Re: [Vo]:comment on Violante data as covered by Steve Krivit
Hi Horace, sorry for the late response, my comments below. 2010/2/7 Horace Heffner hheff...@mtaonline.net: On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote: 2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. I think the experimenters were competent. They knew what they were doing. Using a Faraday constant of 96,485 C/mol, and (conservatively) a valence of 4, n for moles produced, I for current = .2 A, t for time = 1 s, we get: n = I * t / (96,485 C/mol * 4) n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or 3.12x10^17 atoms per second. We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/cm^3 = 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s. If this is correct (highly suspect! 8^), then at a current density of 200 mA/cm^2 we have a thickness of 183 atoms removed per second, or 445 angstroms per second. This would be correct if palladium, when driven as an anode, did dissolve in an alkaline electrolyte (they classically used LiOD in that M4 experiment, according to their original report at http://newenergytimes.com/v2/archives/1998epri/TR-107843-V1.PDF , thanks to Steve Krivit for the link), which it doesn't, see the Pd/H2O Pourbaix diagram at http://www.platinummetalsreview.com/jmpgm/data/datasheet.do?record=532database=cesdatabase which shows that such corrosion only occurs in an acidic electrolyte (pH 3). and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). I guess you mean they are *found* there, couldn't they be both produced on the surface, only with more kinetic energy in the helium nuclei (alphas) than in the tritium nuclei for some reason, so that the helium is implanted more deeply? I find the idea of two different nuclear reaction sites producing different products a bit unlikely. No, most of the 4He reactions occur sub-surface. What do you think produces a volcano? A surface reaction? The volcanos you mention could also be impact craters produced by a local chain reaction on the surface. The typical 4He produced by CF does not have MeV kinetic energy, and is not surface produced. If it were there would be massive alpha counts. There is not sufficient kinetic energy to push alphas that deep into the Pd. You may well have a point here. A ref for those deep alphas would be welcome BTW. This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. Maybe it's just not possible, because you can't make large D fluxes collide head-on Head on collisions, i.e. kinetics, can not possibly account for cold fusion. Not alone I agree, it's more subtle than that, but the Ds do have to meet don't they? I submit that the Ds following/pushing each other down the lattice corridors like fish in a fish swarm have no reasons to experience frequent close encounters. in the bulk, this can only happen at a significant scale on the surface (desorbing vs incident fluxes). In the bulk, it seems to me the deuterons just push and follow each other down the lattice's concentration gradients, and never really collide hard. Also, if Bose Einstein Condensates are involved, they requires cold bosons for their formation. Head-on collisions may be a plausible mechanism for deuteron kinetic energy removal. This would only be the case if the collisions were almost all totally inelastic. Good point, although the combined effect of their respective colleagues pushing from behind could conceivably result in many of the collisions being inelastic. In any case, surface or subsurface, we certainly all agree that something special occurs in the surface region, so the surface plays a determinant role in CF. Maybe we could collaboratively establish a list of what we know is special about the surface, here are a few items for a start: a/ only place where frequent D encounters are possible (as mentioned above) b/ adsorption heat is higher than absorption heat, i.e. the trapping potential for Ds is deeper on the surface than in the bulk (probably due to the surface Pds having dangling bonds) c/ place
Re: [Vo]:comment on Violante data as covered by Steve Krivit
