Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Am 14.09.2011 01:20, schrieb Horace Heffner: On Sep 13, 2011, at 12:55 PM, Peter Heckert wrote: Am 13.09.2011 22:47, schrieb Man on Bridges: Hi, On 13-9-2011 20:44, Horace Heffner wrote: snip calculation of lead shielding Hmmm, is there a way to start and stop a gamma radiation source, as it may be used only to trigger the process? There is no other way than shielding or increasing the distance. Rossi could inside use a shield that is moved electrically or by heat (bimetal). Or he could control the distance to the gamma source. If it is a very small point source the /local/ intensity of radiation could be changed by factor 10^2 or 10^3. Peter The above is incorrect. A 2 cm thick lead shield will only reduce Co-60 gammas by 75%. I = I0 * exp (-0.694 * x) So we want I/Io = 0.01 to achieve 1/100 reduction factor. I/I0 = exp (-0.694 * x) 0.01 = exp (-0.694 * x) ln(0.01) = -0.694*x x = ln(0.01)/(-0.694) = 6.63 It takes 6.6 cm of lead to divide Co-60 gamma intensity by 100. Similarly, it takes about 10 cm of lead (on all sides) to attenuate CO60 gammas by a factor of 1/1000. Maybe I am in error. I understand it this way: A shield cannot alter the wavelength and so it cannot alter the photon energy respective frequency. Only the amount or density of gamma photons can be changed by photon absorption. Now, lets assume the gamma radiator has a volume of 1mm. Then the photon density in 100mm distance must be 4 times weaker as the density directly measured in 0.5 mm distance at the surface of the gamma source. (Inverse square law as in optics) Even without shield we can get a large attentuation factor purely from distance, if the diameter of the source is small. So if the gamma source is in direct contact with nickel, the photon density must be 100 times larger than in 10 mm distance. Is this wrong? Another thought: I think Rossi is naive and will loose if he think he can commercialize a discovery of this magnitude and eternal history changing importance and keep it secret. This is impossible to do, he must go the scientific route, not the commercial route. Also his fans and investors are naive to believe this. As soon as it is totally and unmistakenly clear, this is a nuclear reaction that produces large amounts of energy, law will stop him. And international scientific research will start. You cannot discover the stone of philosophers and commercialize this and keep it secret, this is impossible. This must be done in a scientific way. As soon as large amounts of energy are produced, it must be also scientifically investigated, if this can be abused to build bombs and so on. Rossi says no, this is not possible, but as long as it is a secret he cannot proof it is without dangers. I think no government can tolerate something like this going on and reaching very large dimensions unsupervised. The unknown potential of danger is too high. Only if his customer is NASA or another large scientific and trusted organisation he could have luck selling this. Best, Peter
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Am 14.09.2011 02:17, schrieb Man on Bridges: Hi, On 14-9-2011 1:20, Horace Heffner wrote: snip calculation Just a thought. Let's suppose Rossi is using a gamma radiation source as a catalyzer. Is it then possible to determine the source (catalyzer) of the gamma source, if the following parameters are known? 1. Maximum allowed gamma radiation level which passes safety certification. 2. Maximum lead shielding thickness used around the reactor. No this is not possible if the spatial dimension and size of the gamma source is unknown. The only possibility is to measure the spectrum of gamma radiation. And as verification it would be great if someone could do a gamma spectrum/intensity scan close to the Rossi reactor. Rossi doesnt allow to measure the spectrum. Bianchini was not allowed to measure it.
