Re: [Vo]:Removing All Doubt (READ HIS BLOG)
He does not use protium. He uses ordinary hydrogen. In the cell some of it is broken down into atomic hydrogen. That is what interacts with the nickel. http://www.journal-of-nuclear-physics.com/?p=62cpage=2#comment-273 I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Sat, January 22, 2011 1:32:11 AM Subject: Re: [Vo]:Removing All Doubt On Jan 21, 2011, at 7:06 PM, Man on Bridges wrote: Horace, Based upon natural presence and absence of radiation I would probably go for this on: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 : pres. 3.634 % 1H1 : pres. 99.985 % 63Cu29 : pres. 69.17 % Kind regards, MoB If I recall correctly, Rossi stated that deuterium kills the reaction, and that he uses pure protium. Looking again at the mean atomic weight 63.6 and mean reaction energy 7.2, I calculated as shown below, you can see that these are weighted averages, and are roughly 4 to 1 in favor of 62Ni vs 64Ni, corresponding roughly to their abundances, 3.634% vs 0.926%. They should be roughly equal in lattice half life (as I defined it in my paper), possibly with a small edge for the larger nucleus. Everything posted by me on this is seat of the pants accuracy. I think for casual discussion like this slide rule accuracy is sufficient to demonstrate the principles. I guess my age is showing! 8^) On Jan 21, 2011, at 4:23 PM, Horace Heffner wrote: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Looking at the first two reactions as likely candidates, with mean atomic weight near 63.6, and mean reaction energy about 7.2, we have an estimated energy density of E = (1/(63.6 gm/mol))*Na*7.2 MeV = 1.09x10^10 J/gm Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt (READ HIS BLOG)
On Jan 22, 2011, at 2:06 AM, noone noone wrote: He does not use protium. He uses ordinary hydrogen. In the cell some of it is broken down into atomic hydrogen. That is what interacts with the nickel. http://www.journal-of-nuclear-physics.com/?p=62cpage=2#comment-273 I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. It would be helpful if you could designate in some way which material is quoted and which is yours. I looked at the above reference and found this: Begin quote: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Andrea Rossi April 29th, 2010 at 9:46 AM I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. Get fun, ‘MR BROWN’, and let your sun smile for ever. A.R. p.s. Now, after your lecturing, I want to put you some questions: 1- Who are you? D.Brown is a fake name, so you approached us unonimously, which is not fair, is it? But I know: it’s fun.. 2- which is your profession? What do you do, besides cozy smiling suns? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - end quote. This material you quote says nothing about protium vs deuterium. Protium is an isotope of hydrogen, whether it is in molecular form or not. It is designated 1H1, while deuterium is designated 1H2, or D. The deuterium ion is designated d, but I have avoided that lower case use because I use d for the down quark. Deuterium is an isotope of hydrogen, whether it is in molecular form or not. Same goes for absorbed into a lattice or not. Ordinary hydrogen is a mix of protium and deuterium. The natural abundance of deuterium is about 0.015%. Ordinary hydrogen is mostly H2, but occasionally (again about 0.015%) it is HD, and rarely (2.25 x 10^-8 %) DD. Somewhere I read that Rossi said deuterium poisons the reaction, so he uses pure protium. Perhaps he did, perhaps it is just my bad memory, perhaps just an unfounded rumor. It sounded weird to me because pure protium is hard to come by. I'm not going to spend any time looking this up though. I did state: If I recall correctly, Rossi stated that deuterium kills the reaction, and that he uses pure protium. Perhaps someone here is familiar with this statement by Rossi. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt (READ HIS BLOG)
Sorry for my error The comment about deuterium poisoning the reaction is a comment Piantelli made to individuals. From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Sat, January 22, 2011 6:55:31 AM Subject: Re: [Vo]:Removing All Doubt (READ HIS BLOG) On Jan 22, 2011, at 2:06 AM, noone noone wrote: He does not use protium. He uses ordinary hydrogen. In the cell some of it is broken down into atomic hydrogen. That is what interacts with the nickel. http://www.journal-of-nuclear-physics.com/?p=62cpage=2#comment-273 I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. It would be helpful if you could designate in some way which material is quoted and which is yours. I looked at the above reference and found this: Begin quote: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - * Andrea Rossi * April 29th, 2010 at 9:46 AM * I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. Get fun, ‘MR BROWN’, and let your sun smile for ever. A.R. p.s. Now, after your lecturing, I want to put you some questions: 1- Who are you? D.Brown is a fake name, so you approached us unonimously, which is not fair, is it? But I know: it’s fun.. 2- which is your profession? What do you do, besides cozy smiling suns?- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - end quote. This material you quote says nothing about protium vs deuterium. Protium is an isotope of hydrogen, whether it is in molecular form or not. It is designated 1H1, while deuterium is designated 1H2, or D. The deuterium ion is designated d, but I have avoided that lower case use because I use d for the down quark. Deuterium is an isotope of hydrogen, whether it is in molecular form or not. Same goes for absorbed into a lattice or not. Ordinary hydrogen is a mix of protium and deuterium. The natural abundance of deuterium is about 0.015%. Ordinary hydrogen is mostly H2, but occasionally (again about 0.015%) it is HD, and rarely (2.25 x 10^-8 %) DD. Somewhere I read that Rossi said deuterium poisons the reaction, so he uses pure protium. Perhaps he did, perhaps it is just my bad memory, perhaps just an unfounded rumor. It sounded weird to me because pure protium is hard to come by. I'm not going to spend any time looking this up though. I did state: If I recall correctly, Rossi stated that deuterium kills the reaction, and that he uses pure protium. Perhaps someone here is familiar with this statement by Rossi. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt (READ HIS BLOG)
http://blog.newenergytimes.com/ A rapidly increasing temperature in an enclosed steel container could be a big, big problem. He was afraid. He wondered whether he should leave the building. Instead he called Focardi in Milano—at 2 in the morning—and asked, “What should I do?” This was before Piantelli knew about the poisoning effect of deuterium. But Focardi came up with a workable idea: introduce nitrogen. And it worked. It stopped the uncontrolled temperature rise and killed the experiment. From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Sat, January 22, 2011 6:55:31 AM Subject: Re: [Vo]:Removing All Doubt (READ HIS BLOG) On Jan 22, 2011, at 2:06 AM, noone noone wrote: He does not use protium. He uses ordinary hydrogen. In the cell some of it is broken down into atomic hydrogen. That is what interacts with the nickel. http://www.journal-of-nuclear-physics.com/?p=62cpage=2#comment-273 I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. It would be helpful if you could designate in some way which material is quoted and which is yours. I looked at the above reference and found this: Begin