Re: [Vo]:defkalion rumour (PESN)
Von: Alan J Fletcher a...@well.com An: vortex-l@eskimo.com Gesendet: 1:42 Dienstag, 22.Mai 2012 Betreff: [Vo]:defkalion rumour (PESN) http://pesn.com/2012/05/17/9602095_LENR-to-Market_Weekly_May17/ Sterling's Assessment: ...The scuttlebutt is that while the third party test results have been positive, there have been indications of instability and inconsistency between tests, which doesn't speak well for production readiness. ... Alan, this sounds about right to me. I would never expect any LENR-reaction to be a self-regulating process, except for very low-power/COP systems, where it does not matter if there are some bursts of 10x thermal output on a single-digit Watt-level. But this is different on a say 10kW-level. At least it seems to be self-limiting on the catastrophic end, which is a good thing ofcourse. Summarizing the experiences from the Italian and Greek on the one side and the US/Japanese on the other: a) I/G (Rossi/DGT) are quite bold in their claims of commercialable LENR b) U/J are more on the low end, with the possible exception of Brillouin. Nanospire being on the freaky side. Not to be dismissed as nonsensical. The preliminary conclusion I draw from that is, that high COP/high energy requires GOOD control. With respect to that it seems only logical, that Rossi assembles 1MW as 50x20kW units. ( despite of that, as the rumor goes, people had to run away from a possible out-of-control Rossi-1MW reactor. Sorry to say, but this, plus some common sense is what we have on the issue, and if it aligns, we should consider it.) This is the reason why I lately tend to prefer a spark-plug based excitation over other possibilities (eg magnetrons, classical 13MHz RF) : It allows to cover a wide parameter-space (spark-energy, pulse-shape, repetition-rate) by simple means. It is one thing to have a small fire warming your teapot, another to eventually have a vast wildfire, burning your habitat down. This, I am afraid, eg Rossi , by way of his mindset, never considered systematically. His natural alliant is therefore the US-military, which is used to taking all sorts of freaky risk-taking, where the Rossi variant is only a minor annoyance . All the best Guenther
[Vo]:defkalion rumour (PESN)
I believe a better analogy is trying to warm your Cofee pot uniformly with a 3700 C capable hydrogen blowtorch without melting it. And just to aggravate things the metal pot tends toward nuclear meltown if Rydberg conditions are just right, which happens to be right where you want to operate... The advantage I see in using tungsten spark plugs is the higher melting point of tungsten at 3400 C might have you buying less spark plugs. Langmuir used tungsten electrodes On Tuesday, May 22, 2012, Guenter Wildgruber wrote: -- *Von:* Alan J Fletcher a...@well.com *An:* vortex-l@eskimo.com *Gesendet:* 1:42 Dienstag, 22.Mai 2012 *Betreff:* [Vo]:defkalion rumour (PESN) http://pesn.com/2012/05/17/9602095_LENR-to-Market_Weekly_May17/ http://pesn.com/2012/05/17/9602095_LENR-to-Market_Weekly_May17/ *Sterling's Assessment:* ...The scuttlebutt is that while the third party test results have been positive, there have been indications of instability and inconsistency between tests, which doesn't speak well for production readiness http://pesn.com/2012/05/17/9602095_LENR-to-Market_Weekly_May17/ Alan, this sounds about right to me. I would never expect any LENR-reaction to be a self-regulating process, except for very low-power/COP systems, where it does not matter if there are some bursts of 10x thermal output on a single-digit Watt-level. But this is different on a say 10kW-level. At least it seems to be self-limiting on the catastrophic end, which is a good thing ofcourse. Summarizing the experiences from the Italian and Greek on the one side and the US/Japanese on the other: a) I/G (Rossi/DGT) are quite bold in their claims of commercialable LENR b) U/J are more on the low end, with the possible exception of Brillouin. Nanospire being on the freaky side. Not to be dismissed as nonsensical. The preliminary conclusion I draw from that is, that high COP/high energy requires GOOD control. With respect to that it seems only logical, that Rossi assembles 1MW as 50x20kW units. ( despite of that, as the rumor goes, people had to run away from a possible out-of-control Rossi-1MW reactor. Sorry to say, but this, plus some common sense is what we have on the issue, and if it aligns, we should consider it.) This is the reason why I lately tend to prefer a spark-plug based excitation over other possibilities (eg magnetrons, classical 13MHz RF) : It allows to cover a wide parameter-space (spark-energy, pulse-shape, repetition-rate) by simple means. It is one thing to have a small fire warming your teapot, another to eventually have a vast wildfire, burning your habitat down. This, I am afraid, eg Rossi , by way of his mindset, never considered systematically. His natural alliant is therefore the US-military, which is used to taking all sorts of freaky risk-taking, where the Rossi variant is only a minor annoyance . All the best Guenther
[Vo]:Press release Blacklightpower
See press release and validation reports from Blacklightpower: http://dev.blacklightpower.com/press/052212-2/ http://www.blacklightpower.com/technology/validation-reports/ Peter
[Vo]:BLP News Release re CIHT
http://www.wkow.com/story/18579657/electricity-generated-from-water-blacklight-power-announces-validation-of-its-scientific-breakthrough-in-energy-production Mark Goldes Co-founder, Chava Energy CEO, Aesop Institute 301A North Main Street Sebastopol, CA 95472 www.chavaenergy.com www.aesopinstitute.org 707 861-9070 707 497-3551 fax
Re: [Vo]:Any SLIders out there? I am one.
On Fri, May 18, 2012 at 10:19 PM, William Beaty bi...@eskimo.com wrote: On Fri, 18 May 2012, David Jonsson wrote: What is happening when you turn off the LEDs? Where is that described? It looks like a possible explanation. Three-dollar e-field detectors, see the project page: http://www.amasci.com/emotor/**chargdet.htmlhttp://www.amasci.com/emotor/chargdet.html But if streetlights respond to DC fields, then nearby cars and distant thunderstorms would have enormous effect. Thanks. I will try to build one. Will this transistor do? http://www.newark.com/nte-electronics/nte451/transistor-jfet-n-channel-4ma-i/dp/29C4598 An idea is to build an array of these and measure with a cheap microcontroller. David
Re: [Vo]:Does Taylor diffusion affects heat?
I have been thinking for a while and I think it should because heat conduction is also described as heat diffusion. Can someone please try a simple experiment to check this? Rotate anything, preferably a gas, and check if the radial heat conductivity decreases. David On Sun, May 6, 2012 at 4:56 PM, David Jonsson davidjonssonswe...@gmail.comwrote: Taylor diffusion means that diffusion is affected by Coriolis forces and thus moves in circles and effectively reduces radial diffusion in rotation fluids. Do not mistake this for Taylor dispersion which is an effect which increases diffusion. Since heat flow is a kind of diffused heat I wonder if it also is affected by Taylor diffusion. The heat motion is definitely affected by Coriolis forces. How could this be analyzed? It seems to have some strange consequences in regard to entropy. It seems like entropy doesn't increase as much when rotating but that seems also versy counterintuitive and it seems like a too easy trick to lower increase of entropy. Common reasoning implies that the process is requiring energy which is usually the case to lower entropy increase. Help me solve this. I have always found entropy to be a strange and weak concept. Or maybe the total entropy changes Taylor diffusion and Taylor dispersion balances each other? David David Jonsson, Sweden, phone callto:+46703000370
Re: [Vo]:Press release Blacklightpower
Impressive, however small power, small energy, results some 5 months old- the phase of scale up, intensification, long term functionality has only started.. Science is great but engineering put's the generators on the market. Let's hope the best Peter On Tue, May 22, 2012 at 4:30 PM, P.J van Noorden pjvannoor...@caiway.nlwrote: See press release and validation reports from Blacklightpower: http://dev.blacklightpower.**com/press/052212-2/http://dev.blacklightpower.com/press/052212-2/ http://www.blacklightpower.**com/technology/validation-**reports/http://www.blacklightpower.com/technology/validation-reports/ Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:Press release Blacklightpower
Haynes 242 alloy: http://www.haynesintl.com/pdf/h3142.pdf T
Re: [Vo]:Press release Blacklightpower
Indeed! So small, that it won't remove skepticism from skeptics. They will say it's an error in measurment. And it is extremely ambitious to talk about 1KJ. 6 orders of magnitude of improvement within a few months. 2012/5/22 Peter Gluck peter.gl...@gmail.com Impressive, however small power, small energy, results some 5 months old- the phase of scale up, intensification, long term functionality has only started.. Science is great but engineering put's the generators on the market. Let's hope the best Peter On Tue, May 22, 2012 at 4:30 PM, P.J van Noorden pjvannoor...@caiway.nlwrote: See press release and validation reports from Blacklightpower: http://dev.blacklightpower.**com/press/052212-2/http://dev.blacklightpower.com/press/052212-2/ http://www.blacklightpower.**com/technology/validation-**reports/http://www.blacklightpower.com/technology/validation-reports/ Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com -- Daniel Rocha - RJ danieldi...@gmail.com
RE: [Vo]:Nickel-hydrogen nuclear ash
Eric - perhaps the original post should have been phrased as “zero believable evidence”… instead of zero evidence. The paper does constitute putative “evidence” after all – actually rather convincing if it could be taken at face value. Romodanov is a mystery. If what he was seeing and reporting was accurate (tritium from hydrogen in very significant quantities) – it should have led to a lucrative method for producing an extremely valuable isotope, especially to some countries. Aside from the science involved, this paper has dollar signs (actually Rials) written all over it. Yet the work apparently fizzled after 2003. Also, the paper is almost “too convincing” to be accurate given what Claytor has published (using deuterium). In the ensuing years, there has been no outside replication of Romodanov, or progress which shows up in the public record. Plus, it is no secret that there are thousands of severely underpaid, top-level scientists in Russia who are desperate to move to the West, under almost any pretense … they have little way to show off their wares other than slick papers, especially if they come with an implied threat. In short, a cynic might opine that this is more a feeler for continuing employment in a more hospitable locale, as it is bona fide science. But it would be instructive to know more of the story. From: Eric Walker Tritium is radioactive, so the evidence of radioactivity in the ash of the Ni-H reaction is nonzero…. Romodanov et al., Nuclear reactions in condensed media and X-ray, Seventh International Conference on Cold Fusion, 1998.
