Re: [Vo]:Adding Energy to get Energy
On Sat, Jun 1, 2013 at 10:10 AM, Jones Beene jone...@pacbell.net wrote: Anyway the Farnsworth Fusor is a fusion reactor that many high school level students have built, including Conrad. It involves adding electrical energy in order to achieve LENR reactions. Sound familiar, Joshua? You missed the point. I have no problem adding energy to get energy. The problem I have is when you get back several times more heat than you used to start it, it should be easy to keep it going on its own. It's like combustion. In the Fusor, they haven't done this, plus what they put in is not heat, but real electrical energy to accelerate ions. They don't get that back, so self-sustaining is harder. It's more like trying to close the loop in electrolysis experiments, where you need electricity, but you produce heat. That takes a bigger COP. The mainstream wants to call it hot fusion but it is not. The gainful reactions are fusion but technically not hot or cold, and yes they are definitely low energy - warm not hot. Well, you can play with labels hot and cold, but this is ordinary fusion in the sense that the Coulomb barrier is overcome (or tunneled through) by kinetic energy, the branching ratios are perfectly standard, and everything is completely consistent with scientific generalizations (theory) already accumulated and verified. The published threshold level for D+D fusion is variously listed at around 1.4 MeV up to 2.2 MeV Where are those published? Because from what I've seen (see Bussard's google talk for example, or just wikipedia) the cross-section for D-D fusion peaks around 50 keV, and is still appreciable below 10 keV. The article on fusors says a minimum of about 4 keV is needed to get useful rates. The sun's interior is 15 billion kelvins, corresponding to about 1.3 keV. That makes for a slow fusion rate, and keeps the sun burning. and yet the Fusor average plasma energy level is lessthan 1 eV But in the fusor, it's not the plasma temperature that gives the ions the energy to fuse. The ions are accelerated into the plasma with a few keV energy. In the fusor, the ions are accelerated to several keV by the electrodes, so heating as such is not necessary (as long as the ions fuse before losing their energy by any process). -- Wiki so it truly is LENR on the input side. No, it truly is not. You don't have a clue.
Re: [Vo]:Adding Energy to get Energy
On Sat, Jun 1, 2013 at 11:25 AM, Edmund Storms stor...@ix.netcom.comwrote: We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. Please try to understand what I'm telling you. Cold fusion and hot fusion require different conditions to cause their initiation, they have different nuclear products, and they result at different rates. These are facts and not a matter of arbitrary definition. Cold fusion requires only a few eV for it to be initiated. In contrast, many keV are required to cause hot fusion at the same rate. Cold fusion produces helium while hot fusion produces fragments of helium. What do you mean fragments? Isotopes? The nuclei? Hot fusion produces isotopes of helium, including 4He very occasionally from DD fusion, but commonly from DT fusion, among other products. Cold fusion requires a solid while hot fusion occurs in plasma. Hot fusion also occurs in a solid in neutron sources where they accelerate hydrogen isotopes into palladium deuteride in commercial neutron sources.
Re: [Vo]:Adding Energy to get Energy
Jones, Did he make the background measure and the active run measure with the detector in the same place and same orientation? If he did, then the dip recorded during the active run would mean an _active_ ecat can reduce background radiation. Harry On Sun, Jun 2, 2013 at 12:08 AM, mix...@bigpond.com wrote: In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700: Hi, [snip] Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. That wouldn't surprise me if contained a couple of cm of lead shielding. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Adding Energy to get Energy
From: David Roberson Robin, how would Rossi prevent the lead from melting at the elevated temperatures? Do you suspect that he has it confined within a closed shell of some kind? I do not recall seeing any place for it to hide. Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. That wouldn't surprise me if contained a couple of cm of lead shielding. Lead shielding is not used by Rossi anymore. That tells you something. He has no worry of high energy radiation. The fact that there is no radiation at all detectable (at kW thermal output) from Rossi's device (above a threshold of tens of keV) is rather conclusive that there is no fusion, and essentially no nuclear reaction of any kind in the MeV range. Even if Robin is correct about fusion with fractional hydrogen having no prompt gamma, the occasional spallation neutron and the numerous Augur cascades brought on by fast ions would create reactions which would be easily detectable by Bianchini - and there would be lots of them at this kind of thermal gain level (unless most of the energy comes from another reaction). The gain in Rossi's device can be nuclear, but not from any known reaction which dumps MeV of energy. This would have been seen. The onus is on the proponent of such a theory to demonstrate how this gain does not involve yet another miracle (in addition to the overcoming the threshold of the primary reaction). The most likely explanation, based on all that we know about the Rossi reaction, is that gain happens in the soft x-ray spectrum and/or the EUV spectrum - which is not detectable with Bianchini's equipment. Hagelstein's magic phonons can be ruled out as a local CoE violation (which he admits). I have written Bianchini to ask that - if given another opportunity to test - will he please look specifically for soft x-rays. That would answer many questions, depending on the outcome. Jones attachment: winmail.dat
Re: [Vo]:Adding Energy to get Energy
OK, Jones, let me try to summarize what you propose. You believe CF is like the Mills effect even though CF is known to produce nuclear products and the Mills effect does not. You believe that Rossi made the Ni-H2 system create energy using the Mills effect while everyone else who explored this combination detected evidence of a nuclear process. Even Mills has apparently failed to make his method work this effectively, which seems ironic. You do not accept my theory of how the presence of D, H, or H+D can change the nuclear products from the same mechanism and account for the behavior. Instead, you propose at least two different mechanisms are operating to produce a very strange and rare energy release. You believe that no gamma is emitted by the e-Cat because no gamma is reported to be detected outside the apparatus. You come to this conclusion in spite of gamma being detected on occasion by several studies using light hydrogen and that Celani claimed the e-Cat emitted gamma during startup. Rossi was even concerned enough to put a lead shield in his early design. If Rossi is causing the Mills effect, then his e-Cat is accumulating hydrinos, which should be easy to detect. In addition, I'm asking him to look for deuterium and tritium. The tritium would be easy to detect and would provide unambiguous support for my model and a clear rejection of the Mills effect. This this summary correct? Ed Storms On Jun 1, 2013, at 3:02 PM, Jones Beene wrote: From: Jed Rothwell wrote: Bianchini finds zero radiation over hundreds of hours of careful radiation testing. Most cold fusion experiments produce no measurable radiation over hundreds of hours, including Pd-D ones. Most cold fusion experiments have been milliwatt level and do not use the very sophisticated setup of Bianchini – who after all is measuring kilowatts and is a leading expert at this. Essen finds no radioactivity in the ash. No excess deuterium or tritium have been documented in Rossi. I doubt anyone has looked for deuterium. It would be very difficult to find. Moderately difficult but not “very difficult” - but as a practical matter for a theoretician – is it wise to build a theory on a foundation that depends upon the viability of an extremely rare reaction (P-e-P), unless you have tested the ash in some basic way - and found a skewed H/D ratio or other indication of excess D? In short, the Rossi effect looks very much like the Mills effect. And the Mills effect looks like cold fusion. And that is precisely why it was a mistake to bifurcate the two, circa 1992. So we're back where we started. I agree with Mike McKubre about the conservation of miracles. But cold fusion requires more miracles than Mills, who with his funding has now proved many details. Mills predicts UV lines and finds them – miracle erased. He predicts no gamma and there is none. He predicts and captures the fractional hydrogen as physical atoms, and has the species tested - and it shows up differently from hydrogen in NMR etc. In fact the only problem with Mills in the miracle department is the lack of the commercial product – and if Rossi gets there first due to the high level of a more robust reaction, and especially if AR has accurately predicted Ni-62 then he wins the big prize... Gulp. Three cheers for Rossi, but in the end – it is LENR, and not cold fusion per se as Ed wants to define it. The ultimate source of energy cannot be determined as of now but Rossi’s hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. Which is to say, the Rossi effect is not fusion but can still be a new kind of nuclear reaction if one can be found with no gamma radiation. I expect that all of these effects are either nuclear in something like the conventional sense, or they are Mills superchemical shrinking hydrogen. I doubt there are two unrelated phenomena so similar in nature. Agreed– and there is one common denominator – QM tunneling. Things tend to be unified at some deep level, as are combustion and metabolism (to use Chris Tinsley's favorite example). Exactamundo! There are probably 5-6 similar variations on the theme of quantum tunneling which result in either 1)full fusion (as in the cold fusion of deuterium into helium) 2)some kind of weak force beta decay (W-L or related theory) 3)accelerated decay or internal conversion decay 4)UV supra-chemistry (energy coming from electron angular momentum) 5)QCD strong force effects (quantum chromodynamics) 6)Any combination of the above – even several of them in the same experiment! Any theory which aspires to encompass all of these begins with QM tunneling, but no simpler theory from there on - works. It cannot be true that all excess heat in Ni-H comes from a single kind of reaction, as the result do not allow this. Even in
RE: [Vo]:Adding Energy to get Energy
