Re: [Vo]:MIT Course Day 5 -- NiH Systems
Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that can not be tasted. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. The human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. I do this my making as few assumptions as possible and then limit these to the most basic possibilities. If this approach fits the data, then we have the answer. If the data are not fit, then additional assumptions are added only where absolutely necessary as exceptions. To start, you need to stop thinking of the LENR process as being caused by ordinary nuclear reactions. For example cross-section data have no application. This data is based on use of high energy particles for which a reaction rate is determined as this energy is changed. This process does not happen during LENR. If this process were operating, LENR could not happen. In fact, rejection of the claim results because this kind of thinking is used. We are dealing with a new kind of nuclear reaction. The challenge is to discover the rules that apply to this reaction, not keep using rules that apply to conventional reactions. The rules of conventional reactions make LENR impossible. The data show that Pd and Ni split into smaller parts. This data results from hundreds of studies and is not in doubt. This fact is the starting point for a search for an explanation. The first assumption results from the need to have something cause this result. That event is assumed to be addition of either one or more d or p to the nucleus by some unknown process, followed by fragmentation. Such a process requires the number of p and n in the initial nucleus to equal the total number in the fragments. As a result, if 2d entered the Ni, the fragments would have to contain a total of 30 p. This limits the element combinations that can result. Such calculations can be called nuclear chemistry because the same rule applies to chemical reactions. In the case of nuclear reactions, unlike chemistry, the number of neutrons also has to remain unchanged. Each isotope of an element has a different number of neutrons. Therefore, different isotope combinations are possible. At this point, we need one more assumption. This assumption says the isotope combination must always be non-radioactive, because that is what is observed most of the time. When this assumption is applied, the combinations are further limited, with some isotopes of Ni having many element combinations and some having only a few possibilities. The periodic table can be searched to discover which elements between He and Ni satisfy these two conditions. I have done this and obtained a distribution. This distribution matches what is observed. Therefore, the two assumptions appear to be correct. Once this information is obtained, the energy from each reaction can be calculated along with the frequency of each reaction, with no other assumptions being required. So you ask how the d or p got into the Ni nucleus. This is a separate question requiring different assumptions. First, energy must be available and it must be applied at the time and place where the nuclear event occurs. In addition, this energy must have a form that does not interact with the surrounding chemical structure. This requirement is unique to LENR, unlike what can happen in plasma. I propose a structure forms I call a Hydroton in which the fusion process takes place. This reaction, and only this reaction, has enough energy to overcome the Coulomb barrier for Ni or Pd. This fact further limits what can be proposed to happen. Of course, a person can imagine all kinds of novel quantum process that might operate, but these can not be tested and they all conflict with basic natural laws, which I will not explain here. I can test the consequence of the fusion reaction using the method applied above. I can add one or more d to the Ni or I can add one or more p. It turns out adding 2 d fit the observations. The question is, what kind of fusion reaction can generate two d? This can only happen as a result of a p-e-p reaction. Having 2d enter means the Ni had to be attracted to two Hydrotons, each of which produced and added 1d. Here we have used a few basic assumptions to explain transmutation and to describe the fusion reaction by showing how they are connected. No additional assumptions are required and no novel or untestable processes have to be suggested. This is how, I suggest, LENR be explored. If this approach is used, LENR can be explained and all the previously unexplained behavior makes sense. That is what I'm attempting to do in the book.
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Edmund Storms Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that cannot be tested. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. The human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. But Ed - it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. Essentially this explanation is dead-in-the-water on two fronts - the lack of tritium, which must be there if the reaction can fuse two protons, and the lack of 1+ MeV quanta. Some kind of spin coupling is far preferable to a proposed reaction which cannot happen. Jones
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Jones Beene .it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. Essentially this explanation is dead-in-the-water on two fronts - the lack of tritium, which must be there if the reaction can fuse two protons, and the lack of 1+ MeV quanta. Some kind of spin coupling is far preferable to a proposed reaction which cannot happen. By the way - the S. Jones paper/slide-presentation mentioned last evening does in fact present a plausible method of spin coupling - PLUS he has real data of the RF signature for such coupling.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Ed--Bob Here-- I would note that testing by the manipulation of spin is possible by changing the static magnetic fields or the oscillating fields given known nuclear magnetic resonance parameters. You suggest that energies associated with spin are not found to involve the magnitude of energy involved. Who determined this situation? Is there a reference supporting this conclusion other that mere assertion? I know of Japanese researcher data regarding the formation of various heavier isotopes after forcing D gas through thin films. However, I am not familiar with the data you suggest for the splitting of Pd and Ni. A couple references would be good. When do you expect to finish your book on the subject? If you have a partial bibliography of references, maybe that would give me the pertinent leads. Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Saturday, February 08, 2014 7:12 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that can not be tasted. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. KThe human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. I do this my making as few assumptions as possible and then limit these to the most basic possibilities. If this approach fits the data, then we have the answer. If the data are not fit, then additional assumptions are added only where absolutely necessary as exceptions. To start, you need to stop thinking of the LENR process as being caused by ordinary nuclear reactions. For example cross-section data have no application. This data is based on use of high energy particles for which a reaction rate is determined as this energy is changed. This process does not happen during LENR. If this process were operating, LENR could not happen. In fact, rejection of the claim results because this kind of thinking is used. We are dealing with a new kind of nuclear reaction. The challenge is to discover the rules that apply to this reaction, not keep using rules that apply to conventional reactions. The rules of conventional reactions make LENR impossible. The data show that Pd and Ni split into smaller parts. This data results from hundreds of studies and is not in doubt. This fact is the starting point for a search for an explanation. The first assumption results from the need to have something cause this result. That event is assumed to be addition of either one or more d or p to the nucleus by some unknown process, followed by fragmentation. Such a process requires the number of p and n in the initial nucleus to equal the total number in the fragments. As a result, if 2d entered the Ni, the fragments would have to contain a total of 30 p. This limits the element combinations that can result. Such calculations can be called nuclear chemistry because the same rule applies to chemical reactions. In the case of nuclear reactions, unlike chemistry, the number of neutrons also has to remain unchanged. Each isotope of an element has a different number of neutrons. Therefore, different isotope combinations are possible. At this point, we need one more assumption. This assumption says the isotope combination must always be non-radioactive, because that is what is observed most of the time. When this assumption is applied, the combinations are further limited, with some isotopes of Ni having many element combinations and some having only a few possibilities. The periodic table can be searched to discover which elements between He and Ni satisfy these two conditions. I have done this and obtained a distribution. This distribution matches what is observed. Therefore, the two assumptions appear to be correct. Once this information is obtained, the energy from each reaction can be calculated along with the frequency of each reaction, with no other assumptions being required. So you ask how the d or p got into the Ni nucleus. This is a separate question requiring different assumptions. First, energy must be available and it must be applied at the time and place where the nuclear event occurs. In addition, this energy must have a form that does not interact with the surrounding chemical structure. This requirement is unique to LENR, unlike what can happen in plasma. I propose a structure forms I call a Hydroton in which the fusion process takes place. This reaction, and only this reaction, has enough energy to overcome the Coulomb barrier for Ni or Pd. This fact further limits what can be proposed to happen. Of course, a person can imagine all kinds of novel quantum
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 8, 2014, at 10:13 AM, Jones Beene wrote: From: Edmund Storms Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that cannot be tested. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. The human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. But Ed – it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. But Jones, we do not know this can not work. You are taking the conventional approach that eventually proves that LENR is impossible. I'm proposing a new approach must be used. Suggesting obscure and untestable processes such as spin coupling does not help. The data can be explained without using these processes. Consequently, why insist they be used. Does nature's behavior not have the last word? We know that tritium is made when D and H are present and this can only result from p-e-d fusion. Is it unreasonable to assume p-e-p also occurs? Nevertheless, this proposal shows where to look for the evidence. I'm waiting for someone to find the d and the subsequent tritium when H+Ni is used. Absence of data is not absent of proof, as many people point out including yourself. What would you expect to find if spin coupling were the process? Ed Storms Essentially this explanation is dead-in-the-water on two fronts – the lack of tritium, which must be there if the reaction can fuse two protons, and the lack of 1+ MeV quanta. Some kind of spin coupling is far preferable to a proposed reaction which cannot happen. Jones
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Edmund Storms Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that cannot be tested. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. The human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. JB: But Ed - it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. ES: But Jones, we do not know this can not work. You are taking the conventional approach that eventually proves that LENR is impossible. Not accurate! Let's be clear: I am very much taking an expanded conventional approach - but it is one which says that in order for LENR to be proved, there must be an energetic reaction for gain which does not produce gamma nor does it produce more than minimal transmutation. Spin coupling, for instance - is well known, and has not been ruled out. That does not mean it is correct, but at least it is not ruled out by experiment. Deuterium fusing from protons can be ruled out. I'm proposing a new approach must be used. Suggesting obscure and untestable processes such as spin coupling does not help. They are not obscure at all - and they are testable. You are incorrect on that point. Several of the alternative theories for Ni-H have a good chance even though real fusion as it is known to the mainstream, is not in evidence. We must find a way to convert nuclear mass to thermal heat and yes spin coupling can do that. Your approach, as it applies to Ni-H does not match experiment, and that is the bottom line. We must rule out fusion of protons to deuterium. That says nothing about the fusion of protons to helium in palladium, which can happen in that kind of reaction BUT NOT in Ni-H. The Rossi experiment absolutely rules out P+P - D. Jones
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 8, 2014, at 11:06 AM, Jones Beene wrote: From: Edmund Storms Bob, we are presented with a complex puzzle. A solution requires testing possibilities against what is observed. A solution is made difficult if mechanisms are proposed that cannot be tested. For example, spin coupling can not be tested against what is known and, in addition, it is not found to involve the magnitude of energy involved. The human mind can imagine an infinite number of possibilities. Some way must be used to limit these possibilities. JB: But Ed – it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. ES: But Jones, we do not know this can not work. You are taking the conventional approach that eventually proves that LENR is impossible. Not accurate! Let’s be clear: I am very much taking an expanded conventional approach – but it is one which says that in order for LENR to be proved, there must be an energetic reaction for gain which does not produce gamma nor does it produce more than minimal transmutation. We agree on these two requirements. The mechanism is in question. Spin coupling, for instance - is well known, and has not been ruled out. That does not mean it is correct, but at least it is not ruled out by experiment. Deuterium fusing from protons can be ruled out. How is this ruled out? You only provide assertions. I consider this possibility based on evidence for tritium production and the assumption that a similar process applies to p and d. So far I see nothing that shows this is not a plausible assumption. I'm proposing a new approach must be used. Suggesting obscure and untestable processes such as spin coupling does not help. They are not obscure at all - and they are testable. You are incorrect on that point. Several of the alternative theories for Ni- H have a good chance even though real “fusion” as it is known to the mainstream, is not in evidence. We must find a way to convert nuclear mass to thermal heat and yes spin coupling can do that. So, you claim spin coupling can convert over 24 MeV/event to heat in the case of deuterium and over 11 MeV/event in the case of transmutation. Has anyone actually shown this amount of energy being involved in spin coupling, either by experiment or theory? Your approach, as it applies to Ni-H does not match experiment, and that is the bottom line. We must rule out fusion of protons to deuterium. That says nothing about the fusion of protons to helium in palladium, which can happen in that kind of reaction BUT NOT in Ni-H. The Rossi experiment absolutely rules out P+P - D. OK Jones, I accept we have a different approach. Nevertheless, your statement above is about a fact. Where does this fact come from? I have seen no evidence supporting such a conclusion. As far as I know, no effort has been described about a search for deuterium. Tritium would be produced in such a small amount, it would be missed unless it was sought with care. In addition, Rossi has no reason to acknowledge tritium production for obvious reasons. Please explain why you make the above statement. Ed Storms Jones
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Jones--Bob Cook here-- I saw that mention also and planned to follow up to address some of Ed concerns about it not being possible. Bob - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Saturday, February 08, 2014 9:24 AM Subject: RE: [Vo]:MIT Course Day 5 -- NiH Systems From: Jones Beene .it is far worse to attempt to rationalize a mechanism which we know for sure cannot work, like P+P fusion to deuterium. Essentially this explanation is dead-in-the-water on two fronts - the lack of tritium, which must be there if the reaction can fuse two protons, and the lack of 1+ MeV quanta. Some kind of spin coupling is far preferable to a proposed reaction which cannot happen. By the way - the S. Jones paper/slide-presentation mentioned last evening does in fact present a plausible method of spin coupling - PLUS he has real data of the RF signature for such coupling.
