Rossi: Lugano Test a ‘Gold Mine’ Which Has Improved E-Cat Performance

Today Andrea Rossi reported that the performance of the E-Cat plant currently under production testing has been a ‘terrific surprise’ ‘due to evolution of the charge and to the control system’. I was quite surprised to read that there had been a change in the charge, because I thought that most of the work that his team had been doing at the plant involved the control system.

It seemed from this statement that they must have learned new things about the charge, and so I posted this question on the Journal of Nuclear Physics:

Did the Lugano test report contribute anything to the evolution of the charge that you mention?

Rossi responded:

Andrea Rossi
March 13th, 2015 at 7:51 PM
Frank Acland:
Yes, the charges have been modified studying very carefully the results of the analysis made by the Independent Third Party. For us that report has been a gold mine.
Warm Regards,

I find this very interesting. We have heard plenty of criticism about the Lugano report for various reasons, but Rossi and his team seem to have found it to be invaluable to their work. Of course, they see it in a different context to most of us because of their experience with the E-Cat over the years, and having much more information than we have.

Where Rossi sees a gold mine, we might just see a pile of rocks because of our lack of background — even though we can read the exact same report that he can.

Something in this report has apparently enabled Rossi’s team to seemingly dramatically improve the performance of the E-Cat. We know that following the publication of the report, Rossi said that he had been surprised by the amount of nickel enrichment that the Lugano testers measured. He said at the time, “the results from the test have gone well Beyond what we found before during our internal R&D.”

So it sounds like the Lugano report uncovered information that Rossi and his team had not seen before — perhaps because they had access to better measurement equipment than Rossi had used previously — and has been tremendously valuable for the development of the E-Cat.

So I wonder who else can find the gold that Rossi sees in the Lugano report.

  • bachcole

    I used to say a lot of derogatory things about “Rossi says”, but I believe this. It is very exciting, even if the plant is way too complicated. (:->)

  • Zeddicus Zul Zorander

    I find it very surprising that Rossi didn’t order a thorough fuel analysis himself. Especially as it turns out to be so helpful in increasing the efficiency of the plant. Analyzing the fuel is sort of reverse enginering the theory behind the LENR phenomenon. I think he overestimated his own understanding of the process.

    Very happy to hear about the long ssm periods. I wonder if the same changes would work on the hot cat.

    • Obvious

      Rossi took the rest of the fuel home with him…

      • Nicholas Chandler-Yates

        this is a good point… rossi and his team have almost certainly re-analysed the material in a more thorough way, giving them even more insight than we have.

      • Alan DeAngelis

        Yeah Obvious, the ash is the “Gold Mine”.

    • Omega Z

      Rossi’s R&D focus has been SSM or intermittent power. Probably the continuous 32 day powered operation provided the surprise.

  • Gerard McEk

    So the charges of the E-cats of the plant in operation can be changed one by one and that is clearly being done to imrove the efficiency (COP) and stability? It seems that the plant is also a research object. I wonder how one can guarantee the long term preformance if you make changes during the run.

  • Andreas Moraitis

    Rossi’s latest statements seem to indicate that the Lugano samples were by and large representative, and that the isotopic shifts did not result from simple fractionation. Besides, Rossi has apparently found a theoretical model that allows him to go beyond a purely Edisonian approach. This will speed up development to an unimagined extent, especially if he can refine his theory by own basic research. We might then see a reactor that runs, once started, almost permanently in SSM in the not too distant future.

    • not sure it is representative of the full material, but informative on what was happening.
      if it is confrmed that they improved the charge from those data, this mean that they start to have a theory… maybe just a phenomenological theory, but some that than inspire engineers.

      this is historical.

      add to that what Prometeon said…

      At last something is starting to happen around theory. note that when there is a theory for LENR it will be accepted instantly. theory , lack of theory, is the only cause of LENr rejection.
      even the conspiracy of hot fusion is only a side problem. key is lack of theory.

      • Omega Z

        With Rossi being surprised by the analysis & As the Lugano sample was made of just 10% of the ash, Possibly IH/Rossi analyzed more of the remaining ash & developed a much better picture.

        Note, Even with a theory, certain entities will still be in denial.

        • I suspect IH could not afford so much isotopic analysis…
          or was afraid to do …
          or as you say something never seen happened because the physicists did not run in SSM.

          about theory, from all I’ve read, from acceptance of many facts, I suspect theory is the 99% key to acceptance.

          denial will disappear instantly with theory. today with a good theory you ca invest trillion $ and ruins hope of billions people in claims that dissent with reality.

          • Zack Iszard

            I work for an analytical chemistry company. We have discussed contracting with another similar company that runs an ICP-MS (the gold standard of elemental analysis these days) for about $100 USD per sampple, and they quantify everything. Price point for triplicate or even pentuplicate testing of the ash, especially if Rossi kept his contacts in Raleigh (the “research Triangle”), would likely be negligible for the invaluable information it provides. Five samples each from unspent fuel and ash at the rate I stated above is about $1000 USD. For greater sensitivity and quantifying accuracy, an ICP-MS that has been calibrated with many different standards can resolve sub-ppm-level constituents with appreciable accuracy.

          • ok, so this mean cost is not a factor.
            maybe did something unusual happen, because of Lugano testers…

          • Alan DeAngelis

            Mike McKubre has a MS that can differentiate molecular deuterium from helium! (37:00 min.)

          • Warthog

            It doesn’t cost much at all to do isotopic analysis. High-resolution mass spectroscopy is done routinely by many contract analysis labs. Just send the sample out, and get results back. This also eliminates possible “inventor bias” with the results.

  • Does anyone know what the Dog-Bone Project is doing in the meanwhile?

  • BroKeeper

    It is appearing more to me the longer the E-Cat runs more factors are injected into the processes by the increased isotopes generated during the length of the processes. This would augment the differing theories of the reactions the longer the powder is subjected to changing conditions and transmutations. Perhaps SSM increases with time as isotopes accumulate, decade and/or react with hydrogen to form He(4) (suggested by Alan DeAngelis below). This could even help stabilize control and provide longer operational time from the added heat generated by such added reactions. I doubt any one theory can fit all the increasing scenarios of a changing powder and its resulting reactions. It’s like having its own life growth stages – like ‘Rossi’s Baby’.