Re: [Vo]:Press release Blacklightpower
Peter, I am somehow tired, as we probably all are, of all those annnouncements of 'breakthroughs' . Randall M. probably sees the 'scene' moving away from him and feels challened to announce his n'th breakthrough. Count me unimpressed. Albeit I consider the LENR phenomenon real, quite a lot of Quacks need a decent dose of psychotherapy, it seems. Sorry to say that. After poking into Mill's 2000+ pages theory of everything, I am impressed by his boldness, but not what he has to present. Von: Peter Gluck peter.gl...@gmail.com An: vortex-l@eskimo.com Gesendet: 16:12 Dienstag, 22.Mai 2012 Betreff: Re: [Vo]:Press release Blacklightpower Impressive, however small power, small energy, results some 5 months old- the phase of scale up, intensification, long term functionality has only started.. Science is great but engineering put's the generators on the market. Let's hope the best Peter On Tue, May 22, 2012 at 4:30 PM, P.J van Noorden pjvannoor...@caiway.nl wrote: See press release and validation reports from Blacklightpower: http://dev.blacklightpower.com/press/052212-2/ http://www.blacklightpower.com/technology/validation-reports/ Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:Press release Blacklightpower
I have seen a lot of water vapor, in my bathtub, in my sink, in fog, emitted from a boiler, at high pressure. There was never any excess energy. Yes, I know, Blacklight discovered the hydrino and they have all of them. Frank
Re: [Vo]:Press release Blacklightpower
Von: Daniel Rocha danieldi...@gmail.com An: vortex-l@eskimo.com Gesendet: 16:40 Dienstag, 22.Mai 2012 Betreff: Re: [Vo]:Press release Blacklightpower Indeed! So small, that it won't remove skepticism from skeptics. They will say it's an error in measurment. And it is extremely ambitious to talk about 1KJ. 6 orders of magnitude of improvement within a few months. ## I am just thinking of Zuckerbererg investing a meager 100million into Blacklight Power. Is it 'real' evidence to invest 'virtual' money into 'virtual' evidence? Count me confused. Guenther
RE: [Vo]:Press release Blacklightpower
-Original Message- From: Terry Blanton Haynes 242 alloy: http://www.haynesintl.com/pdf/h3142.pdf Hmmm ... Good eye, Terry. It is probably not coincidental that this is one of the highest available alloys in molybdenum content. Moly is possibly the best fit catalyst in the periodic table, but the unalloyed metal is extremely prone to corrosion. Anyone who reads the BLP experiments with an eye towards best fit can see the 'usual suspects' in Column one and two of the Table. The criterion is to be as close as possible at 27.2 eV - using the confusing Mills rules of engagement, and eliminating 3-body reactions. Molybdenum ions are a good fit at 27.13 which is off by only .07 eV and this for the Mo(2+) ion; plus the miss with moly is on the low side. It seems possible, using the energy hole analogy, that you would want to err on the low side of 27.2 eV if you cannot hit it exactly - since the approaching hydrogen atom carries momentum. Jones attachment: winmail.dat
Re: [Vo]:Nickel-hydrogen nuclear ash
Jones, Not valuable. No market. Who would buy it? What would you use it for? Half-life of tritium (hydrogen-3) is 12.3 yr. Warm Regards, Reliable, a thinking person Jones Beene wrote: Eric - perhaps the original post should have been phrased as “zero believable evidence”… instead of zero evidence. The paper does constitute putative “evidence” after all – actually rather convincing if it could be taken at face value. Romodanov is a mystery. If what he was seeing and reporting was accurate (tritium from hydrogen in very significant quantities) – it should have led to a lucrative method for producing an extremely valuable isotope, especially to some countries. Aside from the science involved, this paper has dollar signs (actually Rials) written all over it. Yet the work apparently fizzled after 2003. Also, the paper is almost “too convincing” to be accurate given what Claytor has published (using deuterium). In the ensuing years, there has been no outside replication of Romodanov, or progress which shows up in the public record. Plus, it is no secret that there are thousands of severely underpaid, top-level scientists in Russia who are desperate to move to the West, under almost any pretense … they have little way to show off their wares other than slick papers, especially if they come with an implied threat. In short, a cynic might opine that this is more a feeler for continuing employment in a more hospitable locale, as it is bona fide science. But it would be instructive to know more of the story. *From:* Eric Walker Tritium is radioactive, so the evidence of radioactivity in the ash of the Ni-H reaction is nonzero…. Romodanov et al., Nuclear reactions in condensed media and X-ray, Seventh International Conference on Cold Fusion, 1998.