quote: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - * Andrea Rossi * April 29th, 2010 at 9:46 AM * I said ‘eventually’ because it is exactly what happens. Of course you know that in English ‘eventually’ means ‘after some time’.We know exactly why and how to make H after the injection of H2 and know exactly how difficult is to use this radical before H2 recombination. This is one of the most important parts of our know how. When we use terms, in this field, we know exactly what we say. We not just made models and calculations, but we made apparatuses which are working from 2 years now. What we are working on is no more an ‘experimental set’, as you wrongly wrote,it is an apparatus which heats up a factory and of which we are organizing the industrialization. I understand you get fun, we don’t: we work on this in a factory totally dedicated to this, and we are pretty good at, as you soon will see. In our team there are Nuclear Physics University professors, with experience from CERN of Geneva, INFN, etc., etc. Your lecturing and sarcastic tone does not qualify you a lot, but we know, you get fun… About the second question, yes, the paper has been peer-reviewed. Get fun, ‘MR BROWN’, and let your sun smile for ever. A.R. p.s. Now, after your lecturing, I want to put you some questions: 1- Who are you? D.Brown is a fake name, so you approached us unonimously, which is not fair, is it? But I know: it’s fun.. 2- which is your profession? What do you do, besides cozy smiling suns?- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - end quote. This material you quote says nothing about protium vs deuterium. Protium is an isotope of hydrogen, whether it is in molecular form or not. It is designated 1H1, while deuterium is designated 1H2, or D. The deuterium ion is designated d, but I have avoided that lower case use because I use d for the down quark. Deuterium is an isotope of hydrogen, whether it is in molecular form or not. Same goes for absorbed into a lattice or not. Ordinary hydrogen is a mix of protium and deuterium. The natural abundance of deuterium is about 0.015%. Ordinary hydrogen is mostly H2, but occasionally (again about 0.015%) it is HD, and rarely (2.25 x 10^-8 %) DD. Somewhere I read that Rossi said deuterium poisons the reaction, so he uses pure protium. Perhaps he did, perhaps it is just my bad memory, perhaps just an unfounded rumor. It sounded weird to me because pure protium is hard to come by. I'm not going to spend any time looking this up though. I did state
Re: [Vo]:Removing All Doubt (READ HIS BLOG)
On Jan 22, 2011, at 3:06 AM, noone noone wrote: Sorry for my error No problem! The comment about deuterium poisoning the reaction is a comment Piantelli made to individuals. Oh, OK, that clears it up. It is my bad memory! It seems to get worse on a daily basis, but I thankfully still have moments of clarity, aided by my Mac. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt
Thank you, now everything depend on-Cu is real, or not! Peter On Sat, Jan 22, 2011 at 12:25 AM, mix...@bigpond.com wrote: In reply to Peter Gluck's message of Fri, 21 Jan 2011 19:31:09 +0200: Hi, [snip] That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter [snip] If all Ni isotopes react equally, and 2/3 of Ni is Ni-58, and we assume single proton fusion, then the primary reaction would be: Ni-58 + H - Cu-59 + 3.42 MeV which then decays rapidly via positron decay according to Cu-59 - Ni-59 + e+ + neutrino + 4.8 MeV (however a considerable portion of this will be lost via neutrinos; say 1/2?). so the total reaction energy is 3.42 + 2.4 = 5.82 MeV / Ni-58. 2/3 *5 kWh / 6 MeV = 1.2E23 Ni-58 reactions, which is 12 gm Ni-58, or about 18 gm Ni altogether (assuming the other isotopes all yield about the same amount of energy / atom). So quite within the realm of possibility. OTOH, if he had 300 gm of Ni, and 1/3 was converted to Cu, then that represents considerably more energy, and one has to wonder where it all went? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
The 59Cu explanation seems to be nonsensical, at least with the conventional reaction explanation implied below, because (1) it does not explain the presence of gram quantities of observable copper (due to the 1.36 minute half-life of 59Cu), (2) the huge flux of 0.5 MeV gammas from positron annihilation would be a major risk to the observers, and proof positive of nuclear reaction, eliminating the need to measure heat, (3) the logarithmic tail-off of positron emission, even given the 1.36 m half-life, should be readily detectable for a long time after power off, (4) and the presence of gram quantities of Ni59, with 76000 y half-life for EC, should be readily detectable in the ash via auger electrons or x-rays. On Jan 22, 2011, at 3:46 AM, Peter Gluck wrote: Thank you, now everything depend on-Cu is real, or not! Peter On Sat, Jan 22, 2011 at 12:25 AM, mix...@bigpond.com wrote: In reply to Peter Gluck's message of Fri, 21 Jan 2011 19:31:09 +0200: Hi, [snip] That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter [snip] If all Ni isotopes react equally, and 2/3 of Ni is Ni-58, and we assume single proton fusion, then the primary reaction would be: Ni-58 + H - Cu-59 + 3.42 MeV which then decays rapidly via positron decay according to Cu-59 - Ni-59 + e+ + neutrino + 4.8 MeV (however a considerable portion of this will be lost via neutrinos; say 1/2?). so the total reaction energy is 3.42 + 2.4 = 5.82 MeV / Ni-58. 2/3 *5 kWh / 6 MeV = 1.2E23 Ni-58 reactions, which is 12 gm Ni-58, or about 18 gm Ni altogether (assuming the other isotopes all yield about the same amount of energy / atom). So quite within the realm of possibility. OTOH, if he had 300 gm of Ni, and 1/3 was converted to Cu, then that represents considerably more energy, and one has to wonder where it all went? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
RE: [Vo]:Removing All Doubt
A close look at the patent application indicates specific mention of copper in the disclosure. The only way to eliminate the expected migration of copper from one part of the reactor to another - is a comparison of the two isotopes: 63Cu and 65Cu. The correct ratio in nature is 69.17 / 30.83 - ergo if testing should show is something like 60/40 then that is proof positive. Case solved ! and let the mass production begin, at the same time as we watch the price of oil careen through the floor. Again - this kind of isotope testing gives away nothing important from Rossi's perspective, if it is limited to only copper isotopes. Anyone communicating with Rossi should press for independent testing of the copper in the spent fuel. Why could Levi not do this? He seems to be trusted by everyone concerned, and no doubt Frascati can do it technically- we have seen their sophisticate mass spec equipment which can distinguish He from D2 - so we know without question that they can do it. Please, Dr. Rossi - let them do it, and before you pack up for India ! Jones From: Horace Heffner The 59Cu explanation seems to be nonsensical, at least with the conventional reaction explanation implied below, because (1) it does not explain the presence of gram quantities of observable copper (due to the 1.36 minute half-life of 59Cu), (2) the huge flux of 0.5 MeV gammas from positron annihilation would be a major risk to the observers, and proof positive of nuclear reaction, eliminating the need to measure heat, (3) the logarithmic tail-off of positron emission, even given the 1.36 m half-life, should be readily detectable for a long time after power off, (4) and the presence of gram quantities of Ni59, with 76000 y half-life for EC, should be readily detectable in the ash via auger electrons or x-rays. On Jan 22, 2011, at 3:46 AM, Peter Gluck wrote: Thank you, now everything depend on-Cu is real, or not! Peter On Sat, Jan 22, 2011 at 12:25 AM, mix...