From: Edmund Storms OK, Jones, let me try to summarize what you propose You believe CF is like the Mills effect even though CF is known to produce nuclear products and the Mills effect does not. Not even close, Ed. I specifically said that I do not address anything to do with cold fusion, as opposed to LENR, and most importantly, this is not an either/or proposition. LENR can have both heat with nuclear products OR heat without nuclear products. And thirdly, we do not need Mills complete theory - but we must borrow parts from his theory to understand Rossi. I have always stated your theory fits Piantelli's experiments, but not Rossi's. You believe that Rossi made the Ni-H2 system create energy using the Mills effect while everyone else who explored this combination detected evidence of a nuclear process. Certainly not everyone else. Ahern's fine replication of Arata finds zero evidence of a nuclear effect and Celani finds none either - basically Piantelli supports the fusion viewpoint, but his work is less convincing Plus - Rossi has possibly advanced the Mills effect - which is now the Rossi effect, by identifying Ni-62 as the active species. BUT in the end - Bianchini has proved that there is NO nuclear products nor nuclear radiation in the Rossi effect. Even Mills has apparently failed to make his method work this effectively, which seems ironic. Mills' proponents, such as Jeff Driscoll think he has proved this. Many others are not convinced. Rossi seems to have gone well beyond Mills, and best of all - by pinpointing the active isotope. You do not accept my theory of how the presence of D, H, or H+D can change the nuclear products from the same mechanism and account for the behavior. Wrong. I do accept that your theory fits the physical evidence for some experiments, like Piantelli, but NOT Rossi's work. You want your theory to cover everything, but unfortunately it does not. Instead, you propose at least two different mechanisms are operating to produce a very strange and rare energy release. Yes. At least five similar mechanisms are present that all involved QM tunneling in one form or another. You believe that no gamma is emitted by the e-Cat because no gamma is reported to be detected outside the apparatus. You come to this conclusion in spite of gamma being detected on occasion by several studies using light hydrogen and that Celani claimed the e-Cat emitted gamma during startup. Rossi was even concerned enough to put a lead shield in his early design. Yes, this is all completely consistent with my hypothesis of multiple related pathways. Rossi no longer uses lead, and the very best testing for radioactivity which has ever been done in LENR finds no radiation in the Rossi effect. I emphasize NONE since there is not the slightest hint of any radiation in Bianchini's results. If Rossi is causing the Mills effect, then his e-Cat is accumulating hydrinos, which should be easy to detect. That could be true - but Rossi has an incentive not to permit this kind of testing. I have also provided a way to partially falsify my hypothesis of soft x-rays. In addition, I'm asking him to look for deuterium and tritium. The tritium would be easy to detect and would provide unambiguous support for my model and a clear rejection of the Mills effect. No. That is not correct. Tritium would have already have been detected by Bianchini if it was there, and it was not there. And it would not reject Mills unless all the complete gain was attributable to fusion, which cannot be the case. In any event, the presence of a small amount of tritium, which is not commensurate with the thermal gain, would bolster my hypothesis of several routes to gain. Jones attachment: winmail.dat
Re: [Vo]:Adding Energy to get Energy
Jones Beene jone...@pacbell.net wrote: No. That is not correct. Tritium would have already have been detected by Bianchini if it was there . . . I do not think so. Tritium would be trapped inside the cell. The decay product is a low energy beta. If a little tritium leaks out of the cell it is not likely to reach the detector, which only covers a small amount of the surface surrounding the cell. The only way Bianchini could detect this would be if Rossi makes a cell with a high quality tube and connectors to the cell contents and allows Bianchini to sample the gas. That is also the only way anyone could detect an increase in deuterium or any other gaseous nuclear product. This is a very difficult and involved thing to do. You have to purge the tube and other hardware. You have to use Swaglok connectors and you have to pay fanatical attention to cleanliness. If you touch any part of metal where the gas will flow, your fingerprint will contain more hydrogen than all of the reaction products from several days of high temperature heat production. Consider this: assuming the ratio of heat to helium is the same as plasma fusion, a Pd-D automobile that runs for a year, producing as much heat as the average gasoline burning automobile, will consume roughly 1 g of D2O. That's 48 million miles per gallon of D2O. - Jed
Re: [Vo]:Adding Energy to get Energy
Jed is correct. Tritium can not be detected by an ordinary detector because the beta is too weak. Unless the required special detector is used, tritium would be totally missed no matter how much is present. That is why tritium is dangerous. Nevertheless, modern methods can detect tritium at a very low level. I suggest the Ni removed from the hot Cat would contain enough tritium to be easily detected if the proper method were used. I have no expectation this effort will be made until the laboratory is found to be contaminated purely by a chance survey done for other reasons. Rossi is playing with fire. Ed Storms On Jun 2, 2013, at 10:20 AM, Jed Rothwell wrote: Jones Beene jone...@pacbell.net wrote: No. That is not correct. Tritium would have already have been detected by Bianchini if it was there . . . I do not think so. Tritium would be trapped inside the cell. The decay product is a low energy beta. If a little tritium leaks out of the cell it is not likely to reach the detector, which only covers a small amount of the surface surrounding the cell. The only way Bianchini could detect this would be if Rossi makes a cell with a high quality tube and connectors to the cell contents and allows Bianchini to sample the gas. That is also the only way anyone could detect an increase in deuterium or any other gaseous nuclear product. This is a very difficult and involved thing to do. You have to purge the tube and other hardware. You have to use Swaglok connectors and you have to pay fanatical attention to cleanliness. If you touch any part of metal where the gas will flow, your fingerprint will contain more hydrogen than all of the reaction products from several days of high temperature heat production. Consider this: assuming the ratio of heat to helium is the same as plasma fusion, a Pd-D automobile that runs for a year, producing as much heat as the average gasoline burning automobile, will consume roughly 1 g of D2O. That's 48 million miles per gallon of D2O. - Jed
RE: [Vo]:Adding Energy to get Energy
You do not need to remove the gas. I know you have heard of Bremsstrahlung, even if the word is almost unspellable to Anglos. Thank heavens for spell checkers and Wiki vids. Here is a little video that tells you why Bianchini would see tritium, if it was there. https://www.youtube.com/watch?v=yYLzarnlcUE It is not the beta decay which is seen - but instead it is the secondary gammas aka Bremsstrahlung . and yes - they would be on the edge of detectability, but a signal should show up above background on his meter - especially when the Rossi device is disassembled, as it is in the Penon report. From: Jed Rothwell Jones Beene wrote: No. That is not correct. Tritium would have already have been detected by Bianchini if it was there . . . I do not think so. Tritium would be trapped inside the cell. The decay product is a low energy beta. If a little tritium leaks out of the cell it is not likely to reach the detector, which only covers a small amount of the surface surrounding the cell. The only way Bianchini could detect this would be if Rossi makes a cell with a high quality tube and connectors to the cell contents and allows Bianchini to sample the gas. That is also the only way anyone could detect an increase in deuterium or any other gaseous nuclear product. This is a very difficult and involved thing to do. You have to purge the tube and other hardware. You have to use Swaglok connectors and you have to pay fanatical attention to cleanliness. If you touch any part of metal where the gas will flow, your fingerprint will contain more hydrogen than all of the reaction products from several days of high temperature heat production. Consider this: assuming the ratio of heat to helium is the same as plasma fusion, a Pd-D automobile that runs for a year, producing as much heat as the average gasoline burning automobile, will consume roughly 1 g of D2O. That's 48 million miles per gallon of D2O. - Jed
Re: [Vo]:Adding Energy to get Energy
Jones, you are simply wrong. I have worked with tritium and I know how it behaves. It cannot be detected using its Bremsstrahlund unless a huge amount is present because this radiation is produced at only a small fraction of the beta and is absorbed very quickly by only a small amount of material in the case of tritium. The video does not show anything about tritium. We simply do not know or are told what is in the supposed light stick or how much tritium is present. To the extent the container holding the tritium containing fluid is thin enough to pass Bremsstrahlund and generate useful light, the amount of tritium would be very dangerous if the container broke. If Rossi had produce enough tritium to be detected this way, everyone in the room would have serious health and legal problems if the tritium got out of the E-Cat. Ed Storms On Jun 2, 2013, at 10:48 AM, Jones Beene wrote: You do not need to remove the gas. I know you have heard of Bremsstrahlung, even if the word is almost unspellable to Anglos. Thank heavens for spell checkers and Wiki vids. Here is a little video that tells you why Bianchini would see tritium, if it was there. https://www.youtube.com/watch?v=yYLzarnlcUE It is not the beta decay which is seen – but instead it is the secondary gammas aka Bremsstrahlung … and yes – they would be on the edge of detectability, but a signal should show up above background on his meter - especially when the Rossi device is disassembled, as it is in the Penon report. From: Jed Rothwell Jones Beene wrote: No. That is not correct. Tritium would have already have been detected by Bianchini if it was there . . . I do not think so. Tritium would be trapped inside the cell. The decay product is a low energy beta. If a little tritium leaks out of the cell it is not likely to reach the detector, which only covers a small amount of the surface surrounding the cell. The only way Bianchini could detect this would be if Rossi makes a cell with a high quality tube and connectors to the cell contents and allows Bianchini to sample the gas. That is also the only way anyone could detect an increase in deuterium or any other gaseous nuclear product. This is a very difficult and involved thing to do. You have to purge the tube and other hardware. You have to use Swaglok connectors and you have to pay fanatical attention to cleanliness. If you touch any part of metal where the gas will flow, your fingerprint will contain more hydrogen than all of the reaction products from several days of high temperature heat production. Consider this: assuming the ratio of heat to helium is the same as plasma fusion, a Pd-D automobile that runs for a year, producing as much heat as the average gasoline burning automobile, will consume roughly 1 g of D2O. That's 48 million miles per gallon of D2O. - Jed