Re: [Vo]:WAY OFF TOPIC Mistaken notions about human populations and longevity
--On Monday, May 21, 2012 5:57 PM -0400 Jed Rothwell jedrothw...@gmail.com wrote: No doubt that is the biological root of the obesity problem. That is why fat people exist. I doubt there are any obese chimpanzees in the wild. Chimpanzees do not have fire and so do not cook their food must devote most of their day to chewing digesting a lot of raw food. Mastering fire cooking food has allowed humans to quickly fuel our large brains using a short digestive system. Ron
Re: [Vo]:WAY OFF TOPIC Mistaken notions about human populations and longevity
Ron Wormus prot...@frii.com wrote: Chimpanzees do not have fire and so do not cook their food must devote most of their day to chewing digesting a lot of raw food. Mastering fire cooking food has allowed humans to quickly fuel our large brains using a short digestive system. True! And we are so dependent on cooking, we could not survive without it. It has altered our very anatomy, including the teeth and stomach. Here is an excellent little book about that: Catching Fire: How Cooking Made Us Human http://www.amazon.com/Catching-Fire-Cooking-Human-ebook/dp/B0028P9BE6/ref=sr_1_1 Some interesting points from that book: Most animals, such as rats and chimps, prefer cooked food. And it tends to make them obese. Chimps that live in Georgia at the Yerkes center have no trouble making and controlling fires. They do not burn themselves or let the fire go out. They use butane lighters to start the fire, but I expect they could use a primitive technique such as rubbing sticks together. So they have the intelligence. They might have discovered fire long ago. If they had discovered it millions of years ago they would not be chimps in their present form, would they? - Jed
Re: [Vo]:Nickel-hydrogen nuclear ash
Jones Beene jone...@pacbell.net wrote: The overage which is in play in this hypothesis is the mystery energy source for Ni-H reactions, whether they be from Mills, Rossi, DGT, Piantelli, Celani, or Thermacore. It is technically nuclear energy, since it comes from a nucleus - but it does not result in rearrangement of the proton nor a new element. I see. Please do not tell Steve Krivit about this. - Jed
[Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I have been researching the cold fusion reaction that is suggested by Rossi and Focardi in their recent paper http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and have a couple of questions. The authors suggest that 3.41 MeV of energy is released by the fusion of a proton with a nickel 58 nucleus into copper 59. I can obtain this value if I calculate the mass difference between a copper 59 atom and a nickel 58 atom plus the mass of the proton and the mass of the extra electron. So far their calculations are in line with mine. The problem arises when I consider the amount of energy required to overcome the coulomb barrier in order to activate the fusion. The two authors seem to overlook this entirely when they calculate the energy available from their proposed reaction. The chart on page 5 of their paper shows that 3.41 MeV is released at the conclusion to the reaction but no allowance is given to the energy needed to initiate it. They do mention the activation energy in their theoretical interpretation on page 7. In this section they calculate that it takes 5.6 MeV to overcome the barrier. The authors use assumed values for the closeness required and thus energy barrier in their example. With these two numbers available I make the assumption that there is a net energy requirement of 5.6 MeV – 3.41 MeV or 2.19 MeV for the fusion. Is there a reason that my calculation is in error? Does the 3.41 MeV have hidden within it the activation energy? I can see no good reason to suspect that this is the case since it would be possible for a device to send high speed protons into a target made of copper. The copper would then shed the 3.41 MeV by some means and that would obviously not repay the debt. Of course I understand that the following beta plus decay would release an additional significant amount of energy as the copper transforms into nickel 59. I calculate this energy as 5.8 MeV when the released positron is annihilated. This value matches that of the two authors which I assume is correct. A recap of the question is: Is the fusion of nickel 58 with a proton and electron into copper 59 an endothermic reaction? Dave
Re: [Vo]:Nickel-hydrogen nuclear ash
I wrote: Now I know how people felt when isotopes were discovered. I meant that isotopes came as a surprise, and people initially questioned the experimental results rather than believe there variations in the weight of an element. It is an interesting episode in the history of science. I read about it decades ago. They were expecting to find that atomic weights are exact integral values starting with hydrogen (1). They got the wrong answers. Quite wrong, in some cases, such as Al, 26.982. As I recall they kept thinking: when instruments improve the results will get better and yield exactly 26.000. This is like the skeptical assertion that as calorimeters improve, the cold fusion effect will go away. The discovery of the neutron cleared up the mystery, but apparently, as mass measurements improve, they have revealed layer of variation below that. More complexity. - Jed
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Your calculation does not take into account the fact that the activation energy barrier releases the energy added to overcome it during the reaction. In this case once coulomb repulsion is overcome, the energy is added back to the system by attractive nuclear force. The 3.41MeV is the change in mass energy balance after the reaction, what happens in between is not important to the calculation. Analogously, If a person descends from the 3rd to second floor of a building, they are just as close to the ground as if they climb from the 3rd to the 53rd floor before climbing back down the the second floor to end their journey. This is the nature of all activation energy barriers. To: vortex-l@eskimo.com Subject: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 13:35:41 -0400 I have been researching the cold fusion reaction that is suggested by Rossi and Focardi in their recent paper http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and have a couple of questions. The authors suggest that 3.41 MeV of energy is released by the fusion of a proton with a nickel 58 nucleus into copper 59. I can obtain this value if I calculate the mass difference between a copper 59 atom and a nickel 58 atom plus the mass of the proton and the mass of the extra electron. So far their calculations are in line with mine. The problem arises when I consider the amount of energy required to overcome the coulomb barrier in order to activate the fusion. The two authors seem to overlook this entirely when they calculate the energy available from their proposed reaction. The chart on page 5 of their paper shows that 3.41 MeV is released at the conclusion to the reaction but no allowance is given to the energy needed to initiate it. They do mention the activation energy in their theoretical interpretation on page 7. In this section they calculate that it takes 5.6 MeV to overcome the barrier. The authors use assumed values for the closeness required and thus energy barrier in their example. With these two numbers available I make the assumption that there is a net energy requirement of 5.6 MeV – 3.41 MeV or 2.19 MeV for the fusion. Is there a reason that my calculation is in error? Does the 3.41 MeV have hidden within it the activation energy? I can see no good reason to suspect that this is the case since it would be possible for a device to send high speed protons into a target made of copper. The copper would then shed the 3.41 MeV by some means and that would obviously not repay the debt. Of course I understand that the following beta plus decay would release an additional significant amount of energy as the copper transforms into nickel 59. I calculate this energy as 5.8 MeV when the released positron is annihilated. This value matches that of the two authors which I assume is correct. A recap of the question is: Is the fusion of nickel 58 with a proton and electron into copper 59 an endothermic reaction? Dave
Re: [Vo]:Spark plugs... thoughts and how-to?
Von: Jojo Jaro jth...@hotmail.com An: Vortex Vortex-l@eskimo.com Gesendet: 7:18 Montag, 21.Mai 2012 Betreff: Re: [Vo]:Spark plugs... thoughts and how-to? Guenter, I believe your bickering is misdirected. ... Did DGT simply make a mistake and accidentally machined an extra hole on both end plates and had to plug it with a spark plug? BTW, machining a spark plug thread is more difficult than machining an ordinary pipe fitting thread, ... Spark Plug CAN deliver High Voltages into the reactor, but NOT High Current, unless it is a highly short-lived transient current spike. Jojo, Sorry if this 'kidding' term comes over as 'bickering'. It was not meant to be such. Just to be funny of sorts. Well. Did not work out, it seems. You hit me, an I feel punished. OK? Actually, in my hypothetical design there are TWO electrodes. One at about 20kV (the HV) , one at 100V (the mesh), but this does not make a difference. For purely practical matters it makes sense to me to use a high-temperature, pressure-tight feedthrough for both. (I doubt whether the DGT design has anything to do with that. My efforts in reverse-engineering greek designs are very limited) Concerning current, I calculate this as follows: Max spark-energy is 300mJ for a duration of 100usec @20kV Which gives approx 7A over that interval, being equivalent to 3kOhm, absorbing all that. I estimate the spark-electrode resistance to be in the mOhm...Ohm range, so it is negligible. (see eg the STM ignition coil driver VBG15NB22T5SP-E or the Fairchild ISL9V3040D3S ecospark) Remember, the secondary (HV-coil) (AC) resistance is in the order of 5-10kOhm, considering Skin-effect and other factors. I did not make a publishable simulation right now, -which You seem to object beforehand- Am unsure whether it makes sense, above common-sense assessments, ie, that the voltage heavily oscillates between +/- kV levels, which is meaningless in conventional ignition, but NOT in a LENR environment, where it has some peculiar effects, which are not lethal, but diminish the efficiency of the whole setup considerably. I attach another pdf to clarify the 'ignition' issue. 'ideal case'. best regards Guenther Visio-my_LENR_ignition-20120522.pdf Description: Adobe PDF document
[Vo]:Another web site about Rossi
See: http://rossifocardifusion.com/
Re: [Vo]:Why Nicola Tesla was the greatest Geek who ever lived.