@bigpond.com wrote: In reply to Peter Gluck's message of Fri, 21 Jan 2011 19:31:09 +0200: Hi, [snip] That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter [snip] If all Ni isotopes react equally, and 2/3 of Ni is Ni-58, and we assume single proton fusion, then the primary reaction would be: Ni-58 + H - Cu-59 + 3.42 MeV which then decays rapidly via positron decay according to Cu-59 - Ni-59 + e+ + neutrino + 4.8 MeV (however a considerable portion of this will be lost via neutrinos; say 1/2?). so the total reaction energy is 3.42 + 2.4 = 5.82 MeV / Ni-58. 2/3 *5 kWh / 6 MeV = 1.2E23 Ni-58 reactions, which is 12 gm Ni-58, or about 18 gm Ni altogether (assuming the other isotopes all yield about the same amount of energy / atom). So quite within the realm of possibility. OTOH, if he had 300 gm of Ni, and 1/3 was converted to Cu, then that represents considerably more energy, and one has to wonder where it all went? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt
In reply to Horace Heffner's message of Sat, 22 Jan 2011 10:52:29 -0900: Hi, The 59Cu explanation seems to be nonsensical, at least with the conventional reaction explanation implied below, because (1) it does not explain the presence of gram quantities of observable copper (due to the 1.36 minute half-life of 59Cu), (2) the huge flux of 0.5 MeV gammas from positron annihilation would be a major risk to the observers, and proof positive of nuclear reaction, eliminating the need to measure heat, (3) the logarithmic tail-off of positron emission, even given the 1.36 m half-life, should be readily detectable for a long time after power off, (4) and the presence of gram quantities of Ni59, with 76000 y half-life for EC, should be readily detectable in the ash via auger electrons or x-rays. [snip] Actually I agree, but the reaction below is the first reaction posited by Rossi-Focardi, which is why I used it. Note however that positron emission may be avoided (or at least severely curtailed) if enhanced electron capture plays a significant role. That may at least explain the lack (or severe shortage) of characteristic gammas. Also, I seem to recall someone being surprised that they got positrons in one run but not in another. A possible explanation for this could be that different size Hydrinos would have different electron capture / positron decay ratios, with smaller Hydrinos enhancing electron capture to a larger degree. I get the distinct impression that the only isotope analysis they have done is with SIMS. [snip] Another possibility exists to explain the copper. If Ed Storms is correct about NAS, and they are fairly large and not destroyed when a fusion reaction occurs, then it's likely that subsequent reactions will occur in the same place, and make use of the metal nuclei in the neighbourhood. That would increase the chances of e.g. Ni-59 being involved in subsequent reactions, and might allow for the series of reactions Rossi-Focardi propose in their paper (See http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
Jones, since you brought this up, I'd like to ask a question about the copper. According to the handy dandy periodic table on my desktop (Kalzium), copper has two stable isotopes, with 34 and 36 neutrons, respectively. Next best is 38 neutrons, with a half life of about 62 hours, and it's downhill from there. The nickel isotopes with 34 and 36 neutrons are also stable, but constitute less than 5% of natural nickel, most of which has either 30 or 32 neutrons. If 30% of a sample of nickel were transmuted into copper by the reaction Ni + H - Cu, the result should, therefore, be highly radioactive, as /most/ of the copper so produced must have either 30 or 32 neutrons, with a half life, respectively, of 82 seconds and about 3.3 hours. In particular, after six months -- or even just one day! -- most of the newly formed copper will have decayed into something else, and won't show up as copper. So, what's the story here? How can the neutron balance work out? How can he have ended up with 30% of the nickel transmuted into (reasonably stable) copper? On 01/21/2011 11:23 AM, Jones Beene wrote: To all concerned, or to anyone harboring lingering doubts about the Bologna demo . There is a surprising simple and extremely convincing way to *remove all doubt* that this device is real. It is so simple that the simple fact that it has not been published yet, is suspicious in itself. (there has been one claim that a test was performed, but not data). Rossi has stated that another long-running device, which was in operation for 6 months continuous, was analyzed and a large percentage of nickel was transmuted to copper. Even if it was less than 30%, it was a lot. This is the key. Based on other disclosures - the amount of transmuted copper recovered from this sample should be in excess of 30 grams and could be as much as a 300 grams. Even without the copper, the nickel from this reactor will have had an isotope shift, so this spent fuel is another key to instant credibility. It can be tested as mixed and there is no need to separate the two metals. In other words, there is no shortage of evidence - either the copper - the ash of the reaction, or the nickel . but the copper is preferable, even in a mixed sample. If he claims the entire sample has been lost, he will lose all credibility in my book. ALL. No one loses such a sample. He is essentially dead in the water, in the eyes of 99% of Physics, if this sample is unaccounted for now. Copper has two isotopes: 63Cu is almost ~69% of the natural ratio. 65Cu is ~31%. That never varies - no matter where the copper came from - Arizona or Chile or Asia. A one gram sample is more than adequate to test - therefore 10 samples sent to 10 labs for isotope analysis should put all doubts to rest, if the ratio varies significantly. With the nickel, the sample should be depleted in 64Ni. There is no rational argument that can account for a ratio which comes from a long standing nuclear reaction being identical to the natural ratio - and if it is identical, then all the copper came from migration from other parts of the device, which is to be expected. Rossi never mentions migration but surely is aware of it. My plea to Ing. Andrea Rossi is to sent samples out soon for testing by University or National Labs. A list can be provided. It is time to put these skeptics of LENR down hard, and you can do that dramatically and very easily, in short order and with minimal effort. They have it coming. You d'man, Andre. show'm your stuff. Jones
RE: [Vo]:Removing All Doubt
From: Stephen A. Lawrence * So, what's the story here? How can the neutron balance work out? How can he have ended up with 30% of the nickel transmuted into (reasonably stable) copper? The short answer is that this percentage must be way off, or there has been a mis-translation. it is possible that they chose a microgram sample which was visually different - and that it had a wildly distorted ratio, for instance, and following that - an incorrect assumption followed. I see now way for such a large ratio over the entire mass of spent fuel, but even one percent is adequate for testing, and any big shift in copper isotopes will be extremely meaningful. Less so with the nickel. Jones
Re: [Vo]:Removing All Doubt
