Re: "free-energy" mechanism with H2O
In a message dated 7/26/04 5:38:10 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: Ironically, the reason that any of this can happen at all, as Fred has been suggesting, is that a high voltage gradient serves to oscillate the ever-present neutrino flux to the degree that they will interact with neutrons. This has been demonstrated. The neutrinos, however, cannot accomplish this accelerated decay thing within the D nucleus, so there has to be QM "partial" tunneling involved first. This is the determining factor on the rate of the reaction. That is why the Fusor is likely a dead-end street as far as viable LENR device is concerned, unless a larger amount of neutrinos can be focused into the device using much larger HV coils, space around the device. Fred threw this out to Earth-Tech, but they didn't bite. Nikola Tesla used flat spiral coils in reverse, and capacitors in resonance, with the condition of synchronisation to capture neutrinos and materialize them into useable energy: " As quoted from Konstantine Meyl, Scalar Waves, Pg. 207: The answer of the potential vortex theory could turn out as follows: If at the transition of the one wire to the wireless transmission the ring-like vortices is purloined the guiding wire the vortices immediately begin to turn around each other, as is observable for flow-technical ring-like vortices. In this way the ring-like vortex the next moment shows it inside. If it before as an electron vortex (fig. 4.3), then it now shows as positron, if it was negatively charged, then it now is positively charged. Following it oscillates back again, etc. Wit that the ring-like vortex on the average has no measurable charge and no mass, because it alternately forms matter and antimatter. Without interaction it has an enormous ability of penetration. In physics such particles are called neutrinos. Tesla thus had, apart from his transmitted energy wave, which turned out to be neutrino radiation, by chance also caught neutrinos which oscillated synchronously. According to the actual level of knowledge do neutrinos penetrate the earth and appear also on the night side. The order of magnitude in every second amounts to approx. 66 billion neutrinos per square centimeter. It is a true bombardment. If we could be able to collect and convert all neutrinos, the won energy would be entirely sufficient to cover the need for energy of the world population (approx. 27 W/m^2). We merely have to materialize them, thus give them mass, charge, and the necessary localization. Tesla was able to do that experimentally! Let's record: The ring like vortices, which Telsa with his transmitter has sent on a journey as electrons with an open vortex center, are neutrinos (fig 7.12). Telsa requests that transmitter and reciever operate in resonance, thus with the same frequency. Under this condition the receiver collects in all oscillating vortices, so that non one is lost. If the neutrinos for instance are just positively charged, when leaving the transmitter electrode, then an electromagnetic force of attraction takes place, if the receiver electrode at the same time is negatively charged. The required operation with the same frequency and opposite phase guarantees that also the next moment, if both, the neutrino and the receiver have changed their polarity, the electromagnetic attraction is preserved. It is obvious that strange neutrinos which fly past and by chance oscillate synchronously are as well attracted. In that way the power collected in the receiver capacitor will increase further and degrees of effectiveness of over 100% are obtainable. Telsa discharges the receiver capacitor timed with the frequency of resonance (fig. 9.9) and points to the difficultly of an exact keeping of the condition of synchronization. .." Baron Von Volsung, www.rhfweb.com\baron, Email: www.rhfweb.com\emailform.html President Thomas D. Clark, Email: www.rhfweb.com\emailform.html, Personal Web Page: www.rhfweb.com\personal New Age Production's Inc., www.rhfweb.com\newage Star Haven Community Services, at www.rhfweb.com\sh. Radiation Health Foundation Trust at www.rhfweb.com Making a difference one person at a time Get informed. Inform others.