Steven, looks like Forbes is reading Vortex also. http://www.forbes.com/sites/alexknapp/2012/05/18/nikola-tesla-wasnt-god-and-thomas-edison-wasnt-the-devil/ T
RE: EXTERNAL: RE: [Vo]:Nickel-hydrogen nuclear ash
Jones, I've always felt there is a relationship between spontaneous emission, pyrophoricity and radioactive materials in a Puthoff atomic model kind of way and that stability is really just a matter of time scales - Forming macro geometries out of Casimir material allows us to modify what Puthoff refers to as the pressure and we can selectively expose gas atoms to higher or lower pressure. I think hydrinos are just regular old hydrogen atoms from their own local perspective and that they appear to crowd into impossibly small pockets from our perspective because of the effect this pressure has on space-time. It might explain the skewed spectroscopy as well because the light is traveling out of the cavities in a Pythagorean relationship with respect to the space time outside the cavities. Almost afraid to hit send for all this thin ice, Fran _ From: Jones Beene [mailto:jone...@pacbell.net] Sent: Monday, May 21, 2012 8:10 PM To: vortex-l@eskimo.com Subject: EXTERNAL: RE: [Vo]:Nickel-hydrogen nuclear ash From: Jed Rothwell Jones Beene wrote: IOW the mass of hydrogen is not a quantum value, and there is no rationale that predicts it will be a single value instead of a range. In fact, mass determination of hydrogen, from various labs in various countries varies all over the place. You are saying the mass varies, and this is not an instrument artifact? As Jon Stewart says, I didn't see that coming. Now I know how people felt when isotopes were discovered. The accepted value for mass of a proton is 938.272013 MeV, but that value (in my hypothesis) is an average of many protons in many situations. Over the years, measurements made in different countries and a different times with different instruments have returned different values (close but different). Some of that is because there can be variation in the feed stock, aside from the instrumentation. In short, hydrogen from natural gas may vary slightly in mass compared to hydrogen from electrolysis of rainwater. This might be the result of the bedrock from which the methane was stored for millions of years having Uranium content which pumped up the non-quark bosons (gluons pions etc). The major hypothesis detail is that the more than half of the proton mass is not quantized, and some of that can be extracted by Coulomb repulsion at close range in IRH (inverted Rydberg hydrogen which is another name for dense hydrogen) - resulting in very fast protons, but only so long there is a usable overage in mass which does not allow quark dispersal. The hypothesis is falsifiable. In short - the average mass can vary to the extent of a fractional percent as either overage or deficit in various sources of hydrogen (say from 937 MeV to 940 MeV). At best, the known value of mass becomes what is really an average based on whatever the most advanced current measurement technique is being used - before recalibration. Everyone recalibrates, as an expedient and so as not to be embarrassed by their instruments. The overage which is in play in this hypothesis is the mystery energy source for Ni-H reactions, whether they be from Mills, Rossi, DGT, Piantelli, Celani, or Thermacore. It is technically nuclear energy, since it comes from a nucleus - but it does not result in rearrangement of the proton nor a new element. Jones
Re: [Vo]:Spark plugs... thoughts and how-to?
What is a spark-plug anyway? Designing an apparatus means knowing as good as possible the essential components and their interaction. If You go to the junkyard and collect your arbitrary ignition electronics, ignition-coil, spark-plug, you are probably in for a surprise. Moderately modern components are evolving and are quite sophisticated. See eg here: http://en.wikipedia.org/wiki/Spark_plug -- Central electrode. And here: http://www.madehow.com/Volume-1/Spark-Plug.html Same with ignition coils and assotiated electronics. For LENR-related purposes it may suffice to say that typical spark-systems develop a bipolar damped voltage at the hot electrode in the range od 10...30msec, where motorbike-systems go up to 20 000 rpm, ie 300Hz, with spark energies of 0.0xJ to 0.3J per spark. The undesirable effect in the LENR-context is the bipolarity of the voltage/ionization direction, but is not killing the effect. It just has to be considered. As said, it diminishes the ionization efficiency by some 50%, which is an annoyance and mainly affects the stability of the mesh-grid-potential, which can be stabilized with a capacitor of sufficient withstand-voltage and capacity, ie several kV several nF, as a first approximation. Note: I am trying to bring some hard parameters in here, and not Rossi said this, DGK done that, Chan exploded a tree, which does a disservice to the effect. Kill the concept, as I try to expose, and You do me a service. Spares me a lot of time. Thank You. Guenther
Re: [Vo]:Spark plugs... thoughts and how-to?
Interesting!!! I take it from your use of a Fairchild IGBT that you are planning to implement a conventional Kettering Ignition Design rather than a Capacitive Discharge Ignition Design. (Since an SCR, rather than an IGBT would be more appropriate in a CDI design.) Yes, there would be severe +/- KV oscillations in a Kettering design. I implemented a CDI design with a heavy diode sink to quickly kill any reverse voltage oscillations. On my design, the resultant LC circuit will have an overdamped harmonic frequency of 11Khz, which would drive havoc inside the reactor. Hence I implement a diode sink to quickly kill the KV oscillations. Another thing, with a Kettering design, you are pretty much limited to a low firing frequency of at most maybe 250 hz. I see that you plan to fire your sparks at 10 hz. At those levels, you will not be able to deliver enough energy into the reaction chamber to initiate any meaningful reaction. It seems you've already calculated your power input to be in the order of 5.5 watts. Seems too low to do anything with it. On my design, I am delivering 260mJ per spark or up to 211 watts at 1000 hz. And my design can be fired up to a practical rate of 10Khz. At this level, I can deliver several KW into my reactor if need be, although I do not plan to go that high. Let me know how your reactor design works out, although I think I can predict severe warping of your electrode and mesh screen. I predict this would be due to severe thermal stesses, electrostatic and electromagnetic attractive forces as well as the sheer turbulence inside that reactor. Once the mesh screen is warped, it would be closer to the other electrode. When that happens, the sparks will follow that closer path all the time. In other words, it will ionize and spark on the same location all the time. That would definitely be detrimental to your nickel nanopowders as that will cook it in short order. In my design, the reactor wall is the anode and I used a larger diameter cathode electrode to prevent warping of the electrodes. Making the reactor wall the anode should repel H+ positive ions into the middle of the reactor chamber where they can hopefully form Rydberg Matter in abundance. Good luck and apologies for biting your head off unnecessarily. Jojo - Original Message - From: Guenter Wildgruber To: vortex-l@eskimo.com Sent: Wednesday, May 23, 2012 1:50 AM Subject: Re: [Vo]:Spark plugs... thoughts and how-to? -- Von: Jojo Jaro jth...@hotmail.com An: Vortex Vortex-l@eskimo.com Gesendet: 7:18 Montag, 21.Mai 2012 Betreff: Re: [Vo]:Spark plugs... thoughts and how-to? Guenter, I believe your bickering is misdirected. ... Did DGT simply make a mistake and accidentally machined an extra hole on both end plates and had to plug it with a spark plug? BTW, machining a spark plug thread is more difficult than machining an ordinary pipe fitting thread, ... Spark Plug CAN deliver High Voltages into the reactor, but NOT High Current, unless it is a highly short-lived transient current spike. Jojo, Sorry if this 'kidding' term comes over as 'bickering'. It was not meant to be such. Just to be funny of sorts. Well. Did not work out, it seems. You hit me, an I feel punished. OK? Actually, in my hypothetical design there are TWO electrodes. One at about 20kV (the HV) , one at 100V (the mesh), but this does not make a difference. For purely practical matters it makes sense to me to use a high-temperature, pressure-tight feedthrough for both. (I doubt whether the DGT design has anything to do with that. My efforts in reverse-engineering greek designs are very limited) Concerning current, I calculate this as follows: Max spark-energy is 300mJ for a duration of 100usec @20kV Which gives approx 7A over that interval, being equivalent to 3kOhm, absorbing all that. I estimate the spark-electrode resistance to be in the mOhm...Ohm range, so it is negligible. (see eg the STM ignition coil driver VBG15NB22T5SP-E or the Fairchild ISL9V3040D3S ecospark) Remember, the secondary (HV-coil) (AC) resistance is in the order of 5-10kOhm, considering Skin-effect and other factors. I did not make a publishable simulation right now, -which You seem to object beforehand- Am unsure whether it makes sense, above common-sense assessments, ie, that the voltage heavily oscillates between +/- kV levels, which is meaningless in conventional ignition, but NOT in a LENR environment, where it has some peculiar effects, which are not lethal, but diminish the efficiency of the whole setup considerably. I attach another pdf to clarify the 'ignition' issue. 'ideal case'. best regards Guenther
Re: [Vo]:Nickel-hydrogen nuclear ash : Rossi -- changed
Rossi --- see #4 Carlo Salvi May 21st, 2012 at 12:59 PM Dear Mr Rossi About the new 600° celsius e-cat: 1)Does it start with the same time of the the first ecat or is it more faster to began to work ? 2)Does it uses the same quantity of Ni/H ? 3)Do you think it still can work for 6 month with one recharge or the new version burns NI/H faster ? 4)Are the ashes still composed with 30% copper or somethings changed ? 5)Do you think this new product will require a different certification from the old version ? 6) when the new product will be released, this will replace the first version or do you think youll sell both products ? Thank you very much, and good luck Mr Rossi. Carlo Salvi Andrea Rossi May 22nd, 2012 at 2:22 AM Dear Carlo Salvi: 1- faster 2- less 3- yes 4- changed 5- yes 6- no: they have different purposes. Warm Regards, A.R.