Re: [Vo]:Adding Energy to get Energy
On Jun 2, 2013, at 10:05 AM, Jones Beene wrote: From: Edmund Storms OK, Jones, let me try to summarize what you propose You believe CF is like the Mills effect even though CF is known to produce nuclear products and the Mills effect does not. Not even close, Ed. I specifically said that I do not address anything to do with cold fusion, as opposed to LENR, and most importantly, this is not an either/or proposition. Please Jones, do not split hairs. You know exactly what I mean by CF and LENR, which I explained before. We are discussing either a nuclear process (CF or LENR) or a nonnuclear process (Mills). This is a clear either/or situation. LENR can have both heat with nuclear products OR heat without nuclear products. No LENR cannot be both. You are simply changing the definition to fit your personal wishes. This is not how the rest of the world defines the word. If you want to make up a different word, please do. This is like calling an apple an orange because you happen to like oranges. Your approach simply causes confusion because we can not discuss the same effect. And thirdly, we do not need Mills complete theory - but we must borrow parts from his theory to understand Rossi. I have always stated your theory fits Piantelli's experiments, but not Rossi's. The Mills theory is a complete and unified model. You can not extract parts that you happen to like. In addition, my theory explains both Rossi and Piantelli. Rossi simply made the effect Piantelli observed stronger. I explain how this might have been done. You might not agree, but nevertheless I have logically explained how this might happen. You have not. I have predicted what is expected to be observed. You have not. I have explained how the Rossi effect must be controlled. You have not. You can accept or reject, but please acknowledge what I claim and discuss the consequences of the idea rather rejecting my ideas by redefining words and proposing ambiguous mechanisms. You believe that Rossi made the Ni-H2 system create energy using the Mills effect while everyone else who explored this combination detected evidence of a nuclear process. Certainly not everyone else. Ahern's fine replication of Arata finds zero evidence of a nuclear effect and Celani finds none either - basically Piantelli supports the fusion viewpoint, but his work is less convincing Of course, some people do not see any effect. This failure is common in this field. In contrast, the effect is clearly seen by other people. Which experience you choose to believe determines how you explain or reject the ideas. I make clear exactly what I accept and reject, and why. Plus - Rossi has possibly advanced the Mills effect - which is now the Rossi effect, by identifying Ni-62 as the active species. BUT in the end - Bianchini has proved that there is NO nuclear products nor nuclear radiation in the Rossi effect. No, Bianchini only failed to detect the energy of radiation his instruments were designed to detect. In addition, he could not demonstrate that radiation was not made inside and being absorbed to below the detection limit. We do know that the light hydrogen system makes low energy radiation that can only result from a nuclear reaction. Whether the proper method was used to detect this radiation emitted from the Ross device is still unknown. Even Mills has apparently failed to make his method work this effectively, which seems ironic. Mills' proponents, such as Jeff Driscoll think he has proved this. Many others are not convinced. Rossi seems to have gone well beyond Mills, and best of all - by pinpointing the active isotope. Rossi claims that Ni62 produces energy because it transmutes to Cu. Mills claims that energy is given off when the electron in a H atom is able to go below the quantum level of 1 by giving this energy to a catalyst. Are you proposing that this catalyst is Ni62? Why would this be the case? Please explain because it makes no sense using the Mills theory. You do not accept my theory of how the presence of D, H, or H+D can change the nuclear products from the same mechanism and account for the behavior. Wrong. I do accept that your theory fits the physical evidence for some experiments, like Piantelli, but NOT Rossi's work. You want your theory to cover everything, but unfortunately it does not. It does not fit everything only because you say it doesn't. I say it does and can predict behavior. We will see who is right when the predictions are tested. Instead, you propose at least two different mechanisms are operating to produce a very strange and rare energy release. Yes. At least five similar mechanisms are present that all involved QM tunneling in one form or another. OK, this is clear.
RE: [Vo]:Adding Energy to get Energy
From: Edmund Storms Jones, you are simply wrong. I have worked with tritium and I know how it behaves. You apparently have not worked with tritium very intuitively, if you cannot understand this simple video. It cannot be detected using its Bremsstrahlund unless a huge amount is present because this radiation is produced at only a small fraction of the beta and is absorbed very quickly by only a small amount of material in the case of tritium. That is not what is being demonstrated before your eyes. Why am I not surprised that you do not want to acknowledge this? Ah. is it because you want tritium to be present in the Rossi reactor when it is not indicated. The video does not show anything about tritium. That is a silly comment, and you know it. We simply do not know or are told what is in the supposed light stick or how much tritium is present. Did you take the time to follow up on the specs? It takes about 5 seconds to find the Wiki site http://en.wikipedia.org/wiki/Tritium_illumination To the extent the container holding the tritium containing fluid is thin enough to pass Bremsstrahlund and generate useful light, the amount of tritium would be very dangerous if the container broke. There are safety concerns, and I would not use this product - but that is not material to the fact that tritium can be detected by its Bremsstrahlung. If Rossi had produced enough tritium to be detected this way, everyone in the room would have serious health and legal problems if the tritium got out of the E-Cat Then it is a good thing that the Rossi effect produces no tritium! But of course, it should if your theory was correct for his device - but it is not correct for the Rossi device. QED Jones
Re: [Vo]:Adding Energy to get Energy
Apparently Jones, I have to be clearer and more emphatic. Tritium can not be detected when it is in a container as massive as the E-cat. THIS IS A FACT. Please at least acknowledge that I might know something about tritium that you do not. The video only shows that some unknown amount of tritium mixed with unknown other radioactive elements was detected in an container of unknown absorption. You are extrapolating this demonstation to conditions that have no relevance to the demonstration. I hope this is clear and we can go on to other subjects. Ed Storms On Jun 2, 2013, at 11:49 AM, Jones Beene wrote: From: Edmund Storms Jones, you are simply wrong. I have worked with tritium and I know how it behaves. You apparently have not worked with tritium very intuitively, if you cannot understand this simple video. It cannot be detected using its Bremsstrahlund unless a huge amount is present because this radiation is produced at only a small fraction of the beta and is absorbed very quickly by only a small amount of material in the case of tritium. That is not what is being demonstrated before your eyes. Why am I not surprised that you do not want to acknowledge this? Ah… is it because you want tritium to be present in the Rossi reactor when it is not indicated. The video does not show anything about tritium. That is a silly comment, and you know it. We simply do not know or are told what is in the supposed light stick or how much tritium is present. Did you take the time to follow up on the specs? It takes about 5 seconds to find the Wiki site http://en.wikipedia.org/wiki/Tritium_illumination To the extent the container holding the tritium containing fluid is thin enough to pass Bremsstrahlund and generate useful light, the amount of tritium would be very dangerous if the container broke. There are safety concerns, and I would not use this product - but that is not material to the fact that tritium can be detected by its Bremsstrahlung. If Rossi had produced enough tritium to be detected this way, everyone in the room would have serious health and legal problems if the tritium got out of the E-Cat Then it is a good thing that the Rossi effect produces no tritium! But of course, it should if your theory was correct for his device – but it is not correct for the Rossi device. QED Jones
RE: [Vo]:Adding Energy to get Energy
Ed, You are not very good at misdirection, try hard as you might - and you are fighting a losing battle in trying to wedge an incorrect theory into the most important LENR experiment out there at present. My advice is to quit before you are completely embarrassed. You theory works in some situations, but it does not work for Rossi's results. Get used to it. Please acknowledge that you read the Penon report and understand that the device was completely disassembled after over 4 megawatt hours of heat was produced, and that no radiation was detected. How much tritium should have been present - if your theory were to be valid? Enough that we would no doubt not be hearing from those guys again. Yet they are still with us and your theory still falls as flat as a pancake, insofar as Rossi is concerned. QED. and yes . let's do move on. Jones From: Edmund Apparently Jones, I have to be clearer and more emphatic. Tritium can not be detected when it is in a container as massive as the E-cat. THIS IS A FACT. Please at least acknowledge that I might know something about tritium that you do not. The video only shows that some unknown amount of tritium mixed with unknown other radioactive elements was detected in an container of unknown absorption. You are extrapolating this demonstation to conditions that have no relevance to the demonstration. I hope this is clear and we can go on to other subjects. Ed Storms On Jun 2, 2013, at 11:49 AM, Jones Beene wrote: From: Edmund Storms Jones, you are simply wrong. I have worked with tritium and I know how it behaves. You apparently have not worked with tritium very intuitively, if you cannot understand this simple video. It cannot be detected using its Bremsstrahlund unless a huge amount is present because this radiation is produced at only a small fraction of the beta and is absorbed very quickly by only a small amount of material in the case of tritium. That is not what is being demonstrated before your eyes. Why am I not surprised that you do not want to acknowledge this? Ah. is it because you want tritium to be present in the Rossi reactor when it is not indicated. The video does not show anything about tritium. That is a silly comment, and you know it. We simply do not know or are told what is in the supposed light stick or how much tritium is present. Did you take the time to follow up on the specs? It takes about 5 seconds to find the Wiki site http://en.wikipedia.org/wiki/Tritium_illumination To the extent the container holding the tritium containing fluid is thin enough to pass Bremsstrahlund and generate useful light, the amount of tritium would be very dangerous if the container broke. There are safety concerns, and I would not use this product - but that is not material to the fact that tritium can be detected by its Bremsstrahlung. If Rossi had produced enough tritium to be detected this way, everyone in the room would have serious health and legal problems if the tritium got out of the E-Cat Then it is a good thing that the Rossi effect produces no tritium! But of course, it should if your theory was correct for his device - but it is not correct for the Rossi device. QED Jones
