Rossi Blog Reader

This page contains all the postings to Andrea Rossi's Journal of Nuclear Physics, with the entries sorted so that Rossi's answers appear under each question (where possible).

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  1. Gherardo

    Dott.Rossi

    Merry Christmas and Happy New Year to you, your co-workers and forum members.

    2014 wishes were fruitfull so this year I hope “may 2015 be the year of the E-Cat”.

    Today I found funny that ECAT acronim is also used by Escambia County Area Transit near Pensacola (Florida).

    All the best, Gherardo

  2. Andrea Rossi

    Gherardo:
    Thank you and same wishes to you, extended, as usual, to all our Readers, from the Team I work in and me,
    A.R.

  3. Andrea Rossi

    Dr Joseph Fine:
    Whereas I cannot comment in positive or negative issues related to the operation of the E-Cat, I totally share the quote from Sir Arthur C. Clarke.
    I too wish you, and again to all our Readers, a healthy 2015, after a peaceful and Merry Christmas.
    A.R.

  4. Jim

    Hi, all is going sound here and of course every one is sharing data, that’s actually fine, keep up writing.

  5. Andrea Rossi

    Jim:
    I take the chance of your comment to remind that all the discussion on course regarding what happens inside the E-Cat during the operation has to be considered totally independent from us. We are publishing the comments as they arrive, with no editing, and the publication of them has nothing to do with any kind of opinion of ours regarding their content. We only spam comments when we deem them unpolite or insulting toward somebody, or when a comment assumes I am sharing any kind of opinion: when I share an opinion I write it. Please note that I never comment on any assumption , in positive or in negative, because I cannot give information of sort about the mechanism of the so called “Rossi Effect”.
    Information about our theoretical bases and operation of the E-Cat, beyond what has already been communicated by us, will be given in due time. Obviously, everybody is welcome to make any assumption and guess he wants.
    Warm Regards,
    A.R.

  6. silvio caggia

    Dear Andrea Rossi,
    Did you dismiss the idea of onion-cat or this technology is embedded within the “dog-bone” hot-cat?
    It really looks like a leek-cat (it.m.wikipedia.org/wiki/Allium_ampeloprasum)… :-)

  7. Andrea Rossi

    Silvio Caggia:
    Ideas are never dismissed, they always work.
    Warm Regards,
    A.R.

  8. Steven N. Karels

    Dear Wlad,

    These are the reactions I would expect in an eCat based on lithium acting on the various other isotopes present in the eCat.

    Start Result Energy Released
    58Ni + 7Li 59Ni+6Li 1.7492963
    59Ni + 7Li 60Ni + 6Li 4.1377298
    60Ni+7Li 61Ni+6Li 0.5702163
    61Ni + 7Li 62Ni + 6Li 3.3465222
    62Ni + 7Li 63Ni + 6Li -0.41204
    64Ni + 7Li 65Ni + 6Li -1.151829

    27Al + 7Li 28Al + 6Li 0.821346
    28Al 28Si + e 4.2961227

    6Li +1H 7Li 0.0069433
    7Li + 1H 4He + 4He 0.0186231

    54Fe + 7Li 55Fe + 6Li 2.0482114
    55Fe + 7Li 56Fe+ 6Li 3.9473333
    56Fe + 7Li 57Fe + 6Li 0.3961208
    57Fe + 7Li 58Fe + 6Li 2.7946144

    Note the negative energy got 62Ni and 64Ni.
    If these are correct, then the following may be observed
    1. The overall energy release is limited by the amount of hydrogen available in the eCat.
    2. Aluminum plays a role in the energy generation as it is converted to silicon.
    3. This would explain why the reaction stops at 62Ni.
    4. It is not clear why The Report indicates 64Ni was consumed.
    5. Why did iron not show a shift in its isotopes (or was it just not reported)?
    6. Is the iron a magnetic catalyst for the other parts of the reactions?

    Thoughts?

  9. Wladimir Guglinski

    Joe,
    concerning your opinion:

    “2. I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.

    you have also to consider that the two nuclei Ni and 3Li7 are very far away one each other (a distance at least two times longer than the distance between 3Li7 nucleus and the orbit 2s1).

    As the magnetic force decreases with the square of the distance, even if the intensity of the magnetic moment of the 2s1 and the magnetic moment of the nucleus Ni was the same, however due to the larger disance the magnetic attraction force between proton and the Ni nucleus would be 4 times weaker than the magnetic attraction force between the proton and the orbit 2s1.

    Besides,
    there is yet the Coulomb attraction between the proton in the newborn 4Be7 and the electron 2s1

    So, concerning the extraction of the proton from the 4Be7, the magnetic moment of the Ni nucleus has not any influence in the mechanism of the extraction.

    regards
    wlad

  10. Wladimir Guglinski

    ERRATA:

    In my comment of December 21st, 2014 at 4:46 AM

    where it is written;

    If this is the case, then the radius of the 2s1 orbit is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.

    the correct is:

    If this is the case, then the radius of the 2s1 orbit in the Ni nucleus is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.

  11. Wladimir Guglinski

    Joe,
    one cold fusion researcher of the Martin Fleischmaan Memorial Project wrote in their page in the Facebook:

    “- We now understand WHY we need the alumina tubes on the feed wires

    However,
    I suspect that Andrea Rossi uses permanent magnets within those alumina tubes. This can be his secret.

    Indeed, let us think about those alumina tubes, as follows:

    1- There is no need them so much long

    2- Why alminium ? Well, because aluminium is not magnetic. By using aluminium the it is easier to hide the secret (for instance, aliminium does not attract small iron pieces).

    3- However aluminium has interesting properties when it interacts with permanent magnets:
    http://terpconnect.umd.edu/~wbreslyn/magnets/is-aluminium-magnetic.html

    By this way, when the electric power is turn off, and the magnetic pulses stop, the excited Ni isotopes continue being aligned toward the axis of the reactor, and this is the reason why the eCat works in self sustained model along some hours. When the 58Ni, 60Ni, 62Ni, 64Ni lose their excitation, the self sustained mode is ended.

    regards
    wlad

  12. Wladimir Guglinski

    Joe,
    actually there is one more step along the transmutation 59Co -> 62Ni:

    59Co + n -> 60Co -> 60Ni ====> 60Ni + n -> 61Ni ===> 61Ni + n -> 62Ni

    I dont know yet why the stable 59Co has reaction with 3Li7, while the stables 63Cu and 65Cu do not have reaction with 3Li7.

    However there is a difference of two protons between 27Co and 29Cu. Then the radius of the 2s1 orbit in 29Cu is larger, and therefore the proton (when it exists the 3Li7) has too much acceleration, and so the neutron is not captured by 29Cu , in spite of the energy of the neutron is not enough to cause the fission of the 29Cu.

    If this is the case, then the radius of the 2s1 orbit is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.

    So, Ni is in the limit for a receptor to have fusion with the emitter 3Li7.

    regards
    wlad

  13. Joe

    Wladimir,

    1. You write,
    “Only 61Ni produces cold fusion, however there is only 1,14% of 61Ni in the fuel, and there is no way to have self sustained mode.”

    So then what is the minimal proportion of 28Ni61 that would allow an SSM? And how do you calculate this proportion?

    2. You write,
    “The magnetic moment of the 2s1 electron is very stronger than the magnetic moment of the nucleus Ni.”

    And that is the problem. The induced magnetic moment of the electron will not allow the weaker magnetic moment of the target nucleus to steal the approaching nucleon from it.

    3. What do you think is the reason for the lack of gamma rays observed in the E-Cat?

    All the best,
    Joe

  14. Joseph Fine

    Andrea Rossi,

    Is it somewhat valid to think of the energy releasing interactions as being glancing blows between two (or more) nuclei? That is, two (or more) nuclei interact at shallow angles such that the heavier nucleus strips off neutrons from the lighter nucleus.

    This is more like atomic etching or removing of material.

    Conceptually (?), this is some type of atomic-scale milling machine where a collection of heavier nuclei (with sufficient kinetic energy) become a machine tool, milling machine or battering ram. In this case, the material removed from the lighter atoms* – which are being battered – are neutrons. Most, if not all, of the removed neutrons are then absorbed (or fused) into the milling machine tool or the heavier nuclei.

    (* Lighter atoms are “damaged” more by collisions with heavier atoms just as lighter cars are damaged more by any collisions with heavier trucks.)

    This is not only miraculous, to quote Sir Arthur C. Clarke:
    “Any sufficiently advanced technology is indistinguishable from magic.”

    Best wishes for a Christmas season full of blessings.

    And also, a happy, healthy New Year to you, your family, your team and your readers.

    Joseph Fine

  15. Wladimir Guglinski

    Joe wrote in December 20th, 2014 at 3:48 PM

    Wladimir,

    1. —————————————–
    Why would the E-Cat not work full time in the SSM just because the fuel might not be 100% 28Ni61? It would produce less power but it should last just as long.
    ——————————————–

    No, Joe,
    when the electric power is turn off, after a time the isotopes 58Ni, 60Ni, 62Ni, 64Ni lose the excitement, and without excitement they have nuclear magnetic moment zero. Therefore they cannot be aligned toward the axis of the reactor, and they do not produce cold fusion.

    Only 61Ni produces cold fusion, however there is only 1,14% of 61Ni in the fuel, and there is no way to have self sustained mode.

    2. ——————————————
    I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.
    ———————————————-

    What you think makes no sense.
    The magnetic moment of the 2s1 electron is very stronger than the magnetic moment of the nucleus Ni.
    The proton of the 3Li7 is extracted by the electron’s orbit, and not by the nuclear magnetic moment of the Ni nucleus.

    The only task of the nuclear magnetic moment of the Ni is to put the z-axis of the Ni aligned toward the axis of the reactor

    .

    3. ———————————————
    The most important question is, at what point in this whole process is excess energy being created?
    ————————————————

    7Li -> 6Li

    58Ni + n -> 59Ni -> 59Co

    59Co + n -> 60Co -> 61Ni ==> 61Ni + n -> 62Ni

    60Ni + n -> 61Ni

    61Ni + n -> 62Ni

    62Ni + n -> 63Cu

    64Ni + n -> 65Ni -> 65Cu

    Note that the isotopes 58Ni(68,1%) + 60Ni( 26,2%) +61 Ni(1,1%) have as a final result the isotope 62Ni.
    So, while 62Ni is transmutting to 63Cu, however 95,4% of the total Ni isotopes are transmutting to 62Ni, and this is the reason why after the 32 days of the eCat working the ash had 98,7% of 62Ni.
    If the eCat continues working , after some months the percentage of 62Ni will be near to zero.

    Also note that the eCat worked only 32 days.

    Andrea Rossi already had reported that in earlier experiments he had found Cu in the ash. This is because the eCat had worked along 6 months, and there was time available for the 62Ni to transmute to 63Cu.
    The transmutation 64Ni -> 65Cu is not representative, since 64Ni has only 0,9% of the fuel.

    In the Lugano Report is said in the page 28:
    “Even if that particular reaction is excluded, since no gammas are observed, we can tentatively use this number for each step towards 62Ni, and the information from ICP-AES that there is about 0.55 gram Ni in the fuel. We find then that there is about 2.2MWh available from the Nickel transformations. Accordingly, from Nickel and Lithium together there is about 3 MWh available, which is twice the amount given away in the test run.

    However,
    these numbers used are referred to hot fusion.
    In the case of cold fusion, the reactions do not give the quantity of energy given in hot fusion.
    So, instead of 2.2MWh available from Ni transformations, for cold fusion actuallthe value must be inferior than 2.2MWh.

    regards
    wlad

  16. Wladimir Guglinski

    orsobubu wrote in December 20th, 2014 at 1:14 PM

    Wlad, I would love you succeeded in explain the Rossi effect without having access to the complete experimental datasets, on the basis of theoretical deduction. I already know mrs. Pamela will reply this way:

    “Hi, Wlad, thanks for the suggestions. We have several experiments in line waiting to be realized. When we have time to test your suggestion I’ll warn you. Hug.”
    ————————————————

    dear orsobubu,
    actually it not is the Pamela’s style.

    In 2009 she sent a reply, concerning her experiment in the US Navy:

    “Subject: RE: absence of gamma-rays in your experiment, and neutron’s background
    Date: Mon, 13 Apr 2009 10:29:47 -0700
    From: pam.boss@navy.mil
    To: wladimirguglinski@hotmail.com
    CC: m_bernstein@acs.org; hestenes@asu.edu; canmarrai@gmail.com

    Like many, we have very few funds and resources. But we will consider your suggestions and see what we can do as time and money permits.

    Regards,

    Pam”

  17. Joe

    Wladimir,

    1. Why would the E-Cat not work full time in the SSM just because the fuel might not be 100% 28Ni61? It would produce less power but it should last just as long.

    2. I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.

    3. The most important question is, at what point in this whole process is excess energy being created?

    All the best,
    Joe

  18. Andrea Rossi

    Felix Rands:
    Thank you for the interesting information
    Warm Regards
    A.R.

  19. Hi Andrea,

    Here is our ragtime-flavored Holiday greeting.

    The very best Christmas wishes for you:

    http://www.youtube.com/watch?v=nKo7Rt027MQ

    -thomas

  20. Andrea Rossi

    Thomas Florek:
    Thank you for this delighting gift that, of course, goes with your and my best wishes for a wonderful Christmas to all our Readers

  21. orsobubu

    Wlad, I would love you succeeded in explain the Rossi effect without having access to the complete experimental datasets, on the basis of theoretical deduction. I already know mrs. Pamela will reply this way:

    “Hi, Wlad, thanks for the suggestions. We have several experiments in line waiting to be realized. When we have time to test your suggestion I’ll warn you. Hug.”

    Good, since you frequent this JONP place you’ve made lots more friends

  22. Dear Andrea Rossi,

    http://www.qnergy.com/products_overview

    I think this could be the right business partner to solve the problem of conversion of hot-cat heat into electricity. The company not only has the necessary technology, but also has sufficient production capacity for starters.

    Best regards and Merry Christmas to you, your family and your partners.
    Felix Rends

  23. Wladimir Guglinski

    ERRATA:

    In my comment of December 20th, 2014 at 5:11 AM

    where it is written:

    As after the reaction 60Ni-3Li7 the 60Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.

    the correct is:

    As after the reaction 61Ni-3Li7 the 61Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.

  24. Wladimir Guglinski

    Email sent to Pamela Mosier-Boss

    From: wladimirguglinski@hotmail.com
    To: pam.boss@navy.mil
    Subject: Your version of Fleischmann-Pons experiment improved by using 105Pd
    Date: Sat, 20 Dec 2014 10:51:55 -0200

    Hi, Pamela

    I have strong reasons in believing that the alignment of the nuclear magnetic moment of the Pd nuclei toward an external magnetic field is responsible for the cold fusion occurrence in the Fleischman-Pons experiment (and of couse also in your version of their experiment. performed by you in the US Navy in 2009).

    Therefore only the stable isotope 105Pd contributes for the cold fusion occurrence in your experiment, since only 105Pd has non-null nuclear magnetic moment.
    All the other isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, have null nuclear magnetic moment, and so they do not contribute for the cold fusion occurrence.

    I had proposed for the cold fusion researchers of the Martin Fleischmann Memorial Project to perform the following experiment, so that to test such hypothesis:

    ========================================================
    THE ALIGNMENT OF NUCLEI AS INDISPENSABLE PRE REQUISITE FOR THE COLD FUSION OCCURRENCE CAN BE TESTED BY EXPERIMENT, as follows:

    1- Fleischmann-Pons experiment will be performed in two different vessels A and B

    2- In both vessels will be used an external source of magnetic field, so that to eliminate the influence of the magnetic fields of the Earth and the Sun

    3 – Magnetic pulses cannot be used in any of the two vessels A and B, in order DO ONT EXCITE the nuclei in both experiments

    4- The vessel A will be filled with the isotope 105Pd and deuterium. As the nucleus 105Pd has nuclear magnetic moment, the 105 Pd nuclei will be aligned toward the vector magnetic field, and the cold fusion must occur .

    5- The vessel B will be filled with the isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, and deuterium. As all those nuclei have null magnetic moment, and as they will not be excited (because there is not magnetic pulses applied), they cannot be aligned toward the vector magnetic field, and therefore COUD FUSION CANNOT OCCUR.
    ========================================================

    However, the researchers of the MFMP responded the following to me:

    “Given that Pure Pd is over $5833 100g, where do you propose one gets affordable 105Pd?

    If you are able to provide the raw materials, we may be able to conduct an experiment.”

    So, they have no money for performing the experiment.

    Then I would like to know:

    May you be interested to make it in the laboratory of the US Navy?

    Can you make it?

    If I am right, the heat produced by using 100% of 105Pd must be 5 times greater than in the experiment made by you in 2009.

    regards
    W Guglinski

  25. Wladimir Guglinski

    eCat full time in self sustained model

    Joe,
    I suspect that by using a fuel composed by 100% of 61Ni the eCat can work full time in the self-sustained-mode.

    The reason why the reactor filled with fuel composed by 58Ni, 60Ni, 61Ni, 62Ni, 64Ni, after some hours stops to work in the self-sustained-mode is because after some time the excited isotopes 58Ni, 60Ni, 62Ni, 64Ni begin to lose their excitation, and so the reactions Ni-3Li7 stop.

    Then there is need to apply magnetic pulses again in the coils of the reactor, in order to excite the isotopes 58Ni, 60Ni, 62Ni, 64Ni, and that’s why there is need to turn on again the electric power, supplying electric current to the coils of the reactor.

    However, the eCat filled with 100% of the isotope 61Ni will work in the self sustained model only while there is yet 61Ni isotopes available within the reactor.

    As after the reaction 60Ni-3Li7 the 60Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.
    So, when the fuel composed by 60Ni is totally converted to 62Ni, there is need to apply again magnetic pulses, in order to excite the 62Ni.

    regards
    wlad

  26. Andrea Rossi

    Wladimir Guglinski:
    As you know, I cannot give information, in positive or in negative, related to the fuel isue, more than I already did.
    Warm Regards,
    A.R.

  27. John Atkinson

    As time passes the e cat gets closer and closer to a world marketed alternative energy source. There will come a time if it has not already accrued,when big oil will threaten or may even sabotage your efforts.The middle East may also try , as illustrated by North Korea’s treats and computer hacks towards Sony . Since a great deal is dependent on computer operation of the e cats when working in unison, have you and Industrial Heat made plans or possibly consulted with the US for such a contingency? It seems at this point it may be the only great treat to your tremendous success and saving this world from itself..

  28. Andrea Rossi

    John Atkinson:
    Than you for your kind attention.
    We have specialists in our Team who deal with this problem.
    Warm Regards,
    A.R.

  29. Wladimir Guglinski

    Andrea Rossi wrote in December 18th, 2014 at 11:41 AM

    Frank Acland:
    It is an interesting idea: under a theoretical and technological point of view, I do not see why not. The issue is in the price: if the electric energy supplied by batteries will be competitive with other sources, the coupling between high efficiency batteries and the E-Cats will be surely possible. Have you an idea of the cost per kWh supplied by this new generation of batteries? I am curious.
    ———————————————-

    I think the eCat spends so much electric power because there is need to excite 98% of the Ni isotopes used as fuel in the eCat.

    The natural abundance of the stable Ni isotopes is the following:

    58Ni = 68,08%

    60Ni = 26,22%

    61Ni = 1,14%

    62Ni = 3,63%

    64Ni = 0,93%

    Only 61Ni has magnetic moment, and so it does not require excitation so that to be aligned by magnetic field along the azis of the reactor, in order to have fusion with 3Li7.

    All the other Ni isotopes have null magnetic moment, and all they require excitation so that to be aligned, in order to have fusion with 3Li7.

    By using a fuel composed by 100% of 61Ni the electric power could be reduced drastically, because there is no need to excite the 61Ni (the electric power for the excitation of the 98% of the other Ni isotopes would be saved).

    However, I do not know how much extra cost would be necessary to obtain a fuel composed of 100% of 61Ni.

    regards
    wlad

  30. Frank Acland

    Dear Andrea,

    There is a great deal of research and development and investment going into battery technology worldwide. I think the trend will be for lower cost per kWh as time goes by.
    I am sure electric vehicles are going to become more and more popular as time goes by. Lack of charging stations for EVs are one obstacle to their widespread use. You should talk with Elon Musk of Tesla. I think together you might be able to work towards a solution!

    Kind regards,

    Frank Aclan

  31. Andrea Rossi

    Frank Acland:
    Very interesting, thank you for the information. I am learning.
    Warm Regards,
    A.R.

  32. Frank Acland

    Dear Andrea,

    You asked about the cost per kWh supplied by the new generation of batteries. I have to thank an E-Cat World reader for finding this. According to the Tesla Motors Web site, they can currently deliver electricity from their lithium-ion batteries at 0.21 Euros per kWh.

    http://www.teslamotors.com/en_CA/goelectric#savings

    Kind regards,

    Frank Acland

  33. Andrea Rossi

    Frank Acland:
    Thank you for this interesting information. As you can se, this is a price good for usual batteries utilization, but very high for us. Nevertheless, the use of batteries as storage of energy is very interesting. By the way: I tested a Tesla, is very funny: same acceleration of a Ferrari, but too short autonomy if you want to get driving fun.
    Warm Regards,
    Andrea

  34. DTravchenko

    Dr Andrea Rossi:
    Gas prices are falling: how are you aware of the drop of the fuels respect the market perspectives for the E-Cat?
    Warm Regards,
    DT

  35. Andrea Rossi

    D. Travchenko:
    Again on the fuels: today this issue has got fuel!
    I do not know what will happen to gas prices and I don’t control them, I only control, together with my wonderful Team, the products for our Customers.
    Prices of fuel will change and the needs of the market will change, this is why we have an excellent business Team to watch the market and ensure we are meeting the Customer’s needs.
    Warm Regards,
    A.R.

  36. Andrea Rossi

    IC Renoir:
    I am not able to provide any information at this time and any information will be shared publicly when appropriate.
    Warm Regards,
    A.R.

  37. Curiosone

    Let me put a hypothetical question: if the oil price will go down enough to make the E-Cat not convenient, what do you think will happen to your and Industrial Heat’s enterprise?
    WG

  38. Andrea Rossi

    Curiosone:
    As this is a hypothetical question, my only response would be to guess, which is not something I wish to be on the record about.
    Warm Regards,
    A.R.

  39. Wladimir Guglinski

    Experiment proposed for the researchers of the Martin Fleischman Memorial Project
    https://www.facebook.com/MartinFleischmannMemorialProject/posts/886318188065548?comment_id=886327334731300&offset=0&total_comments=8&notif_t=feed_comment

    But it seems they have no money so that to perform the experiment.

    ========================================================
    THE ALIGNMENT OF NUCLEI AS INDISPENSABLE PRE REQUISITE FOR THE COLD FUSION OCCURRENCE CAN BE TESTED BY EXPERIMENT, as follows:

    1- Fleischmann-Pons experiment will be performed in two different vessels A and B

    2- In both vessels will be used an external source of magnetic field, so that to eliminate the influence of the magnetic fields of the Earth and the Sun

    3 – Magnetic pulses cannot be used in any of the two vessels A and B, in order DO ONT EXCITE the nuclei in both experiments

    4- The vessel A will be filled with the isotope 105Pd and deuterium. As the nucleus 105Pd has nuclear magnetic moment, the 105 Pd nuclei will be aligned toward the vector magnetic field, and the cold fusion must occur .

    5- The vessel B will be filled with the isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, and deuterium. As all those nuclei have null magnetic moment, and as they will not be excited (because there is not magnetic pulses applied), they cannot be aligned toward the vector magnetic field, and therefore COUD FUSION CANNOT OCCUR.
    ========================================================

    .
    REPLY BY MFMP:

    Martin Fleischmann Memorial Project : Given that Pure Pd is over $5833 100g, where do you propose one gets affordable 105Pd?

    Wlad Guglinski dont you think to be important in order to decypher the cold fusion mystery?

    Wlad Guglinski The 105Pd natural abundance is 22, 23%. In the case my theory is correct, and MFMP decides to perform the experiment, there is need to take care, because in this version of the Fleischmann-Pons experiment the energy created by cold fusion will be 5 times greater

    .

    Comment by me here in JoNP:

    So, anyone who is trying to explain cold fusion thinks that there is need only of proposing the nuclear reactions, and the energy produced in each of the reactions.

    Nobody wishes to realize that there is need to understand the cold fusion mechanisms occurring in the phenomenon.

    It is like to want to understand how the planets are attracted by the Sun, without to know the existence of the gravitation.

  40. JCRenoir

    Dear Dr Rossi,
    We all would like to have more information about the 1 MW plant dlivered to the customer months ago!
    Godspeed,
    JCRenoir

  41. Stefano Landi

    Dear Andrea. Please give us an update on the e-cat / e-mouse system you talked months ago.
    Kind regards
    Stefano

  42. Andrea Rossi

    Stefano Landi:
    As you correctly write, I talked months ago of this issue and I have nothing to add to this matter, in general.
    Warm Regards,
    A.R.

  43. Brandon Hurd

    Dear Andrea Rossi

    Has Industrial Heat informed President Obama about the E-Cat? If so, did you get a formal response?

    If not, don’t you believe the U.S. government will want to be involved with the technology when it is ready for mass roll-out in the market?

    Regards
    Brandon Hurd

  44. Andrea Rossi

    Brandon Hurd:
    I am not involved in this kind of issues, but I am sure that Industrial Heat has given all the due information where it is opportune and proper.
    Warm Regards,
    A.R.

  45. Greg Leonard

    Dear AR and Rodney Nicholson,
    To heat an object, the heat source needs to be a some significant higher temperature.
    I can imagine that a gas powered burner would work fine here.
    What I am unsure about is whether the temperature difference between the ‘heater’ ecat and the ‘heated’ ecat will be enough to get the ‘heated’ ecat into self-sustain mode.

    I imagine the gas powered ecat would be extremely useful when the output requirement is simply heat, and in a place far from the electric grid.
    Greg Leonard

  46. Andrea Rossi

    Greg Leonard:
    Surely the heat production for direct use of heat is the more efficient, for the first and second thermodynamic principle.
    Warm Regards,
    A.R.

  47. Steven N. Karels

    Alessandro,

    I recently purchased a Chevy Volt. It requires 12 Amps of 115VAC input power for 10 hours to completely charge the car’s batteries and this results in a range of about 30 miles. So a likewise continuous charge rate might yield a daily range of about 70 miles. This charging rate is equivalent to an electric output rate of 1.3 kW. Assuming typical Carnot efficiencies means a thermal generation of around 4kW. With a COP of 3, then I would guess a 10kW eCat could provide enough power for one vehicle with a daily range of 70 miles. A substantially larger eCat would be required for continuous long distance traveling. But this says the numbers are close. Getting rid of excess heat might be a problem but I suspect could be engineered to handle it.

  48. Andrea Rossi

    Steven N Karels:
    Yes, I too think that batteries could be a storage .
    Warm Regards,
    A.R.

  49. Steven N. Karels

    Dear Andrea Rossi,

    I think it is more than just energy cost economics. The use of batteries to store energy is more applicable to temporal energy sources such as photovoltaic or wind sources – batteries would store electrical energy during the time when the energy source is not available. Not so with the eCat which, by its nature, can run continuously for months at a time. The eCat is best suited for Baseline (continuous) electrical generation. The only way I see electricity could be used to provide the input energy for eCat operation (besides control) is if the cost of electricity becomes so low as to be much cheaper than natural gas.

  50. Andrea Rossi

    Steven N. Karels:
    All this issue is intriguing, even if not in the immediate future.
    Warm Regards,
    A.R.

  51. Alessandro Coppi

    To complete the idea of Paul, when the car is parking, it will be connected to the grid and upload energy.

  52. Andrea Rossi

    Alessandro Coppi:
    Interesting.
    Warm Regards,
    A.R.

  53. Paul

    Andrea,

    The quickest way to a legal e-cat powered car is to have a suitcase sized e-cat electrical generator (mini-turbine + Generator, Stirling engine + generator, or e-cat direct conversion) in your trunk, recharging the car’s batteries (while at the same time using the batteries to power the e-cat). Since, unlike the car, the e-cat would run 24/7 it could be lower power than the car engine.

    Paul

  54. Andrea Rossi

    Paul:
    Thank you for your idea.
    Warm Regards,
    A.R.

  55. Giovanni

    Dear Dott. Rossi
    interesting development at: Zenn Motor / EEStorrFanFib
    Best regards and happy Christmas and New Year!
    Giovanni

  56. Andrea Rossi

    Giovanni:
    Thank you for the information,
    Warm Regards,
    A.R.

  57. Steven N. Karels

    Frank and Andrea,

    I guess I am not seeing the advantage of high efficiency batteries when generating electricity. Andrea Rossi previously posted that part of the reason of going to gas-power eCats as opposed to self-powering them using produced electricity was the relatively low cost of natural gas compared to the price of generated electricity.

    Even if a high efficiency battery was 100% efficient, it would be no better than using electricity produced by an eCat system to feed-back the power as input to the eCat system. The previous argument on natural gas powered eCats was why consume a precious product such as electricity for heating the eCats when natural gas could do the job at a much cheaper cost per thermal unit. I fail to see why batteries would change this argument.

  58. Andrea Rossi

    Steven N. Karels:
    Good point, but, as I said, this is just matter of energy price.
    Warm Regards,
    A.R.

  59. Frank Acland

    Dear Andrea,

    There is a lot of effort being expended these days into producing better batteries, and other energy storage devices that can be charged by ‘free’ fuels like solar and wind.

    For example, Tesla Motors’ gigafactory (currently under construction) which will manufacture batteries for electric cars is planned to be powered solely by a combination of solar, wind and geothermal.

    What are your thoughts on the possibility having battery driven E-Cats?

    Many thanks,

    Frank Acland

  60. Andrea Rossi

    Frank Acland:
    It is an interesting idea: under a theoretical and technological point of view, I do not see why not. The issue is in the price: if the electric energy supplied by batteries will be competitive with other sources, the coupling between high efficiency batteries and the E-Cats will be surely possible. Have you an idea of the cost per kWh supplied by this new generation of batteries? I am curious.
    Warm Regards,
    A.R.

  61. Regarding:

    Andrea Rossi
    December 17th, 2014 at 3:31 PM
    Koen Vandewalle:
    The gas fueled Hot Cat is a logic evolution of the Hot Cat, due to obvious economic considerations.
    Warm regards
    A.R.

    While it appears some electrical stimulation is essential for the functioning of the Ecat, might it be possible in a multiple unit system for the energy required to heat units not in self-sustain mode to be provided by heat from the units in self-sustain mode? This way, from the heat input point of view, the entire device would be self sufficient and a gas supply would not be necessary? No doubt you have already considered this.

    Rodney Nicholson.

  62. Andrea Rossi

    Rodney Nicholson:
    We are considering this, as you correctly write.
    Warm Regards,
    A.R.

  63. Koen Vandewalle

    Dear Andrea,

    Is the original idea of a Gas-Cat yours, based on an industrial need that you have knowledge of ? Or was it an initiative of one of your (potential) customers ?

    It differs a lot from the original E-Cat and the Hot-Cat.

    Kind Regards,
    Koen

  64. Andrea Rossi

    Koen Vandewalle:
    The gas fueled Hot Cat is a logic evolution of the Hot Cat, due to obvious economic considerations.
    Warm regards
    A.R.

  65. ing. Michelangelo De Meo

    Hello Dr. Rossi ,
    I enclose a very interesting article on one of the first applications that could have the Hot Cat in Italy ie DISTRICT HEATING .
    3 million inhabitants in Italy using heat .
    The Hot Cat would be the ideal system to reduce the cost of heating, cooling and domestic hot water .

    http://www.edilportale.com/news/2014/12/risparmio-energetico-e-sostenibilita/il-teleriscaldamento-in-italia-serve-3-milioni-di-abitanti_43090_27.html

  66. Andrea Rossi

    Ing. Michelangelo De Meo:
    Thank you for the information,
    Warm Regards,
    A.R.

  67. Curiosone

    Dr Rossi, thank you for your answer, as usual, but probably you wrote a typo: photons interact with the electromagnetic force!
    W.G.

  68. Andrea Rossi

    Curiosone:
    Wrong.
    Photons carry electromagnetic force, but they are electrically neutral, therefore cannot interact with the electromagnetic force!
    Warm Regards,
    A.R.

  69. georgehants

    Dear Mr Rossi, you say that you are far ahead of most of your competition.
    Are you taking into account that for many years you had a very small team working on your Research and that now, those who are clever enough to see the potential of your discovery can put many hands to work.
    Would this not mean that the competition could catch you up in a very short time.
    Best wishes

  70. Andrea Rossi

    Georgehants:
    You are right, but we are working at the maximum of our possibilities, independently from what can happen outside. We must think that our competition is as strong as we are, if not better, and act consequently, true or not as it may be.
    Warm Regards,
    A.R.

  71. Curiosone

    Dr Rossi:
    Can you explain which are the forces felt by the different particles?
    W.G.

  72. Andrea Rossi

    Curiosone:
    quarks: Electromagnetism (E), Strong (S) , Weak(W), Gravitation (G) , Higgs (H)
    charged leptons: E, W, G, H
    neutrinos: W, G, H
    photons: G
    gluons: S, G
    W+ W- : E, W, G, H
    Z: W, G, H
    graviton: G
    Higgs: W, G, H
    Warm Regards,
    A.R.

  73. DTravchenko

    Will you attend the ICCF of Padua (Italy) in April ?
    DT

  74. Andrea Rossi

    DTravchenko:
    I will not be able to attend, because in that period I will be in the USA in symbiosis with the 1 MW plant. I take this chance to say that I wish the greatest success to all the scientists of the ICCF.
    Warm Regards,
    A.R.

  75. keV

    Dear Ing. Rossi,

    Concerning the longevity of the fuel – have you tried vibrating the fuel contents during self-sustain mode to see what effect this vibration has on the reaction. For some reason I have this vision of it only being the surface atoms facing the Hydrogen that are reacting (due to the relatively small amount of powder charge actually consumed) and thought that vibrations may bring fresh material to the top. If the vibration does lengthen the overall output time of a single charge, in the 1MW reactor, single charges could be vibrated in sequence (say one individual charge per day) without decreasing the overall real-time output of the whole 1MW plant significantly whilst extending the charge longevity of all the individual e-cat units.

    Of course all this would depend on vibration making a beneficial difference to longevity of charge power output :¬)

    Just a rambling(and no doubt ignorant)thought; a little different from the usual questions you get these days though!

    Regards,
    Kev

  76. Andrea Rossi

    Kev:
    As you know, I cannot give information regarding the fuel, either in positive or in negative.
    Warm Regards,
    A.R.

  77. ing. Michelangelo De Meo

    Hello Dr. Rossi, in a radio important Italian is spoken of ‘ E -cat and the last experiment . You can listen to the interview with Professor Bo Höistad , professor of nuclear physics at Uppsala University , who participated in the experiment .
    Readers who follow her around the world , can read the transcript of what he said the professor on the link below . Congratulations

    http://www.radio24.ilsole24ore.com/player.php?channel=2&idpuntata=gSLAFnX5p&date=2014-12-12&idprogramma=smart-city

    http://22passi.blogspot.it/2014/12/il-ritorno-delle-cat-su-radio24.html

  78. Andrea Rossi

    Ing. Michelangelo De Meo:
    Thank you: very interesting.
    Warm Regards
    A.R.

  79. Giovanni

    Dear Dott. Rossi
    in a previous post of mine, I was pointing to the news (E-Cat world) that the Prime Minister of the Republic of Italy has today (12.12.2014) awarded its High Patronage to the ICCF-19 event. The post has not passed the moderation, perhaps because of the link I have inserted.
    Something is moving…
    My best regards

  80. Andrea Rossi

    Giovanni:
    As I already said, our work of the last 4 years has moved the giants.
    Warm Regards,
    A.R.

  81. Dima Redko

    Dear Andrea,
    How do you think, if the oil prices will continue to fall as fast as they do, will your technology still be competitive when it is finally released, considering the much cheaper oil, gas, and electricity and the high price of 1MW plant 1.5M USD?

  82. Andrea Rossi

    Dima Redko:
    The history of oil prices is a roller coaster…I have not the cristal ball.
    The price of the E-Cat will be adjusted to the market by mass production, in due time.
    Warm Regards,
    A.R.

  83. Herb Gillis

    Andrea Rossi:
    Do you think it would be possible (in principle) to achieve the Rossi Effect in a fully liquid medium [such as a molten metal, metal compound; or molten salt], or is the solid state also a fundamental requirement for the Effect?
    Kind Regards; HRG.

  84. Andrea Rossi

    Herb Gillis:
    Sorry, I cannot answer this kind of questions.
    Warm Regards,
    A.R.

  85. eernie1

    Dear Andrea,
    What amazes me is the fact that you have been able to keep the recipient and the location of your delivered unit a secret this long. Even the Manhattan project, one of the most guarded government secrets, had a Russian spy who was divulging information about the atomic project back to Moscow. You must have a very loyal crew(at least 25 indicated by your blogs)who have not succumbed to the possible temptation of easy money or sharing information with an intimate acquaintance. Almost as miraculous as your machine!
    Silent regards.

  86. Andrea Rossi

    Eernie1:
    You are right: our Team is fantastic, also under this point of view.
    Warm Regards,
    A.R.

  87. Frank Acland

    Dear Andrea,

    Has your customer been able to use any of the heat you have been making for useful purposes yet?

    Kind regards,

    Frank Acland

  88. Andrea Rossi

    Frank Acland:
    All I am authorized to say is that the plant has been delivered. Due information regarding the operation will be given in due time.
    Warm Regards,
    A.R.

  89. orsobubu

    Chi fa la spia non è figlio di Maria non è figlio di Gesù quando muore va laggiù

  90. Steven N. Karels

    Dear Andrea Rossi,

    For those of us who love to do independent analyses — Can you tell us the approximate mass of the fuel going into the 100+ reactors for the 1 MW thermal unit? Can we assume a 1 gram fuel mass per reactor?

  91. Andrea Rossi

    Steven N Karels:
    Due information will be given at the end of the test, with exception of information restricted to those that have the right on it.
    Warm Regards,
    A.R.

  92. Robert Curto

    Dear Wladimir Guglinski,
    I read in some blogs you are complaining that Dr. Andrea Rossi has spammed your comments to sway the attention to the fact that you have discovered how the E-Cat works, you also have accused him of giving wrong information, to not allow anybody to explain how the E-Cat works !
    Excuse me, but all this is ridiculous. First of all Dr. Andrea Rossi has always
    said he does not want , or cannot talk about the mechanism that makes the
    Rossi Effect, so he swayed nothing, just said he cannot give this kind of information, secondly he repeatedly said that he does not agree with your theories and that they have nothing to do with the E-Cat, and also he has repeatedly said he adheres to the Standard Model.
    He always gently hosted your comments and published your articles, that all the other Magazines have always refused. He offered you unlimited space on his blog.
    Now he spams several comments of yours, and you insult him !
    I think that anyone with thinking faculty can understand that he spammed your comments because, as I read on the other blog, you have mixed up your theories and the E-Cat connected theory.
    With the E-Cat he probably wants not to involve your theories.
    I would like to hear from Dr. Andrea Rossi, if I have guessed correctly.
    Robert Curto
    Ft. Lauderdale Florida
    USA

  93. Bob

    Dear Andrea Rossi

    For the 1 MW plant now in operation, can you tell us whether:

    1. Fuel has been added or removed since the plant began operation.

    2. If there has been no fuel added or removed, is there a time before the expiration of the one year operating period when fuel addition or removal is planned.

    3. Whether the quantity of fuel used has met or not met your expectations for fuel consumed.

    Thanks

    Bob

  94. Andrea Rossi

    Bob:
    One of the things we have to test is the duration of a charge under the stress of a 1 MW plant in a long period. We plan not to change the charge until we have a decrease of efficiency, to check which is its real duration under stress. Due information about this issue will be given at the end of the test, probably within one year. Good question.
    Warm Regards,
    A.R.

  95. Robert Curto

    Drs. Joseph Fine and Andrea Rossi, thanks for all your help in getting Roger
    Green’s excellent website on the JoNP.
    I hope if the Readers subscribe to his Newsletter, they will enjoy it as much as I do.
    Thanks to you both,
    (and with a little help from God)
    Robert Curto

  96. Andrea Rossi

    Herb Gills:
    The issue is much more complex than you say; isotopic shifts are caused by reactions and themselves cause further reactions, about which, obviously, I cannot give information, as I wrote many times.
    The role of hydrogen is foundamental. All I meant is just that the main nuclear reactions are not necessarily fusion.
    Warm Regards,
    A.R.

  97. Herb Gillis

    Andrea Rossi:
    Since you believe the energy source in the Ecat is isotopic shifts, can you give us any guidance at all about the role the hydrogen plays? Do you think that at some point the hydrogen could be eliminated?
    Kind Regards;
    HRG.

  98. DTravchenko

    Dear Andrea Rossi:
    Congratulations for your interview with Salvo TV, that I managed to translate with a friend of mine who speaks Italian. About the part in which you talk of the Universities: which university you think is the best in Italy to study Physics?
    Warm Regards,
    DT

  99. Andrea Rossi

    DTravchenko:
    Should a Russian come to Italy to study Physics, I’d suggest him the Alma Mater of Bologna.
    Warm Regards,
    A.R.

  100. JCRenoir

    Do you think that Aether exists ?

  101. Andrea Rossi

    JC Renoir:
    No.
    Warm Regards,
    A.R.

  102. Curiosone

    Dr Rossi:
    What do you suggest, regarding the LENR, to a university student of Phyisics? What would you suggest him to read?
    W.G.

  103. Andrea Rossi

    Curiosone:
    To a University student of Physics I suggest to let alone LENR and study Physics as his Professors teach the matter to him. What a student has to do is to learn as much as possible and as well as possible . Most University Prof of Physics are very good teachers; I have known many of them, even many that think LENR can’t work, and all of them have a very solid knowledge of Physics foundamentals. The period students spend in University is one of the foundamental pillars of their future, and they have not to play with this fact. First of all they have to learn, and to learn they have to study. I would say that to study at least 6 hours per day, plus the time of the lessons is a good rythm. This does not leave much time to make other things seriously. After they will have got the degree, at that point they can look for diversifications. Example: Picasso has been able to become the Picasso we usually refer to after learning to paint as a Raffaello in the Art Academy he attended; that’s how eventually he became Picasso.
    Warm Regards,
    A.R.

  104. Andrea Rossi

    Dr Joseph Fine:
    Thank you for the correction,
    Warm regards,
    A.R.

  105. Wladimir Guglinski

    Joe,
    I think I finally discovered how occurs the capture of the pair electron-positron by the proton when it is accelerated toward the Ni nucleus in the Rossi-Effect.

    The mechanism is caused by a combination between the disturbance in the helical trajectory of the proton and a shrinkage-dilation in the orbit of the electron 2s1.

    It happens as shown in the Figure bellow:
    http://peswiki.com/index.php/Image:Shrinkage_of_the_electron%27s_orbit_in_the_Ni-3Li7.png

    FIG. 1:
    Before the Ni and 3Li7 are coupled along the z-axis, the orbit of the 2s1 electron has a small radius, because the orbit is situated in the 3Li7.

    FIG. 2:
    When Ni and 3Li7 are coupled, the orbit of the 2s1 is shared by the two electrospheres of Ni and 3Li7. As the electrosphere of Ni is larger, the orbit of 2s1 has a dilation.

    FIG. 3:
    The orbit 2s1 begins to attract the proton, and the proton begins to attract the orbit 2s1, and therefore the proton is pulling the orbit toward the 3Li7 nucleus.
    At the same time, the component Ft begins to cause a shrinkage in the orbit 2s1:
    http://peswiki.com/index.php/Image:Acceleration_on_the_proton_by_electron_orbit_in_Rossi-Effect.png
    Therefore the orbit 2s1 has displacement toward the 3Li7 nucleus, while the orbit also experiences a shrinkage.

    FIG. 4:
    The orbit 2s1 continues to have shrinkage, and a displacement along the 3Li7 nucleus. The shrinkage and the displacement occur discretely (proportional to multiples of Planck’s constant)

    FIG. 5:
    Finally, the proton crosses the plane of the orbit 2s1.
    Well, then now the proton is pulling the orbit 2s1 toward the Ni nucleus.
    So immediately the orbit 2s1 experiences a large displacement going to take its initial position, like it had earlier in the FIG. 2, and therefore the orbit experiences a very big dilation.
    The large dilation of the orbit 2s1 captures a pair positron-electron from the aether, the proton captures the electron, and the positron is emitted.

    The phenomenon occurs similarly as happens in the atom, when the electron jumps from an energy level to another one. When the elecron jumps, the atom captures a pair “particle-antiparticle” from the aether (the photon), and the atom emits the photon.

    regards
    wlad

  106. ing. Michelangelo De Meo

    Congratulations, Dr. Rossi , a great interview. I suggest to all the readers of the journal to listen carefully . Force and courage, Rossi.

    http://salvo5puntozero.tv/intervista-chiacchierata-con-andrea-rossi-inventore-e-cat-12122014/

  107. Franco Sarbia

    Dear Dr. Andrea Rossi.

    How much time will still require the procedure to obtain a patent for e-cat and cat hot,valid for the international market?
    Warm Regards.

    Franco Sarbia

  108. Andrea Rossi

    Franco Sarbia:
    There is not a term.
    Warm Regards,
    A.R.

  109. Wladimir Guglinski

    Joe wrote in December 13th, 2014 at 1:58 AM

    Wladimir,

    You write,
    “And when the proton crosses this point where Ft = Fz, the force Ft becomes stronger, and finally the proton decays in a neutron.”

    Must beta+ decay occur exactly at the point in space where Ft = Fz? If so, why? If not, does this mean that Ft = Fz is just a rough estimate?
    ————————————————-

    Joe,
    actually it is more complex than shown in the figure.
    http://peswiki.com/index.php/Image:Acceleration_on_the_proton_by_electron_orbit_in_Rossi-Effect.png

    Because that situation shown in the figure occurs when the proton and the electon are in phase (both them are in the right side regarding the center of the helical trajectory).
    In this case Ft tries to increase the radius of the proton’s helical trajectory.

    But as the proton has acceleration, it means that the proton and the electron are not always in phase.

    When the proton and the electron are out of phase (the proton at right, and the eletron at left side of the center of the helical trajectory).
    In this case Ft tries to shrinkage the radius of the proton’s helical trajectory.

    So,
    as the proton progresses in its motion, actually the radius of the helical trajectory experiences a very fast dilation-shrinkage.

    When the proton hits the plane of the orbit, Fz= 0, and Ft is maximum. And therefore the maximum dilation-shrinkage in the radius of the helical trajectory occurs when the proton is crossing the plane of the orbit.

    I dont know where is the point where the proton captures the pair electron-positron. But probable it is when the proton hits the plane of the electron’s orbit, because it is the point where occurs the maximum dilation-shrinkage of the cross-section of the proton’s orbit around the center of its helical trajectory.

    Joe,
    also note that, as the proton progresses in its motion toward the plane of the orbit, we have:

    1- The component Ft increases

    2- The component Ft is maximum when the proton hits the plane of the electron’s orbit

    It means that while the proton is moving, the radius of the electron’s orbit also is submitted to a shrinkage.
    So, it is reasonable to suppose that the radius of the electron’s orbit experiencies discrete contractions (discrete shrinkages in the radius, multiples of the Planck’s constant).
    The components Fz and Fm contribute for the shrinkage of the radius of the electron’s orbit.

    But when the proton crosses the plane of the orbit, the components Fz and Fm change their action, and now they contribute for the expansion of the radius of the electron’s orbit.

    Therefore, when the proton crosses the plane, the radius of the electron’s orbit can experience a big expansion (multiple of the Planck’s constant), and such expansion on the surface of the electron’s orbit is responsible for extracting the pair electron-positron from the aether (a similar phenomenon as occurs in the atom, when the electron emits photons extracting them from the aether because the electron changed its orbit from one energy level to another in the electrosphere of the atom).

    regards
    wlad

  110. Robert Curto

    Dr. Rossi, this is from Roger Green, I hope your readers will click on:

    ( see the link on the comment of Joseph Fine 2014/12/13 h 03.44 PM)

    Robert Curto
    Ft. Lauderdale Florida
    USA

  111. Andrea Rossi

    Frank Acland, Wladimir Guglinski:
    I forgot to answer to the question 4 of the Frank Acland’s comment, sorry: I answer in seconds while working…
    Answer: as you have read on the Report of the ITP after the Lugano test, energy comes substantially from isotopical shifts, which is not a fusion, at least for what concerns the final results.
    Warm Regards,
    A.R.

  112. orsobubu

    Felix, OMG that’s an amazing, professional elaboration. It is a long time I would propose to Nikolova a movie script of The New Fire, but lacked an actor and director up to the task. I have to say that the face measures do match very well, but the most interesting fact is the matching during the face animated movements, where the mind of the observers automatically fills and compensates the less than perfect details. You could make a great service to art, to the comprension of the head of the inventor and, ultimately, to the comprension of the inner reactor and science itself, if you posted the video elaborated in such a way to hide the left side with a black matter or something. We cannot elaborate on the E-cat features, because they are not disclosed, so we are forced to twiddle with its inventor’s features. I think we’re into something big here.

  113. Wladimir Guglinski

    Frank Acland wrote in December 12th, 2014 at 10:46 PM

    Dear Andrea,
    I have been reading some translated reports of the interview and wonder if you could confirm whether these translated points are correct:

    1. That the 1 MW plant operates with no external drive input for 3/4 of the operation (self sustain mode)
    2. That Industrial Heat will sell heat (not plants)
    3. That you have discussed the E-Cat with the CEO of Volvo
    4. That the E-Cat reaction is not necessarily fusion
    5. Massive sales will begin once the first 1 MW plant is verified consolidated
    ———————————————————————-

    .

    Andrea Rossi wrote in December 12th, 2014 at 11:37 PM

    Frank Acland:
    1- That’s so far our best available situation with E-Cats and Hot Cats.
    2- No, I did not say this. I said that our plants now make heat and that’s the product our Customers can use so far.
    3- Yes, in Göteborg during the year 2012, after an engineer of Volvo had attended a test in our factory in Bologna ( Italy).
    4- ???
    5- What I said is that mass production cannot start before the operation of the first industrial prototype, installed in a factory of a Customer, is consolidated after a long period of continuous operation ( at least one year).
    —————————————————————

    .

    Dear Andrea,
    You forgot to answer the question 4

    regards
    wlad

  114. Wladimir Guglinski

    Joe,
    I think that the proton transmute to neutron in a fraction of second after crossing the plane of the orbit of the elecron, because:

    1- Beyond the Coulomb force Fz acting in the helical trajectory shown in the figure, there is also a magnetic force Fm due to the attraction between the magnetic field of the proton and the magnetic field of the electron’s orbit.
    http://peswiki.com/index.php/Image:Acceleration_on_the_proton_by_electron_orbit_in_Rossi-Effect.png

    2- Therefore the compoment accelerating the proton is Fz + Fm.

    3- The component Ft trying to increse the radius of the helical trajectory is never stronger than Fz + Fm before the proton to hit the plane of the orbit.

    4- In the instant when the proton crosses the plane of the orbit Ft is maximum, and Fz= 0.

    5- When the proton crosses the plane, Fz and Fm change their direction, and so the force trying to increase the radius of the helical trajectory is Fz + Fm + Ft. In this instant the proton captures a pair positron-electron from the aether, and it becomes a neutron emitting a positron.

    regards
    wlad

  115. Joe

    Wladimir,

    You write,
    “And when the proton crosses this point where Ft = Fz, the force Ft becomes stronger, and finally the proton decays in a neutron.”

    Must beta+ decay occur exactly at the point in space where Ft = Fz? If so, why? If not, does this mean that Ft = Fz is just a rough estimate?

    All the best,
    Joe

  116. Frank Acland

    Dear Andrea,

    I am sorry I cannot understand Italian — because you apparently had an interesting interview with Salvo Mandarà today.

    I have been reading some translated reports of the interview and wonder if you could confirm whether these translated points are correct:

    1. That the 1 MW plant operates with no external drive input for 3/4 of the operation (self sustain mode)
    2. That Industrial Heat will sell heat (not plants)
    3. That you have discussed the E-Cat with the CEO of Volvo
    4. That the E-Cat reaction is not necessarily fusion
    5. Massive sales will begin once the first 1 MW plant is verified consolidated

    Thank you!

    Frank Acland

  117. Andrea Rossi

    Frank Acland:
    1- That’s so far our best available situation with E-Cats and Hot Cats.
    2- No, I did not say this. I said that our plants now make heat and that’s the product our Customers can use so far.
    3- Yes, in Göteborg during the year 2012, after an engineer of Volvo had attended a test in our factory in Bologna ( Italy).
    4- What I said is that mass production cannot start before the operation of the first industrial prototype, installed in a factory of a Customer, is consolidated after a long period of continuous operation ( at least one year).
    Warm Regards,
    A.R.

  118. Yuri

    Hello Andrea, I’ve just read a translation into English of your interview on Vessy’s blog, http://www.ecat-thenewfire.com/blog/. When you say to the interviewer that the E-Cat is in ssm for 3/4 of the time do you refer to the low-temperature E-Cat of IH’s customer or to the Hot-Cat?
    Regards,
    Yuri G.

  119. Andrea Rossi

    Yuri:
    Both.
    Warm Regards,
    A.R.

  120. Wladimir Guglinski

    Wladimir Guglinski
    December 12th, 2014 at 11:20 AM

    eernie1 wrote in December 11th, 2014 at 4:56 PM

    wlad,
    In your reply to me I think you meant 2s1 instead of 1s1.How can an electron electrosphere(in the eV range)accelerate the neutron up to 14 MeV?
    ————————————————————–

    Eernie,
    I would like to calculate the energy of the proton when it crosses the plane of the orbit of the electron.
    However the calculation is very complex, because:

    1- The attraction force on the proton increases proportional to the inverse of the square of the distance.
    So, as the proton progresses, the force on it increases quickly.

    2- With the acceleration of the proton, the magnetic field induced by the motion of the proton increases with the growth of its speed. And therefore the attraction force between the magnetic field of the proton and the magnetic field of the electron orbit also increases.

    .

    Therefore the kinetic energy of the proton grows quickly and strongly, and when the proton crosses the orbit of the electron its kinetic energy had a very big increase.

    regards
    wlad

  121. Wladimir Guglinski

    ERRATA:

    Where it is written:

    Consider that the radius of the electron’s orbit is 0,5×10-10m (20 times shorter than the distance 10^-11m). In this case the proton will move 95% of its trajectory with the condition Fz > Ft.

    the correct is:

    Consider that the radius of the electron’s orbit is 0,5×10-12m (20 times shorter than the distance 10^-11m). In this case the proton will move 95% of its trajectory with the condition Fz > Ft.

  122. Wladimir Guglinski

    Joe wrote in December 12th, 2014 at 4:34 PM

    Wladimir,

    A decrease in radius could be expected in the absence of the Coulomb interaction. But since this interaction exists and would be trying to increase the radius of the helical trajectory, what law of Nature are you using to explain how the Coulomb interaction can be countered?
    ————————————————————-

    Joe,
    the attraction force between the proton and the electron is shown in the figure, where it is decomposed in two components Fz and Ft:

    http://peswiki.com/index.php/Image:Acceleration_on_the_proton_by_electron_orbit_in_Rossi-Effect.png

    The component Fz accelerates the proton, and therefore the action of Fz decreases the radius orbit of the helical trajectory of the proton.

    The component Ft tries to increase the radius of the helical trajectory.

    As Fz is very larger than Ft when the proton is far away of the plane of the electron’s orbit, then Ft is not able to increase the radius of the helical trajectory.

    The distance between the nucleus 3Li7 and the plane of the orbit is about 10^-11m.
    Consider that the radius of the electron’s orbit is 0,5×10-10m (20 times shorter than the distance 10^-11m). In this case the proton will move 95% of its trajectory with the condition Fz > Ft.

    But when the proton arrives to a position near to the plane of the orbit, the condition is Ft = Fz (the proton already has travelled 95% of its displacement).
    And when the proton crosses this point where Ft = Fz, the force Ft becomes stronger, and finally the proton decays in a neutron.

    regards
    wlad

  123. orsobubu

    Fantastic interview, I’ve really enjoyed it, congratulations to Mandarà, a journalist I didn’t know. I’m only sorry that not italian readers will have some difficulties, maybe the video can be uploaded on youtube with the machine translation utility

    Andrea, you know that every man has his face divided into two halves, not perfectly symmetrical; well, if you look at the video and cover by the hand the left side of your face, observing only the right side, you will find that you and actor Clint Eastwood are like two peas in a pod! ehheheh

  124. Andrea Rossi

    Orsobubu:
    bang, bang!
    Warm Regards,
    A.R.

  125. Joe

    Wladimir,

    A decrease in radius could be expected in the absence of the Coulomb interaction. But since this interaction exists and would be trying to increase the radius of the helical trajectory, what law of Nature are you using to explain how the Coulomb interaction can be countered?

    All the best,
    Joe

  126. Dear readers (especially those who can understand Italian),

    I own a small web TV and this morning I had the pleasure to have an interview with Andrea Rossi.

    Please find the link to whatch it: http://salvo5puntozero.tv/intervista-chiacchierata-con-andrea-rossi-inventore-e-cat-12122014/

    I hope you will enjoy it.

    Thanks and regards.
    Salvo Mandarà

  127. Greg Leonard

    Dear Silvio Caggia
    What will happen when the AI discover the biggest threat to the human race is … the human race.
    The biggest threat to the planet Earth is … too many humans

    Hopefully it will take some time before we get to that stage.

  128. Wladimir Guglinski

    eernie1 wrote in December 11th, 2014 at 4:56 PM

    wlad,
    In your reply to me I think you meant 2s1 instead of 1s1.How can an electron electrosphere(in the eV range)accelerate the neutron up to 14 MeV?
    —————————————-

    Eernie,
    the proton and the neutron are not accelerated by the energy of the electron orbit, they actually are accelerated by the magnetic field produced by the orbit.

    The proton moves along the distance between the nucleus and the orbit, in the order of 10^-11m, which is a very big distance ( 10.000 times larger than the radius of the nucleus )

    regards
    wlad

  129. Wladimir Guglinski

    eernie1
    December 11th, 2014 at 4:56 PM

    wlad,
    In your reply to me I think you meant 2s1 instead of 1s1.How can an electron electrosphere(in the eV range)accelerate the neutron up to 14 MeV?
    —————————————-

    Eernie,
    as I already had explained with the help of many figures, when two nuclei align their magnetic fields along the z-axis direction, the outter electron orbit is shared by the electrosphere of the two nuclei, and so the orbit takes the position perpendicular to the z-axis.

    In the case Pd-D2, the orbit 1s1 takes the position perpendicular to the z-axis, and so we have:

    1- Firstly the proton has a big acceleration due to its attraction with the orbit 1s1

    2- After the proton transmutation to neutron, the neutron is accelerated due to its repulsion with the orbit 1s1.

    I had shown it to Joe, in the case of Ni-3Li7:

    http://peswiki.com/index.php/Image:Why_neutron_is_repelled_by_the_electron_orbit.png

    regards
    wlad

  130. Wladimir Guglinski

    Joe
    December 12th, 2014 at 2:39 AM

    Wladimir,

    In your model, as the proton approaches the plane of rotation, does the radius of its helical trajectory increase, decrease, or remain constant?
    ————————————————–

    Joe,
    when a particle is accelerated, the radius of the HT decreases (tends to zero when the velocity aproach to the velocity of light).

    when a paritcle is disaccelerated, the radius of the HT increases (tends to infinite when the velocity tends to zero).

    Therefore as the proton approaches the plane of rotation the radius of its helical trajectory decreases.

    regards
    wlad

  131. silvio caggia

    Dear Andrea Rossi,
    In next years we will have:
    A) Free (or near free) energy (e-cat and other competitors)
    B) Artificial Intelligence
    The combination of A) and B) will give always more services and goods with always less human work.
    We have to resign to the idea that work will be limited to only a small percentage of “work-addicted” people.
    The real challenge will be to invent new ways to distribute services and goods to everyone on the earth according to their needs.

  132. Frank Acland

    Dear Andrea,

    You have talked the necessity of bringing down the cost of E-Cat production for competitive reasons and using robotized production lines, which would reduce the role of human employees in the manufacturing process.

    Do you forsee most employment opportunities surrounding the E-Cat coming on the R&D side of your operations? Or other areas?

    Many thanks,

    Frank Acland

  133. Andrea Rossi

    Frank Acland:
    As you well know cars are made with intense use of robots, but still carmakers are strong employers. The areas of employment are extensive, not just in R&D, but also along the production lines and all the infinite series of activity made necessary from the evolving sophistication of a product like this.
    Warm Regards,
    A.R.

  134. alex

    Dear Ing. Rossi,

    In your reply to Daniel G.Zavela you wrote:

    “..it is the milestone that signals the first commercial product based on LENR ..in the free market. The success of this plant goes beyond anything else, and nothing will take a single hour of my work but it from now through the end of 2015″. Does this mean that you are targeting roll out of commercial e-car or hot-cat, or domestic e-cat, for early 2016?

    God be with you.

  135. Andrea Rossi

    Alex:
    I spammed your comment because written in an untranslatable Language. Please resend it in English.
    Warm Regards,
    A.R.

  136. Andrea Rossi

    Dr Peter Forsberg:
    I agree.
    Warm Regards,
    A.R.

  137. Hans-Joachim Müller

    Dear Dr. Rossi,
    to have a job for earning money is very important for the majority of man. But from a general point of view this discussion is a little bit on the surface. What man needs is food, clothing, heat, electricity, transportation, education, information, health care and so on and so on. To generate these things in a higher amount, higher quality and with a smaller amount of costs and work is the goal of every tecnical progress and developement. An also very hard problem is, to give all people, willing to contibute, the chance to contribute and to earn money in this way for her life.
    Nevertheless,I think the developement of E-cat will bring about also large impulses for employment of many people.

    Hans-Joachim Müller

  138. orsobubu

    Andrea, this is a great answer to an important question. You should go and explain something yourself in Frank’s E-cat World site, where he posted another very good article titled “When Robots Replace Human Workers”. They would not listen to me. In average, people commenting the article (who are, without almost any exclusion, big fans of your achievements) are totally unaware of the relationship between work and capital/money.

    They sincerely think various incredible fantasies like this: every future, single, independent worker will free himself from current wage system buying a robot and putting it to work, and he is compensated for it: “People don’t have to have economic value. All they need is credit. If someone has no money, he takes out a loan, just like he is a student. Instead of investing the money in a diploma for which there is no market, he buys robots that stand in a factory somewhere. The robots earn an income, and the income goes to the owner’s bank account. The only limiting factor will be the demand for robots and the products they manufacture.”

    heheheee

  139. Peter Forsberg

    Dear Andrea Rossi,

    Regarding your view about financial specualtion. I could not agree with you more. There are flaws in the western capitalistic model. I am a strong believer in the free market, but in the capital sector there are serious flaws that cause enormous risk and harm.

    Regards

    Peter

  140. Joe

    Wladimir,

    In your model, as the proton approaches the plane of rotation, does the radius of its helical trajectory increase, decrease, or remain constant?

    All the best,
    Joe

  141. Frank Acland

    Dear Andrea,

    What are the kinds of jobs you hope to see created as a result of your work with the E-Cat?

    Many thanks,

    Frank Acland

  142. Andrea Rossi

    Frank Acland:
    Thank you for this important question. The mass production of E-Cats will generate a miscellanea of jobs: blue collars, white collars, chemical engineers, physicists, infomatics, electronic engineers…industrial E-Cats are very complex machines, demanding many integrated disciplines; I hope this work of us will be a game changer in the employment sector. I really hope that we will be able to create jobs in massive measure, directly and indirectly. I have been born as an enterpreneur, and the essence of enterpreneurship is to create jobs. Unfortunately in our present times money is made principally by means of financial speculations and it is not good, because this kind of richness is made without producing things with real value; real value can be conferred only by the work incorporated in it. In this sense we hope to help the game changing. Industry and production of real things redistributes welfare and make shared richness, while financial speculation concentrates richness in few parasitic organizations without making jobs and therefore without sharing richness. Financial speculation is the real liability of contemporary economy; is made by persons that want to earn money without producing anything useful, just making money with money, so that the money loses its original nature, which is the incorporation of work. Banks are using money to make more money in more or less borderline speculations, instead of financing the industries properly.To return to the sound economy based on the production of real things we must invent products worth to be made; the E-Cat, if it works well, can help. In the meantime the law should forbid ( really) to the banks to act as financial speculators, using the money to speculate instead of financing really productive activities that create jobs and produce shared welfare.
    Warm Regards,
    A.R.

  143. eernie1

    Wlad,
    If you were talking about the electron of the 1D2 then 1s1 is the correct nomenclature.
    Regards.

  144. eernie1

    wlad,
    In your reply to me I think you meant 2s1 instead of 1s1.How can an electron electrosphere(in the eV range)accelerate the neutron up to 14 MeV?
    Regards.