That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter On Fri, Jan 21, 2011 at 6:23 PM, Jones Beene jone...@pacbell.net wrote: To all concerned, or to anyone harboring lingering doubts about the Bologna demo . There is a surprising simple and extremely convincing way to *remove all doubt* that this device is real. It is so simple that the simple fact that it has not been published yet, is suspicious in itself. (there has been one claim that a test was performed, but not data). Rossi has stated that another long-running device, which was in operation for 6 months continuous, was analyzed and a large percentage of nickel was transmuted to copper. Even if it was less than 30%, it was a lot. This is the key. Based on other disclosures - the amount of transmuted copper recovered from this sample should be in excess of 30 grams and could be as much as a 300 grams. Even without the copper, the nickel from this reactor will have had an isotope shift, so this spent fuel is another key to instant credibility. It can be tested as mixed and there is no need to separate the two metals. In other words, there is no shortage of evidence - either the copper - the ash of the reaction, or the nickel . but the copper is preferable, even in a mixed sample. If he claims the entire sample has been lost, he will lose all credibility in my book. ALL. No one loses such a sample. He is essentially dead in the water, in the eyes of 99% of Physics, if this sample is unaccounted for now. Copper has two isotopes: 63Cu is almost ~69% of the natural ratio. 65Cu is ~31%. That never varies - no matter where the copper came from - Arizona or Chile or Asia. A one gram sample is more than adequate to test - therefore 10 samples sent to 10 labs for isotope analysis should put all doubts to rest, if the ratio varies significantly. With the nickel, the sample should be depleted in 64Ni. There is no rational argument that can account for a ratio which comes from a long standing nuclear reaction being identical to the natural ratio - and if it is identical, then all the copper came from migration from other parts of the device, which is to be expected. Rossi never mentions migration but surely is aware of it. My plea to Ing. Andrea Rossi is to sent samples out soon for testing by University or National Labs. A list can be provided. It is time to put these skeptics of LENR down hard, and you can do that dramatically and very easily, in short order and with minimal effort. They have it coming. You d'man, Andre. show'm your stuff. Jones
Re: [Vo]:Removing All Doubt
Here's a handy-dandy Table of Nuclides http://atom.kaeri.re.kr/ Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
RE: [Vo]:Removing All Doubt
Peter, The amount of copper found is of low comparative importance. The *isotope shift* from the natural ratio after 6 months is extremely important. This can only be determined by specialized equipment. It is so important to establishing proof of a nuclear reaction, or to changing the (presumed negative) opinion of experts like Director Chu, that nothing else comes remotely close. I cannot express it strongly enough in words than that this is probably the one factor which will make or break the opinions of experts - and without an demonstrable isotope shift, this good work may languish. Jones From: Peter Gluck That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway.
Re: [Vo]:Removing All Doubt
After some digging I think I got close to the source of the 30% copper assertion. The following items are from Rossi's blog. First: Question from William: William January 20th, 2011 at 9:01 AM http://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19862 ... /elided his first three questions .../ 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Rossi's answer: Andrea Rossi January 20th, 2011 at 10:14 AM http://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19868 Mr William: ... 4- No ... Further message from William, apparently in response to this denial: William January 20th, 2011 at 11:30 AM http://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19880 Hello Mr. Rossi, I found the following comment. Dear Pierre, Thank you for your important questions, here are the answers: 1- the Ni powder I utilized were pure Ni, no copper . At the end of the operations in the reactor the percentage of copper was integrally bound to the amount of energy produced. A charge which has worked for 6 monthes, 24 hours per day, at the end had a percentage of Cu superior to 30% 2- About the Ni isotopes: the isotopes after the operations were substantially changed in percentage. We are preparing a campaign of analysys with a Secondary Ions Mass Spectrometer at the University of Padua (Italy), at the end of which the data will be published on the Journal Of Nuclear Physics. Warm Regards, Andrea I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. On 01/21/2011 12:07 PM, Jones Beene wrote: *From:* Stephen A. Lawrence Ø So, what's the story here? How can the neutron balance work out? How can he have ended up with 30% of the nickel transmuted into (reasonably stable) copper? The short answer is that this percentage must be way off, or there has been a mis-translation... it is possible that they chose a microgram sample which was visually different -- and that it had a wildly distorted ratio, for instance, and following that -- an incorrect assumption followed. I see now way for such a large ratio over the entire mass of spent fuel, but even one percent is adequate for testing, and any big shift in copper isotopes will be extremely meaningful. Less so with the nickel. Jones
Re: [Vo]:Removing All Doubt
It is in the same forum. http://www.journal-of-nuclear-physics.com/?p=62 From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Fri, January 21, 2011 1:03:24 PM Subject: Re: [Vo]:Removing All Doubt After some digging I think I got close to the source of the 30% copper assertion. The following items are from Rossi's blog. First: Question from William: William January 20th, 2011 at 9:01 AM ... elided his first three questions ... 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Rossi's answer: Andrea Rossi January 20th, 2011 at 10:14 AM Mr William: ... 4- No ... Further message from William, apparently in response to this denial: William January 20th, 2011 at 11:30 AM Hello Mr. Rossi, I found the following comment. Dear Pierre, Thank you for your important questions, here are the answers: 1- the Ni powder I utilized were pure Ni, no copper . At the end of the operations in the reactor the percentage of copper was integrally bound to the amount of energy produced. A charge which has worked for 6 monthes, 24 hours per day, at the end had a percentage of Cu superior to 30% 2- About the Ni isotopes: the isotopes after the operations were substantially changed in percentage. We are preparing a campaign of analysys with a Secondary Ions Mass Spectrometer at the University of Padua (Italy), at the end of which the data will be published on the Journal Of Nuclear Physics. Warm Regards, Andrea I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. On 01/21/2011 12:07 PM, Jones Beene wrote: From:Stephen A. Lawrence Ø So, what's the story here? How can the neutron balance work out? How can he have ended up with 30% of the nickel transmuted into (reasonably stable) copper? The short answer is that this percentage must be way off, or there has been a mis-translation… it is possible that they chose a microgram sample which was visually different – and that it had a wildly distorted ratio, for instance, and following that – an incorrect assumption followed. I see now way for such a large ratio over the entire mass of spent fuel, but even one percent is adequate for testing, and any big shift in copper isotopes will be extremely meaningful. Less so with the nickel. Jones
