Rossi Congratulates Carl-Oscar Gullström on Theoretical LENR Paper

A few days ago, Carl-Oscar Gullström, a doctoral student in the Department of Physics and Astronomy at Uppsala University, Sweden submitted a paper to E-Cat World for publication here, titled ” “Low radiation fusion through bound neutron tunneling.”

The paper can be read at this link:

https://www.scribd.com/doc/244393652/Low-radiation-fusion-through-bound-neutron-tunneling

Today, I posted a link to the paper, along with the abstract, on the Journal of Nuclear Physics to bring it to Andrea Rossi’s attention. Apparently, it has been well received by Rossi, who posted the following:

Frank Acland:
Please extend to Carl-Oscar Gullstrom my congratulations for his very intelligent paper. We are going through them with attention.
If Carl-Oscar Gullstrom contacts me, we can have an exchange of opinion.
Warm Regards,
A.R.

I will contact Mr. Gullström so he can get in touch with Andrea Rossi.

  • George N

    I only read the abstract of the theory, but it sounds like neutron tunneling is used to get past the columb barrier — neutron tunneling is an accepted concept in physics. May or may not be what is actually happening, but the important thing is that it is a testable theory that is within the confines of mainstream physics — hopefully this allows a few more mainstream scientists to take more interest in LENR and test this new theory. But then again, I am no scientist and I am not sure if the proposed theory is all within accepted concepts of physics…

    • Andreas Moraitis

      I don’t want to be nitpicking, but you should not say „neutron tunnelling“. Neutrons do not need to tunnel. It’s the “bound neutron” (a complex of a neutron and other nucleons, including at least one proton – that is, basically a nucleus) that tunnels into the Coulomb barrier, so that the neutron can be separated from the nucleus. This is also known as a “stripping reaction”. And you are right, this kind of reactions is, in contrast to many others that have been proposed, recognized by mainstream physics. Although there are some open questions (what is the role of the hydrogen, where do the missing neutrons come from…), Gullström’s concept could have a good chance to be accepted. By the way, the involved reactions have already been proposed by others, including Alan de Angelis on this blog.

      • GordonDocherty

        As soon as the word “nucleus” appears, then there are two nuclei – so, talk about “tunnelling” into the Coulomb barrier doesn’t strictly make sense. Both source (with neutron) and target nuclei must each have Coulomb barriers – otherwise all source nuclei would fuse in an instant and, as put eloquently elsewhere, “BOOM!!!”. So, we are back to matching (and possibly merging) Coulomb barriers. Now, I suppose one idea is that two barriers “in synch” now provide a uniform wave mechanism by which the neutron can move across – there is no discontinuous “wave wall” to bounce the neutron back. So:

        1. make a slow-moving (neutron (collapsing Hydrino + energy), bind it with a proton, wait for proton-neutron to approach a larger nucleus (in spin-synch) and, “after a phononic nudge”, the neutron transfers to the larger nucleus. In this case, we would still be left with Hydrogen ions.

        2. make a slow-moving (neutron (collapsing Hydrino + energy), bind it with a larger nucleus (Lithium, for example), let the larger nucleus approach a small nucleus (in spin-synch) and, “after a phonic nudge”, the neutron transfers to the smaller nucleus.

        3. don’t make any neutrons. Instead, use the Hydrogen Ions as carriers to move a neutron from lighter to heavier nucleii (so, Lithium to Nickel / Nickel to Nickel)

        4. combination of 1, 3 and possibly 2 above

        5. somehow allow neutrons to transfer between nuclei fixed in lattices to shake loose from one lattice, travel to another lattice, and there rebind. This last option looks decidedly unlikely in my mind, as there is no apparent transport mechanism or path, just an unstable nucleus that dumps its neutron – perhaps because there are more neutrons than protons (for example, Lithum-7). Back to “BOOM!!!” again… Still, never say never.

        No matter what the exact mechanism, however, it seems all are relying on the following:

        1. spin resonance between nuclei to allow Coulomb fields to at least link up in a stable manner (providing more of a ” ‘carrier’ wave in synch with no discontinuities between nuclei” to allow neutrons to oscillate along, rather than a tunnel)

        2. “vibrational” resonance between nuclei, again to allow Coulomb fields to at least link up in a stable manner (as distance between nuclei at least does not increase and relative (polar) axial alignment is preserved)

        3. some way of “shaking a neutron loose” from one nucleus, allowing it to travel to another “on the wave”

        4. source and target nuclei to be in close enough proximity to allow this to happen, while at the same time being far enough away from other nuclei so as to effect a clean transfer

        Of course, with all this said, where is the excess energy coming from? Is it because of a difference in the binding energies of a neutron between source and target nuclei, so that the net energy released in transmutations is very much greater than zero? Also, is the initial (IR) heat that is applied required to energize the source nuclei enough to allow the release of a neutron? And how does all of this fit in with the Brillouin Model and Brillouin Zones (if, indeed, it does)? Do they provide the potential wells in which transfer nuclei reside? Is, indeed, the purpose of the Hydrogen ions (protons) to allow transfer of neutrons from one, larger nucleus to another, even larger one? Is this why the Lithium-7 is consumed and the abundance of the Nickel-62 isotope increase? In which case, is LENR, at least in part, a “proton-as-transfer mechanism” mechanism within a larger electromagnetic structure (lines of force, perhaps in combination with “filaments of electrons”) that spin-aligns and induces resonance within the nucleii of the still larger crystal structures?

        Complicated, but at least such an explanation (intuitively) covers the experimental data from multiple, independent experiments covering supposedly independent phenomena…

        Anyway, I’d hate to be writing the math for that one (or, as this “problem” is layered, perhaps it is enough to look at each layer in isolation, saying that if the math for layer X holds, then we can assume these underlying conditions supporting layer X+1. After all, when working out the collision of two macro objects (such as two marbles), we don’t first go back to look at the individual atoms and how they were formed into the object. Instead, we take the object as a given… a sort of “object orientation approach” for the physical world…

        • I don’t catch all, but it seems that bounded neutron tunneling, imply that two protons don’t merge in NiH…

          it can explain maybe dd LENR fusion, but for NiH is have to involve a nucleaus containing neutrons?

          • Andreas Moraitis

            Indeed there are studies on „deuteron stripping“ with regard to Pd-D systems. According to Gullström, in Rossi’s reactor the lithium, and partly the 64Ni, do the job (= providing the neutrons).

          • GordonDocherty

            Not (necessarily) merging, rather Coulomb barriers do need to synchronize, to get:

            n n n
            u u u

            not:

            n nXn
            u X u

            that is, a reflective barrier between the Coulomb barriers surrounding the source and target nuclei.

            There is a tendency to draw Coulomb barriers like waves on the sea, so if you get high enough… but, of course, they are perturbations in a shell (think cell membrane). Now, to form a tunnel you would need to merge the two barriers completely (so, no more wave in tunnel), at which point the probability of fusion shoots up. If, on the other hand, you only go as far as two force fields “pulsating in synch”, then a “neutral” neutron can now cross the barrier without any risk of reflection or falling between the gap…

        • GordonDocherty

          One thing we do know. If you have:

          1. No metallic lattice (with correct geometry) – metallic means creation of electron cloud, of course
          2. No lattice saturation (with hydrogen)
          3. No heat
          4. No resonance (phononic and, likely, magnonic)

          you get no reaction…

        • Andreas Moraitis

          There are two nuclei with a repulsive Coulomb force between them, so there is no contradiction, I think. What you say about the lattice is interesting. One might expect that both Li and Ni nuclei/ions should not be able to move as freely as protons or deuterons. There is still a possibility that the reactions occur mainly at the surface, but even then one could especially ask how a 64Ni nucleus would approach a lighter Ni nucleus or vice versa.

        • Herb Gillis

          As a first layer approximation; Has anyone done calculations of the S1 radius of the hydrogen atom in tightly confined “box” (to approximate the conditions in a metal lattice)? Intuition suggests that the atom gets smaller when tightly confined. Is this intuition right? If so; how much smaller does it get? Perhaps small enough to increase probability of fusion? Could a tightly confined H atom be something like a “hydrino”? Seems to me this calculation must have been done somewhere. Its just the “particle in a box” problem- – where the particle is an H atom.

          • tlp

            Hydrino formation could cause part of the excess heat and these isotope transmutations other part (helped with those hydrinos). As Gullström calculated that much more nickel is needed to get that 1.5 MWh.

          • Andreas Moraitis

            I you would convert all atoms in 0.1 g hydrogen (5.9747*10^22) from the ground state to hydrino level 8, only 2.279 kWh of energy (857 eV per atom) would be released. Theoretically there are much deeper levels, but as far as I remember Mills says that they are only seldom reached. Besides, one might expect that the tiny hydrinos would escape quickly through the reactor walls or sealings before they can undergo further transformations.

          • tlp

            How do you predict that level 8? What if it goes all the way to 137? And there may be more hydrogen than 0.1 g. At least some water: ” The

            leads were reconnected and the cap sealed with a mixture of water and alumina powder cement.” (from the report)

          • Andreas Moraitis

            Level 137 should give about 679 kWh for 0.1 g, so theoretically it would be possible to reach the 1.5 MWh with somewhat more hydrogen. One would have to ask Mills if he thinks that transitions to such a deep level are realistic.

          • tlp

            Thanks for this calculation, I was not sure how to calculate it. So tiny part of the heat could come from transmutations, mostly from hydrino formation. And refueling the reactor might mean mainly to add that tiny amount of hydrogen in it. (In form of fuel powder and a little water)

          • Andreas Moraitis

            You can calculate the theoretical energy gain for a complete transition from the ground state to hydrino level n by this (hopefully correct) formula:

            (n^2 -1) x 13.598 eV

            See http://rvanspaa.freehostia.com/Hydrinos_explained.html .

            WolframAlpha (http://www.wolframalpha.com/ ) provides a convenient way to obtain the number of atoms, just type “number of atoms in 0.1 g hydrogen” into the text box. (For individual isotopes, you can use the symbols “1H”, “7Li” etc.)

          • Ophelia Rump

            When galaxies compress together do they get smaller?
            Since they are mostly empty space, they merge and begin to interact in ways which destabilize their normal routines.

  • George N

    I only read the abstract of the theory, but it sounds like neutron tunneling is used to get past the columb barrier — neutron tunneling is an accepted concept in physics. May or may not be what is actually happening, but the important thing is that it is a testable theory that is within the confines of mainstream physics — hopefully this allows a few more mainstream scientists to take more interest in LENR and test this new theory. But then again, I am no scientist and I am not sure if the proposed theory is all within accepted concepts of physics…

    • Andreas Moraitis

      I don’t want to be nitpicking, but you should not say „neutron tunnelling“. Neutrons do not need to tunnel. It’s the “bound neutron” (a complex of a neutron and other nucleons, including at least one proton – that is, basically a nucleus) that tunnels into the Coulomb barrier, so that the neutron can be separated from the nucleus. This is also known as a “stripping reaction”. And you are right, this kind of reactions is, in contrast to many others that have been proposed, recognized by mainstream physics. Although there are some open questions (what is the role of the hydrogen, where do the missing neutrons come from…), Gullström’s concept could have a good chance to be accepted. By the way, the involved reactions have already been proposed by others, including Alan de Angelis on this blog.

      • Mark Szl

        What do you make of the comments made by somebody named Peter Ekström? Post by SamMalle points to it.