    • Alan DeAngelis

      Thanks BroKeeper, I’m going (or I’ve gone) senile. Now I remember. We went over this before:

      2 H(1) + Ni(62) > Ni(60) + He(4) 9.87 MeV (no gamma rays)
      And then:
      2 H(1) + Ni(60) > Ni(58) + He(4) 7.90 MeV (no gamma rays)

      And the Ni-58 wouldn’t be seen. It would be fed back into these reactions.

      Li(7) + Ni(58) > Ni(59) + Li(6) 1.74 MeV

      Li(7) + Ni(59) > Ni(60) + Li(6) 4.13 MeV

      Li(7) + Ni(60) > Ni(61) + Li(6) 0.569 MeV

      Li(7) + Ni(61)) > Ni(62) + Li(6) 3.34 MeV

  • Valeriy Tarasov

    From mine point of view, the improvement of charge
    can be achieved in the case of only 7Li usage, since its fission to alpha particles, resulting from interaction with protons, is source of energy (i.e. 6Li should be removed from the fuel).

  • US_Citizen71

    Perhaps the fuel in the Lugano test was different than what Rossi/IH were normally running in their reactors, missing an ingredient or two, a different particle size for the ingredients possibly or maybe in a different ratio than normal. If they are trying to protect IP then they wouldn’t likely use the secret formula to power a test set up like Lugano since the researchers were allowed before and after samples of the fuel. The mix was likely calculated to give a good result but not as good as the proprietary formula and it likely out performed Rossi/IH’s predictions of output. So they got a data point that made them try a new fuel mixture somewhere in between the Lugano fuel and the secret fuel and it has a better SSM than the previous go to mix. It makes sense and doesn’t require reaching too much to connect the data points.

    • SteveW

      I agree and that’s what I’ve been saying. The e-cat H2 has such a high operating temperature that it makes it difficult to experiment with which is exactly what I would want if I was Rossi. Sure, it glows red hot and looks cool but it’s overkill for electrical generation and just makes engineering difficult.

      • Axil Axil

        The Lagano test may have revealed to Rossi that the function of the nickel micro-particles are transitory. These particles may only be required at startup just to get the LENR reaction going without producing the BANG.

        But the Lagano testers ran the reactor very hot, and many if not most of the nickel particles melted. But after 32 days of operation, the reaction was gaining in effectiveness and vigor even as the nickel particles were being degraded by the high heat. This revelation may have allowed Rossi to rethink his fuel mix strategy. Rossi may have been surprised that the nickel particles showed limited transmutation.

        The nickel particles may only be required to setup a quantum mechanical precondition that allows the LENR reaction to begin gradually and smoothly without a BANG.

        Once startup is achieved, the LENR reaction precedes as a fire would by maintaining the conditions necessary for it continuing progression.

        No LENR reaction has yet to be restarted. Could it be that the nickel particles can only be used once at the initial startup. And once used these particles become ineffective.

        • Veblin

          Hank Mills
          October 8th, 2014 at 4:32 PM
          There are individuals saying that after the test the fuel had been melted – according to the electron microscope images – and the reactor would not have been capable of being restarted because the reaction sites would have been destroyed. Would the reactor have been capable of being restarted?

          Andrea Rossi

          Levi Strauss
          February 7th, 2012 at 5:28 PM
          If using the Ecat for a heat source, can the Ecat be turned off for summer and then restarted in the fall, or once shut down, will a new cartridge have to be installed?

          Andrea Rossi
          Can shut off and restart when you want, no substitution is necessary

          • Axil Axil

            What Hank and Levi is asking Rossi “is your product any good” Of course, Rossi will say YES. But Rossi’s reactor must be under control at all times or else it will fail in one way or another. The Lagano test team abused their reactor and ran it hot. The micrographs of the nickel particles show that. Any yet the reaction was very strong at shut down which is surprising based on the shape of the sample of particles shown.

          • SteveW

            Thanks for finding those statements from Rossi concerning restarting the reactor. My response to these statements is contained in the response to Axil Axil directly above.

        • SteveW

          At the initial start-up, the fine Nickel micro-grain particles are necessary in order to load the Ni lattice with H. Once the Ni is loaded, gas pressure is needed to maintain this loading and as I have said, I don’t believe Hydrogen is a necessary constituent in this pressurizing gas which prevents the Ni lattice from unloading.

          After the Ni is loaded, it doesn’t matter what happens to the Ni. If the Ni melts into a blob, the Ni is still loaded with Hydrogen as long as gas pressure is maintained. The more robust LENR reaction would be expected from the elevated temperature under this premise.

          When Rossi was asked early in 2012 by Levi whether the the e-cat could be shut down and restarted, I believe at that time he thought it could. That was very early in the development. It would take months of testing to determine that it could not and months of trying to design a reactor capable of holding Hydrogen that ultimately ended in failure (IMHO). I believe that Rossi first developed the e-cat H2 with the idea that the Hydrogen gas could be maintained within the reactor but later realized that under the extreme operating conditions, it was just too difficult to prevent the Hydrogen from permeating the reactor vessel for very long and certainly not long enough for a practical life of a reactor charge.

          Rossi then realized (or he may have already known- who knows) that he didn’t have to maintain the Hydrogen within the reactor, to maintain the reaction. He only had to maintain the Hydrogen long enough at start-up to load the Ni lattice. If you can’t design a reactor to hold Hydrogen for the life of a fuel charge, you may has well save the money and just make the reactor capable of holding the Hydrogen just long enough to load the Ni.