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Edmund Storms * Deuterium fusing from protons can be ruled out. How is this ruled out? You only provide assertions. No, I provide two facts from the Rossi experiments. No gamma. No tritium. These are facts, not assertions. Rossi is presently the best hope for the future of LENR and it does little good for anyone to try to confuse the broader field by hopelessly trying to explain another anomaly in another context. The Rossi effect, at this point in time, is worth dropping everything else for - in order to understand it. I consider this possibility based on evidence for tritium production and the assumption that a similar process applies to p and d. Tritium is not seen in Rossi, nor is it seen from protons alone, and it is not relevant to the Rossi effect, except in its absence. There is no evidence that a similar process to Rossi is involved in Pd-D of PF and it is counter-productive to confuse the two. In fact, all the important evidence shows the two cannot be similar in any meaningful way. So, you claim spin coupling can convert over 24 MeV/event to heat in the case of deuterium and over 11 MeV/event in the case of transmutation. You must be kidding, right? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. PF is different from Rossi - end of story ...unless someone can supply real proof of a connection. None has been shown. Ockham is not proof of anything, and has never provided a valid level of understanding to any field of science, especially since parsimony is completely adverse to QM. QM is anti-Ockham. Jones attachment: winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 8, 2014, at 12:11 PM, Jones Beene wrote: From: Edmund Storms * Deuterium fusing from protons can be ruled out. How is this ruled out? You only provide assertions. No, I provide two facts from the Rossi experiments. No gamma. No tritium. But Jones, Rossi detected radiation, made efforts to shield, and this radiation was detected at the time by Celani and other people detect rasdiation using light water. As for tritium, do you actually believe Rossi? I don't! He has given no indication of how or when this detection was made. These are facts, not assertions. Rossi is presently the best hope for the future of LENR and it does little good for anyone to try to confuse the broader field by hopelessly trying to explain another anomaly in another context. The Rossi effect, at this point in time, is worth dropping everything else for - in order to understand it. That is what I'm trying to do, but using information from other sources. Rossi made heat. He has shown no ability to explain the process. He has limited ability to make suitable measurements. And he is a confused source of information. Why would you accept him as an authority about science? I consider this possibility based on evidence for tritium production and the assumption that a similar process applies to p and d. Tritium is not seen in Rossi, nor is it seen from protons alone, and it is not relevant to the Rossi effect, except in its absence. Tritium has been made using normal water in an electrolytic cell. Do you think a different mechanism applies here compared to Ni-H2? There is no evidence that a similar process to Rossi is involved in Pd-D of PF and it is counter-productive to confuse the two. In fact, all the important evidence shows the two cannot be similar in any meaningful way. OK Jones, this is a difference of opinion only nature will resolve. So, you claim spin coupling can convert over 24 MeV/event to heat in the case of deuterium and over 11 MeV/event in the case of transmutation. You must be kidding, right? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. Transmutation, even using Cu production. generates about 6 MeV of energy/event. The products detected by DGT and other people require about 12 MeV/event to be released. This is fact based on the mass change. PF is different from Rossi - end of story ...unless someone can supply real proof of a connection. None has been shown. Ockham is not proof of anything, and has never provided a valid level of understanding to any field of science, especially since parsimony is completely adverse to QM. QM is anti-Ockham. You opened a can of worms I don't have time to kill. Nevertheless, what you said has no justification. Ed Storms Jones winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Sat, Feb 8, 2014 at 11:11 AM, Jones Beene jone...@pacbell.net wrote: No, I provide two facts from the Rossi experiments. No gamma. No tritium. ... These are facts, not assertions. Jones, your analysis is often insightful. But here you're stating facts, and then implying assumptions on the basis of those facts as facts as well. You assume that d+d fusion will result in a gamma, and then when no gamma is seen, you assume that d+d fusion in NiH is not possible. You have assumed away some mechanism that might be fractionating the gamma. And then later you draw upon related arguments to support this assumption. In repeating this line of reasoning, you are as guilty of simple, repetitive assertion of your assumptions as Ed is of his. Simply asserting an assumption to be true, or drawing upon such an assumption implicitly to reason about other things, does not make the assumption true. I suspect d+d fusion is not going on in Rossi's reactor either, but for reasons other than a missing gamma. We have no evidence one way or another about tritium, but no specific reason to believe it is there either. In fact, all the important evidence shows the two cannot be similar in any meaningful way. This is an overstatement. Can we all adopt a more measured tone? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Can you provide a link to the Bianchini report? For some reason I'm having trouble finding it. I assume that this was the appendix provided in connection with the Elforsk test? The only report I'm finding deals with a different subject relating to the E-Cat, in 2010 [1]. In the Elforsk test, no radiation was seen. There were obviously working parameters for the radiation monitor and an upper and lower threshold beyond which it would not have been effective. I do not know what type of monitor was used or what these thresholds were. But what we can deduce from this situation is that no penetrating radiation was escaping the system. It is a nonsequitor to conclude anything about the amount of energy being dissipated, let alone to conclude something about spin coupling as a possible mechanism. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. This is a simple assertion. Can we lay off of these a little? Eric [1] http://e-cataustralia.com/pdf/Levi_Bianchini_and_Villa_Reports.pdf
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 8, 2014, at 12:26 PM, Eric Walker wrote: On Sat, Feb 8, 2014 at 11:11 AM, Jones Beene jone...@pacbell.net wrote: No, I provide two facts from the Rossi experiments. No gamma. No tritium. ... These are facts, not assertions. Jones, your analysis is often insightful. But here you're stating facts, and then implying assumptions on the basis of those facts as facts as well. You assume that d+d fusion will result in a gamma, and then when no gamma is seen, you assume that d+d fusion in NiH is not possible. You have assumed away some mechanism that might be fractionating the gamma. And then later you draw upon related arguments to support this assumption. In repeating this line of reasoning, you are as guilty of simple, repetitive assertion of your assumptions as Ed is of his. Simply asserting an assumption to be true, or drawing upon such an assumption implicitly to reason about other things, does not make the assumption true. I suspect d+d fusion is not going on in Rossi's reactor either, but for reasons other than a missing gamma. We have no evidence one way or another about tritium, but no specific reason to believe it is there either. Eric, no one believes d+d fusion occurs in the Rossi reactor. The d we are discussing results from p-e-p fusion only. I agree with the other comments you make. Ed Storms In fact, all the important evidence shows the two cannot be similar in any meaningful way. This is an overstatement. Can we all adopt a more measured tone? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Can you provide a link to the Bianchini report? For some reason I'm having trouble finding it. I assume that this was the appendix provided in connection with the Elforsk test? The only report I'm finding deals with a different subject relating to the E-Cat, in 2010 [1]. In the Elforsk test, no radiation was seen. There were obviously working parameters for the radiation monitor and an upper and lower threshold beyond which it would not have been effective. I do not know what type of monitor was used or what these thresholds were. But what we can deduce from this situation is that no penetrating radiation was escaping the system. It is a nonsequitor to conclude anything about the amount of energy being dissipated, let alone to conclude something about spin coupling as a possible mechanism. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. This is a simple assertion. Can we lay off of these a little? Eric [1] http://e-cataustralia.com/pdf/Levi_Bianchini_and_Villa_Reports.pdf
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Eric Walker I provide two facts from the Rossi experiments. No gamma. No tritium. ... These are facts, not assertions. [snip] You assume that d+d fusion will result in a gamma, and then when no gamma is seen, you assume that d+d fusion in NiH is not possible. You have assumed away some mechanism that might be fractionating the gamma. Not exactly. I’m glad you brought this particular detail up - because that is not precisely what I am assuming away. What I am stating is that even if “fractionating” some gamma radiation is remotely possible, in principal or as a hypothesis - and there is no proof or even good evidence in physics that this is possible, or can be accomplished at all at high temperature … but even if it can - that problem pales when it is realized that any such mechanism MUST be completely leak-proof or else there will be fatalities. Complete shielding by fractionation would be infinitely more unlikely with highly penetrating radiation than good shielding. We know from cosmology that gamma radiation can escape from neutron stars – yet, give me a break, proponents want to suggest that a few grams of nickel powder will shield for gammas- and not just a little bit but completely 100% shield. Think about the absurdity. The risk/reward situation is such that 99% or four nines for leakage is not nearly good enough. One cannot simply propose the leap that goes all the way from partial fractionation to complete blockage. Do not overlook that fact that at the intense level of thermal output of the Rossi reactor, even a leakage of one part per billion would be fatal to Rossi or anyone else. That is the problem. Not so much that it might work some of the time, but that do be valid as a commercial item - it has to happen all the time with no exceptions. It is not merely the lack of a known phenomenon in mainstream physics for “fractionating gammas” but the fact that for fifty years, billions have been spent by the US and USSR looking for light weight gamma shielding in order to have reactors carried by bombers. There is no indication that had any success at all. Gammas are very penetrating, and that makes it baffling tome - as to how this fractionation thing has gotten any momentum at all.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Sat, Feb 8, 2014 at 12:05 PM, Jones Beene jone...@pacbell.net wrote: The risk/reward situation is such that 99% or four nines for leakage is not nearly good enough. One cannot simply propose the leap that goes all the way from partial fractionation to complete blockage. That is precisely what is being proposed. Whether this suggestion is amenable to you is a different question. But such 100 percent efficiency in fractionation is what is implied in the PdD research. You take deuterium, you place it with a palladium system using electrolysis or gas pressure, and in important experiments you get both excess heat and 4He above background at an order of magnitude correlation with the heat. E.g., on the order of 24 MeV heat leaving the system above what has been put in in terms of input energy per 4He observed. Despite what some journalists might want to believe, that suggests exactly d+d → 4He + Q, with no gamma, brought about through one mechanism or another, direct or indirect. Unless the 4He/heat experiments are to be set aside or dramatically reinterpreted, there's no escaping that Q. The question is what happens to it. Do not overlook that fact that at the intense level of thermal output of the Rossi reactor, even a leakage of one part per billion would be fatal to Rossi or anyone else. Yes -- which is precisely why a mechanism that is thought to intercept any gammas in flight is untenable. Not mentioned is the possibility that the conditions under which such fractionation occurs are a requirement for cold fusion to even happen. This does not seem like too much of a stretch of the imagination. There is no indication that had any success at all. Gammas are very penetrating, and that makes it baffling tome - as to how this fractionation thing has gotten any momentum at all. It's the PdD research and the 4He/heat correlation. You're overlooking this or ignoring it, or assuming as one might that PdD and NiH have nothing in common. But if we accept what the correlation implies, and we are willing to draw weak conclusions from PdD to NiH, as I would venture the large majority of folks are, then it is not a long-shot to assume that something analogous, although perhaps different in components, is also occurring in NiH. Once it's been established that a Q of 24 MeV can be fractionated without penetrating radiation in *some* context, it is not a leap of faith to conclude that similar behavior can be sought in other contexts. Eric
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Eric Walker Jones Beene wrote: The risk/reward situation is such that 99% or four nines for leakage is not nearly good enough. One cannot simply propose the leap that goes all the way from partial fractionation to complete blockage. That is precisely what is being proposed. Then that is precisely why it is wrong. If complete leak-proof gamma shielding is possible - we do not need LENR and we can go directly to subcritical fission, photofission or a small scale hybrid with a desktop accelerator - which is known, proved and reliable. Natural uranium is two orders of magnitude cheaper than deuterium. Who needs deuterium if gammas can be perfectly shielded by grams of loaded metal? Whether this suggestion is amenable to you is a different question. Forget me. Who is it amenable to? Answer: a handful of LENR proponents who started out in PdD and refuse to see that Rossi is very different, or who would love to find something better but have no option? Where is the kilowatt PdD reactor? Cough… cough… Rossi is the future of LENR and it is counterproductive to be lost in the past. PdD is an exercise in futility. But such 100 percent efficiency in fractionation is what is implied in the PdD research. No it isn’t. Lack of gammas ab initio is what is implied in LENR research. The two are completely different, not different ways of saying the same thing. Jones attachment: winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Let me see if I understand your position, Jones. You believe the behavior using deuterium has no relationship to the behavior when H is used. You believe nature has several ways to initiate LENR depending on which isotope of hydrogen is used, with the mechanism for D only working in Pd and the mechanism for H only working in Ni. You accept what Rossi has said without question and apply it only to the Ni-H2 system. Consequently, nothing observed by anyone else applies. You propose the mechanism that works for Ni+H2 is outside of conventional nuclear behavior including not producing the calculated amount of energy the mass change requires. Is that a fair description? Ed Storms On Feb 8, 2014, at 1:56 PM, Jones Beene wrote: From: Eric Walker Jones Beene wrote: The risk/reward situation is such that 99% or four nines for leakage is not nearly good enough. One cannot simply propose the leap that goes all the way from partial fractionation to complete blockage. That is precisely what is being proposed. Then that is precisely why it is wrong. If complete leak-proof gamma shielding is possible - we do not need LENR and we can go directly to subcritical fission, photofission or a small scale hybrid with a desktop accelerator - which is known, proved and reliable. Natural uranium is two orders of magnitude cheaper than deuterium. Who needs deuterium if gammas can be perfectly shielded by grams of loaded metal? Whether this suggestion is amenable to you is a different question. Forget me. Who is it amenable to? Answer: a handful of LENR proponents who started out in PdD and refuse to see that Rossi is very different, or who would love to find something better but have no option? Where is the kilowatt PdD reactor? Cough… cough… Rossi is the future of LENR and it is counterproductive to be lost in the past. PdD is an exercise in futility. But such 100 percent efficiency in fractionation is what is implied in the PdD research. No it isn’t. Lack of gammas ab initio is what is implied in LENR research. The two are completely different, not different ways of saying the same thing. Jones winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
In reply to Eric Walker's message of Fri, 7 Feb 2014 22:05:07 -0800: Hi Eric, [snip] On Fri, Feb 7, 2014 at 9:54 PM, Bob Cook frobertc...@hotmail.com wrote: I think you have the decay scheme for Ni-59 wrong. It has a 76,000 year half life and decays by electron capture as you said. It's good that you seem to know your way around these nuclear transitions. That makes you and Robin and a few others who can keep the rest of us honest. The data I have indicate no gamma activity, in the transition to the Cu-59 nucleus. I'm thinking of this reaction: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-62(P%2CG)29-CU-63%2C%2CSIG If you check the boxes for the A.Simon data, and quick plot on the this page, then click retrieve, you will get a cross section plot for the reaction. The cross sections are in barns, and the proton energy is in MeV. Note that all proton fusion related cross sections happen at high energy, because until the advent of CF, no one was able to fuse protons with heavy nuclei by any other means than high kinetic energy. (And the CF method is still not understood, if it works at all.) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:MIT Course Day 5 -- NiH Systems