      • fact police

        The proposed reaction *is* neutron tunneling. It’s just not tunneling through a Coulomb barrier, but through a nuclear force barrier. The neutron is bound to one nucleus, and requires energy to be freed from it — that’s the barrier. The bound state in the other nucleus is lower, so overall energy is released. Classic tunneling.

        Apart from the question of the probability of this reaction occurring, the problem with the theory is that it does not address the main objections to the observations: the lack of radiation and the absence of intermediate isotopes.

        It also admits problems with neutron and energy economy.

        Finally, the paper does not invoke any special conditions to explain the proposed reactions, and to explain why they have not been observed in the many nuclear experiments in all phases of matter, at all accessible energies, including copious experiments in metal hydrides for nuclear weapons and battery design.

        • Andreas Moraitis

          This usage of „tunnelling“ is new to me, but it may be so. Note that Gullström refers explicitly to the Coulomb barrier. Of course it is true that also the attraction of the strong force must be overcome. An interesting question is how exactly the interaction between the (still bound) neutron and the target nucleus takes part in this process.

          • fact police

            This usage of tunneling is however the oldest usage. The concept was first used to explain alpha decay, in which the barrier was the strong force. There is no Coulomb barrier in alpha decay; it helps along all the way.

            The mention of the Coulomb barrier in the paper is in reference to proton capture or proton fusion.

            The neutron in a bound nucleus has energy, and if a lower state is available in reasonable proximity, there is a finite probability it will tunnel to that state, depending on the barrier height, width, and the energy of the neutron. The real question is how the special LENR conditions influence this probability.

          • Andreas Moraitis

            As far as I understood it, in the case of alpha decay it is as well the Coulomb barrier that would prevent the ejection of the alpha particle. That’s why Gamov introduced the tunnel effect. It’s the same problem as in fusion, only the direction of the movement is opposite: from “inside to outside” instead of “outside to inside”.

          • fact police

            There is no Coulomb barrier in alpha decay. The Coulomb force between the alpha particle and the rest of the nucleus is repulsive. It tries to eject the alpha particle. The force that tries to keep it bound to the nucleus is the strong nuclear force. That’s the barrier it has to tunnel through.

          • Andreas Moraitis

            The range of the strong force is very short, therefore it will not retain the alpha particle any longer as soon as it has been separated. Indeed, the particle would be repelled from the nucleus if it were located at its periphery. But since it is usually positioned inside the nucleus, it will be retained by the surrounding protons – due to the Coulomb force, not the strong interaction. But I do not want to quarrel, I’m not a physicist and of course I may be wrong…

          • fact police

            First, to be clear, the alpha particle *is* on the periphery, and it only feels a repulsive force due to the Coulomb interaction. The strong force holds it in the nucleus, and, as you say is short range, but very strong. That’s why the barrier is narrow, allowing the alpha to tunnel through. But the barrier wall it faces — the attractive force — is the nuclear force. Without the nuclear force, there would be no resistance to decay (no bound state to begin with).

            Alpha decay only happens for nuclei above Z = 82, because that’s when the Coulomb force is large enough to make tunneling feasible.

            In spite of all this, the barrier is often called a “Coulomb barrier” in text books, even for alpha decay, though the careful ones call it a potential barrier. It’s often called a Coulomb barrier, because when you draw the total potential you have a deep well from the nuclear force, a high wall, and then a 1/r drop off from the Coulomb potential.

            Without the Coulomb force, it’s just a well, and particles in the well see an infinitely thick wall, not really a barrier. They have to attain energy equivalent to the well depth to escape.

            But with the Coulomb force, it’s more like a crater at the top of mountain. The walls are still due to the well (nuclear force), but now, on the outside it drops off, so the walls have a finite thickness, and it’s feasible to tunnel through. The Coulomb force is responsible for *reducing* the wall thickness, and that may be why it’s often called a Coulomb barrier, even in alpha decay. Of course, it’s the mountain slope that keeps other charged particles from entering the nucleus, so it makes sense to call it a Coulomb barrier for them.

            This question comes up at physics.stackexchange.com/questions/10863/tunneling-of-alpha-particles, though the answers are not completely clear.

            In any case, it doesn’t matter what interaction produces the potential barrier. The key for a tunneling interaction is that there is an energy state below the peak of the barrier a finite distance away, which is also below the energy of the particle. If that’s the case, there’s a finite probability for a particle to tunnel through the barrier. Those conditions are met for neutrons to tunnel from a Li nucleus to a Ni nucleus. The only question is what that probability is, and how does it compare to a dozen other possible Li + Ni interactions, many with much higher Q-values (none of which are observed in these conditions).

          • Andreas Moraitis

            Thank you for your detailed explanation. It is somewhat disappointing that many sources use the term “Coulomb barrier” apparently in a misleading way. In my own field of research, I have often found that you cannot rely on statements in textbooks, even if they come from the most renowned authors. Once printed, the errors circulate for decades…

          • Hello Officer (“fact police” – great alias!), I think you have won this particular debate but full credit to young Carl-Oscar for not being deterred by too much academic orthodoxy. However, the much more experienced Dr
            Edmund Storms is also not deterred and has just produced an entire
            book on his own theory of LENR, as detailed in the link below:

            http://www.infinite-energy.com/iemagazine/issue116/stormbook.html

            Briefly, Storms makes several basic assumptions and then justifies them in painstaking detail. These are:

            . The LENR process does not take place in a chemical lattice.

            . The LENR process takes place only in cracks of a critically small gap size.

            . All isotopes of hydrogen can fuse by the same basic process, with only the nuclear products being different.

            . The basic process removes energy over a period of time
            as photon emission. Most of this emission does not leave the apparatus.

            . The fusion process causes the transmutation reactions.

            . The overall process is consistent with all natural law and requires introduction of only one new process.

            . Cold fusion and hot fusion are not related in any way.

            I,
            for one, have already bought the Kindle version and, while parts of it are rather heavy going for a non-chemist like me, it really does look like
            Storms has done a Sherlock Holmes – like job and found a series of very non-intuitive paradigm shifts that can, at least, explain all the peer-reviewed observations and measurements on LENR so far published.

            That said, it would be great if you and Carl-Oscar were to
            read and comment on the Storms book and, in turn, if Edmund Storms were
            to read and comment on Carl-Oscar’s paper.

      • GordonDocherty

        As soon as the word “nucleus” appears, then there are two nuclei – so, talk about “tunnelling” into the Coulomb barrier doesn’t strictly make sense. Both source (with neutron) and target nuclei must each have Coulomb barriers – otherwise all source nuclei would fuse in an instant and, as put eloquently elsewhere, “BOOM!!!”. So, we are back to matching (and possibly merging) Coulomb barriers. Now, I suppose one idea is that two barriers “in synch” now provide a uniform wave mechanism by which the neutron can move across – there is no discontinuous “wave wall” to bounce the neutron back. So:

        1. make a slow-moving (neutron (collapsing Hydrino + energy), bind it with a proton, wait for proton-neutron to approach a larger nucleus (in spin-synch) and, “after a phononic nudge”, the neutron transfers to the larger nucleus. In this case, we would still be left with Hydrogen ions.

        2. make a slow-moving (neutron (collapsing Hydrino + energy), bind it with a larger nucleus (Lithium, for example), let the larger nucleus approach a small nucleus (in spin-synch) and, “after a phonic nudge”, the neutron transfers to the smaller nucleus.

        3. don’t make any neutrons. Instead, use the Hydrogen Ions as carriers to move a neutron from lighter to heavier nucleii (so, Lithium to Nickel / Nickel to Nickel)

        4. combination of 1, 3 and possibly 2 above

        5. somehow allow neutrons to transfer between nuclei fixed in lattices to shake loose from one lattice, travel to another lattice, and there rebind.

        This last option looks decidedly unlikely in my mind, as there is no apparent transport mechanism or path, just an unstable nucleus that dumps its neutron – perhaps because there are more neutrons than protons (for example, Lithum-7). Back to “BOOM!!!” again… Still, never say never.

        No matter what the exact mechanism, however, it seems all are relying on the following:

        1. spin resonance between nuclei to allow Coulomb fields to at least link up in a stable manner (providing more of a ” ‘carrier’ wave in synch with no discontinuities between nuclei” to allow neutrons to oscillate along, rather than a tunnel)

        2. “vibrational” resonance between nuclei, again to allow Coulomb fields to at least link up in a stable manner (as distance between nuclei at least does not increase and relative (polar) axial alignment is preserved)

        3. a way of “shaking a neutron loose” from one nucleus and allowing it to travel to another “on the wave between the nuclei”

        4. source and target nuclei being in close enough proximity to allow transfers to happen, while still being far enough away from other nuclei to effect a clean transfer – lattice geometry clearly playing an important part here

        Of course, with all this said, where is the excess energy coming from? Is it because of a difference in the binding energies of a neutron between source and target nuclei, so that the net energy released in transmutations is very much greater than zero? Also, is the initial (IR) heat that is applied required to energize the source nuclei enough to allow the release of a neutron? And how does all of this fit in with the Brillouin Model and Brillouin Zones (if, indeed, it does)? Do the Brillouin Zones provide the potential wells in which transfer nuclei can reside? Is, indeed, the main purpose of the Hydrogen ions (protons) actually to provide the transfer mechanism by which neutrons from one nucleus can transfer to another, larger one? Is this why in the e-Cat tests Lithium-7 was seen to be “consumed” while the ratio of Nickel-62 to other Nickel isotopes was seen to increase? In this case, is LENR, at least in part, using a “proton-as-transfer” mechanism within a larger electromagnetic structure (lines of force, perhaps involving “electron filaments / arcs”) that spin-aligns and induces resonance within the nucleii of the still larger crystal structures?

        True, such an explanation is complicated, but at least it (intuitively) covers the experimental data from multiple, independent experiments from Industrial Heat, Brillouin Energy and BlackLight, covering supposedly independent phenomena…

        Anyway, I would not like to be writing the math for such a complex environment – unless, as this “problem” is layered, each layer were examined and modeled in isolation. In that case, it could be said that if the math for layer X holds, then we can assume the underlying conditions A,B,C,.. hold to support layer X+1. After all, when working out the collision of two macro objects, we don’t first go back to look at the individual atoms and how they were formed into the object. Instead, we take the object as a given. So, maybe LENR+ is best approached and modeled using an “object orientation paradigm” and applying it to the physical world. It would also make writing computer simulations a heck of a lot easier…

        • Mark Szl

          I knew i liked the paper you posted earlier for a reason. Lol.

          Thanks.

        • I don’t catch all, but it seems that bounded neutron tunneling, imply that two protons don’t merge in NiH…

          it can explain maybe dd LENR fusion, but for NiH is have to involve a nucleaus containing neutrons?

          • Andreas Moraitis

            Indeed there are studies on „deuteron stripping“ with regard to Pd-D systems. According to Gullström, in Rossi’s reactor the lithium, and partly the 64Ni, do the job (= providing the neutrons).

          • GordonDocherty

            Not (necessarily) merging, rather Coulomb barriers do need to synchronize, to get:

            n n n
            u u u

            not:

            n nXn
            u X u

            that is, a reflective barrier between the Coulomb barriers surrounding the source and target nuclei.