          On October 8th, 2014 when Rossi was asked again by Hank Mills whether the reactor could have been re-started, Rossi responded with a simple one word answer, “YES”. In Rossi’s mind, that could mean, “Yes, with a new fuel charge and re-sealing the reactor, it could be re-started”. That wouldn’t be a lie- just misdirection. Rossi has admitted to misdirection in the past so why would he not do it here. Why wouldn’t the testing team not try to re-start the reactor unless they were instructed to limit their test to preclude it. And why would they be instructed to preclude it if it could re-start?

  • US_Citizen71

    Maybe the difference between the hotcat and the warmcat is nothing more than the frequency and intensity used to control the reaction. Both signals could be a harmonic of the actual perfect frequency. So as the harmonic gets closer to the actual perfect frequency higher temperature reactions becomes possible due to more reactions per time interval?

  • Mats002

    Yes. This is Classic Human Behavior, when you introduce a new idea, you are the one to perform it.

  • Bob Greenyer

    Frustrated by the incompetence of the USPTO relating to potential reaction paths (not the thermal comments, that was fair) I did a bit of thinking these past two days, and came up with this.

    I’ll try and make a presentation to explain it all later, but wanted to get it out there in its raw form for discussion and before anything similar is wrapped up in patents.

    There is a little more context on our FB

    As I worked, the results kept explaining things observed in an “auto consistent” way.

    I would appreciate anyones comments – I’m tired, so there are likely schoolboy errors.

    • Andreas Moraitis

      Interesting considerations, especially since the role of other elements than Ni and Li has been rarely discussed in this context. However, while the number of hydrogen atoms might suffice for the postulated reactions, there is still the problem of “missing neutrons”, regarding the isotopic shifts in Ni. As has been demonstrated, neither the 7Li nor the 64Ni or the hydrogen (via electron capture) could provide enough neutrons to convert the lighter Ni isotopes to 62Ni. This deficit would even increase if lithium and hydrogen were consumed by other reactions. But after all we do not know how reliable the numbers are, and I guess that the Lugano team and others will provide more data any time soon. So you could still be right with your analysis.

      • Pekka Janhunen

        Agreed. I want to add that appearance of new elements is not a very reliable indicator of nuclear reaction, because all substances (including Al2O3) contain impurities and at high temperature the impurities may start to migrate, come to the surface and form gaseous compounds. If LENR active sites produce soft ionising radiation, they also attract impurities from the gas phase by breaking up molecules into free radicals which are chemically active and react with the nearest solid surface to become sessile again. Observing severe isotopic shifts is a more robust indicator of nuclear activity: in concurrence with the Lugano report authors,I don’t know how to explain the appearance of pure Ni-62 (originally a minor and middle isotope) by means of chemical enrichment.

        One also has to remember that e.g. Parkhomov test was ~100 times shorter than Lugano. Therefore it would be surprising to see major isotopes shifts in Parkhomov ash. If one would see such changes there, it would be a very interesting finding because it would constrain theories a lot.

        • Bob Greenyer

          We do not know what was in Parkhomov fuel or ash precisely, I requested samples of both by Dr Parkhomov said he had committed to a respected and trusted friend that they would do the analysis.

      • Bob Greenyer

        I have added the reaction chains and reasoning for the xNi > 62Ni refinement at the bottom of the document and the same reasoning (in addition with reference to Piantellis patent extension) explains why it might be advantageous to use pure 62Ni with 7Li.

        • Pekka Janhunen

          Criticism: 1) The beta+ radiation (~0.5 MeV) is not observed and radiation from Ni59 was not seen in the ash. 2) If Ni62 doesn’t react, why would it be advantageous?

          • Alan DeAngelis

            To 1): Maybe there is a reaction that causes Ni(59) to disappear that is much faster than the reaction that forms Ni(59). So, its concentration would be essentially zero.

          • Bob Greenyer

            About 59Ni

            Firstly, the half life is 76000 years

            Secondly, since the ash analysis sees near 100% 62Ni, then it would not be an issue.

            I expect that it was necessary to have a long run to ensure that there would be no measurable radiation shortly after the experiment – so that you could permit ash analysis.

            I have explained the 62Ni point in the live document under the title “Why might having pure Ni62 be advantageous?”

          • Andreas Moraitis

            Rossi said some time ago in an interview that they had seen 511 keV gammas. I guess, however, that the alumina casing would not be able to block them. Maybe he controls the reaction meanwhile more precisely, so that the gammas can be avoided.

          • Axil Axil

            In a system where many things are doing the same identical thing and those things can communicate, something called blockade applies. In a system of a trillion occelators all vibrating at the identical same fervency and amplitude, any energy applied to that system no matter how high will be blockaded from causing a hot spot in that collection of oscillators. The oscillators will spread out that energy burst throughout all one trillion oscillators evenly.

            This happens in the Ni/H reactor. All the dipoles in the reactor will become coherent; they will all oscillate at the same fervency and the same amplitude. Any burst of energy that originates inside that coherent system will be blockaded from causing a hot spot. This system is called a BEC – Bose Einstein condensate.



            In the Ni/H reactor, the job of the nickel particles are to initially get all the dipoles to vibrates coherently. After doing this, the BEC will be maintained indefinitely by the dipole blockade.

          • Andreas Moraitis

            Your link does not work for me, but Widom & Larsen advocate a similar hypothesis in one of their patent applications (I guess you know it). Nevertheless, this has to be experimentally examined. Pekka has proposed to put a gamma source behind the running reactor in order to determine if there is a significant shielding effect.

          • Axil Axil

            I fixed the link as follows:


            Why not just add a gamma source to the fuel.

          • Pekka Janhunen

            One could also add the gamma source to the fuel. I just thought that by using an external source behind the reactor one doesn’t have to handle active materials and does not have to modify the charge in any way or even to open the reactor. Whichever is easier, is just a technical question.

        • Andreas Moraitis

          It makes no numerical difference if you convert the protons directly or indirectly. If it is correct that there have been 0.9g Ni and 0.1g LiAlH4 in the fuel, then the number of available nucleons would come to

          6.35*10^21 from H (1 proton available)
          1.436*10^21 from 7Li (1 neutron available)
          1.6956*10^20 from 64Ni (2 neutrons available)
          = 7.956*10^21 nucleons.