On Feb 9, 2010, at 2:09 AM, Michel Jullian wrote: Hi Horace, sorry for the late response, my comments below. 2010/2/7 Horace Heffner hheff...@mtaonline.net: On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote: 2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. I think the experimenters were competent. They knew what they were doing. Using a Faraday constant of 96,485 C/mol, and (conservatively) a valence of 4, n for moles produced, I for current = .2 A, t for time = 1 s, we get: n = I * t / (96,485 C/mol * 4) n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or 3.12x10^17 atoms per second. We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/ cm^3 = 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s. If this is correct (highly suspect! 8^), then at a current density of 200 mA/cm^2 we have a thickness of 183 atoms removed per second, or 445 angstroms per second. This would be correct if palladium, when driven as an anode, did dissolve in an alkaline electrolyte (they classically used LiOD in that M4 experiment, according to their original report at http://newenergytimes.com/v2/archives/1998epri/TR-107843-V1.PDF , thanks to Steve Krivit for the link), which it doesn't, see the Pd/H2O Pourbaix diagram at http://www.platinummetalsreview.com/jmpgm/data/datasheet.do? record=532database=cesdatabase which shows that such corrosion only occurs in an acidic electrolyte (pH 3). It has been pointed out to me privately that hydrogen charge transport has to be accounted for as well, i.e. that hydrogen evolution reduces the effective corrosion current. However, since the reversed current cleaning process was carried out in part to degass the Pd, I expect the hydrogen contribution to the positive surface charge of the Pd anode would be extremely diminished in the latter part of this cleaning process. Well, this is indeed an interesting electrochemical problem. My experience is that nothing, including platinum, totally avoids anodic corrosion if there is a current present. Passification works in part by eliminating the current at the potential at which the passivation is occurring, or less, i.e. by building an insulating layer. I do not think passification of *highly loaded* Pd is possible. The evolving hydrogen would prevent accumulated oxidation of the Pd surface. I have done various passivation experiments (not with Pd though) and my experience has been that passification takes considerable time, even for metals that are not loaded with hydrogen, and once it does occur, the current is highly reduced. Further, if a constant current source is used then the voltage rises to the point where the passified surface is breached. Beyond that, and this is a fairly irrelevant point I know, I think Pd corrodes as an anode in the presence of current in neutral Ph salt electrolytes. The EPRI article states: They accomplished loading with a combination of initial low cathode current densities of ~20-50 mA/ cm2, followed by current ramps up to ~1.0 A/cm2. Current reversals to deload or “strip” the cathodes of D and clean the surface by temporarily making it an anode resulted in high loadings. It seems to me the Pd would be dissolving during the deloading process when the current is reversed. Also, apparently my estimate of 200 mA/cm^2 was too low - it was probably 1 A/cm^2. It would be interesting to actually do an experiment with Pd wire, loading and then reversing the current repeatedly for a long period and then weighing the wire. It seems to me the experimenters would not have gone thorough this procedure if the current reversal did not actually clean the electrode surface, i.e. expose a pure Pd surface. A fully passified surface would not be effective at loading hydrogen as a cathode because it would not even be conductive. If pure Pd was exposed to the electrolyte as an anode it seems to me certain that Pd was being dissolved in the process. One thing I take to be self evident to anyone who has practical experience with electrochemistry experiments. If you have current through an anode then *some* of that anode is going into solution, and that includes platinum. I
RE: [Vo]:comment on Violante data as covered by Steve Krivit
-Original Message- From: mix...@bigpond.com If lattice resonance is a factor, then some depth may be required to build up a strong enough resonance effect that the mechanism can operate. (analogous to adding more dipoles to a TV antenna) Hi Robin, Lattice resonance and depth below the surface could be a factor. However, it is also possible that depth is counterproductive for a certain kind of resonance effect - which only works on the surface layer itself, and only with hydrogen. There is a provocative animation on the Wiki entry: http://en.wikipedia.org/wiki/Infrared_spectroscopy ... which shows an animated visualization (if you have html tuned on) of the many vibration modes that only operate at the surface layer and probably only operate with a very mobile atom like H or D. Simple diatomic molecules like deuterium, which have only one bond, may stretch and contort at high frequencies, but apparently there are large frequency gaps which are forbidden, leading to either super-radiance or sub-radiance. I have lost the citation from a few weeks ago that claimed that below a threshold of about 10 nm, the expected blackbody frequency is upshifted for nanostructures, in general. If anyone has that cite (site) handy please post it. All