-Original Message- From: Edmund Storms Let me see if I understand your position, Jones. You believe the behavior using deuterium has no relationship to the behavior when H is used. No relationship is too strong. After all, both involve hydrogen loaded metal and QM. But the assessment is almost correct if you base it on two different types of tunneling with vastly different outcomes. If PdD does result in fusion to helium, then my answer is almost yes, there is almost no relevance of that dynamic to the Rossi effect of Ni-H, which is not a known fusion reaction, and may not technically be fusion at all. However, I am not convinced that PdD works this way, and frankly - it is a diversion to even bring it up for now, since it detracts from the really important issue - which is the proper understanding of the Rossi effect. The two are almost as unrelated as mainstream fission is from mainstream fusion. You believe nature has several ways to initiate LENR depending on which isotope of hydrogen is used. Absolutely yes - to the degree that the name: LENR includes any thermal anomaly, not necessarily related to a known nuclear reaction, and there are 12 or more distinct routes to thermal gain. You accept what Rossi has said without question LOL. No one can accept Rossi's full account, as it is self-contradictory; but it is the totality of the evidence that matters especially the lack of gammas and that detail comes from experts, not Rossi. You propose the mechanism that works for Ni+H2 is outside of conventional nuclear behavior - including not producing the calculated amount of energy the mass change requires. Exactamundo. This is where one must depart from the original Focardi/Rossi account. The most prevalent active mechanism is not proton addition to nickel, with transmutation to copper. The facts do not support that. There could be a small contribution but the main reaction is different. Like it or not, Steven Jones - and his new slides do support the viewpoint of Bob Cook and many others on spin coupling - due to magnetic field collapse and with real data, and a photon signature in the RF. He also disputes the connection of Helium to excess heat. I would not go that far, but it is sufficient to say the PdD and NiH are very different. attachment: winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Sat, Feb 8, 2014 at 2:07 PM, Jones Beene jone...@pacbell.net wrote: However, I am not convinced that PdD works this way, and frankly - it is a diversion to even bring it up for now, since it detracts from the really important issue - which is the proper understanding of the Rossi effect. How is it a diversion to bring up an apparently well-established conclusion that a large quantum of mass energy can be fractionated without penetrating radiation? That was the point that was at issue. Answer: it's not a diversion. The conclusion may be flawed, the evidence may be flawed, the interpretation may be flawed, and/or the research may be flawed. But a consensus conclusion about the fractionation of a 24 MeV quantum into non-penetrating radiation is something to be addressed in a conversation dealing with the question of whether fractionation is possible. I'm not trying to say that the fractionation conclusion is for sure what is going on, either in NiH or in PdD. Only that it's not out in the wilderness either, as some would tendentiously make it out to be. :) Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Thanks Jones, you make our difference in approach very clear. In contrast, I assume all LENR results from the same process regardless of which isotope of hydrogen is used or which metal lattice contains the NAE. Of course, different nuclear products result from D and H, and different transmutation products result from Pd and Ni. We will see which approach is most useful in making the effect work better and is consistent with the observed behaviors. Let the games begin. :-) Ed Storms On Feb 8, 2014, at 3:07 PM, Jones Beene wrote: -Original Message- From: Edmund Storms Let me see if I understand your position, Jones. You believe the behavior using deuterium has no relationship to the behavior when H is used. No relationship is too strong. After all, both involve hydrogen loaded metal and QM. But the assessment is almost correct if you base it on two different types of tunneling with vastly different outcomes. If PdD does result in fusion to helium, then my answer is almost yes, there is almost no relevance of that dynamic to the Rossi effect of Ni-H, which is not a known fusion reaction, and may not technically be fusion at all. However, I am not convinced that PdD works this way, and frankly - it is a diversion to even bring it up for now, since it detracts from the really important issue - which is the proper understanding of the Rossi effect. The two are almost as unrelated as mainstream fission is from mainstream fusion. You believe nature has several ways to initiate LENR depending on which isotope of hydrogen is used. Absolutely yes - to the degree that the name: LENR includes any thermal anomaly, not necessarily related to a known nuclear reaction, and there are 12 or more distinct routes to thermal gain. You accept what Rossi has said without question LOL. No one can accept Rossi's full account, as it is self- contradictory; but it is the totality of the evidence that matters especially the lack of gammas and that detail comes from experts, not Rossi. You propose the mechanism that works for Ni+H2 is outside of conventional nuclear behavior - including not producing the calculated amount of energy the mass change requires. Exactamundo. This is where one must depart from the original Focardi/ Rossi account. The most prevalent active mechanism is not proton addition to nickel, with transmutation to copper. The facts do not support that. There could be a small contribution but the main reaction is different. Like it or not, Steven Jones - and his new slides do support the viewpoint of Bob Cook and many others on spin coupling - due to magnetic field collapse and with real data, and a photon signature in the RF. He also disputes the connection of Helium to excess heat. I would not go that far, but it is sufficient to say the PdD and NiH are very different. winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric-- I agree with your observation about Jones and Ed. In their give and take just before this message I was not sure who was saying what. The symbol seemed to have no significance as to who was talking. Bob Cook - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Saturday, February 08, 2014 11:26 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Sat, Feb 8, 2014 at 11:11 AM, Jones Beene jone...@pacbell.net wrote: No, I provide two facts from the Rossi experiments. No gamma. No tritium. ... These are facts, not assertions. Jones, your analysis is often insightful. But here you're stating facts, and then implying assumptions on the basis of those facts as facts as well. You assume that d+d fusion will result in a gamma, and then when no gamma is seen, you assume that d+d fusion in NiH is not possible. You have assumed away some mechanism that might be fractionating the gamma. And then later you draw upon related arguments to support this assumption. In repeating this line of reasoning, you are as guilty of simple, repetitive assertion of your assumptions as Ed is of his. Simply asserting an assumption to be true, or drawing upon such an assumption implicitly to reason about other things, does not make the assumption true. I suspect d+d fusion is not going on in Rossi's reactor either, but for reasons other than a missing gamma. We have no evidence one way or another about tritium, but no specific reason to believe it is there either. In fact, all the important evidence shows the two cannot be similar in any meaningful way. This is an overstatement. Can we all adopt a more measured tone? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Can you provide a link to the Bianchini report? For some reason I'm having trouble finding it. I assume that this was the appendix provided in connection with the Elforsk test? The only report I'm finding deals with a different subject relating to the E-Cat, in 2010 [1]. In the Elforsk test, no radiation was seen. There were obviously working parameters for the radiation monitor and an upper and lower threshold beyond which it would not have been effective. I do not know what type of monitor was used or what these thresholds were. But what we can deduce from this situation is that no penetrating radiation was escaping the system. It is a nonsequitor to conclude anything about the amount of energy being dissipated, let alone to conclude something about spin coupling as a possible mechanism. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. This is a simple assertion. Can we lay off of these a little? Eric [1] http://e-cataustralia.com/pdf/Levi_Bianchini_and_Villa_Reports.pdf
Re: [Vo]:MIT Course Day 5 -- NiH Systems
I am not going to try to quote who and what from this thread regarding fractionating gammas (too long of a story line now). What I have come to believe and what I initially missed, and what I think many Vorts may be missing in this, is that the LENR reaction and the fractionating are not two separate processes. Jones (et al) are correct that if there is a fractionating mechanism that is an independent effect, it could not be 100% efficient and some high energy photons would escape as a marker of this inefficiency. The important possibility to realize is that the fractionating and the LENR are both part of the SAME mechanism. There can be no leaks because without the fractionating mechanism operating, there would not be any LENR. On each pair of hydrons, the fractionating mechanism is required to allow the nuclear reaction to occur. This guarantees no leakage, except for secondary effects. So in this scenario, 100% efficient fractionating is possible. Bob Higgins
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Hi Eric, I have made progress and have constructed a new reactor optimized to allow low energy photons to escape. These would be unmistakable signatures of LENR without having to be so optimized to show excess heat to the extent it proves a nuclear source. I have seen transient heat bursts and I want to correlate these with emitted photons. Unfortunately, I am on a temporary hold to get myself and my little lab moved across the US to NM. Bob On Fri, Feb 7, 2014 at 10:49 PM, Eric Walker eric.wal...@gmail.com wrote: On Fri, Feb 7, 2014 at 7:43 AM, Bob Higgins rj.bob.higg...@gmail.comwrote: Rossi has stated that he starts with 10 micron sized particles (since identified as a nickel powder produced from the carbonyl process), adds a catalyst (widely believed to be a nanopowder of some kind), and processes the mix in a way that leads to amplified tubercles on the surface. Thanks for the helpful clarification. I didn't realize that. The main reference I have found is Hank Mills's PESN article [1]. I'm curious where Mills got this information. It sounds like you have made a lot of progress on getting an NiH reactor set up. Have you seen anything interesting? Eric [1] http://pesn.com/2012/01/02/9601998_Defkalion_Claims_No_Problem_with_Revealing_Cold_Fusion_Catalyst/
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric, Jones and Ed--Bob Cook here-- Note that Pam Mosier Boss and Larry (the radiation count specialist consultant for SPAWAR) talked about the CR-39 scheme for monitoring radiation from the Pd-D system they worked with. (This was 2009 at the U of Mo.) They saw evidence of tritium, neutrons, and high energy alphas and He-3. Gamma radiation was also apparent.However there was no apparent gamma radiation associated with the major reaction of 2 D's going to He-4, only the evidence of large melted areas in the Pd electrode and no apparent kinetic energy associated with those alphas. They alphas from the D-D fusion were produced in the Pd electrode, apparently standing, yet there was distribution of the excess energy to the electrode to cause the significant melting of the Pd. They did not see any indication of fission parts of the Pd. . At least if there was any they did not report it. If such fission products were energetic they would have been observed in their CR-39 detector. The reaction (D-D fusion) was real and with no irradiation measured. My assessment is that it happened much like a small nuclear explosion except much faster--instantaneously--once the quantum system was properly stimulated. Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Saturday, February 08, 2014 11:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Feb 8, 2014, at 12:26 PM, Eric Walker wrote: On Sat, Feb 8, 2014 at 11:11 AM, Jones Beene jone...@pacbell.net wrote: No, I provide two facts from the Rossi experiments. No gamma. No tritium. ... These are facts, not assertions. Jones, your analysis is often insightful. But here you're stating facts, and then implying assumptions on the basis of those facts as facts as well. You assume that d+d fusion will result in a gamma, and then when no gamma is seen, you assume that d+d fusion in NiH is not possible. You have assumed away some mechanism that might be fractionating the gamma. And then later you draw upon related arguments to support this assumption. In repeating this line of reasoning, you are as guilty of simple, repetitive assertion of your assumptions as Ed is of his. Simply asserting an assumption to be true, or drawing upon such an assumption implicitly to reason about other things, does not make the assumption true. I suspect d+d fusion is not going on in Rossi's reactor either, but for reasons other than a missing gamma. We have no evidence one way or another about tritium, but no specific reason to believe it is there either. Eric, no one believes d+d fusion occurs in the Rossi reactor. The d we are discussing results from p-e-p fusion only. I agree with the other comments you make. Ed Storms In fact, all the important evidence shows the two cannot be similar in any meaningful way. This is an overstatement. Can we all adopt a more measured tone? There is no high energy event in the Rossi effect, or it would have been seen in the Bianchini radiation monitoring. Can you provide a link to the Bianchini report? For some reason I'm having trouble finding it. I assume that this was the appendix provided in connection with the Elforsk test? The only report I'm finding deals with a different subject relating to the E-Cat, in 2010 [1]. In the Elforsk test, no radiation was seen. There were obviously working parameters for the radiation monitor and an upper and lower threshold beyond which it would not have been effective. I do not know what type of monitor was used or what these thresholds were. But what we can deduce from this situation is that no penetrating radiation was escaping the system. It is a nonsequitor to conclude anything about the amount of energy being dissipated, let alone to conclude something about spin coupling as a possible mechanism. Spin coupling does not apply to the fusion of deuterium into helium. You are intentionally conflating two unrelated effects. This is a simple assertion. Can we lay off of these a little? Eric [1] http://e-cataustralia.com/pdf/Levi_Bianchini_and_Villa_Reports.pdf
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Sat, Feb 8, 2014 at 5:33 PM, Bob Cook frobertc...@hotmail.com wrote: However there was no apparent gamma radiation associated with the major reaction of 2 D's going to He-4, only the evidence of large melted areas in the Pd electrode and no apparent kinetic energy associated with those alphas. The finding of standing/low-energy alphas is very common in the older PdD research. It took me literally years to accept that this was possible. Ed Storms has been insisting on it, and I just didn't see how it was consistent with fusion, so I assumed that the alphas were being stopped in the material and the accompanying Bremsstrahlung screened out by the substrate and the material intervening between the substrate and the measurement device. Realistically, this is improbable, because holding an alpha emitter behind a sheet of palladium will give rise to Bremsstrahlung as the alphas collide with the far side of the sheet, as was seen in a control in one early experiment. I now am willing to accept the experimental finding of alphas being born without kinetic energy. The reason for this is that I think there is something along the lines that Bob Higgins described here [1] going on. It was not until I had some kind of explanation that made sense to me that I was able to go along with what the experimenters were saying. Note that in a lot of the CR-39 experiments (there have been many over the years), there is evidence for ~10 MeV alphas and ~1-3 MeV protons. But the important question is whether they are at a sufficient level to explain what is going on, and I think the consensus is that they are not. All of that is of interest in the context of NiH insofar as it points out the possibility of the gamma energy being dissipated in a benign way. Eric [1] http://www.mail-archive.com/vortex-l@eskimo.com/msg89992.html
Re: [Vo]:MIT Course Day 5 -- NiH Systems