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Am 14.09.2011 08:20, schrieb Peter Heckert: As soon as it is totally and unmistakenly clear, this is a nuclear reaction that produces large amounts of energy, law will stop him. And international scientific research will start. You cannot discover the stone of philosophers and commercialize this and keep it secret, this is impossible. This must be done in a scientific way. As soon as large amounts of energy are produced, it must be also scientifically investigated, if this can be abused to build bombs and so on. Rossi says no, this is not possible, but as long as it is a secret he cannot proof it is without dangers. I think no government can tolerate something like this going on and reaching very large dimensions unsupervised. The unknown potential of danger is too high. Only if his customer is NASA or another large scientific and trusted organisation he could have luck selling this. It will also be impossible to sell this internationally and keep it secret. How to get around customs controls? Rossi was involved in gold smuggle, if it is true, what they write. If he wants to sell internationally then he must produce in these countries where he sells. Best, Peter
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
On Sep 13, 2011, at 10:20 PM, Peter Heckert wrote: Am 14.09.2011 01:20, schrieb Horace Heffner: On Sep 13, 2011, at 12:55 PM, Peter Heckert wrote: Am 13.09.2011 22:47, schrieb Man on Bridges: Hi, On 13-9-2011 20:44, Horace Heffner wrote: snip calculation of lead shielding Hmmm, is there a way to start and stop a gamma radiation source, as it may be used only to trigger the process? There is no other way than shielding or increasing the distance. Rossi could inside use a shield that is moved electrically or by heat (bimetal). Or he could control the distance to the gamma source. If it is a very small point source the /local/ intensity of radiation could be changed by factor 10^2 or 10^3. Peter The above is incorrect. A 2 cm thick lead shield will only reduce Co-60 gammas by 75%. I = I0 * exp (-0.694 * x) So we want I/Io = 0.01 to achieve 1/100 reduction factor. I/I0 = exp (-0.694 * x) 0.01 = exp (-0.694 * x) ln(0.01) = -0.694*x x = ln(0.01)/(-0.694) = 6.63 It takes 6.6 cm of lead to divide Co-60 gamma intensity by 100. Similarly, it takes about 10 cm of lead (on all sides) to attenuate CO60 gammas by a factor of 1/1000. Maybe I am in error. The error I am pointing out is that it does not matter at all how small the source inside the device is - assuming it is centrally located. It could be microscopic, or a couple cm in diameter and this would make no difference at all to the gamma flux measured at the surface of the Rossi device. If the source were even the size of one atom, vs a few mm, or cm, it would make no difference to the intensity measured at the surface of the Rossi device. The intensity is proportional to surface area divided by total counts per minute. The size of the source inside the device is of no consequence provided it is centrally located. Rossi used two counters right up against the device, primarily in coincidence mode, but they would saturate at the count rate expected, so coincindece mode would be irrelevant. Celani measured radiation right near the device before turn on as near background, using two different types of counters. It is not possible to put enough lead in the device to suppress the 1.33 MeV gammas from cobalt to even a non-lethal level - provided there is enough cobalt to sustain a 15 kW reaction at one gamma per LENR reaction. Celani also measured the counts a few meters away. A few meters away, where Celani also measured, is not enough to suppress the counts to background. If a 2 cm thick lead shielded source has even a very modest amount of Co-60 then detectors nearby will detect the gammas - at all times. I showed it would take at least 6x10^11 gammas a second to account for a 12 kW LENR reaction, even assuming 10 MeV per reaction, which is high. Even if Rossi could stuff his source behind a blanket of 6.6 cm of lead on all sides, giving a device radius of 13 cm, leaving no room for water or fuel, that would only reduce the count by a factor of 100, thus outside the reactor a 6x10^9 count per minute (cpm) source would be manifest. At a distance of 6.6 meters, the flux would be reduced by a factor of 6.6/660 = 10^-4, or to 6x10^5 cpm. Celani could not miss this. I understand it this way: A shield cannot alter the wavelength and so it cannot alter the photon energy respective frequency. Yes. Only the amount or density of gamma photons can be changed by photon absorption. That is in practical terms true. Some of the gammas cause positron emission which results in a lower energy gamma, but at CO60 energy levels this is not important. Now, lets assume the gamma radiator has a volume of 1mm. Then the photon density in 100mm distance must be 4 times weaker as the density directly measured in 0.5 mm distance at the surface of the gamma source. (Inverse square law as in optics) This is where the conceptual error occurs. The source is not measured at its radius. It is measured at the radius of the Rossi device, and further. Even without shield we can get a large attentuation factor purely from distance, if the diameter of the source is small. This is irrelevant because the distances at which measurement actually occurred are fixed. So if the gamma source is in direct contact with nickel, the photon density must be 100 times larger than in 10 mm distance. Is this wrong? You are mixing apples and oranges. There is a difference between how the radiation affects the Ni and determining the amount of radiation by counting outside the device. If the Co60 were a nano-sized particle it would provide a high intensity radiation to nano-sized nickel particles at nano-distances from it, but not to all the fuel. A point source does not provide a means to irradiate the entire fuel at the point source flux level. What counts in irradiating the fuel is achieving as nearly as possible a 1-1