Re: [Vo]:Removing All Doubt
Jones Beene wrote: If he claims the entire sample has been lost, he will lose all credibility in my book. ALL. No one loses such a sample. He is essentially dead in the water, in the eyes of 99% of Physics, if this sample is unaccounted for now. I believe the samples have been sent out for analysis. I heard that somewhere, not sure where. I do not think he would claim the sample is lost. Based on my communications with him so far, I think he is likely to say he is sorry but the analysis is confidential because of trade secrets. He says he is keeping many secrets. He is not as bad as Patterson and Reding, and some others in this field, who actively tried to reduce belief in their work. They tried to reduce their own credibility, so that others would not enter the field and compete with them. They wanted more future market share, as Reding put it. I told Reding that he would end up with 100% of zero dollars, whereas with a sane business strategy he might get 1% of a trillion dollars. He did not agree. I was right. I can understand why Rossi is keeping secrets. The Patent Office is partly to blame, because they have embargoed cold fusion. Rossi is also to blame because his patents are . . . peculiar. - Jed
Re: [Vo]:Removing All Doubt
On 01/21/2011 01:43 PM, noone noone wrote: It is in the same forum. http://www.journal-of-nuclear-physics.com/?p=62 Thank you!
Re: [Vo]:Removing All Doubt
It is nuclear, completely nuclear, and only nuclear? Or nuclear- is only a secondary phenomenon?. A Heat balance is a must. The same in classical cold fusion, it is good to believe the helium story but not easy to prove Peter On Fri, Jan 21, 2011 at 7:48 PM, Jones Beene jone...@pacbell.net wrote: Peter, The amount of copper found is of low comparative importance. The **isotope shift** from the natural ratio after 6 months is extremely important. This can only be determined by specialized equipment. It is so important to establishing proof of a nuclear reaction, or to changing the (presumed negative) opinion of experts like Director Chu, that nothing else comes remotely close. I cannot express it strongly enough in words than that this is probably the one factor which will “make or break” the opinions of experts - and without an demonstrable isotope shift, this good work may languish. Jones *From:* Peter Gluck That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway.
Re: [Vo]:Removing All Doubt
Do you have Skype, MSN, Yahoo, etc? Would you like to chat? From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Fri, January 21, 2011 1:51:36 PM Subject: Re: [Vo]:Removing All Doubt On 01/21/2011 01:43 PM, noone noone wrote: It is in the same forum. http://www.journal-of-nuclear-physics.com/?p=62 Thank you!
Re: [Vo]:Removing All Doubt
Stephen A. Lawrence wrote: 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Andrea Rossi January 20th, 2011 at 10:14 AM http://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19868 Mr William: ... 4- No ... Further message from William, apparently in response to this denial . . . I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. I take that to mean No, I cannot confirm that. Meaning I cannot confirm or deny; that's a secret. As I said, he makes no bones about the fact that he keeps secrets. It could also be confusion because of language problems. Or maybe he is contradicting himself . . . - Jed
Re: [Vo]:Removing All Doubt
I discuss with pleasure but chat is incompatible with my multitasking life style. In meantime I am writing my blog (Search No 2/439) But I answer any e-mail asa soon as I can. I have a bad experience with chat. Excuse me. Peter On Fri, Jan 21, 2011 at 8:59 PM, noone noone thesteornpa...@yahoo.comwrote: Do you have Skype, MSN, Yahoo, etc? Would you like to chat? -- *From:* Stephen A. Lawrence sa...@pobox.com *To:* vortex-l@eskimo.com *Sent:* Fri, January 21, 2011 1:51:36 PM *Subject:* Re: [Vo]:Removing All Doubt On 01/21/2011 01:43 PM, noone noone wrote: It is in the same forum. http://www.journal-of-nuclear-physics.com/?p=62 Thank you!
Re: [Vo]:Removing All Doubt
I took the No to mean Currently, there is no independent confirmation. Harry From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com Sent: Fri, January 21, 2011 2:15:45 PM Subject: Re: [Vo]:Removing All Doubt Stephen A. Lawrence wrote: 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Andrea Rossi January 20th, 2011 at 10:14 AM Mr William: ... 4- No ... Further message from William, apparently in response to this denial . . . I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. I take that to mean No, I cannot confirm that. Meaning I cannot confirm or deny; that's a secret. As I said, he makes no bones about the fact that he keeps secrets. It could also be confusion because of language problems. Or maybe he is contradicting himself . . . - Jed
Re: [Vo]:Removing All Doubt
Regardless of the exact amount transmuted, there is an explanation of all this given on Rossi's website. (/When all else fails, read the documentation!/) http://www.journal-of-nuclear-physics.com/?p=62 He says that Ni^x + p - Cu^(x+1) does, indeed, typically produce an unstable result, but it decays back to Ni^(x+1), after which it can pick up another proton, and repeat the process until it ends up as Cu^63, which is stable. He also asserts that the relative proton capture rates of all isotopes of Ni must be identical, as they're determined by electrostatic issues: The capture rate of protons by Nickel nuclei cannot depend on the mass values of different isotopes Finally, he says that they've been testing the ash and it's /not radioactive/: No radioactivity has been found also in the Nickel residual from the process. I don't understand that. If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. In particular, there should probably be a really large fraction of Ni^59 present (31 neutrons), with a 75 ky half-life, and I'd think that would make the sample pretty hot. Or so it seems; I haven't done the calculations to back up the intuition. In any case the text on that page is interesting and certainly worth reading. On 01/21/2011 02:15 PM, Jed Rothwell wrote: Stephen A. Lawrence wrote: 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Andrea Rossi January 20th, 2011 at 10:14 AM http://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19868 Mr William: ... 4- No ... Further message from William, apparently in response to this denial . . . I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. I take that to mean No, I cannot confirm that. Meaning I cannot confirm or deny; that's a secret. As I said, he makes no bones about the fact that he keeps secrets. It could also be confusion because of language problems. Or maybe he is contradicting himself . . . - Jed
Re: [Vo]:Removing All Doubt
On 01/21/2011 01:59 PM, noone noone wrote: Do you have Skype, MSN, Yahoo, etc? Would you like to chat? Got Skype downloaded but never installed it. I'm already spending too much time on this. Thank you very much for the offer. *From:* Stephen A. Lawrence sa...@pobox.com *To:* vortex-l@eskimo.com *Sent:* Fri, January 21, 2011 1:51:36 PM *Subject:* Re: [Vo]:Removing All Doubt On 01/21/2011 01:43 PM, noone noone wrote: It is in the same forum. http://www.journal-of-nuclear-physics.com/?p=62 Thank you!