          To convert all the lighter nickel isotopes (58, 60, and 61) in the fuel to 62Ni you would need 3*10^22 additional nucleons. Considering the minor amounts of 58Ni and 60Ni in the ash does not change much. Either there is at some point an error of more than 250%, or something fundamental has hitherto been overseen.

          • Andreas Moraitis

            Should read „overlooked“…

          • Axil Axil

            In my thinking about what is happening to produce all those neutrons when no free neutrons are seen roaming about, I believe that the magnetic fields produced by SPPs as strong enough (at least 140 MeV) to condense pions out of the vacuum to convert protons from absorbed hydrogen nuclei to neutrons inside the nickel nucleus.

          • Andreas Moraitis

            Why shouldn’t the pions decay to muons, which might then enable ‘legal’ cold fusion?

          • Axil Axil
          • Bob Greenyer

            Perhaps you have not read the spreadsheet, no neutrons are required, just Piantelli’s theory of capture 1H via H- virtual electron in lowest shell (which comes from the ionic LiH, no need for catalytic surfaces) or coulomb based repulsion of the p that did not quite make it in at 6.7MeV that will interact with other xNi and Li / Al etc.

            All the products are short lived if unstable and explains the cascade to 62Ni (even 64Ni to 62Ni) and Piantelli already explained in his patent before Lugano why 7Li would disappear and 6Li would not. It also accounts for Silicon observed and drop in COP after day 8 etc.


            No exotic theories required (well apart from accepting Piantellis experimentally derived observations)

          • Axil Axil

            At the most fundamental level, all the various types of LENR reactions come from the same mechanism. Piantilli’s reaction is the same as Rossi’s and most challenging to explain as common is the Cravens Golden ball reaction as common.
            All Cravens uses in the golden ball is magnetic powder to produce excess heat. No ions are produced there.
            The reason why there are hundreds of LENR theories is that each experiment reveals an emergent property of the LENR reaction. In their own context they are consistent but each are not fundamental. Every type of LENR reaction must be explained through the same fundamental mechanism. That is the challenge
            There are also a handful of LENR miracle’s to explain. Piantell has not done that yet to my knowledge..

          • Pekka Janhunen

            Let’s take it step by step. So if I understand it right, Piantelli is proposing that H- (p+e+e) can act as a “heavy electron” which catalyses fusion in a similar way as muon does in muon catalysed fusion. If some process, for example a lone X ray quantum, kicks out an electron from the lowest shell of Li7 atom, the vacant electron state might get (according to the hypothesis) temporarily filled by H-, after which the proton of H- gets close to Li7 and fusion can occur. If the fusion for some reason proceeds directly to two alpha particles (rather than going through the Be8 intermediate state), one gets two 8.5 MeV alphas, which thermalise and produce some soft bremsstrahlung X-rays of less than 1.1 keV energy.

            A big theoretical problem in this scenario is that the large electric field gradient close to Li7 nucleus should tear apart the electrons from the H- long before the H- can settle itself to orbiting the nucleus, considering also that the orbital radius of the “heavy electron” would be much smaller than the size of the H- itself. However, although the process does not make sense classically, perhaps there is some hope that quantum mechanics might allow for it to work sometimes, although usually the proton would come out empty handed.

            If such process works, there might exist a nonlinear chain reaction or feedback effect if the process requires a seed X-ray to kick out the electron while also producing X-rays itself. That might rhyme with the sometimes bursty nature of LENR phenomena.

            I do not understand the sentence “coulomb based repulsion of the p that did not quite make it in at 6.7MeV”.

          • Axil Axil

            How does pressure oscillation increase that Piantill sites increase the effectiveness of the LENR mechanism that Piantelli proposes?

          • Bob Greenyer

            The molten Li – Al – H with dissolved Ni on solid Ni is a new reaction domain

            Basically, early breakdown of LiAlH4 (or NaAlH4 in E-cat / HCat1) strip oxides from Ni. Then the molten metal mix wets to the surface where LiH or NaH are on ionic – that is they have H- just what is needed for Piantelli reaction pathways.

          • Axil Axil

            How does Piantelli’s theory hold up under the Conservation of Baryon Number, Conservation of Parity, Conservation of Isospin and the Conservation of Lepton Number?

          • Bob Greenyer

            That is above my pay grade (I earn nothing!) however, rest assured, Piantelli takes ALL these things into consideration and is meticulous at checking everything is consistent.

          • Axil Axil

            How does the theory explain how a proton turn into a neutron by showing everything that goes into the reaction and what comes out? Did you see such a layout of the reaction in documentation or the patent?
            LENR cannot violate conservation laws.

          • Bob Greenyer

            Axil, I took the H- capture / p ejection process as published by Piantelli and ran it trough the NNDC calculator, I then looked at the products stability, found the corresponding isotope in Wikipedia and looked to see if it was stable or not. If not, I looked at the decay time, mode of decay and the resultant isotope. Since the reaction zone at high pressures of H2 is ONLY the Nickel bulk or the interface between the Nickel bulk and the molten Li – Al – H – Ni (in solution) then I looked at the phase of the products/decay product to see if they would remain in play.

            If the product was a transition metal, Piantelli would say that it can do 1H capture from H- or receive a p ejectile from the primary Ni interactions.

            Piantelli only lists the initial reactions in his patent, he does not list what the natural decay of unstable ones

            This is what he says in his patent

            “In the case of Nickel, the internal primary nuclear reactions of direct capture, as calculated taking into account the conservation of the spin and of the parity, as well as the Gamow coefficient, can be written:

            1H+58Ni→59Cu+3.417 MeV   {1a}

            1H+60Ni→61Cu+4.796 MeV   {1b}

            1H+61Ni→62Cu+5.866 MeV   {1c}

            1H+62Ni→63Cu+6.122 MeV   {1d}

            1H+64Ni→65Cu+7.453 MeV   {1e}.”