I agree with you Eric, the jury is still out. Ed's way of thinking is more in line with my recent thoughts about a retarding magnetic field effect. He may not agree, but it is easier for me to understand how a process that slows down the snap action associated with the acceleration of the charged particles by the strong force could allow the energy to be dissipated slowly instead of in one large pulse. I visualize forcing the proton(s) to crawl to the nickel nucleus or each other kind of like moving through molasses. After all, it is well known that electromagnetic radiation is generated by the acceleration of charged particles and the rate of that acceleration must determine the spectrum of the radiation emitted. Large magnetic fields have been shown to divert moving charged particles. As I have mentioned previously, DGT has reported the presence of a much larger external magnetic field that anyone would have expected and I assume that they would not have placed that report into the public arena had it been false. I am taking them at their word about this measurement until proven otherwise. A large external magnetic field might well translate into an extremely large internal field at the active sites. Couple that with positive feedback and you get a significant amount of power generation. So far this is the theory that I favor. Dave -Original Message- From: Eric Walker eric.wal...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Sat, Feb 8, 2014 5:25 pm Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Sat, Feb 8, 2014 at 2:07 PM, Jones Beene jone...@pacbell.net wrote: However, I am not convinced that PdD works this way, and frankly - it is a diversion to even bring it up for now, since it detracts from the really important issue - which is the proper understanding of the Rossi effect. How is it a diversion to bring up an apparently well-established conclusion that a large quantum of mass energy can be fractionated without penetrating radiation? That was the point that was at issue. Answer: it's not a diversion. The conclusion may be flawed, the evidence may be flawed, the interpretation may be flawed, and/or the research may be flawed. But a consensus conclusion about the fractionation of a 24 MeV quantum into non-penetrating radiation is something to be addressed in a conversation dealing with the question of whether fractionation is possible. I'm not trying to say that the fractionation conclusion is for sure what is going on, either in NiH or in PdD. Only that it's not out in the wilderness either, as some would tendentiously make it out to be. :) Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Bob Higgins--Bob Cook here-- I agree with your logic regarding 100% efficient fractionating as possible. As I noted in an earlier comment Mosier-Boss etal at SPAWAR saw two separate reactions, the one LENR with no radiation being D-D going to He-4. It was also the dominant reaction that happened in their Pd-D unshielded cell. Bob - Original Message - From: Bob Higgins To: vortex-l@eskimo.com Cc: Bob Higgins Sent: Saturday, February 08, 2014 5:23 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems I am not going to try to quote who and what from this thread regarding fractionating gammas (too long of a story line now). What I have come to believe and what I initially missed, and what I think many Vorts may be missing in this, is that the LENR reaction and the fractionating are not two separate processes. Jones (et al) are correct that if there is a fractionating mechanism that is an independent effect, it could not be 100% efficient and some high energy photons would escape as a marker of this inefficiency. The important possibility to realize is that the fractionating and the LENR are both part of the SAME mechanism. There can be no leaks because without the fractionating mechanism operating, there would not be any LENR. On each pair of hydrons, the fractionating mechanism is required to allow the nuclear reaction to occur. This guarantees no leakage, except for secondary effects. So in this scenario, 100% efficient fractionating is possible. Bob Higgins
RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
IMHO grain size and geometry of these other alloys as powders will have a major effect on their LENR activity. Fran From: Jones Beene [mailto:jone...@pacbell.net] Sent: Thursday, February 06, 2014 5:16 PM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems From: Jed Rothwell Superior for what? Conducting protons? Surely not for loading hydrogen. I have never heard that. Surely you read Ahern's Arata replication for EPRI ? He achieved better loading than the standard of 1:1 with nickel-palladium alloy (at low Pd ratio in the alloy). Many alloys which are tailored for hydrogen storage are in fact better than palladium for that single property (which is the atomic ratio of lattice atoms to hydrogen atoms) This does not meant they will be more active for LENR - only that they will absorb more atoms of hydrogen per atom of lattice. That is what they are designed for. In fact, the alloys which store the most hydrogen are most often NOT anomalous as to energy release, when further stimulated. Unfortunately, the two fields have not been systematically investigated for determining the best of both worlds. Jones
RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Bob, Much discussion regarding micro “tubule” geometry of Rossi powders leads many of us to consider the hair like protrusions as forming nano geometry between the grains as they pack to form a bulk powder. Fran _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: Thursday, February 06, 2014 5:50 PM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems -Original Message- From: Bob Cook Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it … Not sure that I follow this. Although the Rossi patent mentions nanometric and specifically a favored isotope - Rossi himself has identified his nickel supplier, and says the geometry of his powder is micron not nano (at least at that point in time). Metals (as opposed to ceramics) can seldom be reduced below 10 microns by normal Industrial methods such as ball milling - due to surface electric properties aka: “agglomeration.” That is one reason why “nano” is so special and not fully appreciated wrt metals. It simply cannot happen in normal metal processing (except with mixed ceramics like the oxides of nickel). You might do well to talk to the Ni-O “nano” suppliers, like Quantum sphere: http://www.qsinano.com/products_nanomaterials.html They will set you straight on the lack of anything truly “nano” as a metal. It must have a surface oxide. … and may be the reason other researchers do not have very good luck at getting a good reaction. No doubt that Rossi, if we can believe his results, has found something that no one else has yet been able to duplicate. It may be serendipitous, but it is not likely to be “nanometric nickel” per se. Jones
RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Roarty, Francis X Bob, Much discussion regarding micro “tubule” geometry of Rossi powders leads many of us to consider the hair like protrusions as forming nano geometry between the grains as they pack to form a bulk powder. Fran The tubes could be hollow as well as in Enculescu’s image below. Does anyone have the citation for Rossi’s nickel lattice having “tubules”? Cannot find it. But check this out. http://www.science24.com/paper/11457 This is a marvelous image of what can be done, in principal, with nickel nanotubes via electroless deposition. It would not surprise me if Rossi’s supplier of nickel has used a similar technique. This particular paper is Romanian/German and has no connection to LENR that I am aware of. I wonder if Peter Gluck is aware of it? Perhaps a gram or two of this actual material should be tried in LENR, due to the possibility of entrapment of hydrogen in the tubes in one dimension, as we have discussed. As a caveat, this electroless nickel deposition technique apparently involves high phosphorous content, which could be a poison (who knows?) _ From: Jones Beene -Original Message- From: Bob Cook * Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it … Not sure that I follow this. Although the Rossi patent mentions nanometric and specifically a favored isotope - Rossi himself has identified his nickel supplier, and says the geometry of his powder is micron not nano (at least at that point in time). Metals (as opposed to ceramics) can seldom be reduced below 10 microns by normal Industrial methods such as ball milling - due to surface electric properties aka: “agglomeration.” That is one reason why “nano” is so special and not fully appreciated wrt metals. It simply cannot happen in normal metal processing (except with mixed ceramics like the oxides of nickel). You might do well to talk to the Ni-O “nano” suppliers, like Quantum sphere: http://www.qsinano.com/products_nanomaterials.html They will set you straight on the lack of anything truly “nano” as a metal. It must have a surface oxide. * … and may be the reason other researchers do not have very good luck at getting a good reaction. No doubt that Rossi, if we can believe his results, has found something that no one else has yet been able to duplicate. It may be serendipitous, but it is not likely to be “nanometric nickel” per se. Jones attachment: winmail.dat
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
morphing through some kind of process of entropy I think you are right, Vacuum = Entropy = Uncertainty! On Fri, Feb 7, 2014 at 10:00 AM, Eric Walker eric.wal...@gmail.com wrote: On Fri, Feb 7, 2014 at 6:53 AM, Jones Beene jone...@pacbell.net wrote: Does anyone have the citation for Rossi's nickel lattice having tubules? Cannot find it. But check this out. Yes, please. If anyone has a reference to Rossi using nickel with tubules, nanotubules, nanohairs, etc., please provide it. My understanding is that he uses micron-sized nickel powder, treated in some way, and rather than something nano-. There are carbon nanotubes, of course, and Rossi, as far as anyone knows, does not use them. It's remarkably difficult to pin the precise details down and keep them pinned down. They keep on moving around and morphing through some kind of process of entropy. Eric
RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
As Eric realizes, this is a critical issue for anyone wanting to replicated Rossi. In fact, the material shown in the previous image, could indeed be called “micron sized” and one would not be dishonest. However the importance of hollow nickel tube could be the sine qua non of the Rossi scheme. Rossi has a history in his revelations, at least back when he was in full fund-raising mode, of first providing a bit too much information (inadvertently) and then backtracking later to try to minimize the damage. Thus, we often see conflicting statements which can be rationalized if one understands the history of “Rossi-speak”. This “tubule” mystery could be an exemplary example of what I am talking about. But did he actually ever say it? From: Eric Walker Does anyone have the citation for Rossi’s nickel lattice having “tubules”? Cannot find it. But check this out. Yes, please. If anyone has a reference to Rossi using nickel with tubules, nanotubules, nanohairs, etc., please provide it. My understanding is that he uses micron-sized nickel powder, treated in some way, and rather than something nano-. There are carbon nanotubes, of course, and Rossi, as far as anyone knows, does not use them. It's remarkably difficult to pin the precise details down and keep them pinned down. They keep on moving around and morphing through some kind of process of entropy. Eric
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Rossi has stated that he starts with 10 micron sized particles (since identified as a nickel powder produced from the carbonyl process), adds a catalyst (widely believed to be a nanopowder of some kind), and processes the mix in a way that leads to amplified tubercles on the surface. A search of tubules will not find the reference, he used tubercles. I have replicated the growth of tubercles by doing just what Rossi described. Begin with micron scale nickel powder (from the carbonyl precipitate process), add a nanopowder, mix, and heat in an oven with cycling H2, Ar, O2 process gas. The result is a porous structure of tubercles with nanowires growing from the surface. I suspect that both the nanowires and the tubercle structure are indicators that I am using similar processing of the powder mix as Rossi, but are not themselves the LENR NAE. The observation is that when processed in that manner, there are plenty of NAE somewhere. It is easy to believe that this structure (from the SEM pictures) will be rife with nanocracks as Dr. Storms suggests for the NAE. In fact, the NAE are likely to be features you cannot see under the SEM rather than the features you can see. Bob Higgins On Fri, Feb 7, 2014 at 10:12 AM, Jones Beene jone...@pacbell.net wrote: As Eric realizes, this is a critical issue for anyone wanting to replicated Rossi. In fact, the material shown in the previous image, could indeed be called micron sized and one would not be dishonest. However the importance of hollow nickel tube could be the *sine qua non* of the Rossi scheme. Rossi has a history in his revelations, at least back when he was in full fund-raising mode, of first providing a bit too much information (inadvertently) and then backtracking later to try to minimize the damage. Thus, we often see conflicting statements which can be rationalized if one understands the history of Rossi-speak. This tubule mystery could be an exemplary example of what I am talking about. But did he actually ever say it? *From:* Eric Walker Does anyone have the citation for Rossi's nickel lattice having tubules? Cannot find it. But check this out. Yes, please. If anyone has a reference to Rossi using nickel with tubules, nanotubules, nanohairs, etc., please provide it. My understanding is that he uses micron-sized nickel powder, treated in some way, and rather than something nano-. There are carbon nanotubes, of course, and Rossi, as far as anyone knows, does not use them. It's remarkably difficult to pin the precise details down and keep them pinned down. They keep on moving around and morphing through some kind of process of entropy.
RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Ah. tubercles instead of tubules . Thanks Bob From: Bob Higgins Rossi has stated that he starts with 10 micron sized particles (since identified as a nickel powder produced from the carbonyl process), adds a catalyst (widely believed to be a nanopowder of some kind), and processes the mix in a way that leads to amplified tubercles on the surface. A search of tubules will not find the reference, he used tubercles. I have replicated the growth of tubercles by doing just what Rossi described.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni-proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved. No radioactivity has been found also in the Nickel residual from the process. This information attributed to Focardi and Rossi comes from their instructive statements, which suggest the nuclear Ni-proton fusion, in the following paper: A new energy source from nuclear fusion S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna University and INFN Bologna Section, (2)Leonardo Corp. (USA) - Inventor of the Patent, March 22, 2010 (international patent publication N. WO 2009/125444 A1) My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Other Ni-hydrogen materials that have been produced by other experimenters should be carefully checked for both the potential radioactive Ni isotopes---Ni-59 and Ni-63. They should be easy to detect given their well known decay modes and probable gamma emissions. (I will look up this information and put it in a subsequent comment.) I know that both Ni-59 and Ni-63 are problems when it comes to nuclear waste disposal of activated metals.) A null radioactivity essay would be revealing as to the process actually occurring in the Ni-hydrogen reactions. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Thursday, February 06, 2014 7:45 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Thu, Feb 6, 2014 at 2:26 PM, Bob Cook frobertc...@hotmail.com wrote: Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it and may be the reason other researchers do not have very good luck at getting a good reaction. I'm guessing that the purity of Rossi's nickel (in terms of 62Ni and 64Ni) is related to avoiding beta-plus and beta-minus decay, and, with beta-plus decay, the 511 keV positron-electron annihilation photons. Some vorts may enjoy this video of a small cloud chamber [1]. It's remarkable that such a small event can have macroscopic effects. Eric [1] http://www.youtube.com/watch?v=xQVMrkJYShc
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Bob--Bob Cook here Your comments are revealing. I believe quantum systems that are big enough to handle the energy fractionation that Hagelstein identifies in his lectures are a requirement for any solid state nuclear reaction. A thermal conductor to get the heat out is also necessary. These two objectives are probably at the heart of Rossi's design. Of course the Kim BEC theory may occur at discrete locations in the Ni creating new quantum systems during the reactor operation. However maintaining such nice locations for months of operation for the BEC's to form is questionable. Bob - Original Message - From: Bob Higgins To: vortex-l@eskimo.com Cc: Bob Higgins Sent: Friday, February 07, 2014 7:43 AM Subject: Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems Rossi has stated that he starts with 10 micron sized particles (since identified as a nickel powder produced from the carbonyl process), adds a catalyst (widely believed to be a nanopowder of some kind), and processes the mix in a way that leads to amplified tubercles on the surface. A search of tubules will not find the reference, he used tubercles. I have replicated the growth of tubercles by doing just what Rossi described. Begin with micron scale nickel powder (from the carbonyl precipitate process), add a nanopowder, mix, and heat in an oven with cycling H2, Ar, O2 process gas. The result is a porous structure of tubercles with nanowires growing from the surface. I suspect that both the nanowires and the tubercle structure are indicators that I am using similar processing of the powder mix as Rossi, but are not themselves the LENR NAE. The observation is that when processed in that manner, there are plenty of NAE somewhere. It is easy to believe that this structure (from the SEM pictures) will be rife with nanocracks as Dr. Storms suggests for the NAE. In fact, the NAE are likely to be features you cannot see under the SEM rather than the features you can see. Bob Higgins On Fri, Feb 7, 2014 at 10:12 AM, Jones Beene jone...@pacbell.net wrote: As Eric realizes, this is a critical issue for anyone wanting to replicated Rossi. In fact, the material shown in the previous image, could indeed be called micron sized and one would not be dishonest. However the importance of hollow nickel tube could be the sine qua non of the Rossi scheme. Rossi has a history in his revelations, at least back when he was in full fund-raising mode, of first providing a bit too much information (inadvertently) and then backtracking later to try to minimize the damage. Thus, we often see conflicting statements which can be rationalized if one understands the history of Rossi-speak. This tubule mystery could be an exemplary example of what I am talking about. But did he actually ever say it? From: Eric Walker Does anyone have the citation for Rossi's nickel lattice having tubules? Cannot find it. But check this out. Yes, please. If anyone has a reference to Rossi using nickel with tubules, nanotubules, nanohairs, etc., please provide it. My understanding is that he uses micron-sized nickel powder, treated in some way, and rather than something nano-. There are carbon nanotubes, of course, and Rossi, as far as anyone knows, does not use them. It's remarkably difficult to pin the precise details down and keep them pinned down. They keep on moving around and morphing through some kind of process of entropy.