  145. john Atkinson

    I see no down side for e-cat if oil prices continue to fall.I believe the cost of oil is primarilly controlled by OPEC.They have desided to not reduce their oil out put to the world market even though North America and other countries are producing more of their on oil which also furnishs the world oil.Opecs stratigy, I believe , is to allow the world market supply to continue to exceed the market demand for oil and drive the prices lower to the point where it is no longer profitable for the north American continent and other regions in the world to pump it out of the ground.Once it reachs this price and is sustain for some time, oil prices will again come up.Once e-cat is introduced into the market and is proven to be a reliable, clean and cheap energy alternative to oil, the oil prices will decline to the point where its supply will equel its demand which will continually drop. I believe the e-cat will remain much cheaper than oil to produce energy and the e-cat will control the oil prices and not the oil prices control the e- cat. Is this not what all this is about?

  146. Andrea Rossi

    John Atkinson:
    All the energy sources must be integrated at the service of mankind and all must be exploited at the maximum possible. Environmental issues can be resolved with the best available Technologies. We must create jobs, not destroy them.
    Warm Regards,
    A.R.

  147. Wladimir Guglinski

    ERRATA:

    Dear JR,

    in my last post, where it is written:

    While the proton is trying to keep its trajectory along the z-axis, the centripetal force of the electron (transmited via Coulomb attraction) tries to open the helical trajectory of the proton, by increasing the radius Rp.

    the correct is:

    While the proton is trying to keep its trajectory along the z-axis, the centrifugal force of the electron (transmited via Coulomb attraction) tries to open the helical trajectory of the proton, by increasing the radius Rp.

    As you know, I avoid to use the word “centrifugal”, because it is ficticious, in spite of it is the most simple way for explanation

    regards
    wlad

  148. Wladimir Guglinski

    Joe wrote in December 10th, 2014 at 8:53 PM

    2- But the action of the electron 2s1 changes the radius R, and now the orbit has a new radius Rc. However, with that instantaneous velocity V the proton cannot have another different radius. With the velocity V the radius must be kept as R.

    i) By the word “orbit”, do you mean the rotation of the 2s1 electron?
    ————————————————————-

    No, Joe.
    There are two orbits:

    1- The orbit of the electron 2s1. Let’s call its radius Re

    2- The orbit of the helical trajectory of the proton. Let’s call it Rp.

    The rotation of the electron 2s1 in the orbit with radius Re changes the radius Rp of the helical trajectory of the proton, from Rp to Rc. (“c” of change, and not of charge).

    The orbit of 2s1 changes the radius Rp because there is Coulomb attraction between the electron and the proton.
    While the proton is trying to keep its trajectory along the z-axis, the centripetal force of the electron (transmited via Coulomb attraction) tries to open the helical trajectory of the proton, by increasing the radius Rp.
    But as the proton moves is moving with acceleration, its radius Rp actually has tendency to decrease, and not to increase.

    .

    ii) If so, does “Rc” mean “radius of charge”?
    —————————————————

    Explained above

    regards
    wlad

  149. Wladimir Guglinski

    eernie1 wrote in December 10th, 2014 at 9:33 PM

    Wlad,
    As long as we are speculating about Mosier-Boss, how about the following known fusions.
    1D2+1D2=1T3+p(3.02 MeV)
    1T3+1D2=2He4+n(14.1MeV)
    1T3+1T3=2He4+2n(12.9MeV)
    How do you think her reactions achieved the large energies?
    ——————————————-

    Eernie,
    these are reactions by HOT fusion:

    https://www.physicsforums.com/threads/nuclear-fusion-waste-products.121166/

    They are obtained by Tokomak or any other hot fusion reactor
    http://en.wikipedia.org/wiki/Tokamak

    You cannot get 1T3+1D2 = 2He4+n(14.1MeV) directly by cold fusion.

    In Mosier-Boss cold fusion, there is a sandwich Pd-D2-T3

    We have:
    1- The neutron of D2 exits the D2 with 2,2MeV and it is captured by T3. The neutron decays, and T3 transmutes to 2He4.
    2- The proton exits the D2 with 2.2MeV and it is pulled by the orbit 1s1, moving with acceleration toward the 1s1 orbit. When it is crossing the orbit 1s1, the proton transmutes to a neutron, and the neutron is pushed by the orbit 1s1 with acceleration toward the Pd nucleus, getting energy up to 14MeV before to hit the Pd nucleus.

    regards
    wlad

  150. eernie1

    Wlad,
    As long as we are speculating about Mosier-Boss, how about the following known fusions.
    1D2+1D2=1T3+p(3.02 MeV)
    1T3+1D2=2He4+n(14.1MeV)
    1T3+1T3=2He4+2n(12.9MeV)
    How do you think her reactions achieved the large energies?
    Regards.

  151. Joe

    Wladimir,

    You write,
    “2- But the action of the electron 2s1 changes the radius R, and now the orbit has a new radius Rc. However, with that instantaneous velocity V the proton cannot have another different radius. With the velocity V the radius must be kept as R.”

    i) By the word “orbit”, do you mean the rotation of the 2s1 electron?

    ii) If so, does “Rc” mean “radius of charge”?

    All the best,
    Joe

  152. Dan C.

    Dear Andrea,

    I think Greg Leonard may be onto something.

    I recall sometime ago there was talk of an E-cat “TIGER”
    Congratulations to You & Your Team.

    Your R&D is advancing much faster then I could have ever imagined. :-)

    https://www.youtube.com/watch?v=kB5ROD4CGG8

    Regards: Dan C.

    PS,
    U.S. Oil production increasing from about 5M barrels a day to 9M with another 1M coming probably has a little bit to do with the price drop. just a little bit… :-)

  153. Robert Curto

    I would like to share my OPINION with:
    Perter Forsberg and Greg Leonard.
    OPEC does this over and over.
    They try to do as much damage as possible to:

    Natural gas production, they already have enough tree huggers claiming dirty water, earthquakes, etc.

    Alternative Energy, Wind, Solar, etc.

    Fuel made from waste, wood, etc.

    After they do as much damage as possible the price will go back up.

    You have to remember God made the Oil.
    All they have to do is pump it out of the Well for thirty years, at very little cost.
    SOMEDAY E-Cat will make them happy to sell their oil for $10 a barrel, and make a nice profit.

    Robert Curto
    Ft. Lauderdale Florida
    USA

  154. Wladimir Guglinski

    eernie1 wrote in December 10th, 2014 at 11:27 AM

    Wlad,
    In the Mosier-Boss reaction, have you forgotten about the energy used to cause the fusion of the deuterons or the mass change which can be translated as energy that must be accounted for?
    ————————————————–

    Eernie,
    the energy used to cause the fusion of deuterons in cold fusion is small, since cold fusion occurs at low energy of pressure and temperature.

    The mass change is taking in account when we calculate the binding energy 2,2MeV of the deuteron, by using Einstein equation E= mc² applied to the mass defect.

    regards
    wlad

  155. silvio caggia

    @Valeriy Y. Tarasov
    Your theory seems to me very similar to RS2theory forum.rs2theory.org
    but my knowledge of both is so little that I could be wrong…
    I suggest to both guru a cross check, maybe you can cooperate in some way…

  156. orsobubu

    I too believe that the fall in oil prices has nothing to do with the recent developments of LENR on the media. Rather it has to do with the global recession, with new climate change policies, with the price war (the Arabs could plan to make non-competitive the american shale) and with speculation. Oil traders are among the most informed people in the world about everything that happens in the world, in any sector can influence prices. Look at these sites, for example:

    http://www.321energy.com/reports/flynn/current.html

    http://www.321energy.com/archives.php?c=oil

    http://www.insidefutures.com/articles/articles.php?author=12

    If you do a search on 321energy.com + cold fusion, you will only find old articles, a clear sign they know the issue but so far the idea has not made them even remotely tickled. Everything can change tomorrow morning of course, but they are too afraid Wlad receive a tip-off about magnetic prospection oil shortage problems and he could elaborate a new theory, so they won’t even barely touch the topic

  157. Giovanni Guerrini

    In the price of hydrocarbons should be also calculated the cost of global warming,but it is very difficoult because it is a factor in future which we don’t know the amount.
    I hope that some economists,with scientists,make projections to have a real price of oil,gas and coal.
    This is a collective cost and it would be a “real carbon tax” on hydrocarbons.

    Regards G G

  158. Wladimir Guglinski

    Joe wrote in December 10th, 2014 at 3:06 AM
    Wladimir,
    You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.
    —————————————————————————

    Joe,
    let us speak about what the proton enjoys.

    We know from the Newton’s laws that Nature does not appreciate changes:
    1- If a body is at rest, it continues at rest, unless a force is applied on it
    2- If a body movies with constant linear velocity, it continues moving with constant velocity unless a force is applied on it

    A particle moves with helical trajectory (zitterbewegung).
    When the particle is near to be at rest regarding the aether, the radius of its helical trajectory tends to infinitum.
    When the velocity of the particle is near to the speed of light, the radius of its helical trajectory tends to zero

    When a particle is being accelerated (as happens with the proton in the LHC), the radius of its helical trajectory is changing (decreasing).

    Consider the proton being accelerated in the LHC.
    The Nature does not appreciate changes. Therefore, the Nature tries to avoid the change in the radius of the helical trajectory , and (in order to avoid that “disturbance” due to the change in the radius of the helical trajectory) the zitterbewegung of the proton extracts particles of the aether, and emits them in the form of photons (Maxwell’s law).

    In the Rossi-Effect, the proton is accelerated by the orbit of the electron 2s1, and therefore the proton is extracting particles from the aether, and it emits them as photons, in order to avoid the “disturbance” in the radius of its helical trajectory.

    To each absolute velocity of the proton in the aether correspond a specific radius of the helical trajectory.

    Then a second disturbance appears, when the proton is near to cross the plane of the 2s1 orbit: the electron tries to change the radius of the helical trajectory of the proton.
    We have:

    1- in a determined instant “t” the proton has a determined instantaneous velocity V and a radius R of the helical trajectory.

    2- But the action of the electron 2s1 changes the radius R, and now the orbit has a new radius Rc. However, with that instantaneous velocity V the proton cannot have another different radius. With the velocity V the radius must be kept as R.

    3- That discrepancy between the instantaneous velocity V and the new radius Rc requires an extraction of an electron from the aether, so that to transform the proton in a neutron

    regards
    wlad

  159. eernie1

    Wlad,
    In the Mosier-Boss reaction, have you forgotten about the energy used to cause the fusion of the deuterons or the mass change which can be translated as energy that must be accounted for? More than enough to create the energetic particles. the contribution of the strong force is just a part of the overall energy balance.
    I think the 2s1 electron must be more energized so it can sling shot the neutron past the influence of the 2px electron much like the influence of gravity on the space vehicles as they pass a massive body.
    Regards.

  160. Andrea Rossi

    Dr Peter Forsberg:
    You made a good point.
    Warm Regards,
    A.R.

  161. Wladimir Guglinski

    Joe wrote in December 10th, 2014 at 3:06 AM

    Wladimir,

    1. ) —————————————-
    free proton has never been observed to undergo beta+ decay. And by free, we mean not bound to other masses. But obviously free protons are handled by scientists with the use of electric and magnetic fields. And energies associated with these fields can be greater than E = 2*m*c^2, yet with no beta+ decay of the proton ever witnessed.

    The scientists at the LHC are not complaining about protons transmuting into neutrons in their device as the protons are being accelerated by fields. If free protons could undergo beta+ decay, especially as quickly as you imagine that they do, the LHC would never have been built.
    ——————————————————–

    And in the LHC the protons are crossing the orbit of an electron, whose radius is in the order of the Bohr radius 10^-11m ?
    And is the proton moving in a positive elecrosphere of a nucleus ????

    Tell them to improve the following in the LHC, and they will observe the decay+ of the proton no LHC:

    1- They have to create a positive electric field in the LHC accelerator (equivalent to that of the electrosphere of the Ni nucleus), within which the proton must be moving.

    2- They must put in the LHC accelerator an electron moving in an orbit with radius R= 10^-11m, perpendicular to the motion of the proton.

    3- The proton must cross the plane of the electron’s orbit in the center of the orbit.

    .

    1.A) ———————————————–
    Here is an energy comparison:

    1 MeV (1.602×10^−13 J): about twice the rest energy of an electron

    versus

    14 TeV: the designed proton collision energy at the Large Hadron Collider (which has operated at half of this energy since 30 March 2010)
    ————————————————

    In the LHC, what is replacing the electric charge of the electron 2s1, so that to induce a helical trajectory in the proton’s motion?

    The charge of the electron in the Rossi-Effect causes a disturbance in the proton’s zitterbewegung.
    In the LHC, what is causing the disturbance in the proton’s zitterbewegung?

    .

    2. ) —————————————
    You state that the proton has a helical trajectory along the z-axis due to interaction with the aether. You also state that a second helical trajectory caused by the Coulomb interaction between proton and rotating 2s1 electron is superimposed upon the proton, and that this causes a “disturbance” in the proton’s original helical trajectory. You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.
    ———————————————

    And you imagine that within a nucleus with excess of protons, one the protons captures an electron from the aether, and loses its electric field by becoming a neutron in order to avoid the “disturbance’ caused by the excess positive charge of the nucleus?

    .

    i) —————————————–
    If the proton were never to become a neutron, and the “disturbance” therefore never avoided, what would happen to the trajectory of the proton?
    ———————————————

    If a nucleus with excess protons do not capture electrons from the aether, and therefore the disturbance of the nucleis is never avoided by the positron emission, what would happen to the nucleus?

    .

    ii) ———————————————–
    If the two overlapping helical trajectories were to have no negative effect upon the successful crossing of the plane of rotation by the proton, could the strength of the Coulomb interaction between proton and 2s1 electron still prevent the proton from entering the target nucleus by pulling it back to the plane of rotation?
    —————————————————-

    it seems yes

    .

    Joe,
    do you think the acceleration of the beam of protons in the LHC did not have occurred to me earlier?
    I already had reflected about the difference between the conditions in the LHC and in the cold fusion experiments.

    You have to keep in mind that in cold fusion there are many special conditions. You cannot compare them with the ordinary conditions with which the physicists had used to deal with along a century.

    regards
    wlad

  162. Peter Forsberg

    I agree with Andrea Rossi and Robert Cuerto. The fall in the price of oil has nothing to do with ECat. I wish it did. It has more to do with world politics and international economic trends.

    But I’m concerned about the fall in the price of oil, since it makes the ECat less competitive in different ways. If this is a problem depends on how far the price will fall, and for how long it will stay low.

    Regards

    Peter

  163. Greg Leonard

    Dear AR and Robert Curto
    I believe the evidence is against your opinion.
    Fracking has been making an impact for years now. It may well have been part of the reason for the Rockefellers to divest themselves of their oil portfolio – but that did not cause a sharp slide in the oil prices.
    The current slide in oil prices is largely down to the giant hedge funds coming out of oil (I had not realised what influence they wield).
    Again, that slide started exactly when the Lugano report was leaked. This is not coincidence, it is cause and effect.
    regards
    Greg Leonard

  164. Andrea Rossi

    Greg Leonard:
    Thank you for your opinion. I do not agree.
    Warm Regards,
    A.R.

  165. Joe

    Wladimir,

    1. A free proton has never been observed to undergo beta+ decay. And by free, we mean not bound to other masses. But obviously free protons are handled by scientists with the use of electric and magnetic fields. And energies associated with these fields can be greater than E = 2*m*c^2, yet with no beta+ decay of the proton ever witnessed.

    Here is an energy comparison:

    1 MeV (1.602×10^−13 J): about twice the rest energy of an electron

    versus

    14 TeV: the designed proton collision energy at the Large Hadron Collider (which has operated at half of this energy since 30 March 2010)

    The scientists at the LHC are not complaining about protons transmuting into neutrons in their device as the protons are being accelerated by fields. If free protons could undergo beta+ decay, especially as quickly as you imagine that they do, the LHC would never have been built.

    2. You state that the proton has a helical trajectory along the z-axis due to interaction with the aether. You also state that a second helical trajectory caused by the Coulomb interaction between proton and rotating 2s1 electron is superimposed upon the proton, and that this causes a “disturbance” in the proton’s original helical trajectory. You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.

    i) If the proton were never to become a neutron, and the “disturbance” therefore never avoided, what would happen to the trajectory of the proton?

    ii) If the two overlapping helical trajectories were to have no negative effect upon the successful crossing of the plane of rotation by the proton, could the strength of the Coulomb interaction between proton and 2s1 electron still prevent the proton from entering the target nucleus by pulling it back to the plane of rotation?

    All the best,
    Joe

  166. Wladimir Guglinski

    eernie1 wrote in December 9th, 2014 at 1:38 PM

    Wlad,
    In the Mosier-Boss experiment where 2 deuterium atoms fuse to create a tritium atom plus a neutron with approx. 10Mev energy to balance the energy equation, the high energy balance is required because the nucleons are under the influence of the strong force which requires the high energy to balance the reaction.
    —————————————————–

    Eernie,
    this also does not work.
    The binding energy of the deuteron is 2,2MeV (the proton and the neutron are bound with 2,2MeV). A neutron cannot be emitted with energy higher than 2,2MeV when the tritium is formed.

    regards
    wlad

  167. Wladimir Guglinski

    eernie1 wrote in December 9th, 2014 at 1:38 PM

    Wlad,
    In the case of the Halo electron of the 3Li7, the strong force is almost completely neutral because of the distance between the neutron and the nucleon cluster. Therefor the neutron can exit the nucleus without the high energy required by the cluster nucleons and can exit the nucleus as a thermal neutron which can then easily react with the 28Ni nuclei.
    —————————————

    Eernie,
    as I said, before to cross the plane of the 2s1 orbit, the neutron has attraction with the orbit, and it is accelerated.
    After crossing the plane of the orbit, the neutron continues to have attraction, and so it is disaccelerated, and therefore it will not hit the Ni nucleus.

    Your proposal could work if the neutron would be emitted with high energy by the 3Li7 decay, and so the attraction force of the 2s1 orbit would not be able to stop the neutron.
    However the neutron exits the 3Li7 with low energy, since it is weakly bound.

    regards
    wlad

  168. Wladimir Guglinski

    Joe wrote in December 9th, 2014 at 2:12 PM

    Wladimir,

    1. ) ————————————–
    You write,
    ” [...] positive charge was favouring the electron capture by the proton.”

    It is not enough to just simply have a positive charge. Energy states, both initial and final, must be taken into account.

    From Wikipedia, Beta Decay:
    “β+ decay can occur when the mass of the initial atom exceeds that of the final atom by at least twice the mass of the electron.”

    This, of course, is referring to beta+ decay occurring INSIDE the nucleus. But outside the nucleus, beta+ decay has never been observed. The reason given, from the same article, is the following:

    “β+ decay cannot occur in an isolated proton because it requires energy due
    to the mass of the neutron being greater than the mass of the proton.”
    ——————————————————

    yes,
    however such condition do not consider that the proton is crossing the orbit of one electron.

    The energy required is E= 2.mc² , necessary to extract a pair positron-electron from the aether. Such energy can be supplied by the electromagnetic energy of the electron 2s1 moving in its orbit.

    2. ) ——————————————–
    You write,
    “And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.”

    What specifically is that “disturbance”?
    ————————————————–

    As you know, the proton moves by helical trajectory (zitterbewegung), due to the interaction with the aether.

    But suppose the proton is NOT moving with zitterbewegung along the z-axis of the nucleus Ni, going toward the direction of the 2s1 orbit. In this case, when the proton is near to the plane of the electron orbit 2s1, the circular motion of the electron applies a Coulomb attraction force on the proton, and by this way the motion of the electron produces a circular motion of the proton around the z-axis. Combined with the rectilinear motion of the proton along the z-axis, such attraction with the electron produces a helical trajectory in the motion of the proton.
    Remember that here we are considering that the proton does not move with zitterbewegung.

    But the proton has its zitterbewegung. And therefore the helical trajectory induced by the electron 2s1 on the proton causes a disturbance in the proton’s zitterbewegung.

    The best way for the proton to avoid the disturbance in its zitterbewegung is by losing its electric field, so that do not have Coulomb interaction with the electron 2s1. And the way for losing its electric field is by a capture of one electron from the aether, becoming a neutron.

    regards
    wlad

  169. ing. Michelangelo De Meo

    Dear Dr. Rossi,
    Here is another great news, the birth of a Research Institute Energie ” Energiforsk ” announced by Elforsk .

    Magnus Olofson after the publication of Lugano E-cat test report have said that there was now a good reason to launch research efforts around LENR…

    http://www.lenr-forum.com/forum/news/index.php/News/39-Elforsk-announce-the-birth-of-Energiforsk-Energie-Research-institute/

  170. Andrea Rossi

    Ing. Michelangelo De Meo:
    Another giant enters in the LENR R&D. Thank you for the information.
    Warm Regards,
    A.R.

  171. Robert Curto

    Dr. Rossi, I believe you are correct about the oil price drop.
    It has nothing to do with the E-Cat.
    It will be many years before the E-Cat has a good effect on oil prices.
    Robert Curto
    Ft. Lauderdale Florida
    USA

  172. Joe

    Wladimir,

    1. You write,
    ” [...] positive charge was favouring the electron capture by the proton.”

    It is not enough to just simply have a positive charge. Energy states, both initial and final, must be taken into account.

    From Wikipedia, Beta Decay:
    “β+ decay can occur when the mass of the initial atom exceeds that of the final atom by at least twice the mass of the electron.”

    This, of course, is referring to beta+ decay occurring INSIDE the nucleus. But outside the nucleus, beta+ decay has never been observed. The reason given, from the same article, is the following:

    “β+ decay cannot occur in an isolated proton because it requires energy due
    to the mass of the neutron being greater than the mass of the proton.”

    2. You write,
    “And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.”

    What specifically is that “disturbance”?

    All the best,
    Joe

  173. eernie1

    Wlad,
    I am curious about your complication of the Rossi effect by insisting that a neutron must decay to a proton in the 3Li7 nucleus before it could be ejected. Isn’t it more simple to assume that the lightly bound Halo neutron of the 3Li7 is made to exit the nucleus by applying an external force(electron electrosphere of 2p1 electrons created by exciting the 2s1 valence electron). In the Mosier-Boss experiment where 2 deuterium atoms fuse to create a tritium atom plus a neutron with approx. 10Mev energy to balance the energy equation, the high energy balance is required because the nucleons are under the influence of the strong force which requires the high energy to balance the reaction. In the case of the Halo electron of the 3Li7, the strong force is almost completely neutral because of the distance between the neutron and the nucleon cluster. Therefor the neutron can exit the nucleus without the high energy required by the cluster nucleons and can exit the nucleus as a thermal neutron which can then easily react with the 28Ni nuclei.
    Regards.

  174. Wladimir Guglinski

    ERRATA:

    Joe,
    in my last reply to you,
    where I wrote:

    But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the proton has no time available for capturing an electron from the aether.

    the correct is:

    But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the nucleus has no time available for capturing an electron from the electrosphere.

  175. Wladimir Guglinski

    Joe wrote in December 9th, 2014 at 1:18 AM

    Wladimir,

    Your quote states that beta+ decay occurs inside nuclei. The newborn proton that crosses the plane of rotation of 2s1 is obviously not inside a nucleus, or even an entire atom. It is in transit between two atoms, Li and Ni.
    ————————————————-

    Joe,
    because the physicists had noted that beta+ decay occurs inside nuclei does not mean that it cannot occur in the transit between Li and Ni.

    The physicists had never studied the cold fusion phenomenon, we are only in the beginning of the discoveries.

    Besides, of course the beta+ decay must occur within the nuclei IN NORMAL CONDITIONS, because the protons are within the nuclei.

    But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the proton has no time available for capturing an electron from the aether.

    We cannot apply all the rules known in the Standard Model, since cold fusion is impossible to occur from the known rules of that model.

    Note that in NORMAL CONDITIONS the 4Be7 decays in 3Li7 by electron capture (from the electrosphere), It has 53,2 days half-life.

    But there is not time available to capture the electron from the electrosphere, because the proton exits the 4Be7.

    When the proton exits the 4Be7, the 4Be7 will not capture an electron from the electrosphere (as occurs in normal decay of 4Be7), since the proton exited the nucleus, and the excess positive charge disappeared for the 4Be7.

    But the excess positive charge continues, because there is an excess of one proton, and it is moving toward the Ni nucleus.

    Therefore there is favourable conditions for the capture of the electron by the proton (from the aether), because:

    1- When the neutron transmuted to proton in the 3Li7, the electrosphere of the newborn 4Be7 became positive.
    2- Exiting the electrosphere of the 4Be7, the proton would transfer the positive charge excess for the electrosphere of the Ni.
    3- Therefore excess positive charge was favouring the electron capture by the proton. And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.

    regards
    wlad

  176. Dear Readers of Journal of Nuclear Physics,

    Meanwhile the long reviewing process, the presented here article “h-Space Theory” has been updated. The major corrections and more detailed explanations were made in the sections presenting the mechanism of gravitational attraction and the determination of electric charge of elementary particles and ions. To download the updated version of article the following link can be used:

    http://h-theory.narod.ru/h-space_theory_v2.pdf

    Best wishes,
    Valeriy Tarasov

  177. Andrea Rossi

    Christen: good idea.
    Warm Regards,
    A.R.

  178. Greg Leonard

    Dear AR,
    You say that all different power sources will be integrated, but there will be a substantial shift away from the common fossil fuels.
    We have already seen the slide in crude oil prices which started when the Lugano test report was leaked. This adjustment happened in just a few months.

    The lower oil price is causing most of the producer countries to lack the finances to run their economy as they have in the past – which will lead to considerable social upheaval.

    The lower oil price has also affected the economic argument for several wind power systems.

    We live in interesting times.

  179. Andrea Rossi

    Greg Leonard:
    I am pretty sure the shift in the oil price is due to the yield of the fracking shale gas production.
    Warm Regards,
    A.R.

  180. Wladimir Guglinski

    Subject: on the fission of the Pd nucleus in the Mosier-Boss experiment

    Dear Joe,
    what do you think about the mechanism prposed for the acceleration of the proton (and neutron), when applied for the explanation of fission prodution of the nucleus Pd in the Mosier-Boss experiment?

    As I said earlier:
    ==============================================================
    Andrea Rossi already told in the JoNP that he had tested all the sort of combinations, and he concluded that Ni is the best fuel for the eCat. For instance, he said that 46Pd is not a good element. Then probably Andrea Rossi already had tested 50Sn, and concluded that it is also not a good element to react with 3Li7.

    However it is easy to understand why Andrea Rossi did not succeed to get a god COP by using 46Pd and 50Sn combined with 3Li7. We can see it in the Fig. 1 ahead,
    FIG. 1:
    http://peswiki.com/index.php/Image:Fig._1-50Sn_as_RECEPTOR_in_eCat.png

    After the decay 3Li7 -> 4Be7, the proton of the 4Be7 is pulled by the orbit of the electron 2s1, and when the proton crosses the plane of the orbit the proton by electron capture transmutes to neutron, which is now pushed by the orbit 2s1 with a repulsion force.

    So, in the reactor using Ni-3Li7 as fuel, the final acceleration on the neutron is not so high, because the radius orbit of the electron 2s1 is not large.
    But in the reactor using Pd-3Li7 the final acceleration on the neutron is very faster, because the radius orbit of the electron 2s1 is very larger. And so the neutron trespasses the 46Pd nucleus, and is not captured.

    ==============================================================

    The neutron is not captured by the Pd nucleus, and most of them cross the nucleus, and exit it.

    But some of the neutrons hit the Pd causing its fission.

    .

    Did you read about fision in the Mosier-Boss experiment?
    ====================================================================
    III. CONCLUSIONS
    In this communication, the use of microscopic examination, sequential etching, and LET spectrum
    analysis to identify the energetic charged particles and neutrons responsible for the tracks in CR-39 detectors used in Pd/D co-deposition experiments were discussed. These analytical techniques gave complementary results .
    The energetic particles were identified as 2.45 MeV neutrons, 3-10 MeV protons, 2-15 MeV alphas, and 14.1 MeV neutrons.
    [...]
    The only known source of these long range alphas is fission reactions. This suggests that the new elements observed on the cathodes are caused by fissioning of the Pd nucleus

    http://newenergytimes.com/v2/conferences/2012/ICCF17/papers/Mosier-Boss-Its-Not-Low-Energy-Paper-ICCF17-pp.pdf
    ===============================================================

    Do you think the authors of cold fusion theories can find an acceptable theory based on the tunneling mechanism, able to explain the fission?

    regards
    wald

  181. Christen

    Dear Andrea,
    I suggest you, yo avoid misunderstandings, the use of SSM acronym for self sustain mode and SSP acronym for start stop mode of operation in the e-cat system.

  182. Joe

    Wladimir,

    Your quote states that beta+ decay occurs inside nuclei. The newborn proton that crosses the plane of rotation of 2s1 is obviously not inside a nucleus, or even an entire atom. It is in transit between two atoms, Li and Ni.

    All the best,
    Joe

  183. Achi

    Dear Mr Rossi,
    I would like to thank you for all the hard work you and your team are putting in to bring LENR to the world. It feels good knowing that such a revolutionary tech is just on the horizon.
    I would like to ask you about the different modes your ecats have.
    I know of the start stop mode where power is cycled, and self sustained mode where there is no power supplied to drive the device, and I would also assume that there is a mode where power is constantly being supplied.

    Of these modes, which do you believe the LENR effect to be the most powerful? Most able to sustain a load?
    Do you believe that the reaction changes effect with the different modes?
    Do you think that could be more than one type of reaction occuring?
    Are there other “modes” in which the ecat runs?

    I understand if you cannot answer my questions but I appreciate your reading them.

    Thank you,
    Achi

  184. Andrea Rossi

    Achi:
    Thank you for your kind consideration.
    About your questions:
    1 + 2: I will be able to answer after at least 1 year of operation of the 1 MW plant
    3: no, if the control system works properly in the specific phases
    4- I do not think so, but I could be wrong
    5- no
    Thank you for your attention,
    Warm Regards,
    A.R.

  185. Italo R.

    Dear Dr. Rossi, what do you think will happen in the sector of world energy when your 1 MW plant will demonstrate unequivocally that it is capable to generate more energy than it receives in input?
    You said many times that all forms of energy must be integrated among themselves, but I am afraid that the arrival of the E-Cat will tremendously upset the economic balances consolidated till now. Do you think that there will be this derangement, or the effect of the presence of E-Cat in the world will be soft because it will be necessarily introduced gradually?
    Kind Regards,
    Italo R.

  186. Andrea Rossi

    Italo R.:
    Beyond the “Spectre” syndrome, we must understand that all the energy sources will integrate, with a slow shift, gradual in time, toward the most convenient in specific sectors of market. This is in the interest of all, therefore History will tend to this, as always happens: History goes where the common interest is.
    Warm Regards,
    A.R.

  187. Hank Mills

    Dear Andrea,

    So, to avoid confusion and any misunderstanding, that is 2 hours of self sustained operation (constant temperature of the individual reactor) with no electrical power to the drive?

    Thanks!

  188. Andrea Rossi

    Hank Mills:
    Yes. But I must repeat that only after at least one year of regular operation we will be able to know the data of a reliable operation.
    Warm Regards,
    A.R.

  189. silvio caggia

    Dear Andrea Rossi,
    You suggested, more than a time, to read a book written by Norman D. Cook to understand the phisics behind e-cat.
    I red it.
    Are you sure to agree with the author’s ideas about LENR?
    It seems to me that Cook interprets C.F. not as Cold Fusion but as Clean Fission…
    Am I wrong?

  190. Andrea Rossi

    Silvio Caggia:
    Yes, you are.
    Warm Regards,
    A.R.

  191. Wladimir Guglinski

    Charles Jameson wrote in December 7th, 2014 at 7:06 PM

    Dear Wladimir

    Although I am a retired engineer with no experience in nuclear physics I commend you on the simplicity and clarity of your theory. Your diagrams are a big help and I think I do have a basic understanding of what you are proposing.
    —————————————————

    Dear Charles,
    we, engineers,
    we know that we cannot design a machine through mathematical equations.

    first we have to do the design of the the machine by doing a drawing its components, and by doing a drawing showing how the components work all they together.
    If the components of the machine do not work harmoniously, it makes no sense to apply the mathematics on it. First of all, we have to succed in puting all the pieces woring well toghether.
    And later we will use mathematical equations, so that to calculate the diameter of axes, gears, pistons, electric generators, relays, coils, etc, in order to be sure that the machine will not break.

    The atomic nucleus is an mechanical-electromagnetic machine. We cannot describe its working by mathematical equations without to know how the components of the machine are working within it.
    However this is precisely what the physicists are trying to do.
    They know that within the machine there are protons and neutrons. But they do not know how protons and neutrons are bound within the machine, they do not know how protons and neutrons are distributed, and how they move within there. They also do not know exactly how is produced the electric field of each proton, and the shape of this electric field.
    Nevertheless, they are sure that they can describe accurately the working of the machine by mathematical equations.

    That’s why it is impossible to explain cold fusion from the current nuclear models

    regards
    wlad

  192. Wladimir Guglinski

    Joe wrote in December 8th, 2014 at 5:57 PM

    1) —————————————
    I am talking about both the proton and electron entering through the hole of the Ni nucleus together along the z-axis.
    ——————————————

    The electron resulted of the neutron decay cannot be pulled by the orbit 2z1, because there is Coulomb repulsion between two electrons.

    .

    2) —————————————-
    Why do you say that the Coulomb barrier is keeping the proton from entering the nucleus when there is supposed to be no Coulomb barrier at the hole?
    ——————————————-

    The proton can enter through the hole of the barrier only when the z-axis of the nucleus is aligned toward the direction of a vector magnetic field , because when this occurs the nucleus stops to gyrate chaotically, and so the shape of its Coulomb force stops to be spherical.

    In normal conditions (when the conditions for cold fusion occurrence are not satisfied), the z-axis of the nucleus is not aligned toward any direction (the z-axis gyrates chaotically), and so the Coulomb barrier takes the spherical shape. In this case, the proton can enter within the nucleus only via HOT fusion.

    .

    2. ) —————————————-
    You write,
    “the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.”

    This is illogical. The usual electron capture is from the electrosphere.
    ———————————————

    Ok,
    then instead of to call it electron capture, let us call it positron emission:

    “Positron emission or beta plus decay (β+ decay) is a particular type of radioactive decay and a subtype of beta decay, in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron and an electron neutrino”
    .
    http://en.wikipedia.org/wiki/Positron_emission

    The mechamism is the following:
    1- A proton captures an electron from the aether, and is converted into a neutron.
    2- When the electron is captured from the aether, a positron is created together, and is emitted.

    In other words, an electron is captured (created) from the aether, and a positron is emitted.

    .

    3) ———————————————
    What you are describing is the creation of an electron from aether particles. This goes against conservation of charge. When an electron is created, it is always done with an accompanying positron. This is known as “pair creation”.
    ————————————————

    Yes, and while the electron is captured by the proton, the positron exits the nucleus

    regards
    wlad

  193. Hank Mills

    Dear Andrea,

    That is a huge accomplishment! Start Stop Mode is impressive, but Self Sustain Mode is phenomenal beyond measure. I hope you can answer this one question: it would be like a Chrismas present for many of us.

    In tests of the individual reactors that compose the plant, how long is the period of self sustain (in which the reactor maintains a constant or increasing temperature without input) that has been deemed to be safe for use with minimal risk of thermal runaway?

    Thank you for sharing this amazing news!

  194. Andrea Rossi

    Hank Mills:
    The longest period of ssm we got so far with the E-Cats is 2 hours, but only after the end of the test period of the 1 MW plant in the factory of the Customer we will have reliable numbers.
    Warm Regards,
    A.R.

  195. Herb Gillis

    Andrea Rossi:
    Are you presently able to select (entirely at your choice) which of the two SSM modes the reactor will be in when external drive is stopped?
    Kind Regards; HRG.

  196. Andrea Rossi

    Herb Gillis:
    This is one of the tasks of the control system.
    Warm Regards,
    A.R.

  197. Joe

    Wladimir,

    1. In one of my comments, I wrote,
    “I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns [proton and electron] would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.”

    You responded,
    “Joe,
    what you say makes no sense.

    The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.

    The electron has negative charge, and it has not repulsion with the Coulomb barrier, and so it can cross it in any point.”

    I am talking about both the proton and electron entering through the hole of the Ni nucleus together along the z-axis. Why do you say that the Coulomb barrier is keeping the proton from entering the nucleus when there is supposed to be no Coulomb barrier at the hole?

    2. You write,
    “the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.”

    This is illogical. The usual electron capture is from the electrosphere. What you are describing is the creation of an electron from aether particles. This goes against conservation of charge. When an electron is created, it is always done with an accompanying positron. This is known as “pair creation”.

    All the best,
    Joe

  198. Frank Acland

    Dear Andrea,

    Would it be more accurate to define ‘ssm’ as self-sustain mode, or start-stop mode?

    Thank you,

    Frank Acland

  199. Andrea Rossi

    Frank Acland:
    SSM is an acronym that can be used for either definition.
    Warm Regards,
    A.R.

  200. Hank Mills

    Dear Andrea,

    The acronym SMM can represent two different modes of operation: start stop mode and self sustain mode.

    Start Stop mode is when the drive is activated with electrical power, the reactor chamber is heated, the nuclear reactions are simulated, and then the input to the drive (resistances) are turned off. When the drive is turned off, the anonamalous nuclear reactions are still talking place and are contributing to the output of the device. The reactor cools, but it does not drop as quickly as it should if the fuel charge was not in place producing a diminishing quantity of excess power. However, the device as a whole is slowly dropping in temperature and the reactions are dying off until the drive is turned back on. When the drive is off, the reactions are not truly sustaining. They are present, but diminishing.

    Self Sustain Mode is when – after the reactor has been heated to operating temperature – the input to the drive is cut off. However, the reactions have been simulated to such a high level they are self sustaining with no external input. This means the temperature of the reactor is staying the same or increasing. Synonyms for the word sustaining could be: unending, constant, continual, enduring, perpetual, etc. This mode is different than start stop in which the reactions produce excess energy after the input is cut off, but drop in frequency so the reactor (if no more input was applied) would drop to room temperature. The small current drawback is that self sustain mode has a slightly increased chance of thermal run away that is not virtually zero like when the start stop mode is used.

    Please let me know if these two explanations are accurate and which mode is used in the new one megawatt plant.

    Of course, a plant using either of these methods would be far superior to any other energy source on the planet. The lower COP of a plant using Start Stop mode would still be far higher than what hot fusion scientists hope to achieve decades from now with teens of billions of dollars in funding.

  201. Andrea Rossi

    Hank Mills:
    The 1 MW plant can use either of these methods.
    Warm Regards

  202. Dave Roberson

    Dear Mr. Rossi,

    Congratulations to you for the progress that you and your team has made during the recent months. I look forward to the day when your designs become widely available and change the world energy equation in a positive manner.

    I have been following your efforts for several years and believe that I understand how your devices behave, at least to a limited extent. There are plenty of questions concerning the SSM mode of operation and I would like to explore one characteristic of that mode if I may. Is it true that you generally operate in that mode by using pulse width modulation of the drive power source, whether this source is electrical or gas energized? This would be consistent with the first third party test report. The radiated power waveform that they published indicated that the internal heating was not like a resistor load but instead demonstrated an increasing rate of temperature rise during the external heating pulse.

    After a measured application of drive power, your device is then placed into the SSM state as it begins to cool down toward the temperature at which the cycle repeats. I suspect that you refer to the period of time during which the device coasts lower in temperature as the SSM condition. Is this correct? It certainly makes a great deal of sense to call it by that name since the internal core is generating a great deal of additional heat energy during that time frame. Also, I have reason to believe that the newly generated heat power causes the rate of temperature decay to be reduced when compared to the response of a resistive load that has no extra internal power generation. The first third party report contained a graph that clearly suggests this type of activity.

    It appears that people are confused by the definition of the SSM and I hope that you will clarify that operation by answering my questions above. Perhaps after that you will be subject to fewer additional questions concerning the matter.

    Thank you for your attention and I wish you much success in your important endeavor.

  203. Andrea Rossi

    Dave Roberson:
    With ssm we mean that part of operation during which the electric power that drives the E-Cats is turned off. That’s all I can say. The cycles are regulated by the complex control system.
    Warm Regards,
    A.R.

  204. ing. Michelangelo De Meo

    Hello Dr. Rossi,
    increase scientists around the world who are studying the LERN . Reading your last report you notice that you are light years ahead of other competitors .

    http://e-catalyzer.it/e-catalyzer/fusione-fredda-aggiornamenti-sul-nanor

  205. Andrea Rossi

    ing. Michelangelo De Meo:
    Very interesting, this is clearly the development of the work of Brian Ahern, that I always said is the one that is making a good replication. This is positive, because adds evidence to the replicability of our work from third parties, even if with different methods. As for the competition, they are at the point we were 10 years ago, but it is obvious that sooner or later we will have competition. Now, thank to the enormous fight made by us, giants are entering in the battlefield: MIT, NASA, Bill Gates, Lockeed Martin etc etc. When the play gets hard, hard guys go into play, and we are not soft, believe me.
    Warm Regards,
    A.R.

  206. JonJon

    Andrea,
    During ssm (self sustained mode), the Ecat Cop is infinity as the input power is turned off, except for the control electronics?

  207. Andrea Rossi

    JonJon:
    This is a tricky question, because in Physics “infinity” is a nonsense and when an equation involves infinity it is considered wrong.
    To be correct we can say this: during the ssm the consumption of electric power from the grid of the E-Cat is zero. Under a more complex concept, there is a consumption of energy that resolves the infinity issue.
    Warm Regards,
    A.R.

  208. Wladimir Guglinski

    Robert Curto wrote in December 8th, 2014 at 1:51 AM

    Dear JR: We miss your counterpoint to what Wladimir Guglinsky writes.
    We loved your corrections based on the Standard Model and many of us miss you.
    I see that Dr. Andrea Rossi prefers to stay out of theoretical confrontations, even if he repeatedly said he stays inside the Standard Model to explain everything, so we need your counterpoint.
    ————————————————-

    Robert,
    perhaps they prefer stay out because even if a reasonable theory can be found based on the “Low Radiation Fusion Through Bound Neutron Tunneling” proposed by Dr. Oscar Gullstroem , however is very hard to explain with the same theory how alpha emission with 15MeV can be produced in Mosier-Boss experiment (and how can occur the nuclear fission of the Pd nucleus hit by particles with low energy).

    Because there is a paradox in the “Low Radiation Fusion Through Bound Neutron Tunneling” proposed by Dr. Oscar Gullstroem, as follows:

    1) He is proposing a theory for the Rossi-Effect, so that to explain how particles with low energy can surpass a high Coulomb barrier.

    2) However, in Mosier-Boss experiment many particles with very high energy are emitted (as neutrons 14MeV and alphas with 15MeV)

    3) So, the paradox is evident:
    if the solution on the puzzle on how particles with low energy can surpass a high Coulomb barrier is by proposing that low energetic bound neutron tunneling can cross it, then how can those low energy neutrons produce emissions with high energy????

    regards
    wlad

  209. Frank Acland

    Dear Andrea,

    When you say “the ssm should result to be by far the main mode of operation of the 1MW E-Cat” does this mean that for the great majority of the time the input power is turned off?

    If so, this should result in tremendous energy savings for the customer, shouldn’t it?

    Many thanks,

    Frank Acland

  210. Andrea Rossi

    Frank Acland:
    We are aiming at that.
    Warm Regards,
    A.R.

  211. orsobubu

    Buck, thanx for the marvelous article on NYtimes. It shows clearly how economic interests drive the political process in an advanced capitalistic/imperialistic economy. Most interesting of all are 1500 disillusioned readers’ comments, as they suddenly discover that democracy never existed as they imagined. The article also demonstrate how federalism is at work, absolutely not enhancing the possibilities for small communities to vote for their representatives, for citizens rights to breath in a clean environment and for endangered lesser prairie chicken to survive; instead, federalism is the best way, in a complex society, to manage the superior interests of the ruling class in spite of impossible, demagogic and idealistic central government electoral promises.

  212. Wladimir Guglinski

    Joe
    December 7th, 2014 at 8:27 PM

    Wladimir,

    1. ) ————————————
    You write,
    “The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.”

    But you said that the purpose of the “hole” was to allow positively charged particles to bypass the Coulomb barrier.

    Only yesterday you wrote,
    “The proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barrier.”

    So, can a proton not enter a nucleus through the “hole” along the z-axis?
    ——————————————————–

    yes,
    by HOT FUSION in the Sun and any engine workng via hot fusion

    2. ) ——————————————
    You write,
    “The neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.”

    i) What is the electron that is “captured”?
    ———————————————

    the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.

    ii) —————————————
    What is the mechanism that sets the intrinsic spin, and therefore the intrinsic magnetic moment, of the newborn neutron AGAINST the induced magnetic moment of the rotating 2s1 electron in order that the neutron be repelled into the target nucleus?
    ———————————————–

    http://peswiki.com/index.php/Image:Why_neutron_is_repelled_by_the_electron_orbit.png

    regards
    wlad

  213. Robert Curto

    Dear JR: We miss your counterpoint to what Wladimir Guglinsky writes.
    We loved your corrections based on the Standard Model and many of us miss you.
    I see that Dr. Andrea Rossi prefers to stay out of theoretical confrontations, even if he repeatedly said he stays inside the Standard Model to explain everything, so we need your counterpoint.
    Robert Curto
    Ft. Lauderdale Florida
    USA

  214. Tom Conover

    Andrea,

    On November 4th, 2014 you replied to Bob that the maximum temperature an operating e-cat can produce is at peak 1,400°C, and that you had achieved that temperature in e-cat operation. Does that mean that the regular e-cat 1MW masterpiece can perhaps easily achieve temps higher than 1,000°C during normal operation?

    Is the peak temperature still 1,400°C for the e-cat 1MW plant?

    Thank you!

    Tom Conover

  215. Andrea Rossi

    Tom Conover:
    The 1 MW E-Cat is a low temperature plant, it reaches max T= 120°C.
    What I wrote regarding the high temperatures is related to the Hot Cat.
    Warm Regards,
    A.R.

  216. Eric Ashworth

    Readers of the JONP. This is how I see the coulomb barrier enigma. The barrier is formed of 90 degree angles of various field resistance from zero to an achievable maximum. Structure is cubic, formed of four pyramids of a horizontal plane, apex inner most and two pyramids apex inner most on a vertical plane. This represents the geometric layout not the actual.

    These four pyramids have to be set one above another on a vertical plane in a spiral format. Each pyramid occupies a quadrant of a quadrangle and ascends a pyramid structure from its base. Thereby divide a pyramid into four levels. At the base of the top level in one of the quadrants mark (p) for proton zone. At the base of the level beneath but advanced by 90 degrees in a quadrant marked (n) for neutron zone. At the base of the level beneath but advanced by 90 degrees in a quadrant mark (e) for the electron zone and finally at the base of the level beneath mark (a) for aether/electrosphere zone. Thereby to ascend the pyramid which is of a cubic dimension being of four horizontal pyramids you enter at the base level of the base pyramid and ascend on a spiral of four 90 degree angles to get to the next level. To reach the apex of the pyramid from its base a total of sixteen 90 degree angles have to be completed.

    Every 360 degrees of ascent requires a vertical descent back to the negative zone being a south pole of the cube. Upon exiting the charge will ascend upon the exterior and enter the cube at the next level above at the base of the above pyramid and repeat the process until the apex is reached. Consequently, when the circuit is completed the charge will enter the apex of the pyramid being the absolute north gate and exit at the absolute south gate being in a central position of the quadrangle. A positive charge represents a compressed negative and a negative charge represents an expanded positive. Four nickel isotopes represent four horizontal pyramids that in combination represent a cubic dimension of nickel.

    The pyramids revolve to remain neutral but the coulomb barrier is not solid with regards a 360 degree coverage. It may appear to be so due to r.p.ms. of the pyramid. Because of inertia caused by a moving economy flow system which is within every structure/charge, the charge is able to escape from its structure being in an accelerated state as an absolute with regards its charge potential. Thereby, the base of the preceeding pyramid is able to capturte, increase it in charge potential and accelerate it in velocity. Upon negotiating the macro cube of four isotopes with no further gate to cross, transmutation of the fourth isotope takes place.

    Between the apex of one pyramid and the base of a preceeding one is a gate/empty space/almost empty and thereby there are minor gates within a cube and major gates between cubes. Thereby for a charge to complete six circuits of a cube being of six pyramids the seventh circuit requires a higher level of a unique vibration that requires a transmutation of one pyramid into the base of the preceeding cube. Out of one cube into a higher dimension which entails the first pyramid/apex of the lower cube becoming the last/base pyramid of the higher cube and untimately the last becomes the first. Such is evolution. If the geometric activity is as such, then to line up more so the horizontal pyramids would aid in the reaction. To introduce hydrogen into the atmosphere could in theory bring a state of chaos into the activity that would result in the production of heat which would contribute to the process by producing more overall activity. Transmutation, I believe, is the major event because without it evolution would not exist and thereby purpose would be a none factor. I put this as a consideration. Regards Eric Ashworth.

  217. Buck

    Andrea,

    about three weeks ago, you responded to a news article in the (Persian) Gulf News describing LENR phenomena as fact and LENR technology as imminent, stating that “now the competition is very serious.”
    link>> http://gulfnews.com/news/world/india/indian-government-urged-to-revive-cold-fusion-1.1413814

    Would you include the following article from the New York Times “Energy Firms in Secretive Alliance with Attorneys General” as another example of “now the competition is very serious”?
    link>> http://www.nytimes.com/2014/12/07/us/politics/energy-firms-in-secretive-alliance-with-attorneys-general.html?hp&action=click&pgtype=Homepage&module=first-column-region&_r=0

  218. Andrea Rossi

    Buck:
    As I always said, LENR must integrate in the complex world of energy sources.
    Warm Regards,
    A.R.

  219. Joe

    Wladimir,

    1. You write,
    “The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.”

    But you said that the purpose of the “hole” was to allow positively charged particles to bypass the Coulomb barrier.

    Only yesterday you wrote,
    “The proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barrier.”

    So, can a proton not enter a nucleus through the “hole” along the z-axis?

    2. You write,
    “The neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.”

    i) What is the electron that is “captured”?

    ii) What is the mechanism that sets the intrinsic spin, and therefore the intrinsic magnetic moment, of the newborn neutron AGAINST the induced magnetic moment of the rotating 2s1 electron in order that the neutron be repelled into the target nucleus?

    All the best,
    Joe

  220. Wladimir Guglinski

    Charles Jameson wrote in December 7th, 2014 at 7:06 PM

    Dear Wladimir

    Your focus has been on the interaction between Ni and Li but Rossi and others have said that hydrogen is a key element. If I understand correctly your theory does not require it to explain the Rossi Effect.
    —————————————————

    Charles,
    you did understand correctly.

    regards
    wlad

  221. Wladimir Guglinski

    Andrea Rossi wrote in December 7th, 2014 at 6:44 PM

    JC Renoir:
    I am also taking in strong account the 2 papers published in November by Dr Oscar Gullstroem on Ecatworld.
    ————————————————

    I would suggest to Dr. Oscar Gullstroem to apply his theory on “Low Radiation Fusion Through Bound Neutron Tunneling” so that to explain how neutrons with 14MeV , protons with 10MeV, and alphas with 15MeV, can be produced in the Mosier-Boss experiment, since the energy available is only 2,2MeV (the binding energy of the deuteron used in the experiment for the cold fusion Pd-1H2):

    III. CONCLUSIONS
    In this communication, the use of microscopic examination, sequential etching, and LET spectrum
    analysis to identify the energetic charged particles and neutrons responsible for the tracks in CR-39 detectors used in Pd/D co-deposition experiments were discussed. These analytical techniques gave complementary results .
    The energetic particles were identified as 2.45 MeV neutrons, 3-10 MeV protons, 2-15 MeV alphas, and 14.1 MeV neutrons.
    [...]
    The only known source of these long range alphas is fission reactions. This suggests that the new elements observed on the cathodes are caused by fissioning of the Pd nucleus
    .

    http://newenergytimes.com/v2/conferences/2012/ICCF17/papers/Mosier-Boss-Its-Not-Low-Energy-Paper-ICCF17-pp.pdf

    The question is:
    how protons (or neutrons) with low energy 2,2MeV (resulted from the fusion 1H2-1H2) can be able to produce the fission of the Pd nucleus ?

    I can see only one answer:
    the proton produced by the 1H2-1H2 fusion is accelerated, it decays in a neutron which is also accelerated until to get 10MeV, and it hits the Pd nucleus, causing its fission (the neutron is not absorbed by the Pd nucleus, as the neutron is aborbed by the Ni nucleus in the Rossi-Effect, because the energy of the neutron in the Mosier-Boss experiment is too much high for its absorption by the Pd nucleus)

    But perhaps I am wrong.
    And then the question is to know how Dr Oscar Gullstroem can explain it with the theory of the Fusion Through Bound Neutron Tunneling.

    regards
    wlad

  222. Charles Jameson

    Dear Wladimir

    Although I am a retired engineer with no experience in nuclear physics I commend you on the simplicity and clarity of your theory. Your diagrams are a big help and I think I do have a basic understanding of what you are proposing. I do have a question though. In your theory explaining the Rossi Effect what part does hydrogen play. Your focus has been on the interaction between Ni and Li but Rossi and others have said that hydrogen is a key element. If I understand correctly your theory does not require it to explain the Rossi Effect.

    Thanks
    Charles Jameson

  223. Curiosone

    In the 1MW plant operating in the factory of Industrial Heat’s customer will be the self sustaining mode important, or it will not be used, as in the Lugano test of the Hot Cat?
    W.G.

  224. Andrea Rossi

    Curiosone:
    The ssm should result to be by far the main mode of operation of the 1MW E-Cat. This thanks to the new control system and this also is the reason of most of the difficulties we have to overcome. Honestly, our Team is making a masterpiece, among many difficulties.
    Warm Regards,
    A.R.

  225. JCRenoir

    Which books are you helping with to study the theory behind the so called Rossi Effect?
    Thank you,
    JC Renoir

  226. Andrea Rossi

    JC Renoir:
    I am perusing the books of Prof. Nornam Cook “Models of the Atomic Nucleus” ( Springer, 2nd edition 2010) and “Nuclear Models” of Prof. Walter Greiner and Prof. Joachim Maruhn (Springer 1996).
    I am also taking in strong account the 2 papers published respectively on October 25 and on November 18 2014 by Carl-Oscar Gullstroem on Ecatworld:
    ” Low radiation fusion through bound neutron tunnelling” and ” Collective neutron reduction model for neutron transfer reaction”.
    Warm Regards,
    A.R.

  227. Wladimir Guglinski

    ERRATA:

    In my previous comment, where it written:

    In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Mg.

    the correct is:

    In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Hg.

  228. Wladimir Guglinski

    Herb Gillis wrote in December 6th, 2014 at 11:44 AM

    2) —————————————–
    I don’t know what the emitter is. There is no obvious source of hydrogen or deuterium in a fluorescent bulb. Possibly it might be the phosphors used. These are typically rare earth metal iodides (such as those of yttrium or europium). There may be other things in the bulbs as well. Certainly there is glass, and argon.
    ———————————————

    Herb,
    there are other ways to produce cold fusion, without the need of emitters.
    For instance:

    a) fusion p+e -> n.
    This occurs in the Conte-Pieralice experiment (a different version of the Don Borghi experiment)
    In their paper published in Infinite Energy Magazine in 1999, they say:

    “During the electrolytic process, a beam of electrons was forced to hit the aluminium cathode, and so induced a direct interaction of the protons and the electrons. A radioactive source of 90Sr + 90Y was utilized, having activity about 2.39 KBq (overall uncertainty  10%). The radioactive source was a disk having a diameter of 20 mm with implanted 90Sr + 90Y by electrodeposition. It was fixed external to the cell and the electron beam was collimated in order to hit directly the surface of the aluminium cathode with a geometry estimated to be 2 at a good approximation.

    We should add that the aim of the experiment was to measure the possible synthesis of a proton and an electron into a neutron. It was not our purpose to effect calorimetric measurements. It was an unexpected result that we discovered that the aluminium cathode melted in the water during the experiment.
    All previous calculations that we performed, accounting for the radioactive source at such low activity, excluded the possibility of the melting of the cathode. The aluminium cathode melted only when we used the radioactive source and not in the course of the same experiment but without the beta source.
    We were not equiped to follow and measure the temperature rise in the cell. In a repetition of the experiment we are considering measuring parameters to investigate the melting of the cathode.”

    b) Other possibility is: a coherent flux of protons (hydrogen) moving along the magnetic vector of a magnetic field target a nucleus aligned by the magnetic field. A proton can enter within a nucleus by the “hole” in the Coulomb barrier of the nucleus.

    In the case of the flourescent lamps, perhaps the emitter can be due to impurities existing in the Mg.

    regards
    wlad

  229. Andrew1

    Hello Dr. Rossi, I’ve just seen the book trailer of your biography on the Vessela Nikolova’s blog: http://www.ecat-thenewfire.com/blog/book-trailer-ecat-thenewfire/. The music is worthy of your discovery! Best wishes, Andrew

  230. Andrea Rossi

    Andrew1:
    Good luck to Vessela Nikolova’s book; you are right, the music is worth listening.
    Warm Regards,
    A.R.

  231. Wladimir Guglinski

    Joe
    December 7th, 2014 at 1:45 AM

    Wladimir,

    1. ) —————————————–
    You write,
    “Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.”

    As I said in a previous post, this could happen, but it would be a very rare event due to the very precise direction needed for the initial velocity of the newborn proton.
    ———————————————–

    Joe,
    in the decay of the neutron the proton is emitted along the z-axis of the 3Li7, and it is easy to undestand why.
    Look at for instance the structure of the 3Li7:
    FIG. 1:
    http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png

    We have:

    1- The electron gyrates about the proton in the structure of the neutron
    2- When the neutron starts to decay, the radius orbit of the electron moving about the proton starts to increase.
    3- Note in the Fig. 1 above that the orbit of the electron is perpendicular to the z-axis of the 3Li7.
    4- So, the magnetic moment of the electron’s orbit is growing up
    5- Then the proton is emitted along the z-axis, because the proton is forced by the Lorentz force to take the direction perpendicular to the electron’s orbit, which coincides with the z-axis of the 3Li7.
    This is shown in the Figure 2:
    FIG. 2:
    http://peswiki.com/index.php/Image:Fig._2-proton_direction_forced_by_Lorentz_force.png

    .

    2. ) —————————————
    You write,
    “This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.”

    I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.
    ————————————————–

    Joe,
    what you say makes no sense.

    The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.

    The electron has negative charge, and it has not repulsion with the Coulomb barrier, and so it can cross it in any point.

    regards
    wlad

  232. Wladimir Guglinski

    Subject: use of 50Sn as receptor, so that to get COP higher than 3

    Dear Joe,
    50Sn has ten stable isotopes, while 28Ni has only five. So, if one succeeds to get cold fusion with 50Sn, while 28Ni gets a COP of 3 the 50Sn will supply a COP very higher than 3.

    Andrea Rossi already told in the JoNP that he had tested all the sort of combinations, and he concluded that Ni is the best fuel for the eCat. For instance, he said that 46Pd is not a good element. Then probably Andrea Rossi already had tested 50Sn, and concluded that it is also not a good element to react with 3Li7.

    However it is easy to understand why Andrea Rossi did not succeed to get a god COP by using 46Pd and 50Sn combined with 3Li7. We can see it in the Fig. 1 ahead,
    FIG. 1:
    http://peswiki.com/index.php/Image:Fig._1-50Sn_as_RECEPTOR_in_eCat.png

    After the decay 3Li7 -> 4Be7, the proton of the 4Be7 is pulled by the orbit of the electron 2s1, and when the proton crosses the plane of the orbit the proton by electron capture transmutes to neutron, which is now pushed by the orbit 2s1 with a repulsion force.

    So, in the reactor using Ni-3Li7 as fuel, the final acceleration on the neutron is not so high, because the radius orbit of the electron 2s1 is not large.
    But in the reactor using Pd-3Li7 the final acceleration on the neutron is very faster, because the radius orbit of the electron 2s1 is very larger. And so the neutron trespasses the 46Pd nucleus, and is not captured.
    The situation is worst in the case of a fuel composed by Sn-3Li7, because the final velocity of the neutron will be very larger than it was occurred by using as fuel Pd-3Li7.
    That’s why Andrea Rossi did not succeed to get a good COP by using 46Pd.

    However, the acceleration on the proton by the orbit 2s1 must be very higher than the acceleration on the neutron, because:

    A) The proton is accelerated by magnetic interaction, It has magnetic moment 2,793, and also by Coulomb attraction with the electron 2s1.

    B) But the neutron is accelerated only by the magnetic interaction, and its magnetic moment is only 1,913, and it is not pushed by Coulomb repulsion

    Therefore, if one discovers a suitable emitter to be used with the 50Sn, maybe he can succeed to use 50Sn as a receptor, and the COP will be very high than the COP 3 obtained by Ni-3Li7.

    The emitter to be used with 50Sn must have a pair number of protons, in order to have a have a second orbit parallel to the the orbit 2s1, as shown in the Figure 2.
    FIG. 2:
    http://peswiki.com/index.php/Image:Fig._2-50Sn_as_RECEPTOR_in_eCat.png

    Let us analyse what can be the best candidates to be the emitter:

    4Be9 as emitter for 50Sn :
    Abundance: 100%
    With the decay of the neutron, 4Be9 transmutes to 5B9.
    The half-life of 5B9 is 800×10-21 s, and it transmutes to 4Be8 by emtting a proton. 4Be8 is unstable, and decays emitting two 2He4.
    The structures of 4Be9 and 4B10 are shown in the Figure 3.
    FIG. 3:
    http://peswiki.com/index.php/Image:Fig._3-structures_of_4Be9_and_5B9.png

    After the decay of the 5B9, the proton exits the nucleus along the z-axis, and goes to target the 50Sn nucleus, accelerated by the resultant of magnetic forces due to the orbits 2s1 and 2s2.

    As the neutron in the 4Be9 is strongly bound than the neutron in the 3Li7, then a higher oscillatory electromagnetic field must be applied than that used in the Rossi’s Ni-3Li7 fusion.

    .

    6C13 as emitter for 50Sn:
    Abundance: 1,1% . There is need to enrich the fuel with C13 isotopes.

    The structure of excited 6C13 is shown in the Figure 4.
    FIG. 4:
    http://peswiki.com/index.php/Image:Fig._4-structures_of_EXCITED_6C13.png

    The structures of the 6C13 and the 7N13 are shown in the Figure 5:
    FIG. 5:
    http://peswiki.com/index.php/Image:Fig._5-structures_of_6C13_and_7N13.png

    After the decay of excited 6C13 to 7N13 by the decay of the neutron to proton. In normal conditions, the 7N13 has half-life of 9,9 minutes, and it transmutes to 6C13 by beta decay. But before the decay the proton can be captured by the orbit of the electron 1p1, and goes to target the 50Sn nucleus.

    .

    8.O17 as emitter for 50Sn:
    Abundance: 0,2%

    .

    10Ne21 as emitter for 50Sn:
    Abundance: 0,27%

    .

    12Mg25 as emitter for 50Sn:
    Abundance: 10%

    Magnesium 12Mg24 has a structure similar to carbon 6C12 (24Mg = 12C + one complete hexagonal floor).
    Figure 6 shows how 12Mg24 changes its structure when it captures a neutron and it is formed the 12Mg25.
    FIG. 6:
    http://peswiki.com/index.php/Image:Fig._6-structures_of_12Mg24_and_12Mg25.png

    In 12Mg25 the neutron is bound to the deuteron D-2 via spin-interaction in a similar way as the neutron is bound to the deuteron in the 3Li7.

    With the decay of the neutron, the newborn 13Al25 has half-life of 7,18s , and in normal conditions it decays to 25Mg by beta decay.

    So, before 13Al25 to decay, its proton will be extracted by the resultant of the magnetic force due to the orbits 1p5 and 1p6, and the proton goes to target the 50Sn nucleus.

    .

    Joe,
    I think would be of interest to perform experiments, so that to verify if by using 50Sn a higher COP than 3 can be obtained, if the suitable emitter is used by suitable conditions of excitation of the emitter.

    What do you think about?

    Regards
    wlad

  233. Wladimir Guglinski

    eernie1 wrote in December 6th, 2014 at 5:40 PM

    Wlad,
    That is my point. In a normal molecular configuration where the valence electron remains in its 2S1 energy state, no neutron emission is observed. In a LENR reaction, the valence electron is driven into a higher energy state where it can influence the removal of the loose neutron in the 3Li7 nucleus by its more energetic electrosphere. But since the orbit would normally be larger and the influence less strong because of the 1/r^2 rule, we must reduce the radius of the orbit to retain the higher interactive strength.
    —————————————–

    Dear Eernie,
    I and Joe are concluding that there is not emission of neutron in the Rossi-Effect.
    Our conclusion is that the neutron decays in a proton, and the proton exits the nucleus.