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Am 14.09.2011 10:08, schrieb Horace Heffner: It is not possible to put enough lead in the device to suppress the 1.33 MeV gammas from cobalt to even a non-lethal level - provided there is enough cobalt to sustain a 15 kW reaction at one gamma per LENR reaction. Yes this is correct. But this is not what I wanted to say. I think there could be a very small gamma source inside, possibly cobalt 60, with a power of milliwatts or microwatts. This gamma radiation could excite the nickel atom and bring it into resonance in a novel, yet unknown way and could trigger the LENR reaction. May be its only used to start the reaction and then shielded, this could explain the gamma burst at startup. I dont think the reactor itself produces gamma rays in the kilowatt range. Widom Larsen theory says, that not gamma rays are produced, because the gamma photons -if there are any- are downshifted to infrared. Piantelli and Focardi in their papers reported either gamma radiation or energy production mutually exclusive, never both at the same time. And so far I understand, they had no shielding, and so they had no high power gamma radiation. No LENR researcher has yet reported hard gamma radiation or has died from gamma radiation so far I know, but many have reported huge amounts of energy. So, why should the Rossi device produce gamma radiation? My theory was, there might be gamma rays, that act as a catalyzer to start and possibly to sustain the LENR reaction,but I cannot believe, the gamma rays are the reason for the thermal energy. This cannot be, as you have correctly explained and this was never before observed in other LENR experiments. Best, Peter
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
On Sep 14, 2011, at 12:29 AM, Peter Heckert wrote: Am 14.09.2011 10:08, schrieb Horace Heffner: It is not possible to put enough lead in the device to suppress the 1.33 MeV gammas from cobalt to even a non-lethal level - provided there is enough cobalt to sustain a 15 kW reaction at one gamma per LENR reaction. Yes this is correct. But this is not what I wanted to say. OK. My comments were based on assumptions I made from what you wrote. Just to raise the level of understanding, here is the basis of my comments, with some paraphrasing. You said you felt the energy levels of gammas from cobalt decay to nickel may be significant to catalyzing a nickel LENR reaction. This to me implies a 1-1 relation of the stimulating gammas to the stimulated nickel. It implies each stimulating gamma is absorbed by the Ni nucleus it stimulates. I suggested that an upper limit to Ni-H LENR energies is about 10 MeV per LENR reaction. This means a 1.33 MeV photon interacts with a Ni nucleus, or Ni plus hydrogen ensemble, and catalyses a reaction that produces 10 MeV. The gammas to which I referred were 1.33 MeV catalytic gammas, not LENR produced gammas. I did not suggest the reaction produced gammas, or that they would be involved in Celani's pre-test background level measurements. I do, however, think there is reason to expect Ni-H LENR reactions to produce gammas, even as measured momentarily by Celani after the experiment started. BTW, it is notable that there could have been a shielded gamma source located, and momentarily unshielded, in the room Rossi was in. Celani's report says Rossi walked in right after Cealani's gamma measurement was pegged. The source Celani measured would not have had to have been in the device itself. Celani did say there were unexplained anomalies in the readings as he moved around the room he was in. I think there could be a very small gamma source inside, possibly cobalt 60, with a power of milliwatts or microwatts. This gamma radiation could excite the nickel atom and bring it into resonance in a novel, yet unknown way and could trigger the LENR reaction. Well, that is the assumption I made in my calculations - that one gamma stimulates one nickel nucleus. The gamm in the process disappears though. It can not go forth and cause more such reactions. Therefore there is a 1-1 relation. There would be a requirement for a kW of catalytic gammas to create around 10 kW of LENR energy output under that assumption. May be its only used to start the reaction and then shielded, this could explain the gamma burst at startup. Here I have some admitted personal biases. I have posted some suggested reasons why gamma bursts might exists during start-up and shut-down, but that is way outside this discussion. I dont think the reactor itself produces gamma rays in the kilowatt range. Well, if the reactor is producing kW levels of free energy heat then that energy has to come from somewhere. If is coming from LENR then the source is likely nuclear. If the energy produced is photonic, and