Re: [Vo]:Adding Energy to get Energy
Jones Beene jone...@pacbell.net wrote: . . . a signal should show up above background on his meter - especially when the Rossi device is disassembled, as it is in the Penon report. They disassemble it by cutting it in half with a saw, don't they? There is no way you could capture tritium by this method! - Jed
Re: [Vo]:Adding Energy to get Energy
OK Jones, useful discussion has come to an end. I will wait until the proper measurements are made . Then we will talk again. Ed Storms On Jun 2, 2013, at 12:59 PM, Jones Beene wrote: Ed, You are not very good at misdirection, try hard as you might - and you are fighting a losing battle in trying to wedge an incorrect theory into the most important LENR experiment out there at present. My advice is to quit before you are completely embarrassed. You theory works in some situations, but it does not work for Rossi’s results. Get used to it. Please acknowledge that you read the Penon report and understand that the device was completely disassembled after over 4 megawatt hours of heat was produced, and that no radiation was detected. How much tritium should have been present - if your theory were to be valid? Enough that we would no doubt not be hearing from those guys again. Yet they are still with us and your theory still falls as flat as a pancake, insofar as Rossi is concerned. QED… and yes … let’s do move on. Jones From: Edmund Apparently Jones, I have to be clearer and more emphatic. Tritium can not be detected when it is in a container as massive as the E- cat. THIS IS A FACT. Please at least acknowledge that I might know something about tritium that you do not. The video only shows that some unknown amount of tritium mixed with unknown other radioactive elements was detected in an container of unknown absorption. You are extrapolating this demonstation to conditions that have no relevance to the demonstration. I hope this is clear and we can go on to other subjects. Ed Storms On Jun 2, 2013, at 11:49 AM, Jones Beene wrote: From: Edmund Storms Jones, you are simply wrong. I have worked with tritium and I know how it behaves. You apparently have not worked with tritium very intuitively, if you cannot understand this simple video. It cannot be detected using its Bremsstrahlund unless a huge amount is present because this radiation is produced at only a small fraction of the beta and is absorbed very quickly by only a small amount of material in the case of tritium. That is not what is being demonstrated before your eyes. Why am I not surprised that you do not want to acknowledge this? Ah… is it because you want tritium to be present in the Rossi reactor when it is not indicated. The video does not show anything about tritium. That is a silly comment, and you know it. We simply do not know or are told what is in the supposed light stick or how much tritium is present. Did you take the time to follow up on the specs? It takes about 5 seconds to find the Wiki site http://en.wikipedia.org/wiki/Tritium_illumination To the extent the container holding the tritium containing fluid is thin enough to pass Bremsstrahlund and generate useful light, the amount of tritium would be very dangerous if the container broke. There are safety concerns, and I would not use this product - but that is not material to the fact that tritium can be detected by its Bremsstrahlung. If Rossi had produced enough tritium to be detected this way, everyone in the room would have serious health and legal problems if the tritium got out of the E-Cat Then it is a good thing that the Rossi effect produces no tritium! But of course, it should if your theory was correct for his device – but it is not correct for the Rossi device. QED Jones
RE: [Vo]:Adding Energy to get Energy
Tritium is preferentially absorbed into nickel. Most of it would be retained in the nickel powder, if it were present. From: Jed Rothwell . . . a signal should show up above background on his meter - especially when the Rossi device is disassembled, as it is in the Penon report. They disassemble it by cutting it in half with a saw, don't they? There is no way you could capture tritium by this method! - Jed
Re: [Vo]:Adding Energy to get Energy
Jones Beene jone...@pacbell.net wrote: Tritium is preferentially absorbed into nickel. Most of it would be retained in the nickel powder, if it were present. Good point. Still, if you were doing a serious study you would not cut it in half. McKubre devised a complicated way to puncture the Arata cells to collect a sample of gas from them. Something like this might be done. - Jed
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Let me say that almost everyone concerned, other than Andrea Rossi himself - would be delighted if tritium had been found in the spent fuel of the HotCat. If tritium were found in proportion to thermal gain - this would explain the mechanism in accordance with Ed Storm's theory - and not only that: the ash would become a valuable by-product as well. It would make everything clearer and pave the road to commercialization. However, aside from that - the best reason to think that that Rossi has it right this time, and has made a major breakthrough with the HotCat relates to his bombshell patent change to bet the farm on Ni-62. This comes into the tritium discussion through the back door, in a perverse way. There is no obvious reason why tritium would have any connection to Ni-62. It does not. Consequently, if tritium were proved to be responsible for the gain via proton fusion, then it would also indicate that Rossi lost the farm, on his bet, since he would have little IP protection. I apologize to Ed for coming-off as unnecessarily derogatory of his theory as applied to Rossi's results, but it should be clear to all concerned that if Ed is correct, Andrea loses almost everything in the race to market by holding a worthless patent. Most of us are interesting in finding the scientific truth - regardless of who wins the pot-of-gold; but if tritium turns up, Rossi is toast in more ways than one, and there seems to be a bit of unfairness there. There are very different implications for framing a valid theory based around the one isotope. This starts with the realization that Rossi (possibly at the insistence of Focardi) did perform experiment with all of the various nickel isotopes, and out of that effort he is now convinced that he has found the one which is responsible: the smoking gun. This opens up a new avenue for understanding which may be more difficult, but not impossible to navigate. Maybe Rossi deserves some kind of penalty for his antics, but in the 'big picture' it also looks like he may have come around at the perfect time to revive a languishing technology. perhaps making the next age of man the Nickel Age. after all he is. http://www.youtube.com/watch?feature=player_embedded http://www.youtube.com/watch?feature=player_embeddedv=L-4zfsy6rsM v=L-4zfsy6rsM Hope that is not overly dramatic. and . yes, there is the little problem of not yet proved. but if it arrives soon, Rossi may insist that we start the Calendar over next January and call 2014 the year 1 AR. Jones
Re: [Vo]:Adding Energy to get Energy
Jones, I agree with your conclusion about Rossi. However, tritium is not his only problem. His patent will probably not reveal how the Ni can be treated to make it active. Simply adding Ni62 is obviously not the only thing he does to the Ni. Without the ability to replicate the patent by a person skilled in the art, it is worthless. I suspect Rossi does not care about the patent because he intends to keep this a trade secret. The Ni62 is only a distraction. I suspect he expects to use the patent to send researches on a wild goose chase and then use legal distractions when this does not work. He is in an untenable situation. He has no idea how or why the effect works, yet he can make extra energy. He needs to sell the device while pretending he understands how it functions. The patent helps him pursued people that he knows what he is doing - for awhile. He hopes that when the truth be known, he is rich enough to fight the challenge. Meanwhile, he is bringing useful attention to the field, which ironically will encourage people to find his error that much sooner. I would hate to be in his shoes. Ed Storms On Jun 2, 2013, at 2:35 PM, Jones Beene wrote: Let me say that almost everyone concerned, other than Andrea Rossi himself - would be delighted if tritium had been found in the spent fuel of the HotCat. If tritium were found in proportion to thermal gain - this would explain the mechanism in accordance with Ed Storm’s theory – and not only that: the ash would become a valuable by-product as well. It would make everything clearer and pave the road to commercialization. However, aside from that - the best reason to think that that Rossi has it right this time, and has made a major breakthrough with the HotCat relates to his bombshell patent change to “bet the farm” on Ni-62. This comes into the tritium discussion through the back door, in a perverse way. There is no obvious reason why tritium would have any connection to Ni-62. It does not. Consequently, if tritium were proved to be responsible for the gain via proton fusion, then it would also indicate that Rossi “lost the farm,” on his bet, since he would have little IP protection. I apologize to Ed for coming-off as unnecessarily derogatory of his theory as applied to Rossi’s results, but it should be clear to all concerned that if Ed is correct, Andrea loses almost everything in the race to market by holding a worthless patent. Most of us are interesting in finding the scientific truth – regardless of who wins the pot-of-gold; but if tritium turns up, Rossi is toast in more ways than one, and there seems to be a bit of unfairness there. There are very different implications for framing a valid theory based around the one isotope. This starts with the realization that Rossi (possibly at the insistence of Focardi) did perform experiment with all of the various nickel isotopes, and out of that effort he is now convinced that he has found the one which is responsible: the smoking gun. This opens up a new avenue for understanding which may be more difficult, but not impossible to navigate. Maybe Rossi deserves some kind of penalty for his antics, but in the ‘big picture’ it also looks like he may have come around at the perfect time to revive a languishing technology… perhaps making the next “age of man” the Nickel Age… after all he is… http://www.youtube.com/watch?feature=player_embeddedv=L-4zfsy6rsM Hope that is not overly dramatic… and … yes, there is the little problem of “not yet proved”… but if it arrives soon, Rossi may insist that we start the Calendar over next January and call 2014 the year 1 AR… Jones
Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sun, 2 Jun 2013 06:15:39 -0700: Hi, [snip] The fact that there is no radiation at all detectable (at kW thermal output) from Rossi's device (above a threshold of tens of keV) is rather conclusive that there is no fusion, and essentially no nuclear reaction of any kind in the MeV range. Even if Robin is correct about fusion with fractional hydrogen having no prompt gamma, the occasional spallation neutron and the numerous Augur cascades brought on by fast ions would create reactions which would be easily detectable by Bianchini - and there would be lots of them at this kind of thermal gain level (unless most of the energy comes from another reaction). I suspect that most of the energy release is from f/h formation, with a few low level fusion reactions thrown in. The number of such reactions varies and is essentially not a well controlled parameter. This is because it depends on the degree of shrinkage of the f/h. Though, as any given device is used longer, the average shrinkage level should increase, and consequently the number of fusion reactions should also increase. Perhaps another reason why Rossi switches cartridges every 6 months. That also means that as a device ages, the COP should increase (unless it is deliberately kept at a specific level.) BTW the highest energy X-ray you get from Ni is about 8 keV, IOW it's soft. Other substances present (with possible exception of the secret sauce), have lower values. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Adding Energy to get Energy
Jones, please do not confuse hot fusion with cold fusion. The difference is in the products. Cold fusion does not produce neutrons and energetic radiation. Hot fusion produce neutrons and radiation because the conditions require the nuclear product to fragment. This fragmentation does not take place during cold fusion. In addition, cold fusion takes places only in a lattice without any additional energy being applied. Hot fusion occurs in plasma where high energy is available, as is the case with the Farnsworth Fusor. The Farnsworth Fusor produces hot fusion, but at a low level. It works at an apparently low energy because the process is efficient and the real energy of the deutrons is not properly calculated. There is no threshold level. The rate is simply roughly related to the log of the energy and becomes undetectable at low energy. Causing hot fusion is trivial. Anyone can do this with high voltage and some D2 gas. The challenge is to produce more energy than is applied. This has not been done using hot fusion using any of the methods. In contrast, cold fusion has accomplished this on many occasions, although with difficulty and at low level. These are facts and not a matter of opinion. Please try to understand the difference between these two phenomenon. Your opinion is important and needs to be correct. Ed Storms On Jun 1, 2013, at 9:10 AM, Jones Beene wrote: In the category of truth is stranger than fiction here is an amazing story of impersonation on several levels http://trib.com/news/state-and-regional/wyoming-teen-who-built-fusion-reacto r-disqualified-from-science-fair/article_15dda5ab-b68e-5fa7- a13f-7b30d22f850 f.html A Wyoming high school student builds a nuclear reactor in his dad's garage - and then is disqualified from the International Science and Engineering Fair on a technicality. The beginning of a conspiracy theory? LENR suppression? His problem could have been: impersonating Philo - :-) Anyway the Farnsworth Fusor is a fusion reactor that many high school level students have built, including Conrad. It involves adding electrical energy in order to achieve LENR reactions. Sound familiar, Joshua? The mainstream wants to call it hot fusion but it is not. The gainful reactions are fusion but technically not hot or cold, and yes they are definitely low energy - warm not hot. The published threshold level for D+D fusion is variously listed at around 1.4 MeV up to 2.2 MeV and yet the Fusor average plasma energy level is less than 1 eV - so it truly is LENR on the input side. It is definitely NOT in any way hot fusion. Since it is orders of magnitude lower input. Since there are neutrons emitted, no one doubts the reaction is nuclear. Plasma LENR reaction produce neutrons but the same does not happen in condensed matter LENR. BTW Conrad is also a YT! Jockey. His channel is replete with his experiments http://www.youtube.com/watch?v=i4Sjg2aNw6w Jones winmail.dat
RE: [Vo]:Adding Energy to get Energy
-Original Message- From: Edmund Storms Jones, please do not confuse hot fusion with cold fusion. The difference is in the products. Not necessarily. Perhaps that is your definition, but as I stated - the Farnsworth Fusor is LENR on the input side. Same with sonofusion - it is the input that matters most - NOT the output. The Fusor is definitely NOT hot fusion. The average plasma temperature is in the range of neon lighting of CFL or CRTs. Sonofusion produces neutrons, and is generally considered cold and the Fusor is similar. Most observers these days label the Fusor warm but it in reality it is FAR closer to LENR than to hot fusion - and IMO if the device is in no-man's land - then LENR group should claim it, just as with sonofusion. Jones
Re: [Vo]:Adding Energy to get Energy
We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. Please try to understand what I'm telling you. Cold fusion and hot fusion require different conditions to cause their initiation, they have different nuclear products, and they result at different rates. These are facts and not a matter of arbitrary definition. Cold fusion requires only a few eV for it to be initiated. In contrast, many keV are required to cause hot fusion at the same rate. Cold fusion produces helium while hot fusion produces fragments of helium. Cold fusion requires a solid while hot fusion occurs in plasma. Cold fusion does not produce neutrons, while hot fusion produces many neutrons when the same amount of energy is released. The term LENR is used to only describe cold fusion. It was not created for it to be applied to hot fusion. On Jun 1, 2013, at 9:48 AM, Jones Beene wrote: -Original Message- From: Edmund Storms Jones, please do not confuse hot fusion with cold fusion. The difference is in the products. Not necessarily. Perhaps that is your definition, but as I stated - the Farnsworth Fusor is LENR on the input side. Same with sonofusion - it is the input that matters most - NOT the output. The Fusor is definitely NOT hot fusion. The average plasma temperature is in the range of neon lighting of CFL or CRTs. The resulting nuclear products are the important criteria. I can produce hot fusion simply by hitting LiD with a hammer. Therefore the applied power is not important. The amount of applied power only changes the rate, not the resulting nuclear reactions. Because neutrons are made by hot fusion, it can be detected at VERY LOW rates. That is why only a few keV can cause a detectable rate. The applied voltage does not change the kind of nuclear reaction that takes place. We must make a clear distinction between the nuclear reaction that results from hot fusion and the different one that results from cold fusion. Your approach confuses this requirement. Sonofusion produces neutrons, and is generally considered cold and the Fusor is similar. Sonofusion produces hot fusion, i.e, it apparently produces equal amounts of neutrons and tritium, not helium. Yes, I know the tritium has not been reported in this case. The relative applied power is not important. I say again, the applied power ONLY CHANGES THE RATE. It does not change the resulting nuclear reaction. Most observers these days label the Fusor warm but it in reality it is FAR closer to LENR than to hot fusion - and IMO if the device is in no- man's land - then LENR group should claim it, just as with sonofusion. I don't know who you are quoting, but they have no idea what they are talking about. Ed Storms Jones
Re: [Vo]:Adding Energy to get Energy
I thought we agreed to call Muon assisted fusion warm fusion. On Sat, Jun 1, 2013 at 12:25 PM, Edmund Storms stor...@ix.netcom.com wrote: We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. Please try to understand what I'm telling you. Cold fusion and hot fusion require different conditions to cause their initiation, they have different nuclear products, and they result at different rates. These are facts and not a matter of arbitrary definition. Cold fusion requires only a few eV for it to be initiated. In contrast, many keV are required to cause hot fusion at the same rate. Cold fusion produces helium while hot fusion produces fragments of helium. Cold fusion requires a solid while hot fusion occurs in plasma. Cold fusion does not produce neutrons, while hot fusion produces many neutrons when the same amount of energy is released. The term LENR is used to only describe cold fusion. It was not created for it to be applied to hot fusion. On Jun 1, 2013, at 9:48 AM, Jones Beene wrote: -Original Message- From: Edmund Storms Jones, please do not confuse hot fusion with cold fusion. The difference is in the products. Not necessarily. Perhaps that is your definition, but as I stated - the Farnsworth Fusor is LENR on the input side. Same with sonofusion - it is the input that matters most - NOT the output. The Fusor is definitely NOT hot fusion. The average plasma temperature is in the range of neon lighting of CFL or CRTs. The resulting nuclear products are the important criteria. I can produce hot fusion simply by hitting LiD with a hammer. Therefore the applied power is not important. The amount of applied power only changes the rate, not the resulting nuclear reactions. Because neutrons are made by hot fusion, it can be detected at VERY LOW rates. That is why only a few keV can cause a detectable rate. The applied voltage does not change the kind of nuclear reaction that takes place. We must make a clear distinction between the nuclear reaction that results from hot fusion and the different one that results from cold fusion. Your approach confuses this requirement. Sonofusion produces neutrons, and is generally considered cold and the Fusor is similar. Sonofusion produces hot fusion, i.e, it apparently produces equal amounts of neutrons and tritium, not helium. Yes, I know the tritium has not been reported in this case. The relative applied power is not important. I say again, the applied power ONLY CHANGES THE RATE. It does not change the resulting nuclear reaction. Most observers these days label the Fusor warm but it in reality it is FAR closer to LENR than to hot fusion - and IMO if the device is in no-man's land - then LENR group should claim it, just as with sonofusion. I don't know who you are quoting, but they have no idea what they are talking about. Ed Storms Jones