            There is a tendency to draw Coulomb barriers like waves on the sea, so if you get high enough… but, of course, they are perturbations in a shell (think cell membrane). Now, to form a tunnel you would need to merge the two barriers completely (so, no more wave in tunnel), at which point the probability of fusion shoots up. If, on the other hand, you only go as far as two force fields “pulsating in synch”, then a “neutral” neutron can now cross the barrier without any risk of reflection or falling between the gap…

        • GordonDocherty

          One thing we do know. If you have one or more of:

          1. metallic lattice (with correct geometry) – metallic means creation of electron cloud, of course
          2. lattice saturation (with hydrogen)
          3. heat
          4. resonance (phononic and, likely, magnonic)

          missing, you get no reaction…

        • Andreas Moraitis

          There are two nuclei with a repulsive Coulomb force between them, so there is no contradiction, I think. What you say about the lattice is interesting. One might expect that both Li and Ni nuclei/ions should not be able to move as freely as protons or deuterons. There is still a possibility that the reactions occur mainly at the surface, but even then one could especially ask how a 64Ni nucleus would approach a lighter Ni nucleus or vice versa.

        • Herb Gillis

          As a first layer approximation; Has anyone done calculations of the S1 radius of the hydrogen atom in tightly confined “box” (to approximate the conditions in a metal lattice)? Intuition suggests that the atom gets smaller when tightly confined. Is this intuition right? If so; how much smaller does it get? Perhaps small enough to increase probability of fusion? Could a tightly confined H atom be something like a “hydrino”? Seems to me this calculation must have been done somewhere. Its just the “particle in a box” problem- – where the particle is an H atom.

          • tlp

            Hydrino formation could cause part of the excess heat and these isotope transmutations other part (helped with those hydrinos). As Gullström calculated that much more nickel is needed to get that 1.5 MWh.

          • Andreas Moraitis

            I you would convert all atoms in 0.1 g hydrogen (5.9747*10^22) from the ground state to hydrino level 8, only 2.279 kWh of energy (857 eV per atom) would be released. Theoretically there are much deeper levels, but as far as I remember Mills says that they are only seldom reached. Besides, one might expect that the tiny hydrinos would escape quickly through the reactor walls or sealings before they can undergo further transformations.

          • tlp

            How do you predict that level 8? What if it goes all the way to 137? And there may be more hydrogen than 0.1 g. At least some water: ” The

            leads were reconnected and the cap sealed with a mixture of water and alumina powder cement.” (from the report)

          • Andreas Moraitis

            Level 137 should give about 679 kWh for 0.1 g, so theoretically it would be possible to reach the 1.5 MWh with somewhat more hydrogen. One would have to ask Mills if he thinks that transitions to such a deep level are realistic.

          • tlp

            Thanks for this calculation, I was not sure how to calculate it. So tiny part of the heat could come from transmutations, mostly from hydrino formation. And refueling the reactor might mean mainly to add that tiny amount of hydrogen in it. (In form of fuel powder and a little water)

          • Andreas Moraitis

            You can calculate the theoretical energy gain for a complete transition from the ground state to hydrino level n by this (hopefully correct) formula:

            (n^2 -1) x 13.598 eV

            See http://rvanspaa.freehostia.com/Hydrinos_explained.html .

            WolframAlpha (http://www.wolframalpha.com/ ) provides a convenient way to obtain the number of atoms, just type “number of atoms in 0.1 g hydrogen” into the text box. (For individual isotopes, you can use the symbols “1H”, “7Li” etc.)

          • Ophelia Rump

            When galaxies compress together do they get smaller?
            Since they are mostly empty space, they merge and begin to interact in ways which destabilize their normal routines.

  • Ophelia Rump

    This is wonderful!

    Congratulations Carl-Oscar Gullström, I hope you get recognition for contributing to the advancement of science. I especially hope that your contribution turns out to be large.

    • georgehants

      This student must have one of those very rare tutors who actually allow a clever young man to practice open-minded science.
      Think how many other students are being still told that Cold Fusion is junk and telling them to only study the steam engine etc.
      Still lets celebrate this Wonderful example of good science.

      • Ophelia Rump

        Yes, I fully agree. A worthy student is not enough, there must also be a worthy environment. Like planting a seed.

  • Ophelia Rump

    This is wonderful!

    Congratulations Carl-Oscar Gullström, I hope you get recognition for contributing to the advancement of science. I especially hope that your contribution turns out to be large.

    • georgehants

      This student must have one of those very rare tutors who actually allow a clever young man to practice open-minded science.
      Think how many other students are still being told that Cold Fusion is junk and advising them to only study the steam engine etc.
      Still lets celebrate this Wonderful example of good science in action.

      • Ophelia Rump

        Yes, I fully agree. A worthy student is not enough, there must also be a worthy environment. Like planting a seed.

  • SamMalle
    • Ophelia Rump

      I don’t think that man likes any new science. Some people are only cut out to be custodians of the previous work of others. These high janitors of science have a strong sense of obligation to maintain what they have been handed down and dedicate themselves to it’s maintenance and preservation. People who come along and try to leave their mark upon science are viewed as vandals. They particularly disdain to be confronted by the ideas of mere graduate students and patent clerks.

    • Peter Ekström actually seams to accept Gullströms calculations. He just doesn’t accept the E-cat becuase of no radiation.

      • Andreas Moraitis

        100 keV gammas are very weak. They can also be observed in the natural background radiation. If the reactor walls were able to shield most of them, the measurable amount could possibly lie in the inconspicuous range. One would have to do the math in order to demonstrate that, though.

      • tlp

        Swedes are talking (writing) to each other!

        cogullstrom wrote: Hi

        Sorry if the document is a little unclear. But the idea that I have not found any annanatans is to be able to get access to nuclear energy without gamma radiation through the use of bound neutrons. The reason is that a free neutron is about 8 MeV above the ground state while a bound is much closer. A free neutron then goes into a state existed about 8 MeV above the ground state and deexiteras to ground state again that emit gamma rays treatment in MeV region. But the first excited state of such ni62 is 1 MeV above the ground state while a bound neutron from such ni64 only is 0.9 Mev above. only possibility is then that the energy is released as kinetic energy to the nuclei. The problem is the plastic enough nickel cores close to each other. The solution then is to use protons as bridges that move the bound neutrons between nuclei. The energy would then end up with the proton which both can be used now neutronbro o and to distribute the energy. 1 MeV proton goes just mickrometer in nickel and max mm in air. The release additionally emit little energy itaget so that no gamma rays arise.

        Hey Carl-Oscar!

        Welcome to the forum! Nice to talk to someone who does not hide behind a cryptic signature!

        Much of what you say is right! My main objection to this theory is that it is incomplete. It explains not all observations.

        You can consider 7Li + 58Ni -> 6Li + 59Ni as a neutron transfer reaction. Some of the excess energy can then be kinetic energy
        of the cores (ignoring, of course, from the Coulomb repulsion prevents the nuclei to come close enough to each other). But we have
        (at least) four problems that the article does not discuss.

        1 It should excite other states than the ground state giving gamma. There is nothing special about the ground state.

        2 One has to pass several nickel isotopes on the road to 62Ni. These should be in isotopic analysis.

        3 One has to explain why the process stops at 62Ni.

        4 Q of 64Ni + 7Li -> 6Li + 65Ni is negative. How can it be that 64Ni disappear in the ashes?

        Point 3 is easy: Q value for 62Ni + 7Li -> 6Li + 63Ni is negative.

        I use my calculator Reaction on http://nucleardata.nuclear.lu.se/database/masses/
        for calculating Q values​​.

        For non-nuclear physicist, I apologize because I use nuclear physics-speak. I know that the C-O know what I mean.

      • Curbina

        Are you sure that Ekström is not being sarcastic when he talks about a Nobel for Rossi and Gullström? Reading the forum google translated surely I get the sensation that Ekström thinks Gullström is a moron for even delving into the idea of LENR being possible. Perhaps much is lost in the translation, anyway.

        • Curbina

          this quote from Ekström says to me he is just being a smart ass:

          “Since Gullström not say anything about how gamma radiation is to heat is the best rating the article can get: incomplete.

          The article will, even though it has no scientific value, to make the intended benefit. Most will uncritically swallow it with and sinker. There are already plenty of articles out there that allegedly explain LENR. One worse than the other!”

  • SamMalle
    • Mark Szl

      Yup just took a look. Thanks.

    • Ophelia Rump

      I don’t think that man likes any new science. Some people are only cut out to be custodians of the previous work of others. These high janitors of science have a strong sense of obligation to maintain what they have been handed down and dedicate themselves to it’s maintenance and preservation. People who come along and try to leave their mark upon science are viewed as vandals. They particularly disdain to be confronted by the ideas of mere graduate students and patent clerks.

    • Peter Ekström actually seams to accept Gullströms calculations. He just doesn’t accept the E-cat becuase of no radiation. So his evaluation of the report was: “incomplete”

      • Andreas Moraitis

        100 keV gammas are very weak. They can also be observed in the natural background radiation. If the reactor walls were able to shield most of them, the measurable amount could possibly lie in the inconspicuous range. One would have to do the math in order to demonstrate that, though.

      • tlp

        Swedes are talking (writing) to each other!

        cogullstrom wrote: Hi

        Sorry if the document is a little unclear. But the idea that I have not found any annanatans is to be able to get access to nuclear energy without gamma radiation through the use of bound neutrons. The reason is that a free neutron is about 8 MeV above the ground state while a bound is much closer. A free neutron then goes into a state existed about 8 MeV above the ground state and deexiteras to ground state again that emit gamma rays treatment in MeV region. But the first excited state of such ni62 is 1 MeV above the ground state while a bound neutron from such ni64 only is 0.9 Mev above. only possibility is then that the energy is released as kinetic energy to the nuclei. The problem is the plastic enough nickel cores close to each other. The solution then is to use protons as bridges that move the bound neutrons between nuclei. The energy would then end up with the proton which both can be used now neutronbro o and to distribute the energy. 1 MeV proton goes just mickrometer in nickel and max mm in air. The release additionally emit little energy itaget so that no gamma rays arise.

        Peter Ekström wrote:
        Hey Carl-Oscar!

        Welcome to the forum! Nice to talk to someone who does not hide behind a cryptic signature!

        Much of what you say is right! My main objection to this theory is that it is incomplete. It explains not all observations.

        You can consider 7Li + 58Ni -> 6Li + 59Ni as a neutron transfer reaction. Some of the excess energy can then be kinetic energy
        of the cores (ignoring, of course, from the Coulomb repulsion prevents the nuclei to come close enough to each other). But we have
        (at least) four problems that the article does not discuss.

        1 It should excite other states than the ground state giving gamma. There is nothing special about the ground state.

        2 One has to pass several nickel isotopes on the road to 62Ni. These should be in isotopic analysis.

        3 One has to explain why the process stops at 62Ni.

        4 Q of 64Ni + 7Li -> 6Li + 65Ni is negative. How can it be that 64Ni disappear in the ashes?

        Point 3 is easy: Q value for 62Ni + 7Li -> 6Li + 63Ni is negative.

        I use my calculator Reaction on http://nucleardata.nuclear.lu.se/database/masses/
        for calculating Q values​​.

        For non-nuclear physicist, I apologize because I use nuclear physics-speak. I know that the C-O know what I mean.

      • Curbina

        Are you sure that Ekström is not being sarcastic when he talks about a Nobel for Rossi and Gullström? Reading the forum google translated surely I get the sensation that Ekström thinks Gullström is a moron for even delving into the idea of LENR being possible. Perhaps much is lost in the translation, anyway.

        • Curbina

          this quote from Ekström says to me he is just being a smart ass:

          “Since Gullström not say anything about how gamma radiation is to heat is the best rating the article can get: incomplete.