Re: [Vo]:Another web site about Rossi
From Jed, See: http://rossifocardifusion.com/ These guys definitely have a lot of time on their hands! Lots of snazzy graphics! I hope it isn't all just smoke and mirrors. So, Jed. What do you think? Did these guys use WordPress, Drupal, or some other concoction? Inquiring minds want to know. As for me, my current plans are to completely overhaul my orionworks web site using XARA Web Designer Premium. See: http://www.xara.com/us/products/webdesigner/ Being spatially oriented, I'm much more comfortable using a true WYSIWYG format. This software fits the bill for my particular needs. It wouldn't work for you however. In hindsight I sometimes think Drupal might have turned out to have been a better fit for your needs. A higher learning curve in the beginning however. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Thanks for the rapid response! I think that I have a mental block regarding this issue. Perhaps I am looking at the fusion process from the wrong angle. Help me understand where my error is as I do not follow your example below. Let's try a mental experiment. I can see that it would be possible to build a machine that accelerates protons until they reach the required threshold level of 5.6 MeV. It would take at least this much energy to generate the projectile proton. Now, when the bullet proton comes close to the nucleus the fusion takes place. At this point in time it has lost all of its kinetic energy and is absorbed into the nucleus. I assume that very soon thereafter the 3.41 MeV is released as a gamma or some other form of radiation. The tables of nuclides informs me that the copper 59 has a half life of 81.5 seconds before the beta plus decay so the energy associated with that process must be in place but does not exit until later. Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize. Dave -Original Message- From: Finlay MacNab finlaymac...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 1:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Your calculation does not take into account the fact that the activation energy barrier releases the energy added to overcome it during the reaction. In this case once coulomb repulsion is overcome, the energy is added back to the system by attractive nuclear force. The 3.41MeV is the change in mass energy balance after the reaction, what happens in between is not important to the calculation. Analogously, If a person descends from the 3rd to second floor of a building, they are just as close to the ground as if they climb from the 3rd to the 53rd floor before climbing back down the the second floor to end their journey. This is the nature of all activation energy barriers. To: vortex-l@eskimo.com Subject: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 13:35:41 -0400 I have been researching the cold fusion reaction that is suggested by Rossi and Focardi in their recent paper http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and have a couple of questions. The authors suggest that 3.41 MeV of energy is released by the fusion of a proton with a nickel 58 nucleus into copper 59. I can obtain this value if I calculate the mass difference between a copper 59 atom and a nickel 58 atom plus the mass of the proton and the mass of the extra electron. So far their calculations are in line with mine. The problem arises when I consider the amount of energy required to overcome the coulomb barrier in order to activate the fusion. The two authors seem to overlook this entirely when they calculate the energy available from their proposed reaction. The chart on page 5 of their paper shows that 3.41 MeV is released at the conclusion to the reaction but no allowance is given to the energy needed to initiate it. They do mention the activation energy in their theoretical interpretation on page 7. In this section they calculate that it takes 5.6 MeV to overcome the barrier. The authors use assumed values for the closeness required and thus energy barrier in their example. With these two numbers available I make the assumption that there is a net energy requirement of 5.6 MeV – 3.41 MeV or 2.19 MeV for the fusion. Is there a reason that my calculation is in error? Does the 3.41 MeV have hidden within it the activation energy? I can see no good reason to suspect that this is the case since it would be possible for a device to send high speed protons into a target made of copper. The copper would then shed the 3.41 MeV by some means and that would obviously not repay the debt. Of course I understand that the following beta plus decay would release an additional significant amount of energy as the copper transforms into nickel 59. I calculate this energy as 5.8 MeV when the released positron is annihilated. This value matches that of the two authors which I assume is correct. A recap of the question is: Is the fusion of nickel 58 with a proton and electron into copper 59 an endothermic reaction? Dave
Re: [Vo]:Press release Blacklightpower
Frank You have to bathe in Ni nano powder to dilate the space time which the vapor then diffuses into becoming hydrino or inverse Rydberg from our perspective - I would posit gas loading is actually based on temporal distortion where many more atoms can perform Lorentzian contraction to fit into an impossibly small space, radiating huge numbers of spontaneous emissions from our perspective which are also frequency skewed due to their Pythagorean relationship to our frame. Instead of accelerating the object to near C and causing time to slow [ether velocity vs object velocity] you instead use Casimir suppression to segregate the ether velocity into faster and slower pockets through which you selectively diffuse your hydrogen. Fran From: fznidar...@aol.com [mailto:fznidar...@aol.com] Sent: Tuesday, May 22, 2012 10:51 AM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Press release Blacklightpower I have seen a lot of water vapor, in my bathtub, in my sink, in fog, emitted from a boiler, at high pressure. There was never any excess energy. Yes, I know, Blacklight discovered the hydrino and they have all of them. Frank
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
Re: [Vo]:Another web site about Rossi
OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: So, Jed. What do you think? Did these guys use WordPress, Drupal, or some other concoction? To reveal this with Google Chrome you do a right-click and select view page source. And voila, it says: meta name=generator content=WordPress 3.1 / !-- leave this for stats please -- As for me, my current plans are to completely overhaul my orionworks web site using XARA Web Designer Premium. See: http://www.xara.com/us/products/webdesigner/ That costs $99. Probably worth it. It looks similar to WordPress. They use the same jargon for components such as template. - Jed
Re: [Vo]:Press release Blacklightpower
At 07:40 AM 5/22/2012, Daniel Rocha wrote: And it is extremely ambitious to talk about 1KJ. 6 orders of magnitude of improvement within a few months. The Weinberg report : http://www.blacklightpower.com/wp-content/uploads/pdf/WeinbergReport.pdf indicates that it's straight-forward electro-chemical engineering. The PDF's secured, so I can't cut and paste . Page 2-3, and particularly Page 3 starting with A combination of increasing the surface area ...
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
The easiest way to visualize the effect is to imagine the case of a roller coaster. Initially a roller coaster must overcome a gravitational energy barrier by a mechanical mechanism that pulls the cars to the top of the track. That energy is not lost but is regained on the down slope. In the case of a nuclear fusion reaction the kinetic energy of the proton overcomes the repulsion of the positively charged nucleus through its high kinetic energy. Once the proton comes close enough to the target nucleus, the strong nuclear force begins to attract the proton with a strength that dwarfs the electrostatic repusion (the down slope). In this way the kinetic energy of the proton is not lost. Practically speaking a proton does not need to overcome the total energy barrier imposed by electrostatic repulsion, because as it gets close to the nucleus the probability that the proton will tunnel through the barrier becomes large. This quantum effect is similar to what is observed in transistors and MIM diodes, where electrons elastically tunnel through thin dielectrics and semiconductors without losing energy. To: vortex-l@eskimo.com Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 14:23:39 -0400 Thanks for the rapid response! I think that I have a mental block regarding this issue. Perhaps I am looking at the fusion process from the wrong angle. Help me understand where my error is as I do not follow your example below. Let's try a mental experiment. I can see that it would be possible to build a machine that accelerates protons until they reach the required threshold level of 5.6 MeV. It would take at least this much energy to generate the projectile proton. Now, when the bullet proton comes close to the nucleus the fusion takes place. At this point in time it has lost all of its kinetic energy and is absorbed into the nucleus. I assume that very soon thereafter the 3.41 MeV is released as a gamma or some other form of radiation. The tables of nuclides informs me that the copper 59 has a half life of 81.5 seconds before the beta plus decay so the energy associated with that process must be in place but does not exit until later. Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize. Dave -Original Message- From: Finlay MacNab finlaymac...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 1:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Your calculation does not take into account the fact that the activation energy barrier releases the energy added to overcome it during the reaction. In this case once coulomb repulsion is overcome, the energy is added back to the system by attractive nuclear force. The 3.41MeV is the change in mass energy balance after the reaction, what happens in between is not important to the calculation. Analogously, If a person descends from the 3rd to second floor of a building, they are just as close to the ground as if they climb from the 3rd to the 53rd floor before climbing back down the the second floor to end their journey. This is the nature of all activation energy barriers. To: vortex-l@eskimo.com Subject: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 13:35:41 -0400 I have been researching the cold fusion reaction that is suggested by Rossi and Focardi in their recent paper http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and have a couple of questions. The authors suggest that 3.41 MeV of energy is released by the fusion of a proton with a nickel 58 nucleus into copper 59. I can obtain this value if I calculate the mass difference between a copper 59 atom and a nickel 58 atom plus the mass of the proton and the mass of the extra electron. So far their calculations are in line with mine. The problem arises when I consider the amount of energy required to overcome the coulomb barrier in order to activate the fusion. The two authors seem to overlook this entirely when they calculate the energy available from their proposed reaction. The chart on page 5 of their paper shows that 3.41 MeV is released at the conclusion to the reaction but no allowance is given to the energy needed to initiate it. They do mention the activation energy in their theoretical interpretation on page 7. In this section they calculate that it takes 5.6 MeV to overcome the barrier. The authors use assumed values for the closeness required and thus energy
Re: [Vo]:Spark plugs... thoughts and how-to?