comes from the nucleus, then it is by definition called gamma radiation, even if in the low energy range for x-rays. Widom Larsen theory says, that not gamma rays are produced, because the gamma photons -if there are any- are downshifted to infrared. It is notable that their patent provided no test data to show there is actually any screening effect: http://tinyurl.com/47al74f If such a screening effect existed it should be comparatively easy (as CF experiments go) to demonstrate it. Here's what I think of WL theory: http://www.mail-archive.com/vortex-l@eskimo.com/msg38261.html So ... you can see I bring my own bias to the conversation. Piantelli and Focardi in their papers reported either gamma radiation or energy production mutually exclusive, never both at the same time. And so far I understand, they had no shielding, and so they had no high power gamma radiation. This is indeed characteristic of LENR - no or nominal levels of high energy gammas. Low energy gammas and EUV are another thing entirely, but that is outside the scope of our conversation. No LENR researcher has yet reported hard gamma radiation or has died from gamma radiation so far I know, but many have reported huge amounts of energy. So, why should the Rossi device produce gamma radiation? It was measure by Celani. Rossi clearly has something that differs much from prior work - if it is as reported. My theory was, there might be gamma rays, that act as a catalyzer to start and possibly to sustain the LENR reaction,but I cannot believe, the gamma rays are the reason for the thermal energy. Yes, high energy gammas can not be the reason for the excess (ou) thermal energy - if it actually exists, which is still very much in doubt. This cannot be, as you have correctly
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Hi Horrace e.a., On 14-9-2011 23:40, Horace Heffner wrote: In any case, I think there is no reasonable possibility of a Co60 source of any possible significance being hidden behind the 2 cm lead shielding. However, there are various other radioactive materials that very well might be hidden behind a few cm of lead, and which might indeed be catalytic - especially beta producers. This brings me back to why I brought these questions forward. Let's suppose Rossi is using somekind of radiation source (not necessarily Co60) as a catalyzer. Is it then possible to determine the catalyzer, if the following parameters are known? 1. Maximum allowed radiation level which passes safety certification. 2. Maximum lead shielding thickness used around the reactor. Kind regards, MoB
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
On Sep 13, 2011, at 3:10 AM, peter.heck...@arcor.de wrote: Hi, Could it be that Rossi uses a Cobalt60 gamma source as catalyzer? Cobalt 60 decays to Nickel60 and emits gamma rays. The gamma spectrum could be just the right spectrum and energy to excite the Nickel nucleus. Maybe it is mainly the Cobalt60 that needs screening and not the reactor? This would explain the thickness of the lead screening. Focardi has calculated the lead screening. So he must know about the gamma rays and probably knows about the catalyzer. Once he said I dont know and dont want to know... Now I would believe he doesnt know, but I can hardly believe he doesnt want to know ;-) Peter It is notable that if a gamma plus high energy beta source were used for stimulation it could be kept in a container that isolated it from the nickel, and thus it would not be seen in post experiment analysis of the fuel. There are numerous reports of the effectiveness of radiation stimulation of LENR. The problem is that shielding merely attenuates gammas. Some always gets through. This could be detected externally. Almost all (99.88%) C60 decay is 0.32 MeV beta followed by 1.12 MeV gamma, followed by 1.33 MeV gamma. About 0.12% is 1.48 MeV beta followed by 1.3325 MeV gamma. I think the lead shielding was stated by Rossi to be 2 cm thick, but don't have a reference handy. Very roughly, the mass attenuation coefficient in lead for 1.12 MeV gammas is about 0.062 cm^2/gm and for 1.33 gammas about 0.057 cm^2/ gm. The linear attenuation coefficient mu for 1.12 MeV gammas is (0.062 cm^2/gm) * (11.4 gm/cm^3) = 0.707/cm, and for 1.33 MeV gammas is (0.057 cm^2/gm) * (11.4 gm/cm^3) = 0.65/cm. The gamma attenuation, from internal intensity I0 to external intensity I at distance x is given by: I = I0 * exp(-mu * x) so for 1.12 MeV gammas we have: I = I0 * exp(-(0.707/cm)*(2 cm)) = I0 * exp(-1.414) I = I0 * 0.243 for 1.33 MeV gammas we have: I = I0 * exp(-(0.065/cm)*(2 cm)) = I0 * exp(-1.3) I = I0 * 0.88 There is in effect (assuming no calc. errors on my part), with regards to either safety or detection, no practical attenuation offered for cobalt 60 gammas by 2 cm of lead. However, cobalt 60 is not the only possibility given the radioactive source is (and must be to avoid post experiment detection) physically isolated from the nickel. There is no requirement for it to decay into Ni, or Cu, or Zn, which were found in the used fuel. Thus, it is possible to choose an alpha or beta source which does not produce large gamma signatures through 2 cm of lead. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Horace, thank you very much. I dont have the knowledge to calculate this. Only know the very basics. Found this via google: http://itcanbeshown.com/NERS425/Lab5/Shielding%20-%20Final%20Version.pdf There is data about screening. My idea was, it could be a very small and weak source, such as used in schools for physics lectures. These are not too dangerous if shielded. If the gamma source is very small the intensity should also decrase by square of distance. The source could be very close to - or inside the nickel powder core . Then the lead and the distance together could shield it close to natural baseline level. Its just an idea. I dont know if this is possible. It also was an idea of me, that metal hydrides are very well researched. Metal hydride hydrogen storage systems are in use worldwide and they use specially developed alloys also such that use nickel as a component. These sytems are belived to be the most secure devices, melting or explosion or abnormal heating is not reported. Some of these are used with very high pressure and temperature. So there are already thousands if not millions man-years of experience, RD and scientific research done for metal hydride systems. systems. But only the LENR researchers find LENR reactions. Why? If LENR reactions where easily to achieve, then this should have been discovered. The developers try to reduce the thermal hysteresis in the load/unload cycle to get best efficiency. So they search for zero hysteresis. When there is LENR energy production then we should have negative hysteresis and if this is possible by common chemical or physical methods, the countless researchers and scientists should have discovered this method or catalyzer. Now, so the Rossis catalyzer must be something very unusual that nobody would ever try to use for a metal hydride storage system. So we need something that ionizises or atomizes the hydrogen molecules, and something that is very unusual for hydride systems. So I came to the idea it must be a radioactive gamma source or device. And it must be separated from the nickel, but can be very close and very small and can be inside.. Also I think, we should not only think about the energy, but also about frequencies and resonances. If Cobalt60 decays into Nickel60, then the gamma radiation spectrum should contain frequencies that are in tune with the resonance frequencies of the nickel nucleus or the inner electron shells of the nickel atom. That was my idea and how I came to it. Best, Peter Am 13.09.2011 20:44, schrieb Horace Heffner: On Sep 13, 2011, at 3:10 AM, peter.heck...@arcor.de wrote: Hi, Could it be that Rossi uses a Cobalt60 gamma source as catalyzer? Cobalt 60 decays to Nickel60 and emits gamma rays. The gamma spectrum could be just the right spectrum and energy to excite the Nickel nucleus. Maybe it is mainly the Cobalt60 that needs screening and not the reactor? This would explain the thickness of the lead screening. Focardi has calculated the lead screening. So he must know about the gamma rays and probably knows about the catalyzer. Once he said I dont know and dont want to know... Now I would believe he doesnt know, but I can hardly believe he doesnt want to know ;-) Peter It is notable that if a gamma plus high energy beta source were used for stimulation it could be kept in a container that isolated it from the nickel, and thus it would not be seen in post experiment analysis of the fuel. There are numerous reports of the effectiveness of radiation stimulation of LENR. The problem is that shielding merely attenuates gammas. Some always gets through. This could be detected externally. Almost all (99.88%) C60 decay is 0.32 MeV beta followed by 1.12 MeV gamma, followed by 1.33 MeV gamma. About 0.12% is 1.48 MeV beta followed by 1.3325 MeV gamma. I think the lead shielding was stated by Rossi to be 2 cm thick, but don't have a reference handy. Very roughly, the mass attenuation coefficient in lead for 1.12 MeV gammas is about 0.062 cm^2/gm and for 1.33 gammas about 0.057 cm^2/gm. The linear attenuation coefficient mu for 1.12 MeV gammas is (0.062 cm^2/gm) * (11.4 gm/cm^3) = 0.707/cm, and for 1.33 MeV gammas is (0.057 cm^2/gm) * (11.4 gm/cm^3) = 0.65/cm. The gamma attenuation, from internal intensity I0 to external intensity I at distance x is given by: I = I0 * exp(-mu * x) so for 1.12 MeV gammas we have: I = I0 * exp(-(0.707/cm)*(2 cm)) = I0 * exp(-1.414) I = I0 * 0.243 for 1.33 MeV gammas we have: I = I0 * exp(-(0.065/cm)*(2 cm)) = I0 * exp(-1.3) I = I0 * 0.88 There is in effect (assuming no calc. errors on my part), with regards to either safety or detection, no practical attenuation offered for cobalt 60 gammas by 2 cm of lead. However, cobalt 60 is not the only possibility given the radioactive source is (and must be to avoid post experiment detection) physically isolated from the nickel. There is