          The article will, even though it has no scientific value, to make the intended benefit. Most will uncritically swallow it with and sinker. There are already plenty of articles out there that allegedly explain LENR. One worse than the other!”

  • hornster

    Mr. Rossi has humbly stated that he can not or will not disclose the reaction, but from all I can find this is new science. I have also kept a open mind and read the the opinions of those who know the accepted versus what may me.

  • Gerard McEk

    Thank you Carl-Oscar. I very much welcome scientists who seriously think how things can happen which cannot be understood by the regulary laws and understandings of fysics, such as ‘Cold Fusion’. Would it be possible to explain your quite technical paper to lay-men? Thank you

  • Gerard McEk

    Thank you Carl-Oscar. I very much welcome scientists who seriously think how things can happen which cannot be understood by the regulary laws and understandings of fysics, such as ‘Cold Fusion’. Would it be possible to explain your quite technical paper to lay-men? Thank you

  • Grek

    Good work Frank!

  • Andreas Moraitis

    This usage of „tunnelling“ is new to me, but it may be so. Note that Gullström refers explicitly to the Coulomb barrier. Of course it is true that also the attraction of the strong force must be overcome. An interesting question is how exactly the interaction between the (still bound) neutron and the target nucleus takes part in this process.

  • Francisnocab2
    • Francisnocab2

      Briefly, the author of this analysis argues that the waveforms shown in some of the ammeter photographs provided in the last two reports seem inconsistent with what would be expected given Rossi’s description of his setup. Critically, the author found that the waveforms shown would be consistent if a couple of leads had been reversed on one of the ammeters. If such an error had been made in setup, the total electrical power delivered to the apparatus would have been underreported roughly by a factor of three: about the same coefficient of performance claimed in the recent energy catalyzer paper.

      • Obvious

        Where does the power come from then, if it is three times higher than measured coming from the wall outlet?

        • US_Citizen71

          Rossi’s magic powers. All pathoskeptics know about Rossi’s magic powers. When he is the room the laws of physics change so that is why he can’t be allowed anywhere near a test.

          • Obvious

            Apparently. If the incorrectly attached probe “logic” is followed to the end, then Rossi can create a box that creates 3X the supply of electrical power than used for input, but then hides the fact that he is wasting it to make heat by intoning a spell that makes the professors hook the reactor clamp connections incorrectly (every time), and further obfuscates the test by making a device that makes it hard to measure the actual heat production to distract the professors from the magical electrical control box 3X power trick.

          • yes you should write Chuck Norris -style jokes on Rossi…
            could be funny.

            It remind me those cargo cult skeptic who imagined that cold fusion experiments were caused by energy storage…
            those weak mind never tried to file a patent for energy storage at the highest known density … they could be billionaire with their claimed artifacts technology.

          • Obvious

            I think there might be some more fun left in the “measurement problem(s)” conundrums.

            For example, if there WAS 3X more power going in to the resistors, then the heat measurements and calculations must be correct.

            In kind, if the heat measurements/calculations are wrong, then 3X the power input theory cannot be correct.

            Otherwise the new Rossi trick would be to make some amount of energy disappear from the system without a trace.
            This should pit the “clamp to resistor problem” folks against the “heat measurement” problem folks. Let the battle of wits begin…..

          • Omega Z

            So if I understand,
            Your saying they can’t have it both ways.
            Just take all the fun out of it why don’t you. 🙂

            It’s freezing in here. Turn down the heat.

          • bachcole

            I can’t fully appreciate the humor because I deliberately avoid subjecting myself to the moronic maliciousness of the skeptopaths. (:->)

        • AlbertNN

          The power comes from the wall outlet. The hypothesis is that both the measurements of the power into the box and out from the box were made in the same way, with the same error.

          • Obvious

            The same error cannot be repeated at both sides of the control box, since the input AC phases (from the wall) are not split, therefore the probe would simply report a negative current to the meter, but the correct magnitude of current regardless. The resistor part of the “problem” applies only when a true, 120° separated 3 phase power is being measured at a split between the sides of a delta resistor load. The logic of the split delta power measurement cannot be applied to the main power feed measurement. There are only opinions on whether normal, 120° separated, 3 phase power is being fed to the resistors from the control box. If the power from the control box is rectified, phase shifted, or otherwise altered from the “normal” 3 phase AC, then the equations used to find the different values are no longer applicable, and the hypothetical results are just wild guesses.

          • AlbertNN

            I am sorry, but you analysis is not correct. It is the same three phases that are measured both at the input and output of the control box. This gives that the same currents are measured at both points, the only difference is that the voltage will be a continuous sinus at the input, and pulsed at the output.

          • Obvious

            I think I was confusing the argument of the split resistor feed (Kirchhoff law discussion) with your idea. In your version, the probe is connected to the pre-split part of the wiring to the delta resistor (as in the report diagram). Is that right?

          • AlbertNN

            Correct, but also correct for the discussions on Kirchoff’s law that I have seen here. The report states that the current is split equally after the split point C in figure 4, but this is not correct in a 3-phase system, according to Kirchoff’s laws.

          • Thomas Kaminski

            Not so. Your assumptions of three phase are wrong. A three-phased triac pulsed controller with less than 120 degree firing angle will split the currents equally between line and delta load.. There are never three active connections simultaneously. Only two phases are connected at any given time.

          • AlbertNN

            No. If we assume that all three resistors are equal, one path will have a resistance of R, and the other one of 2R when only two phases are connected. This does not give equal currents, either in reality or according to Kirchoff.

          • Obvious

            If I understand this issue correctly, the C2 leads can conduct current from multiple phases, and therefore conduct current more often than the C1 leads. The C2 leads act as extensions to the Delta pattern. So at the maximum phase separation of 120 degrees, the square root of 3 rule applies, but still only affects the calculations of Joule heating of the cables . So the worst case scenario is that the secondary (C2) cables consume 13.97 W, rather than 1.6 W, in the dummy run. Which is still insignificant compared to the resistor heating in the control (unloaded) reactor output.

            The Joule heating of the cables is simply a subtraction from the measured input to the reactor, as transmission loss. The remainder of the power (Control box output minus Joule heating) is considered to be the power consumed by the reactor. So using the maximum C2 heating loss, even tripling the Joule heating values reported for the loaded reactor, the effect is to raise COP slightly.

            So the 1/2 current usage actually is a far more conservative measurement of possible Joule heating loss. And more appropriate when the controller pulses are delivering considerably less than 120 degrees of current flow, as shown in figure 5.

          • Thomas Kaminski

            The square root of three applies only if:

            1). There delta load is symmetric (all legs the same)
            2). The load is “linear” (resistive, inductive, capacitive components)
            3). The three-phase power is sinusoidal and equal in magnitude, but separated in phase by 120 degrees.
            4). The Three phase lines are always attached.

            With a triac controller, you can think of each phase of the power source being “connected” when triac is conducting and “disconnected” when the triac is not conducting. For small conduction angles, only two phases of the supply are connected at any time. The third phase is not connected. Over one cycle of 50 Hertz power with three-phase power lines L1, L2, L3, the following connections occur with a resistive load (assuming the heaters are resistive):

            1) L1 to L2 with a load of R in parallel with 2*R
            2) L2 to L1 with a load of R in parallel with 2*R
            3) L1 to L3 with a load of R in parallel with 2*R
            4) L3 to L1 with a load of R in parallel with 2*R
            5) L2 to L3 with a load of R in parallel with 2*R
            6) L3 to L2 with a load of R in parallel with 2*R

            The instantaneous current that flows is a simple Ohm’s Law calculation at each instant in time. The magnitude of the RMS current will not have a simple square-root-of-three relationship for Line to Phase current. It will be computed by integrating a portion of the sine function over the conduction angles for each of the above cases.

          • Obvious

            Perhaps you can answer this, please.

            If the load is a balanced resistive load, when the amp probe on one phase is reading 19.7 A, and the other two phases also read the same for all three phases, how much average (RMS?) current is entering (being used by) the system:

            With full 3 phase (100% conduction)?
            With a low conduction angle pulse centered on the peak voltage (Maybe 30°)?

            Is it 19.7A (the average of all three phases?).
            19.7 A because what goes in must come out.
            Zero because what goes in must come out.
            59.1 A (the sum of the currents in three phases).
            34.1 A (the square root of the sum of the squares of the three currents)
            34.1 A (19.7 times the square root of three)

          • Thomas Kaminski

            This is not an easy question to answer. I assume that the meters measured 19.7 A RMS. The actual current waveforms are complex, chopped portions of sinusoids. To calculate the power, you also need to know the voltage applied. Power is simply the RMS Volts times the RMS amps. You can either measure the Line-to-Neutral volts times the Line Amps and then multiply by three for the total power. Alternately, you can multiply the Line-to-Line RMS Voltage times the Delta Load (each leg) RMS Current and then multiply by three to get the total power. Both should give you the same answer. What was not reported in the paper was the measured RMS voltage. They reported 19.7 amps (RMS) to calculate the Joule heating in the wires ( R*I^2), but they did not report the RMS voltage of either Line to Neutral or Line-to-Line.

            Here is an Open Office spreadsheet that shows the complex waveforms that happen with a similar triac circuit where the triacs are in series with the Delta load resistors. See if you can identify the RMS voltages and RMS currents.

            https://drive.google.com/file/d/0B0I7pqe_KM9taXNfdUlaaU5GVnc/view?usp=sharing

          • Obvious

            Thanks. The file came in pieces, but I get the gist of it.
            I’ll see if I can open it better.
            The voltage is probably near the maximum for each phase, based on the waveform drawn on the paper in the new bigger Vessy photo.

          • Thomas Kaminski

            Download it and run OpenOffice. Look at the plots. They are interesting.

          • For long this electric discussion made me yawn as we missed too many data, but assuming the I1eff+I2eff=I3eff

            it became clear that more or less I1 and I2 are synchronous, they are the same phase…

            It is not at all 3phase controller driving a 3phase load, but a 3phase battery of 6 dimmers driving 3 e-cat with each 2 coils…

            this explains the photography that recently appeared on Vassy’s site.

            why 2 coil ? maybe to allow some balancing effect to make the cat purr…

            http://www.lenr-forum.com/forum/index.php/Thread/867-Is-Lugano-E-cat-simply-a-single-phase-load-with-two-coils/?postID=1836#post1836

            just an idea…
            basically I trust Wattmeter, and human stupidity only when this defend the consensus and follow the group opinion.

            What make me nearly furious is that the real problem is that E-cat was not correctly calibrated and this cast a doubt…

          • Thomas Kaminski

            Wrong again, though partly correct. The current does split r, 2r, but when conducting L1 to L2 it splits 1 to 2 with one init of current through r and 1/2 unit of current through 2r . When conducting 1 to 3, the currents are 1/2 unit through 2r and 1 through r. L1 sees three “units” of current and both resistors see 1 and 1/2 units of current. Still a 2 to 1 split, just at different times.

          • Obvious

            Please examine the photo of the experimental set up (Lugano Report page 4). It shows the probes connected in the correct direction. I have downloaded and read both the PCE830 manual, and the Compact Fusion Three Phase Power Controller manuals. The correct probe orientation is current flowing from front to back of the PCE830 amp probes. You can even see that the I3 line is connected to the far right load port on the controller by comparing the current probe ports in the PCE830 manual with the colors in the photo (blue). With the clamp opening lever on the left, the front of the probe is up. See pages 1, 11, and 12 of the PCE 830 manual. The load side of the power supply is on the bottom of the Controller box (facing the camera in the page 4 photo). See page 13 of the CCI Compact Fusion Operator Manual. Current flowing from the controller therefore flows from the box to the reactor, so the current is flowing through the probe from front to back (correctly).