    Our conclusio is because if the neutron would exit the 3Li7, there is no way to explain the 10MeV neutrons emitted in the Pamela Mosier-Boss experiment from the fusion 1H2 +1H2 -> 1H3.

    In the Rossi-Effect, the neutron is also accelerated , but with energy lower than 10MeV, because Ni has 28 protons, while Pd has 46 protons, and so the 2s1 orbit in the Pd nucleus has larger radius than in the Ni nucleus. That’s why in Rossi-Effect the neutron is captured by the Ni, while in Mosier-Boss experiment the neutron is NOT captured by the Pd nucleus, because the neutron is moving too must faster than the maximum speed which allows its capture.
    The neutron acceleration is impossible if it exits the 3Li7 in the form of a neutron, because:

    1- when the neutron exits the 3Li7, it is accelerated due to magnetic attraction with the magnetic field induced by the 2s1 orbit.

    2- but after crossing the plane of the 2s1 orbit, the neutron must be attracted by the orbit, and so it will be stoped before to target the nucleus Ni.

    That’s why now Joe and me are considering that the neutron decays in a proton in the 3Li7, the proton is accelerated, and after to cross the 2s1 orbit the proton transmutes to a neutron via electron capture, and the neutron is repelled by the 2s1 orbit, and so the neutron experiences acceleration after to cross the 2s1 orbit.

    So, the neutron can target the Ni nucleus only if it decays in a proton before to exit the 3Li7 in the Rossi-Effect.

    regards
    wlad

  234. Joe

    Wladimir,

    1. You write,
    “Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.”

    As I said in a previous post, this could happen, but it would be a very rare event due to the very precise direction needed for the initial velocity of the newborn proton. The reason for this is due to the Lorentz force which contains velocity, v, as a factor. You must not simply look at the fact that 2s1 might induce a magnetic moment and imagine that a CHARGED particle would be automatically pulled to it. For example, if the initial velocity of the proton were parallel to the plane of rotation of 2s1, it would rotate itself in its own plane and never travel along the z-axis. Here are two figures explaining this:

    https://en.wikipedia.org/wiki/File:Lorentz_force.svg

    https://en.wikipedia.org/wiki/File:Cyclotron_motion.jpg

    2. You write,
    “This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.”

    I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.

    All the best,
    Joe

  235. Wladimir Guglinski

    Joe wrote in December 6th, 2014 at 4:12 PM

    1) —————————————
    If the newborn proton is immediately affected by 2s1, then no 4Be7 could ever be formed. The proton’s initial velocity would ensure the presence of a Lorentz force acting upon the proton, keeping the proton from ever settling in a gravitational flux n(o) and thereby preventing the formation of a 4Be7. (The role of the other electrons is not important since there would be cancellation of effect due to symmetry.)
    ——————————————————–

    Joe,
    the proton has charge, and the unique way it can take so that to exit the Coulomb barrier around the newborn 4Be7 is by passing through the “hole” in the Coulomb barier.
    Therefore, the proton will exit the 4Be7 moving about the z-azis direction.

    This also happens in the alpha decay of 92U238. The experiemnts have shown that 2He4 is emitted in a radial line, i.e. along the z-axis which passes by the center of the nucleus 92U.

    The 2He4 exits the 92U by the “hole” in the Coulomb barrier because its energy is 4MeV, and the Coulomb barier has 8MeV, and so the 2He4 can exist the 92U only trhough the z-axis.

    This happens in all decay of all nuclei when a particle with charge is emitted, as protons and alpha particle.

    Theferore, when the proton exits the 4Be7, it moves along the z-axis, going perpendicular to the plane of the 2s1 orbit, and it hits the center of the orbit.

    regards
    wlad

  236. eernie1

    Wlad,
    That is my point. In a normal molecular configuration where the valence electron remains in its 2S1 energy state, no neutron emission is observed. In a LENR reaction, the valence electron is driven into a higher energy state where it can influence the removal of the loose neutron in the 3Li7 nucleus by its more energetic electrosphere. But since the orbit would normally be larger and the influence less strong because of the 1/r^2 rule, we must reduce the radius of the orbit to retain the higher interactive strength.
    Regards.

  237. Wladimir Guglinski

    eernie1 wrote in December 6th, 2014 at 11:27 AM

    Wlad,
    The normal valence electron of the 3Li7 is a 2s1 electron. However in its normal state within the molecules it is associated with there is no neutron emitted from the nucleus of the 3Li7. I think the bound electrons must be excited in some manner up to at least the 2p1 energy level before the force of the electrosphere can get involved in dislodging the Halo neutron of the 3Li7 nucleus. The orbit of the electron must also shrink due to an external negative field(perhaps from a negative Hydrogen ion)applied to the exterior of the atomic electron configuration.
    ———————————————-

    not necessarily, Eernie

    within the molecules, the nuclei of the atoms are not aligned towards the z-axis as happens in the cold fusion reactors.

    So, in the molecules the electric field of the atoms is spherical, there is not any electron orbit perpendicular to the z-axis, and therefore the neutron and the deuterons of the 3Li7 are not submitted to an oscillation along any specific direction, as occurs in the cold fusion reactors, where they get oscillation toward the z-axis.

    We have to take caution, in order do not get wrong conclusions by the comparison between the behavior of the atoms in normal conditions and the special conditions existing within a cold fusion reactor.
    regards
    wlad

  238. M. Smith

    “Quite recently, a sharp increase of the probability of fusion of various elements was found in accelerator experiments for the cases when the target particles are either imbedded in a metal crystal or are a part of the conducting crystal”
    This is an interesting finding, as well as the common technique in LENR devices of using high surface area powders or sponges, as it amplifies findings in a very different area of “wierd science” called torsion research. Shpilman and other torsion scientists in Russia found that some anisotropic magnetic substances like ferrite (or nickel powder perhaps?) emit “torsion” waves under electrical, magnetic, or rotational stresses. These are radio waves with very odd characteristics, like the capability of promoting transmutation. Sue Benford credibly validated Shpilmans torsion devices as having transmuting effects on silver halide films.
    http://www.journaloftheoretics.com/articles/4-1/Benford-axion.htm
    Radio waves are already used in NMR spectroscopy, along with magnetic fields (which cause precession of the nucleic spins) to alter spin states of nuclei.
    http://en.wikipedia.org/wiki/Nuclear_magnetic_resonance
    It isn’t that much of a stretch, at least to me, to assume that the radio waves that nuclei emit could have additional vectors which affect the strong or weak forces.
    Also, you have the wonderful coincidence of anisotropic (layered, discontinuous) magnetic or diamagnetic substances with the Orgone devices of Wilhelm Reich and others.

  239. Joe

    Wladimir,

    We are not dealing with natural circumstances whereby the 3Li7 may become 4Be7 after a certain amount of time. We are dealing with a technology (E-Cat) that might be artificially decaying the neutron much faster than natural and then having the newborn proton pulled toward a likewise artificially placed 2s1 electron rotation. In this scenario, no 4Be7 could ever be formed since the newborn proton would not naturally place itself in the gravitational flux n(o) as you have pictured it. The way that you imagine it, the proton would behave the same independently of the presence or absence of a rotating 2s1 electron. This is illogical. There is contradiction even between your statements:

    “1- The orbit of 2s1 is not able to extract the proton from the newborn 4Be7.”

    “4- [...] the orbit of the 2s1 electron is pulling the proton, and so the proton exists the 4Be7.”

    If the newborn proton is immediately affected by 2s1, then no 4Be7 could ever be formed. The proton’s initial velocity would ensure the presence of a Lorentz force acting upon the proton, keeping the proton from ever settling in a gravitational flux n(o) and thereby preventing the formation of a 4Be7. (The role of the other electrons is not important since there would be cancellation of effect due to symmetry.)

    All the best,
    Joe

  240. Herb Gillis

    Wladimir Guglinsky:
    In response to your two questions:

    1) Mercury transmutation in CFL bulbs is discussed at the link below, with references to original analytical paper (from 2013):

    http://www.e-catworld.com/2014/12/05/compact-flourescent-lightbulbs-we-are-surrounded-by-lenr-in-our-own-homes-gordon-docherty/

    2) I don’t know what the emitter is. There is no obvious source of hydrogen or deuterium in a fluorescent bulb. Possibly it might be the phosphors used. These are typically rare earth metal iodides (such as those of yttrium or europium). There may be other things in the bulbs as well. Certainly there is glass, and argon.

    All this raises some further questions:
    - What if one added H2 to one of these bulbs?
    - What if nickel and lithium vapors were added?
    - What about a combination of Ni, Li, and H2 vapors:
    – with the original mercury vapor,
    – or without the original mercury vapor?

    Regards; HRG.

  241. eernie1

    Wlad,
    The normal valence electron of the 3Li7 is a 2s1 electron. However in its normal state within the molecules it is associated with there is no neutron emitted from the nucleus of the 3Li7. I think the bound electrons must be excited in some manner up to at least the 2p1 energy level before the force of the electrosphere can get involved in dislodging the Halo neutron of the 3Li7 nucleus. The orbit of the electron must also shrink due to an external negative field(perhaps from a negative Hydrogen ion)applied to the exterior of the atomic electron configuration.
    Regards.

  242. Wladimir Guglinski

    Dear Joe,
    in my theory for the explanation on the Rossi-Effect the ORBITS of the electrons (in both Ni and 3Li7) have a SYMMETRICAL distribution about the z-axes of both Ni and 3Li7.

    Neverthelless,
    you are exhibiting many resctrictions against my theory.

    I am not criticizing you. You actually are doing a good work, because we need to eliminate all the restrictions regarding the theory.

    However, think about the other theories, as those proposed by Stoyan Sarg, Widom-Larsen, Peter Hagelstein, Fedir Mykhaylov, etc.

    From their theories it is simply impossible to explain cold fusion.
    As it is impossible to explain cold fusion from any theory where the Coulomb barrier is spherical

    Even by Stoyan Sarg theory is not possible to explain cold fusion, because:

    a) his Coulomb barrier is non-spherical, however there is not a “hole” in the barrier.

    b) the orbits of the electrons will not have a symmetrical distribution about the z-axis.

    What do you think about ?

    Do you think is possible to explain cold fusion by considering a spherical Coulomb barrier?… as are trying Widom-Larsen, Peter Hagelstein, Fedir Mykhaylov, and others. …

    regards
    wlad

  243. Wladimir Guglinski

    Dear Herb,
    also remember that in order to have cold fusion there is to have a receptor and an emitter.

    In Fleischmann-Pons cold fusion the emitter is supplied by heavy water, which has deuterons. Their experiment does not work with hydrogen

    In the Rossi-Effect, the emitter is 3Li7.

    What would be the emitter in the mercury bulbs?

    regards
    wlad

  244. Wladimir Guglinski

    Herb Gillis wrote in December 5th, 2014 at 9:10 PM

    Wladimir Guglinsky:
    Can your theory explain the apparent transmutation of mercury in fluorescent bulbs? The evidence for transmutation looks pretty strong, and it seems to be a neutron transfer process- – very reminiscent of what was observed in Lugano report (with nickel and lithium).
    If the two processes are indeed similar then it would seem to imply that LENR is not necessarily a solid state phenomenon. Perhaps its time to start looking at ionic liquids??
    ——————————————————-

    Herb,
    I did not know this phenomenon on the transmutation of mercury.
    Is there an analysis of the elements before and after a buld beging to work ?

    regards
    wlad

  245. Wladimir Guglinski

    Joe wrote in December 5th, 2014 at 8:35 PM
    Wladimir,

    1. ) —————————————————
    You have created a further alteration to your model by the introduction of 4Be7. Why would the newly born proton settle to help create a 4Be7 when a 2s1 is acting on it and disrupting any potential creation of a 4Be7?
    ——————————————————–

    I did not understand your question.
    The proton does not help to create the 4Be7.

    The 4Be7 is resulted from the decay of the neutron to proton in the 3Li7.
    The neutron decays because:
    a) it is weakly bound to the deuteron via spin-interaction
    b) the deuteron has oscillation in the 3Li7, due to repulsion with the central 2He4
    c) the oscillatory electromagnetic field applied within the eCat causes the decay of the neutron
    5) So, the 4Be7 is formed due to the decay of the neutron

    The situation is:

    1- The orbit of 2s1 is not able to extract the proton from the newborn 4Be7.
    2- Due to repulsion between the proton and the deuteron in the newborn 4Be7, the proton and the deuteron take the places shown in the figure:
    FIG. 1:
    http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png
    3- In the structure of the 4Be7, the proton begins to oscillate aling the z-axis, due to the acton of the Coulomb attraction with the electrons 1×1 and 1s2:
    FIG. 2:
    http://peswiki.com/index.php/Image:Oscilation_of_the_proton_in_4Be7.png
    4- Finally the amplitude of the oscillation increases to much, because the orbit of the 2s1 electron is pulling the proton, and so the proton exists the 4Be7.

    2. ) ——————————————
    When a neutron decays, the resultant particles separate from each other, so there must be an initial nonzero velocity for the proton. Therefore , the Lorentz force does have its role to play and must be carefully determined in order to ascertain the viability of your model.
    ————————————————

    Yes,
    but with that initial velocity the proton is not able to exit the newborn 4Be7 (otherwise, the 4Be7 would have to decay in fraction of seconds in normal conditions, however it has half-life 53,2 days).

    In the 3Li7, the deuteron-neutron bound via spin-interaction are kept within the 3Li7 thanks to the following equilibrium:
    1- The centrifugal force actuates on the deuteron-neutron
    2- The magnetic force of the proton avoids the deuteron-neutron to be expelled from the 3Li7
    3- When the neutron decays in a proton, due to repulsion the proton and the deuteron take those symmetric positions regarding to the central 2He4.
    4- The centripetal force on the newborn proton than when it was acting on the deuteron-neutron. That’s why the orbit of the proton decreases, and the structure of the 4Be7 gets the equilibrium shown in the Fig. 1.
    5- The newborn 4Be7 has a half-live of 53,2 days, and therefore that initial velocity of the proton due to the decay of the neutron is lost (in the form of heat).
    6- But as already explained, the action of the electrons 1s1 and 1s2 together with the orbit of 2s1 apply an oscilaltory motion on the proton about the z-axis, the amplitude increases, and finally in fraction of seconds the proton exits the 4Be7.

    regards
    wlad

  246. Herb Gillis

    Wladimir Guglinsky:
    Can your theory explain the apparent transmutation of mercury in fluorescent bulbs? The evidence for transmutation looks pretty strong, and it seems to be a neutron transfer process- – very reminiscent of what was observed in Lugano report (with nickel and lithium).
    If the two processes are indeed similar then it would seem to imply that LENR is not necessarily a solid state phenomenon. Perhaps its time to start looking at ionic liquids??
    Regards; HRG.

  247. Joe

    Wladimir,

    1. You have created a further alteration to your model by the introduction of 4Be7. Why would the newly born proton settle to help create a 4Be7 when a 2s1 is acting on it and disrupting any potential creation of a 4Be7?

    2. When a neutron decays, the resultant particles separate from each other, so there must be an initial nonzero velocity for the proton. Therefore , the Lorentz force does have its role to play and must be carefully determined in order to ascertain the viability of your model.

    All the best,
    Joe

  248. Robert Curto

    Dr. Rossi, the second edition is out.
    Google:
    AN IMPOSSIBLE INVENTION SECOND EDITION
    Robert Curto
    Ft. Lauderdale, Florida
    USA

  249. Andrea Rossi

    Robert Curto:
    Thank you for the information,
    Warm Regards,
    A.R.

  250. Wladimir Guglinski

    Joe wrote in December 5th, 2014 at 1:07 PM

    Wladimir,

    So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?
    ———————————————-

    Joe,
    also note that the two electrons 1s1 and 1s2 of the newborn 4Be7 apply a rapid oscillation on the proton along the z-axis, because the Coulomb attraction on the proton changes the direction of its motion, as shown in the figure:

    http://peswiki.com/index.php/Image:Oscilation_of_the_proton_in_4Be7.png

    As the orbit of the 2s1 applies a strong magnetic force on the proton along the z-axis, the amplitude of the oscillation along the z-axis starts to increase, and so quickly the proton exits the 4Be7, going along the z-axis accelerated by the force applied by the orbit 2s1.

    regards
    wlad

  251. Wladimir Guglinski

    ERRATA:

    In my last post, where it is written:

    But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the exits of the 4Be7 nucleus begins to be accelerated along the z-axis, by starting from an initial velocity zero.

    the correct is:

    But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the proton exits the 4Be7 nucleus beginning to be accelerated along the z-axis, by starting from an initial velocity zero.

  252. Wladimir Guglinski

    Joe wrote in December 5th, 2014 at 1:07 PM

    Wladimir,

    The direction of initial velocity of the newly born proton would be the most important factor for accomplishing a nucleon transfer. The reason for this is shown in the Lorentz force:

    F = q ( E + v x B )

    As you can see, it is not just a simple matter of summing electric and magnetic forces, although it does include that. If v is not parallel to the z-axis, the proton may never strike the plane of rotation. So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?
    —————————————————————-

    Joe,
    the initial velocity of the proton is zero, because the half-life of the newborn 4Be7 is 53,2 days.

    So, after the decay of the neutron, the proton stays in the 4Be7.

    Look at the structure of the 4Be7 after the decay of the neutron in the 3Li7:

    http://peswiki.com/index.php/Image:4Be7_structure_after_the_decay_of_3Li7.png

    You can see it also in the Figure 38 at the page 49 of the paper Stability of Light Nuclei:
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    The 53,2 days of the half-life of the 4Be7 occurs in normal conditions.
    But with the z-axis of the 4Be7 aligned with the orbit of the electron 2s1, the proton is pulled by a strong magnetic field induced by that orbit, and the exits of the 4Be7 nucleus begins to be accelerated along the z-axis, by starting from an initial velocity zero.

    regards
    wlad

  253. Joe

    Wladimir,

    The direction of initial velocity of the newly born proton would be the most important factor for accomplishing a nucleon transfer. The reason for this is shown in the Lorentz force:

    F = q ( E + v x B )

    As you can see, it is not just a simple matter of summing electric and magnetic forces, although it does include that. If v is not parallel to the z-axis, the proton may never strike the plane of rotation. So the question is, is it even possible for a newly born proton to have an initial velocity that is parallel to the z-axis?

    All the best,
    Joe

  254. Wladimir Guglinski

    Daniel De Caluwé wrote in December 5th, 2014 at 5:56 AM

    @Wladimir,
    Why does the more losely bound neutron in 3Li7 decays? Due to the excitation by the alternating magnetic field? And does this give enough energy to let it decay? How much energy actually is needed to let that neutron decay in a proton + electron, and are all necessary conditions fulfilled to let it happen?
    ————————————-

    Daniel,
    free neutrons decay in proton+electron in about 15 minutes.

    Within the nuclei the neutrons are stable, because they are bound to deuterons via spin-interaction.
    The spin-interaction between a deuteron and a neutron within the nuclei reinforces the binding energy between the electron and the proton within the structure of the neutron.

    But a neutron weakly bound (as in the 3Li7) is in the verge of having decay (there is very little lacking for it to decay, because it is poorly connected to the deuteron).

    So, with the excitation of the 3Li7 through the oscillatory electromagnetic field, two things may occur:

    a) with NO excitation, the neutron has a normal oscillatory motion about the z-axis in the 3Li7 nucleus, because of magnetic interactions with the other deuterons.

    b) with excitation, the spin-interaction deuteron-neutron becomes weaker, because the oscillation of the neutron about the z-axis increases its amplitude .

    c) the excitation of the neutron exceeds the tolerable limit, and the neutron decays.

    regards
    wlad

  255. Wladimir Guglinski

    Joe,
    in the figure posted by me earlier it is shown only the Coulomb force on the proton:

    http://peswiki.com/index.php/Image:For%C3%A7as_sobre_o_proton_no_3Li7.png

    But there is also a atrong magnetic force applied by the orbit of the electron 2s1, pulling the proton along the z-axis.

    So, the proton acquires a helical trajectory which axis is the z-axis.
    The radius of the helical trajectory (about the z-axis) increases as the proton progresses.

    But the proton travels the distance 10^-11m between the nucleus and the orbit 2s1 in fraction of seconds, and so the growth of the radius of the helical trajectory is despicable when the proton hits the orbit of 2s1.

    Thefore the proton hits the orbit 2s1 practically in the center of the orbit

    regards
    wlad

  256. Andrea Rossi

    TO THE READERS:
    Today has been published on the JoNP the paper ” h- Space Theory” by Dr Valery J. Tarasov.

  257. Steven N. Karels

    Dear Andrea Rossi,

    You posted “Zero emissions with a gas fueled E-Cat is impossible. But due to the ssm and the efficiency the emissions will be of one order of magnitude less than with conventional systems.” An interesting revelation. If we take this literally, then the effective COP of the gas-fired system will be about 10.

    As any housewife can tell you, one of the advantages of a gas stovetop is that the heat is more quickly applied and removed compared to an electric range.

    a. Does this known factor allow you to run a higher effective COP than an electrically heated eCat?
    b. Is there a substantial improvement in SSM when using a gas-fired eCat over that of an electrically heated eCat?
    c. Do you obtain better control with a gas-fired eCat over that of an electrically heated eCat?

  258. Andrea Rossi

    Steven N. Karels:
    a- no, it is not this to make the difference: it is the ssm; this issue is not peculiar to the NG
    b- no
    Warm Regards,
    A.R.

  259. Dan C.

    Dear Mr. Rossi,

    If I may- For those concerned with the CO2
    If the electricity that powers the E-cat comes from a power plant supplied by Natural Gas.

    1. You use the equivalent of 3Kw of NG to produce the 1Kw of electricity that powers the E-cat producing 10Kw heat.
    Or
    2. You can use the equivalent of 1Kw of NG directly to the E-cat to produce 10Kw heat.

    That’s a 2/3rds reduction in NG use. There obviously will be some efficiency differences but, you have reduced CO2 output. If the electricity comes from coal, much more so. And it is cheaper.

    In time if/when E-cats should efficiently produce electricity, this would change the economics & at that point you would use the electricity if it’s cheaper then NG. In the mean time you have options. Gas or Electric. Options encourage faster adaptation.

    I recently read where China has near 1500Gw of installed capacity & want to double that by 2050. 3000Gw times 3 for electrical conversion. 9000Gw of Hot-cats.

    That’s a lot of Hot-cats Mr. Rossi.
    No rest for you & your team.

    work, work, work

    Sincerely, Dan C.

  260. Andrea Rossi

    Dan C.:
    Thank you for your insight. Please also read the answer I gave to Alessndro Coppi. Anyway: the ssm, not applied in Lugano test for the reasons well explained in the Report of the ITP, makes a substantial difference.
    Warm Regards,
    A.R.

  261. Marco Serra

    “due to the ssm and the efficiency the emissions will be of one order of magnitude less than with conventional systems”.
    WWWOOOOWWWW !! What a new !! This is the best Christmas gift I’ve ever received. Really. Thank you.
    Let me elaborate a bit. Because emissions are proportional to power produced, and you are able to get the same power with a tenth of emissions, does it mean that we are talking of a COP > 10 ?

    Please answer yes ;)

    God bless you
    Marco

  262. Andrea Rossi

    Marco Serra:
    Please see the answer I gave to Alessandro Coppi.
    Warm Regards,
    A.R.

  263. Alessandro Coppi

    Hi Andrea,
    In your answer to Patrick Ellul you said:
    “Zero emissions with a gas fueled E-Cat is impossible. But due to the ssm and the efficiency the emissions will be of one order of magnitude less”.
    Therefore the cop is at least 10!

    Congratulations
    Alessandro Coppi

  264. Andrea Rossi

    Alessandro Coppi,
    We will give these data when this cycle of R&D will be finished. So far we must adhere to what is written in the report of the Independent Third Party. The numbers published by the Independent Third Party are the only ones so far verified by a neutral party. The next Third Party will be the Customer, no one else.
    The products in the market will tell us which are the effective data in the effective production world.
    Warm Regards,
    A.R.

  265. Daniel De Caluwé

    @Wladimir,

    I think our previous messages crossed, and maybe the emoticon that I used wrongly gave the impression, but I can assure you that there was no tone of irony in my words. You really gave a good answer. The only question that pops up now, and that maybe I can ask now, is this: Why does the more losely bound neutron in 3Li7 decays? Due to the excitation by the alternating magnetic field? And does this give enough energy to let it decay? How much energy actually is needed to let that neutron decay in a proton + electron, and are all necessary conditions fulfilled to let it happen?

  266. Wladimir Guglinski

    Joe,
    look the forces on the proton when it is exiting the 3Li7

    http://peswiki.com/index.php/Image:For%C3%A7as_sobre_o_proton_no_3Li7.png

    regards
    wlad

  267. Joe

    Wladimir,

    The newly born proton might be able to cross the plane if it were released from the neutron (after decay) in a direction that is parallel to the z-axis. But what is the possibility or probability of such an occurrence?

    All the best,
    Joe

  268. JonJon

    Andrea,
    Could the gas fuelled Hotcat run on biogases,at least they are carbon neutral and can be generated onsite from organic matters,like farm wastes?

  269. Andrea Rossi

    JonJon:
    Yes, but biogas is not a much diffused commodity. Where it is available, there is no reason, under the technological point of view, not to make use of it.
    Warm Regards,
    A.R.

  270. Eric Ashworth

    Wladimir, Been keeping up with your replies and I believe it to be just a matter of time before QRT is taken seriously by the academic community. Well done. I am putting some of my own material together that may tie in with yours. All the best, Regards Eric Ashworth.

  271. Wladimir Guglinski

    Daniel De Caluwé wrote in December 4th, 2014 at 4:09 PM

    @Wladimir,

    Thank you very much for your answer, that is sufficient to me.

    So, your hypothesis for the Rossi-Effect still stands, and probably never was in danger.
    ———————————————

    Daniel,
    I hope there is not a tone of irony in your words.

    Actually any theory is in danger before its confirmation.

    And we know that, even after their confirmation, some theories are in danger, as happened to the Einstein’s Special Theory of Relativity, confirmed by a solar eclipse in 1919. Today we know that his concept of empty space is wrong.

    But at least my theory is according to some experimental evidences. For instance, it explains the 10MeV neutrons in the Mosier-Boss experiment.

    In serious danger is any theory unable to explain the experimental evidences.

    Any cold fusion theory must be able to explain the 10MeV neutrons in Mosier-Boss experiment, and not only the Rossi-Effect.

    regards
    wlad

  272. Patrick Ellul

    Dear Andrea,

    How do you feel about having to give up the “zero emissions” ideal for a gas fuelled e-cat?

    Regards,
    Patrick

  273. Andrea Rossi

    Patrick Ellul:
    Zero emissions with a gas fueled E-Cat is impossible. But due to the ssm and the efficiency the emissions will be of one order of magnitude less than with conventional systems.
    Warm Regards,
    A.R.

  274. Daniel De Caluwé

    @Wladimir,

    Thank you very much for your answer, that is sufficient to me.

    So, your hypothesis for the Rossi-Effect still stands, and probably never was in danger. ;-)

    Kind Regards,
    Daniel.

  275. Wladimir Guglinski

    Dears Joe , Daniel Le Caluwé,
    and further readers of the JoNP

    my theory on cold fusion is the last hopefulness of Physics.

    Do you know other cold fusion theory so beautiful and elegant as the mine?

    My theory is coherent, it is simple, it is logic, and it works by plausible mechanism, without the need of to appeal to strange and doubtful mechanisms neither to ad hoc hypotheses.

    Besides, it gives explanation for phenomena impossible to be explained from other cold fusion theories, as for instance:

    1- The emission of neutrons with 10MeV in the Pamela Mosier-Boss experiment, since only 2,2MeV are available for the neutron before the cold fusion occurence.

    2- Why the scattering experiments detected a spherical Coulomb barrier, while the Coulomb barrier of nuclei is actually non-spherical, and this nuclear property is decisive for cold fusion occurrence.

    Those considerations give me the sure that my theory is the last hope for a coherent explanation for cold fusion.

    If there is other theory capable to explain cold fusion better than the mine, I would like to know it.

    regards
    wlad

  276. Frank Acland

    Dear Andrea,

    You mention that the current plant you are working on (and giving your full attention to) is a normal E-Cat plant.

    Does this mean that development of Hot Cat plants will wait until the current task of perfecting your first 1 MW plant is complete?

    Many thanks,

    Frank Acland

  277. Andrea Rossi

    Frank Acland:
    No, it does not mean that: as I explained, we are also working on the gas fueled Hot Cat, this R&D bringing with itself the Hot Cat evolution.
    Warm Regards,
    A.R.

  278. Wladimir Guglinski

    Joe wrote in December 4th, 2014 at 2:51 AM

    Wladimir,

    1) ——————————————-
    Your latest alteration to your model is intended to eliminate the flip in the magnetic moment (GEP) by replacing the initial magnetic moment by a force of attraction that is electric instead. The main problem with this is that the motion of the proton (which replaces the neutron) would not be easy to determine since the proton has an electric charge.
    Because of this, the proton would suffer a Lorentz force due to the presence of electric and magnetic fields caused by the rotating electron. The result might be that the proton rotates in situ and never crosses the rotation plane. Or the proton might follow a helical trajectory out of the source nucleus again without ever crossing the rotation plane. Your alteration makes the situation of your model more difficult to ascertain.
    ————————————————————-

    Joe,
    what you say can be applied for the proton moving in an electrosphere in normal conditions, where the electrons move in all the directions.

    However, when the 3Li7 and the Ni nucleus have their z-axes aligned, the orbits of the electrons 1s1 and 1s2 is about the z-axis.

    The positions of two electrons 1s1 and 1s2 are always in perfect symmetry regarding the z-axis, and therefore the proton is always attracted by two contrary forces of the same value.
    The Coulomb attraction between the proton and 1s1 is cancelled by the attraction between the proton and 1s2, and the proton goes moving along the z-axis in a helical trajectory whose center-line is the z-axis.

    Therefore the proton crosses the rotation plane of the orbit 2s1 (p1 in my figure).

    regards
    wlad

  279. Herb Gillis

    Anyone:
    A few days ago there was a discussion about the role artificial intelligence (AI)could play in LENR research. The discussion was mostly joking. However; it may not be a joke after all. Advise you see the link below, which is to an AI application that may be uniquely useful in LENR research- – for discovering hidden relationships in large amounts of data (where there is no established theory yet). The Site speaks for itself:
    (http://ccsl.mae.cornell.edu/eureqa)
    Regards; HRG.

  280. Wladimir Guglinski

    Daniel De Caluwé wrote in December 4th, 2014 at 2:38 AM

    My answer: When the weakly bound neutron of the 3Li7 decays in a proton emitting one electron, as you wrote, then what happens with that new born electron? If it is not captured somewhere else in the nucleus, and escapes to the electrosphere, than you get a reorganisation of the electrosphere and also a new atom: 4Be7 (a neutron became a proton, and the electron escaped to the electrosphere ;-) ), with a pair of electrons (with opposing spins!) in the 2s outer subshell of its electrosphere, and in that case, the mechanism you proposed can not work anymore.
    ———————————————————–

    Daniel,
    4Be7 decays in 3Li7 by electron capture.
    http://en.wikipedia.org/wiki/Isotopes_of_beryllium

    Let see what happens:

    1- the neutron decays: n-> p+e, and the first electron exited the nucleus.

    2- the 4Be7 is formed in the form of a ion, and the proton is attracted by the orbit of the electron 2s1.

    3- In fraction of second the proton exits the nucleus of the 4Be7 and so the 4Be7 transmutes to 3Li6. The ion 4Be7 has not time to capture one electron so that to form the 2s outer subshell

    4- In that fraction of second the proton crosses the orbit 2s1 and by electron capture it transmutes to neutron.

    The half-life of the 4Be7 is 53,2 days, but this occurs in normal conditions (when there is NOT an electron orbit 2s1 attracting the proton along the z-axis direction).

    regards
    wlad

  281. Wladimir Guglinski

    Daniel De Caluwé wrote in December 4th, 2014 at 12:20 AM

    1) ———————————————
    My question: At the moment when, according to your model (of the neutron according to your QRT), the neutron decays in a proton + electron, where does that electron goe?
    ————————————————-

    Daniel,
    om beta-decay of a nucleus, the electron is emitted and leaves away the nucleus.

    2) —————————————-
    What happens with it in between the decay of the neutron and its recapturing by the proton?
    ——————————————–

    No, the phenomenon of electron (or positron) capture depends on the charge of the nucleus. Such electron (or postitron) is created (that’s why Dirac proposed his theory on the space filled by virtual electrons)

    3) ————————————————
    Thinking on the Pauli exclusion principle, or your newly GEP (Guglinsky Exlusion Principle), could this newly born electron disturb the mechanism (of acceleration of the proton by the other electron p1 of the 3Li7) that you describe?
    —————————————————–

    No, the newly born electron is extracted from the space by the proton, and they form the neutron.

    The other electron (resulted from the neutron’s decay) exited the nucleus.

    regards
    wlad

  282. Frank Acland

    Dear Andrea,

    Once you get this first plant operating perfectly, like Henry Ford, do you plan to make it your “Model T”, which you can duplicate on a mass production assembly line?

    Kind regards,

    Frank Acland

  283. Andrea Rossi

    Frank Acland:
    Yes, that is the strategy. That is why this first industrial working plant is so important. It will be the “Rosetta Stone” of our industrial know how.
    Warm Regards,
    A.R.

  284. Tommaso di pietro

    Goodmorning ing.Rossi.
    Did you start The test with hot cat to generate electric power?
    If so…when ?

    Thanks in advance

  285. Andrea Rossi

    Tommaso Di Pietro:
    Yes, we are working on that too. At the moment we are oriented toward a classic Carnot Cycle assembly, even if we are looking with curiosity to all the alternatives. The production of electric power with the Carnot Cycle goes together with the R&D we are making with the gas as a fuel, for obvious reasons.
    Warm Regards,
    A.R.

  286. andreino

    Dear Dr. Rossi,
    recently you said that one of you competitor has replicated a lenr device.
    Is that one much more similar to the e-cat or the hot cat or something else? Are the COP similar ?

    Thank you

  287. Andrea Rossi

    Andreino:
    I am referring to the work of Brian Ahern.
    Warm Regards,
    A.R.

  288. Joe

    Wladimir,

    Your latest alteration to your model is intended to eliminate the flip in the magnetic moment (GEP) by replacing the initial magnetic moment by a force of attraction that is electric instead. The main problem with this is that the motion of the proton (which replaces the neutron) would not be easy to determine since the proton has an electric charge. Because of this, the proton would suffer a Lorentz force due to the presence of electric and magnetic fields caused by the rotating electron. The result might be that the proton rotates in situ and never crosses the rotation plane. Or the proton might follow a helical trajectory out of the source nucleus again without ever crossing the rotation plane. Your alteration makes the situation of your model more difficult to ascertain.

    All the best,
    Joe

  289. Daniel De Caluwé

    Dear Wladimir,

    i) You wrote:

    Rossi-Effect:
    Due to the excitation of the 3Li7 by the high frequency of the oscillatory electromagnetic field applied in the reactor, the weakly bound neutron of the 3Li7 decays in a proton emitting one electron.
    Due to the positive charge of the proton, it is strongly attracted by the orbit of the electron p1, and gets a big speed toward the z-axis.
    When the proton crosses the orbit of p1 with very fast speed, by electron capture the proton transmutes to a neutron again.
    The orbit of the electron p1 starts to apply a repulsion force in the newborn neutron, and it is accelerated along the z-axis toward the nucleus Ni.

    My answer: When the weakly bound neutron of the 3Li7 decays in a proton emitting one electron, as you wrote, then what happens with that new born electron? If it is not captured somewhere else in the nucleus, and escapes to the electrosphere, than you get a reorganisation of the electrosphere and also a new atom: 4Be7 (a neutron became a proton, and the electron escaped to the electrosphere ;-) ), with a pair of electrons (with opposing spins!) in the 2s outer subshell of its electrosphere, and in that case, the mechanism you proposed can not work anymore.

    ii) Concerning your article “Cold fusion mystery finally deciphered”, I also have remark, about Figure 6: The outer electron of 3Li7 is NOT a p1 electron, but in fact a 2s1 electron, because it belongs to the 2s subshell of 3Li7, but, of course, as this works in the same way, this probably does not have any effect on your proposed mechanism.

    Kind Regards,
    Daniel.

  290. Daniel De Caluwé

    Dear Wladimir,

    In the case of the Rossi-Effect, you wrote:

    Rossi-Effect:
    Due to the excitation of the 3Li7 by the high frequency of the oscillatory electromagnetic field applied in the reactor, the weakly bound neutron of the 3Li7 decays in a proton emitting one electron.
    Due to the positive charge of the proton, it is strongly attracted by the orbit of the electron p1, and gets a big speed toward the z-axis.
    When the proton crosses the orbit of p1 with very fast speed, by electron capture the proton transmutes to a neutron again.
    The orbit of the electron p1 starts to apply a repulsion force in the newborn neutron, and it is accelerated along the z-axis toward the nucleus Ni.

    My question: At the moment when, according to your model (of the neutron according to your QRT), the neutron decays in a proton + electron, where does that electron goe? What happens with it in between the decay of the neutron and its recapturing by the proton? Thinking on the Pauli exclusion principle, or your newly GEP (Guglinsky Exlusion Principle), could this newly born electron disturb the mechanism (of acceleration of the proton by the other electron p1 of the 3Li7) that you describe?

    Kind Regards,
    Daniel.

  291. Wladimir Guglinski

    Dear Joe,
    as you do not enjoy my Guglinski Exclusion Principle, there is another alternative, as I explain ahead.

    Rossi-Effect:
    Due to the excitation of the 3Li7 by the high frequency of the oscillatory electromagnetic field applied in the reactor, the weakly bound neutron of the 3Li7 decays in a proton emitting one electron.
    Due to the positive charge of the proton, it is strongly attracted by the orbit of the electron p1, and gets a big speed toward the z-axis.
    When the proton crosses the orbit of p1 with very fast speed, by electron capture the proton transmutes to a neutron again.
    The orbit of the electron p1 starts to apply a repulsion force in the newborn neutron, and it is accelerated along the z-axis toward the nucleus Ni.

    .

    Pamela Mosier-Boss experiment:
    After the decay of the deuteron, the charge of the proton is strongly attracted by the orbit of the electron s1 (of the decayed deuteron, now in the new form of a proton).
    When the proton crosses the orbit of the electron s1, it decays in a neutron. This newborn neutron is repelled by the orbit s1, and so the neutron goes moving with acceleration along the z-axis toward the direction of the Pd nucleus.

    regards
    wlad

  292. Wladimir Guglinski

    Dears Joe, Eernie, Orsobubu, Steven, Eric, Daniel Caluwé, Herb Gillis, and those ones interested in the discussion.

    I posted our discussion on my theory for cold fusion in Peswiki, under the title Cold fusion mystery finally deciphered, in this link:

    http://peswiki.com/index.php/Cold_fusion_mystery_finally_deciphered

    regards
    wlad

  293. Wladimir Guglinski

    Joe wrote in December 3rd, 2014 at 2:24 PM:

    Wladimir,

    Even your newly minted Guglinski Exclusion Principle (GEP) (patent pending) can not explain the energy and acceleration of the neutrons in the Mosier-Boss (MB) experiment. As you recently mentioned, the magnetic moments of pairs of electrons within each orbital of an electrosphere cancel. Therefore, there is no net effect upon a neutron traveling through an electrosphere.
    ————————————————

    yes,
    but we can think about other hypothesis.
    The neutron is accelerated by the orbit of the electron s1 of the deuteron, and not by the electrons of the nucleus Pd (the orbit is perpendicular to the z-axes of the two nuclei Pd and 1H2).

    The proton decays in a neutron immediatelly after the decay of the deuteron. When the neutron crosses the orbit of the electron s1, the electron of the neutron changes its spin, and the orbit s1 applies a force of repulsion on the neutron, and it accelerates.

    As the neutron gets a very fast motion, it is not captured by the nucleus Pd.

    This is the unique solution for explaining the 10MeV of the neutrons in the Mosier-Boss experiment.
    If this explanation cannot be able to explain it, the Physics is in a serious problem.

    regards
    wlad

  294. Bernie Koppenhofer

    Dr. Rossi: Good to hear the installation is going good. Is there a possibility you might release results before a year? Our world desperately needs this technology. Thanks.

  295. Andrea Rossi

    Bernie Koppenhofer:
    I am not able to answer, anyway within a year or so we will have consolidated results.
    Warm Regards,
    A.R.

  296. tommaso di pietro

    Dear ing. Rossi,
    Is the megawatt plant actually in operation composed by low temperature e cat inside the big container or by hot cat inside the cube:)?

    Thamks in advance

  297. Andrea Rossi

    Tommaso Di Pietro:
    It is a normal E-Cat, not a Hot Cat. Outside is similar to the 1 MW plant of 2011, inside is completely different.
    Warm Regards,
    A.R.

  298. Joe

    Wladimir,

    Even your newly minted Guglinski Exclusion Principle (GEP) (patent pending) can not explain the energy and acceleration of the neutrons in the Mosier-Boss (MB) experiment. As you recently mentioned, the magnetic moments of pairs of electrons within each orbital of an electrosphere cancel. Therefore, there is no net effect upon a neutron traveling through an electrosphere.

    All the best,
    Joe

  299. Robert Curto

    Dr. Rossi, P2G or PtG is Power to Gas.
    A P2G plant converts electrical power to gas, that can then be stored in Pipelines.
    Google:
    Power to Gas
    Robert Curto
    Ft. Lauderdale Florida
    USA

  300. Andreas Moraitis

    Dear Andrea Rossi,

    There are various civil applications for MagneGas, one of them is metal cutting:

    http://vimeo.com/47166079

    By the way: It was invented by Ruggero Santilli, whom you know well.

    Best regards,
    Andreas Moraitis

  301. Andrea Rossi

    Andreas Moraitis:
    Thank you for your interesting information and congratulations to Dr Ruggero Santilli.
    Warm Regards,
    A.R.

  302. Mark Saker

    Dear Andrea,

    What is the rationale behind 110 computers vs a larger computer with 1+1 redundancy handling all 110 reactors simultaneously.

    Surely the actual computational power required cannot be that great that it would trouble a recent CPU?

    Would that not reduce the cabling?

    thanks

    Mark Saker

  303. Andrea Rossi

    Mark Saker:
    The control system has been designed by our specialists based on the best available technologies related to the controls we need.
    Warm Regards,
    A.R.

  304. Bob

    Dear Andrea Rossi

    1. Does the 1MW plant exchange its heat with

    a. Air

    b. Some other gas

    c. Water

    d. Some other fluid

    2. Can you reveal whether you are obtaining improvements in the Carnot cycle efficiency of the heat exchange apparatus through your own proprietary efforts and if so what order of magnitude are you achieving?

    Thanks

    Bob

  305. Andrea Rossi

    Bob:
    1.c ( Water)
    2. We will give performance data when they will be consolidated.
    Warm Regards,
    A.R.

  306. Richard Hill

    Dr. Rossi,
    You have said that you are working flat out on your customer’s plant. In such projects in the real world there are many things that have to be done, many problems to solve. I’m sure you are solving them one by one. It would be very good for us if you could give us an example of a typical problem (or even a small one) that you have overcome.

  307. Andrea Rossi

    Richard Hill:
    Let me say just this: we have now 110 computers integrated with each other to control the operation of about 10,000 components. And no former statistics and experience. Guess what this means. Luckily, the team is made by top level guys at any working level.
    Warm Regards,
    A.R.

  308. Frank Acland

    Dear Andrea,

    Can you comment on how successful your team has been so far in solving the problems you have encountered with the 1 MW plant?

    Many thanks,

    Frank Acland

  309. Andrea Rossi

    Frank Acland:
    I would say that the main problems have been resolved, but we have to see what happens in a long term operation. We must reach that kind of reliability that Henry Ford reached when he decided to sell the Model T Ford in massive quantities. It takes time. This first plant working in the factory of a Customer that has to get industrial profit from it must reach perfection to make us satisfied.
    Warm Regards,
    A.R.

  310. Wladimir Guglinski

    Joe wrote in December 2nd, 2014 at 11:51 PM

    Wladimir,

    ii) is not part of any orbital within the electrosphere of 3Li7 since it orbits a different entit
    Therefore, the Pauli exclusion principle does not apply here in trying to explain the flip of the spin as the electron crosses the plane. Another reason must be given, or your model is not viable.
    —————————————————–

    Joe,
    then call it Guglinski Exclusion Principle

    It seems the phenomenon occurs in the Pamela Mosier-Boss experiment, because she detected emission of neutrons with 10MeV, while the energy avaikable is only 2,2MeV (binding energy of the deuteron).

    It seems the neutrons are accelerated when they cross some orbits of electrons in the electrosphere of the Pd nuckeus;
    There is no way to exlain the 10MeV by considering the laws of the Standard Model

    regards
    wlad

  311. Wladimir Guglinski

    Joe,
    why do I need to explain an impossible phenomenon(according to the Standard Model) by using the known laws proposed in the Standard Model ?

    If we consider the Standard Model, my model is also no viable from other aspects. For instance, my model of field is different of the model considered in the Quantum Field Theory.

    But you yourself had pointed to us that from the model proposed in QFT it is impossible to explain the null magnetic moment of the even-even nucle with Z=N, because of the monopolar nature of the electric field.

    Besides, my theory for the explanation on cold fusion is also no viable (by considering the Standard Model) because my theory is based on a new nuclear model which works with laws different of those considered in the Standard Nuclear Physics (for instance, the aggregation of nuclei is not promoted by the strong force).

    However,
    the reversion of the magnetic moment of a neutron when it crosses an orbit of one electron can be tested by experiments.

    Besides,it is possible the phenomenon occurs in the Pamela Mosier-Boss experiment, because she detected neutrons with 10MeV, while the binding energy available is only 2,2MeV (the binding energy of the deuteron).

    Therefore, dear Joe,
    we are dealing with questions that cannot be solved by using the same old laws considered in the Standar Model.
    Some news laws must be discovered

    regards
    wlad

  312. Wladimir Guglinski

    Joe wrote in December 2nd, 2014 at 11:51 PM

    1) —————————————-
    Therefore, the Pauli exclusion principle does not apply here in trying to explain the flip of the spin as the electron crosses the plane.
    ——————————————-

    JOe,
    then call it Guglinski Exclusion Principle

    Because a neutron crossing the orbit of an electron was never observed in any experiment, and the physicists do not know such phenomenon.
    As they also did not know cold fusion some years ago.

    2) —————————————–
    Another reason must be given,
    ——————————————–

    Then cold fusion is impossible, Rossi-Effect is a fraud, and he must stop his experiments

    3) —————————————–
    or your model is not viable.
    ——————————————–

    And cold fusion is no viable too

    regards
    wlad

  313. Joe

    Wladimir,

    The Pauli exclusion principle states that for two electrons residing in the same orbital, n, l, and m(l) are the same, so m(s) must be different and the electrons have opposite spins. But the electron orbiting the proton within the QRT neutron

    i) has no defined orbital in a general sense

    ii) is not part of any orbital within the electrosphere of 3Li7 since it orbits a different entity.

    Therefore, the Pauli exclusion principle does not apply here in trying to explain the flip of the spin as the electron crosses the plane. Another reason must be given, or your model is not viable.

    All the best,
    Joe

  314. Andrea Rossi

    Dr Joseph Fine:
    Thank you: sincerely, I did not know MagneGas: interesting, even if unapplicable for civil apparatuses.
    For rockets, maybe…
    Warm Regards,
    A.R.

  315. Wladimir Guglinski

    Andrea Rossi wrote in December 2nd, 2014 at 1:04 PM

    Herb Gillis:
    To alter the composition of a fuel makes practically impossible to get a certification. We are aiming to commercial apparatuses, not of experimental prototypes.
    ———————————-

    Yes, dear Andrea
    but you can work in two ways.
    By one, working so that to improve the technology based on the present fuel, in order to get the certification.

    By the second, to put a small crew so that to test new fuels, with the aim to obtain a higher COP for the eCat.

    Having the certification obtained from the present fuel, in the future you dont need to tell to people who buy the eCat that you changed the composition of the fuel (in the case you discover a new more efficient fuel).

    However, I do not intend to teach a priest how to pray Mass.

    regards
    wlad

    regards
    wlad

  316. Wladimir Guglinski

    Andrea Rossi
    December 2nd, 2014 at 7:50 AM

    Curiosone:
    This fact also casts some doubt about the certainty that what they found there is really the Higgs boson, but after 9 billion euro spent for this quest it is not politically correct to say so.
    ——————————————–

    If the physicists had asked my opinion, I would had told them that a wrong theory never can be confirmed. The theory was developed by considering the empty space, but the space is no empty, and so the theory makes no sense.

    But it is a good lesson for the scientists, in order they learn that the mathematics cannot be applied as corroboration for theories based on stupid conjectures, as that crazy concept of empty space proposed by Einstein, according to which an empty thing (the empty space) could be able to have contraction and to be the supporter of physical entities as the magnetic fields.

    400 years ago Galileo advised the scientists that the Science cannot be divorced to the Logic.

    The scientific method is making jokes with the physicists.
    First, they had interpreted wrongly the Michelson-Morley experiment, and concluded that Einstein was right, because it is impossible to detect the aether by experiments. And as the loyalty to the scientific method prescribes that they cannot accept the existence of something not detectable by experiments, they rejected the obvious: the space cannot be empty.
    Later, they used the scientific method so that to find objections against cold fusion, as to claim that Fleischmann and Pons were wrong, because their experiments were not replicable, and the measurement of the excess heat was wrong, and only because they actually were trying to save their Standard Model, from whose principles cold fusion is impossible.
    Finally, in 2012 they had interpreted as successful an experiment which detected a boson, believing that it was the Higgs boson, in spite of the theory was developed from the concept of empty space, and despite of the fact that in 2011 an experiment published in the journal Nature had shown that the space is not empty, since the space is able to create light.

    In the future the scientists will remember the 20th Century as the most obscure and tenebrous in the History of the Science.

    regards
    wlad

  317. Wladimir Guglinski

    Joe wrote in December 2nd, 2014 at 1:03 AM

    2. I repeat my last question, what is the specific mechanism that flips the spin of the electron when it crosses the plane of the 1p1 rotation?
    ——————————————-

    Pauli’s Principle:

    http://peswiki.com/index.php/Image:PAULIprincipleWHENelectronCHANGESorbit.png

    regards
    wlad

  318. Joseph Fine

    Herb Gillis,

    If Flame temperature were the primary consideration, you might use Magnegas instead of Acetylene. You can also listen to this awful musical background. Perhaps the flame temperature of Magnegas is too high! (You could melt everything!!)

    Independent tests have established that MagneGas™ with a flame temperature of 5,819°C / 10,560°F (Verified by CCNY and the Institue of Ultraspectroscopy http://www.magnegas.com/docs/MG-Flame-report.pdf) is the fastest, most precise and most energy efficient cutting fuel available today.

    http://www.magnegas.eu/metal

    Very warm regards,

    Joseph Fine

  319. Frank Acland

    Dear Andrea,

    Here is an interesting article dealing with your E-Cat from the Dome Magazine in Michigan (talks mainly about public policy issues). The title is ‘The Miraculous Machine’

    http://domemagazine.com/glazer/lg111414

    Kind regards,

    Frank Acland

  320. Andrea Rossi

    Frank Acland:
    Thank you.
    Warm Regards
    A.R.

  321. Herb Gillis

    Andrea Rossi:
    Have you considered the possibility of adding acetylene to the natural gas stream (or even using pure acetylene) in the gas-driven Ecat? Acetylene would give a hotter flame. Acetylene is derived from natural gas and cracking of hydrocarbon feedstocks.
    Regards; HRG.

  322. Andrea Rossi

    Herb Gillis:
    To alter the composition of a fuel makes practically impossible to get a certification. We are aiming to commercial apparatuses, not of experimental prototypes.
    Warm Regards,
    A.R.

  323. Wladimir Guglinski

    ERRATA:

    In my last post, where is written:

    But while 28Ni requires an emitter with pair number of protons, as 50Sn is very biggest than 28Ni then the best emitter for 50Sn can be with odd number of protons.

    the correct is:

    But while 28Ni requires an emitter with odd number of protons, as 50Sn is very biggest than 28Ni then the best emitter for 50Sn can be with even number of protons.

  324. JCRenoir

    How are going the experiments with the gas fueled Ecat?
    JCR

  325. Andrea Rossi

    JC Renoir:
    We are working on this issue very, very, very hard.
    Warm Regards,
    A.R.

  326. Wladimir Guglinski

    Perhaps 50Sn requieres an emitter with Z=pair

    As we have seen, the 4Be9 cannot be an emitter for 28Ni because the size of Ni is not sufficiently large, so that to supply a large difference between the radius of the orbit of p1 and p2, as shown in Figure 7.
    FIG. 7:
    http://peswiki.com/index.php/Image:FIGURE-7-substitute3Li7-28Ni.png

    But while 28Ni requires an emitter with pair number of protons, as 50Sn is very biggest than 28Ni then the best emitter for 50Sn can be with odd number of protons.

    Indeed, as 50Sn has a big electric field, we can use as emitter a nucleus with pair number of protons, as 4Be9. Because of the big size of the 50Sn the radius of the orbit of the electron p1 will be very larger than the radius of the orbit p2. By this way the big magnetic force applied by the orbit p1 will be counterbalanced by a contrary force applied by the orbit p2, and by this a suitable force can be obtained. This shown in the figure ahead:
    http://peswiki.com/index.php/Image:4Be9-USEDasEMITTERfor50sN.png

    Perhaps 4Be9 is not the best emitter to be used with 50Sn, but the best emitter can be found, because it is possible to find the suitable combination of difference in the radii of the orbits , so that to apply the suitable force on the neutron of the emitter.

  327. DTravchenko

    Dear Dr Andrea Rossi:
    Are the licenses that Leonardo Corporation bought back from the former licensees for sale?
    Warm Regards,
    DTravchenko

  328. Andrea Rossi

    D. Travchenko:
    No, they are not.
    Warm Regards,
    A.R.

  329. Curiosone

    To the Readers:
    is somebody able to explain what is in Physics the “Hierarchy problem” ?
    Thanks,
    W.G.

  330. Andrea Rossi

    Curiosone:
    There are two parameters in the Standard Model that are unnatural: the values of the Higgs Field in empty space and of the energy density in empty space, the so called “vacuum energy”: they are much smaller than they ought to be along the mathematical equations, calculating the energy of the virtual particles in the empty space that are necessary to justify the expansion rate of the Universe. The value of the Higgs field should be 10^16 bigger than it has been measured to be by the CERN experiment. This fact also casts some doubt about the certainty that what they found is really the Higgs boson, but after 9 billion euro spent for this quest it is not politically correct to say so. This enormous discrepance between the value the Higgs field should have and the observed one is called the “hierarchy problem”; it is far from being resolved, as well as the one of the energy density in empty space ( the two things seem to be connected).
    Warm Regards,
    A.R.

  331. georgehants

    Dear Mr. Rossi you will be aware that Cold Fusion seems to be very close to some kind of acceptance and the possibility of the 25 years of neglect by main-line science to end.
    Is there anything that yourself or your company can do or say at this time that would help to give it that final push into generally excepted Research.
    For those in need of low cost power to help for instance in the supply of safe drinking water any more delay is life threatening to many people.
    All best wishes.

  332. Andrea Rossi

    Georgehants:
    I agree with you, after the Report of the Independent Third Party ( and the work of many colleagues of mine, like for example Brian Ahern in the MIT) many scientists of the so called mainstream science have changed opinion. Before our work, to show anything regarding the LENR to a mainstream scientist was like to show garlic and crucifix to Dracula; today most accept that it is a field that is worth R&D. The final push can only be given by reliably working plants that produce energy in a profitable way in an industry. This is exactly what our Team is doing by means of the 1 MW plant upon which we are working so hard.
    Warm Regards,
    A.R.

  333. Wladimir Guglinski

    ERRATA:

    The correct structure of the 11Na23 is actually the following:

    http://peswiki.com/index.php/Image:3Li7-and-11Na23-SIMILARITIES-Rev1.png

    I tried several times to upload the correct version by editing it in Peswiki, however the blog does not work.

  334. Wladimir Guglinski

    orsobubu wrote in December 1st, 2014 at 7:18 PM

    Dear Wladimir,

    2 –—————————————–
    Does the book include the arguments treated in your new paper waiting for publishing here on JONP?
    ———————————————

    NO

    3 –—————————————–
    I understood the book will only be published as open access PDF file, by Dr. Prakash Somani publishing house. So there will not be a paper printed edition?
    ———————————————-

    It seems Dr. Prakash wished to publish a small printed edition of about 100 volumes.
    But I dont know his final decision.
    Besides,
    dear Orsobubu,
    I dont understand what is going on.
    Dr. Prakrash told the book would be published until the end of 2014.

    The last exchange of email by us was the following:
    ====================================================================
    On Saturday, September 20, 2014 8:34 AM, Wladimir wrote:

    Hi, Dr. Prakash
    did you give up to publish the books ?
    regards
    wlad

    .

    Date: Sat, 20 Sep 2014 02:53:14 -0700
    From: psomani1@yahoo.com
    Subject: Re: did you give up ?
    To: wladimirguglinski@hotmail.com

    Dear Author,
    Thank you for reaching to me. I will live to my promise. Your books are under typesetting and I will get back to you soon.
    Your books will be published before the end of this year (2014) SURELY. I will arrange to send you the proof – for corrections (if any) soon. We are still working with the Figures of your book.
    With best regards
    Dr. Prakash Somani
    ====================================================================

    4 –——————————————-
    You said that in the book will be something too on plagiarism of your discoveries in shape of nuclei and z-axis. But I remember that there was, here on JONP, an exchange of views with Stoyan Sarg in this regard, where he claimed to have made these discoveries prior to yours, and thus concluding:

    “In 2001 I Submitted My BSM theory to the Canadian Office of Intellectual Property with a claim for discovery of new models of atoms. I received a document in 2001, but to be surer, in 2002 I archived the full electronic version of the theory and the Atlas of Nuclear Atomic Structures in the AMICUS database of the National Library of Canada, where it Obtained and international catalog reference number . Once done, the dated electronic version can not be changed, so it serves as a legal proof of the dates of my intellectual property. The deposited electronic versions are publicly accessible as I show below.All These archives have assign ISBN number like books. In May 2002, I published the first scientific article about my theory in the official physical archive operated by Cornell University http://lanl.arxiv.org/abs/physics/0205052”
    But maybe I missed the follow. Can you summarize what is currently your position on this issue?
    ———————————————-

    yes, I can summarize, dear Orsobubu.
    If the nuclear model by Dr. Sarg is correct, he does not need to be afraid. Nothing can take off the merit of his theory. His prediction of the z-axis will be one among several correct predictions.

    But if his nuclear model is wrong, his prediction on the z-axis loses any merit as prediction.

    The same is applied to my nuclear model.

    I did not discover an z-axis in the nuclei. Actually I have discovered a new nuclear model, where the distribution of protons and neutrons is along the z-axis.

    5 –—————————————–
    One last thing, what do you think of this article:

    http://vixra.org/pdf/1408.0109v4.pdf

    A post by Axil against Standard Model appeared on the blog of Mats Lewan says:

    “From the summary: Mass generation via chiral symmetry breaking and the Higgs becomes irrelevant for two independent reasons. First, in the absence of the weak force there is no need for massive gauge bosons. And second, the chiral impedance is scale invariant, cannot communicate energy but rather only quantum phase, cannot deliver mass. Similarly, mass generation in QCD via dynamic chiral symmetry breaking is seen to be not possible in light of the scale invariance of chiral impedances. In the impedance approach the origin of mass is the energy in the fields of the coupled modes represented in the impedance network and connected by impedance mis-matches. The calculated mass of the electron is correct at the nine signi cant digit limit of experimental accuracy, the muon at a part in one thousand, the pion at two parts in ten thousand, and the nucleon at seven parts in one hundred thousand.

    This says that there is no Higgs field. The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson. I have been postulating EMF condensation into particles as an important mechanism in LENR. This quantum impedance idea supports that belief. Energy gain in LENR is a energy balancing and transfer process in an EMF based energy transfer mechanism. The discovery of the “God particle” has been brought into question recently as a misidentification. I believe that when LENR is accepted, the standard model will need a rework to get rid of the Higgs field.”
    ———————————————————

    Dear Orsobubu,
    Higgs theory was debunked by the detection of the aether by experiments in 2011, published in the journal Nature.
    Higgs developed his theory by believing that the space is empty, and therefore the space cannot have structure, so that to interact with the particles, in order to give them mass.
    His theory is superfluous face to the existence of the aether.

    But it’s good to see that Higgs theory is not satisfactory even from the theoretical viewpoint.
    Even some academic physicists are already questioning the Higgs theory, saying that the boson found in the LHC is not a Higgs-Boson which gives mass to particles, but it is actually only one more particle as many other produced by the collision of protons.

    regareds
    wlad

  335. Joe

    orsobubu,

    Axil writes,
    “The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson.”

    I contend that there is no amount of “condensing” of a field that will form any property of a particle, including mass. Further, no type of manipulation of a field will ever create a particle since a field is of a wave nature. And wave and particle natures are mutually exclusive. I argued this point a little while ago here on the JoNP against Quantum Field Theory.

    All the best,
    Joe

  336. Joe

    Wladimir,

    1. You write,
    “The electron can change its intrinsic-spin simply changing its original position up to a new position upside-down.”

    What you state here is essentially what I stated in my last post. The effect of being “upside-down” is that the movement of the orbit ring is now reversed.

    2. I repeat my last question, what is the specific mechanism that flips the spin of the electron when it crosses the plane of the 1p1 rotation?

    All the best,
    Joe

  337. Wladimir Guglinski

    <b<Efficiency of emitters and receptors in Rossi’s eCat

    In the figures showing the structures of nuclei, the protons and neutrons are not at rest. They oscillate quickly, due to repulsions.
    When the nuclei are aligned along an external magnetic field, and their z-axis is aligned with that field, the protons and neutrons start to oscillate toward the z-axis direction.

    With the application of an oscillatory electromagnetic field, the oscillation of the neutron in the 3Li7 increases its amplitude, and so it helps the magnetic force (applied by the orbit of the electron p1 in the 3Li7) to extract that neutron to send it toward the 28Ni nucleus.

    So, the efficiency of an emitter depends on some conditions of the cold fusion reactor, but the efficiency also depends on some properties, as follows:
    - How the neutron is bound in the emitter
    - The nuclear magnetic moment of the emitter
    - The size of the emitter compared with the size of the of the receptor (because the radius orbit of the electron p1 increases with the size of the receptor)

    The magnetic moment of the emitter helps to get quickly the alignment between the emitter and the receptor.
    3Li7 has magnetic moment +3,26
    5B11 has magnetic moment +2,68
    7N15 has magnetic moment -0,28 , but excited it becomes -2,4

    3Li7 is better than 5B11 from other viewpoint: after the 3Li7 decay, the 3Li6 has small magnetic moment, +0,822. Then suppose 3Li7 supplies a neutron to a nucleus 60Ni, and it transmutes to 61Ni. Having small magnetic moment, the newborn 3Li6 leaves away the 61Ni, and a new 3Li7 couples with the newborn 61Ni, and the 3Li7 supplies a neutron to it. A cascade reaction occurs:
    60Ni + 3Li7 -> 61Ni + 3Li6
    61Ni + 3Li7 -> 62Ni + 3Li6
    62Ni + 3Li7 -> 63Ni + 3Li6

    Unlike, 5B11 after decaying transmutes to 5B10 which has magnetic moment +1,8, and so it will not leave away the newborn 61Ni so easily as 3Li6 does.

    7N15 after decaying transmutes to 7N14 which has magnetic moment +0,403 , weaker than that of 3Li6. Therefore from such viewpoint 7N15 is better than 3Li7.

    3Li7 has also a big unbalance of masses, which can contribute for the extraction of the neutron.

    Other factor wich can increase the efficiency is the quantity of stable isotopes.
    However all the emitters have one unique stable isotope:
    3Li7, 5B11, 7N15, 9 F19, 11Na23, 13Al27, etc.

    But regarding the efficiency of the receptors, the number of stable isotopes can have a strong influence.
    Let us analyse the elements suitable to be receptors, and their stable isotopes.
    The elements marked with “(*)” are those ones which have more than 5 cascades.
    For instance, 22Ti has an uninterrupted cascade from Ti46 until Ti50.