Von: Jojo Jaro jth...@hotmail.com An: Vortex Vortex-l@eskimo.com Gesendet: 21:32 Dienstag, 22.Mai 2012 Betreff: Re: [Vo]:Spark plugs... thoughts and how-to? Interesting!!! Glad jojo no problem. Misunderstandings are quite normal. Anyway. Ignition coils tend to heavily oscillate on the secondary side, which is quite undesirable in the LENR case, because it tends to neutralize the direction of the ion movements. Which is irrelevant in a combustion-motor, but not with LENR. The function of my hypothetical auxiliary mesh-grid can be more easily seen if not used. See my attached sketch. Here you can see that H+ ions tend to oscillate around their point of generation and finally neutralize with high probability. The simple trick seems to be to rectify the potential , such that the H+ ions travel towards the reactant. This can be accompished a) by -well- rectification b) by applying an auxiliary potential via the mesh (a) rectifiication- would do the job , but only for a very short time. By rectification one gains a lot. The 20-10-5..kV pulses then all work in the right direction. (b) -aux mesh potential- on the other hand, only works if the time-interval between sparks is sufficiently large (1:10..1000) compared to the dominant potential (20kV) of the major pulse, which is, say, a couple of usec. So the mesh in the strict sense is not necessary, but only for fine-control or low frequency sparks (say 100msec interval). We are not there yet. best regards Guenther Attached You find a crude graphic representation of said situation. my_LENR_20120522_HV and ion movement.pdf Description: Adobe PDF document
Re: [Vo]:Press release Blacklightpower
Alan J Fletcher a...@well.com wrote: The PDF's secured, so I can't cut and paste . Page 2-3, and particularly Page 3 starting with A combination of increasing the surface area ... Secured PDFs are annoying, aren't they? Here is the text you meant to quote: A combination of increasing the surface area of the supported catalyst and improving/optimizing the supported catalyst could increase the power output of a single CIHT cell by many orders of magnitude. In the former case, using the power density of about 3 mW/cm2 of the Mo-anode cell and a thickness of each cell of a stack of 30 microns, the projected power density is 1 kW/1. Furthermore, the 3 mW/cm2 based on the geometrical surface area of the Mo electrode can be increased by large factors by using textured materials with much larger surface areas than the geometrical surface areas. Moreover, an improvement of five orders of magnitude, which is not unprecedented in the heterogeneous catalysis literature between the first hit and the optimized catalyst . . .
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
Re: [Vo]:Press release Blacklightpower
Alan, If you have firefox you can render most PDFs in your browser without any extra reader. https://addons.mozilla.org/en-US/firefox/addon/pdfjs plain js + html5 You can do all the copying you need then ;-) mic 2012/5/22 Alan J Fletcher a...@well.com: At 07:40 AM 5/22/2012, Daniel Rocha wrote: And it is extremely ambitious to talk about 1KJ. 6 orders of magnitude of improvement within a few months. The Weinberg report : http://www.blacklightpower.com/wp-content/uploads/pdf/WeinbergReport.pdf indicates that it's straight-forward electro-chemical engineering. The PDF's secured, so I can't cut and paste . Page 2-3, and particularly Page 3 starting with A combination of increasing the surface area ...
Re: [Vo]:Another web site about Rossi
From Jed: As for me, my current plans are to completely overhaul my orionworks web site using XARA Web Designer Premium. See: http://www.xara.com/us/products/webdesigner/ That costs $99. Probably worth it. It looks similar to WordPress. They use the same jargon for components such as template. I think it is quite different from Word Press and Drupal. Fundamentally different! IMO, working with XARA's web developer is much more like working with Adobe Illustrator or CorelDraw. Very WYSIWYG layout. You are completely isolated from managing any of the underlying code - HTML, CSS, etc... The lack of access to the underlying code may irk some geeks, but it's fine by me. While XARA is suitable for my personal needs the way the software is currently implemented it isn't really good at building and/or managing a complex web site with multiple sub-directories and lots and lots of pages that might change dynamically. Hopefully, later releases will address some of these matters, especially managing multiple sub-directories under the jurisdiction of a single domain name, and/or sub-domains as well. XARA is very good at whacking out a kick-ass presentation web site in no time at all. It's a great prototype web builder. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
In the case of regular fusion the kinetic energy of the proton would be conserved in the reaction products (i.e. the total energy of the system would be 5.6 MeV + 3.41MeV) these reaction products might be a Cu59 nucleus with 9.01MeV of kinetic energy or more likely some combination of gamma rays, neutrinos, nuclear excitation, and other nuclear reaction products. I am not familiar with the specifics of this reaction. The fusion of Iron inside a star is a different matter. Iron can fuse with hydrogen in an exothermic reaction because hydrogen has zero binding energy due to the fact that it has a single proton nucleus. Iron cannot fuse with itself inside a star because the resultant reaction would be endothermic, this is why stars burn out, not because of H + Fe fusion. To: vortex-l@eskimo.com Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 17:53:30 -0400 That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
[Vo]:Re: [Vo] Offtopic -- pdfjs
At 02:59 PM 5/22/2012, Michele Comitini wrote: If you have firefox you can render most PDFs in your browser without any extra reader. https://addons.mozilla.org/en-US/firefox/addon/pdfjs plain js + html5 I'm on FF10 ... it's installed, but it still opens with the regular reader. I couldn't see any help on their web page.
Re: [Vo]:Press release Blacklightpower
You can always print, scan and OCR it with a cheap all-in-one printer. T
RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
Awkshully, the most stable isotope in the periodic table is not one of iron’s major isotopes. 62Ni has the highest binding energy per nucleon of all isotopes, including iron, even though the average binding energy per nucleon for iron is slightly higher than nickel. Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. From: David Roberson I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question.
[Vo]:New WLT Transmutation : Tungsten (W) to Gold
via Krivit Fascinating Reading: Larsens Latest on LENRs and Gold http://blog.newenergytimes.com/2012/05/21/fascinating-reading-larsens-latest-on-lenrs-and-gold/ http://www.slideshare.net/lewisglarsen/lattice-energy-llc-lenr-transmutation-networks-can-produce-goldmay-19-2012 # Once you've got those slow neutrons, transmutations are easy. Suggests it's been observed experimentally since the 1920's .. and in nature. Could be more profitable than gold-mining. (lenr.qumbu.com -- analyzing the Rossi/Focardi eCat -- and the defkalion hyperion -- Hi, google!)
Re: [Vo]:Bologna Italy 6.3 Earthquake
At 05:09 AM 5/20/2012, Ron Kita wrote: I guess that this will be headline news in a few seconds: http://news.yahoo.com/6-3-magnitude-earthquake-strikes-near-bologna-italy-022116451.html Dear Prof. Brian Josephson, Nobel Prize: Thank you for your attention. The earthquake has hit my house, which is in Ferrara, where the sismic event has been strong, but not our factory, which is in Bologna, where it has hit less, so our work has not been affected. No injuries to my family, thanks to God. Anyway, the people of this geographic area is used to react strongly to adverse events. Again thank you, Warm Regards, A.R.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.com wrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly * borrowed* and then a short time later, the debt is repaid – before the net gain is obvious. *From:* David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
[Vo]:This is how a 1 dimensional nanowire stores charge.