            Also, this nullifies the Kirchhoff argument, since the probes are connected to the pre-split feed to the reactor delta, as is shown in the diagram on page 5 of the Lugano report.

          • AlbertNN

            Then your version of the report is better than mine. The compression artefacts in the version I have makes it impossible to make such a judgement. And we do not know if the PCE830 in figure 5 was connected to the input or output of the control box.

            Kirchoff’s law is equally valid at the input and the output of the control box. The analysis done with it does take the placement according to figure 4 into account.

          • Obvious

            OK, so going through the math so far, using formula in the report, I recalculated the Joule heating of the supply cable secondary leads to 13.97 W, rather than 1.6 W, resulting in a total supply cable Joule heating of 19.07 W. (From the Dummy test part). So that carries over to the page 20 totals, resulting in an “Actual Input” of 467 W instead of the former 479 W value, actually improving the dummy run input to output comparison.

            Now I’m digging through the rest of the math to see where else it affects values downstream.

    • Obvious

      So…. assuming that the clamp on the “resistor” lead was positioned wrong, the apparently magic control box somehow multiplies the input power by 3, since the input power was also measured between the building electrical outlet and the control box, as well as the circuit from the control box to the “resistors”, and these figures agree, minus power consumed by the control box (which is within manufacturer specs)?

      • you make a very good point that not only they did the mistake 2 times,
        they did if after the calibration runs which was rightly done

        some magic allow that to produce a COP of more than 3 despite the tric which gives 3…

        in fact I suspect they understestimate their trick.
        with a V setup as it seems, investing the common phase of the V make power exactly zero… with some swap you can rotate phase (it seems it is their hypothesis)…

        believe me as electronic engineer, once you imagine the testers can invert clamps, forget to merge power, and notice nothing, you’d better forget about science…

        those guys simply imagine a conspiracy of stupidity not top admit they are wrong.
        a classic scheme.

        • AlbertNN

          The only problem is that the photo of the power meter in the report shows that one of the current probes connected to it were inverted when the photo was taken.

          • Obvious

            To me, the image looks like one complete cycle of one phase of AC, heavily clipped by pulse modulation.

          • AlbertNN

            Please read the analysis of the picture that is linked above by Francisnocab2.

          • not necessarily, as it can also be a trigger problem.
            If the photo is the one after the triac, the voltage is not sinusoidal. if before there can be some detail on trigering we ignore…

            moreover whatever you imagine, that mistake is unavoidable… as I say, it is like choosing the wrong seat in a car and trying to drive.

            note also that the conspiracy assume that
            1- the error was done on the two current probes of the front and middle PCE380
            2- it was not done with the dummy, and the two powermeter were OK
            3- which implies that a front PCE380 clamp, which had no reason to me moved was moved in the wrong direction
            4- note also that the controlbox consume some energy and that some weird effect would be visible is clamp were misplaced

            5- it assume that Industrial heat provided devices that don’t work, hoping that an unbelievable incompetence and improbable luck would save them from ridicule and bankruptcy.

            that is extraordinary assumption. you need evidence, and not just photo without details commented by people who are clearly not experienced in managing oscilloscope, powermeter, and even basic power electronics (like assuming sinusoidal for switched controller, assuming no feedback, assuming no resistance change).

          • AlbertNN

            1. Or they did think that this was the correct way to connect the probes
            2. Not necessarily done with the dummy, as we do not know if the heating then was done with three phases. It could only be done with two, omitting the wrongly connected one.
            3. See 2.
            4. Only if you know exactly how much the control box was consuming beforehand.
            5. They might think that it works, see 1.

            What I would appreciate is more documentation, and some comments from the authors on the issue. But they have been very silent this far.
            The analysis linked above is not assuming anything on your list. Or can you point out any error in it?

  • gdaigle

    Could Carl-Oscar Gullström’s proposition be used as a predictive model of which metal lattices might more easily achieve bound neutron tunneling? Now that Pd-D and NiH in lattices have been demonstrated to work, are there other lattices in which the Coulomb Barrier can be even more easily overcome? How about other metal hydrides? Other ferromagnetic materials such as iron? In conditions designed for quantum tunneling (ie. in tunnel diodes)? In Type 1 metal superconductors?

    • Ophelia Rump

      I think you are asking the big question.
      The future turns on those answers.

  • bachcole

    Cool. Very cool. Ultra cool.

  • bachcole

    Cool. Very cool. Ultra cool.

  • GordonDocherty

    According to Carl-Oscar Gullström’s excellent paper “Low radiation fusion through bound neutron tunneling”, and with explicit reference to the Lugano experiment, neutrons are transferred between nuclei such that energy is released as neutrons tunnel between two potential wells created by two nuclei. Here, he states that bound neutron tunneling should be considerably larger than Coulomb barrier tunneling.and give ground state – ground state interaction if the neutron energy levels are about the same in the two nuclei involved in the transition (source and target). He then goes on to consider tunneling probabilities, and put forward an excellent case for what was seen at Lugano.

    The only question then is “what role does the hydrogen play”, or, equivalently, “why does this only work when the lattice is saturated with hydrogen?”. Also, the melting point of Lithium is stated as 1500C – so, how does the Lithium get so hot in the first place? In the case of the e-cat, I believe the answer is in the hydrogen playing a pivotal role in transferring neutrons – neutron carriers, as it were. So, instead of:

    Li -> Ni

    we see

    Li -> H…H -> Ni.

    Of course, as the lithium heats up, it turns first to liquid (at 180.54 C) and then it boils (at 1342 C), so there should be a step change in energy output as the temperature crosses 1342 C – more lithium atoms are now free to interact with the hydrogen AND direct with the nickel. I’m assuming, of course, that the lithium is not instantly heated from solid to gas. Still, the e-cat does start “working” above around 180 – 190 C (which may have something to do with the release of hydrogen from hydride “impurities” in the nickel lattices), but does reflect the operational characteristics of the e-Cat.

    So, no matter what, the hydrogen cannot be written out…

    • ecatworld

      Thank you, Gordon. Can you explain why you say ‘the melting point of Lithium is stated as 1500C’ and then later say ‘Of course, as the lithium heats up, it turns first to liquid (at 180.54 C) and then it boils (at 1342 C), ‘

      Many thanks,

      Frank

      • GordonDocherty

        The paper says 1500 C – other sources say 1342 C (actually 1347 C or 1620 K – didn’t have my glasses on properly), the latter being for pure lithium of course

        • GordonDocherty

          I’m just bothered by the role of the hydrogen – after all, without the hydrogen, no reaction occurs…

          • Axil Axil

            Nickel and lithium act separately. Lithium and hydrogen combine as the major reaction mechanism at these high reactor temperatures.

        • LCD

          You mean boiling point?

          • GordonDocherty

            Yep – wow, a really bad day yesterday

          • LCD

            Lol

        • Ted-X

          Just please consider that there will be a PARTIAL vapor pressure of the Lithium present in the ECAT.

          • GordonDocherty

            True – but, then, this would likely be nowhere near enough to generate the heat seen, and while the partial pressure will increase as the temperature increases, something has to… but, wait, maybe that is what the heaters are for. So, maybe less hydrogen is needed, if there is a plentiful enough supply of “free” Lithium atoms… Was this also the mechanism in all those exploding Lithium batteries…? It still looks likely that Hydrogen is required, however, for a really long-lived energy release and a “decent” COP

          • georgehants

            Igor, sorry have only just seen your comment, so am adding this reply very late.
            I feel sure that only the usual unsavory attacks etc are removed on this site, except those against me, that I do not mind as it is good to show-up some of these unpleasant scientists for what they are.
            Better to moderate than have pages full of crazy scientists denying and debunking everything beyond a steam engine, don’t you agree?

    • Dods

      I also remember reading that the Debye temperature of Nickel might have something to do with the reaction which is 177 C.

    • Bob Greenyer

      On 10.10.14 I said on MFMP FB

      “Rydberg state H helps in Neutron transport from Li7 > Ni”

      That was assuming H was in Rydberg state.

      According to Stoyan, Rydberg state H has 630 times the magnetic moment of H. That is over 2 orders of magnitude more sensitive to any magnetic stimulation.

      H in the lattice may provide screening in the same way that D has been proved to do in palladium.

      • tlp

        It may also be fractional Rydberg state H, called hydrino by Mills. And I think there is more hydrogen than just in the fuel. At least some water: ” The

        leads were reconnected and the cap sealed with a mixture of water and alumina powder cement.”

      • Axil Axil

        Most hydrogen will form hydrogen nano-crystals called Rydberg matter as a condensate of low temperature hydrogen plasma. The atom count of this type of crystal will be less than 100. The core of this crystal will be electron depleted with the exterior covered with orbiting electrons.

    • Ted-X

      The vapor pressure of the lithium in the reactor may still be too low to boil at 1342 deg. C, as the pressure inside the ECAT is higher than atmospheric (due to hydrogen present , presumably from LiAlH4). However, Lithium will have SOME VAPOR PRESSURE starting from 180 deg. C. The onset of the LENR at 180 deg. C seems to be correlated with SOME Lithium being evaporated (having SOME, although relatively low, vapour pressure). The vapor pressure of Lithium is increasing gradually with the temperature and can be estimated.

    • Axil Axil

      The are no neutrons in the hydrogen isotope that the Rossi reactor uses.

  • GordonDocherty

    According to Carl-Oscar Gullström’s excellent paper “Low radiation fusion through bound neutron tunneling”, and with explicit reference to the Lugano experiment, neutrons are transferred between nuclei such that energy is released as neutrons tunnel between two potential wells created by two nuclei. Here, he states that bound neutron tunneling should be considerably larger than Coulomb barrier tunneling.and give ground state – ground state interaction if the neutron energy levels are about the same in the two nuclei involved in the transition (source and target). He then goes on to consider tunneling probabilities, and put forward an excellent case for what was seen at Lugano.

    The only question then is “what role does the hydrogen play”, or, equivalently, “why does this only work when the lattice is saturated with hydrogen?”. Also, [correction applied] “the operating temperature of 1500C is above the boiling point of Lithium” – but, how does the reactor get so hot in the first place? In the case of the e-cat, I believe the answer is in the hydrogen playing a pivotal role in transferring neutrons – neutron carriers, as it were. So, instead of:

    Li -> Ni

    we see

    Li -> H…H -> Ni.

    Of course, as the lithium heats up, it turns first to liquid (at 180.54 C) and then it boils (at 1342 C), so there should be a step change in energy output as the temperature crosses 1342 C – more lithium atoms are now free to interact with the hydrogen AND direct with the nickel. I’m assuming, of course, that the lithium is not instantly heated from solid to gas. Still, the e-cat does start “working” above around 180 – 190 C (which may have something to do with the release of hydrogen from hydride “impurities” in the nickel lattices), but does reflect the operational characteristics of the e-Cat.