    20Ca = Ca40, Ca42, Ca43, Ca44, Ca46
    (*) 22Ti = Ti46, Ti47, Ti48, Ti49, Ti50
    24Cr = Cr50, Cr52, Cr53, Cr54
    26Fe = Fe54, Fe56, Fe57, Fe58
    28Ni = Ni58, Ni60, Ni61, Ni62, Ni64
    30Zn = Zn66, Zn67, Zn68, Zn70
    32Ge = Ge70, Ge72, Ge73, Ge74
    34Se = Se74, Se76, Se77, Se78, Se80
    36Kr = Kr78, Kr80, Kr82, Kr83, Kr84
    38Sr = Sr84, Sr86, Sr87, Sr88
    40Zr = Zr90, Zr91, Zr92, Zr94
    (*) 42Mo = Mo92, 94Mo, Mo95, Mo96, Mo97, Mo98
    (*) 44Ru = Ru96, Ru98, Ru99, Ru100, Ru101, Ru102, Ru104
    46Pd = Pd104, Pd105, Pd106, Pd108, Pd110
    48Cd = Cd106, Cd108, Cd110, Cd111, Cd112, Cd114
    (*) 50Sn = Sn112, Sn115, Sn116, Sn117, Sn118, Sn119, Sn120, Sn122, Sn124
    52Te – Te120, Te123, Te124, Te125, Te126

    As we see, 50Sn has a cascade uninterrupted from Sn115 until Sn120. Beyond the cascade, it has yet 3 isotopes able to have cold fusion: Sn112, Sn122, Sn124.

    Therefore, it seems 50Sn would be the better element to supply the higher COP for the eCat.

    Unfortunately 50Sn has a big quantity of protons, and perhaps it cannot get cold fusion by using 3Li7 as emitter (the orbit of the electron p1 of the 3Li7 will be very large, and the neutron will be thrown with too much energy, and the neutron will trespass the 50Sn nucleus).

    But perhaps 50Sn can get cold fusion with other emitter, as for instance 11Na23.
    The sodium 11Na23 has a structure similar to the 3Li7, as shown in the figure ahead:
    http://peswiki.com/index.php/Image:3Li7-and-11Na23-SIMILARITIES.png

    They both have a big unbalance of mass, which contributes for the expulsion of the neutron.

    They both have spin 3/2 (in Na23 the neutron is also weakly bound, because it is bound via spin-interaction with the deuterons D-2 and D-3, and they are far away.

    11Na23 has also a good magnetic moment: +2,21

    After emitting a neutron, 23Na transmutes to 22Na, which is no stable, and transmutes to 22Ne. 22Ne has magnetic moment zero, and therefore it leaves away the newborn receptor 50Sn(+n), which can start a new cascade with another 11Na23.

  338. Wladimir Guglinski

    Joe wrote in December 1st, 2014 at 5:49 PM

    what mechanism is responsible for reversing this movement of the electron along its own orbit ring?
    —————————————–

    Joe,
    I think you did not understand how the electron changes its spin.

    With the electron does not occur “reversion” of movement, like for instance occurs with the four wheel of a car, when it is moving ahead, and the driver stops the motion, and he puts the car moving in a reverse gear (in this case the four wheels of the car gyrate in reverse).

    The electron can change its intrinsic-spin simply changing its original position up to a new position upside-down.

    regards
    wlad

  339. Wladimir Guglinski

    Joe wrote in December 1st, 2014 at 5:49 PM

    Wladimir,

    You write,
    “The two orbits continue being parallel.

    The intrinsic-spin of the electron in the neutron changes from up to down.”

    In QRT, the intrinsic spin of a particle is defined as its rotation about its central axis. Therefore, what mechanism is responsible for reversing this movement of the electron along its own orbit ring?
    ——————————————–

    Joe,
    the intrinsic-spin of the electron can be changed by several sort of interactions. Even the interaction with the flux n(o).
    See Figure 9 in the page 12 of the paper Stability of the Light Nuclei:
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    regards
    wlad

  340. orsobubu

    Dear Wladimir,

    in March I asked you about the publication of the translation of your book “The Missed U-Turn”, treating the story of the evolution of physics, according to your vision of alternation of Newton and Descartes methods, up to include Rossi’s E-Cat and your Quantum Ring Theory models.

    1 – You said more recent discoveries have been added, about aether, the mass of particles and neutron halo, together with references to plagiarism cases by Nature and European Physical Journal. You said that you do not intend to include proposals explaining LENR. It seems to me that in these last months you elaborated quite a significant amount of material on LENR, so I wonder now if you plan to include something about this.

    2 – Does the book include the arguments treated in your new paper waiting for publishing here on JONP?

    3 – I understood the book will only be published as open access PDF file, by Dr. Prakash Somani publishing house. So there will not be a paper printed edition?

    4 – You said that in the book will be something too on plagiarism of your discoveries in shape of nuclei and z-axis. But I remember that there was, here on JONP, an exchange of views with Stoyan Sarg in this regard, where he claimed to have made these discoveries prior to yours, and thus concluding:

    “In 2001 I Submitted My BSM theory to the Canadian Office of Intellectual Property with a claim for discovery of new models of atoms. I received a document in 2001, but to be surer, in 2002 I archived the full electronic version of the theory and the Atlas of Nuclear Atomic Structures in the AMICUS database of the National Library of Canada, where it Obtained and international catalog reference number . Once done, the dated electronic version can not be changed, so it serves as a legal proof of the dates of my intellectual property. The deposited electronic versions are publicly accessible as I show below.All These archives have assign ISBN number like books. In May 2002, I published the first scientific article about my theory in the official physical archive operated by Cornell University http://lanl.arxiv.org/abs/physics/0205052”

    But maybe I missed the follow. Can you summarize what is currently your position on this issue?

    5 – One last thing, what do you think of this article:

    http://vixra.org/pdf/1408.0109v4.pdf

    A post by Axil against Standard Model appeared on the blog of Mats Lewan says:

    “From the summary: Mass generation via chiral symmetry breaking and the Higgs becomes irrelevant for two independent reasons. First, in the absence of the weak force there is no need for massive gauge bosons. And second, the chiral impedance is scale invariant, cannot communicate energy but rather only quantum phase, cannot deliver mass. Similarly, mass generation in QCD via dynamic chiral symmetry breaking is seen to be not possible in light of the scale invariance of chiral impedances. In the impedance approach the origin of mass is the energy in the fields of the coupled modes represented in the impedance network and connected by impedance mis-matches. The calculated mass of the electron is correct at the nine signi cant digit limit of experimental accuracy, the muon at a part in one thousand, the pion at two parts in ten thousand, and the nucleon at seven parts in one hundred thousand.

    This says that there is no Higgs field. The electromagnetic field condenses under the action of quantum EMF impedance to form the mass of the electron, pion, and meson. I have been postulating EMF condensation into particles as an important mechanism in LENR. This quantum impedance idea supports that belief. Energy gain in LENR is a energy balancing and transfer process in an EMF based energy transfer mechanism. The discovery of the “God particle” has been brought into question recently as a misidentification. I believe that when LENR is accepted, the standard model will need a rework to get rid of the Higgs field.”

    Many thanks.

  341. Joe

    Wladimir,

    You write,
    “The two orbits continue being parallel.

    The intrinsic-spin of the electron in the neutron changes from up to down.”

    In QRT, the intrinsic spin of a particle is defined as its rotation about its central axis. Therefore, what mechanism is responsible for reversing this movement of the electron along its own orbit ring?

    All the best,
    Joe

  342. Wladimir Guglinski

    Daniel De Caluwé wrote in December 1st, 2014 at 7:35 AM

    Dear Wladimir,

    Interesting hypothesis based on your theory, but, unfortunately, except A. Rossi, who signed a non disclosure agreement, there’s nobody else at the moment who can test it.
    —————————————————–

    Dear Daniel
    along years Andrea Rossi had tested many elements, and he knows what emitters do not produce cold fusion.
    So, Rossi already has intuition on the question whether my theory can be correct, or not.

    Obviously he will say nothing, not only because he decided to disclosure nothing about his reactor , but also because if he says the theory can be correct, by this way he is giving a tool for his competitors so that to look for the best combination of emitters and receptors.

    regards
    wlad

  343. Daniel De Caluwé

    Dear Wladimir,

    Interesting hypothesis based on your theory, but, unfortunately, except A. Rossi, who signed a non disclosure agreement, there’s nobody else at the moment who can test it.

    Kind Regards,
    Daniel.

  344. Wladimir Guglinski

    Joe wrote in November 30th, 2014 at 4:02 PM

    Wladimir,

    Are you saying that, before the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is parallel to the orbit of the 1p1 electron; and that after the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is anti-parallel to the orbit of the 1p1 electron?
    —————————————–

    of corse not.

    The two orbits continue being parallel.

    The intrinsic-spin of the electron in the neutron changes from up to down

    regards
    wlad

  345. Wladimir Guglinski

    ERRATA:

    In my last commment, where it written:

    But with the aim of verifying the present theory, would be of interest to test 7N15 as receptor.

    the correct is:

    But with the aim of verifying the present theory, would be of interest to test 7N15 as emitter.

  346. Wladimir Guglinski

    Substitute for 3Li7 and 28Ni in Rossi’s eCat

    The fuel used in the eCat is composed by an emitter 7Li of neutrons together with Ni isotopes receptors of neutrons.

    The emitter 7Li has three electron orbits:
    a) two orbits of electrons s1 and s2, which cancel each other their magnetic fields

    b) one orbit of electron p1, perpendicular to the z-axes of 7Li, whose magnetic field extracts the neutron of the nucleus 7Li
    FIG. 6:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    The radius of the orbit p1 depends on the element used as receptor.
    a) By using a receptor with big quantity of protons, the orbit will be larger, and the force of extraction is stronger.
    b) By using a receptor with low quantity of protons, the orbit will be smaller, and the force of extraction is weaker.

    .

    1- Beryllium 9Be isotope used as emitter

    Beryllium has only one stable isotope, the 4Be9.

    Figure 7 shows the orbits s and p.
    FIG. 7:
    http://peswiki.com/index.php/Image:FIGURE-7-substitute3Li7-28Ni.png

    The orbits s1 and s2 cancell each other their magnetic moments.

    The orbits p1 and p2 also cancell each other.

    CONCLUSION: Cold fusion cannot be obtained by using Be as emitter

    Any element having a pair number of protons cannot be used as an emitter of neutrons in cold fusion reactors.

    .

    2- Boron isotopes used as emitter

    The stable boron isotopes are 5B10 and 5B11. They occur naturally as follows:
    5B10 – 19,9%
    5B11 – 80,1%

    The isotope 5B10 is not of interest to be used as emitter, because it has not a neutron bound with a deuteron via spin-interaction.

    The good candidate to replace 3Li7 is 5B11, because they have similar structures, since they both have a neutron bound with a deuteron via spin-interaction, as shown in the Fig. 8 ahead.
    FIG. 8:
    http://peswiki.com/index.php/Image:FIGURE-8-substitute3Li7-28Ni.png

    The structure of the 5B11 is also shown in the page 60 Fig. 44 of the paper Stability of Light Nuclei
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    The magnetic force for the extraction of the neutron will be due to the orbit of the electron p3, shown in Figure 9.
    FIG. 9:
    http://peswiki.com/index.php/Image:FIGURE-9-substitute3Li7-28Ni.png

    As the neutron in the 5B11 is strongly bound than the neutron in the 3Li7, a stronger magnetic force due to the electron p3 orbit will be required.

    The 28Ni can be replaced by other receptors, in order to verify if the COP of the eCat can be increased.
    The following receptors could be used if their partnership with 5B11 would be able to extract the neutron from the 5B11 emitter:
    20Ca
    22Ti
    24Cr
    26Fe
    30Zn
    32Ge
    34Se
    36Kr
    38Sr
    However, probably the radius orbit of the p3 electron will not be sufficiently large so that to induce a magnetic force capable to extract the neutron.
    There is need to choose elements with higher number of protons.

    We have to expect that a good receptor to work with 5B11 must be found between Z=40 and 50:
    40Zr
    42Mo
    44Ru
    46Pd
    48Cd
    50Sn

    .

    3- Nytrogen isotopes used as emitter

    The two stable isotopes of nitrogen are 7N14 and 7N15.
    The isotope 7N14 is not of interest to be used as emitter, because it has not a neutron bound with a deuteron via spin-interaction.

    The structure of the stable 7N15 is shown in the Figure 10, and it could be used as emitter.
    FIG. 10:
    http://peswiki.com/index.php/Image:FIGURE-10-substitute3Li7-28Ni.png

    Unfortunatelly, 7N15 naturally occurring is only 0,37% , while 7N14 in 99,63%, and so nytrogen is not commercially viable to be used as emitter.

    But with the aim of verifying the present theory, would be of interest to test 7N15 as receptor.

  347. Joe

    Wladimir,

    Are you saying that, before the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is parallel to the orbit of the 1p1 electron; and that after the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is anti-parallel to the orbit of the 1p1 electron?

    All the best,
    Joe

  348. Wladimir Guglinski

    Andreas Moraitis wrote in November 30th, 2014 at 11:48 AM

    Dear Wladimir,

    In your reply to Joe you wrote: „[…] my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).”

    Does the difference between the mass of the neutron and the sum of the masses of the proton and the electron (about 782 keV/c^2) result from the relativistic speed of the electron?
    ————————————————

    Yes,
    and it is calculated in my paper Anomalous Mass of the Neutron, published in JoNP:
    http://www.journal-of-nuclear-physics.com/?p=516#more-516

    regards
    wlad

  349. ing. Michelangelo De Meo

    Hello Dr. Rossi,
    Italy is investing in the construction of expensive hydroelectric power stations in Montenegro (Albania ) for the production of electricity . Via a high-voltage submarine cable , connecting the two shores of the Adriatic Sea , will lead the electric current in Abruzzo and Puglia . Italy continues to squander money on ” clean energy ” . I hope that his plant Hot Cat is the solution to remedy this waste impactful . With the hope that it goes as soon as possible on the world market .

    http://www.primadanoi.it/news/mondo/554816/Elettrodotto-in-Abruzzo–la-lunga.html

  350. Andrea Rossi

    Ing. Michelangelo De Meo:
    Thank you for the interesting information.
    Warm Regards
    A.R.

  351. Andreas Moraitis

    Dear Wladimir,

    In your reply to Joe you wrote: „[…] my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).”

    Does the difference between the mass of the neutron and the sum of the masses of the proton and the electron (about 782 keV/c^2) result from the relativistic speed of the electron?

    Best regards,
    Andreas Moraitis

  352. Wladimir Guglinski

    Joe wrote in November 29th, 2014 at 10:54 PM

    Wladimir,

    In your model, does the neutron ever change the direction of its spin? If so, at what point does it do so in its travel from source nucleus to target nucleus?
    ——————————–

    Joe,
    we are obliged to conclude that yes, the neutron needs to change its spin.

    It changes its spin just after crossing the cross section of the p1 electron orbit.

    I think the neutron do it as consequence of the least action principle.

    As you know, my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).

    The proton has magnetic moment +2,793
    The neutron has magnetic moment -1,913

    So, the orbit of the electron about the proton yields the magnetic moment -(1,913+2,793) = -4,706.

    Before crossing the cross-section of the electron orbit p1, the neutron is moving having its electron moving with a parallel orbit with the orbit p1, and intrinsic spin up.

    When the neutron crosses the cross-section, the electron of the neutron changes its intrinsic-spin to down. So, the proton of the neutron is constrained to also change its spin, and therefore the neutron starts to move with a contrary spin (contrary to the spin the neutron had before to cross the orbit p1).

    Saying the contribution of the least action principle in other words: it is most confortable for the neutron to change its spin after crossing the cross-section of the orbit p1.

    regards
    wlad

  353. Wladimir Guglinski

    Herb Gillis wrote in November 29th, 2014 at 11:09 AM

    Wladimir Guglinsky:
    Would you necessarily need to have an even number of protons if you could otherwise obtain an even number of electrons? For example; many elements form stable ions which have odd numbers of protons but even numbers of electrons (for example; halides like chlorine). Do you foresee the possibility of LENR in an ionic solid (or liquid) containing one of these ions?
    ——————————

    Dear Herb,
    positive ions with even quantity of electrons will probably capture one electron, and they will have odd quantity of electrons. So cold fusion will not occur.

    In the case of negative ions with even quantity of electrons cold fusion would be possible if the excess electron is not expelled
    However we have no guarantee that the excess electron of the ion will not be expelled due to repulsion with the electron p1 of the other nucleus aligned with the ion, since the excess electron is weakly bound to the ion.

    regards
    wlad

  354. Joe

    Wladimir,

    In your model, does the neutron ever change the direction of its spin? If so, at what point does it do so in its travel from source nucleus to target nucleus?

    All the best,
    Joe

  355. Wladimir Guglinski

    Joe wrote in November 29th, 2014 at 3:51 PM

    And even if it would continue successfully past the plane, the neutron will experience the force of attraction when its South is pulled by the North of the electron’s orbit.

    The situation is worse if the neutron’s vector is allowed to change its orientation at any time during its travel. The much stronger magnetic moment of the electron’s orbit will ALWAYS end up PULLING the neutron towards it, therefore effectively preventing any neutron transfer from occurring.
    ———————————–

    Joe
    what you say makes no sense

    If there is force of attraction on the neutron when its spin is in clockwise direction, then if the neutron changes its spin in counter clockwise direction the force cannot continue to be of attraction

    http://peswiki.com/index.php/Image:ATRA%C3%87AO_PROTON_NEUTRON.PNG

    regards
    wlad

  356. Joe

    Wladimir,

    If both vectors, that of the neutron’s magnetic moment and that of the electron orbit’s magnetic moment, always stay in the same direction, there will be attraction at the beginning. But when the neutron is almost halfway across the plane of the electron’s orbit, repulsion will be experienced due to the North of the neutron’s vector approaching the North of the electron’s orbit vector. And even if it would continue successfully past the plane, the neutron will experience the force of attraction when its South is pulled by the North of the electron’s orbit. These two conditions should prevent a successful neutron transfer.

    The situation is worse if the neutron’s vector is allowed to change its orientation at any time during its travel. The much stronger magnetic moment of the electron’s orbit will ALWAYS end up PULLING the neutron towards it, therefore effectively preventing any neutron transfer from occurring.

    All the best,
    Joe

  357. Herb Gillis

    Wladimir Guglinsky:
    Would you necessarily need to have an even number of protons if you could otherwise obtain an even number of electrons? For example; many elements form stable ions which have odd numbers of protons but even numbers of electrons (for example; halides like chlorine). Do you foresee the possibility of LENR in an ionic solid (or liquid) containing one of these ions?
    Regards; HRG.

  358. ing. Michelangelo De Meo

    Dear Dr. Rossi:
    It is necessary to be fast with the delivery of your plants.
    The Global Carbon Project of the Tyndall Centre for Climate Change Research published in the journal Nature the results of a study of 2012. This study shows that among the countries and regions that have emitted more carbon dioxide in 2011 are: China ( with the 28 percent of emissions ) , the US ( 16 percent ) , the European Union ( 11 percent ) and India ( 7 percent ) . It is estimated that , in 2012 , global emissions have increased by 2.6 percent , reaching a record high of 35.6 billion tons . China is the country with the largest amount of emissions , but on a per capita has a relatively low level , with only 6.6 tons per person , well below the 17.2 tons per capita issued by the United States . The European Union has issued 7.3 tons of CO2 per capita .

    Italy wants to buy electricity from Montenegro and take it with a submarine cable .

    Terna is developing plans for new submarine interconnections with the Balkans with the aim of contributing to the diversification of sources and areas of energy supply and the reduction of the price of electricity in Italy and to increase the safety levels of safety Italian electricity system .

    http://www.terna.it/default/Home/AZIENDA/chi_siamo/terna_estero/Terna_nei_Balcani/interconnessioni_balcani.aspx

  359. Wladimir Guglinski

    Joe,
    I think only the nuclei with pair number of protons are suitable to transmute by cold fusion, because they also have pair number of electrons in the electrosphere.

    Because of the pair number of electrons, each pair of electrons cancell each other their magnetic moment, and so they do not influence the penetration of the neutron.

    While the nuclei with odd number of protons, have odd number of electrons, and therefore they always have one unpaired electron, and its magnetic moment do not allow the penetration of the neutron coming from other nucleus.
    For instance, suppose Andrea Rossi had used a fuel composed by 29Cu and 3Li7 in his E-Cat. The neutron after exit the 7Li would not succeed to enter within the 29Cu nucleus, because the magnetic moment due to the orbit of the unpaired do not allow it.

    It also seems that the best nuclei to get cold fusion is those ones with biggest nuclear magnetic moment, because they can have their z-axis aligned more quickly, and so the reactions will occur faster, and by this way a higher COP will be obtained.

    From this viewpoint, 24Cr would be best than 28Ni.
    But 24Cr has 24 protons, and therefore its positive electric field is smaller than that of 28Ni. Therefore the orbit radius of the electron p1 of the 7Li will be shorter (since the orbit of the 7Li is influenced by the electric fields of 24Cr and 7Li working together), and by this reason the magnetic moment of the p1 orbit will not be so efficient to extract the neutron of the 7Li.

    In the case of a fuel 46Pd-7Li be used in the E-Cat, the radius orbit of the p1 electron of the 7Li will be too much large, and the neutron will be captured from the 7Li with too much energy (high velocity). Then, instead of being captured by the 46Pd nucleus, the neutron actually crosses the 46Pd without to be captured, and the 46Pd will have not transmutation.

    Probably this is the reason why, after trying many elements, the tests made by Rossi have pointed that 28Ni is the best fuel to react with 7Li and to get the higher COP

    regards
    wlad

  360. Frank Acland

    Dear Andrea,

    You may be interested to see your work discussed in this article on the Science Blog at the Huffington Post:

    Low Energy Nuclear Reactions: Papers and Patents
    http://www.huffingtonpost.com/david-h-bailey/low-energy-nuclear-reacti_b_6189772.html

    Best wishes,

    Frank Acland

  361. Andrea Rossi

    Frank Acland:
    Thank you for the information,
    Warm Regards,
    A.R.

  362. Wladimir Guglinski

    Other reactions in Fleischmann-Pons and Mosier-Boss experiments

    After the formation of tritium by the first sandwich Pd-1H2-1H2, other sandwiches can be formed, as follows:

    1) Pd-1H2-1H3

    2) Pd-1H3-1H2

    3) Pd-1H3-1H3

    In these sandwiches 2He4 is formed

    As it is not used an oscillatory magnetic field so that to excite the nuclei, 2He4 is never captured so that to forma sandwich, because 2He4 has no magnetic moment

    Pd is not transmutted, because as the proton has charge, it is deviated by the electrons of the Pd electrosphere, and so the proton never hits the Pd nucleus

    In Fleischmman-Pons experiment they did not use an external magnetic field, and so the nuclei were aligned by the Earth’s magnetic field.
    In the days of magnetic storms of the Sun they did not succeed to replicate the results.

    In Mosier-Boss experiment she used an external magnetic field, and so the nuclei were aligned by that field.

    ,

    Regarding Rossi’s E-Cat

    In Rossi’s E-Cat, probably in the first experiments he applied an oscillatory magnetic field unable to excite the nuclei, and therefore only Ni61 had transmutation, because it has magnetic moment -0,75.

    But 58Ni and 60Ni had no transmutation, because they have magnetic moment zero, and so they cannot align their z-axis, and therefore they did not capture 7Li nuclei.
    By this way Rossi obtained a very low COP.

    By improving the oscillatory magnetic field, Rossi did succeed to excite the nuclei, and by this way 58Ni and 60Ni started to have transmutation, and by this way he got to increase the COP of the E-Cat.

    Andrea Rossi could confirm it to us, but as he never gives any information about the reactor, he will tell nothing.

  363. Wladimir Guglinski

    Fleischmann-Pons and Pamela Mosier-Boss experiments finally deciphered

    Fleischmann-Pons experiment and Mosier-Boss experiment are practically the same.
    But Mosier-Boss experiment has a special interest, because in her experiment he measured the energy of the neutrons emitted, in order of 10MeV, and such result is paradoxical, since the binding energy of the deuterons is only 2,2MeV.

    Let us see the mechanism of the cold fusion reactions in the Mosier-Boss experiment.

    1) A Pd nucleus couples with a deuteron, and they align their z-axes.
    Therefore both the Pd and 1H2 lose their spherical Coulomb barrier.

    2) The electron s1 of the deuteron takes an orbit perpendicular to the z-axes of the two nuclei Pd and 1H2

    3) The deuteron has only one proton, and so its positive electric field is small. Therefore the radius orbit of the electron s1 is short. By consequence, the magnetic force produced by the s1 orbit is not able to remove the neutron tied to the proton in the deuteron.

    4) The couple Pd-1H2 is stable, and nothing happens.

    5) Then the couple Pd-1H2 capture a second deuteron, and then a sandwish is made: Pd-1H2-1H2

    6) The electron s1 of the second deuteron takes an orbit perpendicular to the z-axis, between the two deuterons.

    7) Therefore the first proton-neutron of the first deuteron (sandwihed by Pd and the second deuteron) start to have a high oscilatory motion, because:

    7.a) The first deuteron is between two parallel orbits, one orbit of the electron s1 of the first deuterion, and other orbit of the electron s1 of the second deuteron

    7.b) the positive charge of the proton of the first deuteron is attracted by two contrary directions along the z-axis direction

    8) The neutron is tied to the proton in the first deuteron by the spin-interaction, which binding energy is 2,2MeV. When, in the oscillatory motion, the proton stops and changes the direction of its motion in contrary direction, due to the inertia the neutron has the tendency to continue its motion, and so the 2,2MeV binding energy is broken, and the neutron exits the first deuteron, and is captured by the second deuteron, and they form a tritium.

    9) The capture of the neutron causes an unbalance of masses in the newborn tritium. So, the z-axis of the tritium starts to gyrate chaotically, and it gets back its spherical Coulomb barrier, and it leaves away the couple Pd-1H2-1H3

    10) But in the instant when the neutron left away the first deuteron, at that the same time the proton is accelerated toward the contrary directionBut in the instant when the neutron left out The proton gets a high velocity, because free of the link with the neutron the proton is pushed by the repulsion with the positive electric field of the tritium.

    11) The proton enters within the Pd nucleus with 2,2MeV, helped by the orbit of the electron s1 pulling it against the Pd nucleus.

    12) Within the electrosphere of the Pd nucleus the proton captures an electron, and they form a neutron. That electron in the Pd electrosphere was moving with helical trajectory. As the electron loses its helical trajectory when it is captured by a proton and they form a neutron, the energy of the helical trajectory is transferred to the neutron. So, the neutron gets a portion of the energy 10MeV from the helical trajectory of the electron captured by the proton.

    13) When the proton entered within the Pd electrosphere, it was accelerated due to its attraction with the electrons of the Pd nucleus. If the proton was to leave away the Pd electrosphere, it would leave it with the same 2,2MeV, because leaving it its velocity would be decreasing by the attraction with the electrons. However, when the proton becomes a neutron, it loses its charge, and when the neutron starts to leave away the Pd nucleus it is not decelerated by attractions with electrons of the Pd nucleus. Therefore the neutron receives the other portion of the energy 10MeV from the acceleration of the proton before to become a neutron.

  364. Wladimir Guglinski

    Joe wrote in November 28th, 2014 at 7:30 PM

    Wladimir,

    Problem 1.
    If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue.
    ===========================================

    Dear Joe,
    I already explained it to Eric:
    ————————————
    However, in the case of the neutron, when the neutrons is moving toward the Ni nucleus, it can happen the following:

    1) Before to arrive to the cross-section of the electron’s orbit, the magnetic field induced by the electron’s motion applies a force of ATTRACTION on the neutron (the two magnetic vectors point to the the same direction).
    The neutron is PULLED by the electron’s orbit toward the Ni nucleus.

    2) After crossing the cross-section, the magnetic field of the electron starts to apply a force of REPULSION on the neutron (because the two magnetic vectors continue with the same direction).
    And so the neutron is PUSHED toward the Ni nucleus.

    However, I did not mention it because I am not sure if the neutron changes its magnetic field regarding the electron’s orbit after crossing the cross-section of the electron’s orbit (in this case, if the neutron changes the magnetic field vector in the contrary direction, then the electron begins to apply a force of ATTRACTION on the neutron, after it crosses the cross-section, and therefore decreasing the speed of the neutron going by inertia toward the Ni nucleus).
    ———————————-

    ,

    Problem 2.
    Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)
    =====================================

    Joe,
    due to Pauli’s Principle, only one electron can take the orbit taken by the electro p1 of the 7Li in the Figure 6 (orbit perpendicular to the z-axis)
    FIG 6:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    As the inner electron p1 of 7Li is more energetic than the outer electrons of the 58Ni, that orbit in the FIG 6 is taken by the electron of 7Li

    So, the electrons of the 58Ni do NOT take an orbit PERPENDICULAR to the z-axis, and they actually cancell each other their magnetic moments (the vector magnetic moment for the electron p1 has contrary direction of the vector for the electron p2, and the vector for the electron d1 has contrary direction of vector for the electron d2, etc)

    regards
    wlad

    regards
    wlad

  365. Wladimir Guglinski

    Dears Joe and Steven Karels

    58Ni and 60Ni have magnetic moment zero

    http://www.webelements.com/nickel/isotopes.html

    Therefore, the oscillatory magnetic field applied within the reactor is used not only for shaking the nuclei, but also for exciting them, because a nucleus with magnetic moment zero cannot be aligned by a magnetic field

    https://www.google.com.br/search?q=nucleus+excitation&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:pt-BR:official&client=firefox-a&channel=sb&gfe_rd=cr&ei=WzF5VIi6E43dwAT_rIKQDw

    Excited 58Ni has magnetic moment -0,1, while excited 60Ni has magnetic moment +0,2

    regards
    wlad

  366. Franco Sarbia

    Dear Dr. Andrea Rossi,
    still about the Hydrogen fueled Hot Cat on which returned Dr Joseph Fine. There are on the market electrolysers with size of a computer case, secure and low price, which store hydrogen in metal hydride cylinders. In case of using batteries of Hot Cat for cogeneration,it would be enough a hydrogen reserve to start the system.Once in operation the electrolyzer would provide the gas needs to drive it. The whole system would therefore be independent of any supply network of cables or pipes. The strategic interest of this hypothesis is too important because you will not have even thought.
    Warm Regards
    Franco Sarbia.

  367. Andrea Rossi

    Franco Sarbia:
    Thank you for your suggestion. The hydrogen produced that way is too expensive due to the very low efficiency of the electrolysis. I have made production of hydrogen with water electrolysis, therefore know well the issue.
    Warm Regards,
    A.R.

  368. Joe

    Wladimir,

    Problem 1.
    If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue. The same force that pulled the nucleon out of the source nucleus will have a much easier time to stop the nucleon’s progress toward the target nucleus when the nucleon crosses the plane of rotation of the 1p1 electron of 3Li7. And when we add to this the matter of distance, like you mentioned, which makes the attraction to the target nucleus even weaker for the nucleon, then I do not see much success for this model.

    Problem 2.
    Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)

    All the best,
    Joe

  369. Jeff Smathers

    I am trying to determine an approximate specific thrust using a thermally (1500 C ) pressurized gas.

    I am trying to ballpark a given mass for the Rossi Ecat and electrical power source, say about 1000 Kg
    and a liquid gas propellant that has a good expansion Coeffiecent with a contained mass also of 1000 Kg.

    Once in space and turned on, what would be the approximate velocity at the depletion of the heated expanding gas?

    And what is the resulting Specific Impulse value compared to say an ION engine….

    Thanks to those who can calculate on this end of the spectrum.

    Jeff Smathers

  370. Tom Conover

    Dear Andrea Rossi

    Have you heard of this “Low-Cost Microchannel Heat Exchanger”?

    Fabricated electronic cooling and high pressure MCHEX units to prove manufacturing approach:

    Tests showed 400% higher heat transfer rates
    80% reduction in volume
    5,000 psi pressure capability
    High effectiveness > 90%
    80% lower estimated external heat loss
    60% estimated lower cost

    http://energy.gov/sites/prod/files/2014/06/f16/A2%20Poster-Altex%20AMO%20RD%20Project%20Peer%20Review%202014.pdf

    Low-Cost Microchannel Heat Exchanger
    DOE Grant DE-EE0004541
    2013-2014

    Hope this helps,

    Tom

  371. Andrea Rossi

    Tom Conover:
    Very interesting, but not commercial yet.
    It is R&D of the DOE.
    Warm Regards
    A.R.

  372. Dear Mr. Rossi,

    I believe you stated years ago that you tried different metals to create the LENR effect, and only nickel really worked for you. I believe you also tried titanium powder, which has an interestingly high melting point of 1,668°C. That was with your old fuel formula used in the Warm-Cat (joke). Have you retried titanium powder with your new fuel formula in the Hot-Cat? Maybe you could play around with ways to create more nano sized cracks in the titanium lattice? Some scientists claim luck with titanium, and it would be fun to see if it reacted differently with the other elements of the new fuel mixture. Would it work at all? If it did work, would it be more stable?

    Best Regards, Christopher Calder

  373. Andrea Rossi

    Christopher Calder:
    I cannot give this kind of information.
    Warm Regards
    A.R.

  374. Andrea Rossi

    Dr Joseph Fine:
    Prices of fuels are very unstable. So far gas is cheaper than any other fuel, included hydrogen. Besides, the use of hydrogen as a fuel makes more complicated the certification issue. Neverheless, your information is interesting.
    Warm Regards,
    A.R.

  375. Joseph Fine

    Andrea Rossi and Readers:

    Ross Koningstein and David Fork, engineers at Google, described their work on the RE<C project in the Nov 2014 edition of "IEEE Spectrum".

    http://spectrum.ieee.org/energy/renewables/what-it-would-really-take-to-reverse-climate-change

    The RE<C project goal was to find methods to improve usage of Renewable Energy by reducing the cost of Electricity while reducing the amounts of CO2 gas and other pollutants going into the atmosphere by the burning of Coal to produce that Electricity.

    Google shelved the RE<C project in 2011 when they realized that their goals were unrealistic. While renewable energy costs had been reduced, they were not reduced below the costs of burning Coal. Nor would it have had significant effect on Global production of CO2.

    ( Note: Instead, Energy Policy has raised the costs of burning coal rather than reducing costs of other sources. The result has been to increase the cost of Electricity. )

    Google would probably be interested in helping develop applications that use "New FIre". (If they haven't contacted IH or you already.)

    1) Can you run/drive an E-Cat with Hydrogen rather than Natural Gas? If so, 2) Is it cheaper to produce natural gas or Hydrogen?

    Renewed regards,

    Joseph FIne

  376. Steven N. Karels

    Dear Andrea Calaon,

    You responded:
    “I see that silicon is present in the ash. Could the following reaction be possible?
    27Al + 7Li + e -> 28Si + 6Li

    Let me say. Hehehe. You noticed those lines at 28 on Figure 9 (lone) and 11 (with Al as well) of the ITPR.
    Let me first say that the reaction you wrote is impossible…”

    The reaction I was considering was in two steps:

    27Al + 7Li + e -> 28Al + 6Li
    while 27Al is stable, 28Li has a half-life of 2.24 minutes and decays to 28Si.

    So, I think, this might explain the removal of the 27Al from the fuel to the ash.

  377. Andrea Rossi

    Orsobubu, Koen Vandewalle:
    Without that ” damned spam robot” this blog could not work. We receive about one thousand of spam-messages per day: can you imagine how could we check a thousand messages per day to select the spam?
    Warm Regards,
    A.R.

  378. orsobubu

    Koen Vandewalle, you said:

    >PS: Andrea, soon or later, that damned spam-robot wille erase the entire J.O.N.P. and all knowledge about E-cat. I have no idea why you keep that monster.

    hehe because the artificial intelligence of the little monster is under the strict surveillance of the IT guy inside Rossi’s Team, and the IT guy is firmly in charge and he rules he commands he knows everything and he makes tons of backups… I’m sure that, even in case of an ultimate catastrophical event, like the one depicted at the end of the film “Planet of the Apes”, the knowledge of the E-cat could not be erased forever, since the IT guy has organized everything to bring it up held tight in the hand of the Liberty Statue, so that a new Charlton Heston could discover it in this way hehe:

    http://4.bp.blogspot.com/-OIF9h3JRpTM/Ti1kM3kFqQI/AAAAAAAAE8M/GIkgM4zivAY/s1600/planet-of-the-apes-statue-of-liberty-blu-ray-disc-screencap-hd-1080p-05.jpeg

    Anyway, from a human science point of view, to me Artificial Intelligence, robotics, Quantum Computing, Singularity etc are even more fascinating than new energy technologies and also physics in general, but I don’t care what skeptics say, Rossi’s personal saga is unmatched!

  379. Steven N. Karels

    Based on the Lugano Report the fuel consists of three components.

    This analysis speculates on those components.

    Known facts:
    1. Page 43: “Figure 1. Three different types of particles from the fuel material”
    2. Page 43: “Figure 2. SEI of two different types of particles from the ash material”
    3. Page 41: “Note that Li cannot be detected using EDS”

    Analysis

    On page 44 of the report, the EDS analysis of the different fuel particles is shown.

    Particle 1: Mostly nickel with some carbon and oxygen and a small amount of aluminum are seen.

    Particle 2: This particle consists of mostly aluminum with some carbon and oxygen and small amounts of nickel and chlorine.

    Particle 3: This particle contains mostly iron (Fe) with some oxygen and smaller amounts of carbon, silicon, chromium and manganese.

    Particle Sizes:
    From Figure 3, the particles sizes may be estimated. Particle 1 is around 50 to 100 microns; particle 2 is around 75 to 150 microns and particle 3 is 100 to 200 microns.

    Hypothesis

    Particle 1 is the nickel components with some cross contamination from the aluminum from particle 2. The oxygen is assumed to be from the air (oxides). Perhaps a trace of carbon either from the air (natural carbon dioxide) or as a small additive to remove the oxides at higher temperatures.

    Particle 2 is assumed to be LiAlH6 (recall the lithium and hydrogen are not detectable using EDS). The nickel is assumed to be cross contamination from particle 1. Unknown where the chlorine came from but it is small relative to the aluminum – it could be an impurity from the LiAlH4 source.

    Particle 3 is mostly iron with the same observation on the oxygen and the carbon. The silicon could be from the adhesive using to hold the sample while the chromium and manganese might be part of the iron sample (an alloy?).

  380. domenico canino

    dear andrea rossi,
    in Bondeno where you have a factory, the Lega Nord political movement, has won the regional elections. 75% percent. Please come back!!!

  381. Andrea Rossi

    Domenico Canino:
    We go where we are called with acceptable proposals. Politics are not my turf. The sole acceptable proposal we got so far has been from the USA, where I am working with a wonderful Team; thanks to them we are making a masterpiece in the factory of the Customer of our Licensee. This said, I conserve a loving sentiment toward Bondeno, where the first E-Cat has been born and where I worked in EON’s factory together with Prof. Sergio Focardi. I wish to the people of Bondeno everything good . They are laborious people that merits it.
    Warm Regards,
    A.R.

  382. Andrea Rossi

    Paul:
    He,he,he..
    A.R.

  383. Eric Ashworth

    Wladimir, Regards artificial intelligence, they did not use artificial intelligence because of it being artificial. If youn want intelligence you have to use the real thing. It’s called human. Not difficult to understand. By using the Oxford English dictionary it will be revealed that A.I. is not a real intelligence which it is not or you could say what created, what is termed artificial intelligence. When you answer the question all will be revealed. Regards Eric Ashworth.

  384. Wladimir Guglinski

    Joe wrote in November 27th, 2014 at 6:03 PM

    Wladimir,

    Every nucleus that you mentioned has a magnetic moment that is smaller than that of 3Li7. If the 1p1 electron of 3Li7 is able to pull a nucleon away from the nucleus of 3Li7, it should be even easier to pull nucleons out of the target nuclei that you mentioned. The result would be a collision of nucleons from both nuclei – intended source and intended target – right at the centre of rotation of the 1p1 electron.
    ———————————————-

    Joe, two things:

    1- 7Li has a neutron weakly bound

    2- 7Li has 3 protons, its electrosphere is small, compared with the electrosphere of Ni, which has 28 protons.
    Therefore the orbit of the electron p1 in the Fig. 6 is very nearest to the nucleus 7Li and far away from the nucleus Ni, As the magnetic force decreases with the square of the distance, you may realize that the magnetic force on the neutron of 7Li is very stronger than on the neutrons of the Ni.
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    The neutron (or proton) will always exit from the lighter nucleus to the heavier one.

    regards
    wlad

  385. Wladimir Guglinski

    Andrea Rossi wrote in November 27th, 2014 at 1:19 PM

    Wladimir Guglinski:
    We do not give information about our R&D regarding issues inside the reactor.
    ————————————–

    Dear Andrea,
    you did not understand.

    I am not asking any information about R&D regarding issues inside the reactor.

    It was suggested the following experiment:

    1- The reactor will be completely empty. Inside the reactor will be put a fuel composed by 20Ca41 , 7Li and H. The ash will be analysed after some days.

    2- Again the reactor will be completely empty, and will be put a fuel composed by 20Ca43 , 7Li and H. The ash will be analysed after some days.

    3- Again the reactor will be completely empty, and will be put a fuel composed by 20Ca40 , 7Li and H. The ash will be analysed after some days.

    .

    So, the test of the ash is regarding a fuel NOT used in your E-Cat, and therefore it has nothing to do with your E-Cat.

    In another words, your reactor will be used only as a vessel so that to promote cold fusion by using a fuel 20Ca, 7Li, and N (this fuel has nothing to do with your technology).

    This experiment is of the interest of the cold fusion researchers, so that to undertand cold fusion.

    But I understand that you do not waht to perform it.

    I think other cold fusion researchers will be interested to do it later.

    regards
    wlad

  386. Andrea Rossi

    Wladimir Guglinski:
    Let me rephrase it: we do not give information about ANY experiment we do or do not do inside our reactors.
    Warm Regards,
    A.R.

  387. Joe

    Wladimir,

    Every nucleus that you mentioned has a magnetic moment that is smaller than that of 3Li7. If the 1p1 electron of 3Li7 is able to pull a nucleon away from the nucleus of 3Li7, it should be even easier to pull nucleons out of the target nuclei that you mentioned. The result would be a collision of nucleons from both nuclei – intended source and intended target – right at the centre of rotation of the 1p1 electron.

    All the best,
    Joe

  388. Koen Vandewalle

    Dear Orsobubu,
    How “Intelligent” is something that allows its misuse or abuse ?
    I share your opinion.
    Kind Regards,
    Koen

  389. Koen Vandewalle

    Dear Wladimir,

    why didnt they use the A.I. nuclear model so that to discover the Rossi’s secret ?

    Andrea stated that some of his findings are serendipities. No idea if Artificial Intelligence can deal with serendipities.
    A tree that grows on a twig tends to collapse. Maybe the A.I. consumes more energy than E-Cat can produce.

    BTW: I very much like your FIGURES on the Calaon-Guglinski theory.

    Kind Regards,
    Koen

    PS: Andrea, soon or later, that damned spam-robot wille erase the entire J.O.N.P. and all knowledge about E-cat. I have no idea why you keep that monster.

  390. Wladimir Guglinski

    Dear Andrea Rossi,

    my theory of the non-spherical shape of the electric field (Coulomb barrier) of the nuclei can be tested in the E-Cat.
    Ahead I explain how it can be done.

    The nuclei have non-spherical Coulomb barrier, as shown in Fig. 1 for the 2He4.
    FIG 1:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    But due to the chaotic spin of the nuclei, the z-axis shown in Fig. 1 gyrates chaotically, and therefore in average the shape of the Coulomb barrier is spherical, as shown in Fig. 2 for the 2He4 (showing also the electrons of the electrosphere)
    FIG 2:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE2.png

    In order to have cold fusion between two nuclei their z-axes must be aligned by an external magnetic field.
    When the z-axis of a nucleus is aligned toward a direction, the chaotic rotation of the z-axis stops, and two nuclei may align their z-axes as shown in Fig. 5 for the nucleus 7Li.
    FIG. 5:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE5.png

    Therefore, only nuclei with nuclear magnetic moments can have transmutation via cold fusion.

    There is no way to get cold fusion with the following nuclei, because they have null nuclear magnetic moment:

    2He4, 4Be8, 6C12, 8′O16, 10Ne20, 12Mg24, 14Si28, 16S32, 18Ar36, 20Ca40, 22Ti44

    So, suppose that a fuel composed by 20Ca41 , 7Li and H is put in the E-Cat, and after some days the analysis of the ash shows that there was transmutation of the 20Ca41.

    The same is repeated with a fueld composed by 20Ca43, 7Li and H, and suppose after some days the analysis of the ash shows that there was transmutation of the 20Ca43.

    Then the next test will be made with a fuel composed by 20Ca40, 7Li, and H. The analysis of the ash must show that 20Ca40 did not had transmutation.

    The same can be repeated with other nuclei with magnetic moment zero, as for instance 14Si28, by comparing with the results made with its isotope 14Si29.

    If the experiments with nuclei having magnetic moment zero show that they do not transmute within the E-Cat, but their isotopes with non-null magnetic moment have transmutation, this will constitute a strong evidence corroborating my model of non-spherical shape of the Coulomb barrier of nuclei.

    In the case the non-spherical shape of the nuclei proposed in my theory is confirmed by experiments made within the E-Cat, it will be an important discovery for the undestanding of the mechanisms responsible for cold fusion occurrence.

    regards
    wlad

  391. Andrea Rossi

    Wladimir Guglinski:
    We do not give information about our R&D regarding issues inside the reactor.
    Warm Regards
    A.R.

  392. orsobubu

    Koen Vandewalle,

    you hope that the economic model is rendered obsolete by the advent of innovative technologies and scientific theories. This makes sense because, in order to have a revolutionary change, the economic structure must evolve in the first place; in fact these technologies may represent a structural change of economic relations, the same way that the steam engine allowed the passage to an industrial economic structure and the definitive affirmation of capitalist production.

    The problem, though, is that the revolutionary process is not mechanical, not deterministic, the contrary, it is dialectical. So there is a relationship between the economic structure and its superstructures, such the political one. If the change of the structure was enough, we would have already had the revolution, because the current means of production and the economic development that have been established since the beginning of the last century would have been enough to get rid of the old capitalistic system, clearly unfit to take full advantage of the actual potential of the social productive forces, as the world wars demonstrate.

    Unfortunately the political superstructure is still inadequate, and it is still in the hands of a class that manages the system in an anarchist way, with enormous waste, overproductions, useless duplications and destructive competition, crisis, violence, environmental damage, underdevelopment, unemployment. On the other side, the class of the producers, who realizes materially all the values, the use value (goods) and exchange values (money), and gently give them to the ruling class, is not yet conscious of its power and is poorly organized.

    Artificial Intelligence can only be developed by huge means of production, by biggest capitalists, from largest financial concentrations, and by the same capitalist states. So, Artificial Intelligence, in the hands of the class that holds these means of production, has no chance of being able to emancipate workers, for the simple fact that it will not be used to revolutionize the economic model, but rather to strengthen it. Therefore will not be used to overcome the social relationship of wage labor, the only system that produces not fictitious capital, but instead will be used to enlarge exploitation, because this is the only way to increase profits and capital.

    But since the reality is dialectical, in the course of this process the working class is inevitably strengthened, and also more and more crisis are inevitably produced. It is inside this inherent contradiction in social/economic relationships – and thus in political organization – that lies the possibility of change, certainly not in the evolution of the technology itself.

  393. KD

    Mr.Rossi
    I have question.:)
    Did you prepare your Thanksgiving turkey using E-cat technology as part of R&D.:)

  394. Wladimir Guglinski

    Joe,

    I did not find the magnetic moments of Ni58 and Ni60

    But as 61Ni has negative magnetic moment -0,75, and as 7Li has positive +3,24, it means that when they align their z-axis, the 61Ni must be up side down (7Li and 61Ni with contrary nuclear spins).

    20Ca41 has mag. mom. -1,59
    20Ca43 has mag. mom -1,31
    and therefore when they align their z-axes with 7Li, the two nuclei Ca also must have contrary spins of the spin taken by 7Li.

    stable 26Fe57 has mag. mom +0,09 (positive like mag. mom. of 7Li).
    This means that, when 57Fe aligns its z-axis with 7Li, they have to have the same spin (rotation in the same direction).

    I dont know yet what to think about the influence of the spin of the two nuclei, when they are gyrating in the same direction, and when they gyrating in contrary direction.

    regards
    wlad

  395. Wladimir Guglinski

    Koen Vandewalle wrote in November 27th, 2014 at 7:35 AM

    Wladimir,

    concerning your post “Dears Joe, Calaon, Orsobubu, Karrels, Eric…” posted on nov 25.

    I expect that the upcoming Artificial Intelligence will make obsolete most of the standard model and new models, finding the real unified field of the universe(s).

    Do you know of the existence of experiments on A.I. in nuclear science ?
    ————————————-

    Dear Koen,
    why didnt they use the A.I. nuclear model so that to discover the Rossi’s secret ?

    regards
    wlad

  396. Wladimir Guglinski

    Joe wrote in November 27th, 2014 at 1:20 AM

    Since your model must necessarily involve a target nucleus that has a greater magnetic dipole moment than the source nucleus in order that a nucleon be successfully drawn, compare the relative strengths of the magnetic dipole moments of Fe and Co; and of Ca and Sc.
    ———————————————

    Not necessarily, Joe

    7Li magnetic moment is greater than Ni.
    For instance, 7Li has magnetic moment +3,25, while 28Ni61 has magnetic moment -0,75

    What pulles the neutron of the 7Li toward the Ni is the orbit of the electron p1 (shared by 7Li and Ni), which magnetic moment is very greater than that of the 7Li:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    regards
    wlad

  397. Koen Vandewalle

    Wladimir,

    concerning your post “Dears Joe, Calaon, Orsobubu, Karrels, Eric…” posted on nov 25.

    I expect that the upcoming Artificial Intelligence will make obsolete most of the standard model and new models, finding the real unified field of the universe(s).

    Do you know of the existence of experiments on A.I. in nuclear science ?

    I also think that IP and patents, and some economic models will be obsoleted by A.I.

    Kind Regards,

    Koen

  398. Andrea Calaon

    Dear All,
    in these days I do not have much time for participating to the discussion of the JoNP. Please understand me. I will comment/answer you as soon as possible.
    Regards
    Andrea Calaon

  399. Joe

    Wladimir,

    You write,
    “As the authors of the Lugano Report were seeking only for Ni and Li isotopes, perhaps in the ash it can be found (if they look for):

    26Fe + Li7 -> 27Co + 2He4 + n

    20Ca + Li7 -> 21Sc + 2He4 + n

    As the authors were not looking for Co and Sc, perhaps they are in the ash.”

    Since your model must necessarily involve a target nucleus that has a greater magnetic dipole moment than the source nucleus in order that a nucleon be successfully drawn, compare the relative strengths of the magnetic dipole moments of Fe and Co; and of Ca and Sc.

    All the best,
    Joe

  400. Eric Ashworth

    Wladimir, With regards your reply 26th Nov 2014. To me conservation of momentum is displacement of field energy that has a knock on effect regarding a centre of gravity. Your formulae P=Mass/velocity if I am correct inertia comes about by the formuae Push=Mass and velocity/Movement that equals inertia or that no push or attraction upon mass equals no velocity/movement and thereby no inertia. If this is so then we are in full agreement. Evolution comes about by inertia. Really, conservation of energy is the disturbance of a field after the manufacture of a body. When field energy is disturbed it will react, over a period of time, in a specific way to re-establish its field or you could say its comfort zone. This is a natural reaction and only to be expected but could be termed a frustration upon an environmental influence.

    In the case of the neutron moving towards the Ni58 nucleus I believe it to be as stated in case 1) of your explanation.

    With regards the neutron moving towards the Ni58 from Li7 I will put forward my theory when I have time and will look forward to your comments. Regards Eric Ashworth.

  401. Wladimir Guglinski

    Steven N. Karels wrote in November 26th, 2014 at 10:07 PM

    A DC current would generate a magnetic field while a high frequency signal might be used to move or excite the fuel?
    ———————————–

    Probably yes

    regards
    wlad

  402. Wladimir Guglinski

    Steven N. Karels wrote in November 26th, 2014 at 10:07 PM

    Wladimir,

    I also suggest that the ash samples were small because not very much ash was actually produced. Recall the operation was only for 32 days and the majority of the nickel within the reactor was not consumed. So the isotopic analysis was on a very small sample and therefore the lesser occurring elements were not many. At some point, the measurements must become noisy due to a lack of material. So we maybe trying to analyze noise.
    —————————————

    Steven,
    I dont think so.
    32 days is enough for all the reactions to occur.
    Besides, as C, Ca, Cl, Fe, Mg, Mn exist in the fuel, why they do not appear in the ash??

    Let me tell you what I am thinking.

    I think several reactions can occur in the E-Cat.

    For instance, he have:

    9F has 1 complete hexagonal floor + 1 deuteron

    26Fe has 4 complete hexagonal floors

    So, 9F can lose one deuteron. Therefore, instead of 6Li7, actually the best element to react with 26Fe is the 9F:

    9F + 26Fe -> 27Co (the 26Fe captures one deuteron from the 9F)

    On the other hand:

    7N has one complete hexagonal floor minus a deuteron

    Therefore 7N can capture a proton:

    7N + H -> 8.0

    However,
    actually there is a lot of reactions to be considered, and there is no way to discover what are the real reactions occuring in the E-Cat.

    Even Andrea Rossi cannot tell us what is going within his E-Cat, in spite of he is promissing to prove that it is possible to explain its working from the Standard Model.

    There is only one way:

    1) After Andrea Rossi gets the patent of the E-Cat (and therefore his invention will be protected against plagiarists, and he will show everything within his reactor), several experiments will be made, as follows:

    2) Within the E-Cat will be put only 26Fe and 7F

    3) After runing along 30 days, the ash will be analysed

    3) Within the E-Cat will be put only 26Fe and 17 Cl

    4) After runing along 30 days, the ash will be analysed

    5) Within the E-Cat will be put only 26Fe and 6Li7

    6) After runing along 30 days, the ash will be analysed

    7) Within the E-Cat will be put only 20Ca and 7F

    8) After runing along 30 days, the ash will be analysed

    9) and so on, with all the combinations between two elements

    Any combination of two elements is able to react by cold fusion.
    The only difference is: some combinations of two elements have an easier reaction than other two elements.

    The paper Theoretical feasibility of cold fusion according to the BSM by Dr. Stoyan Sarg was published one year ago.
    Now the paper was published again, because the results of the E-Cat were confirmed by the Lugano Report.

    However, the paper was published again not because it gives a good prediction.
    In the Abstract Dr. Sarg says:
    “The analysis also predicts the possibility of another cold fusion reaction based on similarities between the nuclear structures of Ni and Cr.”

    But this is not true.
    Actually cold fusion occurs with any combination of two elements. The difference is because some combinations of two elements, together with some suitable improvements, give a higher COP.

    It seems to me Andrea Rossi used the paper of Dr. Sarg as a strategy to deviate the competitors from the secrets of his E-Cat, puting then in the wrong way.
    As he also told that the fuel of he E-Cat is Ni.
    Andrea Rossi wants to protect his invention, and sometimes he gives wrong informations, so that to deceive his competitors.
    He is playing a cat-mouse game with the competitors.

    So, it’s a waste of time to try to discover what are the reactions within the E-Cat.
    We have to wait Andrea Rossi to get the Patent.

    regards
    wlad

  403. Steven N. Karels

    Wladimir,

    I also suggest that the ash samples were small because not very much ash was actually produced. Recall the operation was only for 32 days and the majority of the nickel within the reactor was not consumed. So the isotopic analysis was on a very small sample and therefore the lesser occurring elements were not many. At some point, the measurements must become noisy due to a lack of material. So we maybe trying to analyze noise.

    The question for Andrea Rossi would be — does the isotopic composition of the eCat reactor change for a much longer run (e.g., a 6 month run)? I would still content that some helium was produced but not captured in their measurements. IMHO.

    I do think their is significance in the three helical wires, nominally for heating the reactor. A DC current would generate a magnetic field while a high frequency signal might be used to move or excite the fuel?

  404. Steven N. Karels

    Wladimir,

    I understand C, carbon, is sometimes added to remove oxides and oxygen (from the initial air in the enclosure) from a heated sample. So perhaps C is added to capture oxygen in the form of CO2. Naturally, I would assume, a more than sufficient amount of carbon would be present to make certain that oxygen and oxides were captured in the operation. So this might explain the presence of carbon. Opinions?

  405. Wladimir Guglinski

    Dears Joe and Calaon,

    I found very interesting the information about the high concentrations of C, Ca, Cl, Fe, Mg, Mn (not found in the ash), and also the elements C, H, O, N, He, Ar and F which cannot be measured quantitatively by the technique used.

    As you know, my new nuclear model is composed by hexagonal floors formed by deuterons, with a central 2He4.
    So, we have:

    8O has 1 complete hexagonal floor

    12Mg has 2 complete hexagonal floors minus 2 deuterons

    18Ar has 3 complete hexagonal floors minus 2 deuterons

    20Ca has 3 complete hexagonal floors

    26Fe has 4 complete hexagonal floors

    28Ni has 4 complete hexagonal floors plus 2 deuterons

    A nucleus with a complete hexagonal floor as 20Ca and 26Fe may have an Accordion-Effect without distortions (while the Accordion-Effect of Ni is distorted).

    So, we have to suppose that 26Fe and 20Ca can get a better alignment of their z-axes with 7Li than 28Ni.

    As the authors of the Lugano Report were seeking only for Ni and Li isotopes, perhaps in the ash it can be found (if they look for):

    26Fe + Li7 -> 27Co + 2He4 + n

    20Ca + Li7 -> 21Sc + 2He4 + n

    As the authors were not looking for Co and Sc, perhaps they are in the ash.

    Dear Calaon,
    what do you think about?

    regards
    wlad

  406. Wladimir Guglinski

    Dears Calaon and Steven Karels,

    probably the nuclear reaction between 7Li and 58Ni occurs easily when the two nuclei are aligned by a magnetic field in the E-Cat, without any additional improvement.

    But probably in the beginning the E-Cat had a very low COP.
    There was need to shake the nuclei within the reactor, in order to get the most high quantity of reactions 7Li+58Ni (and also 7Li+ 60Ni and 7Li+61Ni) by second.

    That’s why along the years Andrea Rossi had improved his reactor, so that to increase the velocity of the quantity of the reactions by second, in order to increase the COP.

    For instance, perhaps the first E-Cat had only one coil in the alumina cylinder.
    By putting 3 coils, if the electric current is AC, inside the reactor occurs an oscilatory magnetic field by the overlap of 3 oscillatory magnetic fields. By this way the nuclei are shaken, and is increases the speed of their interaction.
    With an additional magnetic field induced by a DC current (or a permanent magnet) the nuclei 7Li and 58Ni have their z-axis aligned.

    In order to increase the COP, perhaps Rossi had used a catalyst.

    But in the last page 53 of the Lugano Report there is an intriguing information:

    ————————————————–
    The measured analytes were Ni, Li, and Al. The elements Ni and Al are measured with two independent emission lines to minimize risk for systematic errors. The elements C, H, O, N, He, Ar and F cannot be measured quantitatively by this technique.

    Sample 1 was ash coming from the reactor in Lugano. Only a few granules of grey sample were possible to obtain from the ash and they didn’t look exactly the same. One large and two very small granules were observed.

    Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a very fine powder. Besides the analyzed elements it has been found that the fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these are not found in the ash.
    ————————————————–

    Let us analyse it:

    1- The catalyst could be C, O, N, Ar and F, because perhaps they appear in the ash, but they cannot be measured quantitatively by the technique used.

    2- Aluminium also can be the catalyst

    3- C, Ca, Cl, Fe, Mg, and Mn cannot be the catalyst, since they do not appear in the ash, and therefore they are consumed within the reactor

    4- However, as C, Ca, Cl, Fe, Mg, and Mn do not appear in the ash, then where did they go ???

    5- As they do not appear in the ash, it means that they had transmuted.

    6- Then why the authors of the Report did not speculate about a possible reactions between them ? (for instance, with hydrogen)

    8- And why did not they try to discover what would be the elements resulted from their transmutation?

    9- Besides, as Rossi claims that his E-Cat consumes Ni, why a hell there is a high concentrations of C, Ca, Cl, Fe, Mg and Mn in the fuel ?? (and they are not found in the ash) ???

    regards
    wlad

  407. Wladimir Guglinski

    Bernie Koppenhofer wrote in November 26th, 2014 at 11:53 AM

    Dr. Rossi: Have all the reactors used in the third party testing been returned to you? Thanks again for this site and sharing, and Happy
    ————————————

    Bernie,
    I suppose that, based on the protocol, before the test Andrea Rossi had already agreed that the reactor would be returned, and the test was done with the Professors’ promise to return the reactor. By the way, probably was made a contract-return of the reactor, signed by IH, Leonardo and the Professors of the ITP.

    regards
    wlad

  408. Bernie Koppenhofer

    Dr. Rossi: Have all the reactors used in the third party testing been returned to you? Thanks again for this site and sharing, and Happy Thanksgiving!

  409. Andrea Rossi

    Bernie Koppenhofer:
    The Hot-Cat that I gave to the Professors of the ITP has been given back to me the day after the day in which the reactor has been turned off. The Professors had only one reactor, because the other 2 that I brought to Lugano as spare parts, just in case of breakages, have not been delivered to the Professors, since no breakages happened to the one we gave them to be tested.
    Warm Regards,
    A.R.

  410. Andrea Rossi

    Dr Joseph Fine:
    Thank you for this contribution.
    Warm Regards,
    A.R.

  411. Wladimir Guglinski

    Eric Ashworth wrote in November 25th, 2014 at 10:44 PM

    Wladimir, Your reply to Joe Nov 24th 2014. “As I said in my comment of Nov 21st. 2014. “Due to the inertia, the neutron continues moving and it enters within the Ni58 through the “hole’ in the electrosphere of the Ni58″.

    This reply of yours I believe to be correct but you use the word inertia which I do not think is fully understood as a cause of an effect. For me inertia is a cause of movement but what causes the object to continue to move when the propelling force is removed?.
    ————————————————–

    in absence of force there is conservation of momentum P= m.V . The inertia does not depend on actuation of a propelling force.

    However, in the case of the neutron, when the neutrons is moving toward the Ni nucleus, it can happen the following:

    1) Before to arrive to the cross-section of the electron’s orbit, the magnetic field induced by the electron’s motion applies a force of ATTRACTION on the neutron (the two magnetic vectors point to the the same direction).
    The neutron is PULLED by the electron’s orbit toward the Ni nucleus.

    2) After crossing the cross-section, the magnetic field of the electron starts to apply a force of REPULSION on the neutron (because the two magnetic vectors continue with the same direction).
    And so the neutron is PUSHED toward the Ni nucleus.

    However, I did not mention it because I am not sure if the neutron changes its magnetic field regarding the electron’s orbit after crossing the cross-section of the electron’s orbit (in this case, if the neutron changes the magnetic field vector in the contrary direction, then the electron begins to apply a force of ATTRACTION on the neutron, after it crosses the cross-section, and therefore decreasing the speed of the neutron going by inertia toward the Ni nucleus).

    regards
    wlad

  412. georgehants

    Dear Mr. Rossi, after all your years of hard work, at this time what is the thing that you are most looking to achieve and will give you the most satisfaction in your work?
    Could it be the successful completion of the 1MW plant for your customer?

  413. Andrea Rossi

    Georgehants:
    Absolutely yes: when the contract signed by IH with their Customer for the 1 MW plant will have been totally satisfied, fullfilled and totally paid for, that will be the first plant in history making real energy in an industrial process. That will be the real game changer in the history of the production of energy, like the “New Fire”. This is why I have no time at all for any other issue, in this period. I need maximum focus, because failure is not an option and all the problems that pop up on daily basis have to be resolved properly to consolidate the technology. I want this masterpiece made by our Team to be perfect.
    Warm Regards,
    A.R.

  414. Wladimir Guglinski

    Dear Herb Gillis

    After the announcement by Fleischmann and Pons of their results in cold fusion experiment, along a decade the replicability of the results was a serious problem.

    As nobody knew how cold fusion occurs (because it is impossible by considering the Standard Model, and there was not any new nuclear model compatible with cold fusion) the cold fusion researchers faced the challenge of to replicate the results they claimed to have obtained earlier.

    I had analysed the problem of replicability by considering my new nuclear model, and I had discovered why in some days the researchers did succeed to replicate the results, and in other days they had failed.

    In those experiments the nuclei were aligned by the magnetic field of the Earth. But in some days there are magnetic storms in the Sun, and so the alignment of the nuclei by the Earth’s magnetic field is disturbed by the influence of the Sun’s magnetic field, and that’s why in some days the researchers did not succeed to replicate the results.
    Also, any apparatus inducing magnetic field in some laboratory could have influence in the results. Therefore, a researcher could succeed to replicate the experiment in his laboratory, but when other researcher tried to replicate the experiment in his laboratory he did not succeed to replicate it.

    Then I had submitted to Infinite Energy my paper What is missing in Les Case’s catalytic fusion, and in 2002 the magazine had published it. In the paper I had suggested to use an external source of magnetic field, in order to eliminate the disturbance of the magnetic field of the Sun (and also to replace the magnetic field of the Earth).