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
Fran-- I agree fully. Bob Cook - Original Message - From: Roarty, Francis X To: vortex-l@eskimo.com Sent: Friday, February 07, 2014 5:36 AM Subject: RE: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems IMHO grain size and geometry of these other alloys as powders will have a major effect on their LENR activity. Fran From: Jones Beene [mailto:jone...@pacbell.net] Sent: Thursday, February 06, 2014 5:16 PM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems From: Jed Rothwell Superior for what? Conducting protons? Surely not for loading hydrogen. I have never heard that. Surely you read Ahern's Arata replication for EPRI ? He achieved better loading than the standard of 1:1 with nickel-palladium alloy (at low Pd ratio in the alloy). Many alloys which are tailored for hydrogen storage are in fact better than palladium for that single property (which is the atomic ratio of lattice atoms to hydrogen atoms) This does not meant they will be more active for LENR - only that they will absorb more atoms of hydrogen per atom of lattice. That is what they are designed for. In fact, the alloys which store the most hydrogen are most often NOT anomalous as to energy release, when further stimulated. Unfortunately, the two fields have not been systematically investigated for determining the best of both worlds. Jones
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
I believe that some fractionation must be taking place, but not to phonons. Phonons are contra-indicated by the experimental evidence. Phonons dissipate rapidly to heat with a decay constant that is based on the acoustic velocity. This means that the temperature will be extremely high near the nanoscale NAE, making it much higher temperature than the bulk of the reactor. It suggests that before any useful total heat is realized for the system, the NAE would burn itself out - melt, evaporate, etc. On the other hand, if the output from the NAE was fractionated to lower energy photons, then the decay constant would be based on the speed of light in the material and the deposition to heat would be spread much farther away from the NAE, allowing heat transport out of the NAE without overheating the NAE structure. The micro-explosions that have been reported are on a micron-scale, not on a nano-scale; nanoscale would be expected with phonons. The whole device melt-downs that have been reported can only happen if the NAE is not that much hotter than the bulk of the device. Photons would spread the heat away from the NAE in such a way that the meltdowns and micron-size explosions could occur. Keep in mind that Dr. Hagelstein has PRESUMED coupling to phonons in the formulation of his mathematical experiment. The formulation is not the completely general case with the best solution popping out. The general formulation is too complex to solve today, so simplifying presumptions must be made, and then the solutions are evaluated for consistency with experiment. The simplified formulation just makes it solve-able, not easy to solve. So, in this sense, Dr. Hagelstein is constructing mathematical experiments (the simplifications) and is testing the solutions to see if they match all of the experimental data. If he guesses right in his simplification (didn't leave out something important in his formulation), and finds a match to all of the experimental data, then he has a good theory. It is all based on the same original physics which cannot be solved in purely general form for the complex condensed matter environment. We may not know enough about the NAE to be able to simulate it today because we don't know what simplifications are appropriate. Bob On Fri, Feb 7, 2014 at 12:54 PM, Bob Cook frobertc...@hotmail.com wrote: Bob--Bob Cook here Your comments are revealing. I believe quantum systems that are big enough to handle the energy fractionation that Hagelstein identifies in his lectures are a requirement for any solid state nuclear reaction. A thermal conductor to get the heat out is also necessary. These two objectives are probably at the heart of Rossi's design. Of course the Kim BEC theory may occur at discrete locations in the Ni creating new quantum systems during the reactor operation. However maintaining such nice locations for months of operation for the BEC's to form is questionable. Bob
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows a distribution that is consistent with what is reported and reveals Ni-58 to be the most active isotope for energy production. I will provide much more detail and justification in my book. Meanwhile, you might consider this proposed process. I propose transmutation takes place in the Rossi cell, but he has incorrectly identified its source and incorrectly attributed the energy to transmutation. I propose most energy results from p-e-p=d fusion, with transmutation resulting from fission of Ni adding only a minor amount of energy. If this is the case, focus on Ni is a waste of time. Ed Storms On Feb 7, 2014, at 10:39 AM, Bob Cook wrote: Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni- proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved. No radioactivity has been found also in the Nickel residual from the process. This information attributed to Focardi and Rossi comes from their instructive statements, which suggest the nuclear Ni-proton fusion, in the following paper: A new energy source from nuclear fusion S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna University and INFN Bologna Section, (2)Leonardo Corp. (USA) - Inventor of the Patent, March 22, 2010 (international patent publication N. WO 2009/125444 A1) My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Other Ni- hydrogen materials that have been produced by other experimenters should be carefully checked for both the potential radioactive Ni isotopes---Ni-59 and Ni-63. They should be easy to detect given their well known decay modes and probable gamma emissions. (I will look up this information and put it in a subsequent comment.) I know that both Ni-59 and Ni-63 are problems when it comes to nuclear waste disposal of activated metals.) A null radioactivity essay would be revealing as to the process actually occurring in the Ni- hydrogen reactions. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Thursday, February 06, 2014 7:45 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Thu, Feb 6, 2014 at 2:26 PM, Bob Cook frobertc...@hotmail.com wrote: Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it and may be the reason other researchers do not have very good luck at getting a good reaction. I'm guessing that the purity of Rossi's nickel (in terms of 62Ni and 64Ni) is related to avoiding beta-plus and beta-minus decay, and, with beta-plus decay, the 511 keV positron-electron annihilation photons. Some vorts may enjoy
RE: [Vo]:MIT Course Day 5 -- NiH Systems
Agreed. The issue of a nearly complete lack of transmutation in many types of Ni-H is revealing. It narrows the range of possible energetic reactions which are possible, given that everything else probably conforms to normal physics. In some experiments (Piantelli) has shown far more transmutation than is seen than in other similar reactions, so I have tried to limit this analysis to Rossi, given the fact that he is clearly miles ahead in the race towards commercialization. If we first understand Rossi, then perhaps we will see why he is so far ahead. Piantelli is eating Rossi's photons, as your grandson the video gamer, might phrase it. It the excess heat is a million times more than can be accounted for by a tiny amount of transmutation, then the explanation leans one way. It the excess heat is less than 100 times more than can be accounted for by a much larger amount of transmutation, then the explanation leans another way. The second is Piantelli, the first is Rossi. Both explanation could be accurate for the type of experiment they are doing, due to small variation in reactants - possibly hidden, and unknown even to the experimenter himself. Rossi's reaction is clearly in the trillion plus range of excess heat over any possible transmutation. Transmutation does not happen without measureable levels of radiation, such as would be seen on the meters of Bianchini, with his expert qualifications. That fact is telling - since no excess radiation was seen. Thus no meaningful transmutation. Mills also reports none. Bottom line - the Rossi reaction is most likely a reaction which fundamentally does not involve either high energy photons or transmutation. High energy photons will occasionally leak and produce transmutation. There is no leak proof way to hide 1.1 MeV. Sorry. No high energy photons, no transmutation, then no fusion reaction has occurred which is known to produce high energy quanta. It is as simple as that. From: Edmund Storms Bob and Eric, the issue of transmutation is basic to understanding LENR.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Ed-- One simple question--In all the Ni-H systems has there been a good evaluation of the residuel radioactivity? What do you mean by fragmentation and Ni fission? For example, what are possible fission products? Lighter isotopes which are radioactively stable? Is the fission process like the reaction of a neutron with U-235 producing fragments with kinetic energy, or do the fragments merely stay put. However, If what you suggest happens, i.e. the introduction of d to the Ni nuclei, why not the following reactions?: Ni-58 goes to Cu-60 (radioactive) Ni-60 goes to Cu-62 (radioactive) Ni-61 goes to Cu-63 (stable) Ni-62 goes to Cu-64 (radioactive) and Ni-64 goes to Cu-66 (radioactive). All the radioactive Cu isotopes emit electrons or positrons and additional x-rays or soft gammas to boot, in addition to the .51 mev x-ray associated with positrons-electron reaction. Cu short-lived activity should be seen if the D-Ni reaction occurs. Rossi and Focardi did not appear to advocate such reactions. And I would have estimated that they would have looked for them. Remember they indicated no residual activity and did not mention the P-e-P reaction in their patent application. . Focardi must surely have known about it--the P-e-P reaction. Everything I have heard Focardi say and write has made sense to me and has seemed to be without obfuscation. (I cannot say this for hot fusion advocates and the APS establishment.) However, it would not be the first time I was wrong. A mentor once said it takes $1,000,000 worth of mistakes to make a good engineer, and that was in the late 60's. Luckily I do not have to worry about the issues Hagelstein and others make about my future career. Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows a distribution that is consistent with what is reported and reveals Ni-58 to be the most active isotope for energy production. I will provide much more detail and justification in my book. Meanwhile, you might consider this proposed process. I propose transmutation takes place in the Rossi cell, but he has incorrectly identified its source and incorrectly attributed the energy to transmutation. I propose most energy results from p-e-p=d fusion, with transmutation resulting from fission of Ni adding only a minor amount of energy. If this is the case, focus on Ni is a waste of time. Ed Storms On Feb 7, 2014, at 10:39 AM, Bob Cook wrote: Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni-proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 6:53 AM, Jones Beene jone...@pacbell.net wrote: Does anyone have the citation for Rossi’s nickel lattice having “tubules”? Cannot find it. But check this out. Yes, please. If anyone has a reference to Rossi using nickel with tubules, nanotubules, nanohairs, etc., please provide it. My understanding is that he uses micron-sized nickel powder, treated in some way, and rather than something nano-. There are carbon nanotubes, of course, and Rossi, as far as anyone knows, does not use them. It's remarkably difficult to pin the precise details down and keep them pinned down. They keep on moving around and morphing through some kind of process of entropy. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Ed--Bob Cook here-- Another question is if D is formed in the Ni-H system as you propose, why not the generation of He-4 as in the Pd system without the nasty fragmentation or fission of the Ni? The key to controlling the Rossi process maybe controlling the formation of D. The energy transfer process would be coupled by spin and distribution of angular momentum between the initially excited He-4* at a high spin state and the electrons of the system and maybe the various Ni nuclei in the system. Nuclear-magnetic spin of nuclei is of course coupled to electromagnetic irradiation signals in MRI technology. The math must be well established. Again spin state coupling with appropriate energy transfer should be explored in theory--this is above my head. Do you know if anyone has looked at this? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows a distribution that is consistent with what is reported and reveals Ni-58 to be the most active isotope for energy production. I will provide much more detail and justification in my book. Meanwhile, you might consider this proposed process. I propose transmutation takes place in the Rossi cell, but he has incorrectly identified its source and incorrectly attributed the energy to transmutation. I propose most energy results from p-e-p=d fusion, with transmutation resulting from fission of Ni adding only a minor amount of energy. If this is the case, focus on Ni is a waste of time. Ed Storms On Feb 7, 2014, at 10:39 AM, Bob Cook wrote: Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni-proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved. No radioactivity has been found also in the Nickel residual from the process. This information attributed to Focardi and Rossi comes from their instructive statements, which suggest the nuclear Ni-proton fusion, in the following paper: A new energy source from nuclear fusion S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna University and INFN Bologna Section, (2)Leonardo Corp. (USA) - Inventor of the Patent, March 22, 2010 (international patent publication N. WO 2009/125444 A1) My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Other Ni-hydrogen materials that have been produced by other experimenters should be carefully checked for both the potential radioactive Ni isotopes---Ni-59 and Ni-63. They should be easy
Re: [Vo]:MIT Course Day 5 -- NiH Systems
- Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows a distribution that is consistent with what is reported and reveals Ni-58 to be the most active isotope for energy production. I will provide much more detail and justification in my book. Meanwhile, you might consider this proposed process. I propose transmutation takes place in the Rossi cell, but he has incorrectly identified its source and incorrectly attributed the energy to transmutation. I propose most energy results from p-e-p=d fusion, with transmutation resulting from fission of Ni adding only a minor amount of energy. If this is the case, focus on Ni is a waste of time. Ed Storms On Feb 7, 2014, at 10:39 AM, Bob Cook wrote: Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni-proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved. No radioactivity has been found also in the Nickel residual from the process. This information attributed to Focardi and Rossi comes from their instructive statements, which suggest the nuclear Ni-proton fusion, in the following paper: A new energy source from nuclear fusion S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna University and INFN Bologna Section, (2)Leonardo Corp. (USA) - Inventor of the Patent, March 22, 2010 (international patent publication N. WO 2009/125444 A1) My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Other Ni-hydrogen materials that have been produced by other experimenters should be carefully checked for both the potential radioactive Ni isotopes---Ni-59 and Ni-63. They should be easy to detect given their well known decay modes and probable gamma emissions. (I will look up this information and put it in a subsequent comment.) I know that both Ni-59 and Ni-63 are problems when it comes to nuclear waste disposal of activated metals.) A null radioactivity essay would be revealing as to the process actually occurring in the Ni-hydrogen reactions. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Thursday, February 06, 2014 7:45 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Thu, Feb 6, 2014 at 2:26 PM, Bob Cook frobertc...@hotmail.com wrote: Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it and may be the reason other researchers do not have very
Re: [Vo]:MIT Course Day 5 -- NiH Systems