Re: [Vo]:Removing All Doubt
Read the documentation if you believe it.. It is kind of forced explanation. OK, what is the energy realeased by this reaction? On Fri, Jan 21, 2011 at 9:35 PM, Stephen A. Lawrence sa...@pobox.comwrote: Regardless of the exact amount transmuted, there is an explanation of all this given on Rossi's website. (*When all else fails, read the documentation!*) http://www.journal-of-nuclear-physics.com/?p=62 He says that Ni^x + p - Cu^(x+1) does, indeed, typically produce an unstable result, but it decays back to Ni^(x+1), after which it can pick up another proton, and repeat the process until it ends up as Cu^63, which is stable. He also asserts that the relative proton capture rates of all isotopes of Ni must be identical, as they're determined by electrostatic issues: The capture rate of protons by Nickel nuclei cannot depend on the mass values of different isotopes Finally, he says that they've been testing the ash and it's *not radioactive*: No radioactivity has been found also in the Nickel residual from the process. I don't understand that. If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. In particular, there should probably be a really large fraction of Ni^59 present (31 neutrons), with a 75 ky half-life, and I'd think that would make the sample pretty hot. Or so it seems; I haven't done the calculations to back up the intuition. In any case the text on that page is interesting and certainly worth reading. On 01/21/2011 02:15 PM, Jed Rothwell wrote: Stephen A. Lawrence wrote: 4) I read a comment on another forum claiming that in one of your cells after six months of operation the remaining nickel powder was 30% copper. Can you confirm this? Andrea Rossi January 20th, 2011 at 10:14 AMhttp://www.journal-of-nuclear-physics.com/?p=360cpage=5#comment-19868 Mr William: ... 4- No ... Further message from William, apparently in response to this denial . . . I saw no further response from Rossi on this, and I don't know what the other forum in which his original comment appeared might have been. Google didn't turn it up for me. Make if this what you will; it's certainly not unambiguous -- looks kind of like an assertion followed by a retraction, but other interpretations are possible. I take that to mean No, I cannot confirm that. Meaning I cannot confirm or deny; that's a secret. As I said, he makes no bones about the fact that he keeps secrets. It could also be confusion because of language problems. Or maybe he is contradicting himself . . . - Jed
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
On 01/21/2011 03:55 PM, Roarty, Francis X wrote: Hi Stephan You state If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. But if you take a relativistic approach like that suggested for lead acid batteries you not only have the potential for fusion but potentially rapid aging of the reactants as well... Aside from the obvious question of why this particular bunch of nickel powder should show a shorter half-life than all the other bunches of nickel powder that have been tested in other laboratories, the problem with what you said is that this is just a simple urn problem, with replacement from probability theory. Even if the half-lives were scaled up or down by an order of magnitude it wouldn't make much difference to the conclusion. By the time 30% of the nickel atoms have been chosen by hydrogen atoms /five times/ (which is what it takes to get from Ni^58 up to Cu^63), nearly all the rest will have been chosen /at least once/. And a lot of those are likely to have been chosen /exactly/ once, and the ones which have been selected just once are the ones with the 75 kY half life. Working out the details to get an exact answer would be messy, in part because a bunch of the nickel is actually Ni^60, but the point is that if any significant fraction of the resulting material were Ni^59 there should be measurable radioactivity. And there's not.
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
What about this reaction: Ni (mass 60, 32 neutrons, pres. 26,223%) + Tritium (mass 3, 2 neutrons, pres. synt) -- Cu (mass 63, 34 neutrons, pres. 69,17%) + gamma radiation. On 21-1-2011 22:10, Stephen A. Lawrence wrote: On 01/21/2011 03:55 PM, Roarty, Francis X wrote: Hi Stephan You state If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. But if you take a relativistic approach like that suggested for lead acid batteries you not only have the potential for fusion but potentially rapid aging of the reactants as well... Aside from the obvious question of why this particular bunch of nickel powder should show a shorter half-life than all the other bunches of nickel powder that have been tested in other laboratories, the problem with what you said is that this is just a simple urn problem, with replacement from probability theory. Even if the half-lives were scaled up or down by an order of magnitude it wouldn't make much difference to the conclusion. By the time 30% of the nickel atoms have been chosen by hydrogen atoms /five times/ (which is what it takes to get from Ni^58 up to Cu^63), nearly all the rest will have been chosen /at least once/. And a lot of those are likely to have been chosen /exactly/ once, and the ones which have been selected just once are the ones with the 75 kY half life. Working out the details to get an exact answer would be messy, in part because a bunch of the nickel is actually Ni^60, but the point is that if any significant fraction of the resulting material were Ni^59 there should be measurable radioactivity. And there's not.
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
On 01/21/2011 04:45 PM, Man on Bridges wrote: What about this reaction: Ni (mass 60, 32 neutrons, pres. 26,223%) + Tritium (mass 3, 2 neutrons, pres. synt) -- Cu (mass 63, 34 neutrons, pres. 69,17%) + gamma radiation. Where's the tritium come from? And why the fake name Man on Bridges?
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
Tritium is Hydrogen ;-) No fake name it's an anagram of my name. On 21-1-2011 22:48, Stephen A. Lawrence wrote: On 01/21/2011 04:45 PM, Man on Bridges wrote: What about this reaction: Ni (mass 60, 32 neutrons, pres. 26,223%) + Tritium (mass 3, 2 neutrons, pres. synt) -- Cu (mass 63, 34 neutrons, pres. 69,17%) + gamma radiation. Where's the tritium come from? And why the fake name Man on Bridges?