    In 2003 Dennys Lets and Dennys Cravens had exhibited in the cold fusion ICCF-10 their experiment where they had used an external source of magnetic field, and by this way they had solved the problem of the missing of replicability in cold fusion experiments.

    Probably Andrea Rossi took knowledge on the experiment made by Lets and Cravens, and then Rossi started to use an external magnetic field in his experiments.

    So,
    it is possible cold fusion can occur in the intestines of some animals, because the alignment of the nuclei of the food is produced by the magnetic field of the Earth.
    Of course the cold fusion occurence requires the animal to be at rest, in order to promote the alignment of the magnetic fields along a long period of time. This is the case, for instance, of the bears when they hibernate.
    As the white bears live in the north pole of the Earth, the alignment is easier to occur, because in the poles there is convergence of a big quantity of magnetic lines of the Earth’s magnetic field.
    http://en.wikipedia.org/wiki/Earth%27s_magnetic_field

    Regarding the birds, they do not change the position of their body when they sleep, cold fusion can occur in their intestines during the night (preferably in countries near to the poles of the Earth).

    regards
    wlad

  415. Wladimir Guglinski

    Herb Gillis wrote in November 25th, 2014 at 6:24 PM

    Wladimir Guglinsky:
    The situation you are describing reminds me of nuclear magnetic resonance (NMR), where nuclear spins may be aligned in a strong magnetic field. NMR is an analysis technique widely used in chemistry and medicine. Do you think that an existing NMR machine might be able to tell us something about LENR? For example; by inserting a mixture of Li7, Ni, and H.
    NMR is usually performed in liquid state, but not always. If your theory is correct then do you think LENR devices could be constructed from liquid mixtures in a sufficiently strong external magnetic field? In this way we could do away with the constraints of a solid matrix?
    ———————————————

    Dear Herb
    probably yes.
    But we have to remember that the E-Cat has some special conditions. For instance, Andrea Rossi uses three parallel non-overlapping coils inside the reactor, and so three magnetic fields are produced.
    In the case the current is AC, then of course he uses another source of a magnetic field, so that to produce a resultant vector magnetic field for those three fields induced by the three coils.

    We have also to remember that cold fusion probably occurs in some animals, because 200 years ago it was observed that in the feces of certain birds appear some elements that do not exist in the food they ate.
    Therefore it is reasonable to suppose that when bears hibernate they can also produce cold fusion in their intestine.

    regards
    wlad

  416. Wladimir Guglinski

    Andrea Rossi wrote in November 24th, 2014 at 11:26 AM

    Wladimir Guglinski:
    There is a line of articles and we publish them, after peer reviewing, in the order we receive them. Every Author has his reasons to consider urgent his own paper’s publication and we do not grant privileges to anybody.o the JoNP’s blog.
    ——————————————-

    Dear Andrea

    In 3rd Dec 2013 the JoNP had published the paper Theoretical feasibility of cold fusion according to the BSM-Supergravitation unified theory, by Dr. Stoyan Sarg.

    Now in 2nd Nov 2014 the JoNP is publishing again the same paper by Dr. Sarg, Theoretical feasibility of cold fusion according to the BSM.

    This paper has been published again in the JoNP because in October 2014 the Lugano Report had confirmed the results of the E-Cat.

    This seems to be a privilege.

    regards
    wlad

  417. Eric Ashworth

    Wladimir, Your reply to Joe Nov 24th 2014. “As I said in my comment of Nov 21st. 2014. “Due to the inertia, the neutron continues moving and it enters within the Ni58 through the “hole’ in the electrosphere of the Ni58″.

    This reply of yours I believe to be correct but you use the word inertia which I do not think is fully understood as a cause of an effect. For me inertia is a cause of movement but what causes the object to continue to move when the propelling force is removed?. There can also be an attractive force. What I believe is that there is a displacement of the magnetic fields contained within the object and which then displaces the objects centre of gravity. When these centres of gravity are distorted by their magnetic fields there is an imballance directly related to the energy input required to move the object at a specific velocity. At N.T.P. an object set in motion will regain its original field but it requires a duration of time which can be measured as a distance. At a higher temperature under normal pressure but with an induced force an object can regain its N.T. within normal pressure to reveal an imballance in its centre of gravity and a distorted magnetic field that will respond to its exterior environment because it contains properties of inertia. The neutron, I believe, has a centre of gravity and a field composed of quarks. When distortion sets in because of inertia there is a period of readjustment that can be interpreted as a distance. If the readjustment occurs over a vector then its a mobile inertia. If it occurs over an oscillation then it is a static inertia. This is how I understand inertia. There seems to be an enigma with regards what is referred to as the coulomb barrier and an inability of being able to cross it. Perhaps and this is what I believe, the referrence coulomb barrier should be coulomb barriers. Structure I believe is made up of densities with regards a geometric structure. This I shall put together and try to explain as best I can. Regards Eric Ashworth.

  418. Joseph Fine

    Andrea Rossi & Readers of the JONP:

    When he was at Hughes Aircraft on a Masters’ Degree Work-Study program, John T. Neer took written Notes of Lectures given by (Physicist) Richard Feynman.

    The 5-Volume set of Lecture Notes are partially similar to the contents of the 3 Volume FLP (Feynman Lectures in Physics) series, but cover different topics as well.

    Volume 5 contains a set of Mathematical Techniques which should be useful to anyone interested in Physics or to those who want to brush up on their skills.

    http://www.thehugheslectures.info/about/

    http://www.thehugheslectures.info/the-lectures/

    As far as I know, there were no references to phenomena similar to the “Rossi Effect”, which should not be surprising if everything falls into ‘Standard Physics’.

    Unfortunately, Richard can not be here to ponder the current mysteries for a few months, or even a few days, and then snap his fingers and say ” Ah-ha! So that’s what’s happening! ”

    Best wishes to everyone for a Happy Thanksgiving. Perhaps someone will use these Lecture Notes and have an “Ah-ha” moment.

    Thankful Regards,

    – Joseph Fine

    ////////////////////////////////

    The Hughes Lectures

    Feynman Lecture Notes by John T. Neer

    The Lectures

    These lectures notes run from the fall of 1966 to 1971. Feynman lectured prior to this period and continued on after 1971. With a few exceptions, the actual 2 hours lectures were not dated. However, the volumes in chronological order.

    I want to stress, again, that these are my personal notes and are only a representation of the lectures I attended. They are to the best of my ability my recreation from memory and my original real time notes. No AV recording system was used in the transcription of my raw notes.

    25 MB Download

    Volume 1
    Astronomy, Astrophysics, and Cosmology
    (224 pages)

    Feynman solicited topic input from the scientists and engineers at the Labs for the coming year. New discoveries were being made in astronomy, astrophysics, and cosmology at the time. This 1966-1967 lecture series focused on these subjects. This volume is unique since, as far as I can tell, Feynman did not lecture on this subject matter at CalTech. While much of the material is now dated, what remains is a look into the mind of Feynman as he worked to explain such topics as stellar evolution, nuclear synthesis, cosmology, “black stars” (aka black holes), and general relativity.

    I inserted more current content from the web which relates to the 1966-67 lectures with recent experimental observations and discoveries. While this lecture series has been “eclipsed” by the tremendous theoretical and experimental advancements over the past 45 years, I am sure the reader(s) will find in these lectures the power of Feynman’s insight and ability to have fun with a new subject not touched on by him at CalTech in his “normal” class and research work. I trust others, more specialized in the topics of volume 1, can and will contribute to the additional information to further enrich the notes in the future. This editing will best be done when the notes are moved and dropped in a dynamic and editable platform, yet to be identified.

    The Volume I subject matter was not part of his prior lecture activity, Feynman would talk with some of his CalTech colleagues who worked in the field of astronomy, astrophysics, and cosmology about their work and theories. He would then come to the lecture literally with a (maybe 2 or 3) 3×5 cards and proceed to pour out 2 hours of theory and complex mathematical representations of the topic of the day. This was his genius and almost mystical in his ability to focus his thinking and presentation ability on the most important aspects of a given topic.

    36 MB Download

    Volumes 2
    Relativity, Electrostatics, Electrodynamics, Matter-Wave Interaction
    (209 pages)

    Feynman reflected on how he could teach his original FLP’s volume 2 & 3 differently and better than in his first pass through the subjects five years earlier. The attendees wanted him to lecture a couple years on the subject matter in the original FLP and essentially let him give his revised, enhanced, and expanded lectures. This then led more naturally into QED with a good foundation layer established. Feynman also tailored his lectures more to the level of his audience understanding they were not freshman and sophomore undergraduates but post graduate, doctorate level scientists, employed doing advanced research.

    49 MB Download

    Volume 3
    More on Matter-Wave Interaction, Intro to Quantum Mechanics, Scattering Theory, Quantum Theory of Angular Momentum, Intro to Lie Group, SU 2 & 3 “stuff”, Quantum Electrodynamics (QED), Pair Production
    (314 pages)

    Feynman went on in greater detail to complete his lectures on wave-matter interaction. From there he started into quantum mechanics and his path history formulation. He extended his lectures to include Lie Group theory and the SU 2&3 “Stuff”.

    Feynman diagrams are discussed in Volume 3 at some length as he went deep into QED theory including such topics as quantum scattering. As better understood today, his diagrams represent a visual language of the complex physical processes at the particle interaction level. I have noted recently that with the power of new computers and new concepts the Feynman diagrams have, arguably, run their course. While this is possibly the case, I would assert that bypassing a fundamental understanding of the Feynman diagram concept makes it hard to understand what replaces them. This is like hand held calculators replacing the need to know the fundamental multiplication tables and being able to check what the calculator is telling you. I personally observed in a number of lectures where Feynman would self-check himself as he was working out the math because he could sense that if he kept going he would not get the right physics. This was his true genius at work. That was truly amazing to both watch and try to absorb in real time.

    13 MB Download

    Volume 4
    Molecular Biology
    (65 pages)

    The Molecular Biology lectures started out and then eventually died out as the year progressed. Feynman found the material challenging to get his head around before the lecture and, therefore, very time consuming. He apparently found a CalTech colleague, Seymour Benzer, who changed from physics to biophysics as a person who stimulated Feynman’s interest in this topic.

    By consensus the lecture series ended early. Feynman was deep into his own parton theory which was his version of quark theory. He and Gell-Mann were collegial competitors in those days.

    In preparing these notes for release I decided to include what notes I had of those lectures only to give evidence of Feynman’s interest to explore all the dimensions of science and nature. For those involved in the field these notes will not provide much informational value particularly with all the advancements on research and understanding of molecular biology. The value, I believe, for the reader is how Feynman thought through the subject matter and mentally organized it so he could lecture on it. That might aid teachers in this field to sharpen up their own presentation material. At the end of the volume are my un-transcribed real-time notes that I never got to but I decided to include for those who are into this field.

    6 MB Download

    Volume 5
    Mathematical Methods/Techniques in Physics and Engineering
    (163 pages)

    By some who have seen samples of my notes Volume 5 has been referred to as the “missing lectures” to the FLP “Red Books”. Feynman himself felt that he should have taught the mathematical methods first and then the physics since math is the “language” of physics. Feynman was apparently talked out of starting with a course in math-physics. The attendees at the lab talked him into a year-long lecture on his approach to mathematics as the language of physics.

    I note here also that the math lectures have been referred on the Reddit by someone as “sophomoric” since all physic students must take similar course work and presumably “master” math while learning the physics. In my own case I wanted to learn the physics and minimize the math, or better said, not confused by the physics because the math was too difficult to grasp.

    This is how Feynman approached physics and how he taught himself, at an early age, by developing many shortcuts through the math; “Feynman diagrams” were one clear by product of his self learning process. He did not want to get bogged down and distracted from understanding the physics. This is why and how he got involved in the Manhattan Project; he was their math wizard.

    One story he told of those days: Someone came running into him needing a quick answer to a nuclear decay process that was described by some expansion series like the Sum from 1 to infinity of 1/(1+n^2)[probably not the real one]. Feynman asked how accurately he wanted the answer and the person said 10% would do for now. Feynman said he took a few seconds and said the answer was 1.3 (or something like that); the person was amazed how fast he could give him that and asked how he did it. He said since you told me you only wanted the answer to 10%, it was only necessary to go to the second term in the series expansion and that was good enough for better than 10% accuracy. This story is emblematic of Feynman’s mathematical thinking which is not sophomoric. This is why he made such a contribution to the Manhattan project and ultimately QED. He did indeed “think different”.

    In my own experience I found in my graduate studies that the some of the professors tended to focus more on the math rigor than in teaching the real physics. In Feynman’s world he “felt” the physics and used the math to express that “feeling” and understanding. Language does not necessarily express the essence of the content contained in the idea being described. One must understand both the power and limitations of the language used when discussing a subject. Words don’t always express what one wants to say; so it is for math and physics.

    Lecture Sidebars: Another “feature”, or aspect, of the notes is my attempt to capture “side bar” topics. These special topics or thoughts (including some philosophical ones) added color and currency to the lectures as only Feynman could deliver. He was unconstrained in the lecture environment to take off on a sidebar and the attendees both enjoyed and encouraged him to do so.

    © Copyright 1966 – 2014 John T. Neer.
    ///////////////////////////////////////

  419. Andrea Rossi

    TO ALL OUR AMERICAN READERS:
    THE TEAM OF THE JoNP WISHES YOU ALL A PEACEFUL THANKSGIVING DAY. MAY GOD BLESS YOU ALL!

  420. Steven N. Karels

    Dear Andrea Rossi,

    Any Black Friday specials on eCat reactors?

  421. Andrea Rossi

    Steven N. Karels:
    he,he,he…
    Not yet.
    Warm Regards,
    A.R.

  422. Wladimir Guglinski

    Dear Dr. Stoyan Sarg,

    In the page 2 of your paper Theoretical feasibility of cold fusion according to the BSM it is written:

    From Fig. 3 we see that the nuclear overall shape for elements with 18<Z<86 have not spherical but elongated shape.

    In the Fig. 3 we also se that 10Ne20, with Z=10, has spherical shape.

    However,
    in 2012 by the journal Nature published the paper How atomic nuclei cluster, where it is shown the shape of the 10Ne20 detected by experiments.

    The Fig. 1 of that paper shows that 10Ne20 has non-spherical shape, and so the structure proposed in your paper is in disagreement with the experimental results, since according to your nuclear model the 10Ne20 must have spherical shape.
    http://www.nature.com/nature/journal/v487/n7407/full/nature11246.html

    I think the readers of the JoNP would like you come to explain why your nuclear model predicted wrongly the shape of the 10Ne20.

    I also would like to remember you that in November 3rd, 2014 at 10:33 AM I had posted as comment herein in the JoNP four questions about your models of proton and neutron (to be responded by you) but you did not respond any of them yet.

    regards
    wlad

  423. Herb Gillis

    Wladimir Guglinsky:
    The situation you are describing reminds me of nuclear magnetic resonance (NMR), where nuclear spins may be aligned in a strong magnetic field. NMR is an analysis technique widely used in chemistry and medicine. Do you think that an existing NMR machine might be able to tell us something about LENR? For example; by inserting a mixture of Li7, Ni, and H.
    NMR is usually performed in liquid state, but not always. If your theory is correct then do you think LENR devices could be constructed from liquid mixtures in a sufficiently strong external magnetic field? In this way we could do away with the constraints of a solid matrix?
    Regards; HRG.

  424. JCRenoir

    I have been impressed by the last data you gave of the 1 MW plant: you confirm half cubic meter of reactor to give 1 MW?
    JCR

  425. Andrea Rossi

    JC Renoir:
    Yes: the reactors combined make a volume of half cubic meter to yield 1 MWh/h of Thermal energy. All the rest of the plant is constituted by heat exchangers.
    Warm Regards,
    A.R.

  426. Curiosone

    Dr Rossi:
    Your critics have found a resistance that has no linear resistivity with the temperature, exactly as you said many times. So this drops the accusations made from someone. Comments?
    Godspeed,
    WG

  427. Andrea Rossi

    Curiosone:
    I cannot comment this issue.
    Warm Regards,
    A.R.

  428. Wladimir Guglinski

    Dears Joe, Calaon, Orsobubu, Karrels, Eric…
    … and anybody interested in the subject.

    Dr. Stoyan Sarg is going to pronounce a speech where he says that Coulomb barrier was wrongly interpreted in scattering experiments:
    “At the beginning it discuses the major methodological error in scattering experiments that leads to a tremendously wrong vision about the Coulomb barrier.”
    http://www.e-catworld.com/2014/11/21/dr-stoyan-sarg-to-address-nanotek-expo-2014-on-lenr/

    But Dr. Sarg is wrong.
    There is nothing wrong with the scattering experiments, and the vision about the Coulomb barrier is correct.

    In normal condictions (different of those occurring in cold fusion experiments) the electric field of nuclei (Coulomb barrier) is spherical, as correctly interpreted by the nuclear theorists, because of the following:

    1- The nuclei have non-spherical Coulomb barrier.
    For instance, the figure shows the Coulomb barrier for the 2Her, shown as yellow in the figure.
    FIG. 1:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    2- But the nuclei have chaotic rotation (due to repulsions between protons) and the z-axis of the Figure 1 is changing its direction every time.

    3- As consequence of the chaotic rotation, the electric field takes in average the spherical shape, as shown in the Figure 2 ahead for the 2He4.
    FIG 2:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE2.png

    4- The spherical Coulomb barrier of the Figure 2 was detected in the scattering experiments, and so the vision of the Coulomb barrier by the physicists was correct

    5- However, in cold fusion phenomena occurs the alignment of the two z-axes of two nuclei (as for instance 7Li and 58Ni in the Rossi’s Effect).
    We see it in the Figure 5 ahead.
    FIG 5:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE5.png

    6- The two Coulomb barriers of 7Li and 58Ni take their original non-spherical shape in the Figure 5 because the z-axes of the two nuclei stop to gyrate chaotically. This happens only in the cold fusion experiments (for instance, in the Rossi’s E-Cat the z-axes of 7Li and 58Ni are aligned along the axis of the alumina cylinder, because of the magnetic field created by the electric current in the coils).

    7- As we realize from Figure 1, there is a “hole” in the Coubomb barrier of the nuclei, along the z-axis:
    FIG 1:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    8- When the two z-axes of two nuclei are aligned (as 7Li and 58Ni in the Figure 5), the two “holes” of the two nuclei are aligned, and so it is easier for a particle as a proton or a neutron to exit one of them and to enter within the other.

    regards
    wlad

  429. Steven N. Karels

    Dear Andrea Rossi,

    Have you considered the application of eCat technology to long distance transportation? Assuming you can solve the issue of the eCat startup time, perhaps an eCat system to generate electricity to charge a large battery bank that drives trains?

    According to Wikipedia, each drive wheel requires up to 3000 hp with around 8 drive wheel per locomotive. In total this would be around 20 MW of electrical power or about 50MW of thermal power as a rough estimate.

    The cost economy for long distance hauling would be significant and there is no exhaust when operating in tunnels or inside buildings.

    A similar larger system might apply for marine cargo transportation.

  430. Andrea Rossi

    Steven N. Karels:
    That is a possible application, but not in a short time. Certifications in that area can take ten to twenty years, as I learnt in a meeting with the CEO of an important truck- maker several years ago.
    Warm Regards,
    A.R.

  431. Wladimir Guglinski

    Andrea Rossi wrote in November 24th, 2014 at 11:26 AM

    Wladimir Guglinski:
    There is a line of articles and we publish them, after peer reviewing, in the order we receive them. Every Author has his reasons to consider urgent his own paper’s publication and we do not grant privileges to anybody.
    —————————————-

    Dear Andrea Rossi,
    it is not a question of privileges.

    It seems the shape of the positive electric field (Coulomb barrier) of the nuclei may be the principal cause which makes possible the cold fusion occurrence.

    There are other authors thinking about, as Dr. Stoyan Sarg, who is going to pronounce a speech where he defends the same hypothesis:
    “At the beginning it discuses the major methodological error in scattering experiments that leads to a tremendously wrong vision about the Coulomb barrier.”.
    http://www.e-catworld.com/2014/11/21/dr-stoyan-sarg-to-address-nanotek-expo-2014-on-lenr/

    But…
    the question is to discover what is the correct model for the shape of the Coulomb barrier, and how it allows the cold fusion occurrence, and such subjetct merits a discussion.
    Why do not do it here?

    Here we have some persons interested in the subject, as Joe, Calaon, Eernie, Steven Karels, Eric, Orsobubu, etc.

    My paper Aether Structure for unification between gravity and electromagnetism just proposes how is composed the structure of the electric field of the elementary particles and the nuclei, and therefore it is of interest to discuss it here, since it can give us a better understanding on how the Coulomb barrier is formed around the nuclei.

    So, the publication is of the interest of many readers here, instead of to be a question of previlege.

    After all, you have to think about the privilege of the readers herein, i.e., their privilege of reading papers concerning what is of their interest: to understand the mechanisms which make cold fusion possible to occur.

    regards
    wlad

  432. Andrea Rossi

    Wladimir Guglinski:
    I have already explained and there is nothing to add. The JoNP works that way, and will not change. Again, if you think it is very important that your paper is read in short time, you can send it linked to a comment to the JoNP’s blog. This will not compromise its eventual publication in the Journal.
    Warm Regards,
    A.R.

  433. Bob

    Dear Andrea Rossi

    1. Do you know whether the so-called rossi effect occur:

    a. in a cylinder that is not straight, as those shown in the Lugano report, but in a curved or spiraled shape?

    b. in a cylinder with an angular bend?

    c. in a vessel where the walls are not round like a cylinder, but in any other non-round shape?

    2. Can you tell your readers the size of the largest and smallest e-cats you have constructed which have operated successfully.

    Thanks

    Bob

  434. Andrea Rossi

    Bob:
    1- I cannot give this information
    2- I can say this: the total volume of the reactors of the 1 MW plant is half cubic meter ( 500 liters of volume). All the rest is heat exchangers.
    Is much bigger the control system, entirely designed by our engineers ( 111 computers integrated). I think our team ( electronic engineers, physicists, blue collars, white collars has made a masterpiece.
    Warm Regards,
    A.R.

  435. Andrea Rossi

    Eernie1, Steven N. Karels:
    I expressed myself wrongly: just wanted to say that I totally disagree with the mantras of the fuels doomed by the E-Cat. I agree with you. Sorry for the concision, that produced misunderstanding.
    Warm Regards,
    A.R.

  436. Steven N. Karels

    eernie1 and Andrea Rossi,

    I think you both are arguing the same point. Energy sources will be integrated. If LENR energy is cheaper than other energy sources, it will slowly displace them. But there is significant financial inertia to change, so the change will be slow to occur.

  437. Bob

    Dear Andrea Rossi

    Can the so-called rossi effect occur

    in a cylinder that is not straight, as those shown in the Lugano report, but curved?

    2. Can the so-called rossi effect occur in a cylinder that is bent?

    3. Can the so-called rossi effect occur in a vessel where the walls are not curved like a cylinder, but straight?

  438. eernie1

    Dear Andrea,
    I do not understand your disagreement. The necessary time required to integrate LENR with other sources will only increase the time that fossil fuels will be required. Can you elaborate?
    Fond regards.

  439. eernie1

    Dear Andrea Rossi,
    I have read many blogs that predict that a successful and non-refutable showing of your device would immediately make the fossil fuel industry disappear or at least non-profitable. However there are many markets where the fuels will be required for many years after introduction of LENR devices.
    One market which is obvious is the gasoline driven devices now in use. There are hundreds of millions of autos, diesel engines, aircraft, power stations and other devices that depend on fossil fuels. These devices have cost their owners Trillions of dollars and have usage lifetimes of decades. Especially in the case of autos, the majority of owners would not be able to discard their present models and purchase LENR driven devices. So at least in this case there would be a market for fuels for many years, especially in less advanced societies. The same case can be made for the other markets(think of the cost of new passenger aircraft).
    LENR will eventually replace the older devices but perhaps not before the cheap sources of fossil fuels will be depleted. The first positive effect of your device will be the decrease in cost of these fuels to people who can least afford them.
    Fond regards.

  440. Andrea Rossi

    Eernie 1:
    I totally disagree.
    All the existing energy sources have to be integrated.
    Warm Regards
    A.R.

  441. Wladimir Guglinski

    Dear Andrea Rossi

    I had submitted my article Aether Structure for unification between gravity and electromagnetism 6 months ago to the JoNP.

    I think would be of interest to publish it, so that I could talk about the formation of the electric fields with Joe, Eernie, and Andrea Calaon, and others.

    In the case the electric field indeed has non-spherical shape as proposed in my theory, this property of the electric field can be connected with the cold fusion occurrence.
    Therefore I think it is of interest to discuss it.

    May you ask the reviewers to speed up the revision of the article?

    regards
    wlad

  442. Andrea Rossi

    Wladimir Guglinski:
    There is a line of articles and we publish them, after peer reviewing, in the order we receive them. Every Author has his reasons to consider urgent his own paper’s publication and we do not grant privileges to anybody. If you want to speed up, you can just link your paper to a comment to the JoNP. In this case you get it immedialetly published here, where rules are totally different.
    Warm Regards,
    A.R.

  443. Wladimir Guglinski

    Joe wrote in November 24th, 2014 at 3:32 AM

    Wladimir,

    If the induced magnetic dipole of the rotating 1p1 electron of 3Li7 can attract the valence neutron of 3Li7, why would that neutron not stop and settle at the centre point of the rotation rather than proceed further and into the 28NiXX nucleus?
    ———————————————–

    As I said in my comment of November 21st, 2014 at 7:07 PM:

    “Due to the inertia, the neutron continues moving, and it enters within the Ni58 through the “hole” in the electrosphere of the Ni58.”

    regards
    wlad

  444. tommaso

    Dear Andrea,
    Do you know the studies of Ruggero Santilli?
    If so,what do you think about it?

  445. Andrea Rossi

    Tommaso:
    I prefer not to comment on the work of our competitors; I know that Ruggero Santilli has to be respected, though. His work is interesting.
    Warm Regards,
    A.R.

  446. Joe

    Wladimir,

    If the induced magnetic dipole of the rotating 1p1 electron of 3Li7 can attract the valence neutron of 3Li7, why would that neutron not stop and settle at the centre point of the rotation rather than proceed further and into the 28NiXX nucleus?

    All the best,
    Joe

  447. Wladimir Guglinski

    Andrea Calaon wrote in November 23rd, 2014 at 5:49 PM

    ————————————–
    You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.
    ======================================

    With figures is easier to understand.
    The flux-n(o) of the 2He4 is shown in the figure:
    http://peswiki.com/index.php/Image:Fig._3.JPG

    In the inner side of DOUGLAS the neutron has magnetic moment -1,913 , because it is crossed by a flux-n(o)-up.
    The neutron in the outer side of DOUGLAS has magnetic moment +1,913, because it is crossed by a fluz-n(o)-down.

    All the other nuclei are formed by the capture of deuterons and neutrons by the flux-n(o) of the 2He4.
    In the figure ahead the 3Li7 is formed by the central 2He4 and the deuteron-neutron captured by the flux-n(o)
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE3.png

    The positive field of the proton is similar to the positive field of the 2He4:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    The field is non-spherical
    But as the proton is made by quarks, and there is repulsion between the quarks up, then the body of the proton has chaotic spin, then in average the field of the proton becomes spherical.

    regards
    wlad

  448. Eric Ashworth

    Andrea Calon, With regards your reply to Wladimir 23rd November 2014. As you are no doubt aware I do not know the academic teachings of nuclear physics. However, I think nuclear physics has some basic simple law to it that escapes many people who are at a complex level before grasping an introductory ABC level (proton, neutron, electron). You stated in your reply to Wladimir, ‘I cannot agree because the electric field symetry is a basic feature of electromagnetism. As a consequence the so called coulomb barrier is identical in all directions. Andrea, think about structure. Are you within a structure or are you outside of a structure?. From what I am aware nobody is outside of a structure and structure has a point (central position) and a periphery being the outer limit. Therefore your reply regards spherical symetry cannot be unless it’s at the centre of a system. This is why the Earth rotates and also nutates. If it was at at a centrifugal position it would not but it is at a centripetal position and it does what it does because it is not symetrical within its field. I believe that nothing can be absolutely spherical in our position within the solar system. Have I missed something?. Regards Eric Ashworth

  449. Wladimir Guglinski

    Andrea Calaon wrote in November 23rd, 2014 at 5:49 PM

    1) ———————————————-
    You seem to say that the shape of the “electrostatic” field (intensity in different directions) is influenced by the kinetics of the particles. Let me say that, if I understood correctly, I can not agree, because the electric field symmetry is a basic feature of electromagnetism. As a consequence the so called Coulomb barrier is identical in all directions.
    ==================================================

    No, you did not understand.

    The field of particles and the field of the nuclei is composed by two spherical fields. Look for instance the two fields of the 2He4:

    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    But due to the chaotic rotation the fields takes in average the spherical shape:

    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE2.png

    This explains why the Coulomb barrier is identical in all directions

    .

    2) ————————————–
    You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.
    =========================================

    My theory is developed from the concept of non-empty space (aether), composed by elementary particles of the aether as electricitons, magnetons, gravitons, permeabilitons.

    The concept of field in my theory emeges from the formation of physical fields composed by electricitons, magnetons, gravitons, and permeabilitons.

    All the nuclei are composed by a central 2He4, which produces a gravity flux named flux-n(o). Each particle as proton and neutron is captured by such flux.
    Due to the laws of interactions between the gravitons of the flux-n(o) and the spin of particles (protons and neutrons), the magnetic moment of those particles can change their sign depending on the direction of the flux-n(o).
    If a flux-n(o)-up crosses the neutron, its magnetic moment is positive.
    If a flux-n(o)-down crosses the neutron, its magnetic moment becomes negative.

    3) ——————————
    Wladimir, let me insist that what I proposed is not going against any consolidated law of physics, not to mention the so called Standard Model.
    =================================

    According to the Standar Model nuclear reactions cannot occur via electromagnetism, and the resaon is easy to be understood: nuclear reactions need to be promoted by particles bound via strong nuclear force.
    Your theory requires a model in which protons and neutrons are bound via electromagnetism.
    Therefore your theory is going against the Standard Model.

    As I already said , the Coulomb barrier in the distances of 2fm within the nuclei is 100 times stronger than the electromagnetism interaction.
    There is need a new nuclear model so that to explain how protons and neutrons can be bound via electromagnetism.

    The Lugano Report is showing that the results obtained from Rossi’s E-Cat are incompatible with nuclear reactions occuring via strong force, as you did point out.

    Therefore Rossi’s Effect is incompatible with the Standard Model, based on the hypothesis of protons and neutrons bound via strong force within the nuclei.

    regards
    wlad

  450. Andrea Calaon

    Dear Wladimir Guglinsky,
    thank you for the appreciation: “Calaon is an expert in nuclear and chemical reactions, a field in which he is working along years”. Wladimir, I have to admit that I am not an expert of nuclear reactions, nor a chemist. I only studied the subjects during University and keep doing it whenever I have time. I have but friends working in the fields of nuclear chemistry, chemistry and physics. However my proven involvement in Physics is only VERY marginal. I participated in the numerical simulation of details of ITER and Wendelstein 7-X, inclusive cracks, but never in the Plasma part. I am only a humble thermo-mechanical numerical simulations specialist, just very fond of Physics.
    What I think helps me is that in my work as a Researcher I am used to strive for discerning between opinion and proven and “reproducible” fact. In the years I managed to debunk a series of misconceptions that had encrusted for long times.

    Back to the LENR.
    As you probably noticed, I have already “withdrawn” the idea of an actual coupling between Li nucleus and an electron, because it seems to me impossible that the coupling can cross the two 1s electrons protecting the Li+ ion.
    I have to premise that I haven’t analysed your theory in detail. But there are some features I don’t understand or probably just do not agree with.
    You seem to say that the shape of the “electrostatic” field (intensity in different directions) is influenced by the kinetics of the particles. Let me say that, if I understood correctly, I can not agree, because the electric field symmetry is a basic feature of electromagnetism. As a consequence the so called Coulomb barrier is identical in all directions.
    My “theory”, which is Dallacasa’s in this respect, explains the “non-sphericity” of the nuclear attractive force among nucleons with the strong dependence of the attractive potential with the reciprocal orientation of the magnetic moments (i.e. spins) (and their phasing). Nothing exotic, just basic electromagnetism.
    You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.
    I am trying to follow the suggestion of our Italian “bon-ton Guru” orsobubu putting a bit more “pepper” in the discussion.
    But you know that actually I would never criticize someone’s work without a reason, and would never offend (consciously) anyone for having a different opinion.

    Wladimir, let me insist that what I proposed is not going against any consolidated law of physics, not to mention the so called Standard Model. I am sure that there is no need to contradict any main evidence of physics to explain LENR. So far I just took the possible validity of the nuclear potential of Dallacasa to its extreme consequences.

    Dear eerinie1,
    it is possible that part of the phenomenology of the electron capture is actually due to the potential of Dallacasa, through what I proposed.

    I would like to say something I think should guide anyone trying to understand the LENR and the device of Andrea Rossi.
    The plethora of all LENR experiments, starting from the inception of F&P, depends on some unusual “mechanism” that is necessarily the same at work in the Hot-Cat. The probability of two different mechanisms at work is nil.
    There must be a single explanation for all manifestations of excess energy. The variations must reside only in “common physics” details. It seems Andrea Rossi has found a way to make that mechanism much more efficient than any other known experiment.
    As a consequence it makes no sense to think about any special mechanism that has the chance to work only with the conditions that the report of the 8th of October seem to have revealed.
    The mechanism should depend critically on the presence of deuterium or hydrogen in a metal matrix.

    It would be extremely interesting to work with all data of the How-Cat: frequencies, polarizations, correct stoichiometry, effects of missing elements or composition shortenings, … but it is impossible. So I am now trying to get suggestions from the works of Iwamura and Tadahiko Mizuno, who publish most of the data.

    Regards to all
    Andrea Calaon

  451. Wladimir Guglinski

    eernie1 wrote in November 23rd, 2014 at 12:47 PM

    1) ————————————
    However using the p electron of the Li, although very volatile, to create a neutron bridge between the 3Li7 and the 28Ni58 etc. seems to be more difficult and the energy balance more complex than just pulling out the loosely bound neutron in the nucleus of the 3Li7 which may be easily done with an external field. Since the neutron is a low energy(thermal)neutron it has a large cross section for reacting with the 28Ni58 and a reasonable life time to accomplish this.
    =======================================

    Dear Eernie,
    cold fusion is not an easy process, because if it was we were seeing cold fusion occurring every time in the nature.
    So, discarding a harder process in favour of another easier one does not seems to be a strong reason.

    Besides, even in the case of the Calaon theory, perhaps it is yet missing the resonance contribution, which I did not mention in my interpretation of his theory via my nuclear model, in order to give the most simplest explanation of the fundamental mechanism due to the electron’s contribution.

    Also, note that Calaon needs to change a litle his initial version.

    2) ————————————-
    I was inspired to think of this approach by your theory of the distortions occurring in even perfect spherical nuclei by the geometry of internal energy fields and spin considerations.
    ========================================

    Dear Eernie,
    actually I dont know what is the correct process.

    But as I said, I think cold fusion does not occur via the most easy way, in spite of I can be wrong and perhaps the Nature uses the mechanism proposed by you instead of the mechanism proposed by Calaon.

    regards
    wlad

  452. Andrea Calaon

    Dear Steven N. Karels,
    You asked:
    “I see that silicon is present in the ash. Could the following reaction be possible?
    27Al + 7Li + e -> 28Si + 6Li
    …”
    Let me say. Hehehe. You noticed those lines at 28 on Figure 9 (lone) and 11 (with Al as well) of the ITPR.
    Let me first say that the reaction you wrote is impossible, because 28Si is equal to 27Al plus a proton, not a neutron.
    In the light of what I said in my last post, I think that direct exchanges of neutrons between nuclei with more than one proton are impossible (He4 is excluded, so from Li upwards).
    The possible sequence that would lead to Si28 is:
    1 : p + e -> pe
    2 : Al27 + pe -> Si28 + e
    The impression, also looking at other LENR experiments is that, once Hydronion (pe) is formed, it can reach any nucleus present, especially those with high magnetic moments:
    Li7: +3.26 [muN]
    Al28 : +3.64 [muN]
    So I would not be surprised to see that Si28 is formed in the Hot-Cat.
    One curiosity is the strong line at 43, because without ions, 43 is only Ca43, a relatively rare isotope.

    Regards
    Andrea Calaon

  453. Franco Sarbia

    Dear Dr. Andrea Rossi, the fuel of the gas fueled Hot Cat could be hydrogen in the future?
    Warm regards.
    Franco Sarbia.

  454. Andrea Rossi

    Franco Sarbia:
    In the future all is possible, just associated to a probability percentage that I am not able to evaluate now on the specific issue you are asking for.
    Warm Regards,
    A.R.

  455. eernie1

    Dear Wlad,
    I have been an advocate of chemo-nuclear reactions causing the Rossi effect ever since he first revealed his device, since the electrons are more easily manipulated and can cause disruptions in the nucleus of atoms as shown by Fermi etal. I understand the process you and Andrea C. are proposing. However using the p electron of the Li, although very volatile, to create a neutron bridge between the 3Li7 and the 28Ni58 etc. seems to be more difficult and the energy balance more complex than just pulling out the loosely bound neutron in the nucleus of the 3Li7 which may be easily done with an external field. Since the neutron is a low energy(thermal)neutron it has a large cross section for reacting with the 28Ni58 and a reasonable life time to accomplish this. I was inspired to think of this approach by your theory of the distortions occurring in even perfect spherical nuclei by the geometry of internal energy fields and spin considerations.
    Fond regards.

  456. Wladimir Guglinski

    orsobubu wrote in November 23rd, 2014 at 9:04 AM

    >Then Calaon needs to decide what he prefers to do

    Yes, he can decide, but think twice about it, sounds like a low-profile compromise to me, I really don’t like where this is going

    What have we become, a Boy Scouts blog? all those battles, the threats, the abuses, the taunts, and now all friends ending up singing kumbaya? and then, what will we do here alone? Please please Sarg, JR, Joe, what the hack are you waiting, it’s just two people after all, this thing can not be heading this way, add a bit of fuel to the fire, come on! For example, what’s this story about Calaon-Guglinsky, who tells me that Guglinsky-Calaon wouldn’t be much better?
    ——————————————

    Dear Orsobubu
    Calaon is an expert in nuclear and chemical reactions, a field in which he is working along years.

    His theory trying to explain Rossi’s Effect is compatible with my nuclear model, since Calaon starts from the hyphotesis that the interactions occuring in the phenomenon are not promoted by strong force.

    I have a nuclear model which can help him to understand the mechanisms involved in the phenomenon.

    Therefore, I think it is of interest to help one each other, since he is collecting data about nuclear and chemical reactions, and he showing evidences that Rossi’s Effect must be due to electromagnetic reactions (compatible with my nuclear model).

    This is not a dispute.
    It is actually an attempt so that to try to understand and to explain the mechanisms involved in cold fusion and Rossi’s Effect.

    I cannot propose to work together neither with an author with a theory incompatible with my nuclear model neither with the author of a new nuclear model (competitor to my nuclear model, as Dr. Sarg).

    If a good work results from a Calaon-Guglinski theory, the benefit is for the science’s advancement.

    regards
    wlad

  457. orsobubu

    >Then Calaon needs to decide what he prefers to do

    Yes, he can decide, but think twice about it, sounds like a low-profile compromise to me, I really don’t like where this is going

    What have we become, a Boy Scouts blog? all those battles, the threats, the abuses, the taunts, and now all friends ending up singing kumbaya? and then, what will we do here alone? Please please Sarg, JR, Joe, what the hack are you waiting, it’s just two people after all, this thing can not be heading this way, add a bit of fuel to the fire, come on! For example, what’s this story about Calaon-Guglinsky, who tells me that Guglinsky-Calaon wouldn’t be much better?

  458. Curiosone

    The report of the ITP is very hard to read. Can you explain in simple words, as you are always able to do, how has been measured toe electric power consumed?

  459. Andrea Rossi

    Curiosone:
    The electric power consumed has been measured with two wattmeters PCE 830, installed one between the control system and the reactor, one between the control system and the plug of the electric power of the laboratory. This set up has been made to check if the control system was able to modify someway the measurement of the wattmeter. The values of the two Wattmeters coincided perfectly, and this gave evidence of the fact that the control system was not able to influence someway the measurement made by the Wattmeter.
    Warm Regards,
    A.R.

  460. Steven N. Karels

    Dear Andrea Rossi,

    In the eCat reactor used in the Lugano Report, the interior chamber of the reactor has three helixes of heating wire. The result of applying a current through these wires was both generation of thermal energy to heat the reactor and generation of a magnetic field due to the current.

    Have you preformed measurements with an applied magnetic field independent of the heating wires (e.g., a permanent magnet or a second electromagnet with either a continuous current or a variable current as part of your control system)?

    You will probably decline to directly answer the question but it is an experiment your team needs to perform if it has not already done so.

  461. Andrea Rossi

    Steven N. Karels:
    Sorry, I can’t answer regarding our internal R&D.
    Warm Regards,
    A.R.

  462. Wladimir Guglinski

    Joe wrote in November 22nd, 2014 at 11:37 PM

    1. ) ———————————–
    Why do you have only the outer electron of 3Li7 involved in the process of neutron transfer? Why are the outer (3d, 4s) electrons of the 28NiXX not involved at all in the Calaon-Guglinski neutron transfer process?
    =========================================

    Joe,
    perhaps they are also involved, since the Ni also changed the shape of its positive field due to the nucleus, and so the electrons in the electrosphere change their orbits, and the outer electrons of the Ni have interaction with the positive field of the 7Li.

    However, in order to simplify the explanation, I had explained only what happens with the outer electron of the 7Li.

    .

    2. ) ——————————–
    Why would the valence neutron at 7fm prefer exiting along the z-axis in which direction it has no momentum than along the xy-plane in which it has angular momentum?
    =====================================

    First of all, the neutron is not at 7fm, actually it is at a distance of 2,391fm.

    The neutron exits along the z-axis because the orbit of the electron p1 in the Figure 6 induces a magnetic dipole moment vector along the z-axis:
    http://en.wikipedia.org/wiki/Magnetic_moment

    FIG. 6:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    As the magnetic moment vector of the neutron is also along the z-axis, then the neutron is pulled by the magnetic moment of the electron p1 toward the z-axis.

    .

    3. ) ———————————–
    (Remember that Andrea Calaon has the two nuclei with their z-axes parallel rather than collinear as per your view.)
    =========================================

    Calaon has not a new nuclear model so that to allow him to understand the physical mechanism involved in the Rossi’s Effect.

    Note that he even did not respond to my question, when I asked to him how to solve that puzzle regarding the nuclear models which do not consider the strong force as the cause of the agglutination of the nuclei: as the electromagentism is 100 times weaker than the Coulomb repulsions in the distances of 2fm within the nuclei, how can the electromagnetis to be responsible for the nuclei aggregation?

    So, he is trying to understand what happens (as everybody) by considering what he knows from the known models (in which the electrosphere of the nuclei is spherical and unalterable).

    Then Calaon needs to decide what he prefers to do.
    He can either keep his initial version or to adopt the new way I am suggesting to him.

    regards
    wlad

  463. Andrea Calaon

    Dear Joe,
    You asked: “Are you saying that an electron can orbit beneath the ground state of an atom?”.
    In a sense yes. It does sound VERY unplausible …, but …
    Probably the p/d/t-e bound state can be interpreted as something beneath the 1s ground state of an hydrogen atom. A bare hydrogen nucleus (any isotope) has no electron orbitals that could interfere with an incoming electron and I proposed that under special conditions the electron couples with the naked nucleus. The bound electron should not have a wave function with the the classical electron orbital, but it would be a bound.
    This bound state is most probably not stable, in the sense that as soon as a photon with the right energy interacts with it, the coupling is destroyed and the electron can either remain bound to the nucleus in a standard orbital or become completely unbound.
    As I already mentioned I suspect that the spectra measured by Randell Mills at al. (if real) are the emissions at the formation of this probably unstable bound state in a plasma.
    For a Lithium ion (+1), which is surrounded by the fully occupied spherical 1s(2) orbital, for me it is still difficult to imagine a mechanism that arrives to the Li-e coupling.

    Think now to the experiments of Yasuhiro Iwamura of Mithsubishi Heavy Industries, where deuterium seems to be able to “enter” into very different nuclei: Ca, W, Ba, Sr, Cs, … plus all intermediate nuclei involved in the large isotopic shifts measured. The only way a positive charge (the deuteron) can reach another nucleus, which is protected by both a negative “sticky” barrier (the inner electron orbitals) and a positive “repelling” barrier (the positive charge of the nucleus), is becoming “picometrically neutral”, at least for a while. If the p/d/t-e pseudo particle is stable enough it can travel through the two barriers and “grab itself” to other nuclei, through Dallacasa’s potential. What follows has been described by me a number of times.

    Another problem with the reaction
    Li7 + e + Nixx -> Li6 + e + Nixx+1
    is that, even if it can form, the Li7-e would need to reach the nucleus of Ni. And this nucleus is protected by its electron shells: 1s2 2s2 2p6 3s2 3p6 4s2 3d8. It seems to me extremely implausible that something of the size of a 1s orbital (the Li+ ion electronic inner shell) can penetrate all those Ni shells. Resuming: a direct neutron exchange between Li7 and Nixx is impossible.
    Something more plausible is that a pe (let us call it “Hydronion”) forms first, then Li7 abandons a neutron, …:
    1 : p + e -> pe
    2 : Li7 + pe -> Li6 + pen
    3 : pen -> de (“Deuteronion”)
    4s: Nixx + de -> Nixx+2 + neutrino
    4c: Nixx +de -> Nixx+1 + pe

    The pen pseudo-particle reacts immediately becoming Deuteronion (de). Deuteronion would be what actually penetrates the Ni electron shells.
    There are two possibilities for reaction 4: 4s (stopping) and 4c (catalytic). If the main reaction chain that realizes the neutron exchange is (1,2,3,4c), the Hydronion (pe) would act as a catalyst, re-entering the chain at reaction 2.
    Deuterium could have not been detected because at 1,400 [C] it would immediately find a way to escape in the gas during the LENR.

    Reagards

    Andrea Calaon

  464. Andrea Calaon

    Dear Joe,
    You asked: “Are you saying that an electron can orbit beneath the ground state of an atom?”.
    In a sense yes. It does sound VERY unplausible …, but …
    Probably the p/d/t-e bound state can be interpreted as something beneath the 1s ground state of an hydrogen atom. A bare hydrogen nucleus (any isotope) has no electron orbitals that could interfere with an incoming electron and I proposed that under special conditions the electron couples with the naked nucleus. The bound electron should not have a wave function with the the classical electron orbital, but it would be a bound.
    This bound state is most probably not stable, in the sense that as soon as a photon with the right energy interacts with it, the coupling is destroyed and the electron can either remain bound to the nucleus in a standard orbital or become completely unbound.
    As I already mentioned I suspect that the spectra measured by Randell Mills at al. (if real) are the emissions at the formation of this probably unstable bound state in a plasma.
    For a Lithium ion (+1), which is surrounded by the fully occupied spherical 1s(2) orbital, for me it is still difficult to imagine a mechanism that arrives to the Li-e coupling.

    Think now to the experiments of Yasuhiro Iwamura of Mithsubishi Heavy Industries, where deuterium seems to be able to “enter” into very different nuclei: Ca, W, Ba, Sr, Cs, … plus all intermediate nuclei involved in the large isotopic shifts measured. The only way a positive charge (the deuteron) can reach another nucleus, which is protected by both a negative “sticky” barrier (the inner electron orbitals) and a positive “repelling” barrier (the positive charge of the nucleus), is becoming “picometrically neutral”, at least for a while. If the p/d/t-e pseudo particle is stable enough it can travel through the two barriers and “grab itself” to other nuclei, through Dallacasa’s potential. What follows has been described by me a number of times.

    Another problem with the reaction
    Li7 + e + Nixx -> Li6 + e + Nixx+1
    is that, even if it can form, the Li7-e would need to reach the nucleus of Ni. And this nucleus is protected by its electron shells: 1s2 2s2 2p6 3s2 3p6 4s2 3d8. It seems to me extremely implausible that something of the size of a 1s orbital (the Li+ ion electronic inner shell) can penetrate all those Ni shells. Resuming: a direct neutron exchange between Li7 and Nixx is impossible.
    Something more plausible is that a pe (let us call it “Hydronion”) forms first, then Li7 abandons a neutron, …:
    1 : p + e -> pe
    2 : Li7 + pe -> Li6 + pen
    3 : pen -> de (“Deuteronion”)
    4s: Nixx + de -> Nixx+2 + neutrino
    4c: Nixx +de -> Nixx+1 + pe

    The pen pseudo-particle reacts immediately becoming Deuteronion (de). Deuteronion would be what actually penetrates the Ni electron shells.
    There are two possibilities for reaction 4: 4s (stopping) and 4c (catalytic). If the main reaction chain that realizes the neutron exchange is (1,2,3,4c), the Hydronion (pe) would act as a catalyst, re-entering the chain at reaction 2.
    Deuterium could have not been detected because at 1,400 [C] it would immediately find a way to escape in the gas during the LENR.

    Regards

    Andrea Calaon

  465. Joe

    Wladimir,

    1. Why do you have only the outer electron of 3Li7 involved in the process of neutron transfer? Why are the outer (3d, 4s) electrons of the 28NiXX not involved at all in the Calaon-Guglinski neutron transfer process?

    2. Why would the valence neutron at 7fm prefer exiting along the z-axis in which direction it has no momentum than along the xy-plane in which it has angular momentum? (Remember that Andrea Calaon has the two nuclei with their z-axes parallel rather than collinear as per your view.)

    All the best,
    Joe

  466. Eric Ashworth

    Regards Calaon – Guglinski material. Yes you are correct. For me what you are showing is a chain system. The sun sits within an interregnum where the north of one system connects with the south of another system. The atomic interaction which you show is a simple state of a more complex solar interaction. Whether energy interacts on a micro or macro scale the outcome is an energy interaction. The reason for an interregnum is because I believe there is a law connected to gravity that pulls energy back with regards a state of loss at the centre of a system.
    Well done, Eric Ashworth

  467. Wladimir Guglinski

    Daniel De Caluwé wrote in November 22nd, 2014 at 7:23 PM

    @Wladimir Guglinski,
    @Andrea Calaon,

    Wow, I’m impressed! Calaon-Guglinski very convincing to me!
    ——————————-

    Daniel,
    it seems I have now two Andreas in my life

    regards
    wlad

  468. DTravchenko

    Dr Rossi:
    How is going the R&D of the gas fueled E-Cats? News?
    Warm Regards,
    DTravchenko

  469. Andrea Rossi

    DTravchenko:
    We are making important progress on this issue. Soon will start to test the first prototypes.
    Warm Regards,
    A.R.

  470. Wladimir Guglinski

    eernie1 wrote i November 22nd, 2014 at 12:09 PM

    Dear Andrea C and Wlad.
    Fermi, Alvarez and Wick have shown both theoretically and experimentally that the injection of an electron into the nucleus occurs naturally in some of the heavier atoms, causing them to be radioactive, emitting Beta particles. These electrons(usually s or p level) are considered present in the nucleus either field wise or particle wise dependent upon whether the investigator treats them as particles or an EM field.
    ———————————————

    Dear Eernie,
    it is not the case.

    Fermi, Alvarez, and Wick were speaking about absorption of electrons of the inner levels s and p of some heavy nuclei by those own nuclei.

    In the case of the Rossi’s Effect, Calaon and I are not speaking about the electrons of the inners levels s and p of the heavier nucleus Ni being absorbed by the own Ni.

    We are speaking about the contribution of the inner levels s and p of the lighter 7Li in the transmutation of the heavier Ni.

    As you know, dear Eernie, in 2006 was published my Quantum Ring Theory where I had predicted that even-even nuclei with Z=N have non-spherical shape. People used to call me mad, because I had the audacity of defy a dogma considered untouchable along 80 years by the nuclear theorists, according to which those nuclei have spherical shape.
    But in 2012 the journal Nature published a paper showing that my prediction was correct: even-even nuclei with Z=N have non-spherical shape:
    http://www.nature.com/nature/journal/v487/n7407/full/nature11246.html

    More recently, since 1989 the nuclear theorists have considered along more than 25 years that cold fusion is impossible.
    They supposed cold fusion to be impossible because according to the Standard Nuclear Physics the positive electrosphere of the nuclei has spherical shape.
    So, they believed that, if a particle positively charged will enter within a nuclei, it must win the Coloumb barrier under conditions of high conditions of pressure and temperature (hot fusion), because the Coulomb barrier involves spherically the whole nucleus.

    But they are wrong.
    The shape of the positive electric field of the nuclei is non-spherical, as the nuclear theorists believed along 80 years.

    However, due to the chaotic rotation of the nuclei, in average the positive electric field of the nuclei takes the spherical shape. And this is the reason why hot fusion occurs needs to occur in the Sun.

    Soon or later, the nuclear theorists will realize that Rossi’s Effect must be explained via the consideration that the positive electrosphere of the nuclei is non-spherical, and this nuclear property is responsible for the occurrence of cold fusion.

    And soon or later, the nuclear theorists will realize that, again, I am right.

    Eernie,
    perhaps we are witnessing the birth of a new theory, to be known in the future as Calaon-Guglinski theory.

    regards
    wlad

  471. Daniel De Caluwé

    @Wladimir Guglinski,
    @Andrea Calaon,

    Wow, I’m impressed! Calaon-Guglinski very convincing to me!

    Kind Regards,
    Daniel.

  472. JCRenoir

    The main stream media of the world are beginning to take seriously your work. Does this help your work?
    JCR

  473. Andrea Rossi

    JC Renoir:
    The only thing that can help our work are well working products. Mass media go with the wind: masses of matter make the wind.
    Warm Regards,
    A.R.

  474. Bernie Koppenhofer

    I am sure you have read the articles about Bill Gates being briefed on LENR. Do you agree, with more money for research, there could be for example, 10 or more pilot installations and research projects which would speed the introduction of the Rossi effect?

  475. Andrea Rossi

    Bernie Koppenhofer:
    The fact that we moved the mountains with our hard work is positive.
    Warm Regards,
    A.R.

  476. eernie1

    Dear Andrea C and Wlad.
    Fermi, Alvarez and Wick have shown both theoretically and experimentally that the injection of an electron into the nucleus occurs naturally in some of the heavier atoms, causing them to be radioactive, emitting Beta particles. These electrons(usually s or p level) are considered present in the nucleus either field wise or particle wise dependent upon whether the investigator treats them as particles or an EM field. The process is called Reverse Beta, conversion electrons, or just Beta decay since the electron presence is subsequently ejected form the nucleus along with a Beta+ or Beta- particle, a Neutrino and a photon whose energy depends on the angle of entrance of the electron. Once the influence of the electron is felt in the nucleus, its spin and its field energy can play havoc with the equilibrium of the resting nucleus resulting in perhaps some strange outcomes. The inner electrons can also be induced to enter the nucleus by imposing an outer negative field on the electron sphere(perhaps a negative Hydrogen ion?).
    Energetic regards.

  477. Gio51

    Dear Dott. Rossi
    Underdeveloped countries need DESPERATELY your devices…!!!!! Pleaase, please, hurry up..!!!
    Gio

  478. Andrea Rossi

    gio51:
    Our Team is working as hard as possible and resolving problems.
    Warm Regards,
    A.R.

  479. Robert Curto

    Dr. Rossi, they have a Machine that will melt snow. They have 4 Machines.
    The smaller one, number two, consumes 40 to 60 gallons of diesel per hour. The fuel tank holds 550 gallons.
    Could an E-Cat supply heat at a lower cost ?
    Google:
    Snow Dragon
    Robert Curto
    Ft. Lauderdale Florida
    USA

  480. Andrea Rossi

    Robert Curto:
    This is one of the possible applications.
    Warm Regards,
    A.R.

  481. Joe

    Andrea Calaon,

    Are you saying that an electron can orbit beneath the ground state of an atom?

    All the best,
    Joe

  482. Wladimir Guglinski

    Andrea Calaon wrote in November 20th, 2014 at 7:33 AM :

    Let me spend a few words to advertise my theory .
    In the reactions above I explicitly added (+e) because I believe that the “secret” of the LENR is a coupling with the electron. Li7 in a uncommon “physical-chemistry” event in the metal matrix, couples with one electron becoming a sort of “new particle”: Li7e. Then this pseudo-particle, which is neutral already at picometric scales, can easily couple (through the same mechanism) with a Ni isotope: Li7eNixx. Li7 and Nixx become forced to travel inside the “circular” electron potential well. Soon they reach “nuclear contact” (at 2-3 [fm]) with very low excess kinetic energy, and can exchange the neutron because it is energetically convenient and probably Li7 offers it on the plane orthogonal to its magnetic moment, right where Nixx can easily “grab it”.

    —————————————————————————

    Dear Calaon,
    I have analysed your idea on the “new particle” Li7e, taking in consideration my nuclear model, and I have arrived to some interesting conclusions.
    Let me explain it.

    Figure 1 ahead shows the nucleus 2He4 with its positive electric field, produced by the two protons.

    FIG. 1:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE1.png

    The nucleus 2He4 has spin about the z-axis shown in the Figure 1.
    However, the two protons have residual repulsion (not in that magnitude of the repulsion considered in the Standard Model, because the electric fields of the protons are immersed within the electric field of the 2He4), and due to the repulsion the two protons have oscillations (zig-zag motion), and since the neutrons are bound to the protons via the spin-interaction, the neutrons also oscillate.

    Due to the oscillation of the two protons and two neutrons, the z-axis is changing its direction all the time. By this reason in average the positive field of the 2He4 is spherical, as shown in the Figure 2, and the two electrons in the electrosphere take the levels s1 and s2.

    FIG. 2:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE2.png

    Now consider the 3Li7 nucleus, shown in the Figure 3.

    FIG. 3:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE3.png

    The magnetic field of the 3Li7 is shown by North-South (blue-pink).
    The magnetic force which links the deuteron to the central 2He4 is induced by the rotation of the proton. The neutron has no charge, and therefore it does not induce magnetic force. The centrifugal force tries to expel the neutron, but it is bound to the deuteron due to spin-interaction.

    The radius of the orbit of the deuterion is 0,405fm, while the radius of the orbit of the neutron is 2,391fm. The two values are calculated in the paper Stability of Light Nuclei published in JoNP, based on the equilibrium between magnetic force on the proton and the centrifugal force on the deuteron-neutron, and I had used the magnetic moment of the 3Li7 measured in the experiment so that to calculate the values 0,405 and 2,391.
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    As the neutron in the 3Li7 is bound to the deuteron via the spin-interaction, but the radius orbit of the neutron is very big (2,391fm), it means that the neutron is weakly bound to the deuteron (and it is the deuteron that avoids the neutron to be expeled by the action of the centrifugal force).

    As happened in the case of the 2He4, the three protons of the 3Li7 are submitted to oscillations due to repulsions, and as the neutron is bound to the deuteron, also the neutron has oscillation.
    So, in spite of the deuteron-neutron move about the z-axis, however the z-axis has a chaotic motion, changing its direction all the time.
    Therefore, in average the positive electric field of the 3Li7 due to the three protons is spherical, and the distribution of the electrons in the positive electrosphere of the Li7 takes the levels shown in the Figure 4.

    FIG. 4:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE4.png

    In the page 3 of the Lugano Report it is said:
    ”Three braided high-temperature grade Inconel cables exit from each of the two caps: these are the resistors wound in parallel non-overlapping coils inside the reactor.”

    Therefore the electric current in the coils induces an internal magnetic field inside the alumina cylinder of the reactor, and when a nucleus 3Li7 approaches to a nucleus 28Ni58 and they couple chemically, both the Li7 and the Ni58 align their nuclear z-axis toward the axis of the alumina cylinder of the E-Cat.

    Being the two z-axis of both Li7 and 58Ni aligned toward the same direction, the two nuclei couple their nuclear magnetic moment, and by this way both the nuclei of 3Li7 and 58Ni stop to gyrate chaotically, and so the nuclear z-axis of the 3Li7 and 58Ni stops of changing their direction: their nuclear z-axis keep the same direction of the axis of the alumina cylinder.

    As the two nuclei stopped to gyrate chaotically, then the two positive electrospheres of 7Li and 58Ni lose the spherical shape they had when they were gyrating chaotically. In other words, both nuclei of 7Li and 58Ni get back the shape of electrosphere shown in the Figure 1 for He4 and Figure 3 for Li7.
    This changing in the electrosphere of the 7Li is shown in the Figure 5, where we see that the electrons s1, s2, and p1 change their orbits.

    FIG. 5:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE5.png

    But note that the electron of the level p1 occupies an unstable level, because its negative charge is attracted not only by the positive electrosphere of the Li7, but it is also attracted by the positive electrosphere of the Ni58.
    So, the electron of the level p1 is attracted by the electrosphere of the Ni58, and then the electron p´1 changes its orbit, taking the place shown in the Figure 6, between the nuclei Li7 and Ni58.

    FIG. 6:
    http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

    The orbit of the electron of the level p1 works now like a coil inducing a strong magnetic moment toward the direction of the two nuclear z-axis of the two nuclei 7Li and 58Ni.

    As the neutron in the 7Li is weakly bound, and it has a big magnetic moment (-1,913), it will be pulled by the magnetic field of the electron p1 toward the direction of the nucleus Ni58.
    Due to the inertia, the neutron continues moving, and it enters within the Ni58 through the “hole” in the electrosphere of the Ni58.

    NOTE: look at the Figure 3 of the paper Stability of Light Nuclei the magnetic moment of the neutron within the nuclei depends on the position of the neutron. When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.
    FIG. 4 of the paper Stability of Light Nuclei:
    http://peswiki.com/index.php/Image:Fig._3.JPG

    Therefore, the 7Li transmutes to 6Li, and 58Ni transmutes to 59Ni.
    The same happens with the isotopes 60Ni, 61Ni, 62Ni.

    I think your theory has chance to be correct, dear Calaon. But it seems there is no way to conciliate your theory with the Standard Model.
    I think your theory requires my nuclear model so that to explain the Rossi’s Effect.

    Regards
    Wlad

  483. Steven N. Karels

    Andrea Calaon,

    I see that silicon is present in the ash. Could the following reaction be possible?

    27Al + 7Li + e -> 28Si + 6Li

    This would effectively remove the aluminum from the fuel and leave the silicon in the ash. Note the 28Si is major isotope in naturally occurring silicon.

    If the above reaction is possible, then the reaction within the eCat could be explained as well as the total energy output for one gram of fuel estimated and the percentage of fuel expended computed.

  484. Herb Gillis

    Andrea Calaon:
    In your response to my last question you said:

    “The LENR do not take place between ANY of the nuclei present in a reacting powder/surface/whatever. The access to the reaction is controlled by chemical properties, not nuclear properties. So even admitting that any neutron rich isotope (apart from Li7) could work as a donor, then you would face the problem of having it react in the LENR.”

    1) Do you have any ideas on what the relevant chemical properties are (for accessing LENR)?

    2) If it turns out that only Nickel [or Ni/Li] has these chemical properties; then do you think it might be possible to use Ni [or Ni/Li] as a matrix alloy (ie. solid state “solvent”) for promoting neutron transfer reactions between other combinations of nuclei?

    Thank you for your insightful remarks.
    Regards; HRG.

  485. Andrea Calaon

    Dear Joe,
    electron, proton and neutron are not points, they have intrinsic sizes.
    The wave function of all s orbitals overlaps significantly with the nucleus. But electrons do not fall into the nuclei of atoms. Fortunately :)
    If you would like to visualize an electron (I am not offering you the perfectly canonical description of the electron … ) imagine a point charge that runs along a helical trajectory at the speed of light. The diameter of the trajectory is fixed: 386 [fm]. And the frequency of the circular component of the motion is fixed as well: about 2.47E20 turns per second. Very quick indeed! The nature of the particle has to do with these fixed parameters. So that you can not have an electron without them. Now the radius of the hydrogen atom (as the most probable distance between the proton and electron in a hydrogen atom in its ground state) is 52.9 [pm]. Therefore the size of the electron is about 0.36% that size. Not a point that would fall onto the nucleus, nor something as big as the orbital.
    The best equations we have for the electron describe how the plane of the intrinsic rotation evolves (for dynamical conditions) or how is distributed on average in stationary conditions (like an atomic orbital).
    The problem of the precise size of the proton arose for an experiment where the size of the proton is estimated thanks to the interaction between an orbital and the nucleus. The experiment uses muons instead of electrons only because they, having a mass 207 times that of the electron, form orbitals that are 207 times more tight around the nucleus than an electron does. And the ratio between the size of the orbital and that of the nucleus is smaller.
    See for example: http://phys.org/news/2013-01-physicists-surprisingly-small-proton-radius.html

    Best Regards
    Andrea Calaon

  486. Robert Curto

    Dr. Rossi, GENeco is a company in the UK that has a Plant that can convert food waste, and human waste, to provide fuel to power 8,500 homes, as well as to provide fuel for a Bio-Bus.
    With one tank of fuel the Bio-Bus can travel 200 miles, and emit 30% fewer emissions then a Diesel Bus.
    Google:
    GENeco
    Click on:
    GENeco
    Robert Curto
    Ft. Lauderdale Florida
    USA

  487. Andrea Calaon

    Dear Herb Gillis,
    The only energetically possible neutron swap reaction with Ni64 acting as a donor is this:
    Ni64 (+ e) + Ni61 -> Ni63 (+ e) + Ni62 + 0.94 [MeV].
    I think however that Li7 acts as a donor in the LENR because it has very special nuclear properties, not found elsewhere.