          • Axil Axil

            Piantelli does not list the neutrinos coming out of the reaction. He would need to list electron neutrinos to account for the fermion conservation law with two fermions going into the reaction and one coming out. When an electron is captured by a proton, a neutron is produced along with a neutrino. Since a proton is changed to a neutron during electron capture, the number of neutrons in the nucleus increases by 1. In all the reactions listed, I see a proton added.
            In addition, those reactions would produce radioactive isotopes. No radiation byproducts are ever detected. How does Piantelli explain this lack of radioactive activation?
            No copper is seen is Rossi’s ash.

          • Bob Greenyer

            Did you look at the spreadsheet?

            there is only captured protons or proton projectile interactions. All progression is via recognised decay.

            The products are all very short lived isotopes (except one that is very long but gets converted anyhow)

            Testing my Piantelli based hypothesis, I suggest deliberately blowing up a reactor and dropping the sintered core immediately into a cloud chamber.


          • Axil Axil

            I will study the spread sheet as time permits.

          • Alan DeAngelis

            Wouldn’t copper in its excited state have a much large cross section than the other nuclei (a bigger target)?


            1H+65Cu*→66Zn*→62Ni+4He (no gamma rays)

          • Alan DeAngelis

            Copper(1) Cu+ is a soft acid by HSAB. H- (hydride) is
            a very soft base by HSAB. They would form a very strong polarizable COVALANT
            bond. It’s not just a pile of bricks as physicists would have you believe.

          • Alan DeAngelis

            This might help explain HSAB theory. See 9:40 min.

          • Bob Greenyer

            Neither does wikipedia list neutrinos coming out of the decay of isotopes when, for instance, there is a B+ emission, perhaps because that people think they have little interactions (Parkhomov would disagree, one of his proudest achievements is a study since 1998 of how neutrinos can affect the half life of unstable elements)

            Piantelli has published that he has seen radioactive isotopes. But as per my sheet, they are predominantly short lived and no neutrons. No radiation measurable outside the reactor, and nothing of consequence after a full burn and a bit of decay time.

          • Axil Axil

            All of Piantill’s observations and theory are immigrant. He must prepare his nickel surface is just the right way or his observations and theory doesn’t work. He must apply heat is just the proper way or things don’t work. The secret sauce must be used…not just anything will work. The pressure must be in range or things don’t work. There is something about the pressure, the heat, the secret sauce. the size of the particles, the metal that the particles are made from, and the surface that is at the essence of LENR. This is what I term fundamental. Piantelli’s theory does not address these things.

          • Bob Greenyer

            Also, as per this section in the spreadsheet

            “Explaining Nickel Transmutation (inspired by Piantellis published patent extension)”

            where I have only partially done the reaction chains, but I have taken them as far as removal of all copper. Interestingly, the result of running it this far is that there is now Gallium in the products, and it is an iterative process, so very difficult to remove all Gallium.

            Gallium seen in Tof-Sims, but then it is used in the process.

          • Pekka Janhunen

            The problem with reactions 1a-1e is that they have only one resultant particle (the copper nucleus), hence to satisfy conservation of energy and momentum, the energy should be liberated by a hard gamma quantum which is not observed. All radiation-free nuclear reactions must have at least two heavy output particles.

            The equations might qualify to this, if one assumes that there are two or more protons on the left-hand side which react with the nickel nucleus simultaneously. Then the extra proton which formally does not participate in the reaction could take the energy away, and they would be Piantelli’s 6.7 MeV protons.

          • Bob Greenyer

            I want your pay grade!

            I would think you need reactions that exceed 6.7 MeV (like 1H + 7Li) to result in a gain.

          • Pekka Janhunen

            Could you tell, how much can we trust that number, 6.7 MeV? If the number is accurate, it’s a very useful datapoint. But if it’s not accurate, then making conclusions from it is not fruitful. After all we need experimental data more than theories at the moment. Theories will follow once reliable data exist. Whereas theories cannot be tested without data.

          • Bob Greenyer

            It is in his patent he says

            “Expelled protons 35″ have an energy of 6.7 MeV. This calculated value is experimentally confirmed by cloud chamber measurements.”


            Here is the process from his original patent (which does not detail the energy)


            “After such adsorption step, the H- ions interact with the atoms of the clusters, provided that a second activation threshold is exceeded, which is higher than the first threshold. By exceeding this second threshold, in accordance with the Pauli exclusion principle and with the Heisenberg uncertainty principle, the conditions are created for replacing electrons of metal atoms with H- ions, and, accordingly, for forming metal-hydrogen complex atoms. This event can take place due to the fermion nature of H- ion; however, since H- ions have a mass 1838 times larger than an electron mass, they tend towards deeper layers, and cause an emission of Auger electrons and of X rays. Subsequently, since the H- ion Bohr radius is comparable with the metal core radius, the H- ions can be captured by the metal core, causing a structural reorganization and freeing energy by mass defect; the H- ions can now be expelled as protons, and can generate nuclear reactions with the neighbouring cores. More in detail, the complex atom that has formed by the metal atom capturing the H- ion, in the full respect of the energy conservation principle, of the Pauli exclusion principle, and of the Heisenberg uncertainty principle, is forced towards an excited status, therefore it reorganizes itself by the migration of the H- ion towards deeper orbitals or levels, i.e. towards a minimum energy state, thus emitting Auger electrons and X rays during the level changes. The H- ion falls into a potential hole and concentrates the energy which was previously distributed upon a volume whose radius is about 10’12 m into a smaller volume whose radius is about 5×10 15 m. At the end of the process, the H- ion is at a distance from the core that is comparable with the nuclear radius; in fact in the fundamental status of the complex atom that is formed by adding the H- ion, due to its mass that is far greater the mass of the electron, the H- ion is forced to stay at such deep level at a distance from the core that is comparable with the nuclear radius, in accordance with Bohr radius calculation. As above stated, owing to the short distance from the core, a process is triggered in which the H- ion is captured by the core, with a structural reorganization and energy release by mass defect, similarly to what happens in the case of electron capture with structural reorganization and energy release by mass defect or in case of loss of two electrons, due to their intrinsic instability, during the fall process towards the lowest layers, and eventually an expulsion of the the H- ion takes place as a proton, as experimentally detected in the cloud chamber, and nuclear reactions can occur with other neighbouring cores, said reactions detected as transmutations on the active core after the production of energy.