    So, no matter what, the hydrogen cannot be written out…

    • Frank Acland

      Thank you, Gordon. Can you explain why you say ‘the melting point of Lithium is stated as 1500C’ and then later say ‘Of course, as the lithium heats up, it turns first to liquid (at 180.54 C) and then it boils (at 1342 C), ‘

      Many thanks,

      Frank

      • GordonDocherty

        1342 C (actually 1347 C or 1620 K – didn’t have my glasses on properly) is the boiling point. Now, rereading the paper, it says “operating temperature of 1500C is above the boiling point of Lithium”) – so, many thanks for pointing this out – another reason why it’s a good idea to discuss things out in the open 🙂 The main point remains, however – the system first has to heat above 1347C if the Lithium atoms are to become mobile… unless, of course, as a liquid, lithium atoms come into contact with the Nickel lattice. Still, this does leave the role of the hydrogen unanswered. Is it simply to “fill the holes”? Such a role seems a bit passive. Also, if it is just a case of Lithium ions sitting next to Nickel ions, why does the Nickel have to be formed in the way it is – wouldn’t a flat surface do just as well? I understand that the nickel, as formed, presents a larger surface area, but the idea of Nickel and Lithium atoms just ending up side-by-side and interacting is clearly not right – there is clearly more going on than that, as the paper states when mentioning spin-orbit and 3-D properties for tunneling probabilities.

        • GordonDocherty

          I’m just bothered by the role of the hydrogen – after all, without the hydrogen, no reaction occurs…

          • Axil Axil

            Nickel and lithium act separately. Lithium and hydrogen combine as the major reaction mechanism at these high reactor temperatures.

        • LCD

          You mean boiling point?

          • GordonDocherty

            Yep – wow, a really bad day yesterday

          • LCD

            Lol

        • Ted-X

          Just please consider that there will be a PARTIAL vapor pressure of the Lithium present in the ECAT.

          • GordonDocherty

            True – but, then, this would likely be nowhere near enough to generate the heat seen, and while the partial pressure will increase as the temperature increases, something has to… but, wait, maybe that is what the heaters are for. So, maybe less hydrogen is needed, if there is a plentiful enough supply of “free” Lithium atoms… Was this also the mechanism in all those exploding Lithium batteries…? It still looks likely that Hydrogen is required, however, for a really long-lived energy release and a “decent” COP

    • Dods

      I also remember reading that the Debye temperature of Nickel might have something to do with the reaction which is 177 C.

    • Bob Greenyer

      On 10.10.14 I said on MFMP FB

      “Rydberg state H helps in Neutron transport from Li7 > Ni”

      That was assuming H was in Rydberg state.

      According to Stoyan, Rydberg state H has 630 times the magnetic moment of H. That is over 2 orders of magnitude more sensitive to any magnetic stimulation.

      H in the lattice may provide screening in the same way that D has been proved to do in palladium.

      • tlp

        It may also be fractional Rydberg state H, called hydrino by Mills. And I think there is more hydrogen than just in the fuel. At least some water: ” The

        leads were reconnected and the cap sealed with a mixture of water and alumina powder cement.”

      • Axil Axil

        Most hydrogen will form hydrogen nano-crystals called Rydberg matter as a condensate of low temperature hydrogen plasma. The atom count of this type of crystal will be less than 100. The core of this crystal will be electron depleted with the exterior covered with orbiting electrons.

    • Ted-X

      The vapor pressure of the lithium in the reactor may still be too low to boil at 1342 deg. C, as the pressure inside the ECAT is higher than atmospheric (due to hydrogen present , presumably from LiAlH4). However, Lithium will have SOME VAPOR PRESSURE starting from 180 deg. C. The onset of the LENR at 180 deg. C seems to be correlated with SOME Lithium being evaporated (having SOME, although relatively low, vapour pressure). The vapor pressure of Lithium is increasing gradually with the temperature and can be estimated.

    • Axil Axil

      The are no neutrons in the hydrogen isotope that the Rossi reactor uses.

  • gc

    it is also very interesting that the original FP palladium deuterium publication used an
    electrochemical cell with an LiOD solution – the papers seemed to focus on the deuterium
    loading into palladium, but any role of Lithium in solution seems to have been overlooked?

  • gc

    it is also very interesting that the original FP palladium deuterium publication used an
    electrochemical cell with an LiOD solution – the papers seemed to focus on the deuterium
    loading into palladium, but any role of Lithium in solution seems to have been overlooked?

  • Has he submitted it to a physics journal for peer review? Maybe they will print it.

    • Zephir

      In the presented form? No way.

  • Has he submitted it to a physics journal for peer review? Maybe they will print it.

    • Zephir

      In the presented form? No way.

  • Obvious

    Sorry, but I cannot jump on this bandwagon. The process proposed might be akin to what happens sometimes in the reactor, but the theory calculations fall short of describing the reaction energy (under reports) by a significant margin. Unless I have messed something up, the calculated specific energy should read 3.5 GJ/kg, or 3.5 MJ/g (page 6) ( better: 3.5e3 MJ/kg), requiring only 1.6 grams of nickel fuel (rather than ~1600, using the reported 3.5 MJ/kg figure). This seems to fix the math somewhat (1.6 grams reported as 2 grams due to significant figures). But then the reaction rates reduce the probability of burning nickel to Ni62 by 50 fold, requiring 100 times more fuel to obtain the calculated 1.6 g (~2 g) burned fuel amount. So in essence the theory predicts a result that is 1% of the empirical results. Either this means that 99% of the energy comes from another reaction, or the theory is flawed.
    If anyone can find errors that I may have made, please feel free to straighten me out.

    • GordonDocherty

      As I wrote above, maybe the hot-cat is an e-cat with Lithium-Nickel neutron transfer
      piggy-backing off the more fundamental H-based e-cat nuclear reaction,
      as the heat from the e-Cat (which also provides the necessary
      environment for neutron transfer) first melts, then boils the Lithium,
      with the whole breaking down should the temperature cause the Nickel
      globally in the reactor to melt… (1455 C). So, run the hot cat reactor
      up to 1400 C and keep it there (+/- 50 degrees C – range of boiling
      point of Lithium to melting point of Nickel) and all should be well.
      Anyway, just something to consider…

    • Omega Z

      Theories are but a tool & always subject to change on new evidence.
      I appreciate his attempt verses those who just say there is no theory, thus I will not even look.

      • Obvious

        The bare bones of the proposed bound neutron tunneling mechanism are intriguing. However, this is one more example showing that the conversion process to Ni62, even if stealing a neutron from Li7, is not the actual main heat production method, because it falls significantly short of producing the energy obtained.
        What were the earlier “warm” cats doing, before Li was added? I suspect that Li6 is both a catalyst and a product of the reaction, assembled from deconstructed H atoms (bypassing He), and has swamped the original Li7 content. We have no idea how much Li went into the reactor: it seems that some was already inside (and did not react on its own). The Ni 62 is a critical part of the reaction, but is built up from the process, which becomes more efficient when all the nickel becomes Ni62 (and possibly Ni64). But these are guesses, I’ll humbly admit. The trick is to fit all the available scraps of info into a theory that supports all of the observations, rather than finding a mechanism, and forcing some numbers into it to fit some results whilst ignoring inconvenient data.

  • Obvious

    Sorry, but I cannot jump on this bandwagon. The process proposed might be akin to what happens sometimes in the reactor, but the theory calculations fall short of describing the reaction energy (under reports) by a significant margin. Unless I have messed something up, the calculated specific energy should read 3.5 GJ/kg, or 3.5 MJ/g (page 6) ( better: 3.5e3 MJ/kg), requiring only 1.6 grams of nickel fuel (rather than ~1600, using the reported 3.5 MJ/kg figure). This seems to fix the math somewhat (1.6 grams reported as 2 grams due to significant figures). But then the reaction rates reduce the probability of burning nickel to Ni62 by 50 fold, requiring 50 times more fuel to obtain the calculated 1.6 g (~2 g) burned fuel amount. So in essence the theory predicts a result that is 2% of the empirical results. Either this means that 98% of the energy comes from another reaction, or the theory is flawed.
    If anyone can find errors that I may have made, please feel free to straighten me out.

    • GordonDocherty

      As I wrote above, maybe the hot-cat is an e-cat with Lithium-Nickel neutron transfer
      piggy-backing off the more fundamental H-based e-cat nuclear reaction,
      as the heat from the e-Cat (which also provides the necessary
      environment for neutron transfer) first melts, then boils the Lithium,
      with the whole breaking down should the temperature cause the Nickel
      globally in the reactor to melt… (1455 C). So, run the hot cat reactor
      up to 1400 C and keep it there (+/- 50 degrees C – range of boiling
      point of Lithium to melting point of Nickel) and all should be well.
      Anyway, just something to consider…

    • Omega Z

      Theories are but a tool & always subject to change on new evidence.
      I appreciate his attempt verses those who just say there is no theory, thus I will not even look.

      • Obvious

        The bare bones of the proposed bound neutron tunneling mechanism are intriguing. However, this is one more example showing that the conversion process to Ni62, even if stealing a neutron from Li7, is not the actual main heat production method, because it falls significantly short of producing the energy obtained.
        What were the earlier “warm” cats doing, before Li was added? I suspect that Li6 is both a catalyst and a product of the reaction, assembled from deconstructed H atoms (bypassing He), and has swamped the original Li7 content. We have no idea how much Li went into the reactor: it seems that some was already inside (and did not react on its own). The Ni 62 is a critical part of the reaction, but is built up from the process, which becomes more efficient when all the nickel becomes Ni62 (and possibly Ni64). But these are guesses, I’ll humbly admit. The trick is to fit all the available scraps of info into a theory that supports all of the observations, rather than finding a mechanism, and forcing some numbers into it to fit some results whilst ignoring inconvenient data.

  • bitplayer

    Getting way ahead of the current state, should Gullström’s paper stand up to review, does it provide any direction for MFMP with respect to tuning the frequency of their heating coils? Seems like Brillouin’s model of compression waves might bridge the two areas, but I could be whistling in the wind here.

    • Bob Greenyer

      At least myself and Bob Higgins are reading it now.

    • Omega Z

      “unfounded speculations under the implication of factual knowledge”
      That pretty well covers all of us. Does it not? 🙂

  • Obvious

    So…. assuming that the clamp on the “resistor” lead was positioned wrong, the apparently magic control box somehow multiplies the input power by 3, since the input power was also measured between the building electrical outlet and the control box, as well as the circuit from the control box to the “resistors”, and these figures agree, minus power consumed by the control box (which is within manufacturer specs)?

    • you make a very good point that not only they did the mistake 2 times,
      they did if after the calibration runs which was rightly done

      some magic allow that to produce a COP of more than 3 despite the tric which gives 3…

      in fact I suspect they understestimate their trick.
      with a V setup as it seems, investing the common phase of the V make power exactly zero… with some swap you can rotate phase (it seems it is their hypothesis)…

      believe me as electronic engineer, once you imagine the testers can invert clamps, forget to merge power, and notice nothing, you’d better forget about science…

      those guys simply imagine a conspiracy of stupidity not top admit they are wrong.
      a classic scheme.

  • Iggy Dalrymple

    Hasn’t that 90 yr old engineer, eernie, been peppering Rossi with “tunnelling” theories for years?

    • Omega Z

      Iggy
      That 90 yr old engineer has knowledge he’s probably not allowed to discuss. In 1 of his exchanges he mentioned working with Nano Technology.

      Nothing spectacular about that, Except, He was talking about the mid 50’s. In fact, much of what he causally mentions is considered state of the art today. I have no doubt, he had direct involvement with the SR-71 blackbird design & construction. I would imagine most of the F-35 state of the art stealth fighter is probably antiquated technology to him.

      Of course, that was the days when 4 people could spend the weekend in a hotel room & could come up with a massive aircraft design change. Using nothing but a slide rule, pencils & poster board. Then present it with a reasonable fixed price.