    Ni64 is only 0.9% of all natural Nickel atoms. So in any case its role can only be minor both energetically and isotopically.

    The LENR do not take place between ANY of the nuclei present in a reacting powder/surface/whatever. The access to the reaction is controlled by chemical properties, not nuclear properties. So even admitting that any neutron rich isotope (apart from Li7) could work as a donor, then you would face the problem of having it react in the LENR.
    As far as I know the Hot Cat is the first device that seem to be based on a neutron swap mechanism activated by the LENR.
    Regards,
    Andrea Calaon

  488. Andrea Calaon

    Dear Steven N. Karels,
    I agree with you, probably some grains or some other parts of the powder reacted fully, some others, not measured, much less.
    If 6Li can be turned into 7Li:
    Li6 + e + p -> Li7 + neutrino + (max) 6.47 [MeV]
    then hydrogen has a role, and turns into a neutron (together with one electron) first in this reaction. Then the neutron is transferred to xxNi.
    Therefore possibly it is not necessary to have all 7Li in the fuel powder at the beginning.
    Best Regards
    Andrea Calaon

  489. Giuliano Bettini

    Dear Andrea,
    extremely interesting the news that some lab was able to replicate the Rossi Effect (even in a minimal part) .
    A question, if I may:
    1. which lab?;
    2. is it “excess heat” (generally speaking)? or
    3. specifically what you call “Rossi Effect”?
    Thanks, kind regards,
    Giuliano Bettini

  490. Andrea Rossi

    Giuliano Bettini:
    As you know, in our laboratory we have analysed all the claims of the competitors and reproduced their apparatuses. We found one that works. I already spoke of it, but it is not correct that I speak on his behalf.
    I suppose publications will follow. For now, I just have to say, honestly, that this competitor of us has made a good job.
    Warm Regards,
    A.R.

  491. JYD

    Dear Andrea

    It could be the best friend for a Spatial HOT-Cat
    http://www.techno-science.net/?onglet=news&news=13372

    Futuristic regards
    JYD

  492. Andrea Rossi

    JYD:
    Thank you for this interesting information.
    Warm Regards,
    A.R.

  493. Frank Acland

    Dear Andrea,

    In case you are not aware, there is a new paper published by Carl-Oscar Gullstrom titled “Collective Neutron Reduction Model for Neutron Transfer Reaction”.

    He writes by way of introduction:

    “So I have improved the neutron transfer theory. In my first attempt the radiation was still a bit high but it is solved now. The trick is to not have high energy protons to drag out the neutrons but instead neutrons that are so low in energy that they can’t enter the nucleon but at the same time they could drag out more neutrons. If it is of interest I attached a document with some simple calculations.”

    Link:

    http://www.scribd.com/doc/247067779/Collective-Neutron-Reduction-Model-for-Neutron-Transfer-Reaction

    Kind regards,

    Frank Acland

  494. Andrea Rossi

    Frank Acland:
    Of course I know also this paper that I received last week from Oscar Gullstroem (I write Gullstroem because I have not the dieresis to put on the “o”). I am studying it since I received it. It is worthwhile the time to be studied carefully.
    Warm Regards,
    A.R.

  495. Steven N. Karels

    Andrea Calaon,

    I have not had a chance to go over your numbers in detail. Given the problem of producing too much energy than the measured test energy is a better scenario than the situation of not being able to produce the measured amount of energy with the measured or estimated components in the fuel.

    Perhaps Dr. Storms concept of a Nuclear Active Environment (NAE) is applicable and the ash was from such an environment and all the nickel at that site was converted.

    My original posting that you responded to asked whether the produced and naturally occurring 6Li could be transformed into 7Li. I understand you said you think it could be so transformed. If this is correct, then the supply of 7Li is only limited by the amount of hydrogen present.

    The Laguno report did not say all the fuel was consumed nor give any indication that the reactor was nearing fuel exhaustion. So the measurement that the ash was fully transformed to Ni62 only tells us what happened at the local site where the ash was produced.

  496. Joe

    Andrea Calaon,

    How can an electron get so close to a nucleus, in order to form a pseudo-particle, without the electron being forced to enter the nucleus due to electrostatic attraction?

    All the best,
    Joe

  497. Wladimir Guglinski

    orsobubu wrote in November 20th, 2014 at 8:59 AM

    Wladimir, is this interesting, about strong force?

    http://press.web.cern.ch/press-releases/2014/11/lhcb-experiment-observes-two-new-baryon-particles-never-seen
    ——————————————–

    Dear orsobubu
    many new unstable particles can be created.

    However, they represent NOTHING for the working of the universe.
    By using the properties of the particles (baryon number, lepton number, parity, strangeness, etc), it is possible to predict new particles, because those properties of the particles is decurrent from the laws of intereaction for the formation of new particles, composed by the agglutination of the elementary particles of the aether (electricitons and magnetons).

    Strong force must be actually a kind of dynamic gravity (the magnitude of the strong force interactions depends on the velocity of the particles which are interacting).

    In spite of the strong force (dynamic gravity) can be responsible for the agglutination of the quarks in order to form the proton and the electron, it does not means that the nuclei are bound via the strong force.

    regards
    wlad

  498. Herb Gillis

    Andrea Calaon:
    Thanks for responding to my question in such detail.
    As a possible alternative explanation: Do you think it possible that the Ni64 may be acting instead as a neutron donor to one of the lighter Ni isotopes (ie. 58, 60, 61) via the same mechanism as Li7? If this is true then perhaps LENR reactions can be achieved between any pairing of a relatively neutron poor nucleous and a relatively neutron rich nucleous?
    Regards; HRG.

  499. Andrea Calaon

    Dear Steven N. Karels,
    the reaction
    6Li + e + p -> 7Li
    is possible, for what I know. And, given that Lithium7 is able to couple with the electron in the stimulated Hot-Cat powder, because we know it most probably reacts with Nixx, I would guess that Li6 should react as well. The magnetic dipole moment of Li6 is +0.8220.. [muN] whereas that of Li7 is 3.2564… [muN], therefore my theory would suggest that in the same conditions, Li6 should react significantly slower than Li7. And Li7 should do something like:
    7Li (+ e) + p -> 2He4 (+ e) + 16.84 [MeV]
    Another point is the abundance of protons in the Nickel metal structure. Is their number high enough to make this reaction “visible” among the neutron swap?
    The experimental results seem to suggest that Li can play the role of an interstitial like the proton. A LiH substructure in the Nickel? I really do not know.

    Checking today the data of my “energy analysis” of yesterday, I noticed a mistake in summing the number of atoms of Ni in the isotopic shift chain.
    I will not repeat the whole thing, but just give the (hopefully) right and important numbers:
    As a reference one gram of natural Nickel contains:
    6.985E21 nuclei of Ni58
    2.691E21 nuclei of Ni60
    1.170E20 nuclei of Ni61
    The total number of single one neutron shifts for a complete forward shift to Ni62 in one gram of Ni is 6.1377E22.
    The experimental average energy of a unitary Nickel forward shift reaction, would be around 1 [MeV]. Far too low.
    These data, together with the energies of the Ni isotopic shifts obtained via neutron swap with Li7 given yesterday, say that:
    A complete isotopic forward shift of Ni58, 60 and 61 to Ni62 of 0.55 [g] of Nickel would liberate 3.757 [MWh]. It is 2.5 times the energy measured during the test.
    For a 1.5 [MWh] are enough 0.22 [g] of natural Nickel, plus 0.17 [g] of natural Lithium.
    The minimum ratio between the weight of Lithium and the weight of Nickel in the powder for guarantying a complete isotopic shift of Ni is 77.4%.
    These corrected data say that the discrepancy between the energy measured and the isotopic and weight analysis is even wider than guessed yesterday. Possibilities:
    The shifts are due to a different reactions with an even lower energy. Does it exists?
    The estimated quantity of Ni in the charge was wrong. Ni was slightly more that 0.22 [g] and it underwent an almost complete isotopic shift.
    The sample showed a complete isotopic shift, but the value was not representative for the whole ash. In reality only 40% of the Nickel particles shifted completely while the others did not react. The non-reacted part was not present in the analyzed grains.
    What do you think?

    Best Regards

    Andrea Calaon

  500. Andrea Calaon

    Dear Herb Gillis,
    Ni64 can not be given a neutron from Li7 because the reaction
    Ni64 (+ e) + Li7 + 1.15 [MeV] -> Ni65 + Li6 (+ e)
    requires 1.15 [MeV] to take place and apparently there are no such energetic photons around.
    Neither it is possible for Ni64 to shift “back” and lose one neutron to a Li6:
    Ni64 (+ e) + Li6 + 2.41 [MeV] -> Ni63 + Li7 (+ e).

    Therefore the disappearance of this Nickel isotope must happen in a different way.

    Let me spend a few words to advertise my theory :) .
    In the reactions above I explicitly added (+e) because I believe that the “secret” of the LENR is a coupling with the electron. Li7 in a uncommon “physical-chemistry” event in the metal matrix, couples with one electron becoming a sort of “new particle”: Li7e. Then this pseudo-particle, which is neutral already at picometric scales, can easily couple (through the same mechanism) with a Ni isotope: Li7eNixx. Li7 and Nixx become forced to travel inside the “circular” electron potential well. Soon they reach “nuclear contact” (at 2-3 [fm]) with very low excess kinetic energy, and can exchange the neutron because it is energetically convenient and probably Li7 offers it on the plane orthogonal to its magnetic moment, right where Nixx can easily “grab it”.

    Back to Ni64
    Remember that Andrea Rossi for a certain time (about 2011-2013) stressed that he had a way to enrich isotopically Ni in number 62 AND 64. Now we know something of what that process is. However that process in its present form seems to consume Ni64 as well.
    Since I believe the mechanism at the base of LENR is always the electron coupling, Ni64 couples with an electron that is itself coupled with another nucleus. In this case most probably the other nucleus is a proton, and the most likely result is:
    Ni64 (+ e) + p -> Cu65 (+ e) + 6.94 [MeV]
    In a small percentage of the reactions the electron crosses the two nuclei right while they are reacting and takes part in the nuclear reaction. This leads to the production of Ni65, which decays beta with a half life of about 2.5 [h]:
    Ni64 + e + p -> Ni65 + neutrino + 5.32 [MeV]
    Ni65 -> Cu65 + e + antineutrino + (max) 2.138 [MeV] (beta decay)

    With the tiny charge the presence of the second process and its beta decay should almost be undetectable (but I haven’t done the numbers).

    I will later answer to Steven N. Karels as well.
    Thank you all for these interactions

    Regards
    Andrea Calaon

  501. Marco Serra

    Dear Andrea,
    you said that “We cannot feed more information to our competition, which now is very powerful”. My question is: how can any competitor be powerful without knowing the core effect that drive the ECat ? Do you know of any lab that succeded in replication of the Rossi Effect even in a minimal part ?

    God bless you
    Marco

  502. Andrea Rossi

    Marco Serra:
    Yes.
    Warm Regards
    A.R.

  503. Andrea Rossi

    Italo R.:
    Thank you for the information.
    Warm Regards
    A.R.

  504. Steven N. Karels

    Andrea Calaon,

    An interesting analysis. Thank you. Please consider if the reaction
    6Li + e + p -> 7Li
    is possible.

    If this were to occur, would not the available amount of 7Li increase?

    Perhaps the rate equations favors the 7Li + Ni over the 6Li -> 7Li to establish the ash lithium isotropic ratio?

    So what would happen is the 7Li fuses with the Ni to become 6Li and the next Ni isotope. Then occasionally an hydrogen nucleus fuses with a 6Li to replenish the 7Li nuclei. And the reaction continues until the hydrogen is depleted. Since there are four times as many hydrogen atoms in LiAlH4 than lithium atoms, the “fuel” is larger than you assumed.

    Thoughts?

  505. Andrea Rossi

    Dr Joseph Fine:
    Thank you: fantastic translation. Happy Thanksgiving (in advance) to you and all our American Readers.
    Warm Regards,
    A.R.

  506. Joseph Fine

    Andrea Rossi, Silvio Caggia,

    The Navajo message translates as “Thanks, Warm Regards.”

    ” Jo, jo, jo. ”

    Happy Thanksgiving (in advance),

    Joseph Fine

  507. Frank Acland

    Dear Andrea,

    You mention Leonardo Corporation bought back some licensing contracts from licensees. Going forward, are there any licensees that will continue with their licensee status?

    Many thanks,

    Frank Acland

  508. Andrea Rossi

    Frank Acland:
    We offered to all our Licensees to buy back from them the licenses. Some of them have accepted the offer, and signed an Agreement that is under NDA, some preferred to hold the licenses: obviously, the licensees that preferred to hold the licenses have continued their Licensee status.
    Warm Regards,
    A.R.

  509. Steven N. Karels

    Dear Andrea Rossi,

    Can you comment on the accuracy of The Report regarding the fuel composition?

    a. Was LiAlH4 used as the source of hydrogen?
    b. Was the nickel content in the fuel 55% by weight?
    c. Was the lithium content of the fuel higher than The Report estimated?

  510. Andrea Rossi

    Steven N. Karels:
    No, I cannot comment.
    Warm Regards,
    A.R.

  511. Bernie Koppenhofer

    Dr. Rossi: It has been suggested safety certification will be a lot easier for the Gas Cat than for the Electric Cat. Is this true?

  512. Andrea Rossi

    Bernie Koppenhofer:
    No, it is not true. The point is not the fuel.
    Warm Regards,
    A.R.

  513. Wladimir Guglinski

    Andrea Calaon wrote in November 19th, 2014 at 10:35 AM

    I am convinced that these are the reactions Andrea Rossi is investigating (remember his excitement at the article speaking about nucleus tunneling between Lithium7 and Nickel?). Note that, if the nuclear binding energy is not related to the nuclear force (as suggested by many and as in my theory), these reactions do not even involve the Weak interaction: They are purely electromagnetic.
    Here are the reactions and the energies:
    Ni58+Li7 ->Ni59+Li6 + 1.749 [MeV]
    Ni59+Li7 ->Ni60+Li6 + 4.138 [MeV]
    Ni60+Li7 ->Ni61+Li6 + 0.570 [MeV]
    Ni61+Li7 ->Ni62+Li6 + 3.346 [MeV]

    =================================================

    Dear Andrea Calaon
    Actually we have to be astonished with the question: why did not the nuclear theorists realize 80 years ago that strong nuclear force cannot be responsible for the nuclear binding energy of the nuclei??? Because if the strong force was interaction which responsible for the attraction proton-neutron and neutron-neutron, then the dineutron would exist in the nature.
    Two neutrons linked by the strong force cannot be separated by the isospin proposed by Heisenberg, because only a force of repulsion would be able to win the attraction by the strong force between two neutrons, and an abstract mathematical concept as the isospin cannot create a force of repulsion.

    So, from a simple question of logic, the strong force must be discarded as the responsible for the attraction proton-neutron and neutron-neutron within the nuclei.

    But the question is not so easy as it seems.
    By considering the Coulomb repulsion in the distances of 1fm between protons and neutrons within the nuclei, the electromagnetic interaction is not able supply the necessary force for the agglutination of the stable nuclei. There is need an interaction 100 times stronger than that promoted by the electromagnetic interaction in a distance of 1fm, and this is the reason why the nuclear theorists had discarded 80 years ago the electromagnetism as the promoter of the nuclear binding energy, and they had adopted the strong nuclear force, which interaction is 100 times stronger than the electromagnetism in the distance of 1fm.

    And now finally the E-Cat is showing what the logic was suggesting to us, when we had faced the obvious: as two neutrons do not form the dineutron, then the strong nuclear force cannot promote the agglutination of the nuclei.
    And the consequence: the nuclear theorists will be obliged to accept this unavoidable fact.

    Nevertheless a problem arises: as the nucleus is not bound via the strong nuclear force, but in the distances of 1fm there is need a force 100 times stronger than that promoted by the electromagnetism, how can the nuclei be bound via the electromagnetism?

    Obviously an acceptable new nuclear model must be able to explain such paradox, and the nuclear theorist will discard the theories which do not solve the puzzle.

    You said: “as suggested by many and as in my theory”.
    However, how do you (and the many) explain it ?

    .

    In the nuclear model proposed in my Quantum Ring Theory the puzzle is solved as follows:

    1) The nuclei are surrounded by an electric field.
    See Figure 1:
    http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png

    2) Suppose a proton fuses with a nucleus. The fusion occurs as explained ahead.

    3) The proton must perforate the electric field of the nucleus, so that to be captured by the nucleus.

    4) When the proton perforates the electric field of the nucleus and they have fusion, the electric field of the nucleus has no repulsion with the electric field of the proton, because the two electric fields fuse by forming one unique electric field, surrounding the nucleus and the proton. By this way, the proton is not submitted to that Coulomb repulsion considered in the Standard Nuclear Model, in the order of 100 times stronger than the electromagnetic interaction.

    5) The equilibrium of the newborn nucleus formed by the proton+(original nucleus) is promoted via the equilibrium between the action of the centrifugal force trying to expel the protons and neutrons of the newborn nucleus and the magnetic force actuating on the protons.

    6) The stability of the light nuclei via equilibrium between magnetic force and centrifugal force is shown and calculated in my paper Stability of Light Nuclei, published in the JoNP:
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    .

    As the nuclear theorists are now accepting the reality of the cold fusion, thanks to the performance of the E-Cat, sure that they will realize that there is need a new nuclear model for explaining the results obtained by Andrea Rossi.

    And as the results obtained by the E-Cat are pointing out that the strong nuclear force cannot be the responsible for the agglutination of the nuclei, then facing the question on what will be the new nuclear model to be chosen (between the many new nuclear models proposing the electromagnetism as the cause of agglutination of the nuclei) of course they will consider only those models capable to explain the puzzle:
    how can the electromagnetism to promote the agglutination of the nuclei, since there is need a force 100 times stronger than that promoted by the electromagnetism?

    The best new nuclear model able to solve the puzzle will be chosen.

    Regards
    wlad

  514. Andrea Calaon

    Dear All,
    if Andrea Rossi allows me I will shortly abuse of the JoNP for a personal message.
    Immediately after my last post, despite the grammatical and lexical mistakes, many peopled re-accessed the site where I have my theory written down.
    I would like to stress that the documents in my theory-site are not up-to-date with the latest “news and changes”. Still new things appear on a daily basis, therefore I will wait first the ideas to settle a bit and then I will write them.
    When a complete review of my theory will be ready I will communicate it in one of my messages to the JoNP.
    Best Regards,
    Thank you Andrea

  515. Herb Gillis

    Andrea Calaon:
    How does your proposed reaction mechanism explain the observed reduction in the concentration of Ni64?
    Regards; HRG.

  516. Joseph Fine

    Andrea Rossi, Silvio Caggia.

    I have not translated the earlier comment made using the Navajo Codetalkers ‘dialect’.

    http://asecuritysite.com/challenges/nav

    Best regards,

    Joseph Fine

  517. Andrea Calaon

    Dear Readers of the JoNP,
    Andrea Rossi a few days ago, commenting the recent exchange of ideas about reactions that can justify the large isotopic shifts in Nickel and Lithium, said:
    “The contradictions or errors possibly emerging from such kind of comments or articles cannot be commented by me”.
    In the “comments” of the Readers (like myself) for sure there are mistakes. But what is Andrea referring to with the word “articles”?
    My guess is that if he could he would have criticized the conclusions of the report regarding the total content of Li and Ni in the powder and in the ash (may be not only these …).
    The ICP-AES sample was 0.21% of the total powder and ash. Since the powder is a mixture of grains of different origin, the sample could well be not representative of the whole population. And so the real total content of Li7 could have been much more than 0.0117 [g] (1.17% of the charge powder), or conversely the Nickel content could have been much less than 0.55 [g].

    In the last few days I eventually extended the study of the reactions to the energy they produce, comparing it to the measured energy released in the experiment. Well, …. I am a bit late, now is one and a half month after the publication of the report. But better late than never. The conclusions I arrive at contradict some of my guesses so far. Fortunately they also strongly point towards interesting conclusions.
    Two starting points:
    - 1.5 [MWh] are equal to 3.37E22 [MeV],
    - The number of Nickel nuclei 58, 60 and 61 (all the forward shifting) present in the 0.55 [g] of Nickel (if the w% estimation is correct) are respectively: 3.84e21, 1.48e21, 6.433e19. (You have to actually to do some numbers from those in the report to get these). These values are obtained using the isotopic ratios.

    Thus the average energy of a unitary Nickel forward shift reaction, FORGETTING LITHIUM, would be around 1.83 [MeV] (you have to consider that the number of reacting nuclei of Ni60 in a complete shift is actually equal to Ni58 plus all initial nuclei of Ni60, and that Ni58 reacts twice since it has to shift by two neutrons, …). If Lithium were added as a separate shift the average value would decrease. 1.83 [MeV] is a particularly low value for isotopic shifts.
    Let us now compare this average energy with the energies that would be released by Nickel shifting separately from Lithium, and starting from lowest masses: father nucleus plus proton plus electron. Here they are:

    Ni58+e+p ->Ni59+neutrino+ (max) 8.22 [MeV]
    Ni59+e+p ->Ni60+neutrino+ (max) 10.61 [MeV]
    Ni60+e+p ->Ni61+neutrino+ (max) 7.04 [MeV]
    Ni61+e+p ->Ni62+neutrino+ (max) 9.81 [MeV]

    Well, it is not necessary to check that the average (in the sense explained above) energy of all these reactions is far above the average experimental value. An this is only for the Nickel isotopic shift, than you should add the reactions that would shift Lithium. Conclusion: in the Hot Cat used in the first 32 days these reactions are not taking place.

    Then let us look at the reactions in which Li7 swaps a neutron with the Ni isotopes. These should be the reactions providing the lowest possible energy, since they realize the two isotopic shifts without further father particles (energy).
    I am convinced that these are the reactions Andrea Rossi is investigating (remember his excitement at the article speaking about nucleus tunneling between Lithium7 and Nickel?). Note that, if the nuclear binding energy is not related to the nuclear force (as suggested by many and as in my theory), these reactions do not even involve the Weak interaction: They are purely electromagnetic.
    Here are the reactions and the energies:

    Ni58+Li7 ->Ni59+Li6 + 1.749 [MeV]
    Ni59+Li7 ->Ni60+Li6 + 4.138 [MeV]
    Ni60+Li7 ->Ni61+Li6 + 0.570 [MeV]
    Ni61+Li7 ->Ni62+Li6 + 3.346 [MeV]

    All these reactions, in the case of a complete forward isotopic shift of 0.55 [g] of natural Nickel would have liberated a total energy 2.248 [MWh]. This value is about 1.5 times the measured energy (with an almost complete isotopic shift towards Ni62. But is as near as I can get.
    I think there are no realistic reactions with lower released energy.

    My conclusions are now these:
    - The reactions that take place in the Hot Cat are those in which the neutron is exchanged between Li7 and Nickel.
    - Hydrogen would seem unnecessary, apart from its possible lattice distortion effects.
    - The reaction of Ni61 to Ni62 should be more “efficient” than the others because experimentally there was absolutely no Ni61 left, whereas all other shifts lead to Ni61 before the final jump to Ni62. In fact the theory I am proposing says that this reaction should be quicker than the others, since Ni61 is the only Nickel isotope that has a magnetic dipole moment.
    - I think that the sampling of the powder for the ICP-AES was not lucky and led to numbers that can be misleading. The almost total isotopic shift of Ni and the analysis above tell me that most probably the charge powder contained a lower % of Ni: more or less 0.37 [g] for each gram. The corresponding Li (with natural isotopic ratio) for a complete Nickel shift would be 0.156[g] for each gram of fuel. Of which Li7 would be 0.144 [g] and Li6 0.0118 [g]. Is looks like the quantity that has the weight indicated in the report (0.0117 [g]) is precisely Li6, and not Li7. The other 47.4% in weight of the charge is made of the other nuclei as described in the Report.

    I am convinced that who is studying this phenomenon for commercial purposes did similar analyses immediately after the report arriving to similar conclusions, independently from the theory applied.

    Best Regards

    Andrea Calaon

  518. Andrea Rossi

    Andrea Calaon:
    I appreciate your efforts. Obviously I cannot comment.
    Warm Regards,
    A.R.

  519. Silvio Caggia

    Dear Andrea Rossi,
    Than-zie Lin Wol-la-chee Nesh-chee Klizzie-yazzi Dibeh,
    Gloe-ih Wol-la-chee Gah Na-as-tso-si Gah Dzeh Klizzie Wol-la-chee Gah Be Dibeh!

  520. Andrea Rossi

    Silvio Caggia:
    Ugh!
    A.R.

  521. Andrea Rossi

    Daniele Passerini (blogger of “22 Passi”)
    You asked me few days ago about why some of our commercial Licensees have cancelled their websites. The reason is that we decided to offer to all our commercial Licensees to buy back their licence at a price, obviously, superior to the price they paid for it. Some of our Licensees have accepted our proposal and sold us back their license.
    The details of the agreements are covered by NDA ( Non Disclosure Agreement).
    We maintained with our former Licensees a friendly and collaborative relationship, open to the possibility of future collaboration upon specific issues.
    Warm Regards,
    A.R.

  522. Buck

    The news about CF has reached a critical point as far as the Oil Industry is concerned.

    Gulf News, the largest English language newspaper in the Gulf region (UAE, Dubai, Oman, Bahrain, Qatar, Kuwait, Yemen, and Saudi Arabia), with a daily circulation of about 110,000, has just reported that India is moving towards getting back into CF research. Your work, the China Nickel Energy connection in Baoding, and Bill Gates’ recent visit with Vittorio Violante were cited in a factual positive fashion.

    To me, the tone takes on an alarmist quality as it presents CF phenomena as fact and CF technology as imminent.

    LINK>> http://gulfnews.com/news/world/india/indian-government-urged-to-revive-cold-fusion-1.1413814

  523. Andrea Rossi

    Buck:
    As I wrote on this blog one hour ago, now the competition is very serious. Thanks to the work of my Team, LENR, that 5 years ago were very “low”, not only in temperature, but also in global consideration, have gained momentum at high level. My Team merits recognition for this: our action and our fight have been the real game changer.
    Warm Regards,
    A.R.

  524. silvio caggia

    Dear Andrea Rossi,
    I understand that you can use electric current in order to “communicate” with the e-cat reactor in a sort of morse-code like:
    ———- (Prepare)
    ………. (Activate)
    ———- (Deactivate)
    But with gas e-cat what will you use? “Smoke signals” like native Cherokees? :-D

  525. Andrea Rossi

    Silvio Caggia:
    Navajo dialect: between Navajo and Cherokees cultural exchanges were frequent. Please keep this a jealously guarded secret: must remain strictly between you and me.
    Warm Regards,
    A.R.

  526. Franco Sarbia

    Dear Andrea Rossi
    1) In the gas fueled Hot Cat, the gas feeding performs a function of “starter” required for
    a) bring the system to operating temperature and make it independent from any network or external power supply?
    b) or also for operating at the regime?
    2) This means that, in both cases, the Hot Cat functioning at regime through its own generator can feed itself the reaction and its electronic control?
    3) The total autonomy in each stage would make safe and completely self-sufficient its cogeneration capacity for the production of electricity and heat even in hostile and remote environments, without any connection to wired networks, eliminating the supply of powerful, expensive, and dirty accumulators to start a Hot Cat System equipped with a battery of hundreds of modules, necessary to provide electricity and heat at homes and production activities of a small town or an urban neighborhood, both in developed countries and in the underdeveloped countries not yet equipped with wired electrical infrastructure. Can you confirm this strategic purpose?

  527. Andrea Rossi

    Franco Sarbia:
    Gas fuel will substitute electric energy to activate the reactor and drive it; I cannot give more particulars until we will have a product ready for the market. We cannot feed more information to our competition, which now is very powerful. We need to reach extreme commercial competitivity before leaking more information. When we will have reached the necessary economy scale our prices will discourage any competition, but before that phase we must be aware of the fact that our Competitors are eating voraciously any single bit of information we are leaking.
    Warm Regards,
    A.R.

  528. Eric Ashworth

    Dear Andrea, Your reply November 17th to Steven N. Karels. You say if the temperature reaches the safety limit the reactors turn off by a law of nature, whatever the source of the heat that causes a rise of the temperature.

    I have several questions regarding your answer:-

    1. If the reactors turn off by a law of nature/response mechanism can you be absolutely positive that a limit of safety has actually been reached?.

    2. Would you consider this law of nature to be a hindrance to the performance of LENRs?.

    3. Your ongoing R&D with regards the e-cat, is this with regards (a) the possible applications of the e-cat?. (b) An attempt to overcome this law of nature?. or
    (c) Both?.

    Regards Eric Ashworth.

  529. Andrea Rossi

    Eric Ashworth:
    1- yes
    2- no
    3- among others, a, not b.
    Warm Regards,
    A.R.

  530. Wladimir Guglinski

    Errata:

    In my last comment,

    instead of the relative kinetic energy of the deuteron regarding the 3Li7 is E= 0,5.m.(V² – v²)

    actually the relative kinetic energy is E= 0,5.m(V-v)²

    regards
    wlad

  531. Wladimir Guglinski

    How cold fusion may contribute for the solar nucleosynthesis

    • Rafael wrote in November 12th, 2014 at 8:23 AM :

    Maybe the sun is the product of a LENR, why not you try to mix the same chemical elements that has in the sun to see if you not create an artificial sun or get electricity or make a nuclear fusion propellant with less chemical elements, we already know what the sun is made of, just see on the wikipedia. Do not forget that the sun also has chromium nickel and calcium.
    —————————————————–

    Dear Rafael
    All the current theories of Modern Physics had been developed from the concept of empty space. So, the concept of field considered in the QFT–Quantum Field Theory does not take in consideration any structure for the space.

    That’s why in the Standard Nuclear Physics the electric field of the particles as the proton, the electron, and also the electric fied of the nuclei, is considered as a spherical field involving the particles, or the nucleus.

    This electric field of the nuclei considered in the Standard Nuclear Physics is a homogeneous sphere (it means that in any point of the field the value of the electric vector is always the same). Therefore, when a particle as the proton is forced to enter within a nucleus because it is submitted to very high pressure and temperature, the energy necessary to win the Coulomb repulsion is always the same (because no matter where is the point of the electric field of the nucleus where the proton enters, since the energy necessary is the same in any point of the electric field of the nucleus).

    Such concept of field adopted in the QFT introduced several puzzles in the Standard Nuclear Physics.
    For instance, when an alpha particle (2He4) exits the nucleus 92U238, it leaves out with an energy lower than the energy necessary to put an alpha particle within the nucleus 92U. Such paradox was solved by Gamow. However his solution is not acceptable, because he introduced another paradox in his solution. Besides, if the 2He4 should exit the 92U as proposed by Gamow, because the electric field of the 92U is spherical the 2He4 would have to exit the 92U with a tangential line (because of the rotation of the 92U). But the experiments show that the 2He4 exits the 92U with a radial line.

    Other puzzle is the emission of solar neutrinos by the Sun, as we see from the paper published by the journal Nature in 1984:
    Solar neutrinos and other problems and their relation to energy production in the Sun:
    http://www.nature.com/nature/journal/v312/n5991/abs/312254a0.html
    Models of the solar interior, based on the usual physical assumptions, predict a neutrino flux several times greater than that observed in the Davis 37Cl experiment. If, as is widely accepted, this discrepancy represents a ‘flaw’ in the standard solar model one would expect this flaw to manifest itself in other ways also. Here we point out some less well known discrepancies between theoretical predictions of the standard solar model and relevant observations. ”.

    The problem is not solved yet, as we realize from the last updated on 23 September 2013 version of the paper The 3He(α,γ)7Be reaction for big bang and stellar nucleosynthesis:
    http://www.york.ac.uk/physics/research/nuclear/nuclear-astrophysics/big-bang/

    ”…; looking for physics beyond standard model of particle physics. Naturally, the reaction attracted early attention of experimentalists and theoreticians alike in the 1950’s. Surprisingly, even today much work needs to be done via nuclear physics experiments to understand this reaction and provide information to the colleagues working on big bang nucleosynthesis, standard solar model and standard model of particle physics.

    Perhaps the puzzle of the rate of solar neutrinos cannot be solved via the Standard Model because they are not considering the cold fusion in the Sun. The emission of neutrinos from cold fusion reactions occurs in a rate very lower than that occurring in hot fusion.

    Probably some steps in the nucleosynthesis of some elements in the Sun occurs via cold fusion. And therefore it is impossible to conciliate any theory developed from the Standard Model with the experimental astronomical observations.

    .

    An experiment published in 2011 proved that space is no empty
    The experiment was published in the journal Nature. Light was produced by the space. And therefore the space cannot be empty. It must have a structure, so that to be able to emit light.
    A vacuum can yield flashes of light
    http://www.nature.com/news/a-vacuum-can-yield-flashes-of-light-1.12430

    Therefore the concept of electric field adopted in the Quantum Field Theory must be wrong.

    A new concept of electric field, based on the concept of a space having a structure is proposed in the Quantum Ring Theory.
    And here a crucial point emerges: there is no way to propose a spherical shape of electric field by considering the space with structure.

    Therefore the concept of field adopted in Quantum Field Theory cannot be correct.

    Other fundamental puzzle impossible to be solved by considering the concept of field adopted in QFT is concerning the null magnetic moment of even-even nuclei with equal quantity of protons and neutrons, as 2He4, 4Be8, 6C12, 8O16, 10Ne20, etc. Due to the monopolar nature of the electric charge. For instance, the 2He4 has two protons. As the nucleus has rotation, the rotation of the two protons has to induce a magnetic moment. So, by considering the model of field adopted in QFT, it is impossible to explain the null magnetic moment of the 2He4, and all the other even-even nuclei with equal quantity of protons and neutrons.
    I had challenged several nuclear theorists for coming to Rossi-Focardi blog Journal of Nuclear Physics-(JoNP) so that to explain how such puzzle could be solved according to the Standard Model. No one nuclear theorist did come.

    The reason why the model of field adopted in Quantum Field Theory cannot explain the null magnetic moment of the even-even nuclei with equal quantity of protons and neutrons is because in QFT it is adopted the mono-field model. In my paper Aether Structure for unification between gravity and electromagnetism, submitted for publication in the JoNP, it is shown that the null magnetic moment of those nuclei can also be explained via a double-field model (an outer electric field concentric with an inner central field composed by gravitons), adopted in Quantum Ring Theory.

    The Fig. 1 ahead shows the two concentric fields of a proton, as proposed in the paper Aether Structure for unification between gravity and electromagnetism.
    FIG. 1
    http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png

    As the radius of the electric field has the magnitude of the Bohr’s radius 10^-11m, and the radius of the nucleus is 10^-15m, of course the Fig. 1 does not show the real proportion between the fields. The Fig. 2 show a better proportionality (but of course not real yet):
    FIG. 2
    http://peswiki.com/index.php/Image:FIGURE_2-_3_fields_in_real_proportionality.png

    As seen in the Figure 2, there are two “holes” in the electric fields of the particles, and also in the electric fields of the nuclei.
    Under suitable condictions of low pressure and temperature, a nucleon as a proton or a deuteron can enter within a nucleus through that hole by having lower energy than that necessary if the nucleon is forced to enter via any other point of the electric field of the nucleus.

    By considering that nucleons also may exit a nucleus via the hole in the electric field of the nuclei, we eliminate two paradoxes of the Standard Nuclear Physics:

    1) The unacceptable paradox introduced by Gamow, proposed for explaining how a 2He4 can exit the 92U with energy lower than the necessary to cross the Coulomb barrier of the electric field of the 92U

    2) Why the 2He4 exits the 92U by a radial trajectory as detected by experiments (impossible to explain from the Gamow theory based on the Standard Model).

    .

    How a nucleon may enter within a nucleus
    Cold fusion may occur by two ways:

    1) Via resonance within vessels with conditions of low pressure and temperature, as occurs in the Rossi’s E-Cat.

    2) Via kinetic energy in vessels with conditions of very high pressure and temperature, as the Sun. Let us see how it may occur:

    a) In the Sun, more than 99,999% of the fusions occur via high nuclear reactions.

    b) Cold fusion occurs in less than 0,001% of the nuclear fusions

    c) It is very hard to occur cold fusion in the Sun, because the nucleons (for instance a proton) have very high kinetic energy in the star. So, when a proton enters within a nucleus via the “hole” in the electric field of the nucleus, the fusion does not occur (the nucleus cannot capture the proton) because due to the very high kinetic energy the proton simply trespass the nucleus, exiting the nucleus in the other “hole” opposite to the “hole” where the proton had entered.

    d) But a cold fusion reaction may occur as follows:

    d.1) Within the Sun all the nuclei are moving very fast, and every time changing the direction of their motion due to the collision with other nuclei.

    d.2) But suppose that a nucleus (for instance 3Li7) in an exact instant is moving along the x-axis with speed “v”, with the “hole” of its field aligned toward the x-axis. And consider that a deuteron with speed “V” (moving in the same direction along the x-axis) in that exact instant collides against the 3Li7, because the speed V is faster than v. In that condition the relative kinetic energy of the deuteron regarding the 3Li7 is E= 0,5.m.(V² – v²), where “m” is the mass of the deuteron. Therefore the kinetic energy of the deuteron in some very rare conditions is suitable low so that, when the deuteron enters within the 3Li7, the deuteron is captured, and 3Li7 transmutes to 4Be9.

    .

    The difference between cold fusion and hot fusion
    There are three basic differences between hot fusion and cold fusion:

    1) Hot fusion occurs when a nucleon enters within a nucleus (only under extreme conditions of high pressure and temperature) when the nucleon succeeds to perforate the electric field of the nucleus.

    2) Cold fusion occurs when the particle enters within a nucleus via the “hole” existing in the electric fields of the nuclei. It can occur either in low or in high conditions of temperature and pressure.

    3) In order to enter within a nucleus via hot fusion, a particle needs to have a very big kinetic energy. So, when the particle enters within the nucleus, due to the very high kinetic energy of the nucleon the nucleus is excited, and this is the reason why gamma photons are emitted. Unlike, in the case of cold fusion, as the particle enters with low energy the nucleus is not excited, and gamma rays are not emitted. So, also the tax of neutrinos emission in cold fusion is lower than in the case of hot fusion.

    Therefore, such property of cold fusion of emitting lower quantity of neutrinos can be response for the question why from the Standard Nuclear Physics there is no way to conciliate the hot fusion reactions in the Sun with the flux of neutrinos emitted by that star.

    Conclusion
    As we may realize, beyond the challenge of finding a theory capable to explain cold fusion by keeping the principles of the Standard Nuclear Physics there are many other challenges in Nuclear Physics impossible to be solved via the Standard Nuclear Physics.

    Up to now the nuclear theorists refused to think about a New Theory based on new fundamental concepts missing in the Standard Nuclear Physics, because of two reasons:
    1) Cold fusion is impossible to occur by considering the Standard Model
    2) Therefore they were sure it would be possible to solve the unsolved questions by keeping the Standard Model.

    But now cold fusion is a reality: Rossi’s Effect was confirmed by three universities of the Europe. And the nuclear theorists worlswide are beginning to accept this new reality, as by one of the top level nuclear phusicist of Russia, Dr Uzikov:
    http://www.proatom.ru/modules.php?name=News&file=article&sid=5595

    Therefore a reasonable person must realize that it makes no sense to continue trying to explain cold fusion via the old Standard Model. Because the problem is beyond the challenge of finding a theory for explaining cold fusion, actually now the cold fusion became the way for solving the unsolved questions.

    Some nuclear theorists have the hope to explain cold fusion via the Standard Model. When we reply to them that cold fusion is impossible to occur from the principles of the Standard Model, they say that we don’t know in deep the Standard Model. However, we may reply to those nuclear theorists: actually you don’t know in deep the structures existing in the Nature. They are very different of that considered in the Standard Nuclear Physics, by beginning from the structure of the space, and as consequence the structure of the electric field of the nuclei, responsible for the difference between hot fusion and cold fusion.

    Regards
    wlad

  532. Steven N. Karels

    Dear Andrea Rossi,

    I can understand your reluctance for further public testing. Too much exposure now can mean too much immediate demand which is bad for a developing business as well as feeding your competition at the expense of your R&D.

  533. Andrea Rossi

    Steven N. Karels:
    Correct.
    Warm Regards,
    A.R.

  534. eernie1

    Dear Andrea,
    Since you are understandably in your quiet period with respect to the mechanics of your device, perhaps we can speculate together about the physics involved and how the analysis of the ashes by the professors can lead us. From your previous remarks, I get the feeling that you are not completely satisfied with your present theory and that the ash analysis surprised you to a certain extent.
    One thing that stands out for me is the apparent neutron involvement indicated by the reported isotope content. At the energies involved for instance any free produced neutrons must be slow(thermal)neutrons, which exhibit therefore a much larger cross section for interaction with other isotopes within the structure. You of course have a much better knowledge of starting materials both quantity and quality so you can more readily pass judgment on any proposed theories. If it is possible, you can save us time and effort if you can rule out various proposals. For example, one theory requires 4He to be generated. Have you ever measured an increase of this isotope when your device is operating? Are the reported ash ratios of isotopes consistent with the starting amounts? I understand that some of the questions may involve confidential information, but if you can provide some answers perhaps we can provide possible reaction directions.
    Mutual regards.

  535. Andrea Rossi

    Eernie1:
    I cannot give this information now.
    In due time we will give information about the theory we see behind the so called Rossi Effect.
    Warm Regards,
    A.R.

  536. gillana

    Dear A.Rossi
    First of all congratulations for visibility gained in the media following your last ITP, in Italy Panorama (n.47) made a good piece.
    Here I would suggest that this visibility is now of paramount importance for the success of the E-Cat and I would continue to do public short demonstrations for promotional purposes.
    Please could you briefly summarize the progress of the new 1MW plant compared to the equipotent plant originally projected.
    Best regards

  537. Andrea Rossi

    Gillana:
    Visibility in this phase of our development is not very important. What is important is that we stabilize the commercial breakthrough pulled by means of the 1 MW plant making profit in the factory of our Customer. All the rest is secondary. We will not make any other public test, because from now on we will be focused exclusively on the market and the public tests will be made totally useless by the regular operation of our commercial plants. Our R&D will be maintained confidential until the commercialization of new products related to it. We cannot give further advantages to the competition, that is eating voraciously every information we feed it with.
    R&D, obviously, will continue in our factory, but it will not be finalized to public tests, it will be finalized to the manufacturing of products.
    Warm Regards,
    A.R.

  538. Steven N. Karels

    Dear Andrea Rossi,

    You have probably already considered this. On the gas-fired eCat reactors, I assume the gas flame will be applied to the exterior surface of each eCat reactor. Therefore, the heat fluid (e.g., water) being heated should be running coaxially inside the eCat reactor. Depending on the requirements you may want to run all of these in parallel or use a series plumbing scheme so that some of the eCat reactors run at a lower temperature and some at a higher temperature range (single stage, two stage or multi-stage designs).

    On the maintenance side, I would strongly suggest that the location of each eCat be readily recognizable (e.g., numbered or lettered) and that the individual eCat reactors be monitored so that fuel lifetime, temperature history and maintenance actions are noted and retrievable. A user-friendly graphics display system should show the overall system status and then be able to “drill-down” to a specific eCat reactor module. A database of the overall system and each eCat reactor should be automatically maintained both at the customer location and at your corporate location, although the information stored may be different. Some type of internet interrogation and control should be considered, with proper security implementations. Some type of built-in-test (BIT) is needed.

    Some type of load-averaging control should be implemented so when the system is running at partial output power, the fuel consumption of the eCat reactors is managed to maximize operational time between maintenance actions (i.e., don’t run one specific eCat reactor all the time forcing it to consume all of its fuel when most of the surrounding eCat reactors have plenty of fuel left).

    Some thoughts,

  539. Gio51

    Dear Dott. Rossi
    If I am not wrong, you stated some time ago that in a reasonable timelapse you would have been be able to open a customer installation to visitors, therefore disclosing the customer’s name. Am I wrong? Which is the situation right now?
    My best regards
    Giovanni

  540. Andrea Rossi

    Gio51:
    Be kind, read all my comments on this issue from some time ago through now.
    Thank you for your attention,
    Warm Regards,
    A.R.

  541. Andrea Rossi

    Orsobubu: again thank you, very appreciated also from all the Team.
    Warm Regards,
    A.R.

  542. BroKeeper

    Dear Readers and Andrea,

    Thank you Andrea, and thanks to Dan C’s help, for making this video available.
    I felt, from reading company PR material and articles on Tom Darden’s ideology in recovering brownfields into sustainable, environmental compatible and profitable properties, there was much more to the story in selecting Cherokee Investments.

    This remarkable 2011 YouTube of UNC Kenan-Flagler business school presentation made it quite clear the reasons why Andrea has entrusted the E-Cat into Tom Darden’s/IH capable hands. Obviously the integrity and purposeful sincere intent projected in this video by Tom and William McDonough is to construct a better world under ‘right’ business models framed with ethical practices. Practices, WM says, founded on Thomas Jefferson’s three directives in the Declaration of Independence “with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness”. With our liberty, McDonough stresses, we should provide for everyone opportunity to the Pursuit of Happiness including the poor with clean water and accessible housing.

    This, in my opinion, is a must view video to those who wish to understand what is behind Andrea and Tom Darden’s philosophical core values and their vision to incorporate the “New Fire” for all generations.
    Thank you, Andrea.
    BroKeeper

  543. Andrea Rossi

    BroKeeper:
    Thank you very much,
    Warm Regards,
    A.R.

  544. orsobubu

    … dedicated also to the whole wizardry Team, of course!

  545. Andrea Rossi

    Orsobubu:
    Thank you. Inspiring for persons who perspire: look, how idealism gears up with materialism!
    Warmest Regards,
    A.R.

  546. orsobubu

    >It appears we moved the mountains!

    No, dear Andrea, this time you didn’t move any mountain, THEY were obliged to go to *your* mountain, instead. And I want to dedicate this energetic and inspired song to your personal story, because you are the man on the silver mountain, “show them the way, the light and the fire, they will scream your name, and make you holy again”

    https://www.youtube.com/watch?v=czybZ-J_X9g

    RAINBOW – Man On The Silver Mountain

    I’m a wheel, I’m a wheel
    I can roll, I can feel
    And you can’t stop me turning
    Cause I’m the sun, I’m the sun
    I can move, I can run
    But you’ll never stop me burning
    Come down with fire
    Lift my spirit higher
    Someone’s screaming my name
    Come and make me holy again

    I’m the man on the silver mountain
    I’m the man on the silver mountain
    I’m the day, I’m the day
    I can show you the way
    And look I’m right beside you
    I’m the night, I’m the night
    I’m the dark and the light
    With eyes that see inside you
    Come down with fire
    Lift my spirit higher
    Someone’s screaming my name
    Come and make me holy again

  547. Giovanni Guerrini

    Yes,dear Dott Rossi,you with Prof Focardi,have moved mountains.
    Before your appearance on the scene,few people knew that “cold fusion” is real,now the world knows it.
    I’d call it “Rossi Focardi effect” !

    Thank you all,again.

    Regards G G

  548. Andrea Rossi

    Giovanni Guerrini:
    Thank you: yes, it is true: before our fight the LENR were mostly forgotten. The fight ( F-I-G-H-T ) we made in these last 4 years has definitely changed the game.
    Warm Regards,
    A.R.

  549. Italo R.

    About William McDonough and Thomas Darden – UNC Kenan-Flagler:

    The correct link is this:

    https://www.youtube.com/watch?v=OfQHvmYEOVI

  550. Andrea Rossi

    Italo R.:
    Thank you, very useful !!!
    Warm Regards,
    A.R.

  551. Andrea Rossi

    You are right, I already sent to the IT Guy the issue.
    Warm Regards,
    A.R.

  552. Dan C.

    Dear Mr. Rossi

    The link you provided to Brokeeper is unavailable. Has a typo
    I believe this is the video,
    The link you provide “v=OfO” should be “v=OfQ”

    William McDonough and Thomas Darden – UNC Kenan-Flagler
    http://www.youtube.com/watch?v=OfQHvmYEOVI

  553. Neri B.

    Dear Andrea,
    referring to the gas E-CAT i have a question.
    Is the gas needed only for reaching an activation point of the reaction and then you switch off the gas and mantain an eletric driver or the gas is burned continously?
    And if so which is the power (in electric Watts) needed to drive the reaction?
    Why the electric driver was so high in the test of TPR2?
    Thank you

  554. Andrea Rossi

    Neri B.:
    I cannot answer. When and if the gas- fueled E-Cat will be available, due information will be given.
    Warm Regards,
    A.R.

  555. BroKeeper

    Dear Andrea,

    Thank you very much for your added response. However, clicking on the YouTube link gives:
    “This Video is Unavailable. Sorry about that” :)
    BroKeeper

  556. Andrea Rossi

    BroKeeper:
    I inform immediately the IT guy about this.
    Warm Regards,
    A.R.

  557. Steven N. Karels

    Dear Andrea Rossi,

    My previous question was in the event of a building fire, not caused by your equipment, would the eCat reactor

    a. pose the same danger level as a tank of compressed hydrogen (because the Rossi Effect might be triggered even if the eCat was powered down) or
    b. would the internal temperature slowly rise (due the building fire and/or the Rossi Effect) until the nickel withing the eCat reactor melted and the Rossi Effect was no longer possible?

  558. Andrea Rossi

    Steven N. Karels:
    I already answered you: the intrinsic safety system of the E-Cat stops the reactor whatever the source of heat, which means also in case of whatever external source of heat, even a fire.
    Warm Regards,
    A.R.

  559. Andrea Rossi

    Frank Acland, Pietro F., George:
    Very interesting!
    It appears we moved the mountains!
    Warm Regards,
    A.R.

  560. tommaso di pietro

    Dear Dr Rossi,
    how long is the operating one megawatt plant?
    it is reasonable to think that the year of which you speak will expire by the first half of next year?

    Thanks in advance

  561. Andrea Rossi

    Tommaso Di Pietro:
    As I said, it is not chronometry.
    Warm Regards,
    A.R.

  562. Bob

    Dear Andrea Rossi

    You recently reported the maximum operating temperature of the e-cat to be 1400C

    Can you say whether this maximum temperature is due to:

    1. the nature so-called Rossi effect, or

    2. the current state of the art of the e-cat technology.

    Thank you.

    Bob

  563. Andrea Rossi

    Bob:
    Both.
    Warm Regards,
    A.R.

  564. Steven N. Karels

    Dear Andrea Rossi,

    When you field your 1MW eCat system, what are the dangers of a building fire? While electricity would be off, the internal temperature of the building could rise to high levels in an uncontrolled structure fire. What implications does this have for the eCat reactors? Could they start to produce power and eventually explode? Obviously the eCat system is not expected to work after such a fire scenario, but are there dangers to firefighters, etc.?

  565. Andrea Rossi

    Steven N. Karels:
    Our industrial plants have obtained the safety certification because they are intrinsecally safe: if the temperature reaches the safety limit the reactors turn off by a law of nature, whatever the source of the heat that causes a rise of the temperature. Besides, we have put all the safety systems imposed by the certification companies. By the way, the case of a building fire is quite unlikely, because the E-Cat plants are put in proper environment, made with not flammable materials. We have to respect all the requirements already existing for thermal energy industrial plants.
    Warm Regards,
    A.R.

  566. George

    Some top nuclear scientists are urging India’s new government to revive research on “cold fusion”, saying it has the potential to provide answers to the country’s energy problem.

    http://www.business-standard.com/article/news-ians/modi-government-urged-to-revive-cold-fusion-114111700763_1.html

  567. Pietro F.

    ….forse sarebbe il caso di sbrigarsi un po!! :)
    ….perhaps it would be appropriate to hurry up a bit!!! :)

    http://www.business-standard.com/article/news-ians/modi-government-urged-to-revive-cold-fusion-114111700763_1.html

    Buon lavoro

  568. Frank Acland

    Dear Andrea,

    I think you will find this article interesting. Nuclear physicists are urging the new Indian government to revive a new LENR program, apparently inspired by your work.

    http://www.business-standard.com/article/news-ians/modi-government-urged-to-revive-cold-fusion-114111700763_1.html

    Kind regards,

    Frank Acland

  569. BroKeeper

    Dear Andrea,

    From your philosophical and spiritual insights in past comments indicate you have a higher objective other than acquiring riches and/or self aggrandizement but sincerely projected concern for the physical relief to this world’s unanswerable sufferings. My question is what is within your core values to select Cherokee and their leadership as your partner to lead this imperative endeavor? In other words what are Industrial Heat’s core values that complimented your vision of a better path for all societies? Could you share your thoughts with your loyal followers of IH’s attended directions to achieve these noble hopeful goals?

    Thank you for your kind attention, Bro

  570. Andrea Rossi

    BroKeeper:
    To better understand the core values of Industrial Heat, you can go to
    http://www.youtube.com/watch?v=OfOHvmYEOVI
    and you also can search
    William Mc Donough and Thomas Darden-UNC- Kenan-Flagler
    Warm Regards
    A.R.

  571. Heath

    Dear Mr. Rossi,
    From your response to Dr Joseph Fine, a quick question. Has that one year with the Customer already begun or is there more problem resolution before to be resolved before that period starts? I am excited about what you and Industrial Heat are working toward, thus my curiosity.

  572. Andrea Rossi

    Heath:
    More problems resolution is on course. But I am an optimist.
    Obviously the time scheduling in such cases is not chronometry: one year is an approximative term.
    Warm Regards,
    A.R.

  573. Steven N. Karels

    Dear Andrea Rossi,

    I think I can understand the decision to proceed toward a natural gas fired eCat.

    The ultimate goal is the production of useful forms of energy (heated water, steam or electricity). I had assumed that the ideal solution would be an electrically heated eCat fed by a portion of its own generated electricity, essentially running for free, using only the cost of the hydrogen, nickel and other fuel components.

    But if we separate the input energy from the output and realize that electricity is currently and probably will always be more expensive than natural gas, we realize that diverting a portion of the generated electricity back to heating the eCat reactors is using precious (more costly) electricity that could be sent on the grid to sell to others.

    So even if the COP goes very high, it makes more economic sense to sell the electricity than to use it to heat eCats. Is this essentially correct?

  574. Andrea Rossi

    Steven N. Karels:
    yes
    Warm Regards
    A.R.

  575. eernie1

    Andrea C.
    There is no way to predict accurately the composition of the fuel content as stated previously because of the variety of possible starting ingredients. Andrea R. has made that abundantly clear warning that any attempt may be misleading. Because the tube is not sealed and allows 4He to escape is the reason that the 4He could be detected by sniffing the surrounding atmosphere. A sealed tube, unless the probe is inserted inside the tube, would not allow measurement. 7Li may only be the starter catalyst, allowing other ingredients to provide the necessary neutrons. The variety and form of the inserted energies may trigger other sources.
    Regards and please continue your investigations.

  576. Dear Andrea Rossi,

    Why not add movable infrared reflectors around the HotCat reactor (of the Lugano type) and make the control system regulate them. When the reactor overheats, the control system would open the reflectors so that thermal radiation can escape more freely. When the reactor cools down too much, the control system would close them again. It seems to me that it should be possible to reach high COP in this way, without changing the core itself and without the core even being “aware” that he is running at higher COP.

    Of course, one might consider moving parts to be risky, but that is probably a solvable engineering problem. For example the reflectors could be bimetallic fins which change shape in reaction to temperature, so that the control system would be passive; effectively a thermostat.

    The fact that you are pursuing a gas-cat makes me think that there may be some reason why this kind of construct does not work. However, I do not see what such reason could possibly be.
    regards, pekka

  577. Andrea Rossi

    Pekka Janhunen:
    The reasons why we are trying gas is very simple:
    1- to make 1 kWhe you need 3 kWht
    2- gas price is very cheap and will become cheaper
    Thank you for your idea, we have worked on it , but I cannot give information related to it.
    Warm Regards,
    A.R.

  578. Andrea Rossi

    Robet Curto:
    Thanks for the information,
    Warm Regards,
    A.R.

  579. silvio caggia

    Dear Andrea Rossi,
    another question about ITPR2, another guess with friends:
    When professors write 19.7 A line current what they mean?
    A) 19.7 Ampere RMS
    B) 19.7 Ampere conventional average
    C) 19.7 Ampere Peak

  580. Andrea Rossi

    Silvio Caggia:
    In the Report you can find clear explication regarding the amperage measurements. I can only take notice of what is written in the Report, as you can too.
    Warm Regards,
    A.R.

  581. Andrea Rossi

    Dr Joseph Fine:
    We will publish a report of the 1 MW plant used by the Customer of IH for his industrial purposes after 1 year of regular operation, when we will be able to give evidence ( if so) of the real profitability of the technology, beyond the laboratory tests: this is the obvious next step of our evolution.
    I do not know what the Professors of the ITP will make, because they are totally independent from us.
    Warm Regards,
    A.R.

  582. Andrea Calaon

    Dear Steven N. Karels,
    Thank you for your suggestion! So far I had not considered the stoichiometric aspects in details. But you are right: it is necessary.
    Eernie1 explained exactly my thoughts regarding the neutron transfer. I had proposed two reactions that consume Li7 and Li6 at different rates. But here there must be something more to have that kind of shift.
    If I made the numbers right the 0.011 grams of Li7 donating one neutron can be responsible for only 22.4% of a one neutron shift of the sole Ni58 (Ni with natural isotopic ratios). And it would be only the 4.7% of the total Nickel forward shift (58,60,61 ->62). Eernie1: may be the 0.011 [g] is not that accurate, but we are speaking about different orders of magnitude. So Ni must be invested by a series of reactions that have little to share with Li.

    Dear Eernie1,
    About He4 I guess it would have been impossible to detect any, since the reactor was not Helium tight (like the system of Tadahiko Mizuno), and the isotopic analyses were done only on the powders.

    Dear All,
    About the hydrides in the charge: If I had to provide a source of hydrogen as constant as possible with temperature I would have used a mixture of reversible metal hydrides:
    Mg(AlH4)2, KAlH4, Na2LiAlH6, K3AlH6, K2LiAlH6, K2NaAlH6, LiGaH4, Ca(AlH4)2, …
    I am not an expert in mass spectra, but the lines on pages 50 and 52 (and others) I think do not contradict this.
    For a Hot Cat powder however there is not much difference since the temperature are so high that any hydride would have released its hydrogen much before.
    Still the Carbon and Oxygen present in the charge are obscure to me.

    The fact that the Nickel powder does not sinter at 1,400 [C] for 32 days is for me quite remarkable.

    Only the surface of the active grains can produce the LENR. In the test almost all Nickel reacted, instead of only the active volume on the surface of the particles. I think the reason is the fact that at 1,400 [C] the metal grains of which the active particles are made of undergo not only grain growth, but also recrystallization, activated by the gamma and the “not too fast” daughters of the LENR. So that sooner or later all Ni nuclei are invested by the LENR.

    Best Regards

    Andrea Calaon

  583. Andrea Rossi

    Andrea Calaon:
    Your comment has been retrieved casually from the spam: next time change the address and send again your comment.
    Warm Regards,
    A.R.

  584. Eric Ashworth

    Dear Andrea, The following immediate information, I am guessing, probably will be of no value to yourself but maybe some of the readers of the journal will find it interesting. I have added some further information that you may find interesting regarding a static barrier that I believe is part of the energy equation and could be a consideration.

    Anyway, when I refer to geometry I am referring to three basic shapes or you could say symbols that make up a stable structure. These being cube, pyramid and sphere or in diagram format square, triangle and circle. Six pyramids form a cube, four on the horizontal and two on the vertical. The two on the vertical are associated with polar activity whereas the four on the horizontal are associated with structural activity that produce the poles, rotation of the structure sets the precedent. The sphere represents the neutral of the structure. Being neutral it does not extend outside of the cube. The space that extends outside of the cube within the base of the pyramid is considered a part of the Absolute volume. The neutral is always between two Absolutes. It’s the apexes of the four pyramids that combine to form a super gravitational focus of force (the apex of a pyramid should have some distinguishing feature when associated in connection with its cube). The volumes at the base of the pyramids form octahedral cavities with values of gravity when six or more neutrals are packed together. On the face of each of the four flats, central position, base of each pyramid is one vortex that responds to the polar activity, central position of the two vertical pyramids. Thereby, four loops of quarks enter and exit the two vertical pyramids at their bases around and on the pole. These loops, as previously described, open and close the vortexes that contain the economy flows in accordance to the exterior environment

    I therefore theorize that when the exterior environment is made-up of many neutrals that structure a lattice, containing many octahedral cavities, the inner neutrals will respond differently from those at the outer edge. To explain why, I will have to refer to the static barrier that all structures possess (the mechanism produces one). The static barrier is directly related to the internal kinetic energy. As previously dealt with. When a structure increases in its kinetic energy the vortexes close because the loops increase in flow and force. This is a self preservation response action. All created structures travel from the base of a pyramid from where it is formed to its apex where the super gravity exists and where higher energy is able to penetrate structure and eventually cause it to fall to pieces so as to be recycled (there are structures that create and those created within). The vortex has an oscillating motion to it responding to two values of gravity maintained by an integral motion of a flow with a force. This oscillating motion creates around the structure a variable static barrier of none flow that is dependent upon its kinetic content. A week static barrier will allow penetration of quarks at the vortex. A slow oscillation is a long stroke. A fast oscillation is a short stroke with regards the vortex and its internal dimension which when closed down produces a short stroke and a strong static barrier.

    An octahedral cavity contains a four static barrier value of a static force. Its the static barrier that provides a resistance to flow. The exterior of a lattice is thereby surrounded by a static barrier of less resistance than that within the interior composed of octahedral cavities (the atom could be considered to have a primitive conscious response mechanism). If what I theorize is correct then LENR should peak and then fall in thermal output. Could I be correct?. I do not want to probe into what could be confidential. Regards, Eric Ashworth

  585. Steven N. Karels

    Dear Andrea Rossi,

    This is in regards to the Andrea Calaon possible nuclear reaction theory.

    Analysis #1

    Based on the Lugano Report, an estimate of the fuel composition may be attempted

    Known facts:
    1. Fuel sample had a mass of 1 gram
    2. Page 29: “From the analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and ICP-AES analysis we find there is about 0.011 grams of 7Li in the 1 gram fuel.”
    3. Page 29: “… the information from ICP-AES that there is about 0.55 gram NI in the fuel.”
    4. Page 28: “From all combined analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and H in addition to the Ni.”
    5. Page 28: “… from the ICP-AES analysis which shows the mass ratio between Li and Al is compatible with a LiAlH4 molecule.”
    6. Page 28: “…natural composition, i.e. 6Li 7% and 7Li 93%”
    7. Page 28: “We remark in particular that hydrogen but no deuterium was seen by SIMS.”

    Analysis

    The average mass of the lithium atoms are 0.07*6 + 0.93*7 = 6.93 amu. Amount of 6Li = 0.011 grams * 6/94 = 0.0007 grams. Total lithium was 0.0117 grams.
    Aluminum atoms have a mass of 27 amu while hydrogen atoms have an average mass of 1.
    So the molecular weight of the LiAlH4 must be 6.93 + 27 + 4 = 37.93 amu.
    Therefore, the amount of LiAlH4 must be 0.011 grams * 37.93 / 6.93 = 0.06 grams and the amount of aluminum must be 0.043 grams. The amount of hydrogen in the LiAlH4 must be 0.006 grams.
    The iron mass must therefore be 1.0 grams (total) – 0.55 grams (Ni) – 0.043 grams (Al) – 0.011 grams (Li) – 0.006 grams (H) = 0.39 grams of iron.
    Element % by Weight
    Nickel 55.0
    Iron 38.9
    Aluminum 4.3
    Lithium (total) 1.2
    Hydrogen (no Deuterium) 0.6
    Total 100.0
    LiAlH4 6.1

    It is also likely that the LiAlH4 was prepared using hydrogen depleted of deuterium.