            According to the above, the actual process cannot be considered as a fusion process of hydrogen atoms, in particular of particular hydrogen isotopes atoms; instead, the process has to be understood as an interaction of a transition metal and hydrogen in general, in its particular form of H- ion.”

          • Pekka Janhunen

            OK, thanks. I could buy the idea that H- ion might act as a “heavy electron” (with strong caveats I explained in another posting), and that with those caveats this might be a way to get a proton (in the form of H-) close to a metal nucleus which might then enable a nuclear reaction. From that point onwards, however, what Piantelli writes does not make sense to me. I do not understand how the proton could be expelled (where would the energy come from and what would happen to the electrons), and anyway, if there is net energy release, there must be a nuclear reaction happening somewhere.

            That said, it is highly interesting if he observes energetic protons and has been able to measure their energy (6.7 MeV). I assume that it’s the maximum energy that he has seen; the proton slows down quickly in matter so one would see a continuum from low values to the max.

          • Bob Greenyer

            He measured it from distance travelled in cloud chamber in addition to calculating it.

            ejected protons are ones that didn’t quite make it to the core “shrinking trojan horse style” but were repelled by the coulomb force.

          • Pekka Janhunen

            To make it work there must be two (or more) protons around the nickel nucleus. One fuses and is not seen again, the other takes the fusion energy and is ejected out. Energy conservation works and one gets an explanation for the absence of radiation. The hard problem is then how to get two protons at once near the nucleus and how they overcome the Coulomb barrier. The protons need not be similar. One could be high energy and one could be low energy, for example. Or it could be a diproton, which is a short-lived unstable state that might exist temporarily. (I find the diproton explanation unlikely, but anyway.)

            Does Piantelli see copper getting formed from nickel?

          • Bob Greenyer

            You will see that the products are short half lives, better to look at the ash after a long run experiment.

            Piantelli wants an experiment to do real-time analysis, this would settle the debate. I understand DGT built one.

          • Pekka Janhunen

            I guess what we are asking is that what is the reaction that Piantelli is proposing? Once the reaction equation is written down, all the conservation laws and more can be checked by the relevant experts.

          • Bob Greenyer

            Piantelli says in his published Patent extension that only TRANSITION metals can perform the primary reaction, NOT Li. Nickel is his preferred metal.

            Based on experiment following hypothesis grounded in non-contoversial physics, he says that H- (of which there is an abundance in the ionic molten LiH + Al directly on the Ni as we showed in our SEM images from ‘Bang!’) gets REALLY close to Ni nucleus, because of charge shielding, yes, in a kind of muon way, in his patent he says fermion.

            Then one of two outcomes occur

            – strong force pull, resulting in capture of 1H
            – didn’t quite make it and coulomb repulsion of p

            under his hypothesis, the calculated energy of the ejected p is around 6.7MeV and this was observed in a cloud chamber also. This is published.

            Elemental transmutations etc observed in his published experiments can be explained by application of this process.

            I didn’t set out to do what I did over the last 2 days, I just did a Parkhomov, I progressed on the basis that Piantelli process was real, in part because the USPTO could not see where the needed p projectile came from to cause the 1H + 7Li reaction quoted by levi… and this was rejecting Piantelli at the same time.

            The p is an ejectile.

            The only hypothesis I make, is that 62Ni is so happy, it only acts as a springboard for p for further reactions, rather than taking on further 1H from the H-.

          • Pekka Janhunen

            If a negative particle (H-) falls deep into a potential well of a Ni nucleus, energy is liberated in the process. (How is this energy liberated?)

            Once fallen into the potential well, the H- has negative total energy. To disintegrate it and to move any of the particles (p,e,e) outside the system would therefore consume energy, not liberate it. The mechanism you describe sounds like both having the cake and eating it not only once, but twice.

            If Piantelli sees 6.7 MeV protons, why couldn’t they be direct products of some nuclear reaction. Do you know how accurate is the number 6.7 MeV? Is it like 6.7+-0.1 ?

          • Bob Greenyer

            I am just working off his published works, like Parkhomov did with the Fuel analysis, I am starting from a foundation that a scientist, basing their statements on empirical evidence and have not set out to deceive. Then I am just using nothing more than the NNDC Q-Calc and Wikipedia Isotopic decay tables, I have not gone any deeper than that, it is just that by running the exercise AFTER

            1. the MFMP had empirically determined the real likely temperatures in the Lugano report


            2. the MFMP had demonstrated empirically for the first time what the reaction environment was likely to be via SEM (thanks to Ed Storms and Kiva Tech) following ‘Bang!’

            it was possible to understand the system better and things just started making sense. Of course, I may be wrong on many counts, the purpose of showing the scribble was to invite comment!

            What I can say is that having spent nearly 2 weeks with Piantelli, the man does not leave anything to chance, he studies, calculates, re-calculates, hypothesises, builds exacting experiments designed to robustly test a hypothesis , tests and tests again, then analyses and analysis again. He would be mortified to publish anything he did not whole-heartedly stand behind. More than that, he is willing to say that despite all his thoroughness, he may still be wrong. The man is a true, broad ranging scientist and invites respect. I only conducted the exploration because the USPTO was effectively dismissing Piantellis patent by inadvertently quoting two reactions in it that were independently cited by Levi as being possible explanations for the Lugano reactors output.

            I imagine some x-rays are lost, do these have a stimulating effect on the rest of the system? I don’t know.

          • Bob Greenyer

            You are ignoring Al and specifically the production of alphas from the second reaction in the top of the spreadsheet.

            I have added the alpha + 58Ni > 62Ni (via short lived unstable intermediaries)

    • Alan DeAngelis

      I see in your Facebook:
      “…there is no evidence in the corpus of nuclear science to substantiate the claim that nickel will spontaneously ionize hydrogen gas and thereafter “absorb” the resulting proton.”

