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  1. 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

  2. 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

  3. 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

  4. DTravchenko

    Dr Rossi:
    How is going the R&D of the gas fueled E-Cats? News?
    Warm Regards,
    DTravchenko

  5. Andrea Rossi

    DTravchenko:
    We are making important progress on this issue. Soon will start to test the first prototypes.
    Warm Regards,
    A.R.

  6. 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

  7. Daniel De Caluwé

    @Wladimir Guglinski,
    @Andrea Calaon,

    Wow, I’m impressed! Calaon-Guglinski very convincing to me!

    Kind Regards,
    Daniel.

  8. 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.

  9. Andrea Rossi

    Bernie Koppenhofer:
    The fact that we moved the mountains with our hard work is positive.
    Warm Regards,
    A.R.

  10. 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?

  11. 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.

  12. Gio51

    Dear Dott. Rossi
    Underdeveloped countries need DESPERATELY your devices…!!!!! Pleaase, please, hurry up..!!!
    Gio

  13. Andrea Rossi

    gio51:
    Our Team is working as hard as possible and resolving problems.
    Warm Regards,
    A.R.

  14. JCRenoir

    The main stream media of the world are beginning to take seriously your work. Does this help your work?
    JCR

  15. Andrea Rossi

    Robert Curto:
    This is one of the possible applications.
    Warm Regards,
    A.R.

  16. Joe

    Andrea Calaon,

    Are you saying that an electron can orbit beneath the ground state of an atom?

    All the best,
    Joe

  17. 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

  18. 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

  19. 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.

  20. 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.

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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.

  26. 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

  27. Andrea Rossi

    JYD:
    Thank you for this interesting information.
    Warm Regards,
    A.R.

  28. 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

  29. 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.

  30. 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.

  31. 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

  32. 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

  33. 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

  34. 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.

  35. 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

  36. 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

  37. Andrea Rossi

    Marco Serra:
    Yes.
    Warm Regards
    A.R.

  38. Andrea Rossi

    Italo R.:
    Thank you for the information.
    Warm Regards
    A.R.

  39. 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?

  40. 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

  41. Joseph Fine

    Andrea Rossi, Silvio Caggia,

    The Navajo message translates as “Thanks, Warm Regards.”

    ” Jo, jo, jo. ”

    Happy Thanksgiving (in advance),

    Joseph Fine

  42. Andrea Rossi

    Dr Joseph Fine:
    Thank you: fantastic translation. Happy Thanksgiving (in advance) to you and all our American Readers.
    Warm Regards,
    A.R.

  43. 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.

  44. Andrea Rossi

    Steven N. Karels:
    No, I cannot comment.
    Warm Regards,
    A.R.

  45. Andrea Rossi

    Bernie Koppenhofer:
    No, it is not true. The point is not the fuel.
    Warm Regards,
    A.R.

  46. 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?

  47. 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

  48. 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

  49. 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?

  50. 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

  51. Herb Gillis

    Andrea Calaon:
    How does your proposed reaction mechanism explain the observed reduction in the concentration of Ni64?
    Regards; HRG.

  52. 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

  53. Andrea Rossi

    Andrea Calaon:
    I appreciate your efforts. Obviously I cannot comment.
    Warm Regards,
    A.R.

  54. 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

  55. Andrea Rossi

    Silvio Caggia:
    Ugh!
    A.R.

  56. 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!

  57. 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.

  58. 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

  59. 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.

  60. 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.

  61. 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.

  62. 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?

  63. 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

  64. Andrea Rossi

    Eric Ashworth:
    1- yes
    2- no
    3- among others, a, not b.
    Warm Regards,
    A.R.

  65. 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.

  66. 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

  67. 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

  68. Andrea Rossi

    Steven N. Karels:
    Correct.
    Warm Regards,
    A.R.

  69. 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.

  70. 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.

  71. 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.

  72. 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

  73. 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.

  74. 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,

  75. 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

  76. 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.

  77. Andrea Rossi

    Orsobubu: again thank you, very appreciated also from all the Team.
    Warm Regards,
    A.R.

  78. 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

  79. Andrea Rossi

    BroKeeper:
    Thank you very much,
    Warm Regards,
    A.R.

  80. orsobubu

    … dedicated also to the whole wizardry Team, of course!

  81. Andrea Rossi

    Orsobubu:
    Thank you. Inspiring for persons who perspire: look, how idealism gears up with materialism!
    Warmest Regards,
    A.R.

  82. 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

  83. 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.

  84. Andrea Rossi

    Italo R.:
    Thank you, very useful !!!
    Warm Regards,
    A.R.

  85. Andrea Rossi

    You are right, I already sent to the IT Guy the issue.
    Warm Regards,
    A.R.

  86. 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

  87. Italo R.

    About William McDonough and Thomas Darden – UNC Kenan-Flagler:

    The correct link is this:

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

  88. 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

  89. 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.

  90. 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

  91. Andrea Rossi

    BroKeeper:
    I inform immediately the IT guy about this.
    Warm Regards,
    A.R.

  92. 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.

  93. 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

  94. 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?

  95. Andrea Rossi

    Frank Acland, Pietro F., George:
    Very interesting!
    It appears we moved the mountains!
    Warm Regards,
    A.R.

  96. Andrea Rossi

    Tommaso Di Pietro:
    As I said, it is not chronometry.
    Warm Regards,
    A.R.

  97. 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

  98. Andrea Rossi

    Bob:
    Both.
    Warm Regards,
    A.R.