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Hi, On 13-9-2011 20:44, Horace Heffner wrote: snip calculation of lead shielding Hmmm, is there a way to start and stop a gamma radiation source, as it may be used only to trigger the process? Why I'm asking, well I remembered an older message earlier this year from Jed. Sorry, for the long text which I dug up from my personal mail-archive, unfortunately it seems not to available in the vortex-mail-archive. Kind regards, MoB On 16-2-2011 20:48, Jed Rothwell wrote: Here is a revised version of the message I sent the other day. Villa reported no gamma emissions or other radiation significantly above background from the Rossi device. Celani, however, said that he did detect something. Here are the details he related to me at ICCF16, from my notes, with corrections and additions by Celani. Celani attended the demonstration on Jan. 14. The device did not work at first. He and others were waiting impatiently in a room next to the room with the device. He estimates that he was around 6 m from the device. He had two battery-powered detectors: 1. A sodium iodide gamma detector (NaI), set for 1 s acquisition time. 2. A Geiger counter (model GEM Radalert II, Perspective Scientific), which was set to 10 s acquisition time. Both were turned on as he waited. The sodium iodide detector was in count mode rather than spectrum mode; that is, it just tells the number of counts per second. Both showed what Celani considers normal background for Italy at that elevation. As he was waiting, suddenly, during a 1-second interval both detectors were saturated. That is to say, they both registered counts off the scale. The following seconds the NaI detector returned to nomal. The Geiger counter had to be switched off to delete overrange, which was 7.5 microsievert/hour, and later switched on again. About 1 to 2 minutes after this event, Rossi emerged from the other room and said the machine just turned on and the demonstration was underway. Celani commented that the only conventional source of gamma rays far from a nuclear reactor would be a rare event: a cosmic ray impact on the atmosphere producing proton storm shower of particles. He and I agreed it is extremely unlikely this happened coincidentally the same moment the reactor started . . . Although, come to think of it, perhaps the causality is reversed, and the cosmic ray triggered the Rossi device. Another scientist said perhaps both detectors malfunctioned because of an electromagnetic source in the building or some other prosaic source. Celani considers this unrealistic because he also had in operation battery-operated radio frequency detectors: an ELF (Extremely Low Frequency) and RF (COM environmental microwave monitor), both made by Perspective Scientific. No radio frequency anomalies were detected. I remarked that it is also unrealistic because the two gamma detectors are battery powered and they work on different principles. The scientist pointed to neutron detectors in an early cold fusion experiment that malfunctioned at a certain time of day every day because some equipment in the laboratory building was turned on every day. That sort of thing can happen with neutron detectors, which are finicky, but this Geiger counter is used for safety monitoring. Such devices have to be rugged and reliable or they will not keep you safe, so I doubt it is easy to fool one of them. Celani expresses some reservations about the reality of the Rossi device. Given his detector results I think it would be more appropriate for him to question the safety of it. When Celani went in to see the experiment in action, he brought out the sodium iodide detector and prepared to change it to spectrum mode, which would give him more information about the ongoing reaction. Rossi objected vociferously, saying the spectrum would give Celani (or anyone else who see it), all they need to know to replicate the machine and steal Ross's intellectual property. Celani later groused that there is no point to inviting scientists to a demo if you have no intentions of letter them use their own instruments. (Note, however, that Levi et al. did use their own instruments.) Jacques Dufour also attended the demonstration. He does not speak much Italian, so he could not follow the discussion. He made some observations, including one that I consider important, namely that the outlet pipe was far too hot to touch. That means the temperature of it was over 70°C. That, in turn, proves there was considerable excess heat. McKubre and others have said the outlet temperature sensor was too close to the body of the device. Others have questioned whether the steam was really dry or not. If the question is whether the machine really produced heat or not, these factors can be ignored. All you need to know is the temperature of the tap water going in (15°C), the flow rate and the power input (400 W). At that power level the outlet
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
So we need something that ionizises or atomizes the hydrogen molecules, and something that is very unusual for hydride systems. If the Rossi reaction turns out to be centered on highly excited hydrogen atoms... The bumpy surface of a nickel lattice will “field-ionized” the Rydberg atoms in a highly excited hydrogen envelope that hug the surface of the nano-particle. This phenomenon may be visualized as arising from the interaction of the Rydberg atom with the electric fields due to its electrostatic “image.” Compared to a hydrogen atom in the ground state, a Rydberg atom has an enhanced susceptibility to these fields. This is because the Rydberg electron experiences a greatly reduced electric field from the ion core due to their larger average separation. Polycrystalline metal surfaces of the nickel lattice of large micro particles will generate inhomogeneous “patch” electric fields outside its surface. These electrostatic fields also influence Rydberg atoms, potentially causing both level shifts and ionization and competing with the more intrinsic image charge effects. In general, patch fields arise from the individual nano-grains of a polycrystalline lattice surface exposing