      The same task today takes 20,000 people with mainframe access & years to develop something that requires redevelopment many times over many years. With a Fixed price that will no doubt triple & quadruple before the first prototype takes flight. Amazing how far we have come. NOT…

      • Axil Axil

        My guess is that the plasmons (sheets of electrons) developed by supersonic speeds of stealth aircraft at play to disrupt, absorb and frequency shift radar signals much in the same way that gammas are disrupted and downshifted in the Ni-H reactor.

  • Iggy Dalrymple

    Hasn’t that 90 yr old engineer, eernie, been peppering Rossi with “tunnelling” theories for years?

    • Omega Z

      Iggy
      That 90 yr old engineer has knowledge he’s probably not allowed to discuss. In 1 of his exchanges he mentioned working with Nano Technology.

      Nothing spectacular about that, Except, He was talking about the mid 50’s. In fact, much of what he causally mentions is considered state of the art today. I have no doubt, he had direct involvement with the SR-71 blackbird design & construction. I would imagine most of the F-35 state of the art stealth fighter is probably antiquated technology to him.

      Of course, that was the days when 4 people could spend the weekend in a hotel room & could come up with a massive aircraft design change. Using nothing but a slide rule, pencils & poster board. Then present it with a reasonable fixed price.

      The same task today takes 20,000 people with mainframe access & years to develop something that requires redevelopment many times over many years. With a Fixed price that will no doubt triple & quadruple before the first prototype takes flight. Amazing how far we have come. NOT…

      • Axil Axil

        My guess is that the plasmons (sheets of electrons) in the air/flight-surface interface developed at supersonic speeds of stealth aircraft at play to disrupt, absorb, and frequency shift radar signals much in the same way that gammas are disrupted and downshifted in the Ni-H reactor.

  • Obvious

    Where does the power come from then, if it is three times higher than measured coming from the wall outlet?

    • US_Citizen71

      Rossi’s magic powers. All pathoskeptics know about Rossi’s magic powers. When he is the room the laws of physics change so that is why he can’t be allowed anywhere near a test.

      • Obvious

        Apparently. If the incorrectly attached probe “logic” is followed to the end, then Rossi can create a box that creates 3X the supply of electrical power than used for input, but then hides the fact that he is wasting it to make heat by intoning a spell that makes the professors hook the reactor clamp connections incorrectly (every time), and further obfuscates the test by making a device that makes it hard to measure the actual heat production to distract the professors from the magical electrical control box 3X power trick.

        • yes you should write Chuck Norris -style jokes on Rossi…
          could be funny.

          It remind me those cargo cult skeptic who imagined that cold fusion experiments were caused by energy storage…
          those weak mind never tried to file a patent for energy storage at the highest known density … they could be billionaire with their claimed artifacts technology.

          • Obvious

            I think there might be some more fun left in the “measurement problem(s)” conundrums.

            For example, if there WAS 3X more power going in to the resistors, then the heat measurements and calculations must be correct.

            In kind, if the heat measurements/calculations are wrong, then 3X the power input theory cannot be correct.

            Otherwise the new Rossi trick would be to make some amount of energy disappear from the system without a trace.
            This should pit the “clamp to resistor problem” folks against the “heat measurement” problem folks. Let the battle of wits begin…..

          • Omega Z

            So if I understand,
            Your saying they can’t have it both ways.
            Just take all the fun out of it why don’t you. 🙂

            It’s freezing in here. Turn down the heat.

          • Obvious

            I think I was confusing the argument of the split resistor feed (Kirchhoff law discussion) with your idea. In your version, the probe is connected to the pre-split part wiring to the delta resistor (as in the report diagram). Is that right?

          • AlbertNN

            Correct, but also correct for the discussions on Kirchoff’s law that I have seen here. The report states that the current is split equally after the split point C in figure 4, but this is not correct in a 3-phase system, according to Kirchoff’s laws.

          • Obvious

            Thanks. The file came in pieces, but I get the gist of it.
            I’ll see if I can open it better.
            The voltage is probably near the maximum for each phase, based on the waveform drawn on the paper in the new bigger Vessy photo.

      • bachcole

        I can’t fully appreciate the humor because I deliberately avoid subjecting myself to the moronic maliciousness of the skeptopaths. (:->)

      • Andreas Moraitis

        Thank you for your detailed explanation. It is somewhat disappointing that many sources use the term “Coulomb barrier” apparently in a misleading way. In my own field of research, I have often found that you cannot rely on statements in textbooks, even if they come from the most renowned authors. Once printed, the errors circulate for decades…

  • Andreas Moraitis

    The range of the strong force is very short, therefore it will not retain the alpha particle any longer as soon as it has been separated. Indeed, the particle would be repelled from the nucleus if it were located at its periphery. But since it is usually positioned inside the nucleus, it will be retained by the surrounding protons – due to the Coulomb force, not the strong interaction. But I do not want to quarrel, I’m not a physicist and of course I may be wrong…

  • morse

    We need an ambassador for LENR, someone that can ignite the passion to the public, something like this short movie:

    http://www.youtube.com/watch?v=32vlOgN_3QQ#t=104

    • bachcole

      We could call the movie “Perserverance”.

  • morse

    We need an ambassador for LENR, someone that can ignite the passion to the public, something like this short movie:

    http://www.youtube.com/watch?v=32vlOgN_3QQ#t=104

    • bachcole

      We could call the movie “Perserverance”.

  • Freethinker

    Kudos to Carl-Oscar Gullström.

    Kudos to him, not only for trying to put some theory and numbers to the bound neutron tunneling in perspective with regards to LENR and ECAT.

    No, kudos to him for he is a PhD student at the Nuclear Physics group at the department of Physics and Astronomy at Uppsala University and that he is publicly attempting, with an open mind, to contribute in understanding what makes LENR works.

    Kudos to him, for he has the audacity to curse aloud in church of science.

    Kudos to him for that he does this, in spite what it may do to his future career possibilities in science.

    Carl-Oscar Gullström, I tip my hat for you.

    Oh, by the way. Need a grant? Maybe you should send ELFORSK a letter. They just might want to pay you to explore your ideas further.

    • Bob Greenyer

      I doff my hat also, and I love it that he is putting it out their warts and all for review – Yay for Live Open Science!

  • Freethinker

    Kudos to Carl-Oscar Gullström.

    Kudos to him, not only for trying to put some theory and numbers to the bound neutron tunneling in perspective with regards to LENR and ECAT.

    No, kudos to him for he is a PhD student at the Nuclear Physics group at the department of Physics and Astronomy at Uppsala University and that he is publicly attempting, with an open mind, to contribute in understanding what makes LENR works.

    Kudos to him, for he has the audacity to curse aloud in church of science.

    Kudos to him for that he does this, in spite what it may do to his future career possibilities in science.

    Carl-Oscar Gullström, I tip my hat for you.

    Oh, by the way. Need a grant? Maybe you should send ELFORSK a letter. They just might want to pay you to explore your ideas further.

    • Bob Greenyer

      I doff my hat also, and I love it that he is putting it out their warts and all for review – Yay for Live Open Science!

  • Obvious

    The same error cannot be repeated at both sides of the control box, since the input AC phases (from the wall) are not split, therefore the probe would simply report a negative current to the meter, but the correct magnitude of current regardless. The resistor part of the “problem” applies only when a true, 120° separated 3 phase power is being measured at a split between the sides of a delta resistor load. The logic of the split delta power measurement cannot be applied to the main power feed measurement. There are only opinions whether normal, 120° separated, 3 phase power is being fed to the resistors from the control box. If the power from the control box is rectified, phase shifted, or otherwise altered from the “normal” 3 phase AC, then the equations used to find the different values are no longer applicable, and the hypothetical results are just wild guesses.

  • Obvious

    To me, the image looks like one complete cycle of one phase of AC, heavily clipped by pulse modulation.

  • not necessarily, as it can also be a trigger problem.
    If the photo is the one after the triac, the voltage is not sinusoidal. if before there can be some detail on trigering we ignore…

    moreover whatever you imagine, that mistake is unavoidable… as I say, it is like choosing the wrong seat in a car and trying to drive.

    note also that the conspiracy assume that
    1- the error was done on the two current probes of the front and middle PCE380
    2- it was not done with the dummy, and the two powermeter were OK
    3- which implies that a front PCE380 clamp, which had no reason to me moved was moved in the wrong direction
    4- note also that the controlbox consume some energy and that some weird effect would be visible is clamp were misplaced

    5- it assume that Industrial heat provided devices that don’t work, hoping that an unbelievable incompetence and improbable luck would save them from ridicule and bankruptcy.

    that is extraordinary assumption. you need evidence, and not just photo without details commented by people who are clearly not experienced in managing oscilloscope, powermeter, and even basic power electronics (like assuming sinusoidal for switched controller, assuming no feedback, assuming no resistance change).

    • AlbertNN

      1. Or they did think that this was the correct way to connect the probes
      2. Not necessarily done with the dummy, as we do not know if the heating then was done with three phases. It could only be done with two, omitting the wrongly connected one.
      3. See 2.
      4. Only if you know exactly how much the control box was consuming beforehand.
      5. They might think that they work, see 1.

      What I would appreciate is more documentation, and some comments from the authors on the issue. But they have been very silent this far.
      The analysis linked above is not assuming anything on your list. Or can you point out any error in it?

  • GordonDocherty

    This looks very interesting – a paper from 2004 by D.V. FILIPPOV and L.I.URUTSKOEV that predicts the results now being seen :

    https://www.mail-archive.com/vortex-l@eskimo.com/msg99443.html

    particularly interesting:

    – input and output energies per initial/final atom are small and do not exceed 10 keV per synthesized atom;
    – there is no release of free neutrons;
    – there is no remnant radioactivity, i.e. non-stable isotopes are not synthesized

    Wow…

  • GordonDocherty

    This looks very interesting – a paper from 2004 by D.V. FILIPPOV and L.I.URUTSKOEV that predicts the results now being seen :

    https://www.mail-archive.com/vortex-l@eskimo.com/msg99443.html

    particularly interesting:

    – input and output energies per initial/final atom are small and do not exceed 10 keV per synthesized atom;
    – there is no release of free neutrons;
    – there is no remnant radioactivity, i.e. non-stable isotopes are not synthesized

    Wow…

    they also say at the end of their paper:

    “Though the mechanism of transformation of nuclei is not understood yet, it is natural to assume that there is some catalyst which may unite nuclei in a cluster and thus create a conditions for a resonance, which might initiates an exchange of nucleons. We consider the
    George Lochak’s magnetic monopole to be one of possible candidates for such a catalyst. Magnetic monopole, proposed by Lochak, is a lepton, i.e. participates in the electroweak interactions and can be treated as a magnetically excited state of neutrino. Such a monopole is massless (or nearly mass-less), very light (from the viewpoint of energy scale) and can be born, for example, in electromagnetic phenomena in a condensed matter”

    So, clusters of nuclei (Rydberg matter clusters, or clusters of clusters?), resonance, and magnetic monopoles – or rather, from Wikipedia, “condensed matter systems contain effective (non-isolated) magnetic monopole quasi-particles, or contain phenomena that are mathematically analogous to magnetic monopoles” – “flux tubes”.

    Again, from Wikipedia :

    “A flux tube is a generally tube-like (cylindrical) region of space containing a magnetic field, such that the field at the side surfaces is parallel to those surfaces. Both the cross-sectional area of the tube and the field contained may vary along the length of the tube, but the magnetic flux is always constant. … Placing a thin superconductor over a magnet will result in flux tubes passing through the small cracks of it.”