of these issues could overlap, based on geometry and super-radiance. In the animation above, the atoms in a CH2 group can vibrate in six different ways: stretching, scissoring, rocking, wagging and twisting. What the animation does not show is the distinct possibility of coherence (semi-coherence) in vibration - such as if these atoms (deuterons) were moving together in a tuning fork analogy. That situation would be expected to self-reinforce. If the actual frequency of vibration were to exceed the blackbody rate at the surface layer, then we might expect it to be coherent. Don't ask me why yet, but it relates to super-radiance and seems to involve atoms that are inactive in the parts of the IR spectrum unless they are stimulated by external agents. I found some published information that hydrogen fits the bill and will not emit in the IR in certain ranges - absent special circumstances. It is Russian and may not be accurate information, since it does not sound logical to me yet. More on that later. Jones
Re: [Vo]:comment on Violante data as covered by Steve Krivit
On 02/08/2010 11:41 AM, Jones Beene wrote: I have lost the citation from a few weeks ago that claimed that below a threshold of about 10 nm, the expected blackbody frequency is upshifted for nanostructures, in general. If I understand you, and if this is true, then it's a violation of the second law of thermodynamics. It's provable by a direct, simple second law argument that all thermal radiators of the same temperature have the same spectrum, with the caveat that radiation is reduced proportionally at frequencies at which the object is reflective or transparent. Here's the argument: When two objects at the same temperature are placed next to each other in a uniformly hot oven (with everything at the same temperature to start with), and a dichroic filter is placed between them, if one radiates more strongly at the filter's peak reflection frequency than the other, their temperatures will change. (The one which radiates more strongly at the mirror's reflection frequency will see more radiation coming in than the other object, and so will get warmer.)
Re: [Vo]:comment on Violante data as covered by Steve Krivit
2010/2/2 Abd ul-Rahman Lomax a...@lomaxdesign.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel
Re: [Vo]:comment on Violante data as covered by Steve Krivit
Two things to consider: (1) reversing the current *does* dissolve the Pd surface, and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote: 2010/2/2 Abd ul-Rahman Lomax a...@loma xdesi gn.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). I guess you mean they are *found* there, couldn't they be both produced on the surface, only with more kinetic energy in the helium nuclei (alphas) than in the tritium nuclei for some reason, so that the helium is implanted more deeply? I find the idea of two different nuclear reaction sites producing different products a bit unlikely. This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. Maybe it's just not possible, because you can't make large D fluxes collide head-on in the bulk, this can only happen at a significant scale on the surface (desorbing vs incident fluxes). In the bulk, it seems to me the deuterons just push and follow each other down the lattice's concentration gradients, and never really collide hard. Also, if Bose Einstein Condensates are involved, they requires cold bosons for their formation. Head-on collisions may be a plausible mechanism for deuteron kinetic energy removal. Michel On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote: 2010/2/2 Abd ul-Rahman Lomax a...@loma xdesi gn.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote: 2010/2/7 Horace Heffner hheff...@mtaonline.net: Two things to consider: (1) reversing the current *does* dissolve the Pd surface, True, but extremely slowly I believe. A Pd anode is known to dissolve relatively fast in acidic electrolytes such as D2SO4, but I don't think that's what they used. It is doubtful whether they reverted the current long enough to dissolve more than a few atomic layers. I think the experimenters were competent. They knew what they were doing. Using a Faraday constant of 96,485 C/mol, and (conservatively) a valence of 4, n for moles produced, I for current = .2 A, t for time = 1 s, we get: n = I * t / (96,485 C/mol * 4) n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or 3.12x10^17 atoms per second. We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/cm^3 = 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s. If this is correct (highly suspect! 