RE: [Vo]:Adding Energy to get Energy
-Original Message- From: Edmund Storms We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. That may be true regarding cold fusion. You are free to stick with that antiquated term if you want to, but do not pretend to speak for the broader field of LENR. I am NOT talking about cold Fusion. Period. LENR is much more than cold fusion in 2013. The two are not synonymous. I have followed what is now called LENR for 23 years too from a different perspective which does not require deuterium - and I believe that the proper definition of LENR must include sonofusion, the Farnsworth Fusor, the Mills effect and the Rossi effect, in addition to cold fusion. In fact- doing so will make understanding the LENR field less confusing, not more - since there is plenty of overlap and we have moved well beyond deuterium. Please try to understand what I'm telling you. I understand what you are saying - but I reject completely your contention that the definition of LENR is somehow fixed by the old days when cold fusion was the only game in town, and fractional hydrogen was considered taboo to cold fusion practitioners. You have overlooked Mills' excellent experiments from the start and continue to overlook his contributions, despite his publications, patents and success in fund-raising - or to consider the newer offshoots of CQM. Mills is NOT cold fusion in any relevant way - but can be included under the broader definition of LENR, especially since many of us have adapted parts of his theory to a nuclear perspective. In short, Mills work is more relevant to understanding Rossi than were PF. In a nutshell - Ed this is our disagreement: You are lost in fading reminiscence of cold fusion of palladium and deuterium - which is going nowhere as of 2013 - now that Nickel-hydrogen is showing an ability to provide kilowatts in contrast to the milliwatts of most cold fusion efforts. Please do not confuse the two. Jones
Re: [Vo]:Adding Energy to get Energy
You can call it what you want. Jones called the muon reaction cold fusion before he applied the term was applied to the F-P effect. Nevertheless, the products are those that result from hot fusion, i.e. equal amounts of neutron and tritium that result from fragmentation of the resulting helium nucleus. In discussions, these two different nuclear processes MUST be described accurately and not confused with each other. To make the description clear, hot fusion is the term applied to the reaction that fragments the helium and cold fusion is applied when helium itself is produced. Of course, the two different reactions have many other important differences. We just need to describe them so that we are clear about which reaction is being described. Ed Storms On Jun 1, 2013, at 10:57 AM, Terry Blanton wrote: I thought we agreed to call Muon assisted fusion warm fusion. On Sat, Jun 1, 2013 at 12:25 PM, Edmund Storms stor...@ix.netcom.com wrote: We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. Please try to understand what I'm telling you. Cold fusion and hot fusion require different conditions to cause their initiation, they have different nuclear products, and they result at different rates. These are facts and not a matter of arbitrary definition. Cold fusion requires only a few eV for it to be initiated. In contrast, many keV are required to cause hot fusion at the same rate. Cold fusion produces helium while hot fusion produces fragments of helium. Cold fusion requires a solid while hot fusion occurs in plasma. Cold fusion does not produce neutrons, while hot fusion produces many neutrons when the same amount of energy is released. The term LENR is used to only describe cold fusion. It was not created for it to be applied to hot fusion. On Jun 1, 2013, at 9:48 AM, Jones Beene wrote: -Original Message- From: Edmund Storms Jones, please do not confuse hot fusion with cold fusion. The difference is in the products. Not necessarily. Perhaps that is your definition, but as I stated - the Farnsworth Fusor is LENR on the input side. Same with sonofusion - it is the input that matters most - NOT the output. The Fusor is definitely NOT hot fusion. The average plasma temperature is in the range of neon lighting of CFL or CRTs. The resulting nuclear products are the important criteria. I can produce hot fusion simply by hitting LiD with a hammer. Therefore the applied power is not important. The amount of applied power only changes the rate, not the resulting nuclear reactions. Because neutrons are made by hot fusion, it can be detected at VERY LOW rates. That is why only a few keV can cause a detectable rate. The applied voltage does not change the kind of nuclear reaction that takes place. We must make a clear distinction between the nuclear reaction that results from hot fusion and the different one that results from cold fusion. Your approach confuses this requirement. Sonofusion produces neutrons, and is generally considered cold and the Fusor is similar. Sonofusion produces hot fusion, i.e, it apparently produces equal amounts of neutrons and tritium, not helium. Yes, I know the tritium has not been reported in this case. The relative applied power is not important. I say again, the applied power ONLY CHANGES THE RATE. It does not change the resulting nuclear reaction. Most observers these days label the Fusor warm but it in reality it is FAR closer to LENR than to hot fusion - and IMO if the device is in no- man's land - then LENR group should claim it, just as with sonofusion. I don't know who you are quoting, but they have no idea what they are talking about. Ed Storms Jones
Re: [Vo]:Adding Energy to get Energy
On Jun 1, 2013, at 11:16 AM, Jones Beene wrote: -Original Message- From: Edmund Storms We are taking about two different phenomenon of nature. Trying to use the same concepts and words to describe both results in confusion. Those of us who have studied cold fusion for the last 23 years have a definition of CF that is not up for discussion. That may be true regarding cold fusion. You are free to stick with that antiquated term if you want to, but do not pretend to speak for the broader field of LENR. Jones, I think I'm in a better position to speak for the field than you are. I am NOT talking about cold Fusion. Period. LENR is much more than cold fusion in 2013. The two are not synonymous. Cold fusion was the term first applied to the phenomenon. Then transmutation was observed, which required the term not be focused on fusion. Consequently, several additional terms were tried and LENR stuck. LENR includes the phenomenon called cold fusion and the reaction producing transmutation. I have followed what is now called LENR for 23 years too from a different perspective which does not require deuterium - and I believe that the proper definition of LENR must include sonofusion, the Farnsworth Fusor, the Mills effect and the Rossi effect, in addition to cold fusion. That is not what is accepted or is accurate. The phenomenon that is called cold fusion produces helium and tritium without neutrons. The phenomenon called hot fusion produces no helium and equal numbers of neutrons and tritium, examples of which are sonofusion, the Farnsworth Fusor, and muon fission. The Mills effect is a different phenomenon all together. His effect is not nuclear, as he admits. The Rossi effect follows from the cold fusion phenomenon when H is used instead of D. I have shown exactly how the D and H systems are related to the cold fusion phenomenon and why tritium is produced without neutrons. I hope you have followed the discussion of my explanation on Vortex. In any case, this has no relationship to the difference between cold fusion or LENR and hot fusion. In fact- doing so will make understanding the LENR field less confusing, not more - since there is plenty of overlap and we have moved well beyond deuterium. Please try to understand what I'm telling you. I understand what you are saying - but I reject completely your contention that the definition of LENR is somehow fixed by the old days when cold fusion was the only game in town, and fractional hydrogen was considered taboo to cold fusion practitioners. Please note what I said above. Your comment has no relationship to what I'm saying. You have overlooked Mills' excellent experiments from the start and continue to overlook his contributions, despite his publications, patents and success in fund-raising - or to consider the newer offshoots of CQM. Mills is NOT cold fusion in any relevant way - but can be included under the broader definition of LENR, especially since many of us have adapted parts of his theory to a nuclear perspective. In short, Mills work is more relevant to understanding Rossi than were PF. That is simply not true. The Rossi effect is claimed to produce a nuclear product. I think the product is wrong, but the focus has been on detecting the product. In addition, the Ni-H2 system produces radiation that CAN NOT result from a Mills reaction. In a nutshell - Ed this is our disagreement: You are lost in fading reminiscence of cold fusion of palladium and deuterium - which is going nowhere as of 2013 - now that Nickel-hydrogen is showing an ability to provide kilowatts in contrast to the milliwatts of most cold fusion efforts. Apparently you have not read my book, or any of my papers or followed the discussion on Vortex. I have no loyity to deuterium. I have made cear that ANY isotope of hydrogen can fuse as a result of the cold fusion (LENR) process. In contrast, Jones, you are mixing applies and oranges and producing confusion. Please read my book. cold fusion using deuterium produces more than milliwatts of power. Rossi has made the Ni-H2 system more active than it ever was, but this does not change the nature of the reaction. Ed Storms Please do not confuse the two. Jones
RE: [Vo]:Adding Energy to get Energy