    So back to Casimir spaces again, looking now at them (also) as flux tubes.

    Again, if one or more of:

    1. metallic lattice (with correct geometry) – metallic means creation of electron cloud, of course
    2. lattice saturation (with hydrogen)
    3. heat
    4. resonance (phononic and, likely, magnonic)

    are missing from the e-Cat, there is no reaction… and Casimir spaces (in cracks in / faults with / “cabbage head” form of the lattice) do appear to play a vital role, as do magnetic field lines, electrons and Brillouin Zones in the metal lattice.

    One more thought. Maybe the hot-cat is an e-cat with Lithium-Nickel neutron transfer piggy-backing off the more fundamental H-based e-cat nuclear reaction, as the heat from the e-Cat (which also provides the necessary environment for neutron transfer) first melts, then boils the Lithium, with the whole breaking down should the temperature cause the Nickel globally in the reactor to melt… (1455 C). So, run the hot cat reactor up to 1400 C and keep it there (+/- 50 degrees C – range of boiling point of Lithium to melting point of Nickel) and all should be well. Anyway, just something to consider…

    • Mark Szl

      Here is a big favor to ask but if one doesn’t then ….

      Gordon Dorcherty and anyone else!

      Could someone give an easy to read review (i.e. plain English which an 18 year old could understand) of the possible theoretical basis of the Ecat that everyone has been discussing so far. That would help us newbies get up to speed. That would include criticism and counter-criticism which always helps in understanding an ideas progress.

  • georgehants

    Below is an example of the worst in science where these scientific traitors are suggesting that Research and Wonderful work being done by people like Carl-Oscar Gullström above, be hidden away to protect Hot Fusion.
    Is their anyone on page that believes that these scientific manipulators should not be removed from any contact with future scientific progress, publicly.
    A perfect example of the cheap, nasty people that inhabit the scientific community.
    ———
    From Cold Fusion Now
    Cold fusion defended against ITER at Channeling Conference
    —–
    During the Oct. 8 roundtable session, Dr. Tsyganov has reported
    that “some of the participants suggested that we avoid rushing to
    promote cold fusion and, therefore, prevent any interference with the
    implementations of international tokamak ITER.”
    —-
    Tsyganov explained, “… it is difficult to ignore the cold fusion
    process because it is much less expensive and much more practical than
    traditional thermonuclear fusion.”
    http://coldfusionnow.org/cold-fusion-defended-against-iter-at-channeling-conference/

  • georgehants

    Below is an example of the worst in science where these scientific traitors are suggesting that Research and Wonderful work being done by people like Carl-Oscar Gullström above, be hidden away to protect Hot Fusion.
    Is their anyone on page that believes that these scientific manipulators should not be removed publicly from any contact with future scientific progress, for example, teaching.
    A perfect example of the cheap, nasty people that inhabit parts of the scientific community.
    Well done Mr. Tsyganov for being a good honest scientist.
    ———
    From Cold Fusion Now
    Cold fusion defended against ITER at Channeling Conference
    —–
    During the Oct. 8 roundtable session, Dr. Tsyganov has reported
    that “some of the participants suggested that we avoid rushing to
    promote cold fusion and, therefore, prevent any interference with the
    implementations of international tokamak ITER.”
    —-
    Tsyganov explained, “… it is difficult to ignore the cold fusion
    process because it is much less expensive and much more practical than
    traditional thermonuclear fusion.”
    http://coldfusionnow.org/cold-fusion-defended-against-iter-at-channeling-conference/

    • Igor B

      Dearest georgehants! Be sure that the “scientific manipulators” are already safely removed at least from this forum. My comments are screened one after another, and believe – no single word of critics pops up in this thread. Do you guys really think that censoring is a way to argue? Does it make you feel more confident?

      • georgehants

        Igor, sorry have only just seen your comment, so am adding this reply very late.
        I feel sure that only the usual unsavory attacks etc are removed on this site, except those against me, that I do not mind as it is good to show-up some of these unpleasant scientists for what they are.
        Better to moderate than have pages full of crazy scientists denying and debunking everything beyond a steam engine, don’t you agree?

  • CimPy

    In case you missed, some more discussions on that paper are also in comments here

  • Obvious

    Please examine the photo of the experimental set up (Lugano Report page 4). It shows the probes connected in the correct direction. I have downloaded and read both the PCE830 manual, and the Compact Fusion Three Phase Power Controller manuals. The correct probe orientation is current flowing from front to back of the PCE830 amp probes. You can even see that the I3 line is connected to the far right load port on the controller by comparing the current probe ports in the PCE830 manual with the colors in the photo (blue). With the clamp opening lever on the left, the front of the probe is up. See pages 1, 11, 1and 12 of the PCE 830 manual. The load side of the power supply is on the bottom of the Controller box. See page 13 of the CCI Compact Fusion Operator Manual. Current flowing from the controller therefore flows from the box to the reactor, so the current is flowing through the probe from front to back.

    • AlbertNN

      Then your version of the report is better than mine. The compression artefacts in the version I have makes it impossible to make such a judgement. And we do not know if the PCE830 in figure 5 was connected to the input or output of the control box.

      Kirchoff’s law is equally valid at the input and the output of the control box. The analysis done with it does take the placement according to figure 4 into account.

      • Obvious

        OK, so going through the math so far, using formula in the report, I recalculated the Joule heating of the supply cables secondary leads to 13.97 W, rather than 1.6 W, resulting in a total supply cable Joule heating of 19.07 W. (From the Dummy test part). So that carries over to the page 20 totals, resulting in an “Actual Input” of 467 W instead of the former 479 value, actually improving the dummy run input to output comparison.

        Now I’m digging through the rest of the math to see where else it affects values downstream.

  • Obvious

    If I understand this issue correctly, the C2 leads can conduct current from multiple phases, and therefore conduct current more often than the C1 leads. The C2 leads act as extensions to the Delta pattern. So at the maximum phase separation of 120 degrees, the square root of 3 rule applies, but still only affects the calculations of Joule heating of the cables . So the worst case scenario is that the secondary (C2) cables consume 13.97 W, rather than 1.6 W, in the dummy run. Which is still insignificant compared to the resistor heating in the control (unloaded) reactor output.

    The Joule heating of the cables is simply a subtraction from the measured input to the reactor, as transmission loss. The remainder of the power (Control box output minus Joule heating) is considered to be the power consumed by the reactor. So using the maximum C2 heating loss, even tripling the Joule heating values reported for the loaded reactor, the effect is to raise COP slightly.

    So the 1/2 current usage actually is a far more conservative measurement of possible Joule heating loss.

    • Thomas Kaminski

      The square root of three applies only if:

      1). There delta load is symmetric (all legs the same)
      2). The load is “linear” (resistive, inductive, capacitive components)
      3). The three-phase power is sinusoidal and equal in magnitude, but separated in phase by 120 degrees.
      4). The Three phase lines are always attached.

      With a triac controller, you can think of each phase of the power source being “connected” when triac is conducting and “disconnected” when the triac is not conducting. For small conduction angles, only two phases of the supply are connected at any time. The third phase is not connected. Over one cycle of 50 Hertz power with three-phase power lines L1, L2, L3, the following connections occur with a resistive load (assuming the heaters are resistive):

      1) L1 to L2 with a load of R in parallel with 2*R
      2) L2 to L1 with a load of R in parallel with 2*R
      3) L1 to L3 with a load of R in parallel with 2*R
      4) L3 to L1 with a load of R in parallel with 2*R
      5) L2 to L3 with a load of R in parallel with 2*R
      6) L3 to L2 with a load of R in parallel with 2*R

      The instantaneous current that flows is a simple Ohm’s Law calculation at each instant in time. The magnitude of the RMS current will not have a simple square-root-of-three relationship for Line to Phase current. It will be computed by integrating a portion of the sine function over the conduction angles for each of the above cases.

      • Obvious

        Perhaps you can answer this, please.

        If the load is a balanced resistive load, when the amp probe on one phase is reading 19.7 A, and the other two phases also read the same for all three phases, how much average (RMS?) current is entering (being used by) the system:

        With full 3 phase (100% conduction)?
        With a low conduction angle pulse centered on the peak voltage (Maybe 30°)?

        Is it 19.7A (the average of all three phases?).
        19.7 A because what goes in must come out.
        Zero because what goes in must come out.
        59.1 A (the sum of the currents in three phases).
        34.1 A (the square root of the sum of the squares of the three currents)

      • For long this electric discussion made me yawn as we missed too many data, but assuming the I1eff+I2eff=I3eff

        it became clear that more or less I1 and I2 are synchronous, they are the same phase…

        It is not at all 3phase controller driving a 3phase load, but a 3phase battery of 6 dimmers driving 3 e-cat with each 2 coils…

        this explains the photography that recently appeared on Vassy’s site.

        why 2 coil ? maybe to allow some balancing effect to make the cat purr…

        http://www.lenr-forum.com/forum/index.php/Thread/867-Is-Lugano-E-cat-simply-a-single-phase-load-with-two-coils/?postID=1836#post1836

        just an idea…
        basically I trust Wattmeter, and human stupidity only when this defend the consensus and follow the group opinion.

        What make me nearly furious is that the real problem is that E-cat was not correctly calibrated and this cast a doubt…

  • hornster

    I am optimistic but have a problem with some claims of the e-cat that should be answered. Not nuclear physics, just applying heat to the reaction.
    1- why is AC needed, but DC is a problem?
    2-why complicate the heating with 3 phase AC measurement complexity. The heating required should be minimal (I hope) and easily measured. 3 phase is only needed for high drains.
    3. I can understand why gas heating is less expensive than electric, but again why is DC heating a problem?
    4. if such a magnificent breakthrough, why is the heat source even an issue?

    • 1- it can simply be that making an AC dimmer is more easy at kW level, and it plugs well in industrial installation
      2- it can be simply (it seems since Vessy’s photo of a 3 core test was 3 monophase) that industrial installation are in 3phase. E-cat in target application is MW.
      3- it may simply be that AC is what the electric e-cat is made for. I bet that it works on DC, if you change the control box
      4- probably because the COP is low, and because plugging a turbine is really hard

      We probably interpret simple engineering and financial decision as “LENR in E-cat requires…” whil it is simply “with triac you need AC” or “for 1MW you need 3phase or gas”

  • hornster

    I am optimistic but have a problem with some claims of the e-cat that should be answered. Not nuclear physics, just applying heat to the reaction.
    1- why is AC needed, but DC is a problem?
    2-why complicate the heating with 3 phase AC measurement complexity. The heating required should be minimal (I hope) and easily measured. 3 phase is only needed for high drains.
    3. I can understand why gas heating is less expensive than electric, but again why is DC heating a problem?
    4. if such a magnificent breakthrough, why is the heat source even an issue?

    • 1- it can simply be that making an AC dimmer is more easy at kW level, and it plugs well in industrial installation
      2- it can be simply (it seems since Vessy’s photo of a 3 core test was 3 monophase) that industrial installation are in 3phase. E-cat in target application is MW.
      3- it may simply be that AC is what the electric e-cat is made for. I bet that it works on DC, if you change the control box
      4- probably because the COP is low, and because plugging a turbine is really hard

      We probably interpret simple engineering and financial decision as “LENR in E-cat requires…” whil it is simply “with triac you need AC” or “for 1MW you need 3phase or gas”