      I don’t know if it absorbs a proton or not BUT if it did, wouldn’t this make it “non-obvious’ and therefore patentable? I just don’t get patent law.

      • Bob Greenyer

        I think that is Piantelli’s point, he showed these effects empirically (by emissions and ash isotopic study) having built experiments designed to test a hypothesis.

        • Alan DeAngelis

          Thanks Bob, I just read that sentence and got distracted by thoughts about Giordano Bruno.

    • Obvious

      A quartz particle was found in the earlier SEM ash/fuel report.

  • GreenWin

    This makes the “critique” from Pomp & Circumstance all the more amusing.

    • psi2u2

      Pomp and Circumstance have nine lives.

      Just you wait……! 😉

  • GreenWin

    Thoughtful comment Poly. You say, “Disapproval by the corporate class, whose main concern is lower taxes for themselves, comes in the form of lack of investment.” Let’s not forget the corporate class includes giant government agencies (EPA, DOD, DOE, NOAA, NASA, etc.) AND the elite corporate climate gang – e.g. Algore, Sierra Club, Audubon, Greenpeace, Earth First. These agents and puppeteers are terrified they are losing control of energy. Fact is, they are. And with it the cash cows that have fed (and permanently corrupted) their agendas. Oh, well.

    LENR raises the standards of living for literally billions. When standards rise – birthrates drop. Hello climate/population gang?? And erm, no need to pave the deserts with PV or windmills, or blow away mountain tops, or drill the Arctic, or crisscross habitat with 100kV transmission lines, or burn the rainforest for fuel or… let the polar bears drown. CF is not like Guy Fawkes – he didn’t actually create anything. It’s more literally a New Fire; able to propel humanity forward just as old fire did.

    • Timar

      Greenpeace under control of the corporate class!? What about the Illuminati?

      • Manuel Cruz

        Greenpeace is on the bankroll of Rockefeller. It’s corrupt because it only attacks certain targets, which are the economic interests of its competitors.

      • GreenWin

        Timar, IF Greenpeace was truly interested in protecting the Arctic etc., they would pay attention to viable alternative energy sources. In the case of LENR, even a modicum of research indicates viability (NASA’s LArc Chief Scientist Dr. Dennis Bushnell, for example.)

        The entire climate consortium has refused to support LENR. And this Der Spiegel article points to Greenpeace loss of credibility due to financial scandal involving currency trading: IF climate doom disappears, how will these corporate “environmentalists” collect their huge salaries??

        • I don’t know for GP, but I interacte with climate skeptic and pro-nuke wamrmist…
          I was surprised that they are open to LENR, asking technology readiness evidences, if not enthusiast already…

          I feel that the line of demarkation in peoples mind is not between climate opinion, environmental concern, fossil, nuke, renewable…

          the line of demarkation is between
          and humanist/protechnology

          the said climate skeptic community is split in many different positions and disagreement…
          some accuse bad science (fraud, budget oriented science), politicized science (accuses communism), some accuse alarmism , or bad solutions (eg promote nuke, or prefer adaptation)… in fact many claimed skeptic are believers but not concluding the same.

          as LENR solve all problems with nuke, climate, oil, coal, pollution, population, third world development, the only opponents will be the malthusians-luddites, this mean most political enviros like GP, excluding real environmentally concerned (take bjorn lomborg, judith curry, claude allègre).

          if you see someone oppose LENR not for being unreal, but evil, it is for political reason.

          be ready, it is a politic war, a religious war in fact, but religion is politic, and vice versa.

  • I bet as soon as the replication attemps spread and spread Rossi/IH is put in the spot.
    He has to deliver something to stay in the game. Either press releases or a working device.

    • psi2u2

      I bet they IH is ready.

  • Alberonn

    Nice but slightly depressing analisys Avatar Polymorph, but Hope returned with the obvious ease with which Alexander Parkhomov replicated the hotcat. For sure we’re no longer dependent on the depressing strategy of IH and AR to try and do things the ‘proper’ way and stall till they are ready to launch under their COMPLETE CONTROL.
    “Parkhomovs” and their kin are bound to rise all around the globe : out of control of repressive western science, paralyzing legislation of “sophisticated” nations and your corporate elite (including Greenwin’s entities, see below :-). Once it hits China and/or India the battle is won…

  • Alan DeAngelis

    Yeah Pedro, I actually agree with you. This may be a local
    hot spot. See my comment.
    I also think there may be other reactions taking place.
    2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV (no gamma rays)

    2 H(1) + Ni(62) > Ni(60) + He(4) 9.87 MeV (no gamma rays)

    2 H(1) + Ni(60) > Ni(58) + He(4) 7.90 MeV (no gamma rays)

    2 H(1) + Ni(58) > Ni(56) + He(4) 5.82MeV

  • Bob Greenyer

    I have added what the ECat, HCat 1 might be and why the ECat is much easier to work with and control.

    • Dr. Mike

      Although I will wait to see experimental results confirming LENR theory, I believe your hypothesis for the various reactions can be verified after you get the basic reactor running. I am certainly looking forward to your results. One thing that I didn’t see in your spread sheet is how all of the Ni in the Lugano reactor was converted to Ni62? Should there be enough hydrogen in a reactor to not only convert all of the Ni to Ni62, but also to supply hydrogen for all other proposed reactions?
      Dr. Mike

      • Bob Greenyer

        I make the hypothesis that 62Ni does not capture more 1H from H- but rather just ejects p as per Piantellis other option.

        Although I have not done all iterations, I have done enough to exclude Cu effectively from the ash (but not Gallium – yet)

        you can see the Piantelli path from xNi>62Ni under the section in the sheet marked

        Explaining Nickel Transmutation (inspired by Piantellis published patent extension)

        It is at the bottom as I only set out to find what would happen to the Aluminum.

        Decay derived Alpha particles may also interact as suggested by Piantelli in his patent extension. An example of how 58Ni becomes 62Ni is given.