8^), then at a current density of 200 mA/cm^2 we have a thickness of 183 atoms removed per second, or 445 angstroms per second. and (2) previous work has shown that helium production takes place near but below the surface (order of microns), while tritium production tends to take place on or very close to the surface (within a few atomic widths). I guess you mean they are *found* there, couldn't they be both produced on the surface, only with more kinetic energy in the helium nuclei (alphas) than in the tritium nuclei for some reason, so that the helium is implanted more deeply? I find the idea of two different nuclear reaction sites producing different products a bit unlikely. No, most of the 4He reactions occur sub-surface. What do you think produces a volcano? A surface reaction? The typical 4He produced by CF does not have MeV kinetic energy, and is not surface produced. If it were there would be massive alpha counts. There is not sufficient kinetic energy to push alphas that deep into the Pd. This has been a classic problem with CF, converting the process into a bulk effect instead of a surface effect for all practical purposes. Maybe it's just not possible, because you can't make large D fluxes collide head-on Head on collisions, i.e. kinetics, can not possibly account for cold fusion. in the bulk, this can only happen at a significant scale on the surface (desorbing vs incident fluxes). In the bulk, it seems to me the deuterons just push and follow each other down the lattice's concentration gradients, and never really collide hard. Also, if Bose Einstein Condensates are involved, they requires cold bosons for their formation. Head-on collisions may be a plausible mechanism for deuteron kinetic energy removal. This would only be the case if the collisions were almost all totally inelastic. The only way that can happen is if they are fusions. Michel On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote: 2010/2/2 Abd ul-Rahman Lomax a...@loma xdesi gn.com: ... A single SRI experiment has been published that made strong efforts to recover all the helium, and it came up with, as I recall, about 25 MeV. That experiment was discussed in the paper submitted by Hagelstein, McKubre et al to the DOE in 2004: http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf They flushed helium out by simply desorbing and reabsorbing deuterium several times, by varying the cell current, which they reversed in the end to get all the D out. It seems to me that if they actually managed to extract all the helium this way, which their resulting Q value suggests (104±10 % of 23.8 MeV), the reaction can't possibly happen in the bulk. Not even subsurface. It has to happen exactly on the surface, with some (about half) of the produced helium nuclei going slightly subsurface. If the reaction itself was subsurface, surely about half of the produced helium couldn't be recovered without more radical means such as the one you suggested below. ... 2. Recovery of *all* the helium -- except perhaps for minor and unavoidable leakage, which should, of course, be kept as small as possible. What occurs to me is to dissolve the cathode. This seems a good idea. I forget the best acid to use, but I do know that palladium can be dissolved. As I recall, Aqua Regia is the best for Pd. Michel Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:comment on Violante data as covered by Steve Krivit
In reply to Horace Heffner's message of Sun, 7 Feb 2010 05:52:36 -0900: Hi, [snip] No, most of the 4He reactions occur sub-surface. What do you think produces a volcano? A surface reaction? The typical 4He produced by CF does not have MeV kinetic energy, and is not surface produced. If it were there would be massive alpha counts. There is not sufficient kinetic energy to push alphas that deep into the Pd. [snip] ...or alternatively fast alphas are produced, but only so deep in the Pd that they don't make it to the surface. If lattice resonance is a factor, then some depth may be required to build up a strong enough resonance effect that the mechanism can operate. (analogous to adding more dipoles to a TV antenna). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
[Vo]:comment on Violante data as covered by Steve Krivit