Re: [Vo]:Removing All Doubt
In reply to Stephen A. Lawrence's message of Fri, 21 Jan 2011 14:35:44 -0500: Hi, [snip] If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. In ...which I effectively already said several days ago. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
On 01/21/2011 04:50 PM, Man on Bridges wrote: Tritium is Hydrogen ;-) Tritium is a vanishingly rare isotope of hydrogen. There is essentially none in the gas injected into the reactor. So, where do you think sufficient quantities of tritium come from to play an interesting role here? No fake name it's an anagram of my name. On 21-1-2011 22:48, Stephen A. Lawrence wrote: On 01/21/2011 04:45 PM, Man on Bridges wrote: What about this reaction: Ni (mass 60, 32 neutrons, pres. 26,223%) + Tritium (mass 3, 2 neutrons, pres. synt) -- Cu (mass 63, 34 neutrons, pres. 69,17%) + gamma radiation. Where's the tritium come from? And why the fake name Man on Bridges?
Re: [Vo]:Removing All Doubt
On 01/21/2011 04:53 PM, mix...@bigpond.com wrote: In reply to Stephen A. Lawrence's message of Fri, 21 Jan 2011 14:35:44 -0500: Hi, [snip] If a tiny fraction of the nickel is transmuted each second, and if nearly all the transmutation events produce unstable copper which eventually decays back to (higher weight) nickel, and if it takes multiple steps to get to stable copper, then by the time we've got a lot of stable copper running around, nearly all the nickel must have been transmuted at least once, and the whole lot should be radioactive. In ...which I effectively already said several days ago. Oh. Oops. Well, hmm -- I guess we agree, eh? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
In reply to Stephen A. Lawrence's message of Fri, 21 Jan 2011 14:35:44 -0500: Hi, [snip] In particular, there should probably be a really large fraction of Ni^59 present (31 neutrons), with a 75 ky half-life, and I'd think that would make the sample pretty hot. Or so it seems; I haven't done the calculations to back up the intuition. [snip] Ni-59 decays almost completely by electron capture directly to the ground state (hence no gammas), and all the energy is take by the neutrino, so this decay is not detectable (however about 1/25000 decays are via positron). See http://atom.kaeri.re.kr/cgi-bin/decay?Ni-59%20EC Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: EXTERNAL: Re: [Vo]:Removing All Doubt
Hello Stephen, I know it's rare, that's why it says pres. synt; while the scientists claim they need to refuel every six months. Kind regards from the Netherlands, Rob Dingemans (a.k.a. man on bridges ;-) On 21-1-2011 22:55, Stephen A. Lawrence wrote: On 01/21/2011 04:50 PM, Man on Bridges wrote: Tritium is Hydrogen ;-) Tritium is a vanishingly rare isotope of hydrogen. There is essentially none in the gas injected into the reactor. So, where do you think sufficient quantities of tritium come from to play an interesting role here? No fake name it's an anagram of my name. On 21-1-2011 22:48, Stephen A. Lawrence wrote: On 01/21/2011 04:45 PM, Man on Bridges wrote: What about this reaction: Ni (mass 60, 32 neutrons, pres. 26,223%) + Tritium (mass 3, 2 neutrons, pres. synt) -- Cu (mass 63, 34 neutrons, pres. 69,17%) + gamma radiation. Where's the tritium come from? And why the fake name Man on Bridges?
Re: [Vo]:Removing All Doubt
In reply to Peter Gluck's message of Fri, 21 Jan 2011 19:31:09 +0200: Hi, [snip] That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter [snip] If all Ni isotopes react equally, and 2/3 of Ni is Ni-58, and we assume single proton fusion, then the primary reaction would be: Ni-58 + H - Cu-59 + 3.42 MeV which then decays rapidly via positron decay according to Cu-59 - Ni-59 + e+ + neutrino + 4.8 MeV (however a considerable portion of this will be lost via neutrinos; say 1/2?). so the total reaction energy is 3.42 + 2.4 = 5.82 MeV / Ni-58. 2/3 *5 kWh / 6 MeV = 1.2E23 Ni-58 reactions, which is 12 gm Ni-58, or about 18 gm Ni altogether (assuming the other isotopes all yield about the same amount of energy / atom). So quite within the realm of possibility. OTOH, if he had 300 gm of Ni, and 1/3 was converted to Cu, then that represents considerably more energy, and one has to wonder where it all went? Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
In reply to Jones Beene's message of Fri, 21 Jan 2011 09:48:47 -0800: Hi, [snip] The *isotope shift* from the natural ratio after 6 months is extremely important. This can only be determined by specialized equipment. It is so important to establishing proof of a nuclear reaction, or to changing the (presumed negative) opinion of experts like Director Chu, that nothing else comes remotely close. [snip] There may be a slight spanner in the works here. If cluster fusion is involved, and it's a relatively slow (in nuclear terms) reaction, then the reaction itself may take place in such a way as to preferentially produce stable isotopes. IOW it may take as many protons and electrons from the cluster as may be required to produce a stable isotope, or perhaps, it fuses the entire cluster, then rearranges internally into a stable isotope, and spits out whatever is left over, carrying the energy of the reaction. Note however that in this case I would not expect to see positron decay with consequent electron annihilation gammas. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
On Jan 21, 2011, at 8:31 AM, Peter Gluck wrote: That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter There are some very fundamental issues, and mysteries involved. The fundamental questions relate to exactly what reactions are involved. Some do not produce copper, so the new copper content only establishes a lower bound on energy at best. Further, the mechanisms involved may not be fixed or even energy conservative, so there is difficulty establishing even a lower bound based on copper production. Generally, LENR has not been found to produce detectable high energy signatures. It also has not been found to produce radioactive products, especially neutrons. If weak reactions are eliminated, especially signature creating weak reactions that have more than femtosecond order, half lives, then what is left as feasible are strong force reactions without radiative products. Such reactions for nickel can be found starting on page 16 of: http://www.mtaonline.net/~hheffner/RptB.pdf which is described in http://www.mtaonline.net/~hheffner/dfRpt as noted earlier. Note that a lot more output possibilities are feasible than just copper, but let's get on with assuming copper is the only output. Those aneutronic strong force copper producing reactions involving 4 or fewer proton fusions with Ni are: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 + 2 p* -- 63Cu29 + 1H1 + 6.122 MeV [-10.582 MeV] (B_Ni:33) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 2 p* -- 65Cu29 + 1H1 + 7.453 MeV [-9.080 MeV] (B_Ni:65) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Note