There was a throw-away line in McKubre's interview with Sterling Allen -- he pointed across the lab and said he was doing a fundamental experiment on phonon interactions with Hagelstein. Maybe there'll be some real data points.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 7, 2014, at 12:54 PM, Bob Cook wrote: Ed-- One simple question--In all the Ni-H systems has there been a good evaluation of the residuel radioactivity? Bob, evidence shows that when Pd or Ni experience transmutation, the resulting nucleus breaks into two parts. These two parts are not radioactive. In other words, the system tries to dissipate all the energy while producing nuclei that have no residual energy, i.e. are not radioactive. Addition of 2(p-e-p) to Ni results is a distribution of stable products, with O, Mg, Si, S Ca, and Ti being the most frequent. Si is matched with S. Prompt alpha emission also occurs leaving behind Ni. This process results from the normal rules of nuclear chemistry. NO copper isotopes are formed. The detected copper has normal isotopic composition, which is not possible to produce from transmutation. I suspect copper results from contamination by materials in the cell. If copper formed, the nucleus would have no way to dissipate the energy, which is essential. What do you mean by fragmentation and Ni fission? For example, what are possible fission products? Lighter isotopes which are radioactively stable? Is the fission process like the reaction of a neutron with U-235 producing fragments with kinetic energy, or do the fragments merely stay put. However, If what you suggest happens, i.e. the introduction of d to the Ni nuclei, why not the following reactions?: Ni-58 goes to Cu-60 (radioactive) Ni-60 goes to Cu-62 (radioactive) Ni-61 goes to Cu-63 (stable) Ni-62 goes to Cu-64 (radioactive) and Ni-64 goes to Cu-66 (radioactive). All the radioactive Cu isotopes emit electrons or positrons and additional x-rays or soft gammas to boot, in addition to the .51 mev x-ray associated with positrons-electron reaction. Cu short-lived activity should be seen if the D-Ni reaction occurs. Rossi and Focardi did not appear to advocate such reactions. And I would have estimated that they would have looked for them. Remember they indicated no residual activity and did not mention the P-e-P reaction in their patent application. . This is true. They clearly have no understanding of nuclear chemistry. They saw transmutation produced and from this observation ASSUMED that heat resulted from transmutation because they found the p-e-p reaction impossible to explain. My approach is to violate as few basic laws as possible and to find an internally consistent process. That goal involves d-e-d, d-e-p and p-e-p type reactions. In addition, transmutation requires energy that is only available from the fusion reaction. These conclusions lead logically to a model that can explain all observations without ad hoc assumptions or using novel processes. Unfortunately, the justification and details require a book to explain, so don't expect a proof here. Ed Storms Focardi must surely have known about it--the P-e-P reaction. Everything I have heard Focardi say and write has made sense to me and has seemed to be without obfuscation. (I cannot say this for hot fusion advocates and the APS establishment.) However, it would not be the first time I was wrong. A mentor once said it takes $1,000,000 worth of mistakes to make a good engineer, and that was in the late 60's. Luckily I do not have to worry about the issues Hagelstein and others make about my future career. Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Feb 7, 2014, at 1:42 PM, Bob Cook wrote: Ed--Bob Cook here-- Another question is if D is formed in the Ni-H system as you propose, why not the generation of He-4 as in the Pd system without the nasty fragmentation or fission of the Ni? That answer is too complicated to explain here. That is why the book is required. Take my word for the present that I have a good reason for this model. The key to controlling the Rossi process maybe controlling the formation of D. The energy transfer process would be coupled by spin and distribution of angular momentum between the initially excited He-4* at a high spin state and the electrons of the system and maybe the various Ni nuclei in the system. Nuclear-magnetic spin of nuclei is of course coupled to electromagnetic irradiation signals in MRI technology. The math must be well established. These fragmentation products release about 11 MeV/fragment. Please tell me how spin state coupling can transfer this amount of energy. Even adding a p to Ni requires about 6 MeV be dissipated. I know of no example of this much energy being transferred by any kind of coupling mechanism. Do you? Ed Storms Again spin state coupling with appropriate energy transfer should be explored in theory--this is above my head. Do you know if anyone has looked at this? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two different results is the challenge. In the case of Ni+H, I propose the p-e-p fusion process deposits the resulting d in the Ni nucleus, resulting in fragmentation of the product in order to dissipate the excess mass-energy. I believe 2d enter all isotopes of Ni when the fusion reaction is operating. As a result, the 1.9 MeV obtained from the p-e-p reaction is added to any energy resulting from occasional transmutation. When the Ni fissions, it must conserve n and p, which produces a distribution of products that can be calculated. This calculation shows a distribution that is consistent with what is reported and reveals Ni-58 to be the most active isotope for energy production. I will provide much more detail and justification in my book. Meanwhile, you might consider this proposed process. I propose transmutation takes place in the Rossi cell, but he has incorrectly identified its source and incorrectly attributed the energy to transmutation. I propose most energy results from p-e-p=d fusion, with transmutation resulting from fission of Ni adding only a minor amount of energy. If this is the case, focus on Ni is a waste of time. Ed Storms On Feb 7, 2014, at 10:39 AM, Bob Cook wrote: Eric-- Your bring up some interesting questions about the Rossi reactor. The information I have included come from Rossi and Focardi's international patent application noted below. 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. The natural isotopic abundances are: Ni-58, 68.08%; Ni-59, 0%--its radioactive with 1/2 life of 80,000 years; Ni 60, 26.22%; Ni-61, 1.14%; Ni-62, 3.63%; Ni-63, 0%--its radioactive with 1/2 life of 92 years; Ni-64, 0.93%. I would pick Ni-60 because it is more than one transmutation (Ni- proton fusion) away from a radioactive residue. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? Rossi and Focardi seem to contradict themselves with the statements below: ...we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved. No radioactivity has been found also in the Nickel residual from the process. This information attributed to Focardi and Rossi comes from their instructive statements, which suggest the nuclear Ni-proton fusion, in the following paper: A new energy source from nuclear fusion S. Focardi(1) and A. Rossi(2)--(1)Physics Department Bologna
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Ed--Bob Cook here Thanks for you response. I need a little more time to think about your ideas. I need to look at the respective products you identify and the likely other respective fission pieces to see if I agree with what you say makes sense. Just roughly thinking, I would expect a neutron or 2 and maybe an alpha in such a fission process. Is there a good reference on what you call rules of nuclear chemistry? Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 2:23 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Feb 7, 2014, at 12:54 PM, Bob Cook wrote: Ed-- One simple question--In all the Ni-H systems has there been a good evaluation of the residuel radioactivity? Bob, evidence shows that when Pd or Ni experience transmutation, the resulting nucleus breaks into two parts. These two parts are not radioactive. In other words, the system tries to dissipate all the energy while producing nuclei that have no residual energy, i.e. are not radioactive. Addition of 2(p-e-p) to Ni results is a distribution of stable products, with O, Mg, Si, S Ca, and Ti being the most frequent. Si is matched with S. Prompt alpha emission also occurs leaving behind Ni. This process results from the normal rules of nuclear chemistry. NO copper isotopes are formed. The detected copper has normal isotopic composition, which is not possible to produce from transmutation. I suspect copper results from contamination by materials in the cell. If copper formed, the nucleus would have no way to dissipate the energy, which is essential. What do you mean by fragmentation and Ni fission? For example, what are possible fission products? Lighter isotopes which are radioactively stable? Is the fission process like the reaction of a neutron with U-235 producing fragments with kinetic energy, or do the fragments merely stay put. However, If what you suggest happens, i.e. the introduction of d to the Ni nuclei, why not the following reactions?: Ni-58 goes to Cu-60 (radioactive) Ni-60 goes to Cu-62 (radioactive) Ni-61 goes to Cu-63 (stable) Ni-62 goes to Cu-64 (radioactive) and Ni-64 goes to Cu-66 (radioactive). All the radioactive Cu isotopes emit electrons or positrons and additional x-rays or soft gammas to boot, in addition to the .51 mev x-ray associated with positrons-electron reaction. Cu short-lived activity should be seen if the D-Ni reaction occurs. Rossi and Focardi did not appear to advocate such reactions. And I would have estimated that they would have looked for them. Remember they indicated no residual activity and did not mention the P-e-P reaction in their patent application. . This is true. They clearly have no understanding of nuclear chemistry. They saw transmutation produced and from this observation ASSUMED that heat resulted from transmutation because they found the p-e-p reaction impossible to explain. My approach is to violate as few basic laws as possible and to find an internally consistent process. That goal involves d-e-d, d-e-p and p-e-p type reactions. In addition, transmutation requires energy that is only available from the fusion reaction. These conclusions lead logically to a model that can explain all observations without ad hoc assumptions or using novel processes. Unfortunately, the justification and details require a book to explain, so don't expect a proof here. Ed Storms Focardi must surely have known about it--the P-e-P reaction. Everything I have heard Focardi say and write has made sense to me and has seemed to be without obfuscation. (I cannot say this for hot fusion advocates and the APS establishment.) However, it would not be the first time I was wrong. A mentor once said it takes $1,000,000 worth of mistakes to make a good engineer, and that was in the late 60's. Luckily I do not have to worry about the issues Hagelstein and others make about my future career. Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting
Re: EXTERNAL: RE: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 7:43 AM, Bob Higgins rj.bob.higg...@gmail.comwrote: Rossi has stated that he starts with 10 micron sized particles (since identified as a nickel powder produced from the carbonyl process), adds a catalyst (widely believed to be a nanopowder of some kind), and processes the mix in a way that leads to amplified tubercles on the surface. Thanks for the helpful clarification. I didn't realize that. The main reference I have found is Hank Mills's PESN article [1]. I'm curious where Mills got this information. It sounds like you have made a lot of progress on getting an NiH reactor set up. Have you seen anything interesting? Eric [1] http://pesn.com/2012/01/02/9601998_Defkalion_Claims_No_Problem_with_Revealing_Cold_Fusion_Catalyst/
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 9:39 AM, Bob Cook frobertc...@hotmail.com wrote: 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. I can only imagine. I'm not sure how one would go about enriching select isotopes of nickel. Perhaps they have sufficiently different properties to make separation straightforward? (E.g., maybe the spin-0 claim one hears occasionally in connection with some isotopes, which is something I know nothing about, can be made use of.) Hank Mills reports that Rossi has found a cheap way to enrich the nickel, although I do not have an opinion about this [1]. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? I was thinking of radioactive species of copper and zinc, primarily. By contrast, I believe 62Ni and 64Ni would go to stable isotopes of copper after proton capture. In natural abundance, 58Ni is the most prevalent, at 68 percent: p + 58Ni → 59Cu + ɣ + Q (2.9 MeV) Here 59Cu is an unstable species which will beta-plus decay to 59Ni, which will then transition to 59Co via electron capture. I believe it will be accompanied by an Auger cascade, so there will be lots of activity taken together. By contrast, p + 62Ni → 63Cu + ɣ + Q (5.6 MeV) Here 63Cu is a stable isotope. If one assumes the ɣ is somehow being fractionated as a large set of lower-energy photons through some as-yet discovered mechanism, as I suspect is happening (I'm rooting for an interaction with the electronic structure, here), then you want 62Ni and 64Ni, because there will be no activity with these isotopes afterwards. Using nickel in its natural isotopes will be like banging the keys on a piano -- there will be lots of noise leaving the system. My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Yes, very much so. This is one of those mutating details, subject to a mysterious law of entropy, where one doesn't know what to believe. In a related connection, I recall an anecdote of an experiment by one of the Italian researchers, perhaps Piantelli, where some nickel that had been undergoing a reaction was placed in a cloud chamber and all kinds of activity was seen. If there is proton capture happening at a significant level, and there is no activity, my guess is that this would be primarily because Rossi has succeeded in enriching the nickel to suitable isotopes to a high degree. But my understanding is that it is also the case that in PdD experiments, transmutations are often seen to stable isotopes, so there may be something inherent to cold fusion that leads to stable isotopes, mitigating perhaps the need for enrichment to very high levels. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 11:35 AM, Jones Beene jone...@pacbell.net wrote: Agreed. The issue of a “nearly complete lack” of transmutation in many types of Ni-H is revealing. Rossi has claimed this: THE AMOUNT OF COPPER WE FIND AFTER 6 MONTHS OF OPERATION IS OF ORDERS OF MAGNITUDE MORE THAT THE IMPURITIES IN THE 99. Ni WE USE. [1] This is from an article by Matts Lewans: Analyses of the nickel powder used in Rossi’s energy catalyzer show that a large amount of copper is formed. Sven Kullander considers this to be evidence of a nuclear reaction. [2] There are many other similar, suggestive statements out there. I think it is hard to justify the conclusion that there is almost no transmutation being seen in Rossi's device. It the excess heat is a million times more than can be accounted for by a tiny amount of transmutation, then the explanation leans one way. This is a finding from (probably low-gain) PdD research. There is little to conclude about NiH, as far as I can tell. Keep in mind that in PdD d+d fusion may be more energetically favorable than transmutation, by way of whatever magic is happening that is causing cold fusion. Transmutation does not happen without measureable levels of radiation, such as would be seen on the meters of Bianchini, with his expert qualifications. This is a reasonable expectation, but I think the conclusion is too pat. It is possible that a combination of enrichment on Rossi's part and a natural mechanism that favors stable isotopes in LENR are sufficient to keep the activity down if you do your prep work correctly. Bottom line – the Rossi reaction is most likely a reaction which fundamentally does not involve either high energy photons or transmutation. Sure -- everyone is entitled to his or her opinion, and we grant you the same. Eric [1] http://www.journal-of-nuclear-physics.com/?p=395cpage=1#comment-20859 [2] http://www.nyteknik.se/nyheter/energi_miljo/energi/article3144827.ece
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Ed--Bob Cook here Spin states of a quantum system reflect the angular momentum of the system and hence the energy associated with that angular momentum. High spin quantum numbers reflect the higher energy of the system. The allowable states are quantized. In magnetic fields the direction of the spin is controlled more or less depending upon the field strength. The allowable number of states is reduced from the situation where there is no magnetic field. Resonant magnetic oscillating fields input to a nucleus with a magnetic moment and non-zero spin state for its ground state, can add energy to the quantum system by changing the spin number of the quantum system. This is the basis for the MRI technology which is an accounting of the energy absorption at a given resonance frequency at well determined locations, identifying the nucleus with the specific resonance frequency absorption . If there is spin coupling, (a basic assumption is that spin is conserved in any nuclear reaction at the end of the reaction) a coupling between various particles subject to integer, J, quantum seems probable. Thus, any He-4* with a high spin integer J quantum number and excess energy--say 10 mev--would distribute this high angular momentum to electrons or other particles in the quantum system--all the many electrons and particles at the same time. The electrons (and other particles) in turn would distribute their excess spin energy (angular momentum) to the lattice as electromagnetic field oscillations or radiation and hence lattice heat. In the end the net spin would be what it was to start with. The reaction would be fast and cause results of the distribution of quantum angular momentum and lattice motion instantaneously. No energetic (kinetic energy) particles are involved, only angular momentum with its corresponding rotational energy. The rotational energy may actually be rotating electric and or magnetic fields associated with the particle with the high spin quantum state. Again I do not understand the details of spin coupling, the actual timing nor the most likely fractionation of the spin/angular momentum among the particles of the quantum system. The basic idea is that the energy associated with the mass loss first shows up as angular momentum or spin of the newly found He-4* and this spin is distributed to the rest of the system. Bob Cook - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 2:34 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Feb 7, 2014, at 1:42 PM, Bob Cook wrote: Ed--Bob Cook here-- Another question is if D is formed in the Ni-H system as you propose, why not the generation of He-4 as in the Pd system without the nasty fragmentation or fission of the Ni? That answer is too complicated to explain here. That is why the book is required. Take my word for the present that I have a good reason for this model. The key to controlling the Rossi process maybe controlling the formation of D. The energy transfer process would be coupled by spin and distribution of angular momentum between the initially excited He-4* at a high spin state and the electrons of the system and maybe the various Ni nuclei in the system. Nuclear-magnetic spin of nuclei is of course coupled to electromagnetic irradiation signals in MRI technology. The math must be well established. These fragmentation products release about 11 MeV/fragment. Please tell me how spin state coupling can transfer this amount of energy. Even adding a p to Ni requires about 6 MeV be dissipated. I know of no example of this much energy being transferred by any kind of coupling mechanism. Do you? Ed Storms Again spin state coupling with appropriate energy transfer should be explored in theory--this is above my head. Do you know if anyone has looked at this? Bob - Original Message - From: Edmund Storms To: vortex-l@eskimo.com Cc: Edmund Storms Sent: Friday, February 07, 2014 10:32 AM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose the hydrogen fusion process provides the required energy and dissipates much of the excess mass-energy. In other words, transmutation can not occur unless fusion is taking place at the same time and place in the material. We now know that two kinds of transmutation occur. Iwamura shows that D can be added to a target resulting in a stable heavier product. Most other claims for transmutation are based on fragments of Pd being found. Explaining these two