  • Alan DeAngelis

    Aluminum to Silicon
    Coupled reactions?

    2H(1) + Ni(n) > Ni(n-2) + He(4) (MeV alpha)

    Al(27) + He(4) > Si(30) + H(1) 2.3722 MeV

    The MeV alpha from the above reaction can cause the below reaction and then the MeV proton from the below reaction can cause the above reaction.

    • Bob Greenyer

      Piantelli described his understanding as more like nucleosynthesis… capture, re-organisation etc. but NOT ‘Cold Fusion’! Any helium is a product of emission and D2 does not work with Nickel (as Celani has found)

      What is happier than Ni62?

      • Alan DeAngelis

        Not deuterium two protons. H(1). For example:
        H(1) + Ni(62) > Cu(63)* Step1
        H(1) + Cu(63)* > Ni(60) + He(4) Step 2
        Over all
        2 H(1) + Ni(62) > Ni(60) + He(4) 9.87 MeV

        And others:
        2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV (no gamma rays)

        2 H(1) + Ni(60) > Ni(58) + He(4) 7.90 MeV (no gamma rays)

        2 H(1) + Ni(58) > Ni(56) + He(4) 5.82MeV

        • Alan DeAngelis

          I know this is a sketchy thought but I think the chemistry sets up the reaction. See the comments of this link for an even sketcher explanation.

          • Bob Greenyer

            Sketchy thoughts are fine by me, thats where I was on Friday. Now it is a full on scribble!

          • Alan DeAngelis

            Would ingesting the lithium help?

          • Bob Greenyer

            Quite possibly!

          • Alan DeAngelis

            “There is only one difference between a madman and me. The madman thinks he is sane. I know I am mad.”
            Salvador Dali

        • Bob Greenyer

          It’s ok, I know you did not mean Deuterium! For the Ni+H system, I was just clarifying what works and what it is not.

          It is my understanding that Nickel can only capture 1H front the H- in low shell at a time – but there is the chance that an ejected p from an unsuccessful capture could impinge on the same atom at the same time as it was capturing 1H from H-, but that may be statistically improbable.

          • Alan DeAngelis

            Hi Bob, just one last thing.

            Its nickel hydride NiH2 (with two very soft polarizable covalent bonds).

            H(1)~~Ni(n)~~H(1) > Cu*~~H(1) > Zn* > Ni(n-2) + He

            It’s not like the hot fusion banging billiard balls together in a plasma approach.

            There’s a typo in that link. Should be: infrared (not inferred) stretching.
            I first talk about F&P but then I talk about protium-nickel hydride bonds.
            And HSAB is Hard Soft Acid Base theory (Ralph Pearson).

          • Bob Greenyer

            Protide is the key, the ion (H-) of hydrogen, two electrons on one proton… this is in abundance with the ionic hydride LiH liquid, which when liquid is in dynamic equilibrium with the H2 in the reactor.

          • Bob Greenyer

            I have added

            “Magnetic / electrostatic field requirement as per Piantellis patent”

            and its relationship to all coil configurations of hot cats to the sheet

          • Alan DeAngelis

            Yeah Bob, you are right. “The near-stoichiometric NiH is unstable and loses hydrogen at pressures below 340 MPa”.

          • Alan DeAngelis

            But the resulting copper in its excited state, Cu*, would have a much large cross section than the other nuclei (a bigger target), a much high probability of undergoing another nuclear reaction than the other nuclei.

          • Alan DeAngelis

            But then again, nickel has a valance of +2. So I’m not so sure if one can assume the structure of bulk NiH from its stoichiometry. I’m not so sure that nickel hydride can be thought of as a discrete molecule of NiH. Bulk NiH may have some domains of [Ni(0)Ni(ii)H2]. I really don’t know. Maybe someone can fill me in on this.

      • Andreas Moraitis

        Successful H- capture by a nucleus could of course be called “cold fusion”. Consider that this term has initially been coined with regard to muon-catalyzed fusion, which works, as you know, in a very similar way.

        • Bob Greenyer

          Kinda. Hot/Cold Fusion was meant to be isotopes of H fusing. This involves H but with anything other than an isotope of H.

  • Is the project dog bone dead?

  • Alan DeAngelis

    Maybe the phosphorous comes from the reducing agent hypophosphorous acid, H3PO2.

    • Alan DeAngelis

      Oh, I just got it through my thick head. Natural phosphorous has only one isotope. So, we weren’t taking about the fuel. We were speculating about a mechanism that might involve the formation of phosphorous. A little too much free association. Having said that, hypophosphorous acid, H3PO2 might be a good reducing agent (a source of hydride) to add to the fuel.
      It can be used to make copper(i) hydride.

  • Ted-X

    Please consider the toroidal shape of the reactor. I posted about it before, the principle would be to induce high currents (rather than eddy currents in a linear reactor), a sort of a welding transformer, where the nickel powder will behave as a single, short-circuited coil, with the heating coil also inducing the current in the nickel powder. Email me privately if you need more details or if you want more comments.

    • Bob Greenyer

      Thanks, I did consider it, we need to first get

      1. a reactor that shows evidence of working
      2. one that works for a reasonable amount of time

      Potentially understanding derived from our experimental work this year already in relation to published patents will help us achieve this.

  • Bob Greenyer

    What temperature should the Hot Cat start operating?

    []=Project Dog Bone=[]

    Added to the live doc.

  • Bob Greenyer

    I have added proposed turn on and operating temperature ranges for the E-Cat, E-Cat HT1 & HT2 based on the use of NaAlH4, Piantelli theory and the new reaction environment that was derived from the SEM images of the ‘Bang!’ reactor.

  • Bob Greenyer

    More clarity added to section

    “The Peak “just at the end” of Lugano”

  • Bob Greenyer

    I have JUST for the first time discovered/read the Focardi/Rossi paper from March 22, 2010 – it shows lack of understanding and serendipity in equal measure, but supports my hypothesis and adds more detail on reactions.

    I have added it to the live sheet.