This is how a 1 dimensional nanowire stores charge. Excess charge electrons: the delocalized electrons, those not associated with specific atoms, follow circular orbits around the tube circumference. As in an atom, this motion causes one orientation of the electron spin to have lower energy than the opposite orientation. http://physics.aps.org/articles/v5/57 Focus: Electron Spin Influences Nanotube Motion Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. You can see that the maximum charge storage capacity would be very large for a macro sized nanotube. An excess charge concentration of just 2000 electrons would be enough to lower the coulomb barrier at the superconductive tip of the tube. Cheers: Axil
Re: [Vo]:New WLT Transmutation : Tungsten (W) to Gold
In reply to Alan J Fletcher's message of Tue, 22 May 2012 16:20:19 -0700: Hi, [snip] via Krivit Fascinating Reading: Larsens Latest on LENRs and Gold http://blog.newenergytimes.com/2012/05/21/fascinating-reading-larsens-latest-on-lenrs-and-gold/ http://www.slideshare.net/lewisglarsen/lattice-energy-llc-lenr-transmutation-networks-can-produce-goldmay-19-2012 # Once you've got those slow neutrons, transmutations are easy. Suggests it's been observed experimentally since the 1920's .. and in nature. Could be more profitable than gold-mining. This ignores the fact that the energy released during the transmutation is worth more than the gold (at least at current energy prices). In the comments Tom Andersen manages to stuff it up completely :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
In reply to Axil Axil's message of Tue, 22 May 2012 22:33:09 -0400: Hi, [snip] Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. I'm not sure that free electrons even have spin. Consider the following and show me where I'm wrong. :) Draw an ellipse on a piece of paper. Cut it out. Stick a pin through one of the foci. The ellipse as a whole can be rotated about the pin. This is l (quantum number). The movement of the electron around the circumference of the ellipse is s. No closed orbit - no s. If someone can find a reference to paper that shows the spin of free electrons (i.e. not attached to atoms), I'd love to see it. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:New WLT Transmutation : Tungsten (W) to Gold
Larsen has all of the slow neutrons. Mills has all of the hydrinos. I have all of the mega-hertz meter Puthoff has all of the ZPE It amazing. Frank
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
Its an old experiment. Stern and Gerlich. Frank -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 11:06 pm Subject: Re: [Vo]:This is how a 1 dimensional nanowire stores charge. In reply to Axil Axil's message of Tue, 22 May 2012 22:33:09 -0400: Hi, [snip] Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. I'm not sure that free electrons even have spin. Consider the following and show me where I'm wrong. :) Draw an ellipse on a piece of paper. Cut it out. Stick a pin through one of the foci. The ellipse as a whole can be rotated about the pin. This is l (quantum number). The movement of the electron around the circumference of the ellipse is s. No closed orbit - no s. If someone can find a reference to paper that shows the spin of free electrons (i.e. not attached to atoms), I'd love to see it. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
In reply to mix...@bigpond.com's message of Wed, 23 May 2012 13:06:16 +1000: Hi, [snip] BTW there are also other ways you can rotate an ellipse. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I understand how your analogy operates now. It is helpful to be able to compare everyday experiences with the phenomenon that one is trying to get a handle upon. What is the possibility that the barrier energy that is supplied by my experimental proton accelerator at approximately 5.6 MeV has become converted into mass of the copper 59 nucleus? According to the charts of nuclides there is a reduction of mass of the copper 59 that equals a modest 3.41 MeV as the proton fuses with the nickel 58. This stage of the reaction is followed by the release of 5.82 MeV in a half life of 81.5 seconds via beta plus decay and positron-electron destruction. The beta plus decay results in the formation of nickel 59. A further calculation that is obtained by taking the same initial nickel 58 and adding a neutron to it to arrive at nickel 59 yields a mass loss equal to 8.99 MeV. This is the value shown in WL charts as well when you follow their reaction path. If you add the two energy releases associated with the beta plus aka the Rossi reaction you get a net of 9.23 MeV. This compares quite well with the value according to the WL process at 8.99 MeV. The difference is less than half the energy equivalent of an electron. I suspect that this error can be discovered with enough digging. In these scenarios the same starting and end results are obtained: Nickel 58 seed results in Nickel 59 product. Also, approximately the same energy is released provided the original input proton energy of 5.6 MeV is assumed to be unimportant as you suggest. Dave -Original Message- From: Finlay MacNab finlaymac...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:23 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? The easiest way to visualize the effect is to imagine the case of a roller coaster. Initially a roller coaster must overcome a gravitational energy barrier by a mechanical mechanism that pulls the cars to the top of the track. That energy is not lost but is regained on the down slope. In the case of a nuclear fusion reaction the kinetic energy of the proton overcomes the repulsion of the positively charged nucleus through its high kinetic energy. Once the proton comes close enough to the target nucleus, the strong nuclear force begins to attract the proton with a strength that dwarfs the electrostatic repusion (the down slope). In this way the kinetic energy of the proton is not lost. Practically speaking a proton does not need to overcome the total energy barrier imposed by electrostatic repulsion, because as it gets close to the nucleus the probability that the proton will tunnel through the barrier becomes large. This quantum effect is similar to what is observed in transistors and MIM diodes, where electrons elastically tunnel through thin dielectrics and semiconductors without losing energy. To: vortex-l@eskimo.com Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? From: dlrober...@aol.com Date: Tue, 22 May 2012 14:23:39 -0400 Thanks for the rapid response! I think that I have a mental block regarding this issue. Perhaps I am looking at the fusion process from the wrong angle. Help me understand where my error is as I do not follow your example below. Let's try a mental experiment. I can see that it would be possible to build a machine that accelerates protons until they reach the required threshold level of 5.6 MeV. It would take at least this much energy to generate the projectile proton. Now, when the bullet proton comes close to the nucleus the fusion takes place. At this point in time it has lost all of its kinetic energy and is absorbed into the nucleus. I assume that very soon thereafter the 3.41 MeV is released as a gamma or some other form of radiation. The tables of nuclides informs me that the copper 59 has a half life of 81.5 seconds before the beta plus decay so the energy associated with that process must be in place but does not exit until later. Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize. Dave -Original Message- From: Finlay MacNab finlaymac...@hotmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 1:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Your calculation does not take into account the fact that the activation energy barrier releases the energy added to overcome it during the reaction. In this case once coulomb repulsion is overcome, the energy is added back to the system by attractive nuclear force. The 3.41MeV is the change in mass energy balance after the reaction,
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
I also suspect that the reaction is a bit more complex than a single hydrogen fusion. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 9:44 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.com wrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly borrowed and then a short time later, the debt is repaid – before the net gain is obvious. From: David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
The process suggested by Rossi has major issues as you point out. I do think that one could force a reaction like this with very high energy protons impacting nickel, but it seems quite unlikely to me that this will happen at the low temperatures encountered within Rossi's reactors. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 6:50 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Awkshully, the most stable isotope in the periodic table is not one of iron’s major isotopes. 62Ni has the highest binding energy per nucleon of all isotopes, including iron, even though the average binding energy per nucleon for iron is slightly higher than nickel. Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. From: David Roberson I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question.