    Analysis #2

    Andrea Calaon suggested the following reactions on JONP (14 Nov 2014):
    Li7 + p + e +Ni58 -> Li6+Ni60 + neutrino + (max) 12.35MeV
    Li7 + p + e + Ni60 -> Li6+Ni62 + neutrino + (max) 10.39MeV

    From the Laguno Report we know that 5825 MJ of energy was produced.
    5825MJ = 3.635E+22MeV

    Relative abundance of nickel (fuel): Ni58: 68%; Ni60: 26%

    What amount of atomic ingredients are needed to produce the net energy observed and to support transmutation of the Ni58 and Ni60 isotopes into Ni62?

    Let X be the total number of atoms of Ni-58 undergoing transmutation.
    Ni58 reaction:

    X = (.68 / (.68 + .26)) * 3.635E+22 MeV / (12.35 MeV + 10.39 MeV) = 1.15637E+21 atoms of Ni58
    Let Y be the total number of atoms of Ni-60 undergoing transmutation.
    Ni60 reaction:

    Y = (.26 / (.68 + .26)) * 3.635E+22 MeV / (10.39 MeV) = 9.67686E+20 atoms of Ni60
    The amounts of hydrogen (H) and Li7 (L) needed to support these reactions are:

    H = X * 2 + Y = 3.28042E+21 atoms of atomic hydrogen
    L = X * 2 + Y = 3.28042E+21 atoms of Li7

    Converting number of atoms to grams:
    Avogadro constant: 1 amu = 1.661E-24 grams
    Hydrogen: 0.005 grams of 0.006 grams present
    Li7: 0.038 grams of 0.011 grams present
    Ni58: 0.111 grams
    Ni60: 0.096 grams
    Total nickel = (0.111 grams + 0.096 grams) / (0.68 + 0.26) = 0.222 grams of 0.55 grams present

    Conclusion: Insufficient lithium is present to generate the reported energy. Perhaps the aluminum reactions might provide some insights?

  586. Andrea Rossi

    Steven N. Karels:
    I cannot comment anything regarding this issue. Everything you or other Readers write about this issue is totally out of the reach of my possibility to answer positively or negatively. Any comment is welcome, but I have precise limitations regarding the IP divulgation.
    The contradictions or errors possibly emerging from such kind of comments or articles cannot be commented by me.
    Warm Regards,
    A.R.

  587. Joseph Fine

    Andrea Rossi,

    From May 13th 2013, when the first ITP report was published, to October 6, 2014, when the second ITP report was published, is an interval of about 17 months. The recent report described the operation of the E-Cat for 32 days.
    Has there been additional ITP work on starting up (or completing) a report on a lengthier operational experiment? Is it possible that such a report might become available in the near future?

    Continuing regards,

    Joseph Fine

  588. Robert Curto

    Dr. Rossi, lower cost electricity with natural gas, with zero emissions.
    Google:
    Net Power
    Robert Curto
    Ft. Lauderdale Florida
    USA

  589. Steven N. Karels

    Andrea Calaon,

    There is another problem. If you review my post of 9 Nov, I calculated and estimated the fuel composition (i.e., Ni 55%, Fe 39%, Al 4.3%, Li 1.1% and H 0.6%).

    To convert to the number of atoms of each element divide by that elements atomic weight (e.g., Li = 1.1/7).

    There is insufficient hydrogen and/or lithium to convert all of the Ni (excluding 62Ni). Perhaps the aluminum had a similar role to lithium?

  590. Andrea Rossi

    Steven N.Karels:
    All the assumptions that you are making regarding the charge can create a lot of confusion, because they are wrong, but I cannot answer, due to the fact that, as you well know, I cannot give any information on the matter.
    Warm Regards,
    A.R.

  591. BroKeeper

    Congratulations Andrea. Your low morning input prayers resulted in very high output, perhaps COP=Infinity. :)

  592. Andrea Rossi

    BroKeeper:
    IH and I share the same values and our Team is perfectly aware of the importance of our work. This is the reason why we are dedicating our lives to this endeavour, to make it integrable with all the sustainable energy sources.
    Warm Regards,
    A.R.

  593. Giuliano Bettini

    Dear Andrea,
    did Bill Gates met with you also?
    Ciao,
    Giuliano Bettini.

  594. Andrea Rossi

    Giuliano Bettini:
    I cannot give information about whom I meet, either positive or negative.
    Warm Regards
    A.R.

  595. Eric Ashworth

    Koen Vandewalle, You are correct and I am aware that without drawings it is extremely difficult to understand. I presume you are referring to the information posted November 13th 2014. This later information is theorized from an understanding of the mechansim that demonstrates a method of controlling an induced flow so as to form a unified field. The design of the mechanism came as an understanding of a unit of energy referred to as a cubic neutral but as you are aware it would be impossible to describe in words the intricate nature of the machines make-up. It is a vast subject that I have been involved with over the past 20 years and I am still learning. Answer to your first question. No you would not build a thruster or a propeller from my latest information. To build a proper thruster that is silent because of no turbulence/disturbance you would build the mechanism as shown with propellers in multiples of four depending upon application. You mention about heat strengthening the vortex. The vortex is a self regulating control mechanism that responds to its environment. You can’t strengthen a vortex in nature. The vortex either increases an internal gravity value of the structure or it reduces it in response to its environment. In the mechanism the vortexes are continually being structured in compliance with the r.p.ms. of the overlapping propellers. Atoms and particles are self contained constructed units containing a specific value of gravity even though they are comprised of quarks and particles, aether, I believe, being not a structure contains no gravity. I refer to it as an Absolute of size with regards that of an Absolute of volume. Thereby, these two Absolutes make and contain everything that exists. I believe there are two positions where gravity exists, one position being centripetal and one position being central of centrifugal force due to the fractional make-up of any structure. Maybe, if I am able to explain the three basic symbols in a concise way some light will be cast on this nebulous subject.
    Regards Eric Ashworth

  596. eernie1

    Steven,
    If the ash results are correct, there is a large increase in 6Li,or a large decrease in 7Li or a combination of both. Also a large decrease in 58Ni, or a large increase in 62Ni or a combination of both. In all cases, there is no change in the atomic number of the resultant materials which implies that if protons are involved there must be a revision back to a neutron in the reactions. I would bet without further data and of course speculation, on 7Li to 6Li and 58Ni to 62Ni with a transfer of neutrons. There is also no report of 4He. I think that if more 7Li is required it can easily be added in some form and since it is approx. 1/9th the Atomic weight of 58Ni would add little to the overall sample size weight wise. There are so many variables that can be manipulated, an estimate of sample content is less certain than the content of the various isotopes.
    Wlad submitted an interesting data point when he showed that there is much more distortion in the nucleus of the 7Li than in the nucleus of the 6Li. This would indicate that the 6Li is much more stable than the 7Li.
    Regards.

  597. Frank Acland

    Dear Andrea,

    You might find this interesting: Bill Gates recently attended a meeting at ENEA in Frascati where he was briefed by Dr. Vittorio Violante about LENR.

    More information here: http://www.e-catworld.com/2014/11/14/bill-gates-addressed-at-enea-by-lenr-research-coordinator-what-does-he-know-about-lenr/

    Kind regards,

    Frank Acland

  598. Andrea Rossi

    Frank Acland:
    Clearly our work is giving his consequences.
    Warm Regards
    A.R.

  599. Koen Vandewalle

    Eric Ashworth,
    what you write remains hard to understand without a drawing.

    Is it possible to build a thruster or a propeller with it ?
    Is it possible to apply some heat e.g. IR radiation, somewhere to strenghten the vortex ?
    A sort of man-made, heat driven tornado that pulls or pushes load.

    Kind Regards,
    Koen

  600. Wladimir Guglinski

    Steven N. Karels
    November 13th, 2014 at 8:51 PM

    Wladimir,

    Just because the percent of 6Li increased does not necessarily mean that 6Li was created. This would only be true if the total amount of lithium in the fuel and the ash were exactly known. We only know the ratio changed.
    —————————————

    you’re right, Steven

    Then there is need to know the % of total lithium in the fuel and in the ash, so that to know whether the 7Li decays in 4He + 4He, or whether it transmuttes to 6Li.

    If the % of total lithium is the same in the fuel and in the ash, then the 7Li is transmutted to 6Li.

    This information is very important, so that to give us the answer on what reaction of 7Li we need to consider

    regards
    wlad

  601. Dear eernie1, Andrea and JoNP Readers,
    Eernie1 You said it! A neutron must pass from Li to Ni. I agree. But IMO there is no tunnelling.

    The more I look at the data, the more it seems to me that the reactions responsible for both the isotopic shift of Ni58, 60, 61 to Ni62, and the change in Lithium isotopic ratio are either these:

    1: Li7 + e + Ni58 -> Li6 + e + Ni59 + 2.26 [MeV]
    2: Li7 + e + Ni59 -> Li6 + e + Ni60 + 4.65 [MeV] … and following

    or these:

    3: Li7 + p + e + Ni58 ->Li6+Ni60+neutrino+(max) 12.35 [MeV]
    4: Li7 + p + e + Ni60 ->Li6+Ni62+neutrino+(max) 10.39 [MeV].

    In all these “electron mediated nuclear reactions” (as in the theory I proposed) one of the electrons does not take part in the actual nuclear reaction and serves only as a coupling mechanism that allows two nuclei to approach down to 2 or 3 femtometers. In my initial theory I did not consider this type of reaction, which could for example lead directly to 2He4 + e from d + e + d.
    If two nuclei have the nuclear properties that allow them to react without the participation of the electron, this type of reaction should actually be possible.
    In the reactions 1 and 2 there is essentially a neutron that moves from Lithium 7 to a Nickel isotope, thanks to the electron that locks both nuclei inside its Zitterbewegung (ZBW) trajectory.
    But if Hydrogen has a any role in the LENR shown in this test of the Hot Cat, than the correct reactions can only be number 3 and 4.
    In reactions 3 and 4 there is a chain of two electron couplings and Nickel can acquire two neutrons. One electron and one proton add to the Ni nucleus together with another neutron coming from Li7.
    I think that reactions 3 and 4 take place in two stages: First an electron captures a proton and becomes the mysterious state of matter that someone imaginatively called Hydrino. In this case however the photon emission is at discrete frequencies, and not continuous, because this time it happens inside a metal matrix. Then the “Hydrino” adds to the Nickel nucleus together with the one neutron of Li7. I will use the symbol ep for the proton captured inside the electron ZBW (the “Hydrino”). Reaction 3 would become:

    3-1: p + e -> ep
    3-2: Li7 + ep + Ni58 -> Li6 + Ni60

    ep has no net charge, therefore it should efficiently couple to the other nuclei.
    The mass analysis of the ashes shows that during the 32 days of the test (that I think is just the Hot Cat “charge priming”) there is no significant development of deuterium and tritium. This should keep the gamma emissions very low.
    During this phase the quantity of hydrogen inside the metallic Ni is probably not high enough for what I call the “classical” LENR reactions (1-5 of my “theory”).
    I guess that exceeding the agreement of max 35 days of continuous testing would have led to the ignition of the “classical LENR” with the associated soft gamma radiation that Andrea Rossi confirmed in various occasions (the lead shield …). Needless to say that the time constraint imposed was a clever move that allowed to show to the world a reactor without any significant gamma radiation and with dramatic isotopic shifts, instead of some ephemerous He4 and may be some inconvenient Tritium.

    If the “non-nuclear” part of my “theory” is right, it seems that there is a way to make Lithium behave as the interstitial hydrogen.

    I am now trying to update the theory I proposed with the possibility of the electron not taking place in the nuclear reaction (as in all cases above), plus some considerations about the transmutations of Iwamura and the Hydrino mystery. I hope I will have time this weekend.

    Best Regards

    Andrea Calaon

  602. Steven N. Karels

    Dear Andrea Rossi,

    While the COP of the electricity power eCat is known from previous postings to be around the 3 or higher, the intrinsic cost of electricity generation is about 3, due to Carnot and other efficiency limitations. This discussion ignores the greater inefficiency of electricity generation and focuses only on heat generation by the eCat reactor.

    My thoughts will be obviously wrong or incomplete as I lack the direct access to the information and ideas that AR’s development team has access to and holds. That said, it may be interesting and informative to express what I contemplate on a gas-fired eCat versus an electricity heated eCat.

    Assume we are interested in only a heat output, e.g., generating warm water or low temperature steam for community heating, etc.
    The current Cat and Mouse design may no longer be needed. AR found better stability in a two stage implementation where a smaller unit, running at a COP of around 3 is used to provide the thermal input to a larger eCat reactor. But one of the cooking benefits of a gas fired stove is better control over heat.

    1. I suggest that the two stage control may not be needed with a gas-fired eCat.

    The major benefit of a gas-fired eCat is the relatively low cost of the fuel compared to electricity. However, keep in mind the electricity is 100% efficient in being turned into heat while the gas-fired approach is around 50% to 70% efficient (some of the heat is lost going up the chimney). There is also the carbon loading issue which I will ignore for now.

    The heat geometry for an electric powered eCat reactor is generally fixed by the heating element geometry. The number of wires entering the eCat reactor is limited. While conceptually a gas-fired eCat could have multiple controlled gas nozzles along the reactor. This may be a unimportant option but it is a feature of the gas-fired design.

    There is no magnetic field being applied in the process of adding heat using the gas-fired approach. This may not be important but a magnetic field could affect the Rossi Effect. A magnetic field or “vibrations” could still be applied as part of the control but conceptually at lower power than for the purposes of heating the eCat to its operating temperature.

    2. No changing magnetic fields with the gas-fired eCat – could be an advantage.

    There is added complexity in the gas-fired eCat. Gas plumbing is added, nozzles, a source of ignition but it is also a well-understood technology. So it may be worth the complexity.

    3. Increased complexity but the energy source is less expensive.

    While the gas-fired approach has chimney losses – heat waste – a well-designed thermal system may be able to recover the majority of the heat loss. This is also a well-understood and mature technology and should be implemented in any production system.

    4. Implement heat waste recovery.

    Some thoughts – Steve

  603. Tom Conover

    Dear Andrea Rossi,

    Dr Vessela Nikolova regarding her book “The New Fire” have inspired many of us already.

    I was happy to see that IH’s Patent #61821914 reports COP 11.07 when measured using water calorimetry. (see page #11 of 13 of the patent).

    More information can be found @ http://www.ecat-thenewfire.com/blog/industrial-heat-patent-cop-11/

    “It seems a very important document. The most relevant part probably regards the COP reached by a Hot-Cat.

    In the ‘experimental results’ of the patent application, describing instead a test on a multiple reactor device, we read that a COP of 11.07 was measured using water calorimetry.

    The cat is out of the bag, indeed!

    Tom

  604. Steven N. Karels

    Wladimir,

    Just because the percent of 6Li increased does not necessarily mean that 6Li was created. This would only be true if the total amount of lithium in the fuel and the ash were exactly known. We only know the ratio changed.

  605. Wladimir Guglinski

    eernie1 wrote in November 12th, 2014 at 6:01 PM

    Dear Andrea,
    How much reliability do you place on the report of the ash contents included in the TP3? If they are correct the Rossi effect must involve the removal from one of the reaction atom nuclei, of a neutron, and the capture of that neutron by another of the involved atoms. Most likely IMO, the 7Li passing a neutron in steps to the 62Ni. The Hydrogen through its spin energy absorbed from an imposed RF field can cause the neutron emission by interacting and destabilizing a neutron rich nucleus such as the 7Li.
    ——————————————

    Eernie1,
    there is a strange conclusion in the Report, because:

    1) In the page 29 they say:
    “One can speculate about the nature of such reactions. Considering Li and disregarding for a moment from the problem with the Coulomb barrier the depletion of 7Li might be due to the reaction p + 7Li = 8Be = 4He + 4 He. The momentum mismatch in the first step before 8Be decays can be picked up by any other particle in the vicinity. In this case the large kinetic energy of the 4 He (distributed between 7 and 10 MeV ) is transferred to heat in the reactor via multiple Coulomb scattering in the usual stopping process. “

    2) If that was be the case, then there would be only decrease of the 7Li, since it is totally transmutted to 4He + 4He.

    3)However, by looking at the composition of the ash in the page 42, we see:
    a) 6Li : the original fuel had 8,6% and the ash has 92,1%
    b) 7Li : the original fuel had 91,4% and the ash has 7,9%

    So, we realize that actually there was transmutation of 7Li to 6Li.

    The structures of 6Li and 7Li according to the nuclear model proposed in QRT is shown ahead:

    6Li:
    http://peswiki.com/index.php/Image:Fig._18.png

    7Li:
    http://peswiki.com/index.php/Image:Fig._19.png

    .

    Those two structures are agree to the quadrupole moment of the two nuclei:

    a) 6Li has Q = -0,0008
    b) 7Li has Q = -0,04

    From the structure of the 7Li we realize that it is submitted to a big unbalance of mass. So, the excitation of the 7Li may cause the rupture of the binding between the deuteron and the neutron, and the 7Li transmutes to 6Li, with the emission of the neutron.

    However, there is need to have a special condition of resonance between 7Li and 58Ni, otherwise the phenomenon would have to occur also easily between 7Li and many other nuclei as it occurs between 7Li and 58Ni.

    It seems the excitation of 58Ni resonates with the excitation of 7Li.

    regards
    wlad

  606. Steven N. Karels

    Peter,

    We had a death in the family so I was away. I missed reading your postings. What day did you post?

  607. Andrea Rossi

    Steven N. Karels:
    Sorry to hear that.
    Please accept my condolences.
    Andrea

  608. Peter Forsberg

    Steven

    Sorry to hear that! My post was on the third of November.

    Regards

    Peter

  609. Eric Ashworth

    Dear Andrea, Thanks for your information regarding spam and what to do if any information inadvertently becomes spam. I do understand that sometimes you cannot comment on certain topics because of the nature of your work. My comments are merely to share information that could be of interest to both yourself and your readers. I must say I have learnt a lot about atomic physics from your journal. Many thanks for the opportunities your journal provides. Regards Eric Ashworth.

  610. Dear Doctor Rossi.

    Does your experimental “Gas Cat” also require small electrical input for controllability, etc?

    Congratulations for the many international awakenings on your wonderful e-cat discoveries…

    Neil Taylor,
    Long Time Follower

  611. Andrea Rossi

    Neil Taylor:
    Yes, it does.
    Thank you for your kind attention.
    Warm Regards,
    A.R.

  612. Peter Forsberg

    Dear Steven N. Karels,

    Did you not understand the exchange between me and Andrea Rossi a few days ago regarding COP and modCOP?

    If you understand that, you know why Andrea is pursuing the Gas-Cat.

    Regards

    Peter

  613. Andrea Rossi

    Italo R.:
    Thank you for your kind attention,
    Warm Regards,
    A.R.

  614. Eric Ashworth

    Dear Andrea, To explain a theory I shall use water as an example of how I see molecular interaction. What is water with regards density, well we all know it can be either steam as a vapour, water as a liquid or ice as a solid. Here is why I think these states are able to exist. Considering each state is comprised of two hydrogen units and one unit of oxygen. Thereby three fields of activity that interact in each state.

    Therefore, draw three circles and overlap them slightly put x in the two overlaps where the vortex forces occur, label steam. Draw three other circles and overlap them a bit more, put y in the two overlaps, label it water. Do the same again but overlap these by a third and put z in the two overlaps, label it ice.

    The overlaps produce two vortexes of a flow force (I believe vortexes are able to gravitate external aether. I do not think aether exists as a single entity. I think it has to be a triplicity and thereby exist as a charge, even when in a free state. An obvious question is, what dictates the adjustment of the x state to the y state and visa versa?.

    Water as steam contains lots of kinetic energy, a week economy flow system and a strong outer static barrier that provides it with a loose bind unlike water and especially ice (This barrier provides atoms/structure with identity. Identity is what puzzled Einstein. Why don’t atoms flow together?. The reason why structures do not flow together is, I believe, because of their internal oscillations of the internal vortexes. Kinetic energy refers to the velocity of the quarks and the vibrational pitch/resonance of the vortexes).

    Economy flow is produced by the vortex. Thereby, less overlap provides less economy, less economy provides less quality/solidity and less economy provides for more kinetic energy and more kinetic energy privides more identity which allows a structure more independence. Water molecules that become more independent form a vapour.

    This reference to a static barrier is fully demonstrable by the previously mentioned mechanism.

    Water as ice contains a degree of latent energy because more overlap provides more economy, more economy provides more Quality/solidity and more economy provides more latent energy. More latent energy is responsible for less identity and thereby a reduced static barrier value creates more solidy between the molecules because they are able to get into closer proximity.

    If this correct then the peculiar characterists of Browns Gas can be accounted for if aether is brought into the equation regarding the expanded molecular state of water and its contracted state back to water. It could be that when quarks enter a vortex they could in theory because of the motion induced on the quarks become more latent in their charge potential resulting in an eventual releasing of a charge with an added kinetic potential.

    Back to the obvious question: In my attempt to answer this question it could be as follows:-

    Draw three lines (the three circles represent a top view, the three lines represent a side view) that overlap by a third. The vortexes are formed in the overlaps. In the space surrounding the molecule are quarks as free aether. Quarks to me represent electrical charge. Heat could be said to be electrical in content. So a lot of quarks in the environment would fuel the vortexes. Now through the areas that represent a vortex mark y and make loops that enter and exit each vortex. As will be evident two loops will flow away at the base and two loops will flow towards at the top (As is demonstrable in the mechanism). This flow is positive at its face and negative at its base with regards the individual quarks, according to the geometry of a cubic neutral or I should say a piece of a cubic neutral. Details of this can follow). As we know unlike particles attract and like particles repel. I believe it is the repulsion caused by the increased strengthy of a field that causes the vortexex to move out. When the environment contains few quarks the vortexes move in and create a stronger economy flow system. The economy flow system is a self protecting mechansim with regards the gravity value of the environment in competition with that of the molecule/structure.

    Andrea could this activity of what I have described in some way help with the workings of your e-cat. I do not know the technicalities with regards nickel but I think the octahedral cavities with regards a possible distortion could play a part. As you are aware this subjects involves causes and effects. Regards Eric Ashworth.

  615. Andrea Rossi

    Eric Ashworth:
    As you know, I cannot comment on this kind of issues. About the theory behind the so called Rossi effect I will be explicit after conclusive reconciliation of the science we got from the work of the ITP.
    By the way: this comment of yours has been casually retrieved by me from the spam, where the robot had relegated it: I saw it because in the first page of the spam section. This makes me think some comment is lost in the spam, therefore I repeat: if a Reader finds his comment not published, he is kindly invited to inform us writing to
    info@journal-of-nuclear-physics.com
    resending the comment as an attachment.
    Warm Regards,
    A.R.

  616. Andrew

    Dear Dr. Rossi,

    I’d like to bring to your attention the following news article published this week on Panorama, major Italian weekly magazine:

    http://www.ecat-thenewfire.com/blog/book-e-cat-on-the-main-italian-magazine/

    Kind Regards
    Andrew

  617. Andrea Rossi

    Andrew:
    Thank you; the article, as correctly suggested from your link, has been generated from an interview to Dr Vessela Nikolova regarding her book “The New Fire”.
    Warm Regards,
    A.R.

  618. silvio caggia

    Dear Andrea Rossi,
    I have made a guess with some friends about figure 5 of ITPR2. According to your e-cat experience, when was this PCE-830 photo made?
    1) During the dummy phase
    2) During the test phase
    3) During an un-documented e-cat experiment
    4) It’s a swedish joke, nothing to do with e-cat

  619. Andrea Rossi

    Silvio Caggia:
    The photo #5 of the Report of the Independent Third Party is very important and has been made on purpose from the Professors. They explained to me that the photo has been taken during the set up of the measurement stuff and they were controlling that the PCE830 was surely able to read perfectly the waves also in extreme conditions: for this reason , as surely have understood the experts and the reviewers to whom the Professors have given the report before the publication, the photo shows the wave also when the system has been put in overload; you can understand it from the acronym “OL” that you can read on the display, while the wave is perfectly described by the instrument.
    Thank you for the intelligent question.

  620. Italo R.

    Dear Dr, Rossi,
    I have found this russian web site where they talk positively about the last test.
    It is written in russian language but easily translating with Google.

    http://www.proatom.ru/modules.php?name=News&file=article&sid=5595

    Kind Regards,
    Italo R.

  621. Andrea Rossi

    Italo R.:
    Yes, I got it. I translated it to my Team.
    Thank you,
    Warm Regards,
    A.R.

  622. Gherardo

    Dott.Rossi,
    my compliments for your interview published on Panorama. It says nothing new for us but it would had been unbelivable one year ago.
    Gherardo

    PS: the text in italian… http://www.astampa.rassegnestampa.it/GruppoTotoAc/View.aspx?ID=2014111928829069

  623. Andrea Rossi

    Gherardo:
    Thank you, very nice! Panorama arrives also here in the USA in the bookshop close to the factory I am working in.
    Warm Regards,
    A.R.

  624. Steven N. Karels

    Dear Andrea Rossi,

    What are the reasons for going to a gas-fired eCat reactor:

    a. Cheaper heat source?
    b. Faster means of applying heat energy?
    c. Variable heating over the reactor?
    d. Better energy coupling (source to input to the eCat)?
    e. Other advantages?

  625. Andrea Rossi

    Steven N. Karels:
    a., obviously multiplied by the fact that making thermal energy by means of thermal energy we have not the factor 3 to turn thermal energy into electric energy.
    Warm Regards,
    A.R.

  626. Gianluca

    it would be interesting to know what the COP of ECAT in a hypothetical function as a domestic boiler (70-80 ° C). The use of low temperature increases the COP?
    Thanks

  627. Andrea Rossi

    Gianluca:
    The temperature chosen for the secondary fluid of a heat exchanger depends on its flow rate and does not affect the efficiency of the generator.
    Warm Regards
    A.R.

  628. Andrea Rossi

    Dr Joseph Fine:
    Thank you for your inspiring comment.
    Warm Regards,
    A.R.

  629. Joseph Fine

    Dear Andrea Rossi,

    This morning, on the Vortex-l Website, I saw a well written – and poetic – article titled ‘The Pale Blue Dot’ (in part to honor the memory of Carl Sagan).

    https://www.mail-archive.com/vortex-l@eskimo.com/msg99693.html

    With your permission, I have copied the text below.
    —————————————————

    The pale blue dot:

    Axil Axil Wed, 12 Nov 2014 13:32:08 -0800

    One of the sticking point that lodges deeply in the gullets of “real”
    science is that LENR is just too perfect to be believed. They are wrong. In
    point of fact, it is beyond too perfect, it is absolutely perfect. The
    corruption of the mind that is our legacy inherited from the mindless
    primitive from which we evolved rebels against the concept of such
    perfection. Such perfection cannot exist in this life. Such perfection can
    only exist and be truly enjoyed in the next. From the pride and prejudice
    born deep within that primordial dark place, mankind does not deserve to
    drink fully this sweet ambrosia of the immortals.

    LENR goes way beyond a great way to produce energy, it is a doorway to a
    new science whose implications when fully appreciated and developed will
    lift mankind up to trod upon brave new worlds spread like dust before
    eternity. A door for humankind will open to savor the power and the
    prerogatives of the gods. When man is wise enough to step through this
    doorway past the impossible that LENR lays open into timeless and unending
    existence, mankind will spread like a rising tide throughout the universe.

    This perfection of LENR is its own threat to its credibility and its
    science is here 1000 years before its proper time. What aborigines from the
    dawn of our past corruption would rightly understand the wonders of our
    present civilization without quaking with fear at the reality of such
    wonders? The science that LENR will reveal and the future that it portends
    it just too awesome to contemplate.

    Carl Sagan explained the emotion behind our current science and cosmology
    when he wrote Pale Blue Dot: A Vision of the Human Future in Space. Sagan
    played for high stakes in this attempt to “de-deify” our entire species.
    His beautiful, secular psalm dedicated to our demotion is unsurpassed. In
    Psalm 8, King David described us as only a little lower than the angels
    while in Pale Blue Dot, Sagan takes great pains to obliterate any sense of
    cosmic significance.

    Sagan says of that picture taken from by a spacecraft from a viewpoint far
    out in space: “We succeeded in taking that picture and, if you look at it,
    you see a dot. That’s here. That’s home. That’s us. On it everyone you
    know, everyone you love, everyone you’ve ever heard of, every human being
    who ever was, lived out their lives. The aggregate of all our joys and
    sufferings, thousands of confident religions, ideologies and economic
    doctrines. Every hunter and forager, every hero and coward, every creator
    and destroyer of civilizations, every king and peasant, every young couple
    in love, every hopeful child, every mother and father, every inventor and
    explorer, every teacher of morals, every corrupt politician, every
    superstar, every supreme leader, every saint and sinner in the history of
    our species, lived there–on a mote of dust suspended in a sunbeam”.

    With the help of LENR, this view claustrophobic view of human existence is
    about to change.

    —————

    Well done, Axil Axil!

    Best regards,

    Joseph Fine

  630. JCRenoir

    Prof. Bert Abbing:
    I congratulate with you for your open letter. Of course I agree with you.
    JCR

  631. Andrea Rossi

    JC Renoir:
    I spammed your comment for reasons you easily can understand, but I will respond to the part concerning the questions emerged on some blog related to the Report of the ITP: you ask when I will answer to the questions put here and there. As a matter of fact, all the questions have been answered in this blog, directly from me or from other expert Readers. It is true that notwithstanding this fact, somebody continues to put again and again the same questions, but the intention of these guys is not to make clear obscure points, but to try to pull us in a discussion where they get confidential information; obviously there are also the agenda-motivated guys: our policy with them is just to ignore them, after the answer has been already given regarding the issue they raise.
    To all the intelligent and honest questions we have answered .
    Warm Regards,
    A.R.

  632. clauzon pierre

    Dear friend Andrea,

    > You have to take a look on the nice paper given in french by Contrepoints about the ecat (Third report). I think that there is a real change in the minds now.

    >

    > http://www.contrepoints.org/2014/10/13/184364-fusion-froide-le-chat-e-cat-est-enfin-sorti-de-sa-boite

    >

    > and also this note taken from Linkedin below:

    >

    Christian Wiesner Director Global Sales / Co-Founder / Shareholder of PWH Plasmawerk Hamburg GmbH and ROTOKINETIK UG

    What Rossi is doing is absolutely OK, every small to medium sized company has to act this way. Don’t forget that every patent is also giving clear instructions to your competitors on how you are doing things, it’s like a manual.

    It’s therefore common practice for smaller companies to only patent technological aspects AROUND your key technology, preventing other people from making direct copies of your products, rather than patenting the core technology itself.

    If one of the giants, like GE, SIEMENS or ABB (why not AREVA), will finally jump on the case, with hundreds of patent lawyers behind, there will be no patent strong enough so these guys can’t find their way around !

    Well done Mr. Rossi, well done !

    Warmest regards and good luck for the 1MW demo to come

    Pierre

  633. Andrea Rossi

    Clauzon Pierre:
    By the way, I want to add that I am grateful to Centrepoint/ Science for the publication and also to Christian Wiesner of PWH Plasmawerk for his kind words.
    Warm Regards,
    A.R.

  634. Rafael

    but maybe the temperature of the sun come from a nuclear reaction and we could call HENR, high energy nuclear reaction. can you tell me where the sun’s energy come from than a nuclear reaction?

  635. Andrea Rossi

    Rafael:
    Please Google ” Wikipedia nuclear reactions on the sun” for a quick answer. All this has nothing to do with LENR. If you are interested to understand the basics of nuclear physics, you can buy a high school physics manual : from your language I think this can be a good start for you.
    Warm Regards,
    A.R.

  636. Jorge Alberto

    Dear Rossi,

    A website of CNN says that LENR is Obama’s secret weapon.

    http://ireport.cnn.com/docs/DOC-1187686?ref=feeds%2Fnewsiest.

    Jorge

  637. Andrea Rossi

    Jorge Alberto:
    Thank you! I did not know that CNN was talking about us!
    Warm Regards,
    A.R.

  638. Paul Calvo

    Hi Dr Rossi

    take a look at this water heater tech, I can see this looking like your home cat – they have a video demonstration.

    http://myheatworks.com/technology.php

    Regards,

    Paul Calvo

  639. Andrea Rossi

    Paul Calvo:
    Thank you for the information,
    Warm Regards,
    A.R.

  640. eernie1

    Dear Andrea,
    How much reliability do you place on the report of the ash contents included in the TP3? If they are correct the Rossi effect must involve the removal from one of the reaction atom nuclei, of a neutron, and the capture of that neutron by another of the involved atoms. Most likely IMO, the 7Li passing a neutron in steps to the 62Ni. The Hydrogen through its spin energy absorbed from an imposed RF field can cause the neutron emission by interacting and destabilizing a neutron rich nucleus such as the 7Li.
    Regards.

  641. Andrea Rossi

    Eernie1:
    The analysis made by the scientists of the third independent party that made the test have been made correctly, with the most sophysticated methodologies and the best available technologies. The results are what they are. Our duty now is to reconcile them with the Standard Model, and this is the work we are making. When we will have conclusive information about this issue, if such information will be publicable, we will give due information. Before that, is totally useless to make inconclusive assumptions. We are working upon not easy equations and I am optimist about the output.
    Warm Regards,
    A.R.

  642. Peter Wolstenholme

    Pete Frimmel:
    Water heats up in a water-fall, so is slightly hotter at the bottom
    than at the top, for reasons of conservation of energy. A volcano
    presumably shows the same effect, because the lava will be stirred up
    as it flows downhill. Heating by internal friction. But I have not
    seen the figures.
    Peter W.

  643. Paul Calvo

    Dr Rossi, Take a look at this water heater – they have a video demonstration.

    http://myheatworks.com/technology.php

    Regards,

    Paul Calvo

  644. Andrea Rossi

    Clauzon Pierre:
    I am delighted to receive this important information from you.
    Warm Regards
    A.R.

  645. Rafael

    Maybe the sun is the product of a LENR, why not you try to mix the same chemical elements that has in the sun to see if you not create an artificial sun or get electricity or make a nuclear fusion propellant with less chemical elements, we already know what the sun is made of, just see on the wikipedia. Do not forget that the sun also has chromium nickel and calcium.

  646. Andrea Rossi

    Rafael:
    The sun, obviously, is not a product of LENR. The temperatures necessary to remake what happens in the sun are in the order of millions °C, which is not properly a “low energy”…
    Warm Regards,
    A.R.

  647. carloluna

    Pete,Wladimir

    The Earth’s mante is made up largely of piezoelectric perovskites that produce heat (mysterious source responsible for half of the heat which keeps the temperature of the Earth) and electricity because of the enormous pressure. The mante flows on the core for the Coriolis force, so the electricity magnetizes the core. Therefore I believe that the Earth’s magnetic field is created.

  648. Wladimir Guglinski

    Pete Fimmel wrote in November 11th, 2014 at 7:20 PM

    Prior to the course of the lava flow turning towards the town, it flowed from the volcano into the sea.

    Temperatures of the lava immediately before it reached the sea were found to be higher than those of the lava emerging from the volcano!

    An interesting source of unexplained heat.
    —————————————-

    Pete,
    nobody knows what is a mysterious source responsible for half of the heat which keeps the temperature of the Earth.

    Half of the 44 TW heat lost to space is due to radioactive decay.
    http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat

    The other source is unknown.

    Perhaps the Nature already had discovered the Rossi’s Effect before Andrea Rossi.

    regards
    wlad

  649. Robert Curto

    Dr. Rossi, I liked your response to Pete Fimmel.
    I am also not an expert on Volcanos.
    However after 10 minutes on Google, I may be able to explain the ‘unexplained heat’
    There are 4 chemical types of Lava.
    Felsic, sometimes >950 C.
    Intermediate, 750 to 950 C.
    Mafic, >950 C.
    Ultramatic, 1,600 C.
    What was the sequence out of the Volcano ?
    Ultramatic followed by Intermediate ???
    Robert Curto
    Ft. Lauderdale Florida
    USA

  650. Pete Fimmel

    Prior to the course of the lava flow turning towards the town, it flowed from the volcano into the sea.

    Temperatures of the lava immediately before it reached the sea were found to be higher than those of the lava emerging from the volcano!

    An interesting source of unexplained heat.

  651. Andrea Rossi

    Pete Fimmel:
    I am not an expert of volcanos. Cannot comment at all.
    Warm Regards,
    A.R.

  652. Steven N. Karels

    Dear Andrea Rossi,

    You posted “Nevertheless, I must repeat that presently the focus of our R&D is restricted to the 1 MW industrial plant and the gas fueled Hot Cat.” Does that mean that no one is working on electricity generation from an eCat?

  653. Andrea Rossi

    Steven N. Karels:

    No, I didn’t mean that.
    Warm Regards,
    A.R.

  654. Frank Acland

    Dear Andrea,

    You mention that you don’t think LENR effects takes place spontaneously on earth. Do you think they occur naturally anywhere in the universe?

    Kind regards,

    Frank

  655. Andrea Rossi

    Frank Acland:
    In Physics nothing is absolutely impossible: everything is associated to a certain percentage of probability.
    Warm Regards,
    A.R.

  656. Franco Sarbia

    Caro Andrea Rossi.
    L’ecat sottoposto a test, ha caratteristiche di dimensione, potenza, temperatura d’esercizio, che sembrano ormai adattarsi a diverse applicazioni quali mini generatori turbo elettrici, capaci di motorizzare: automobili, motoveicoli, piccole imbarcazioni, e aerei leggeri. Sistemi modulari complessi a turbina ed elettrici ad alto grado di sicurezza potrebbero essere propulsori di grandi navi, sottomarini, aerei, e veicoli spaziali. State già lavorando a simili applicazioni?
    Cordiali saluti.
    Franco Sarbia

    Dear Andrea Rossi.
    The ECAT tested, has characteristics of size, power, operating temperature, which now seem to adapt to different applications such as mini electric turbo generators, capable of powering: cars, motorcycles, small boats, and light aircraft. Complex modular systems, turbine and electrical high degree of security may be engines of large ships, submarines, aircraft, and spacecraft. Are you already working in similar applications?
    Best regards.
    Franco Sarbia

  657. Andrea Rossi

    Franco Sarbia:
    Well, the applications you are looking for belong to the future of the Hot Cat, possibly. Nevertheless, I must repeat that presently the focus of our R&D is restricted to the 1 MW industrial plant and the gas fueled Hot Cat.
    Warm Regards,
    A.R.

  658. Steven N. Karels

    An Estimate of the Hydrogen Pressure within the eCat Reactor

    Assume the ideal gas law

    Amount of available hydrogen (from the LiAlH4) is 0.006 grams
    eCat Reactor operating temperature of 1500 K
    Interior eCat cavity dimensions assumed to be 0.5” diameter by 20cm in length
    Ideal Gas Law: PV = nRT where R = 0.08205 L * atm / ( mole * K)

    V = L * pi * D * D / 4 = 20cm * 3.1415927 * 1.27cm * 1.27cm / 4 = 25.34 cc = 0.02534 liters
    n = 0.006 grams / 2 grams per mole (diatomic hydrogen gas in the reactor interior volume) = 0.003 moles
    P = nRT / V = 0.003 moles * 0.08205 L * atm / (mole * K) * 1500K / 0.02534 = 14.57 atm

    This would be the maximum pressure. Actual pressure would be reduced by hydrogen adsorbed into the nickel

  659. Andrea Rossi

    Orsobubu: I retrieved this comment of yours from the spam: fortunately it was in the first spam page in good evidence. Beware next time you don’t find your comment published is because it contains something taken as advertising by the robot. In that case, just inform me sending an email to info@leonardocorp1996.com
    Warm Regards,
    A.R.

  660. Boss

    Dear Rossi ,
    Unibo published the tprII
    here : http://amsacta.unibo.it/4084/

    Is there a reason in your opinion for the e-cat recent developments being discussed and confined to online blogs and forums?
    Why do you think the media are not talking about It ?
    Regards

  661. Andrea Rossi

    Boss:
    We did not make press conferences because we deem it premature. It is necessary to see in operation the 1 MW plant for a long enough time to be sure of a commercial breakthrough before it is worth to make a diffused communication. For the same reason so far our publications are limited to scientific and technological context.
    Warm Regards,
    A.R.

  662. georgehants

    Dear Mr. Rossi, understanding that you have delayed your small domestic E-Cat because of the certification problems, are you still working on getting it ready for market as it is the device most suitable to use in villages etc. for water purification etc. and is very urgently needed in these situations.
    Beat wishes

  663. Andrea Rossi

    Georgehants:
    At this moment our focus is on the 1 MW plant and the gas fueled Hot Cats.
    Warm Regards,
    A.R.

  664. Pete Fimmel

    An item of news of interest for the last 3 weeks or more has been the volcanic lava flow in Hawaii. I haven’t noticed any discussion on why it continues to glow red and does not solidify.

    Surely this would be of interest to the LENR community. Perhaps someone could offer an explanation for it remaining at such a high temperature for weeks on end.

    If it were molten aluminium at 700˚C its temperature would drop 400˚C in 12 minutes from contact with ambient temperature surface soil.

    Looks like ‘natural’ LENR to me.

  665. Andrea Rossi

    Pete Fimmel:
    The time necessary to the solidification of the lava is not a number, but a complex system of integrals, related to the heat source power, the heat conduction, convection, irradiation, the heat exchange surface, the mass of the lava to be cooled etc. Without these calculations it is impossible to give an answer, but, honestly, I think that no LENR are happening there. I know how complex is the mechanism to get the so called “Rossi effect” and I do not think it can happen spontaneously.
    Warm Regards,
    A.R.

  666. Wladimir Guglinski

    orsobubu wrote in November 10th, 2014 at 9:20 AM

    More doubts over the Standard Model:

    http://www.techtimes.com/articles/19802/20141108/shocking-cern-may-not-have-discovered-elusive-higgs-boson-particle-after-all.htm

    http://www.techtimes.com/articles/19806/20141109/researchers-claim-higgs-boson-particle-still-elusive-what-did-cern-discover-then.htm
    ——————————————-

    Andrea Rossi wrote in November 10th, 2014 at 10:35 AM

    Orsobubu:
    Your last comment is interesting also in general in the following sense: an experiment made by the creme of the scientists of all the world, with funding of tens of billions of Euros, raises doubts: this makes normal and understandable that also the E-Cat science can raise doubts among the scientific echelons.
    Nobody is immune from doubts. Respect them, I have an advantage, though: with a commercial breakthrough I can make futile any kind of doubt.
    =========================================================

    Dears Orsobubu and Andrea Rossi

    the brain of the scientist is one among the most mysterious and paradoxical things of the universe.
    Because although the scientist is known as the most rational of the beings, however the most scientists become irrational when the theories they believe is threatened by some experiment.

    The Higgs theory was conceived 50 years ago by considering the empty space, because he was led by the need of explaining (from the concept of the empty space), from where the particles get their mass.

    But in 2011 a new experiment had proven that the space is not empty, because an empty space cannot create light (the empty space cannot get energy from nothing):
    A vacuum can yield flashes of light
    http://www.nature.com/news/a-vacuum-can-yield-flashes-of-light-1.12430

    Therefore,
    the Higgs theory lost its merit and lost its sense, because he had proposed it with the aim of solving the paradox: how particles can get mass from the empty space?.

    The mass of the particles comes from their interaction with the aether, which existence was proven by the experiment published in the journal Nature.

    And so the physicists would have to realize that there is need to develop a New Physics, based on the concept of the aether:
    a new model of the atom
    a new model of the nucleus
    a new model of elementary particles

    But instead of to face this unavoidable situation requiring a New Physics, the physicists persist in keeping their old theories developed under the hypothesis of the empty space.

    Probably, after the publication of the paper by Nature in 2011, the own Higgs had said to himself:
    “What a hell… I developed my theory by supposing the space as empty, and now this experiment proves the empty space is a myth. Oh, my God, in this case… probably my theory is wrong… there is no need any boson to give mass to the particles, they can get mass from the interaction with the non-empty space…”

    Suppose that an alien scientist from another planet comes here (probably getting energy for his spacecraft from the Rossi’s Effect), and we tell him that scientists in the Earth continue believing that there is need a kind of Higgs boson so that to supply mass to the particles, in spite of in 2011 a new experiment detected that the space is no empty, as Higgs supposed for proposing his theory.

    Well,
    immediately the alien scientist would reply to us:

    “This is unbelievable. What sort of scientists do you have in this planet?”

    regards
    wlad

  667. orsobubu

    Andreas Moraitis, Peter Forsberg, StudentG, Silvio Caggia:

    I can give you a well done link on Husserl

    http://www.filosofico.net/husserl.htm

    translated:

    http://translate.google.com/translate?hl=en&sl=auto&tl=en&u=http%3A%2F%2Fwww.filosofico.net%2Fhusserl.htm

    This is perhaps the busiest italian site for students of philosophy. I must say that, although Husserl departs from Berkeley’s classical idealism and then try to solve the problem of solipsism (the material non-existence of other men), anyone who is preparing to such studies should necessarily compare him with a general critique of modern idealism, as for example the powerful Lenin’s one in “Materialism and Empiriocriticism”, which faces the same problems on the crisis of science and in general of the life of humanity today, resolved of course – in the works and actions by the author – not through subjectivism and phenomenology (the reflection on the subject itself, You can seek an answer exclusively in yourself, as said by Rossi), but through the material intervention of the man in society and economy, ultimately by class struggle. I’m sure that among the 11 professors examining Rossi’s doctoral thesis (he’s constantly under a commission of Professors), Geymonat, who was critic with Husserlian fenomenology, would have liked to hiddenly shoot him in the back with a flat out 0/110 :)

    In fact, what interests most to me is to remember the figure of Rossi’s Professor of Relativity, Ludovico Geymonat, the knowledge of which could also be useful here to non-italian readers, like Peter Forsberg (hi!), when he considers empirically evident the failure of alternatives to the capitalist mode of production; when it is rather obvious (and it had to be especially to a master of logic as Geymonat) that those systems were based on absolutely capitalistic social relationships (market, money, wage labor, banks, etc.), although a state controlled market instead of a free market.

    Geymonat had a degree in mathematics and philosophy, he was the most important epistemologist (philosopher of science) we’ve had in Italy, and one of the foremost in the world. Together with Feyerabend, he demolished the thesis of Karl Popper, whose critique vs. dialectical materialism/marxism had very little scientific foundation, dictated mostly by geopolitical needs in a mccarthyistic political climate, depending on the creation of an anti-communist ideology in anti-soviet key.

    Among the students of this master of scientific philosophy there are contemporary important scholars in Italian and international cultural life, which often followed different paths from the thought of their teacher, and it gives me great satisfaction to know that Andrea Rossi has also been part of this group. This fact alone is sufficient to me to ridicule any critics by detractors that use “established science arguments” against him.

    I badly translate here a few sentences about Geymonat and science, from the site by philosopher Diego Fusaro:

    http://www.filosofico.net/geymonat.htm

    translated:

    http://translate.google.com/translate?hl=en&sl=auto&tl=en&u=http%3A%2F%2Fwww.filosofico.net%2Fgeymonat.htm
    —–
    In the course of his work, the reflection comes, in particular, to two conclusions: to reject a formalist-conventional interpretation of knowledge (which can and should relate to the real facts and truth), and to value analysis as not static and abstract but dynamic and practical (everything is “dynamic”, everything is “motion”); analysis must take into serious consideration “the complex dialectic, both theoretical and technical-experimental, pushing the scientist to ever more daring generalizations of his results”.

    For Geymonat scientific theories are “an act essentially historical, indissolubly linked to the level of human civilization and therefore at a level of our instruments of knowledge and action”, restoring in this way the thesis expressed by Lenin in Materialism and Empiriocriticism, and dialectical materialism by Engels … He believed in science as a content of truth, albeit temporary. But he did not believe in the neutrality of science. Science is a powerful tool, the most powerful that has given the man. It is not indifferent to the social group which holds it: if science is the prerogative of the ruling classes, it becomes a powerful tool of coercion. If science becomes the preserve of the lower classes as well, then it becomes the most powerful tool of liberation and social progress … This view, class, politics and science had, in lucid and coherent speech by Geymonat, two specific consequences. The socialization of scientific discourse, resulting in attention to the communication of science to the general public. And the social commitment of the scientist … But Geymonat is not only uncomfortable for academics and politicians. Nor for his students and his readers. It ‘also inconvenient, and perhaps especially for scientists, who Geymonat calls for a more stringent commitment: to recognize that their science is not neutral, that knowledge they produce have enormous effects on society. And, therefore, scientists can not think of “focus their activities on pure research without being distracted” by other concerns.

    They must focus on the tormentors of the society “to reveal the social truth, just as Galileo had pointed his telescope to the sky to reveal the physical truth. Scientists must engage “with the utmost seriousness to address the urgent problem to make sense of human philosophical, ethical and political science. “Because if science “will not be able to broaden and deepen its duties, if it fails to take the position of high responsibility that competes in today’s world, if it is unable to spread throughout the critical spirit, will eventually betray their mission. In that case, will soon become a factor not of progress, but of genuine ruin of increasingly dangerous dehumanization of society, “as a tool of emancipation of the whole society to an instrument of power for small oligarchies … The force of arms, the laws, the propaganda that this would be the best of the possible states, are the means of defense of “the Order of Things” and is called Power. Geymonat writes: “If we ask a revolutionary the things which he would change in this Order he shall answer: I want to change everything. But if we turn to a conservative, more or less said, he will say: I want to put some modifications, apply some reform. That is simply not possible and so it is as if to say: I do not want to change anything. A simple examination of what has happened and is happening continuously in the development of society, shows that efforts to reform the order existing have to invest all of that Order or fail. ”
    —-

    Finally, I want to add something that few people know. Throughout his academic and scientific life, Geymonat was always admittedly Marxist, but also hostile to the positions of Lenin. In the late 70s there was his shocking “Leninist turn” – harshly criticized by all colleagues – in which he exerted self-criticism of previous positions. Of course I may be wrong, but I think the early repudiation of Lenin was the result of a kind of residue of “personal interest”, in the sense that italian academics and politician in the 60s and 70s were very close to soviet ideology and far from the critics of Leninist wing. The latter, of course, was deeply adverse to the pro-Soviet official Communist party (remember that, since the 30s, Stalin had worldwide killed all the Leninist opposition to his regime, and yet in the Hungary repression, eg., they did not have better luck). It’s true that Geymonat was a profound critic to the academy, but it was always, in the end, what made him a living, and he could have reasonably expected the retirement to fully express his revolutionary spirit (to the contrary, Rossi will not express, surelys he will not retire either). Coincidently, most of italian Rossi’s “scientific” enemies belong to that same leftist political and academic area fought by Geymonat.

    I don’t want to make an apology, but I must recognize that although Rossi, like other Geymonat’s students, has followed a different philosophy, inspired by confidence in the market economy, however he distinguished his life according to the teachings of the illustrious Professor, working on the concrete, economic reality of the society, which he achieves by combining in himself the characteristics of the scientist, of the inventor, of the entrepreneur and of the epochè (in the end I must to add a stupidity at any cost)

  668. Alessandro Coppi

    Making a paragon between the e-cat development and early radio experiments, we could say that Fleischmann e Pons finding can be compared with Hertz’s resonator, the ITP tests represent the shot of rifle behind the hill, the last e-cat device the first electronic tube, the 1 MW plant represents the Poldhu station.
    From this point in forward we can imagine a fast and global affirmation of the e-cat technology.
    I have not casually spoken of Guglielmo Marconi, there are many common points between you and him, and your work, seriously risks to overcome his radio for importance in the mankind history.

    ···—·—
    Alessandro Coppi

  669. KeithT

    Dear Andrea Rossi,

    For a E-Cat within a circulating thermal fluid circuit, the fluid could be heated via gas, electric or even another E-Cat, once the primary E-cat was up to temperature then producing excess heat it would be a case of extracting the excess heat via a heat exchanger. If you have multiple E-Cats within a circuit it only requires the initial heat source to get the circuit up to temperature, after this it becomes a question of thermal heat balance and control of the individual E-Cat thermal cycling to get overlap.

    Apologies for taking too simplistic a view i am trying top understand the potential of the E-Cat.

    Regards,
    Keith Thomson

  670. Andrea Rossi

    Keith T.:
    Thank you for your suggestion.
    Warm Regards,
    A.R.

  671. Frank Acland

    Dear Andrea,

    Can you tell us:

    1. Who made the reactor that was used in the Lugano report?

    2. Who prepared the powder that was used in the reactor?

    3. What your role was (if any) in preparing the above items?

    Many thanks,

    Frank Acland

  672. Andrea Rossi

    Frank Acland:
    1- The reactor that has been used in the Lugano test has been manufactured in the factory of Industrial Heat, in Raleigh, North Carolina.
    2- The charge has been prepared by Industrial Heat, as all the charges are now, obviously upon the instructions I delivered with the know how.
    3- I had no role in the preparation of the reactor and of the charges, because I trust my magnificent Team. After months of rehearsing, under my direction, the Team of IH is able to manufacture everything without my help. For example, the 1 MW for the Customer of IH has been completely manufactured by them. The reactor used in Lugano is just one out of many of them manufactured in the factory of IH by their workers, directed by their engineers. The charges are made by the top level persons that have access to them.
    Warm Regards,
    A.R.

  673. Steven N. Karels

    Andrea and KeithT,

    A gas-fired heat source has some temperature that it operates at. I would assume there is an optimal temperature for control of the eCat. The eCat ideal operating temperature is probably lower than the gas-fired heat source (or else the gas-fired heat source would not be a good heat supplier). Therefore, the molten salt might act as an efficient mechanism to change the gas-fired temperature to an optimal temperature that is needed for the eCat? Maybe the eCat can output energy to heat the molten salt to improve effective COP? Thoughts?

  674. Andrea Rossi

    Steven N. Karels:
    No, it does not work like that.
    and there is no point to put a heat exchanger between the heat source and the reactor.
    Warm Regards,
    A.R.

  675. Eric Ashworth

    Ran out of space.

    Could these suggestions help solve some enigmas with regards atomic physics?. I would also like to think that maybe Andrea could use some of the information.

    There is more interesting information with regards the geometry of a cubic neutral of energy but I need to cut this short and I apologize to Andrea for taking up so much space on his blog. Regards Eric Ashworth.

  676. Andrea Rossi

    Eric Ashworth:
    I use all the information I receive.
    Thank you.
    Warm Regards,
    A.R.

  677. Eric Ashworth

    Koen Vandewalle & Greg Leonard, I feel it necessary to put forward further information with reference to the previous site of mine referring to the unifying field oscillation mechanism. This information when understood and considered could explain the anomalous mass of the neutron and how gravity features into the enigma. This technical so please bear with me.

    These findings and theories are from my understanding and observations with regards the mechanism previously referred to a a ‘unifying field oscillator’ which came about as an understanding I perceived as the internal dynamics of the atom. I believe the origin of geometry and math is related to an understanding of energy with regards atomic make-up.

    The mechanism design is based upon a binary interaction to produce and control a flow. This being one circle referred to as the field which is divided into four quadrants. Midway between the centre of the field and its periphery on each of the dividers which now become diameters are drawn four circles that extend from centre to periphery. Thereby a one third overlap is achieved by each circle with its neighbour. The mechanism is now designed using four rotors, two blades per rotor and a necessary timing sequence. Rotors are spaced accordingly to allow for a necessary vacant central space. The static baffle of the assembly (being a frame) contains a central divider/mantle between inner and outer field and upon which rotors are fixed at their strategic points. This reference to mantle comes from an observation in nature and referes to the division between the chambers of the frame that occupy the inner and outer positions of the field

    The field now has a central vacant space (super/major gravitational), four vacant spaces at the periphery of the field(minor gavitational fields), four gavitational at each rotor, four inner vortex flows (vortex gravitational), four outer vortex flows (vortex gravitational). The vortex flows are the economy flows within the mechanism that bind the flow structure together (binding force). The mechanism contains two gates of the super gravitational force which includes all the other gravitational forces and three types of flow, curvature, linear and static. The linear flow refers to an exterior macro curvature flow that responds accordingly to the two gates. Reference to my previous material will explain flows. There is a horizontal and a vertical plane to the mechanism as can be seen, four divisions of the horizontal between two gates of the super gravitational. Four divisions become four dimensions, two gates become two dimensions and thereby six dimensions in all. Consequently I refer to the mechanism as being able to generate a cubic neutral of energy. This mechanism is thereby able to prove a theory of unification, it is also able to be studied so as to investigate what I term energy interaction. Mobile energy (flows), static energy (atomic substance in structure) and the interaction between these two states which I term (the static and mobile mechanics of energy interaction). The static and mobile mechanics with regards the mechanism is in reference to the static baffle and its interaction with the generated flows.

    So in consideration with regards the mechanism I shall extend my thoughts to particle structure and attempt to explain stable and unstable structure and therefore the following is guesswork. To be stable a structure has to be a cubic neutral composed of six parts that compliment one another. The proton being of three quarks is unstable and so too is the neutron but together they form a cubic neutral of energy in the same way that two propellers overlap in the mechanism. The overlap of two particles is dependent upon their construction within a field and nothing can be formed outside of a field that conforms to the field. Every field contains its super/major gravitational force and its four minor gravitational forces. Aminor gravitational force is a black hole on the periphery of the field, this being one quarter strength of the super gravitational force. Fusing of structure (the construction of a vortex) occurs from the moment it is formed at the periphery from vortex gravitational force. Vortex forces formed at the periphery of a macro field will produce solid atomic substance due to the super gravitational force acting progressively on the structure as it traverses the field. Vortexes not formed at the absolute periphery will be formed of a gravity value less than that of a quarter and will always be less dense in their structural make-up because of this. Three distinct densities with a fourth being less distinct thereby exist because of a divisional relationship within the field.

    The economy flow system of vortex forces within a structure provide the quality value pertaining to the structure. Good quality provides solidity, poor quality is responsible for week structure. Metals have good structure, Hydrogen has week structure. This is important: good structure has to have a good overlap between its particles to form a good economy flow system whereas, particles that barely overlap or oscillate to form a partial overlap with regards time and space i.e. time in/time out have very week strurcture. It is dependent upon the gravitational forces of the particles (particle gravitation forces depends upon their origin of manufacture) that dictates the overlap that is responsible for the vortex forces of the economy flow systems.

    In the mechanism I see four particles that overlap by one third, providing good structure. Each rotor can represent a particle. Thereby imagine three quarks (each quark composed of three aethers) rather than two blades rotating around a central vortex (I believe geometry dictates energy whether it be atomic or charged particles) within a field. Three evenly spaced quarks in a binary interaction with three other quarks of another particle (three quarks form an unstable neutral particle) form a stable neutral cube of energy. To picture this situation in the mechanism it will be seen that two quarks of the neutron would be travelling out to the periphery on an arc and one travelling into the central field whereas with regards the proton its activity would be visa versa. To travel into a field the vortex closes producing less inner gravity and less kinetic energy, to travel outwards of a field opens the vortex producing more inner gravity and more kinetic energy. Each quark composed of three aethers in vortex make-up. If so a quark could be likend to a mini heat pump with regards its charge potential as it completes a circuit.

    I shall now refer to flows and the mechanism. As I previously mentioned two curvature vortex flows, produced by the overlapping propellers, upon contact spin out to form economy flows from the produced linear flow. This linear flow encapsulates the cubic neutral of generated energy. The linear flow represents for me the electron flow force produced by a proton neutron activity. However, when the proton and the neutron are unable to produce a linear flow because of a week interaction then the neutron because of its quark activity responds according to the field of its manufacture, its two negative quarks become more negative by position and the proton acts visa versa. This results in a cubic neutral of charge energy which is not the same as a cubic neutral of atomic energy made up of particles. Consequently, hydrogen could be said to be comprised of an expanded neutron that circumvents a contracted proton in an oscillating motion. The interaction between two these particles could be oscillatory in activity because of the week bond. If what I say is correct then could this explain the enigma with regards P+E=N. The loss of .789 could be due to the formation of an economy flow system or it could be said that .789 has something to do with a binding force energy. The subject does intrigue me.

  678. KeithT

    Dear Andrea Rossi,

    If an E-Cat can be gas powered, could it be powered by a heated thermal fluid / molten salt.

    Regards,
    Keith Thomson

  679. Andrea Rossi

    Keith T:
    To what purpose? Any heat exchange implies a loss of efficiency.
    Warm Regards,
    A.R.

  680. ing. Michelangelo De Meo

    The Independent Third Party Report Has Been Posted by Professor Levi Following the site :
    http://amsacta.unibo.it/4084/
    Also, the Report of the Independent Third Party has been published on Google Scholar:
    http://scholar.google.it/scholar?q=E.Foschi%2C+H.Essen&btnG=&hl=it&as_sdt=0

  681. Andrea Rossi

    Ing. Michelangelo De Meo:
    Thank you, I am delighted to read that.
    Warm Regards,
    A.R.

  682. KeithT

    Dear Andrea Rossi,

    Regarding the gas E-Cat, is a pure gas powered E-Cat possible, or would you still require electrical controls. There is still many remote locations on Earth that do not have an electrical supply.

    Regards,
    Keith Thomson

  683. Andrea Rossi

    KeithT:
    Electric controls are necessary, but they consume a small amount of energy, easy to backup.
    Warm Regards,
    A.R.

  684. StudentG

    Dear Mr. Rossi,

    Even though I was just googleing Husserl and don’t know much about him and his work, it seemed to me that his ideas are kind of in line with core teachings of the Buddha as interpreted by Theravada Buddhism. So I was just wondering if you know about Theravada or have interest in this teaching and see parallelities to Husserl. To clarify, where this question is coming from, I’m a student in Nanoengeneering from Germany and following this blog for a while now with much interest. So my interest is not religious, but as you seem to be really open minded and experienced in the combination of scientific science and philosophy science, im seeking for an answer in this perspective, if even possible.

    Best greetings and Good luck

  685. Andrea Rossi

    StudentG:
    Thank you very much for your interesting comment. In my Philosophy studies I also have studied Buddhism, and your innuendo is intriguing. I leave to you your subjective interpretation. You can seek an answer exclusively in yourself. Whatever I could say is not important. By the way: I am sure you will make a great engineer in nanotechnologies and I am sure I will hear of you in the near future. As you know better than me, in this period of your life the most important thing is what you learn from your Professors. Leave all the rest in a second place.
    If it is necessary to study 8 hours per day, you have to study for 12 hours.
    Good luck!
    Andrea Rossi

  686. Giovanni Guerrini

    Dr. Bert Abbing

    Thank you.

    Regards G G

  687. Andrea Rossi

    Orsobubu:
    Your last comment is interesting also in general in the following sense: an experiment made by the creme of the scientists of all the world, with funding of tens of billions of Euros, raises doubts: this makes normal and understandable that also the E-Cat science can raise doubts among the scientific echelons.
    Nobody is immune from doubts. Respect them, I have an advantage, though: with a commercial breakthrough I can make futile any kind of doubt.
    Warm Regards,
    A.R.

  688. Curiosone

    I read the comment of Bert Abbing: there is the difference between a true physicist and a bunch of guys with an agenda

  689. DTravchenko

    What impressed me from the open letter of Bert Abbing is the fact that your enemies qualify themselves as big chemists and physicists, but are privy of elementary knowledge of chemistry and physics.
    Warm Regards,
    DTravchenko

  690. Robert Curto

    Dr. Rossi, excellent post by Dr. Bert Abbing.
    I was happy you gave him a nice pat on the back, and added a fact about
    the 1 MW Plant.
    Robert Curto
    Ft. Lauderdale Florida
    USA

  691. Andrew

    Dear Dr. Rossi,

    i’d like to bring to your attention this interesting interview:
    http://www.ecat-thenewfire.com/blog/lenr-seen-from-nuclear-engineer/

    Kind Regards
    Andrew

  692. Andrea Rossi

    Andrew:
    Thank you for the link to this interview. This is another important endorsement to LENR coming from the top level Science.
    The considerations of the nuclear eng. Piero Andreuccetti of CESI ( Italian R&D leading center belonging to ENEL) are very intelligent. I also agree with him about the work on course at the MIT directed by Dr Brian Ahern.
    Warm Regards,
    A.R.

  693. Andrea Rossi

    Dr Bert Abbing, Physicist:
    Thank you for your substantially correct insight; now, the operation of the 1 MW plant in the factory of IH’s Customer will be the commercial breakthrough.
    Warm Regards,
    A.R.

  694. orsobubu

    >I got it: you got stuck seat in the theater…

    ahhaha this was very good

    …and I learned about the film, many thanks to you sir, I was really going to miss it!

  695. BertAbbing

    Bert Abbing PhD, Physicist

    Open Letter

    To whom it may concern,

    Dear sirs I’m writing to you because I’m ashamed by the biased, negative, unscientific behavior of some blogs regarding the Rossi Ecat topic.
    Not even what is written in these blogs is completely without any scientific foundation but also the comments or contributions that contrast with their opinions, are harshly attacked with a behavior that is very far from what is a normal scientific debate.
    Making a more in depth search one can find that the same persons, with same nicknames, write the same, or similar, comments on “different” blogs, revealing a network of organized disinformationist voted only to diffuse doubt, suspicion and a kind of “conspiracy theory” among common people and persons who eventually have to take decisions but are not trained in physics or other scientific fields.
    This network is what I would call a “negationist mafia”, and is not surprising to realize that many of those people have economic interests.
    As an example I will examine here pages against the TPR2 that are a good example of a general behavior. This pages are self published in blog by a company called “StepChange Innovations GmbH”, that is self defining “a technology development and consulting firm based in Germany”, and appears to operate mainly in the chemical industry field. The articles, as others also against Rossi, are all signed by Dr. Christian Schumacher the company CEO who has no background as a physicist presenting himself as having “ 20 years of experience in the chemical industry with global players such as Hoechst AG and DyStar Textilfarben GmbH as head of R&D, senior regional business manager Asia Pacific, head of e-commerce, head of marketing services, new product development manager and R&D chemist”.
    With such a background he made several mistakes in the article writing and appears clear that his main interest is not scientific.