different crystal faces of the individual nano-crystals. Each of these faces has a different work function due to differing surface dipole layers. For example, Singh-Miller and Marzari have recently calculated the work functions of the (111), (100), and (110) surfaces of gold and found 5.15, 5.10, and 5.04 eV, respectively. These differing work functions correspond to potential differences just outside the surface beyond the dipole layer. Consequently, charge density must be redistributed on the surface to satisfy the electrostatic boundary conditions, producing macroscopic electric fields. While patch fields were first discussed extensively in the context of thermionic emission they are present near polycrystalline metal structures of any type, including electrodes and electrostatic shields. A bumpy nickel lattice surface provides Rydberg atoms with the same spill over ionization function that palladium does for ground state H2 atoms and it keeps the ionization localized on the surface of the nickel lattice. If you are interested in this subject read this paper for more theoretical background: *http://arxiv.org/PS_cache/arxiv/pdf/1008/1008.1533v3.pdf* To amplify the production of Rydberg atoms, I would use potassium or lithium catalysts as a dopant in the hydrogen envelope. Rossi has put together many different mechanisms that all work together to amplify the cold fusion process. The secret catalyst is only one of his tricks. It will not function on its own hook unless optimally combined with all the other mechanisms; the nano surface of the micro particle catalyst surface preparation being just one. By the way, Rossi has said many times that no radioactive materials are used in his reactor. Best regards, Axil On Tue, Sep 13, 2011 at 3:50 PM, Peter Heckert peter.heck...@arcor.dewrote: Horace, thank you very much. I dont have the knowledge to calculate this. Only know the very basics. Found this via google: http://itcanbeshown.com/**NERS425/Lab5/Shielding%20-%** 20Final%20Version.pdfhttp://itcanbeshown.com/NERS425/Lab5/Shielding%20-%20Final%20Version.pdf There is data about screening. My idea was, it could be a very small and weak source, such as used in schools for physics lectures. These are not too dangerous if shielded. If the gamma source is very small the intensity should also decrase by square of distance. The source could be very close to - or inside the nickel powder core . Then the lead and the distance together could shield it close to natural baseline level. Its just an idea. I dont know if this is possible. It also was an idea of me, that metal hydrides are very well researched. Metal hydride hydrogen storage systems are in use worldwide and they use specially developed alloys also such that use nickel as a component. These sytems are belived to be the most secure devices, melting or explosion or abnormal heating is not reported. Some of these are used with very high pressure and temperature. So there are already thousands if not millions man-years of experience, RD and scientific research done for metal hydride systems. systems. But only the LENR researchers find LENR reactions. Why? If LENR reactions where easily to achieve, then this should have been discovered. The developers try to reduce the thermal hysteresis in the load/unload cycle to get best efficiency. So they search for zero hysteresis. When there is LENR energy production then we should have negative hysteresis and if this is possible by common chemical or physical methods, the countless researchers and scientists should have discovered this method or catalyzer. Now, so the Rossis catalyzer must be something very unusual that nobody would ever try to use for a metal
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Am 13.09.2011 22:47, schrieb Man on Bridges: Hi, On 13-9-2011 20:44, Horace Heffner wrote: snip calculation of lead shielding Hmmm, is there a way to start and stop a gamma radiation source, as it may be used only to trigger the process? There is no other way than shielding or increasing the distance. Rossi could inside use a shield that is moved electrically or by heat (bimetal). Or he could control the distance to the gamma source. If it is a very small point source the /local/ intensity of radiation could be changed by factor 10^2 or 10^3. Peter
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
On Sep 13, 2011, at 11:50 AM, Peter Heckert wrote: Horace, thank you very much. I dont have the knowledge to calculate this. Only know the very basics. Found this via google: http://itcanbeshown.com/NERS425/Lab5/Shielding%20-%20Final% 20Version.pdf There is data about screening. Yes. They provide an *average* linear attenuation coefficient for Co60 of 0.694 cm−1, which is 0.694/cm. I provided estimates of .707/ cm and 0.65/cm for the two differing types of gammas. My estimates were based on numbers I pulled off a graph of attenuation by mass coefficients, so have some error. Still, I am right on the range provided by the paper you reference. The bottom line is that 2 cm lead provides practically no shielding for Co60 gammas. Using the attenuation coefficient from your referenced article of 0.694/cm, we have: I = I0 * exp(-(0.694/cm)*(2 cm)) = I0 * exp(-1.388) I = I0 * 0.25 This means 25 percent of Co60 gamma radiation would get through the Rossi lead shielding. My idea was, it could be a very small and weak source, such as used in schools for physics