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 2:23 PM, Edmund Storms stor...@ix.netcom.com wrote: NO copper isotopes are formed. The detected copper has normal isotopic composition, which is not possible to produce from transmutation. I suspect copper results from contamination by materials in the cell. You may have jumped to a conclusion on this one on the basis of some low-gain nickel research. I do not think we can conclude that no copper is ever formed, if Rossi's statements are to be taken at face value. In addition, transmutation requires energy that is only available from the fusion reaction. This is an assumption. My intuition tells me that an arbitrarily high potential can form during a transient between two electrically insulated grain boundaries as well, and some interesting things can happen in that connection, although I am not yet able to characterize this system. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 2:34 PM, Edmund Storms stor...@ix.netcom.com wrote: Even adding a p to Ni requires about 6 MeV be dissipated. I know of no example of this much energy being transferred by any kind of coupling mechanism. Do you? That is the question that once answered will win someone the Nobel prize. I assume that something like this is happening. I assume that the person who wins the Nobel prize will be a mainstream physicist who repackages something that was haphazardly mentioned on this list without giving or perhaps even knowing to give credit, long after people have forgotten about this list. Eric
RE: [Vo]:MIT Course Day 5 -- NiH Systems
Yes Rossi did promote Ni-Cu at one time, but it is not his current spiel. AFAIK – since the death of Focardi, Rossi no longer strongly promotes any theory for gain - but it may still be on his blog. Kullander also admitted the copper found was of natural isotope distribution. Copper is well-known to migrate by thermal diffusion rapidly by when in contact with other metals and heated. Rossi probably now realizes this. Rossi’s old reactors were made largely of copper alloy. Any copper found would of necessity have to be radioactive, if from transmutation – and since it was not radioactive, nor in an anomalous isotope distribution – then it had to be from migration instead. From: Eric Walker Jones Beene wrote: Agreed. The issue of a “nearly complete lack” of transmutation in many types of Ni-H is revealing. Rossi has claimed this: THE AMOUNT OF COPPER WE FIND AFTER 6 MONTHS OF OPERATION IS OF ORDERS OF MAGNITUDE MORE THAT THE IMPURITIES IN THE 99. Ni WE USE. [1] This is from an article by Matts Lewans: Analyses of the nickel powder used in Rossi’s energy catalyzer show that a large amount of copper is formed. Sven Kullander considers this to be evidence of a nuclear reaction. [2] There are many other similar, suggestive statements out there. I think it is hard to justify the conclusion that there is almost no transmutation being seen in Rossi's device. It the excess heat is a million times more than can be accounted for by a tiny amount of transmutation, then the explanation leans one way. This is a finding from (probably low-gain) PdD research. There is little to conclude about NiH, as far as I can tell. Keep in mind that in PdD d+d fusion may be more energetically favorable than transmutation, by way of whatever magic is happening that is causing cold fusion. Transmutation does not happen without measureable levels of radiation, such as would be seen on the meters of Bianchini, with his expert qualifications. This is a reasonable expectation, but I think the conclusion is too pat. It is possible that a combination of enrichment on Rossi's part and a natural mechanism that favors stable isotopes in LENR are sufficient to keep the activity down if you do your prep work correctly. Bottom line – the Rossi reaction is most likely a reaction which fundamentally does not involve either high energy photons or transmutation. Sure -- everyone is entitled to his or her opinion, and we grant you the same. Eric [1] http://www.journal-of-nuclear-physics.com/?p=395cpage=1#comment-20859 http://www.journal-of-nuclear-physics.com/?p=395cpage=1 [2] http://www.nyteknik.se/nyheter/energi_miljo/energi/article3144827.ece attachment: winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Bob, I like your approach. In june I whimsically imagined all the resulting energy of fusion being transformed into He with linear momentum. This would preserve conservation energy but violate conservation of momentum. At the time it did not occur to me that the energy could be transformed into angular momentum and thereby preserve conservation of momentum. Harry On Fri, Feb 7, 2014 at 11:35 PM, Bob Cook frobertc...@hotmail.com wrote: Ed--Bob Cook here Spin states of a quantum system reflect the angular momentum of the system and hence the energy associated with that angular momentum. High spin quantum numbers reflect the higher energy of the system. The allowable states are quantized. In magnetic fields the direction of the spin is controlled more or less depending upon the field strength. The allowable number of states is reduced from the situation where there is no magnetic field. Resonant magnetic oscillating fields input to a nucleus with a magnetic moment and non-zero spin state for its ground state, can add energy to the quantum system by changing the spin number of the quantum system. This is the basis for the MRI technology which is an accounting of the energy absorption at a given resonance frequency at well determined locations, identifying the nucleus with the specific resonance frequency absorption . If there is spin coupling, (a basic assumption is that spin is conserved in any nuclear reaction at the end of the reaction) a coupling between various particles subject to integer, J, quantum seems probable. Thus, any He-4* with a high spin integer J quantum number and excess energy--say 10 mev--would distribute this high angular momentum to electrons or other particles in the quantum system--all the many electrons and particles at the same time. The electrons (and other particles) in turn would distribute their excess spin energy (angular momentum) to the lattice as electromagnetic field oscillations or radiation and hence lattice heat. In the end the net spin would be what it was to start with. The reaction would be fast and cause results of the distribution of quantum angular momentum and lattice motion instantaneously. No energetic (kinetic energy) particles are involved, only angular momentum with its corresponding rotational energy. The rotational energy may actually be rotating electric and or magnetic fields associated with the particle with the high spin quantum state. Again I do not understand the details of spin coupling, the actual timing nor the most likely fractionation of the spin/angular momentum among the particles of the quantum system. The basic idea is that the energy associated with the mass loss first shows up as angular momentum or spin of the newly found He-4* and this spin is distributed to the rest of the system. Bob Cook - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Friday, February 07, 2014 2:34 PM *Subject:* Re: [Vo]:MIT Course Day 5 -- NiH Systems On Feb 7, 2014, at 1:42 PM, Bob Cook wrote: Ed--Bob Cook here-- Another question is if D is formed in the Ni-H system as you propose, why not the generation of He-4 as in the Pd system without the nasty fragmentation or fission of the Ni? That answer is too complicated to explain here. That is why the book is required. Take my word for the present that I have a good reason for this model. The key to controlling the Rossi process maybe controlling the formation of D. The energy transfer process would be coupled by spin and distribution of angular momentum between the initially excited He-4* at a high spin state and the electrons of the system and maybe the various Ni nuclei in the system. Nuclear-magnetic spin of nuclei is of course coupled to electromagnetic irradiation signals in MRI technology. The math must be well established. These fragmentation products release about 11 MeV/fragment. Please tell me how spin state coupling can transfer this amount of energy. Even adding a p to Ni requires about 6 MeV be dissipated. I know of no example of this much energy being transferred by any kind of coupling mechanism. Do you? Ed Storms Again spin state coupling with appropriate energy transfer should be explored in theory--this is above my head. Do you know if anyone has looked at this? Bob - Original Message - *From:* Edmund Storms stor...@ix.netcom.com *To:* vortex-l@eskimo.com *Cc:* Edmund Storms stor...@ix.netcom.com *Sent:* Friday, February 07, 2014 10:32 AM *Subject:* Re: [Vo]:MIT Course Day 5 -- NiH Systems Bob and Eric, the issue of transmutation is basic to understanding LENR. First of all, transmutation has a very high barrier requiring an explanation of how this can be overcome. Second, the resulting energy has to be dissipated in ways known to be possible. I propose
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric--Bob Cook here-- It may be that laser radiation at selected frequencies could activate organic Ni compounds selectively based on the mass of the Ni isotope. The activation may be ionization or other molecular changes to increase or decrease solubility and the opportunity for separation. Selective activation may also be possible via the magnetic properties of the respective Ni isotopes with oscillating magnetic fields I think you have the decay scheme for Ni-59 wrong. It has a 76,000 year half life and decays by electron capture as you said. The data I have indicate no gamma activity, in the transition to the Cu-59 nucleus. This is unusual situation that the new nucleus is not formed in an excited state. This is a nice feature of Ni-59, and it should cause no problems. Ni-59 does have a neutron activation cross section, and checking for it in reactor residue should not be to difficult. One would use neutron activation with gamma evaluation of the activated product.One would need to use a research reactor for a source of neutrons like the U of Missouri has. Bob Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Friday, February 07, 2014 8:17 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Fri, Feb 7, 2014 at 9:39 AM, Bob Cook frobertc...@hotmail.com wrote: 1. Is Rossi separating Ni isotopes for the Ni he uses in the reactor? This would be expensive. I can only imagine. I'm not sure how one would go about enriching select isotopes of nickel. Perhaps they have sufficiently different properties to make separation straightforward? (E.g., maybe the spin-0 claim one hears occasionally in connection with some isotopes, which is something I know nothing about, can be made use of.) Hank Mills reports that Rossi has found a cheap way to enrich the nickel, although I do not have an opinion about this [1]. 2. Is there radioactive ash (Ni-59 or Ni-63) left in the spent reactors? I was thinking of radioactive species of copper and zinc, primarily. By contrast, I believe 62Ni and 64Ni would go to stable isotopes of copper after proton capture. In natural abundance, 58Ni is the most prevalent, at 68 percent: p + 58Ni → 59Cu + ɣ + Q (2.9 MeV) Here 59Cu is an unstable species which will beta-plus decay to 59Ni, which will then transition to 59Co via electron capture. I believe it will be accompanied by an Auger cascade, so there will be lots of activity taken together. By contrast, p + 62Ni → 63Cu + ɣ + Q (5.6 MeV) Here 63Cu is a stable isotope. If one assumes the ɣ is somehow being fractionated as a large set of lower-energy photons through some as-yet discovered mechanism, as I suspect is happening (I'm rooting for an interaction with the electronic structure, here), then you want 62Ni and 64Ni, because there will be no activity with these isotopes afterwards. Using nickel in its natural isotopes will be like banging the keys on a piano -- there will be lots of noise leaving the system. My final observation is that the Rossi-Focardi comment that there is no radioactivity in the residue needs to be checked. Yes, very much so. This is one of those mutating details, subject to a mysterious law of entropy, where one doesn't know what to believe. In a related connection, I recall an anecdote of an experiment by one of the Italian researchers, perhaps Piantelli, where some nickel that had been undergoing a reaction was placed in a cloud chamber and all kinds of activity was seen. If there is proton capture happening at a significant level, and there is no activity, my guess is that this would be primarily because Rossi has succeeded in enriching the nickel to suitable isotopes to a high degree. But my understanding is that it is also the case that in PdD experiments, transmutations are often seen to stable isotopes, so there may be something inherent to cold fusion that leads to stable isotopes, mitigating perhaps the need for enrichment to very high levels. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Fri, Feb 7, 2014 at 9:54 PM, Bob Cook frobertc...@hotmail.com wrote: I think you have the decay scheme for Ni-59 wrong. It has a 76,000 year half life and decays by electron capture as you said. It's good that you seem to know your way around these nuclear transitions. That makes you and Robin and a few others who can keep the rest of us honest. The data I have indicate no gamma activity, in the transition to the Cu-59 nucleus. I'm thinking of this reaction: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-62(P%2CG)29-CU-63%2C%2CSIG What data are you using? Do they include proton capture cross sections? Up to now I have only been able to work out the Q values but have had no insight into the cross sections. The Exfor cross section data are hard to make sense of. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
I wrote: I'm thinking of this reaction: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-62(P%2CG)29-CU-63%2C%2CSIG Sorry, that should have been: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-58(P%2CG)29-CU-59%2C%2CSIG Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric-- I am looking at the Knolls Atomic Power Laboratory, Chart of the Nuclides, Thirteenth Addition Revised as of July 1983. This chart does not include proton capture cross sections. I do not believe I have seen proton capture cross sections for any isotopes. The cross section would have to be a function of the proton energy. The thermal neutron cross section of the proton is 0.333 barns and its integral cross section is 0.150 barns. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Friday, February 07, 2014 10:05 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Fri, Feb 7, 2014 at 9:54 PM, Bob Cook frobertc...@hotmail.com wrote: I think you have the decay scheme for Ni-59 wrong. It has a 76,000 year half life and decays by electron capture as you said. It's good that you seem to know your way around these nuclear transitions. That makes you and Robin and a few others who can keep the rest of us honest. The data I have indicate no gamma activity, in the transition to the Cu-59 nucleus. I'm thinking of this reaction: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-62(P%2CG)29-CU-63%2C%2CSIG What data are you using? Do they include proton capture cross sections? Up to now I have only been able to work out the Q values but have had no insight into the cross sections. The Exfor cross section data are hard to make sense of. Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Eric--Bob here I looked at the link and have now seen a list of cross sections for the Ni-59, P reaction. I must study the protocol for measuring the specified cross sections to understand the sig and dsig data. Off hand I do not understand these labels. My guess is that the energies listed are the average of the data within a 1 sigma band of all the data (and also a 2 sigma band of all the data) at a specified incident proton energy. Bob - Original Message - From: Eric Walker To: vortex-l@eskimo.com Sent: Friday, February 07, 2014 10:07 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems I wrote: I'm thinking of this reaction: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-62(P%2CG)29-CU-63%2C%2CSIG Sorry, that should have been: https://www-nds.iaea.org/exfor/servlet/X4sSearch5?reacc=28-NI-58(P%2CG)29-CU-59%2C%2CSIG Eric