    Let us review just few of the main errors in the article and also in some of the comments.
    First of all we note an annoying repetition of old arguments who were already answered in the past but, like “gutta cavat lapidem”, probably in the author mind there is the strategy that repeating a false argumentations will at least win over the new readers.
    One of the arguments is clearly wrong and demonstrate all the ignorance in the field. The fact that the emissivity of the IR camera was set to 1 ( Black Body ) in the TPR1 have in reality set a LOWER LIMIT to the energy measurements because the temperatures read by any IR camera will be the minimum possible in this case. This point was already discussed in the past.
    Also calling IR imaging for just a “qualitative method” demonstrate the author inexperience
    in measurements and laboratory methods. If one has the humility to search in the literature can find that Thermographic Calorimetry is commonly used in a variety of fields ( e.g. medical, but also military and aerospace ) where a fluid calorimeter is not usable.
    A fluid calorimeter is a complex device to design and at the temperatures of the hotcat is even more complex requiring careful calibration of fluid flow ( forced air ? ) and fluid temperature. Probably in this kind of device a heat exchanger ( eg. air to water ) that would require also an extremely careful calibration.
    All this means high costs and long times and also would be of no interest because the only interest of the Professors that conducted the test, was to have a lower limit of the produced heat and avoid any possible error that could increase the measured value. They achieved that calibrating the apparatus using the dummy, ( actually the same reactor without the charge ) and taking always the most conservative scenario. They specified that many times in both reports.
    This point was completely missed, probably purposely or by inexperience, by the blog author.

    Thermographic recording was the method of choice because was sufficiently simple, direct, and economic to obtain the desired result.
    .
    It is good scientific practice when writing to be precise, true and documented so finding phrasing like:
    “There is still criticism about the way of energy input, vaguely described and not ruling out manipulations by wiring tricks, inaccurate measurement of the output energy, and lack of proper calibration.”
    Immediately let the educated reader to discard all the article as unscientific and just disinformation. The energy input is described in the report with an electric diagram, there was also a calibration, the point of measurement were two giving coherent figures. So the main content of this affirmation is FALSE, the other is also vague and belongs to the realm of hypothesis and paranoid suspects not to reality.

    The rest is just a sad reading. All “Miracles” about wires was already clarified in the report on page 19 where is written “ All the characteristics of these resistors, however, such as their geometric dimensions and the exact makeup of the alloy they are made of, are covered by trade secret. ” So is a fact and not an hypothesis, as written by a certain “Thomas Clarke” in a comment against “KJ”, that that was a special custom alloy made especially for the reactor and NOT standard Inconel. Without these data all mumbling about fusion of wires is just nonsense.

    The author of the blog page, probably pressed by the agenda to attack the report fails even in what it should be his own field: chemistry. He tries to transmit the idea that LiH4 is so unstable that would not be usable. The fact LiH4 does effectively have a violent reaction with water is written even in the Wikipedia page (http://en.wikipedia.org/wiki/Lithium_aluminium_hydride) from which he has copied the chemical formulas, but the reaction with moisture of air, is much more slower. Even in the reference he cites, copying part of the abstract without understanding it, is written that “The water absorption up to 11.7% due to exposure to air for 1 h does not change in any drastic way the hydrogen desorption rate of ball milled LiAlH4 “ . This means that the reaction is slow with a low release of Hydrogen in air, so low that from the same abstract we learn that this hydride , “ does not self-ignite on contact with air but can only be ignited by scraping ”, and from Wikipedia page : “Aged, air-exposed samples often appear white because they have absorbed enough moisture to generate a mixture of the white compounds lithium hydroxide and aluminium hydroxide.”
    Aged means long term exposure and not that “would decompose rapidly upon storage in air, in presence of moisture”.
    So we have the FACT that the author of this blog article is deliberately manipulating scientific data to follow an agenda.
    The other reference cited by the author is a downloadable Open Access publication that almost nothing has to do with the examined topic.

    As a final note regarding this first article let me comment about the ridiculous thesis of reversed clamp cited in the comments. If we enter in the domain of the hypothetical and paranoid suspect every scientific paper on earth is questionable so this hypothesis have no foundation but is only “ conspiracy theory “, with no value.

    The second part of the article is even worse, on a quality point of view, than the first.
    We find a mix of unrelated facts. Rossi Patent, any declaration from him, or any other theory has nothing to do with the measurements, as the fact that this or that isotope is commercially available. Also what happened in 2011 is not of any interest here because does not regard this paper and also no result was published from that sample.
    If the group had not mentioned Cu for example is because there was NOT. This means that fortunately for them the reaction p+Ni is excluded. But this does NOT mean that ALL reactions are excluded.
    The only fact that remains is that from the samples collected there is an evidence of isotope shift. It is also NOT true that the fuel was exhausted if we read the data with carefully.
    Part of the Li7 was still there and I would not be surprised if the main source of energy would be the Li7.
    As regarding sampling and handling of the powders we should presume that the group have this abilities. One of the members of the group is even specialist in Forensic Chemical Analisys.
    For obvious reasons Rossi participated to reactor loading and powder extraction ( in the presence of at least a member of the committee ).
    It was his reactor. But is not written in the report that he participated in sampling and handling that are others phases. Again even in that part of the article we find a deliberate distortion of facts in order to demonstrate predetermined thesis.
    Nuclear Physics and Quantum Mechanics are not field for lay persons so I would invite the author of that blog to refrain to comment or criticize Theoretical Physics works.

  696. Andreas Moraitis

    Dear Andrea Rossi,

    I understand your attitude – although I am convinced that the 110/110 would not have been granted without reason. Anyway, the subject of your thesis is one of the most interesting (and challenging) subjects that one could choose for a philosophical study. Maybe one day when you have more time you might want to summarize your thoughts on it.
    Unfortunately, while everybody has at least heard of Einstein’s General Relativity, Husserl’s work is almost unknown to the wider public. Viewing both concepts in context must be an effective “catalyzer” for unbiased thinking – perhaps one of the most urgent needs in a rapidly evolving world.

    Best regards,
    Andreas Moraitis

  697. Andrea Rossi

    Andreas Moraitis:
    The reason why Husserl’s importance for the approach to any scientific issue ( remember he was originally a physicist and a mathematician) is not well known is due to the extreme difficulty of his texts. To read Husserl is one of the most difficult tasks you can imagine under the intellectual point of view, also because his German language is very difficult to translate, and usually translations lose part of the meaning he wants to carry with his words. I had to take lessons of German language with a specialist, to study in German the “Ideen zur einer reinen phaenomenologie und phaenomenologischen philosophie” ( I improperly wrote ‘phaenomenologie’ because I have not the dieresis to put on the ‘a’) and only reading in German I could understand what he wrote. Also in this case, even if in the exam of Filosofia Teoretica I gave on it Prof. Enzo Paci granted me a 30/30 cum laude, I was and am convinced that I have got only a fraction of it. It is immense but very difficult; you have to stay hours on every page, otherwise its content flows on the surface of your brain like water on granite.
    P.S.
    The translation in Italian of the “Ideen” has been made by Prof Enzo Paci: when during the exam I told him that to understand Enzo Paci I had to read Edmund Husserl he laughed like crazy.
    Warm Regards,
    A.R.

  698. Steven N. Karels

    Dear Andrea Rossi,

    Determining the Fuel Composition for an eCat Reactor

    Based on the Lugano Report, an estimate of the fuel composition may be attempted

    Known facts:
    1. Fuel sample had a mass of 1 gram
    2. Page 29: “From the analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and ICP-AES analysis we find there is about 0.011 grams of 7Li in the 1 gram fuel.
    4. Page 29: “… the information from ICP-AES that there is about 0.55 gram NI in the fuel.
    5. Page 28: “From all combined H in addition to the Ni.
    6. Page 28: “… from the ICP-AES analysis which shows the mass ratio between Li and Al is compatible with a LiAlH4 molecule.
    7. Page 28: “…natural composition, i.e. 6Li 7% and 7Li 93%
    8. Page 28: “We remark in particular that hydrogen but no deuterium was seen by SIMS.

    Analysis

    The average mass of the lithium atoms are 0.07*6 + 0.93*7 = 6.93 amu.
    Aluminum atoms have a mass of 27 amu while hydrogen atoms have an average mass of 1.
    So the molecular weight of the LiAlH4 must be 6.93 + 27 + 4 = 37.93 amu.
    There for the amount of LiAlH4 must be 0.011 grams * 37.93 / 6.93 = 0.06 grams and the amount of aluminum must be 0.043 grams. The amount of hydrogen in the LiAlH4 must be 0.006 grams.
    The iron mass must therefore be 1.0 grams (total) – 0.55 grams (Ni) – 0.043 grams (Al) – 0.011 grams (Li) – 0.006 grams (H) = 0.39 grams of iron.

    Element % by Weight
    Nickel 55.0
    Iron 39.0
    Aluminum 4.3
    Lithium 1.1
    Hydrogen (no Deuterium) 0.6
    Total 100.0
    LiAlH4 6.0

    It is also possible that the LiAlH4 was prepared using hydrogen depleted of deuterium.

  699. Andrea Rossi

    Steven N. Karels:
    Obviously I can add nothing to what has been published in the Report of the ITP.
    Warm Regards,
    A.R.

  700. Greg Leonard

    Dear Eric Ashworth
    I do not believe any sort of separate vanes can counteract the torque generated by the motor on the body of the vehicle.
    A pair of contra-rotating rotors may be necessary.
    Incidentally, the Entecho web site appears to have been hacked this morning!!
    Greg Leonard

  701. Andrea Rossi

    Andrea Moraitis:
    I do not think it is worth. I studied Relativity in the ” Università Degli Studi di Milano” with Prof. Ludovico Geymonat , who was my Professor of “Filosofia della Scienza” and in my doctoral thesis, that I made with Prof. Enzo Paci as the doctoral advisor and my teacher of Husserl’s Phenomenology, made of Relativity a paradigmatic example of a work that has been born by means of an epochè of all the consolidated pre-existing knowledge. Honestly, is an immature work that is not worth to be published. At those times I already was working, designing, manufacturing and selling incinerators with energy recovery and had to work at least 10-12 hours per day so I had to study during the night. My thesis got a 110/110 from the 11 Professors that examined me, but I valued it 60/110 then, much less now. It is superficial. I had not the time and the focus necessary to make it as deep as I should have done. Forget it.
    Warm Regards,
    A.R.

  702. Wladimir Guglinski

    What would be the very small change in the Laws of Physics supposed by E. Teller ?

    Dear readers of the JoNP,
    I was thinking about what McKubre told Edward Teller had said to him:

    “He didn’t think cold fusion was a reality, but said if it were he could account for it with a very small change in the laws of physics.”
    http://archive.wired.com/wired/archive/6.11/coldfusion_pr.html

    What sort of very small change in the Laws of Physics could be occuring to Edward Teller, from which would be possible to explain cold fusion?

    Well,
    probably Edward Teller knew the Don Borghi experiment, from which neutrons are obtained from fusion proton-electron at low energy. And as this is impossible from the Laws of the Standard Model, maybe occurred to him that one of the mysterious mechanisms that rule the nuclear reactions in cold fusion is concerning the fusion proton-electron at low energy: p+e = n.

    Perhaps he also would be thinking about a new sort of gravity Planck’s constant acting within the nuclei, as I propose in my paper Anomalous Mass of the Neutron
    http://www.journal-of-nuclear-physics.com/files/Anomalous%20mass%20of%20the%20neutron.pdf

    The academic physicists are finally surrendering themselves to the reality of cold fusion, as we see from what was said by the Russian nuclear physicist Dr Vitaly Uzikov:
    http://www.proatom.ru/modules.php?name=News&file=article&sid=5595

    As the fusion proton-electron may have an important contribution for the nuclear reactions occuring in cold fusion, one can hope that the next step is the acceptation of the Don Borghi experiment.
    So, one may even hope that Don Borghi experiment will finally be accepted by the academics, and his experiment will be repeated in the universities.

    After all, as the cold fusion is being finally accepted by the academics, there is no reason anymore for the refusal of the Don Borghi experiment.

    regards
    wlad

  703. orsobubu

    >Silvio Caggia:
    >Dear Andrea Rossi,
    >a philosophic question:
    >which is your preferite philosopher?

    >Edmund Husserl.
    >Warm Regards,
    >A.R.

    Silvio Caggia, I cant’ believe it, I’m sure there was a slip a lapse a typo a virus or some other sort of connection bug or internet malfunction and the name of his preferite starts with M and ends with X, I know it better because Andrea Rossi not only is fond of the cold fusion and the epochè but he likes the permanent revolution very much and things like that also

  704. Andrea Rossi

    Orsobubu:
    I got it: you got stuck seat in the theater looking permanently at “Interstellar”, until, resisting the roller coaster vibrations of the spaceship when it crossed at the speed of light the barrier of the wormhole, you reached, through the same wormhole, the planet “Alltrue”, so called because in it all you imagine gets true.
    Warm Regards
    A.R.

  705. Andreas Moraitis

    Dear Andrea Rossi,

    I have been tempted to ask you this for a long time, but for some reason I did not dare to do it: Would it be possible that you publish online your doctoral thesis on the relationship between the philosophy of Edmund Husserl and Albert Einstein’s theory of relativity? I would be very interested in reading it. Of course, I would not hesitate to improve my sparse Italian for this purpose.

    Best regards,
    Andreas Moraitis

  706. Eric Ashworth

    Koen Vandewalle & Greg Leonard, Because of your interest in this specialiosed subject I shall provide further information to help explain and so that you can give it further consideration. As you are aware Wladimir has technical in depth knowledge of the atomic nucleous which is an extremely specialised subject with technical jargon. Language is only useful when you can understand it and this can cause a problem. However, I shall continue to put forward my considerations of energy interactions and hope that Wladimir can glean some useful information from what I think based on my observations of the mechanism involved in my research. Regards Eric Ashworth.

  707. Curiosone

    Did you see the movie Interstellar?
    I saw it yesterday in New York. They say can teach the basics of relativity: is it true?

  708. Andrea Rossi

    Curiosone:
    Who wants to read where Fantasy meets Science, can read Asimov.
    Who wants to get fun, sort of ‘Donald Duck goes stellar’ , can go to see this movie, which is a lot more Fanta than Scientific. The return of the protagonist still young to meet his grown old daughter is conceptually wrong and misleading in terms of Relativity, because speed of light cannot be overcome, therefore time cannot playback; the scene where the astronauts go at the speed of light suffering for vibrations similar to a roller coaster is tragicomic. Who goes to see this movie must let alone Science and forget Relativity.
    It is a funny cartoon.
    Warm Regards
    A.R.

  709. Eric Ashworth

    Dear Greg Leonard, Regards your reply November 6th pertaining to the torque generated on the Entecho craft by the centrifugal fan. The torque I believe is compensated by a series of blades securred to the inner rigid body of the craft. My suggestion is to syncronise the angle of attack of the fixed blades to that of the flow generated by the centrifugal fan by computer control of the fixed blades with regards r.p.m.. As you know R&D is ongoing and the Entecho technology is still in this stage. Regards Eric Ashworth.

  710. Steven N. Karels

    Dear Andrea Rossi,

    Reading The Report, I noticed on page 52 (Appendix 4) there was presence of aluminum in the fuel but it disappeared in the ash. The Report proposes that the aluminum came from Lithium Aluminum Hydride (page 28) which they assume was employed to release hydrogen gas at elevated temperatures. So what happened to the aluminum?

  711. Andrea Rossi

    Steven N. Karels:
    We are working on the reconciliations after the results of the Report. We will publish the conclusions when we will have resolved all the equations involved.
    Warm Regards
    A.R.

  712. JCRenoir

    Congratulations for the peer reviewing of Dr Uzikov published on Proatom.
    Great achievement.
    JCT

  713. Andrea Rossi

    JC Renoir:
    Thank you,
    Warm Regards,
    A.R.

  714. DTravchenko

    Dr Andrea Rossi:
    Did you see the article on Proatom written by Dr Vitaly Uzikov?
    http://www.proatom.ru/modules.php?name=News&file=article&sid=5595
    This is a peer reviewed nuclear physics magazine and Dr Uzikov is a preminent figure of the Russian Nuclear Physics world. Congratulations to the Professors of the ITP, this is an important endorsement from the mainstream Russian scientific environment. If you come in Russia you will find friends of much higher level than you can even imagine.
    From Russia, with love and with the regular Warm Regards,
    D. Travchenko

  715. Andrea Rossi

    D. Travchenko:
    This paper is becoming viral and I am very honoured of what you say: yes, I agree totally with you in regard of the inportance of this publication.
    Warm Regards,
    A.R.

  716. ing. Michelangelo De Meo

    Dear Dr. Rossi to announce that the prestigious Russian site
    http://www.proatom.ru/index.php is praising his story
    signed by VAUzikov leading progettazionetecnologo , NIIAR .
    He concluded ” … that Rossi is back in his marathon winner scientific .. ” .
    Dr. Rossi continues to run without looking back and win for all of us .
    Thank you for your extraordinary work .

    http://www.proatom.ru/modules.php?name=News&file=article&sid=5595

  717. Andrea Rossi

    Ing. Michelangelo De Meo:
    Thank you, your link is very important because it also has the English translation. The position of Dr Vitaly Uzikov is a milestone in the evolution of the scientific mainstream science. Now we have to reconcile the results of the Report of the ITP with the Standard Model.
    Warm Regards,
    A.R.

  718. Silvio Caggia

    Dear Andrea Rossi,
    a philosophic question:
    which is your preferite philosopher?
    I try a guess: Jeremy Bentham?

  719. Andrea Rossi

    Silvio Caggia:
    Edmund Husserl.
    Warm Regards,
    A.R.

  720. Italo R.

    Dear Dr. Rossi:
    assuming that on the beginning of the reaction the charge is formed with various components mixed homogeneously, and that during the reaction there are inhomogeneous variations within the charge. A sample taken at one point of the mixture is not representative of what has happened in the rest of the charge.
    Are we confident that the sample was taken significantly?
    After all, sampling is a science on its own.

    Kind Regards
    Italo R.

  721. Andrea Rossi

    Italo R.:
    Thank you for this very important link.
    Dr Uzikov is a top level nuclear phusicist of Russia and his attention is extremely important, also as a recognition coming from the top levels of the mainstream scientific world community.
    Warm Regards,
    A.R.

  722. Wladimir Guglinski

    Dears readers of the JoNP

    Sincerely… I dont like to call cold fusion by LENR.

    More than 20 years ago, there was a very strong resistance against cold fusion, because the phenomenon is impossible from the Laws of the Standard Physics. To speak about cold fusion was a tabu in the scientific community.

    With the aim to lubricate the oposition of the academic physicists against cold fusion, and soften the resistance against the acceptation of the phenomenon, Edmund Storms had proposed to call it LENR.

    But today we dont need to call it LENR, because today cold fusion is a reality. It was confirmed by 3 universities in Europe, and Andrea Rossi is putting his cold fusion E-Cat in the market.
    So, we dont need to be afraid of any resistance anymore. Cold fusion is a reality.

    Storms got his 15 minutes of fame. And we dont need to use the word LENR. We have to call it just by the correct name: “cold fusion”, because nuclear fusion occurs in the phenomenon, as stated by the report published by the 3 universities.

    I think to call things by the correct word they mean is the correct way to call them.

    regards
    wlad

  723. Wladimir Guglinski

    JR wrote in November 7th, 2014 at 4:40 PM

    Daniel,

    FYI, it also fundamentally makes no sense to listen to Wlad’s characterization of other people’s statements, theories, etc…, as he has little understanding of them to begin with and even less interest in trying to actually understand more.
    ——————————–

    of course
    I will never understand why a chord can never be broken by the fast rotation of a stone tied to its end, and I also cannot understand why the strong nuclear force can never be won by the rotation of a proton moving with very high speed, as you claim.

    Do you understand it, Daniel ??

    Unfortunatelly I cant, dear Mr. JR.
    I suggest you to propose a New Physics based on Crazy Laws, and then I hope we will be able to understand your arguments.

    As I also do not understand why you have no shame to say nonsenses here, where everbody may realize your lack of understanding in fundamental questions in Physics.

    regards
    wlad

  724. JR

    Daniel,

    FYI, it also fundamentally makes no sense to listen to Wlad’s characterization of other people’s statements, theories, etc…, as he has little understanding of them to begin with and even less interest in trying to actually understand more.

  725. Wladimir Guglinski

    JR wrote in November 5th, 2014 at 7:31 AM

    Dear Daniel,

    If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.
    ———————————————

    Dear readers,
    despite Mr. JR has already lost the discussion here,
    we have to take the opportunity,
    so that to show that,
    AGAIN,
    Mr. JR uses the inversion of the causality in his arguments.

    What Mr. JR said is equivalent to say:

    A stone tied to the end of a chord and moving circularly very fast cannot cause the rupture of the chord, because the centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the force of the chord in this case.

    Therefore,
    according to Mr. JR,
    a chord tied to a stone moving in circular orbit can NEVER be broken.

    According to Mr. JR, a very fine nylon thread for fishing (0.1mm in diameter) can easily keep a mass 100.000kg moving in circular trajectory, because the line will never break, because the line is subjected to the centrifugal force, which is ficticious.

    Those ones who believe that a fine nylon thread will never be break by a mass of 100.000kg moving in circular trajectory tied to the end of the nylon, they can believe in what Mr. JR says here in the JoNP.

    regards
    wlad

  726. Wladimir Guglinski

    Gell-Mann versus Edward Teller

    Mallove speaking about Gell-Mann:
    From the principles of Quantum Mechanics cold fusion occurrence is impossible to occur, as stated by the Nobel Laureate Murray Gell-Mann at a public forum (lecture at Portland State University in 1998): “It’s a bunch of baloney. Cold fusion is theoretically impossible, and there are no experimental findings that indicate it exists” 3.
    3- E. Mallove, CSICOP: “Science Cops? at War with Cold Fusion, Infinite Energy, V. 4, No. 23, 1999

    E. Teller:
    McKubre was summoned by Edward Teller. “He didn’t think cold fusion was a reality, but said if it were he could account for it with a very small change in the laws of physics.”
    http://archive.wired.com/wired/archive/6.11/coldfusion_pr.html

    So,
    two two laureates Nobel Prize with different opinions on cold fusion.
    Why?

    Simple.
    Gell-Mann was sure the cold fusion is impossible because he was sure the Standard Nuclear Physics was developed under unchanging and correct Laws of Physics.

    While Teller knew that something is missing in the Laws of the Standard Model.

    And there is no need to be a genius like Teller to realize that the Standard Model is not the final theory, since there are unsolved puzzle in Nuclear Physics, and some nuclear properties cannot be explained by the Standard Model.

    Teller also knew that by keeping the Laws of the Standard Model would be impossible to explain cold fusion, and that’s why he said that at least a small change in the Laws of Physics is needed.

    However, even a “small” change in the Laws of Physics always represent a big changing in the Physics.

    regards
    wlad

  727. Mark Saker

    Dear Andrea,

    1) I know you are now looking into the result of the nickel analysis.

    Even if you cannot determine the cause of Ni 90% transmutation in one month, would an easy fix be to add six grams of material to give you a 6month usage cycle (which is still not very much). Would this have any adverse effects?

    Or..
    WE’LL SEE

    2) Would the reactor continue to work regardless of the change in Nickel isotope (you probably cannot answer this ) :)
    EXACTLY

    3) Also, is there video footage of the test at the times you were involved such as emptying the reactor, etc. Of course I fully believe in the e-cat, I’m just thinking of the pathological skeptics. I’m interested to see how far they will go before they convert :) I’m guessing very far!
    THERE IS NO ANY VIDEO FOOTAGE REGARDING THE LUGANO TEST; IF SOME IS AROUND, IT IS A FALSE PRODUCTION. THE CHARGE HAS BEEN PUT AND EXTRACTED BY THE COMMETTEE

    4) Could you request the ITP authors release some more photos to quench our thirst for new stuff….or you can release a picture of the 1MW device. hehe
    NO. PHOTOS OF THE 1 MW PLANT WILL SURELY BE AVAILABLE IN DUE TIME

    5) Are you aware of IH giving any media announcements related to the Ecat in the near term or will they not talk until the 1MW planty has been running for a year? Surely you must be in contact with them?
    NO NEWS UNTIL THE R&D AND TEST UPON THE PLANT SUPPLIED TO THE CUSTOMER WILL BE COMPLETED

    6) Any plans to come to England, I’ll be glad to cook you a meal
    I TAKE NOTICE OF THIS

  728. Andrea Rossi

    Mark Saker:
    In the gas- fueled E-Cat, which is under R&D so far, the use of a thermoelectric device to produce electricity for intrinsic use is not opportune, due to its very low efficiency.
    Warm Regards,
    A.R.

  729. Giuliano Bettini

    Dear Andrea,
    about the Gas-Cat.
    The current cat (TPR2) seems to be stable (constant input heat to maintain a constant reaction).
    Why the self-produced heat is not enough?
    It seems that, for the reaction, not only heat it’s needed.
    (But… with a very very small amount of electric energy, I must suppose….).
    Have I missed something?
    PS: I know that you do not comment what is happening in the reactor, as a matter of fact I know you do not comment anything. :)
    Self Sustaining Regards,
    Giuliano Bettini.

  730. Andrea Rossi

    Giuliano Bettini:
    You wrote the answer yourself.
    Warm Regards,
    A.R.

  731. Frank Acland

    Dear Andrea,

    You asked yesterday for assistance tracking down references to LENR in the Manhatten Project. There is now a thread about this topic on E-Cat World. Some readers have come up with some references you might find useful. You can heck out the comments here:

    http://www.e-catworld.com/2014/11/06/rossi-asks-for-help-cold-fusion-in-manhattan-project/

    Best wishes,

    Frank Acland

  732. Andrea Rossi

    Frank Acland:
    Thank you for this useful link.
    Warm Regards,
    A.R.

  733. Curiosone

    General Emilio Spaziante has pleaded guilty for corruption and has been sentenced to serve 4 years in prison: he is the officer of the Guardia di Finanza that closed Petroldragon and all the other factories of yours twenty years ago: any comment?

  734. Andrea Rossi

    Curiosone:
    No comment.
    Warm Regards,
    A.R.

  735. Wladimir Guglinski

    Daniel De Caluwé wrote in November 6th, 2014 at 3:41 AM

    Dear Wladimir,

    And as I’m not a nuclear physisist, I did not now that the (or at least some) nucleï rotate so fast. Thank you for the explanation and the interesting links!
    —————————————

    Daniel,
    and the situation becomes worst when some nuclei are excited. Their rotation is so fast that it causes the deformation of the nucleus.

    See the figure at left in the title Recent physical results of the article High-spin physics ISP.
    In the figure we see that a spheric nucleus is deformed when it is excited getting a high-spin, and there is a changing in the spheric shape: the nucleus takes the shape of an elipsoid, under the action of the centrifugal force:

    Recent physical results
    First evidence of magnetic rotation in nuclei around mass A = 80

    “The conventional concept of nuclear rotation is based on the existence of a deformed mass distribution of the nucleus (see left figure).”
    https://www.hzdr.de/FWK/MITARB/rs/highspin.html

    Therefore,
    under very fast nuclear rotation, the magnitude of the centrifugal force on protons and neutrons is very larger than that of the strong nuclear force (as calculated by me here in the JoNP).
    Note that the nucleus is deformed due to the fast rotation, and therefore the excited high-spin nuclei would have to be desintegrated under the action of the centrifugal force, if protons and neutrons were bound via the strong nuclear force.

    Nevertheless, those excited high-spin nuclei survive, and it means that the strong nuclear force is not the responsible for the nucleus aggregation.

    regards
    wlad

  736. JCRenoir

    1-Are you sure you will be able to reconcile the results of the ITP, regarding the isotopes shift, remaining in the Standard Model ?
    2- How is going your new invention of the gas fueled E-Cat?
    JCR

  737. Andrea Rossi

    JC Renoir:
    1- yes
    2- working on it: very promising
    Warm Regards,
    A.R.

  738. Tom Conover

    Andrea Rossi,

    I started here —
    https://www.google.dk/?gws_rd=cr&ei=_59bVJi6B-nB7Aaw5oD4Cw#q=Edward+Teller+fusion

    .. then checked here:
    http://www.britannica.com/EBchecked/topic/586350/Edward-Teller
    Although the Los Alamos assignment was to build a fission bomb, Teller digressed more and more from the main line of research to continue his own inquiries into a potentially much more powerful thermonuclear hydrogen fusion bomb.

    .. and then found more here:
    http://www.worldscientific.com/worldscibooks/10.1142/p337

    Edward Teller Lectures
    Lasers and Inertial Fusion EnergyWith a Foreword by E M Campbell
    Edited by: Heinrich Hora (University of New South Wales, Australia), George H Miley (University of Illinois, Urbana, USA)

    About This Book E-Book Reviews Supplementary
    How to achieve unlimited, safe, clean and low-cost energy by laser- or beam-driven inertial nuclear fusion has preoccupied all winners of the Edward Teller Medal since its inception in 1991. This book presents their findings, meeting discussions, and personal insights from Edward Teller himself. Expect discussion of important advances anticipated in the future such as multi-billion dollar fusion research projects (NIF), and new schemes such as the petawatt-picosecond laser-plasma interactions evoking new physics and coupling mechanisms.

    .. hoping this enough to start your review of “fishing for fusion”

    Warm Regards
    Tom

  739. Andrea Rossi

    Tom Conover:
    Thank you, interesting .
    Warm Regards,
    A.R.

  740. “Today I have been informed from an Indian nuclear physicist that, during the Manhattan Project, Oppenheimer and Teller expressed the opinion that cold fusion was a possibility.”

    Dear Andrea Rossi,

    Perhaps the colleague refers to the Oppenheimer Phillips Process:

    http://goo.gl/h01H3z
    http://goo.gl/9n5ZNm
    http://en.wikipedia.org/wiki/Oppenheimer%E2%80%93Phillips_process

    An exothermic stripping reaction (like for example the Oppenheimer-Phillips process), would explain the transmutations taking place and solve the problem with the coulomb barrier.

    Same Days ago I have had the same idea in the LENR Forum:
    http://www.lenr-forum.com/forum/index.php/Thread/442-Is-Calcium-Rossi%E2%80%99s-Secret-Catalyst/?postID=1811#post1811

    Best regards
    Felix Rends

  741. Andrea Rossi

    Felix Rends:
    Thank you very much, this is very interesting. This could give further evidence that LENR have right of citizenship in the Standard Model system.
    Warm Regards,
    A.R.

  742. Andrea Rossi

    To the Readers, request for help:
    Today I have been informed from an Indian nuclear physicist that, during the Manhattan Project, Oppenheimer and Teller expressed the opinion that cold fusion was a possibility. If true, this is important under a historical point of view, but I have not been able to find a reference of this. Is any of our Readers able to inform us about similar reference?
    Warm Regards,
    A.R.

  743. Koen Vandewalle

    Greg Leonard, Eric Ashworth,
    The simulated device creates multiple vortex structures because of the shapes of the blades of rotor and stator, in combination with exact flow, viscosity and density of the fluid.
    The clue may be that the torque on the rotor that is mechanically connected with the outer casing, is possible because the blades on the casing “recoil” and “bump” on the vortices that are internally created, and act, as Eric says, as solid objects.
    Awsome !!!

    You should really cooperate with Wladimir and his knowledge on aether structure. With that simulation software, and a “fluid” that has proper electric and magnetic properties, and some “black-hole-stuff” you might see the appearance of nuclearish familiar stuff.

    But…. it looks expensive.

    Kind Regards,
    Koen

  744. Magico Lipton

    Dear Andrea,
    now we know that, during the reaction, new Isotopes are forming.
    Is the corresponding COP:
    1. constant? or
    2. it changes (linearly, or in some more complex way)?
    Many thanks,

  745. Andrea Rossi

    Magico Lipton:
    1- the COP raises with the temperature, as explained in the Report of the ITP
    2- see above
    Warm Regards,
    A.R.

  746. Greg Leonard

    Dear Eric Ashworth
    Thanks for the interesting post and the web sites you point to.
    A quick look at the Entecho idea leaves me wondering how they are planning to deal with the torque acting on the fan.
    Surely the craft will spin until the air resistance matches the rotor torque!
    What have I missed?
    Greg Leonard

  747. Daniel De Caluwé

    Dear Wladimir,

    You wrote: Daniel, we are not speaking about atoms. We are speaking about nuclei. The rotation of the nucleus is independent on the rotation of the atom (electrosphere).

    My answer: Yes, I agree, we spoke about the rotation of the nucleus of the atom and not of the atom itself. I realised my mistake after I posted my previous message, but I did not correct it anymore. So my remark about the rotation of the nucleus when the atom is chemically bound (via electrosphere), was a stupid one, because the nucleus still rotates within the atom even when it is chemically bond (via the electrosphere) isn’t it? And as I’m not a nuclear physisist, I did not now that the (or at least some) nucleï rotate so fast. Thank you for the explanation and the interesting links!

    Kind Regards,
    Daniel.

  748. Wladimir Guglinski

    Daniel De Caluwé
    November 5th, 2014 at 6:28 PM

    Dear Wladimir,

    It’s not necessary to explain to me, because I understood you well, from the beginning, when you first calculated the tearing apart of the atom, due to the centrifugal (and not ‘centripetal’ ;-) ) force, caused by the fast motion (around its axis) of the atom, but the only question that remains is this: does the individual atom really rotates so fast??? (Certainly not when it is chemically bond ;-)
    ——————————————–

    Daniel
    we are not speaking about atoms.
    We are speaking about nuclei.

    The rotation of the nucleus is independent on the rotation of the atom (electrosphere).

    On the Rotation of the Atomic Nucleus
    http://journals.aps.org/pr/abstract/10.1103/PhysRev.53.778

    Wikipedia:
    Therefore there are several possible answers for the nuclear magnetic moment, one for each possible combined l and s state, and the real state of the nucleus is a superposition of them. Thus the real (measured) nuclear magnetic moment is somewhere in between the possible answers.
    http://en.wikipedia.org/wiki/Nuclear_magnetic_moment

    A simple model for nuclear rotation at high angular momenta
    Abstract
    A simple solvable model of particles coupled to a rotor is introduced. The solutions illustrate some properties of the nucleus rotating with high angular momentum.
    http://www.sciencedirect.com/science/article/pii/0370269371900578

    Single-Particle and Collective Aspects of Nuclear Rotation
    The spectra of rapidly rotating nuclei reveal two distinct components in the build up of the total angular momentum, corresponding to collective rotation and alignment of orbital angular momentum of individual particles. Various aspects of the interplay of these two mechanisms are discussed. The pattern of collective excitations built upon an yrast state of aligned particle motion is analyzed on the basis of a simple model. For the strongly deformed nuclei, the relative contribution of alignment and collective rotation is characterized by two different moments of inertia referring to the yrast envelope and the collective bands. The behaviour of these moments in the transition region from superfluid to normal phase is considered. Finally, some of the consequences of the build up of angular momentum by alignment and collective rotation are considered for the region of the highest spins, where pair correlations are expected to play a minor role.
    http://iopscience.iop.org/1402-4896/24/1B/001

    Chirality of nuclear rotation
    FIG. 1. The discrete symmetries of the mean field of a rotating triaxial reflection symmetric nucleus (three mirror planes). The axis of rotation (z) is marked by the circular arrow. It coincides with the angular momentum~J.
    http://arxiv.org/pdf/nucl-th/0001038.pdf

    regards
    wlad

  749. Daniel De Caluwé

    Dear Wladimir,

    It’s not necessary to explain to me, because I understood you well, from the beginning, when you first calculated the tearing apart of the atom, due to the centrifugal (and not ‘centripetal’ ;-) ) force, caused by the fast motion (around its axis) of the atom, but the only question that remains is this: does the individual atom really rotates so fast??? (Certainly not when it is chemically bond ;-)

    Kind Regards,
    Daniel.

  750. Wladimir Guglinski

    On the ficticious centrifugal force

    Dear Daniel De Caluwé

    As the centrifugal force is ficticious, how can it cause the rupture of a string ?

    Let me explain it.

    Suppose you wishes to cause the rupture of a string A with your two hands. So, we have to apply on the string A two contrary forces with your hands, in order to cause its rupture.

    Now let us to do an analogy with the case of a stone moving in circular orbit tied to the end of a string B, while you hold the other end with your hand.

    With analogy to the rupture of the string A with your two hands (where two contrary forces are applied), it seems that two contrary forces must be applied on the ends of the string B. One force is applied by your hand, and the other force is applied by the stone (the centrifugal force, acting in contrary direction of the force applied by your hand).

    Before the rupture of the string B, the force applied by the stone and the force applied by your hand must be equal (since the string was not disrupted).

    But let us analyse it by applying Newton’s law. As the string B is submitted to two contrary and equal forces, the resultant on the string B is zero, and therefore it must be at rest (or to move in rectilinear motion).
    So,
    the centrifugal force does not exist, it is ficticious.

    There is only one force: it is the force applied by your hand. And what is done by this force?
    Well, in each fraction of time such force applied by your hand changes the direction of the stone motion. In other words, you need to apply a force (transmited by the string B to the stone) in order to change every time the direction of the motion of the stone.

    In general, when we have to analyse a phenomenon in which a body has circular motion, the use of the centrifugal force simplifies the analysis and the explanation of the phenomenon. In other words, in spite of we know that the centrifugal force is ficticious, however we use to consider its action, so that to simplify the analysis and the explanation of the phenomenon.

    But sometimes, along a discussion, often we find people like Mr. JR, and they adopt the strategy of refuting our arguments, by claiming that the centrifugal force is ficticious. In this case, there are two situations:

    1- The person uses this sort of argument because he does not understand the discussion

    2- He uses this sort of argument with bad intent, in order to cause confusion to peoples who are reading the debate. By this way, by claiming that the centrifugal force is ficticious and does not exist, he tries to convince the listeners that he is right, and his opposer is wrong.

    So, each reader here has to conclude himself what is the case of our friend Mr. JR.
    I simply wash my hands.

    regards
    wlad

  751. George

    Dr. Rossi gets compliments Elforsk seriously. Read the article on page 13 -14
    Isotopic change indicate ”Cold” nuclear reaktion -Elforsk follows the development
    New test on Andrea Rossis Ecat show clear signs of isotope change in the fuel.
    The results indicate that the cause could be nuclear reaction in cold
    temperatures.

    http://issuu.com/elforsk_/docs/elforsk_perspektiv_2_2014?e=7916266/10018941

  752. Andrea Rossi

    George:
    Thank you for the link to Elforsk, very important.
    Warm Regards,
    A.R.

  753. WaltC

    Dear Andrea,
    When you spoke to Neri B. of “Dangerous situations”, are we talking:
    1) Dangerous- as in “BOOM!”,
    2) Dangerous- as in it breaks, stops working and needs to be replaced (but no “BOOM!”),
    3) or something else?

    Thanks, WaltC

  754. Andrea Rossi

    Waltc:
    …and the right answer is…2!
    Warm Regards,
    A.R.

  755. Steven N. Karels

    Dear Andrea Rossi,

    Have you considered the possibility that the 62Ni lone isotope in The Report was not nickel at all but a molecule with a mass of 62?

  756. Andrea Rossi

    Steven N. Karels:
    I stay on the SIMS analysis made by the Swedish Institute you can see in the Report of the ITP.
    Warm Regards,
    A.R.

  757. Eric Ashworth

    Koen, in reply to your November 4th reply. I too am a graphical person. I can provide you with two web sites that deal with fluid dynamics. One of these, a close associate, displays a centrifugal fan that operates in a unique concept. It is a www. site belonging to entecho.com.au/contact.php As you know when semi solid objects spin they appear solid. The other concept is on a http//: site belonging to triteckindustries.ca This concept is similar to the entecho one but does not feature the centrifugal fan even though it could in its design which involves propellers and a baffle arrangement. Both of these technologies require movement to induce and control a flow which is able to be studied and measured. See what you think. Regards Eric Ashworth.

  758. Wladimir Guglinski

    Daniel De Caluwé wrote in November 5th, 2014 at 3:53 AM

    @JR,

    Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).
    —————————————–

    Daniel,
    let me explain it by an easy words.

    Suppose you take a string and you tie a stone with mass “m” in its end.
    And you put the stone moving with circular trajectory with speed V and radius R around your hand.

    The stone tries to escape, applying a force on the string. So, you have to apply a force on the string, otherwise the stone go away with the string.

    The force of the stone is given by Fc = m.V²/R.
    Let us call it centrifugal force (note that, in spite of it is ficticious, however it is able to cause the rupture of a string, because it is actually due to the inertia of the stone, and the inertia is no ficticious).

    The string is able to support a force Fs.

    If the speed of the stone increases so much, the centrifugal force Fc will be stronger than the force Fs of the string, and the string will have a rupture, and the stone will go away.

    Now let us apply it to the nucleus, as follows:

    1- The stone plays the role of a proton

    2- The string plays the role of the strong nuclear force

    The strong nuclear force must be stronger than the centrifugal force Fc , otherwise the proton will move away, leaving the nucleus.

    I showed by calculation that with a speed 10% of the light speed, the centrifugal force Fc on the proton is 500N, while the Coulomb force is 50N.

    In the distance of 2fm (the radius of the nucleus 2He4), the Coulomb repulsion has the same magnitude of the strong nuclear force ( 50N ).

    Therefore we conclude that the centrifugal force on the proton is 10 times stronger than the strong nuclear force.

    regards
    wlad

  759. Hi Dan C.:

    “When Rossi says all energies will be integrated, he is just being realistic. it will take many decades to transition. ”

    I have to very much agree with this. If you take the energy content of the entire current annual consumption of crude oil, natural gas and coal, and add present nuclear power generation which will be phased out eventually as the plants exceed their useful lives; then calculate how many one megawatt units are required to displace it, you will see it will take a long time.

    Even assuming initial production of 10,000 one megawatt units a year and doubling production each year for a few years it will still take a long time, simply because the current usage of coal, natural gas and oil is so huge. Having said this, though, it didn’t take very long to install optical fiber and related equipment across the entire world when the internet appeared.

    The critical factor permitting the rapid expansion of *any* business is that the product be sold at a price which allows profits to be high enough so that they can be reinvested in the construction of larger and larger manufacturing plants quickly. If a company is making only small profits it will not have the resources to expand quickly, if at all. LENR will be no different.

    Charging a high enough price will be critical in determining how long it takes for LENR technology to achieve maturity. If E-cat units are sold at prices sufficient to finance rapid expansion, then it will certainly not take 100 years.

    Of course the profits must also be high enough to finance both research into new products and the improvement of existing ones, to stay ahead of potential competitors. Also, when competition eventually appears, in order to continue to thrive it will be critical for the company to maintain a very strong balance sheet so it can take on competition likely to come from some very large and very well financed companies in the energy and transportation industries.

    Rodney.

  760. Wladimir Guglinski

    JR wrote in november 4th, 2014 at 8:38 PM

    The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming
    ————————————————————–

    The radii quoted by Bethe due to the rotation of the nucleus as a whole, as if it were a solid body? Be it as it may, numerous measurements had been made by 1936.

    The magnetic moments of nuclei
    Bethe explains that the magnetic moments of numerous are known, thanks to a remark made by Pauli, who showed that if the nucleus behaves as a small magnet, it can somewhat perturb the observed atomic spectral lines caused by electrons making transitions from one quantum state to another. This was called the “hiperfine structure” of the spectral lines, and the magnetic moment of the nucleus could be deduced from it. The magnetic moments of about thirty nuclei were known in 1936.

    page 347
    http://books.google.com.br/books?id=IJa4afSc-MsC&pg=PA347&lpg=PA347&dq=nucleus+rotation+hans+bethe&source=bl&ots=_QY1zPUjBj&sig=XYZNgCuUdsHRJw_biDW6M_mvqEM&hl=en&sa=X&ei=0IVaVPaBGYGUNtaqgPgI&ved=0CEgQ6AEwCA#v=onepage&q=nucleus%20rotation%20hans%20bethe&f=false

    So,
    the nuclei have rotation, and since it influences even the atomic spectral lines, the rotation must be very fast, because the magnetic moment of the electron is 9284 (x10^-27 J/T) , while the magnetic moment of the proton is only 14 (x10^-27 J/T).
    Without a large angular momentum the protons and neutrons could not influence the spectral lines (the distance between the electrons in the electrosphere and the protons in the nucleus is 10^-11m , while the size of the proton is 10^-15m, and so the size of the proton is despicable regarding its distance to the electrons). And the magnetic moment decreases with the square of the distance.

    Therefore there are two alternatives:

    1- Mr. JR does not know the nuclear properties of nuclei

    2- Mr. JR knows them, however he tries to deceive people by lying

    regards
    wlad

  761. Frank Acland

    Dear Andrea,

    Regarding the heat measurements in the Lugano report (leaving alone the isoptopic shifts reported): were these measurements in line with what you would have expected based on your own R&D experience with the E-Cat?

    Many thanks,

    Frank Acland

  762. Andrea Rossi

    Frank Acland:
    Yes, taking in account the strong conservative mode maintained by the Professors.
    Warm Regards,
    A.R.

  763. Neri B.

    Dear Andrea,
    in TPR 1 we saw 3 tests: in the first the reactor melted, in the others two tests the COP was 5.6 and 2.9.
    Recently you stated that someone has experienced the cat could become a tiger.
    Can you please tell us which is the highest COP you ever achieved in your internal test for a reasonable period of time and at what temperature?
    Thank you

  764. Andrea Rossi

    Neri B.:
    When I said the Cat can become a Tiger it was referred ironically to a totally different issue…anyway we reached very high COPS, but in very Dangerous situations, so it is not proper to talk of them. That was extreme R&D
    Warm Regards,
    A.R.

  765. Wladimir Guglinski

    Daniel De Caluwé wrote in November 5th, 2014 at 3:53 AM

    @JR,

    Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).
    —————————-

    Dear Daniel
    I avoided to call it centrifugal force because the centrifugal force does not exist, I was sure Mr. JR would use it so that to refuse my argument.

    .

    =============================================
    JR wrote in November 5th, 2014 at 7:31 AM

    Dear Daniel,

    If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.
    ——————————————

    Therefore, according to Mr. Jr,
    when a Formula 1 driver makes a turn too fast, instead of being vented out of the curve, the car is pulled towards the center of curvature … ha ha ha

    My God …
    … then the designers of Formula 1 tracks are crazy, because they put barricades and tires on the outside of the track, to protect against car crashes.

    According to Mr. JR, the designers had to put these barricades on the inside of the runway, so cars do not be thrown into the center of the trajectory …
    ha ha ha

    Dear Mr. JR
    your lack of knowledge of elementary physics is awesome.

    In spite of the centrifugal force is a fictitious force, however due to the rotation of the nucleus the protons and neutrons are submitted to the tendency to be expelled from the nucleus, due to the INERTIA of their motion.

    The protons and neutrons try to continue in a straight TANGENTIAL trajectory, and the strong nuclear force on the protons and neutrons have to avoid they be expelled from the nucleus by such INERTIA

    This tendency due to the INERTIA is vulgarly known as centrifugal force. Within the nuclei the INERTIA is contrary to the strong nuclear force.

    The value of such INERTIA tendency is Fc = m.V²/R, and (as I have shown here) it is at least 10 times of magnitude stronger than the strong nuclear force.

    Therefore, the strong nuclear force cannot avoid the protons and neutrons to be expelled from the nucleus, because the action of the INERTIA on them is 10 times stronger.

    regards
    wlad

  766. Wladimir Guglinski

    JR wrote in November 4th, 2014 at 8:38 PM

    1) ——————————-
    The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming
    ———————————-

    You are wrong.
    the 4H3 nucleus has spin zero because one proton has spin up and the other proton has spin down, while one neutron has spin up and the other neutron has spin down. The total spin is zero.

    However all the nuclei have rotation, and so each nucleon (proton or neutron) is submitted to the centripetal force.

    2) ———————————-
    a centripetal force CAUSES binding, the centripetal force IS the nuclear force, having the centripetal force get larger than the coulomb doesn’t matter much since the coulomb effect is small compared to the binding from the strong force, etc…
    ————————————-

    No, according to the Standard Model, which causes binding is the strong nuclear force (and it is not centripetal, i.e., it does not point out to the center of the nucleus, since there is also attraction between two neighbors nucleons).

    3) ———————————-
    The closest you come to a true statement is when you *assume* that 11Be can’t be bound by the strong force and then make the bold and daring conclusion that, if there is no binding, then it would not be bound. Bravo.

    I don’t think you could have been more wrong. Actually, you could have been more wrong (and certainly will) simply by saying more. I eagerly await to see what simple ideas you screw up next…
    ————————————-

    Bravo, Mr. JR, you are using the Heisenberg phantasmagoric method, so that to explain how the cluster of the 11Be can keep a halo neutron without any sort of attraction force between them.

    Heisenber awarded the Nobel Prize with his phantasmagoric method.
    There is a good chance you may get the Nobel Prize too.

    regards
    wlad

  767. Wladimir Guglinski

    Dan C. wrote in November 5th, 2014 at 12:31 PM

    Dear Wladimir,

    You say:
    “However Andrea Rossi does not want to speak about, because the best is do not put angry the owners of the other energy sources existing in the planet nowadays.”

    I respectfully disagree.
    And Mr. Rossi can correct me if I’m wrong.

    When Rossi says all energies will be integrated, he is just being realistic. it will take many decades to transition.
    —————————————-

    just what I did mean to say. I said in the future, and the future does not mean tomorow, or next year. The future means after some decades.

    regards
    wlad

  768. Dan C.

    Dear Wladimir,

    You say:
    “However Andrea Rossi does not want to speak about, because the best is do not put angry the owners of the other energy sources existing in the planet nowadays.”

    I respectfully disagree.
    And Mr. Rossi can correct me if I’m wrong.

    When Rossi says all energies will be integrated, he is just being realistic. it will take many decades to transition. We will need all those other forms of energy during this period.

    On many blogs, you see people proclaiming LENR will obliterate fossil fuels within a few short years. It will take more then a few short years just to get started. While many think this will happen fast, I hope it happens fast enough to preserve the fossil resources for all our other needs.

    People greatly underestimate the magnitude of the task of transitioning to LENR energy. If it could be mostly accomplished in 50 years, it would be a great feat for society. In this respect, Rossi is right to say all energies will be integrated. At least for the foreseeable future.

    With respect,
    Dan C.

  769. Boss

    Dear Andrea Rossi,
    I have a really simple question for you, but I want to be sure you won’t spam it as a tedious comment coming from that class you refer to as ‘snakes’ (which in fact I’m not).
    We are talking about science, not religion, and therefore, big claims have to be supported by huge proofs. Faith has nothing to do with the subject.
    In the TPRI some doubts emerged, (such as DC current not excluded) , and eventhough I was totally sure there wasn’t any DC hidden some where, the faith in your honesty couldn’t be enough for science.
    In fact in the TPRII DC presence was checked and excluded.
    At this point some other doubts have been raised, and just like as for TPRI, I am preatty sure no clamp was inverted. But belief goes along with faith, hence it’s not scientifical.
    One picture of the setup would be enough and you would also give evidence that out there it’s full of ‘snakes’.
    The TPRII was intended to prove that the Cat works without any reasonable doubt. Then prove it.
    Thanks for your attention, always cheering for you.
    Regards

  770. Andrea Rossi

    Boss:
    This situation of the “changed position of the clamps” is very funny, while it is also an evidence of the correctness of the work of the Professors. Lacking real reasons to make a serious critic, these persons make “assumptions”: they “assume” that the clamps of the two PCE830 have been changed of position, and upon this “assumption” are writing all their lectures. I make you a simple example of what is going on: you are driving your car correctly, respecting all the laws related to driving, but suddenly a policeman stops you and says: ” I assume you were going overspeed, so you have to pay a fine”. No evidence at all that you have violated the speed limit, but, based on his assumption, he wants to fine you.
    This situation is exactly the same.
    THE SET UP OF THE EXPERIMENT, INCLUDED THE SET UP OF THE TWO PCE830 HAS BEEN DONE BY THE PROFESSORS, NOT BY ME. THE PROFESSORS CONTROLLED EVERY DAY THE CORRECTNESS OF ALL THE CONNECTIONS. ONE OF THEM (PROF ROLAND PETTERSON) WAS SPECIFICALLY DEDICATED TO THIS TASK. THE CLAMPS HAVE NEVER BEEN DISCONNECTED, EXCHANGED, DISPLACED OR ANYTHING LIKE THAT.
    Obviously the persons that have an agenda finalized to try to say negative things, not having serious things or citics to make, now fish in the lake of “assumptions” and “hypotesis”. From this lake you can fish out all the monsters you want, being just “assumptions”.
    Photos: I was not allowed to make photos and therefore I do not have any photo. The Professors know perfectly how the clamps have been put and know perfectly that no displacement or changement has been done.
    The level of this critic is so low, that it is not worth the time of an answer, so, as you rightly wrote, the temptation to spam it has crossed my brain, but you are always so kind that I decided to answer.
    Warm Regards,
    A.R.

  771. Nicola Cortesi

    Dear Andrea,

    today on the webpage of the popular italian journal “Il Fatto Quotidiano”, an article appeared on the recent nomination of Fabiola Gianotti to director of CERN. At the end of the article, the journalist Andrea Von Flue, himself a physician, guesses if the new director’ll start a research project also on LENR. You can read the article in italian on:

    http://www.ilfattoquotidiano.it/2014/11/04/grazie-fabiola-gianotti-politica-perso-unaltra-occasione-per-tacere/1191758/

    Something has changed!

    Hot Regards,

    Nicola

  772. Andrea Rossi

    Nicola Cortesi:
    thank you for the important information. This is an accomplishment due, obviously, to the Report of the ITP.
    Warm Regards,
    A.R.

  773. JR

    Dear Daniel,

    If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.

  774. Daniel De Caluwé

    @JR,

    Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).

    Kind Regards,

  775. Peter Forsberg

    Dear orsobubu,

    I can understand that many people can think that communism is a good way to rule man. Truly there are some distinctly bad concepts in the incarnation of lesse faire economy known as capitalism. But the empirical evidence is against communism. It works extremely badly. But this is not a political blog, so I will spare everyone the details.

    Regards

    Peter

  776. Joe

    Koen,

    Tapping into a free energy flow would be the best. The E-Cat would then behave like a wheel at the bottom of a waterfall. Back in 2011, some people were asking Dr Rossi if the E-cat performed differently at various times of the day. They were implying that the Sun was perhaps responsible for providing the excess energy to the E-Cat. This would have been in the form of a neutrino flux. The question remains though, what is the source of the overunity found in the E-Cat?

    All the best,
    Joe

  777. JR

    Wlad said:

    “Dears Mr. Joe , Mr. JR , and dr. Stoyan Sarg

    As is known, the centripetal force on the protons and neutrons is not considered in the Standard Nuclear Physics.”
    ———————–
    Wrong. That was an easy one!

    Of course, even if it were true, almost everything else you said following this was wrong. The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming, a centripetal force CAUSES binding, the centripetal force IS the nuclear force, having the centripetal force get larger than the coulomb doesn’t matter much since the coulomb effect is small compared to the binding from the strong force, etc…

    The closest you come to a true statement is when you *assume* that 11Be can’t be bound by the strong force and then make the bold and daring conclusion that, if there is no binding, then it would not be bound. Bravo.

    I don’t think you could have been more wrong. Actually, you could have been more wrong (and certainly will) simply by saying more. I eagerly await to see what simple ideas you screw up next…

  778. Joe

    Daniel,

    Thank you for your kind words, and for reading the article by Prof Meyl.

    I think tunneling is a strong candidate for the central process occurring in the E-Cat.
    Two reasons are the following:

    1. The hot spots on the E-Cat resemble vortex losses in a capacitor as explained by Prof Meyl. (Compare Fig 12 in the article by Prof Meyl with images of the large vertical E-Cat when not running.) And electric capacitance can be used instead of magnetic induction to create an oscillating charge density and near-field waves, to answer your question.

    2. The lack of radiation from the E-Cat can be explained by re-absorption which is typical behaviour in near-field radiation. Some radiation does emerge though as EM waves.

    All the best,
    Joe

  779. Wladimir Guglinski

    Why is centripetal force neglected in Standard Nuclear Physics?

    Dears Mr. Joe , Mr. JR , and dr. Stoyan Sarg

    As is known, the centripetal force on the protons and neutrons is not considered in the Standard Nuclear Physics.

    However, a simple calculation shows that centripetal force within the nuclei can have a higher magnitude than Coulomb repulsion . Let us see the calculation.

    Units used:
    Charge of the proton: 1,6×10^-19 C
    Mass of the proton and neutron: 1,7×10^-27 kg
    K= 9×10^9 Nxm²/C²

    I will consider the velocity of the protons 3% of the speed of light c=3×10^8m/s, and so their speed is v= (3×10^-2)x(3×10^8) = 9×10^6m/s.

    Actually the speed of protons due to the rotation of the nuclei cannot be lower than 10% of the light speed, but we will be conservative, and so let consider only 3%.

    Let us consider the nucleus 2He4, by considering the two protons with a distance of 2fm between them (2fm = 2×10^-15m).

    1- Coulomb repulsion between the two protons

    Fe = K.q²/R² = 9×10^9 x (1,6×10^-19)²/(2×10^-15)² = 50N

    2- Centripetal force on each proton

    Fc = 1,7×10^-27 x (9×10^6)²/2×10^-15 = 70N

    .

    So, by considering the speed of protons to be 3% of the light speed (which is an underestimated value), the centripetal force on each proton within the 2He4 has the same magnitude of the Coulomb repulsion force between the two protons.

    If we consider the velocity of protons in the order of 10% of the light speed, we get Fc = 510N (one order of magnitude stronger than the strong nuclear force in a distance of 2fm).

    .

    Obviously the influence of the centripetal force is stronger in other nuclei, as for instance 11Be, where there is a halo neutron moving with radius R=7fm about the cluster. As the strong nuclear force does not actuate in a distance of 7fm, the halo neutron in the 11Be would have to be quickly expelled from the nucleus 11Be, due to the centripetal force on it, since the centripetal force increases with the radius: Fc = m.w².R , where “w” is the angular velocity.
    In the 11Be the centripetal force on the halo neutron is 145N, while the strong nuclear force is practically zero.

    And the situation becomes worst, because the neutron decays in a proton, and the 4Be11 transmutes to 5B10 with a halo proton with orbit radius R=7fm.
    So, beyond the 145N due to centripetal force there is the actuation of a Coulomb force a little weaker than 50N, while the strong nuclear force is practically zero.

    The halo proton would have be expelled quickly from the newborn 5B10, and so 5B10 would have to decay. But this not happens, because the proton actually goes back to the cluster, and the 5B10 becomes stable.

    .

    So, the question is:
    Why the nuclear physicists neglect the centripetal force on the protons and neutrons????

    I hope to hear a good explanation from Mr. JR , or any nuclear theorist he wishes to invite come here to explain it to us.

    Regards
    wlad

  780. Andrea Rossi

    Wladimir Guglinski:
    I confirm what I wrote answering to Peter Forsberg. When I want to say a thing I say it.
    Warm Regards
    A.R.

  781. Wladimir Guglinski

    Peter Forsberg
    November 4th, 2014 at 2:22 AM

    Dear Wladimir,

    You say that the secondary E-cat could produce 1kWh of electricity. I have yet to hear Rossi say that he can produce electricity with good enough efficiency. Do you have other information?
    —————————————–

    Dear Peter
    of course in the future there will be only three forms of energy sources used by mankind: the cold fusion, the magnetic generators, and the hidroelectric plants, because they are 100% clean ecologically speaking (but new hidroelectric plants will not be built).

    However Andrea Rossi does not want to speak about, because the best is do not put angry the owners of the other energy sources existing in the planet nowadays.

    regards
    wlad

  782. Wladimir Guglinski

    Joe wrote in November 4th, 2014 at 5:08 PM

    Wladimir,

    1. You say,
    “I dont know, since I dont know the charge of the electriciton.”

    Should not the charges on your aether particles be UNIT by definition? Otherwise, you would need even more fundamental particles to explain your aether particles.
    ———————————–

    Yes,
    but the unit of charge in Physics is the electron charge e =1,6×10^-19C, and I cannot change what was established in Physics.
    The charge of an electriciton will be something like 10^-30.e

    Besides, the gravity flux of particles as the proton and electron agglutinate electricitons in the form of rings, and the gravity flux passes by withing the rings. These rings of electricitons move with the speed of light, inducing the fields of electricitons.

    2. You say,
    “As I dont know how many electricitons form the bodies of the quarks up and down.”

    Do the magnetons, or any other of the QRT aether particles, also form bodies?
    ——————————————–

    No,
    magnetons are agglutinated in the form of magnetic fields induced by the flux of electricitons

    3. You say,
    “Perhaps you also would like to claim that Heisenberg’s abstract mathematical concept of Isospin is able to create a force of repulsion between two neutrons.”

    a) My concept of a neutron-next-to-a-proton-asymetrically-attenuating-the-electric-field-thus-causing-a-nonzero-Q(b), simply known as the NNTAPAATEFTCANQ(b) conjecture, for the deuteron is not an abstract mathematical concept but a concrete physical reality… maybe.
    —————————————–

    Joe,
    we are talking about the puzzles of the Standard MOdel, and not the solutions proposed in your theory.

    4. You say.
    “If this is the case, there would not be necessary 33 theories proposed along 66 years. One unique theory would solve the puzzle.”

    But maybe my NNTAPAATEFTCANQ(b) theory can be that theory. I shall copyright it. If you have free time, you can be my lawyer as I can not afford a real one.
    —————————————

    So, you are my competitor, since I solve the puzzle with the model of neutron n=p+e (and the quarupole moment of the deuteron is calculated in my paper Anomalous Mass of the Neutron)

    So, I would not be a good lawyer for you… rsss

    regards
    wlad

  783. Joe

    Wladimir,

    1. You say,
    “I dont know, since I dont know the charge of the electriciton.”

    Should not the charges on your aether particles be UNIT by definition? Otherwise, you would need even more fundamental particles to explain your aether particles.

    2. You say,
    “As I dont know how many electricitons form the bodies of the quarks up and down.”

    Do the magnetons, or any other of the QRT aether particles, also form bodies?

    3. You say,
    “Perhaps you also would like to claim that Heisenberg’s abstract mathematical concept of Isospin is able to create a force of repulsion between two neutrons.”

    a) My concept of a neutron-next-to-a-proton-asymetrically-attenuating-the-electric-field-thus-causing-a-nonzero-Q(b), simply known as the NNTAPAATEFTCANQ(b) conjecture, for the deuteron is not an abstract mathematical concept but a concrete physical reality… maybe.

    b) It is force that is the abstract mathematical concept. Force has never been measured, it has always been calculated. It is a holdover from when we humans believed that spirits moved things by applying a force to them. The only concrete things that can be measured are distance (rod) and time (clock).

    4. You say.
    “If this is the case, there would not be necessary 33 theories proposed along 66 years. One unique theory would solve the puzzle.”

    But maybe my NNTAPAATEFTCANQ(b) theory can be that theory. I shall copyright it. If you have free time, you can be my lawyer as I can not afford a real one.

    All the best,
    Joe

  784. Bob

    Dear Andrea Rossi

    1. Do you know what is the maximum temperature an operating e-cat can produce?

    2. Have you achieved that temperature in an e-cat operation?

    3. Are there any other e-cat applications you are presently working on in addition to plant for manufacturing and possible aircraft engine use?

    Thanks

    Bob

    P.S. Thanks to all the nuclear theorists who are posting to this blog. The exchange of knowledge is truly remarkable.

  785. Andrea Rossi

    Bob:

    1- at peak 1,400°C
    2- yes
    3- presently we are focused on the 1 MW plant; apart from this, we are making R&D mainly for gas driven E-Cats; obviously I am also studying on the reconciliations of the results of the measurements made by the ITP.
    Warm Regards,
    A.R.

  786. orsobubu

    Peter Forsberg,

    thank you for the link, this part was very interesting:

    “Probably more inhibiting than anything else is a feeling of responsibility. The great ideas of the ages have come from people who weren’t paid to have great ideas, but were paid to be teachers or patent clerks or petty officials, or were not paid at all. The great ideas came as side issues.

    To feel guilty because one has not earned one’s salary because one has not had a great idea is the surest way, it seems to me, of making it certain that no great idea will come in the