that equations (B_Ni:83) and (B_Ni:65) the extra proton involved merely plays a catalytic role, holding the nucleus together for a longer period and in an initially much more de-energized state. So, excluding weak reactions, and reactions involving large clusters of protons, the most likely candidate reactions producing CU are: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Looking at the first two reactions as likely candidates, with mean atomic weight near 63.6, and mean reaction energy about 7.2, we have an estimated energy density of E = (1/(63.6 gm/mol))*Na*7.2 MeV = 1.09x10^10 J/gm The production of 50,000 kWh then produces, using the above two reactions and considering Ni abundances, roughly produces a mass of copper M: M = (50,000 kWh)/(1.09x10^10 J/gm) = 16.5 gm We are left with some obvious questions. What about the other isotopes of nickel? Shouldn't they be involved? What prohibits radioactive nuclei from forming? We have involved the naturally occurring 58Ni, 60Ni, 61Ni 62Ni, and 64Ni, as well as trace amounts of 59Ni, as well as the other unknown and intentionally not disclosed ingredients. Given that 58Ni has 68% natural abundance, it is of interest as to why we do not see: 58Ni28 + p* - 59Cu29 which normally decays into 59Ni38 quickly, or possibly, given the deflation fusion scenario, the involvement of an apparently instantaneous electron capture: 58Ni28 + p* - 59Ni28 which has a 76000 y half life. Apparently, neither this nor any other radioactive material shows up in the output, however. Not a surprise, as few, or at least no confirmed, heavy LENR reactions produce radiative byproducts, except possibly tritium. Tritium production, is from a different process, tunneling of hydrogen to a cloaked hydrogen location, not tunneling of cloaked hydrogen to lattice nuclei locations which is responsible for heavy transmutation LENR. It seems a reasonable premise then that no radioactive material is *ever* produced in Rossi's experiment.Why this happens in general in LENR needs an answer. Nothing will be fully understood until why this happens is answered. A clue as to what might be happening is offered in pp 20-24 of: http://www.mtaonline.net/~hheffner/CFnuclearReactions.pdf n ( 939.57 MeV/c2) + e - lambda0 (1115.7 MeV/c2) + K0 ( 497.6 MeV/c2) + e p (938.27 MeV/c2) + e - lambda0 (1115.7 MeV/c2) + K0 ( 497.6 MeV/ c2) + antineutrino p (938.27 MeV/c2) + e - sigma+ (1189.3 MeV/c2) + K0 ( 497.6 MeV/ c2) + e These are sub-reactions, that are confined within the boundaries of the new heavy composite nucleus, except possibly for the escape of the K0. Given the extended stay of the electron in the de-energized nucleus, the probability of strange quark pairs in the vicinity increases, as does the above three reactions. These reactions, due to the catalytic effect of the nuclear electron,
Re: [Vo]:Removing All Doubt
On 22-1-2011 2:23, Horace Heffner wrote: Note that a lot more output possibilities are feasible than just copper, but let's get on with assuming copper is the only output. Those aneutronic strong force copper producing reactions involving 4 or fewer proton fusions with Ni are: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 + 2 p* -- 63Cu29 + 1H1 + 6.122 MeV [-10.582 MeV] (B_Ni:33) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 2 p* -- 65Cu29 + 1H1 + 7.453 MeV [-9.080 MeV] (B_Ni:65) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Note that equations (B_Ni:83) and (B_Ni:65) the extra proton involved merely plays a catalytic role, holding the nucleus together for a longer period and in an initially much more de-energized state. So, excluding weak reactions, and reactions involving large clusters of protons, the most likely candidate reactions producing CU are: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Horace, Based upon natural presence and absence of radiation I would probably go for this on: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 : pres. 3.634 % 1H1 : pres. 99.985 % 63Cu29 : pres. 69.17 % Kind regards, MoB
Re: [Vo]:Removing All Doubt
On Jan 21, 2011, at 7:06 PM, Man on Bridges wrote: Horace, Based upon natural presence and absence of radiation I would probably go for this on: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 : pres. 3.634 % 1H1 : pres. 99.985 % 63Cu29 : pres. 69.17 % Kind regards, MoB If I recall correctly, Rossi stated that deuterium kills the reaction, and that he uses pure protium. Looking again at the mean atomic weight 63.6 and mean reaction energy 7.2, I calculated as shown below, you can see that these are weighted averages, and are roughly 4 to 1 in favor of 62Ni vs 64Ni, corresponding roughly to their abundances, 3.634% vs 0.926%. They should be roughly equal in lattice half life (as I defined it in my paper), possibly with a small edge for the larger nucleus. Everything posted by me on this is seat of the pants accuracy. I think for casual discussion like this slide rule accuracy is sufficient to demonstrate the principles. I guess my age is showing! 8^) On Jan 21, 2011, at 4:23 PM, Horace Heffner wrote: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) 64Ni28 + 3 p* -- 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83) Looking at the first two reactions as likely candidates, with mean atomic weight near 63.6, and mean reaction energy about 7.2, we have an estimated energy density of E = (1/(63.6 gm/mol))*Na*7.2 MeV = 1.09x10^10 J/gm Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Removing All Doubt
In reply to Man on Bridges's message of Sat, 22 Jan 2011 05:06:04 +0100: Hi, [snip] Based upon natural presence and absence of radiation I would probably go for this on: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 : pres. 3.634 % 1H1 : pres. 99.985 % 63Cu29 : pres. 69.17 % The problem with choosing a reaction based on 62Ni28 is that it is less than 4% of the Ni present, so that even if all of it reacted, it wouldn't explain 30% Cu. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:Removing All Doubt
On Jan 21, 2011, at 10:03 PM, mix...@bigpond.com wrote: In reply to Man on Bridges's message of Sat, 22 Jan 2011 05:06:04 +0100: Hi, [snip] Based upon natural presence and absence of radiation I would probably go for this on: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 62Ni28 : pres. 3.634 % 1H1 : pres. 99.985 % 63Cu29 : pres. 69.17 % The problem with choosing a reaction based on 62Ni28 is that it is less than 4% of the Ni present, so that even if all of it reacted, it wouldn't explain 30% Cu. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Yes, indeed. My impression is the 30% number is not very solid though. Nothing seems really solid to me for that matter. That's life out here in the peanut gallery! 8^) I think we'd all be very satisfied to see a commercially viable MW plant. Hopefully it won't take as long to get there as it has taken BLP. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