-Original Message- From: Edmund Storms The Mills effect is a different phenomenon all together. His effect is not nuclear, as he admits. Yes, but that is not relevant to understanding Rossi. Many other researchers, including Miley have incorporated major parts of Mills theory into a nuclear version for Ni-H - using the important Rydberg energy details - like IRH (inverted Rydberg hydrogen). It is easily possible that Mills' theory, like your own (and everyone else's) is partly correct and partly wrong. It is a major mistake to be ignorant of Mills experiments when analyzing Rossi. The Rossi effect follows from the cold fusion phenomenon when H is used instead of D. No, it doesn't. Just the opposite, in fact. There is no evidence of any cold fusion effect in the Rossi results. You are intentionally conflating with Piantelli. Bianchini finds zero radiation over hundreds of hours of careful radiation testing. Essen finds no radioactivity in the ash. No excess deuterium or tritium have been documented in Rossi. In short, the Rossi effect looks very much like the Mills effect. The Rossi effect is claimed to produce a nuclear product. Many inaccurate claims have been made about the Rossi effect, but no nuclear product has been documented by anyone including Focardi, who is responsible for that detail. Testing of the copper showed natural isotope balance, indicating metal migration - not transmutation ash. In addition, the Ni-H2 system produces radiation that CAN NOT result from a Mills reaction. Piantelli alone has made claim this claim, but we are talking about the Rossi effect and Mills. Piantelli is irrelevant to Rossi. Bianchini finds zero radiation over hundreds of hours of careful radiation testing of Rossi's results on three separate occasions. Celani saw something on startup but nothing at all during operation. Rossi may have used a startup isotope, but there is NO radioactivity at all during operation. Your theory may work for Piantelli's results, which have trade secrets that make it different from Rossi - but your theory is incompatible with Rossi's actual results. Again, Rossi see no radiation during operation and no nuclear ash, like Mills. Rossi uses potassium catalyst, like Mills (this has been documented in the spectrographic data). Rossi see long term gain, like Thermacore. Rossi has no radioactivity in the ash or in the process itself. Once again, the Rossi effect bears every resemblance to Mills, and none to Piantelli at all, or to cold fusion or as you chosen to define it. Apparently you have not read my book, or any of my papers or followed the discussion on Vortex. I have no loyalty to deuterium. I have read your book and other material but continue to reject the notion that it is relevant to Rossi's actual results. You know that I have stated several times that your theory may well apply to Piantelli's results, but not to Rossi's yet you continue to conflate Rossi and Piantelli because your theory falls flat with Rossi. Jones attachment: winmail.dat
Re: [Vo]:Adding Energy to get Energy
Jones Beene jone...@pacbell.net wrote: Bianchini finds zero radiation over hundreds of hours of careful radiation testing. Most cold fusion experiments produce no measurable radiation over hundreds of hours, including Pd-D ones. Essen finds no radioactivity in the ash. No excess deuterium or tritium have been documented in Rossi. I doubt anyone has looked for deuterium. It would be very difficult to find. You could not look for it with any of the Rossi cells I have seen. You need something smaller, tightly sealed with high grade Swaglok fittings. I do not know if they have looked for tritium either. I'll bet it would escape. In short, the Rossi effect looks very much like the Mills effect. And the Mills effect looks like cold fusion. So we're back where we started. I agree with Mike McKubre about the conservation of miracles. I expect that all of these effects are either nuclear in something like the conventional sense, or they are Mills superchemical shrinking hydrogen. I doubt there are two unrelated phenomena so similar in nature. Things tend to be unified at some deep level, as are combustion and metabolism (to use Chris Tinsley's favorite example). - Jed
RE: [Vo]:Adding Energy to get Energy
From: Jed Rothwell wrote: Bianchini finds zero radiation over hundreds of hours of careful radiation testing. Most cold fusion experiments produce no measurable radiation over hundreds of hours, including Pd-D ones. Most cold fusion experiments have been milliwatt level and do not use the very sophisticated setup of Bianchini - who after all is measuring kilowatts and is a leading expert at this. Essen finds no radioactivity in the ash. No excess deuterium or tritium have been documented in Rossi. I doubt anyone has looked for deuterium. It would be very difficult to find. Moderately difficult but not very difficult - but as a practical matter for a theoretician - is it wise to build a theory on a foundation that depends upon the viability of an extremely rare reaction (P-e-P), unless you have tested the ash in some basic way - and found a skewed H/D ratio or other indication of excess D? In short, the Rossi effect looks very much like the Mills effect. And the Mills effect looks like cold fusion. And that is precisely why it was a mistake to bifurcate the two, circa 1992. So we're back where we started. I agree with Mike McKubre about the conservation of miracles. But cold fusion requires more miracles than Mills, who with his funding has now proved many details. Mills predicts UV lines and finds them - miracle erased. He predicts no gamma and there is none. He predicts and captures the fractional hydrogen as physical atoms, and has the species tested - and it shows up differently from hydrogen in NMR etc. In fact the only problem with Mills in the miracle department is the lack of the commercial product - and if Rossi gets there first due to the high level of a more robust reaction, and especially if AR has accurately predicted Ni-62 then he wins the big prize... Gulp. Three cheers for Rossi, but in the end - it is LENR, and not cold fusion per se as Ed wants to define it. The ultimate source of energy cannot be determined as of now but Rossi's hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. Which is to say, the Rossi effect is not fusion but can still be a new kind of nuclear reaction if one can be found with no gamma radiation. I expect that all of these effects are either nuclear in something like the conventional sense, or they are Mills superchemical shrinking hydrogen. I doubt there are two unrelated phenomena so similar in nature. Agreed- and there is one common denominator - QM tunneling. Things tend to be unified at some deep level, as are combustion and metabolism (to use Chris Tinsley's favorite example). Exactamundo! There are probably 5-6 similar variations on the theme of quantum tunneling which result in either 1)full fusion (as in the cold fusion of deuterium into helium) 2)some kind of weak force beta decay (W-L or related theory) 3)accelerated decay or internal conversion decay 4)UV supra-chemistry (energy coming from electron angular momentum) 5)QCD strong force effects (quantum chromodynamics) 6)Any combination of the above - even several of them in the same experiment! Any theory which aspires to encompass all of these begins with QM tunneling, but no simpler theory from there on - works. It cannot be true that all excess heat in Ni-H comes from a single kind of reaction, as the result do not allow this. Even in the same experiment, one could see three similar but different pathways to thermal gain that all share QM tunneling as the starting point, but differ on everything else. Ockham be damned ! Don't forget that appeals to parsimony were used by skeptics to argue the wrong side of many past issues - against DNA for instance, as the carrier of genetic information. There is a long list of Ockham failures and the workable LENR theory will be on the next one.
Re: [Vo]:Adding Energy to get Energy
Jones Beene jone...@pacbell.net wrote: Most cold fusion experiments have been milliwatt level and do not use the very sophisticated setup of Bianchini . . . Fleischmann and Pons ran hundreds of tests with boiling cells, at 20 to 100 W. They has sophisticated detectors. They found nothing as far as I know. I doubt anyone has looked for deuterium. It would be very difficult to find. ** ** Moderately difficult but not “very difficult” Impossible with any of the Rossi reactors I know of. You need complicated Swaglok connections, as I said, to extract the sample of gas without contamination. It could be done. It should be done. But it has not been done as far as I know. You would have to design that particular experiment around this goal. - Jed
Re: [Vo]:Adding Energy to get Energy
On Sat, Jun 1, 2013 at 2:02 PM, Jones Beene jone...@pacbell.net wrote: The ultimate source of energy cannot be determined as of now but Rossi’s hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. I don't exclude the possibility that there's something Millsean going on here, but I will take pleasure in quoting this post at a future point in time if research were to come out and show that d+p fusion is happening. Eric
RE: [Vo]:Adding Energy to get Energy
Eric, I have dined on crow before and prefer mine well-charred with a nice Pinot Noir… The ultimate source of energy cannot be determined as of now but Rossi’s hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. I don't exclude the possibility that there's something Millsean going on here, but I will take pleasure in quoting this post at a future point in time if research were to come out and show that d+p fusion is happening. Eric
Re: [Vo]:Adding Energy to get Energy
On Sat, Jun 1, 2013 at 5:33 PM, Jones Beene jone...@pacbell.net wrote: I have dined on crow before and prefer mine well-charred with a nice Pinot Noir… Foul! Fowl demands a white, say chardonnay,
Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sat, 1 Jun 2013 14:33:22 -0700: Hi, [snip] Eric, I have dined on crow before and prefer mine well-charred with a nice Pinot Noir The ultimate source of energy cannot be determined as of now but Rossis hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. Some clean reactions: (Type I) 64Ni + H2 = 62Ni + 4He + 11.8 MeV (no gammas) 62Ni + H2 = 60Ni + 4He + 9.88 MeV (no gammas) 60Ni + H2 = 58Ni + 4He + 7.9 MeV (no gammas) Alternatives: (Type II) 64Ni + H2 = 65Cu + H (fast) + 7.45 MeV (no gammas) 62Ni + H2 = 63Cu + H (fast) + 6.12 MeV (no gammas) Dirty: 58Ni + H2 = 56Ni (radioactive) + 4He + 5.83 MeV 60Ni + H2 = 61Cu (radioactive) + H (fast) + 4.8 MeV Decay reactions: 56Ni = 56Co (radioactive) = 56Fe 61Cu = 61Ni However, if you start with 62Ni ( 64Ni) then you need to wait a very long time before enough 56Ni shows up to produce significant radioactivity. The no means no (few) primary gammas. A very small percentage of the fast particles will produce the occasional spallation neutron, which will in turn, over a long period of time, result in the production of some radioactive isotopes. However the rate of production may be below the rate at which they decay, so the overall level of radioactivity may remain very small. The H2 is of course f/H molecules. I have no idea whether Type I or Type II reactions would predominate. Note also that in the Type II case, follow on Type II reactions would result in stable Zn isotopes. Nevertheless, I suspect that indeed the primary source of energy in his reactor is the formation of f/H. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Adding Energy to get Energy