Re: [Vo]:New WLT Transmutation : Tungsten (W) to Gold
You guys need to share with the rest of us. Dave -Original Message- From: fznidarsic fznidar...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 11:07 pm Subject: Re: [Vo]:New WLT Transmutation : Tungsten (W) to Gold Larsen has all of the slow neutrons. Mills has all of the hydrinos. I have all of the mega-hertz meter Puthoff has all of the ZPE It amazing. Frank
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
On Tue, May 22, 2012 at 3:50 PM, Jones Beene jone...@pacbell.net wrote: ** Another reason that Focardi/Rossi’s claim of nickel going to copper is brain-dead. Or a gambit intended to divert attention from what is really going on. Eric
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
http://arxiv.org/pdf/0909.3060.pdf Electric charge enhancements in carbon nanotubes On page 4 of the reference, three of my recently made assertions are demonstrated. 1. Charge is concentrated at the tip of the tube, 2. Tube Charge will be attracted to and amplified by the contract with the lattice where the charge will accumulate. 3. The lattice will respond to the concentrated charge from the tube with an induced counter charge that is capable of lowering the coulomb barrier. In an electron rich hydrogen envelope where there is a large excess of degenerate electrons produced by spark discharge, the figures for charge density listed in the reference will be greatly exceeded. The degenerate electron will pack onto the surface of the nanotubes. This theory is applicable to the Chin type reaction. In addition, remember that excess degenerate electrons will also lower the coulomb barrier broadly by long range opposite charge induction. *If someone can find a reference to paper that shows the spin of free electrons* * * *(i.e. not attached to atoms), I'd love to see it. :)* Look into dirac cones on the surface of three dimensional topologic insulators. The spin will move in synchrony with the circular path of the electron around the cone. I believe this is called strong spin orbit coupling. http://online.kitp.ucsb.edu/online/topomat11/hasan/pdf/Hasan_TopoMat11_KITP.pdf On Tue, May 22, 2012 at 11:06 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 22 May 2012 22:33:09 -0400: Hi, [snip] Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. I'm not sure that free electrons even have spin. Consider the following and show me where I'm wrong. :) Draw an ellipse on a piece of paper. Cut it out. Stick a pin through one of the foci. The ellipse as a whole can be rotated about the pin. This is l (quantum number). The movement of the electron around the circumference of the ellipse is s. No closed orbit - no s. If someone can find a reference to paper that shows the spin of free electrons (i.e. not attached to atoms), I'd love to see it. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
In reply to fznidar...@aol.com's message of Tue, 22 May 2012 23:08:53 -0400 (EDT): Hi, [snip] Its an old experiment. Stern and Gerlich. IIRC that deals with Silver ions, not free electrons. Frank -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 11:06 pm Subject: Re: [Vo]:This is how a 1 dimensional nanowire stores charge. In reply to Axil Axil's message of Tue, 22 May 2012 22:33:09 -0400: Hi, [snip] Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. I'm not sure that free electrons even have spin. Consider the following and show me where I'm wrong. :) Draw an ellipse on a piece of paper. Cut it out. Stick a pin through one of the foci. The ellipse as a whole can be rotated about the pin. This is l (quantum number). The movement of the electron around the circumference of the ellipse is s. No closed orbit - no s. If someone can find a reference to paper that shows the spin of free electrons (i.e. not attached to atoms), I'd love to see it. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
The hard part is making the coulomb barrier go away; after that the fusion is easy. On Tue, May 22, 2012 at 11:24 PM, David Roberson dlrober...@aol.com wrote: I also suspect that the reaction is a bit more complex than a single hydrogen fusion. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 9:44 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? There are a number of assumptions at issue in this tread that I would like to counter. I believe that a cooper pair of Protons fuses with the nickel nucleus. The reaction products of this type fusion are listed in the Kim paper http://www.physics.purdue.edu/people/faculty/yekim/BECNF-Ni-Hydrogen.pdf The cooper pair of protons speculation assumes a superconductive surface on the nano nickel powder which must be hot at 4ooC to thermalize the gamma radiation from the fusion reaction. His thermalization is done by averaging the gamma ray energy over N coherent atoms(thermal energy value = gamma energy/N). The coulomb barrier is greatly lowered by Rossi’s catalyst (aka secret sauce). Penetration of this greatly weakened or nonexistent barrier consumes no reaction energy. Cheers: Axil On Tue, May 22, 2012 at 6:27 PM, David Roberson dlrober...@aol.comwrote: I just reviewed the wikipedia article on stars. They support the idea that iron is the last element that can be fused before the process becomes endothermic. It is an interesting read and I should be kicked in the rear for not reading it before asking my question. Of course they might not be entirely accurate as is sometimes the case, but on this occasion my calculations and their article suggests otherwise. As I write this I am wondering if the wikipedia model assumes iron fusing with iron versus iron fusing with hydrogen. I guess I should pursue this a bit further to see what the implications are if both of the reactants are iron. I appreciate the inputs that have been presented and I will think about them carefully as I try to understand the issue. Dave -Original Message- From: David Roberson dlrober...@aol.com To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:53 pm Subject: Re: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? That idea crossed my mind but I still do not know where the 5.6 MeV of energy imparted upon the proton wound up. If the path were exothermic I would expect to be able to recover(or at least locate) all of the 5.6 MeV as well as some extra energy. I recall reading an article years ago that suggested that fusion energy was possible within stars until the final product was iron. The star would then collapse under the influence of gravity due to the lack of extra heat. Could this be the effect that I am calculating? It does seem to add up in the numbers. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Tue, May 22, 2012 5:22 pm Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic? Are you discounting QM and quantum tunneling? One could say that the in tunneling - threshold energy is briefly * borrowed* and then a short time later, the debt is repaid – before the net gain is obvious. *From:* David Roberson Could you help me understand how the 5.6 MeV is recovered or released? Is there extra energy released into the copper crystal structure that equals this magnitude? I am having a difficult time trying to get back the 5.6 MeV to make the next proton energetic enough for the next reaction. Forgive me for being ignorant about this mechanism, but it truly is difficult to visualize.
Re: [Vo]:This is how a 1 dimensional nanowire stores charge.
What I was describing is the spin hall effect in a one dimensional topological insulator… see page 8 of last reference. On Wed, May 23, 2012 at 12:39 AM, Axil Axil janap...@gmail.com wrote: http://arxiv.org/pdf/0909.3060.pdf Electric charge enhancements in carbon nanotubes On page 4 of the reference, three of my recently made assertions are demonstrated. 1. Charge is concentrated at the tip of the tube, 2. Tube Charge will be attracted to and amplified by the contract with the lattice where the charge will accumulate. 3. The lattice will respond to the concentrated charge from the tube with an induced counter charge that is capable of lowering the coulomb barrier. In an electron rich hydrogen envelope where there is a large excess of degenerate electrons produced by spark discharge, the figures for charge density listed in the reference will be greatly exceeded. The degenerate electron will pack onto the surface of the nanotubes. This theory is applicable to the Chin type reaction. In addition, remember that excess degenerate electrons will also lower the coulomb barrier broadly by long range opposite charge induction. *If someone can find a reference to paper that shows the spin of free electrons* * * *(i.e. not attached to atoms), I'd love to see it. :)* Look into dirac cones on the surface of three dimensional topologic insulators. The spin will move in synchrony with the circular path of the electron around the cone. I believe this is called strong spin orbit coupling. http://online.kitp.ucsb.edu/online/topomat11/hasan/pdf/Hasan_TopoMat11_KITP.pdf On Tue, May 22, 2012 at 11:06 PM, mix...@bigpond.com wrote: In reply to Axil Axil's message of Tue, 22 May 2012 22:33:09 -0400: Hi, [snip] Many delocalized electrons would orbit the diameter of the nano-tube in a cooper paired counter rotational spin up spin down couplet. The electron motion would be superconductive and the total excess charge would be proportional to the length of the nanotube. I'm not sure that free electrons even have spin. Consider the following and show me where I'm wrong. :) Draw an ellipse on a piece of paper. Cut it out. Stick a pin through one of the foci. The ellipse as a whole can be rotated about the pin. This is l (quantum number). The movement of the electron around the circumference of the ellipse is s. No closed orbit - no s. If someone can find a reference to paper that shows the spin of free electrons (i.e. not attached to atoms), I'd love to see it. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:New WLT Transmutation : Tungsten (W) to Gold
No neutrons please. The fusion of one or two protons will transmute tungsten(N) into elements in the platinum group(n+1) or (N+2). The test is easy. There will be no beta decay and no tungsten isotopes. On Tue, May 22, 2012 at 7:20 PM, Alan J Fletcher a...@well.com wrote: via Krivit Fascinating Reading: Larsen’s Latest on LENRs and Gold http://blog.newenergytimes.com/2012/05/21/fascinating-reading-larsens-latest-on-lenrs-and-gold/ http://www.slideshare.net/lewisglarsen/lattice-energy-llc-lenr-transmutation-networks-can-produce-goldmay-19-2012 #http://www.slideshare.net/lewisglarsen/lattice-energy-llc-lenr-transmutation-networks-can-produce-goldmay-19-2012 Once you've got those slow neutrons, transmutations are easy. Suggests it's been observed experimentally since the 1920's .. and in nature. Could be more profitable than gold-mining. ** ** (lenr.qumbu.com -- analyzing the Rossi/Focardi eCat -- and the defkalion hyperion -- Hi, google!)
Re: [Vo]:Nickel-hydrogen nuclear ash
On Tue, May 22, 2012 at 7:41 AM, Jones Beene jone...@pacbell.net wrote: Eric - perhaps the original post should have been phrased as “zero believable evidence”… instead of zero evidence. The paper does constitute putative “evidence” after all – actually rather convincing if it could be taken at face value. You forced me. :) Ni + K2CO3 + H2O: tritium 26 * background. Notoya et al., Tritium generation and large excess heat evolution by electrolysis in light and heavy water-potassium carbonate solutions with nickel electrodes, Fusion Technology, 26,179, 1994; Alkali-hydrogen cold fusion accompanied by tritium production on nickel, Trans. Fusion Technology, 26, 205, 1994. Ni + K2CO3 + H2O: tritium 10-100 * background. Notoya, Alkali-hydrogen cold fusion accompanied by tritium production on nickel, in the proceedings of the Fourth International Conference on Cold Fusion, 1993. Ni + K2CO3 + D2O, H2O: tritium 339 * background. Srinivasan et al., Tritium and excess heat generation during electrolysis of aqueous solutions of alkali salts with nickel cathode, in the proceedings of the Third International Conference on Cold Fusion, 1992. Ni + Li2CO3 + H2O: tritium 145 * background. Srinivasan et al., op cit. Please confirm either that these references do not meet your evidentiary standards or that the Ni-H2O electrolytic system is different in some basic way from the Ni-H2 system when considering the question of radiation. Eric