lectures. These are not too dangerous if shielded. If a 2 cm thick lead shielded source has even a very modest amount of Co-60 then detectors nearby will detect the gammas - at all times. If the gamma source is very small the intensity should also decrase by square of distance. The source could be very close to - or inside the nickel powder core . Then the lead and the distance together could shield it close to natural baseline level. Its just an idea. I dont know if this is possible. This is not possible if the source produced enough radioactivity to have any effect. Cosmic rays are very detectable and energetic enough to have an effect on LENR if a near background level of Co-60 can have an effect. A a background level of radiation the radiation would have to trigger a significant chain reaction to produce measurable heat. Cosmic rays should trigger such a chain reaction too. It is also notable that Co60 gammas have enough energy to trigger positron-electron pair generation. Coincidence counters were placed up close to the experiment (one on each side) and detected none. It also was an idea of me, that metal hydrides are very well researched. Metal hydride hydrogen storage systems are in use worldwide and they use specially developed alloys also such that use nickel as a component. These sytems are belived to be the most secure devices, melting or explosion or abnormal heating is not reported. Some of these are used with very high pressure and temperature. So there are already thousands if not millions man-years of experience, RD and scientific research done for metal hydride systems. systems. But only the LENR researchers find LENR reactions. Why? Except for rare explosions, LENR researchers for the most part have had difficulty reliably measuring the effects, they are typically so small. Metal hydride storage systems usually require heating systems, involve large temperature variations, and large reaction enthalpies. LENR effects would not even be noticed unless very robust, which is highly unlikely. Secondly, (relatively) very little LENR research has been done on ordinary hydrogen. Most research has been done on deuterium based systems. If LENR reactions where easily to achieve, then this should have been discovered. LENR reactions are *not* easy to achieve at this point (unless of course Rossi is on to something that makes it easy.) The developers try to reduce the thermal hysteresis in the load/ unload cycle to get best efficiency. So they search for zero hysteresis. When there is LENR energy production then we should have negative hysteresis and if this is possible by common chemical or physical methods, the countless researchers and scientists should have discovered this method or catalyzer. Pretty difficult to say, not knowing what Rossi's method is, or even if it works as advertised. Now, so the Rossis catalyzer must be something very unusual that nobody would ever try to use for a metal hydride storage system. Agreed. So we need something that ionizises or atomizes the hydrogen molecules, and something that is very unusual for hydride systems. So I came to the idea it must be a radioactive gamma source or device. And it must be separated from the nickel, but can be very close and very small and can be inside.. There have been numerous reports of limited success with radiation stimulation of loaded lattices. Mostly these involved betas (electrons from electron microscope guns or accelerators), alphas, or neutrons. Also I think, we should not only think about the energy, but also about frequencies and resonances. If Cobalt60 decays into Nickel60, then the gamma radiation spectrum should contain frequencies that are in tune with the resonance
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
On Sep 13, 2011, at 12:55 PM, Peter Heckert wrote: Am 13.09.2011 22:47, schrieb Man on Bridges: Hi, On 13-9-2011 20:44, Horace Heffner wrote: snip calculation of lead shielding Hmmm, is there a way to start and stop a gamma radiation source, as it may be used only to trigger the process? There is no other way than shielding or increasing the distance. Rossi could inside use a shield that is moved electrically or by heat (bimetal). Or he could control the distance to the gamma source. If it is a very small point source the /local/ intensity of radiation could be changed by factor 10^2 or 10^3. Peter The above is incorrect. A 2 cm thick lead shield will only reduce Co-60 gammas by 75%. I = I0 * exp (-0.694 * x) So we want I/Io = 0.01 to achieve 1/100 reduction factor. I/I0 = exp (-0.694 * x) 0.01 = exp (-0.694 * x) ln(0.01) = -0.694*x x = ln(0.01)/(-0.694) = 6.63 It takes 6.6 cm of lead to divide Co-60 gamma intensity by 100. Similarly, it takes about 10 cm of lead (on all sides) to attenuate CO60 gammas by a factor of 1/1000. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Rossi e-cat catalyzer, Gamma rays
Hi, On 14-9-2011 1:20, Horace Heffner wrote: snip calculation Just a thought. Let's suppose Rossi is using a gamma radiation source as a catalyzer. Is it then possible to determine the source (catalyzer) of the gamma source, if the following parameters are known? 1. Maximum allowed gamma radiation level which passes safety certification. 2. Maximum lead shielding thickness used around the reactor. And as verification it would be great if someone could do a gamma spectrum/intensity scan close to the Rossi reactor. Kind regards, MoB