I'm correcting this comment as to the Violante data using more accurate numbers as provided by Violante and inferred from that. The substance of this remains the same. http://newenergytimes.com/v2/news/2010/34/345revisions.shtml We have learned, through a better understanding of their paper, that the authors did not perform calorimetry. Rather, they used the helium measurements to back-calculate the excess heat they would have expected from the amount of helium they measured, assuming the hypothesis of a D+D 4He + 23.8 MeV (heat) reaction. That statement appears to be radically incorrect. If it were true, the green dots would be right on the helium actually measured! You have misunderstood the chart, and you are directly contradicting the article. The chart plots, for three experiments, the numbers of helium atoms found, with error bars. This is total helium, and it appears that background helium is included. There are, however, some problems with the presentation. On the one hand, the experiment that shows a green dot on the money, is the noisiest point, it's actually a low excess helium measurement, obscured by plotting total helium including background. I doubt that the intention was obfuscation, though, rather it seems a bit sloppy to me. But it was only a conference paper! You state that they did not perform calorimetry. On the contrary, they describe their calorimetry in the paper, http://newenergytimes.com/v2/library/2005/2005Apicella-SomeResultsAtENEA.pdf, in detail, and they give the data in the text, and I have converted to MeV using the NASA energy calculator at http://heasarc.gsfc.nasa.gov/cgi-bin/Tools/energyconv/energyConv.pl Laser 2: 23.5 kJ, 1.47 x 10^17 MeV Laser 3: 3.4 kJ, 2.12 x 10^16 MeV Laser 4: 30.3 kJ, 1.89 x 10^17 MeV If we expect 24 MeV/He-4, these figures would translate to Laser 2: 0.612 x 10^16 atoms Laser 3: 0.088 x 10^16 atoms Laser 4: 0.787 x 10^16 atoms If background is to be added, 0.55 x 10^16 per the chart (and from Violante direct data), this becomes expected measurement: Laser 2: 1.162 x 10^16 atoms Laser 3: 0.638 x 10^16 atoms Laser 4: 1.337 x 10^16 atoms And these are the green dot positions (read from the chart): Laser 2: 1.20 x 10^16 atoms Laser 3: 0.72 x 10^16 atoms Laser 4: 1.27 x 10^16 atoms It appears that they took the energy, divided it by 24 MeV/He4, and plotted that as the green dots for reference. However, the positions aren't exact, so they have made some approximation or there is some other factor they have not disclosed. Nevertheless, the green dots are *approximately* what they say they are: measured energy converted to expected helium at 24 MeV. (Note that Violante does acknowledge some sloppiness in the plots, but this does not greatly affect the presentation.) For reference, here is the helium data taken from the chart: Laser 2: 0.80 x 10^16 to 0.97 x 10^16 atoms, increase over background: 0.245 - 0.415 x 10^16, midpoint 0.330 Laser 3: 0.68 x 10^16 to 0.79 x 10^16 atoms, increase over background: 0.125 - 0.235 x 10^16, midpoint 0.180 Laser 4: 0.94 x 10^16 to 1.18 x 10^16 atoms, increase over background: 0.385 - 0.625 x 10^16, midpoint 0.505 Numbers reported by Violante in correspondence with Krivit, helium atoms, increase over background: Laser 2: 0.35 x 10^16 Laser 3: 0.10 x 10^16 Laser 4: 0.50 x 10^16 Calculated Q factors from the energy/helium (from my reading of the chart): Laser 2: 35 - 60 MeV, midpoint 45 MeV Laser 3: 9 - 17 MeV, midpoint 12 MeV Laser 4: 30 - 49 MeV, midpoint 37 MeV Laser 3 certainly looks like an outlier. From the better data provided by Violante: Laser 2: 42 MeV Laser 3: 21 MeV Laser 4: 38 MeV I'd have been much happier with statements of the actual measured values, or series of values, but this kind of specific and detailed data is often omitted. The round numbers are very clearly claimed. Then there are the green dots. These are not presentations of raw data, but of the raw energy data (stated explicitly as numbers) interpreted as helium on the hypothesis of 24 MeV/He-4. But there is an unfortunate problem. They do not state how they correlate measured helium with total helium, and they are not clear on whether or not the data in the chart is measured helium including background, the caption implies that it is the increase, but the caption could be interpreted merely to indicate that an increase over background is shown, and, from the calculations above, the figures are for total helium, i.e., background plus increase. However, the variation in the background is not stated. Do the error bars include that? It is quite unfortunate that they did not present the data clearly! They did do calorimetry, they are explicit about that. Those are the measured energy figures given, and those figures were not simply extrapolated from helium measured as you claimed: were it so, the green dots would be meaningless, but they also
Re: [Vo]:comment on Violante data as covered by Steve Krivit
Abd ul-Rahman Lomax wrote: I'm correcting this comment as to the Violante data using more accurate numbers as provided by Violante and inferred from that. Provided where? When? To you in private correspondence, or did you find the data elsewhere? I can poke around and see if I have some unpublished data . . . - Jed
[Vo]:comment on Violante data as covered by Steve Krivit