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Alan-- I watched the Hagelstein 5th day lecture last night. With respect to the NiH system some of his optic and sonic coupling arguments went over my head. I did understand the electron shielding argument associated with overcoming the coulomb repulsion issue in the Ni matrix. Its not apparent how this shielding would function at a surface, however. I thought that the solubility of H in nano Ni particles may be considerably higher than it is in bulk Ni. In the Defkalion system there is apparently a Ni matrix which would have low H concentration compared to the nano Ni with its high surface area to volume ratio. This would help focus the energetic reaction in the nano particles and preserve the integrity of the base Ni matrix, keeping the fuel in tact. Hagelstein did not significantly address spin conservation and coupling between the H, D and electrons or spin of the Ni or Pd atoms themselves. This coupling may be buried in his equations and operators--I'm not sure. I think spin coupling is important, particularly with whatever magnetic fields exist within the respective systems. Finally, I do not think Hagelstein addressed the electron-positron reaction with its 0.51 MEV gammas that Rossi and Focardi have identified associated with Cu isotope production, nor other radiation observed in various experiments on the NiH system done by Focardi and others. Check out: Focardi, S., Gabbani, V., Montalbano, V., Piantelli, F. and Veronesi, S., Focardi, S., et al. Evidence of Electromagnetic Radiation From Ni-H Systems, Proceedings of the Eleventh International Conference on Condensed Matter Nuclear Science, Marseille, France, (2004) I would be surprised that Focardi did not monitor He-3 and/or H-3, for the same reason Hagelstein indicated interest in He-3 production in the NiH experiments. Bob - Original Message - From: Alan Fletcher a...@well.com To: vortex-l vortex-l@eskimo.com Sent: Wednesday, February 05, 2014 6:40 PM Subject: [Vo]:MIT Course Day 5 -- NiH Systems All of Ruby Carat's/Jeremy Ry's videos are now up http://coldfusionnow.org/2014-cold-fusion-101-video-lectures/ Particularly day 5 Hagelstein http://www.youtube.com/watch?feature=player_embeddedv=Al7NMQLvATo From my cryptic notes (H:M) : 1:25 : he disagrees with Ed Storms, because you need the electron cloud/Gamow factor for the reaction rates. (Gives up 10 orders of magnitude) 1:29 NiH Talks about H2 clustering in Ni Keywords are Fukai phase and elevated vacancy formation 1:50+- Phonon/Accoustic coupling should be about 8Thz -- compare with the recent discussion about Bushnell's 5-30Thz stimulation (Actually I couldn't see Bushnell saying that) Says Piantelli encountered charge generation -- compare Rossi EMF and Defkalion Magnetic effects (I think it comes from He3 creation) 2:05 Briefly discusses Rossi and Defkalion. Says that their COP from Ni powder is in line with Piantelli's rod. Says they should NOT be dismissed out of hand. My thoughts : since H doesn't easily diffuse into Ni (Unlike D in Pd) it's more likely to be a surface effect.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Thu, Feb 6, 2014 at 3:32 PM, Bob Cook frobertc...@hotmail.com wrote: I would be surprised that Focardi did not monitor He-3 and/or H-3, for the same reason Hagelstein indicated interest in He-3 production in the NiH experiments. Bob Hagelstein said that detecting a He-3 signal with a mass spectrometer is difficult because it might be confused with a HD signal depending on resolution of the spectrometer. Harry
RE: [Vo]:MIT Course Day 5 -- NiH Systems
-Original Message- From: Bob Cook My thoughts : since H doesn't easily diffuse into Ni (Unlike D in Pd) it's more likely to be a surface effect. Perhaps - but misleading. Pure nickel is not a great proton conductor- and one must pay dearly to get pure nickel. But why? It takes only a small amount of selected other metals, as alloying agents for nickel, to far exceed palladium. For instance, 95% nickel and 5% palladium is superior to palladium, at a fraction of the cost. There is a wealth of data on hydrogen storage alloys which tends to be overlooked as candidate alloys for LENR. Jones
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Thu, Feb 6, 2014 at 4:48 PM, Jones Beene jone...@pacbell.net wrote: -Original Message- From: Bob Cook My thoughts : since H doesn't easily diffuse into Ni (Unlike D in Pd) it's more likely to be a surface effect. Perhaps - but misleading. Pure nickel is not a great proton conductor- and one must pay dearly to get pure nickel. But why? It takes only a small amount of selected other metals, as alloying agents for nickel, to far exceed palladium. For instance, 95% nickel and 5% palladium is superior to palladium, at a fraction of the cost. There is a wealth of data on hydrogen storage alloys which tends to be overlooked as candidate alloys for LENR. Jones I think Swartz said in Friday's 2014 MIT video that his lastest NANOR composed of Ni and Pd. harry
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Jones Beene jone...@pacbell.net wrote: Perhaps - but misleading. Pure nickel is not a great proton conductor- and one must pay dearly to get pure nickel. But why? It takes only a small amount of selected other metals, as alloying agents for nickel, to far exceed palladium. For instance, 95% nickel and 5% palladium is superior to palladium, at a fraction of the cost. Superior for what? Conducting protons? Surely not for loading hydrogen. I have never heard that. - Jed
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Harry-- Its not so difficult if you suspect HD--you need to dissociate the HD molecule first and then do a mass spec test on the gas coming out of the system. Neither the H nor the D have a atomic weight (AW) of 3 and a charge of +2. H-3 would be the other most likely AW of 3 and it would be radioactive. It could be gettered from the gas stream with a hydrogen getter. Hagelstein wished during the lecture 2 or 3 times that he could get funding to check for He-3--he commented on loosing funding once for the He-3 testing. He implied that the funding entities did not want him to find He-3. Bob - Original Message - From: H Veeder To: vortex-l@eskimo.com Sent: Thursday, February 06, 2014 1:47 PM Subject: Re: [Vo]:MIT Course Day 5 -- NiH Systems On Thu, Feb 6, 2014 at 3:32 PM, Bob Cook frobertc...@hotmail.com wrote: I would be surprised that Focardi did not monitor He-3 and/or H-3, for the same reason Hagelstein indicated interest in He-3 production in the NiH experiments. Bob Hagelstein said that detecting a He-3 signal with a mass spectrometer is difficult because it might be confused with a HD signal depending on resolution of the spectrometer. Harry
RE: [Vo]:MIT Course Day 5 -- NiH Systems
From: Jed Rothwell Superior for what? Conducting protons? Surely not for loading hydrogen. I have never heard that. Surely you read Ahern's Arata replication for EPRI ? He achieved better loading than the standard of 1:1 with nickel-palladium alloy (at low Pd ratio in the alloy). Many alloys which are tailored for hydrogen storage are in fact better than palladium for that single property (which is the atomic ratio of lattice atoms to hydrogen atoms) This does not meant they will be more active for LENR - only that they will absorb more atoms of hydrogen per atom of lattice. That is what they are designed for. In fact, the alloys which store the most hydrogen are most often NOT anomalous as to energy release, when further stimulated. Unfortunately, the two fields have not been systematically investigated for determining the best of both worlds. Jones
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Jones-- I agree that there are a number of alloys that do better at hydrogen solubility than Ni. However, they may not have the body centered crystal array and may actually have differing phases, some of which hold the hydrogen better than others in the same alloy. The simple crystal structure of pure Ni may be of an advantage in the LENR business. Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it and may be the reason other researchers do not have very good luck at getting a good reaction. If you want to be careful about how you stimulate a quantum system with fixed input frequencies, various crystals and impurities may not help. Bob - Original Message - From: Jones Beene jone...@pacbell.net To: vortex-l@eskimo.com Sent: Thursday, February 06, 2014 1:48 PM Subject: RE: [Vo]:MIT Course Day 5 -- NiH Systems -Original Message- From: Bob Cook My thoughts : since H doesn't easily diffuse into Ni (Unlike D in Pd) it's more likely to be a surface effect. Perhaps - but misleading. Pure nickel is not a great proton conductor- and one must pay dearly to get pure nickel. But why? It takes only a small amount of selected other metals, as alloying agents for nickel, to far exceed palladium. For instance, 95% nickel and 5% palladium is superior to palladium, at a fraction of the cost. There is a wealth of data on hydrogen storage alloys which tends to be overlooked as candidate alloys for LENR. Jones
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Thu, Feb 6, 2014 at 5:00 PM, H Veeder hveeder...@gmail.com wrote: On Thu, Feb 6, 2014 at 4:48 PM, Jones Beene jone...@pacbell.net wrote: -Original Message- From: Bob Cook My thoughts : since H doesn't easily diffuse into Ni (Unlike D in Pd) it's more likely to be a surface effect. Perhaps - but misleading. Pure nickel is not a great proton conductor- and one must pay dearly to get pure nickel. But why? It takes only a small amount of selected other metals, as alloying agents for nickel, to far exceed palladium. For instance, 95% nickel and 5% palladium is superior to palladium, at a fraction of the cost. There is a wealth of data on hydrogen storage alloys which tends to be overlooked as candidate alloys for LENR. Jones I think Swartz said in Friday's 2014 MIT video that his lastest NANOR composed of Ni and Pd. harry uh sorry his older versions of NANOR use Ni and Pd. Harry
Re: [Vo]:MIT Course Day 5 -- NiH Systems
Jones Beene jone...@pacbell.net wrote: Surely you read Ahern's Arata replication for EPRI ? He achieved better loading than the standard of 1:1 with nickel-palladium alloy (at low Pd ratio in the alloy). Hmmm . . . I ascribe that to the small particle size. I assume the hydrogen is sticking to the surface, not being absorbed the way it is with bulk palladium. I could be wrong. Also, I wonder if that ratio is measured reliably. With a small mass of metal it can be difficult to measure loading accurately. - Jed
RE: [Vo]:MIT Course Day 5 -- NiH Systems
-Original Message- From: Bob Cook * Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it … Not sure that I follow this. Although the Rossi patent mentions nanometric and specifically a favored isotope - Rossi himself has identified his nickel supplier, and says the geometry of his powder is micron not nano (at least at that point in time). Metals (as opposed to ceramics) can seldom be reduced below 10 microns by normal Industrial methods such as ball milling - due to surface electric properties aka: “agglomeration.” That is one reason why “nano” is so special and not fully appreciated wrt metals. It simply cannot happen in normal metal processing (except with mixed ceramics like the oxides of nickel). You might do well to talk to the Ni-O “nano” suppliers, like Quantum sphere: http://www.qsinano.com/products_nanomaterials.html They will set you straight on the lack of anything truly “nano” as a metal. It must have a surface oxide. * … and may be the reason other researchers do not have very good luck at getting a good reaction. No doubt that Rossi, if we can believe his results, has found something that no one else has yet been able to duplicate. It may be serendipitous, but it is not likely to be “nanometric nickel” per se. Jones attachment: winmail.dat
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Thu, Feb 6, 2014 at 2:26 PM, Bob Cook frobertc...@hotmail.com wrote: Also I suspect that the nano Ni that is produced is pretty pure. That may be why Rossi uses it and may be the reason other researchers do not have very good luck at getting a good reaction. I'm guessing that the purity of Rossi's nickel (in terms of 62Ni and 64Ni) is related to avoiding beta-plus and beta-minus decay, and, with beta-plus decay, the 511 keV positron-electron annihilation photons. Some vorts may enjoy this video of a small cloud chamber [1]. It's remarkable that such a small event can have macroscopic effects. Eric [1] http://www.youtube.com/watch?v=xQVMrkJYShc
Re: [Vo]:MIT Course Day 5 -- NiH Systems
I wrote: Hmmm . . . I ascribe that to the small particle size. I assume the hydrogen is sticking to the surface, not being absorbed the way it is with bulk palladium. I mean it is adsorbed rather than absorbed. Further, I meant that palladium particles will also adsorb large amounts -- I think. - Jed
Re: [Vo]:MIT Course Day 5 -- NiH Systems
1:50+- Phonon/Accoustic coupling should be about 8Thz -- compare with the recent discussion about Bushnell's 5-30Thz stimulation (Actually I couldn't see Bushnell saying that) I viewed it again, and couldn't hear a specific frequency mentioned.
Re: [Vo]:MIT Course Day 5 -- NiH Systems
On Wed, Feb 5, 2014 at 6:40 PM, Alan Fletcher a...@well.com wrote: Particularly day 5 Hagelstein http://www.youtube.com/watch?feature=player_embeddedv=Al7NMQLvATo Interesting video. I have always been a little mystified by Peter Hagelstein's theory. My understanding is that in its current form it involves two receivers which form semi-independent oscillators with the 24 MeV donor. In one oscillator, phonon modes are weakly coupled with the donor, and in the other oscillator, the nuclear degrees of freedom are strongly coupled with it. These two oscillators work in conjunction to fractionate the 24 MeV quantum into relatively low-energy phonon modes, which dissipate energy from the system as heat. The reason for two oscillators instead of one is that you have to have a strongly coupled partner to interact with the 24 MeV donor, and the phonon modes can't provide this kind of coupling. Mono-vacancies are also important in the theory, for in a mono-vacancy in palladium, there will still be significant electron charge density, and this charge density will have the effect of screening the reactants somewhat with the negative charge of the electrons, thereby reducing Coulomb repulsion between reacting nuclei. At 1:10:20, Hagelstein addresses how his model does not fall into the trap set by Huizenga's three miracles. In connection with two of those miracles, Hagelstein does not believe that his theory requires that Coulomb repulsion be altered, nor that it requires the branching ratios for t/3He/gammas to be changed in the case of d+d fusion. I had a hard time understanding him when he explained why it was that these two miracles were avoided. It might have been something along the lines of Coulomb repulsion being overwhelmed by the large number of coherently coordinated actors in the system, and the branching ratios being applicable to incoherent fusion, whereas we're dealing with a coherent system, so they do not apply. It's likely that I misunderstood one or both of these points. As an uninformed bystander, there are several challenges that I have with Hagelstein's theory. The first is the expectation that in a 700 C system there will be any kind of coherent coordination of phonons, let alone a coordination of phonon modes sufficient to fractionate on the order of 10E12 reactions per second into heat. Although Hagelstein's theory is focused on PdD, which has typically been operated at lower temperatures, he also seeks to apply it to NiH, which is often operated at higher temperatures. If I have understood him, he notes that the coherence of the phonons has to be more than just local and must extend across a significant portion of the system [1]. A second difficulty I have is the notion that phonons can be coupled to a nuclear reaction that is underway. I can imagine electromagnetic coupling, e.g., the coupling of a [dd]* intermediate state with the positive charges of the nuclei and the negative charges of the electrons, but it seems too abstract to say that a reaction can directly couple with phonons. This probably just goes back to a deficiency in my understanding of quantum mechanics. Note that electromagnetic coupling with lattice sites would lead to phonons as a side-effect, and electromagnetic coupling with electrons would lead to photons. There's good reason to think that Hagelstein is correct in assuming that plain old fusion is going on (d+d fusion in the case of PdD), and in wanting to fractionate the resulting mass-energy of the source across a large number of sinks, instead of trying to devise a way to catch a 23 MeV gamma or fast t and 3He in flight. What I don't understand yet is why he does not consider electromagnetic coupling with electrostatic charges in the Coulomb rich environment. Perhaps this is because it might imply that in such an environment the branching ratios would change, depending on how you look at the matter. Eric [1] I have a similar difficulty with BECs and hydrotons -- how do such delicate creatures form and survive in something as chaotic as a metal at high temperatures?