-Original Message- From: mix...@bigpond.com Hi Robin, The H2 is of course f/H molecules. Still three body reactions - no way Nevertheless, I suspect that indeed the primary source of energy in his reactor is the formation of f/H. Yup. By a large factor. There is actually an easy way to reinforce but not falsify this conclusion, and without much ado. I've suggested that the SiC and SiN tubes at ~800 C will be partially transparent to soft x-rays/EUV. If so, the emissivity will be in excess of one, although only slightly so and only if the inner SS tube allows some soft x-ray radiation through (not known). That should be relatively easy to confirm, and could have been done in previous testing. Since there is no other reasonable explanation for emissivity in excess of one, it would be worth the small effort. Jones
Re: [Vo]:Adding Energy to get Energy
On Sat, Jun 1, 2013 at 5:10 PM, Eric Walker eric.wal...@gmail.com wrote: On Sat, Jun 1, 2013 at 2:02 PM, Jones Beene jone...@pacbell.net wrote: The ultimate source of energy cannot be determined as of now but Rossi’s hundreds of hours of operation at kilowatt levels with no gammas clearly indicates NO fusion. I don't exclude the possibility that there's something Millsean going on here, but I will take pleasure in quoting this post at a future point in time if research were to come out and show that d+p fusion is happening. Eric hydrinos could be hydrotons that didn't get close enough to fuse. Harry
Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sat, 1 Jun 2013 17:27:32 -0700: Hi, [snip] -Original Message- From: mix...@bigpond.com Hi Robin, The H2 is of course f/H molecules. Still three body reactions - no way No, these are all two body reactions, because the f/H is bound in a pico/femto molecule, and approaches the target nucleus as a single (composite) entity. Upon arrival, several things can happen: 1) Two neutrons tunnel out of the target nucleus into the molecule, producing 4He. (note that tunneling of a single neutron is out of the question, as it is energetically forbidden). This option has the great advantage that only neutrons tunnel, which is, I suspect, more likely than protons tunneling in.) The energy is shared as kinetic energy between the lighter Ni nucleus and the new alpha particle. or 2) The molecule tunnels into the nucleus producing an exited nucleus that then decays through particle emission. That emission can take many forms. It may be a proton, a neutron, an alpha particle, or the nucleus of a lighter element, i.e. a fission reaction. (Strictly speaking all these possibilities are fission reactions.) or 3) Only one proton of the pair tunnels into the target nucleus resulting in a transmutation reaction, the energy of which is shared as kinetic energy between the newly transmuted nucleus and the remaining free proton. (This option is I suspect much more probable than #2, because it involves the concurrent tunneling of only a single proton, rather than both of them. Comparison with #1 is more difficult because of the ease of tunneling for neutrons compared to protons). 4) Any of the above, where at least some of the energy is shared with one or both of the shrunken electrons too. 5) A repeat of all of the above, where however the original molecule is a magnetically bound composite of at least 2 f/H molecules, containing at least 4 f/h atoms. In this case the energy released when all 4 are involved in a transmutation reaction, as in #2 above, is much larger, so the fission channel becomes much more likely. (This is beginning to sound like a patent application. ;) BTW each of the fast particles produced can breed hundreds of new f/H molecules, provided that sufficient normal Hydrogen is within reach. Each of these can then produce another transmutation reaction, resulting in the localized micro explosions that are responsible for the craters that are detected. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Adding Energy to get Energy
-Original Message- From: mix...@bigpond.com Hi Robin, The H2 is of course f/H molecules. Still three body reactions - no way No, these are all two body reactions, because the f/H is bound in a pico/femto molecule, and approaches the target nucleus as a single (composite) entity. What is the separation distance between the two protons? They may be relatively close, but it is hard to imagine that this is not three body. Why are spallation neutrons not produced?
RE: [Vo]:Adding Energy to get Energy
Jones, Interesting concept..[snip] No, these are all two body reactions, because the f/H is bound in a pico/femto molecule,[/snip] how about combining it with Naudt's paper on relativistiv hydrogen, the hydrogen has an equivalent negative acceleration of relativistic proportion from suppression geometry which is breaking the isotropy into a tapestry of different values..could the covalent bond hold / oppose the transition to different fractional values within the tapestry such that it collides with a non fractional or lesser fractional hydrogen from a temporal angle? It would be a Lorentzian contracted molecule approaching the normal molecule .. I am questioning if a covalent bond can drag an inertial frame with it into a different frame and collide on some relativistic vector that reduces the columb barrier? Time dilation might gain something but more than that I wonder if the barrier is lower on the time axis. Fran Jones http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=from:%22Jones+Be ene%22 Beene Sat, http://www.mail-archive.com/search?l=vortex-l@eskimo.comq=date:20130601 01 Jun 2013 19:35:51 -0700 -Original Message- From: mix...@bigpond.com Hi Robin, The H2 is of course f/H molecules. Still three body reactions - no way No, these are all two body reactions, because the f/H is bound in a pico/femto molecule, and approaches the target nucleus as a single (composite) entity. What is the separation distance between the two protons? They may be relatively close, but it is hard to imagine that this is not three body. Why are spallation neutrons not produced?
RE: [Vo]:Adding Energy to get Energy
Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. -Original Message- From: mix...@bigpond.com Hi Robin, Why are spallation neutrons not produced?
Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sat, 1 Jun 2013 19:35:11 -0700: Hi, [snip] -Original Message- From: mix...@bigpond.com Hi Robin, The H2 is of course f/H molecules. Still three body reactions - no way No, these are all two body reactions, because the f/H is bound in a pico/femto molecule, and approaches the target nucleus as a single (composite) entity. What is the separation distance between the two protons? That depends on the model. In mine, it can be as small as a few fm. They may be relatively close, but it is hard to imagine that this is not three body. Three body reactions are usually rare, because it is unlikely that all three bodies will be in the same place at the same time. However if two of the three are already bound together, and hence guaranteed to both be present, then the chances are the same as for a two body reaction. Why are spallation neutrons not produced? The energy needed to remove a neutron for each of the isotopes is: 64Ni: 9.66 MeV 62Ni: 10.6 MeV 60Ni: 7.82 MeV 58Ni: 12.217 MeV For the three reactions: 64Ni + H2 = 62Ni + 4He + 11.8 MeV (no gammas) 62Ni + H2 = 60Ni + 4He + 9.88 MeV (no gammas) 60Ni + H2 = 58Ni + 4He + 7.9 MeV (no gammas) there isn't a lot of difference between the energy of the alphas and the spallation energy, which means that such an alpha has to get really lucky and hit a Ni nucleus close to where it was formed, before it loses too much energy ionizing other atoms. Since very few will be so lucky, few spallation neutrons will be created. BTW note also that alphas and other nuclei are both positively charged, so they repel one another making direct hits even more unlikely. Note also that *pure* 62Ni isn't capable of creating any at all (9.88 MeV 10.6 MeV). However once a significant amount of 60Ni had accumulated, it would be able to. OTOH, the 60Ni may get gobbled up by the H2 reaction before any significant spallation could occur. (Perhaps a reason for Rossi's patent application? ;) So it doesn't look like neutron spallation is going to be much of a problem. The proton spallation energy is: 64Ni: 12.55 MeV 62Ni: 11.14 MeV 60Ni: 9.53 MeV 58Ni: 8.17 MeV Obviously not a problem for the top two. However for the bottom two it is possible. Not a problem for 60Ni, because 59Co is stable, however 57Co is radioactive. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Adding Energy to get Energy
In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700: Hi, [snip] Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. That wouldn't surprise me if contained a couple of cm of lead shielding. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Adding Energy to get Energy
Robin, how would Rossi prevent the lead from melting at the elevated temperatures? Do you suspect that he has it confined within a closed shell of some kind? I do not recall seeing any place for it to hide. Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Jun 2, 2013 12:08 am Subject: Re: [Vo]:Adding Energy to get Energy In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700: Hi, [snip] Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. That wouldn't surprise me if contained a couple of cm of lead shielding. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Adding Energy to get Energy
In reply to David Roberson's message of Sun, 2 Jun 2013 00:41:42 -0400 (EDT): Hi, [snip] Robin, how would Rossi prevent the lead from melting at the elevated temperatures? Do you suspect that he has it confined within a closed shell of some kind? I do not recall seeing any place for it to hide. Dave You are correct. :) I was confused with the earlier versions that used lead shielding. However any solid will provide *some* shielding. -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, Jun 2, 2013 12:08 am Subject: Re: [Vo]:Adding Energy to get Energy In reply to Jones Beene's message of Sat, 1 Jun 2013 19:59:52 -0700: Hi, [snip] Let me add that in the appendix to the Penon report, David Bianchini finds not only no significant radiation over background, but actually the peak radiation counts are slightly less during the experiment than background, indicating the apparatus shields the detector from cosmic rays slightly. That wouldn't surprise me if contained a couple of cm of lead shielding. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