http://newenergytimes.com/v2/news/2010/34/345revisions.shtml We have learned, through a better understanding of their paper, that the authors did not perform calorimetry. Rather, they used the helium measurements to back-calculate the excess heat they would have expected from the amount of helium they measured, assuming the hypothesis of a D+D 4He + 23.8 MeV (heat) reaction. That statement appears to be radically incorrect. If it were true, the green dots would be right on the helium actually measured! You have misunderstood the chart, and you are directly contradicting the article. The chart plots, for three experiments, the numbers of helium atoms found, with error bars. This is total helium, and it appears that background helium is included. There are, however, some problems with the presentation. On the one hand, the experiment that shows a green dot on the money, is the noisiest point, it's actually a low excess helium measurement, obscured by plotting total helium including background. I doubt that the intention was obfuscation, though, rather it seems a bit sloppy to me. But it was only a conference paper! You state that they did not perform calorimetry. On the contrary, they describe their calorimetry in the paper, http://newenergytimes.com/v2/library/2005/2005Apicella-SomeResultsAtENEA.pdf, in detail, and they give the data in the text, and I have converted to MeV using the NASA energy calculator at http://heasarc.gsfc.nasa.gov/cgi-bin/Tools/energyconv/energyConv.pl Laser 2: 23.5 kJ, 1.47 x 10^17 MeV Laser 3: 3.4 kJ, 2.12 x 10^16 MeV Laser 4: 30.3 kJ, 1.89 x 10^17 MeV If we expect 24 MeV/He-4, these figures would translate to Laser 2: 0.612 x 10^16 atoms Laser 3: 0.088 x 10^16 atoms Laser 4: 0.787 x 10^16 atoms If background is to be added, 0.555 x 10^16 per the chart, this becomes expected measurement: Laser 2: 1.167 x 10^16 atoms Laser 3: 0.643 x 10^16 atoms Laser 4: 1.342 x 10^16 atoms And these are the green dot positions: Laser 2: 1.20 x 10^16 atoms Laser 3: 0.72 x 10^16 atoms Laser 4: 1.27 x 10^16 atoms It appears that they took the energy, divided it by 24 MeV/He4, and plotted that as the green dots for reference. However, the positions aren't exact, so they have made some approximation or there is some other factor they have not disclosed. Nevertheless, the green dots are *approximately* what they say they are: measured energy converted to expected helium at 24 MeV. For reference, here is the helium data taken from the chart: Laser 2: 0.80 x 10^16 to 0.97 x 10^16 atoms, increase over background: 0.245 - 0.415 x 10^16, midpoint 0.330 Laser 3: 0.68 x 10^16 to 0.79 x 10^16 atoms, increase over background: 0.125 - 0.235 x 10^16, midpoint 0.180 Laser 4: 0.94 x 10^16 to 1.18 x 10^16 atoms, increase over background: 0.385 - 0.625 x 10^16, midpoint 0.505 Calculated Q factors from the energy/helium: Laser 2: 35 - 60 MeV, midpoint 45 MeV Laser 3: 9 - 17 MeV, midpoint 12 MeV Laser 4: 30 - 49 MeV, midpoint 37 MeV Laser 3 certainly looks like an outlier. I'd have been much happier with statements of the actual measured values, or series of values, but this kind of specific and detailed data is often omitted. The round numbers are very clearly claimed. Then there are the green dots. These are not presentations of raw data, but of the raw energy data (stated explicitly as numbers) interpreted as helium on the hypothesis of 24 MeV/He-4. But there is an unfortunate problem. They do not state how they correlate measured helium with total helium, and they are not clear on whether or not the data in the chart is measured helium including background, the caption implies that it is the increase, but the caption could be interpreted merely to indicate that an increase over background is shown, and, from the calculations above, the figures are for total helium, i.e., background plus increase. However, the variation in the background is not stated. Do the error bars include that? It is quite unfortunate that they did not present the data clearly! They did do calorimetry, they are explicit about that. Those are the measured energy figures given, and those figures were not simply extrapolated from helium measured as you claimed: were it so, the green dots would be meaningless, but they also would be consistent, i.e., all three experiments would show green dots right on the money. The only experiment that shows that ratio, roughly, is the one with the lowest energy production, and the error bars in the helium measurement would make this not as important as it might seem. In any case, nobody with any sense would look at the series of three experiments and think that it was some kind of definitive confirmation of 24 MeV/He-4. It's one data point that looks like that, that's all, and two data points, less down in the noise, that look like there is missing helium, the same as with about everyone else. Definitely,