Re: "free-energy" mechanism with H2O
Hi Robin, > Actually I was trying to come up with a way of measuring the ratio of "stripping" > reactions to fusion reactions, but upon further consideration I realize that since > both would produce lone neutrons, this isn't a good measure anyway. That measurement can be done with a neutron spectrometer. In fact this has been done and most neutrons appear to be 2.5 MeV, which is as expected for D+D fusion. This is the main reason that Fusorites don't buy the (formerly flawed) hypothesis for stripping. I think that the earlier hypothesis which neglected the immediate n decay is also what you have been referring to in these posts. But that is not the way this *present version* of stripping should be understood. This is a new twist. I shouldn't even call it a hypothesis yet - as it must be tested, which can be done. Consequently, if what I am proposing happens in fact, the split-neutron will decay too fast to be noticed as a free neutron, and it should not have ever been picked up by any neutron spectrometer anyway, which it wasn't. So far, so good. The most noticeable thing, and the test that would prove my hypothesis, is a magnetic plasma divertor which samples a small portion of the bulk plasma spectrographically. This is probably way too expensive for most small inventors to manage - plus - none of the ones I know are convinced of this hypothesis anyway. Side note: the "not-invented-here" mentality extends even to the garage inventor group... IF protons are seen near 500 KeV and beyond - then yes, I think my suggestion is the only reasonable explanation. These would NOT be the proton which is left over from stripping - that would be much lower energy. Let me make that clear - we are not talking about the proton left over when a deuteron is split into a proton and a neutron. That proton is cold - the putative "hot" proton needed to drive the secondary reaction (D+D) is the new twist on stripping. Instead, these hot protons would be the decay protons from accelerated neutron decay. These are the very particles (along with the betas) that provide the "missing" energy necessary to cause real D+D fusion and real 2.5 MeV neutrons. Make no mistake - real D+D fusion occurs, it just cannot be caused by 20 kv of input, unless there is an intermediary reaction which has heretofore gone unnoticed. That reaction is not stripping, per se, it is accelerated neutron decay. It is even possible that stripping will NOT occur frequently in circumstances where immediate decay is not favored by other circumstances. As stated earlier, for every 2.5 MeV neutron seen, I expect that you will find a multiple of "hot" protons - maybe in a ratio of from 2-1 up to 10-1. These would be your "real" power source, and would be the reason that it doesn't help to raise the voltage of the Fusor, nor to increase the plasma density. In the end, this is a QM reaction and it has nearly maxed-out at ~10^10 decays per second per 10^20 deuterons present (very rough approx). Another way to prove/disprove this hypothesis is to run the Fusor for a few hours, shut it down and run everything through a high precision mass-spec. I believe that the ratio of hydrogen to tritium, or hydrogen to 3He, will be on the order of 20-1 up to 50-1. If you started the run with 100% D2 as fuel, how would you explain this otherwise? Side Note: if the protons which I am suggesting are present were a bit hotter, one might find that boron atoms in a Fusor would fuse. Some might anyway. It should be tried. Proton-Boron fusion (fission) p + 11 B --> 3 alphas + 8.7 MeV is the reaction which probably offers the ultimate energy resource on the earth, because, the fuels are ubiquitous on the earth, cheap and fast neutrons are not generated. It has been shown, however, that bremsstrahlung power losses are too large to satisfy the ignition condition in a solenoid because the atomic number of boron is so large but as to a Fusor, well, let's say it should be tried. The "real" power source of the Fusor is IMHO not the 20 kilovolt input, nor a recycling of the D+D energy (as the neutron carries most of that off anyway) NOR is the "real" power source "stripping" per se." IMHO, the real power source is the nearly instantaneous beta decay of the neutron which has been slightly freed from the proton (QM tunneling effect) and is then subject to accelerated decay. As stated, it is even possible that stripping will NOT occur frequently in circumstances where immediate decay is not favored by other circumstances. Ironically, the reason that any of this can happen at all, as Fred has been suggesting, is that a high voltage gradient serves to oscillate the ever-present neutrino flux to the degree that they will interact with neutrons. This has been demonstrated. The neutrinos, however, cannot accomplish this accelerated decay thing within the D nucleus, so there has to be QM "partial" tunneling involved first. This is the de
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Mon, 26 Jul 2004 07:26:26 -0700: Hi, [snip] >> So there are lots of 700 keV electrons detected? > >How would you detect these in a Fusor? By putting a small instrument portal in the wall? > >If the question is, "are secondary gammas detected" the answer is yes. Actually I was trying to come up with a way of measuring the ratio of "stripping" reactions to fusion reactions, but upon further consideration I realise that since both would produce lone neutrons, this isn't a good measure anyway. A better alternative might be to measure the ratio of neutron production to that of He3 production. Since, if I'm not mistaken, the reaction you propose would result in D splitting into N + P, while a fusion reaction would result in N + He3. If the ratio of He3 to N is small, then we can assume that the D -> N + P reaction predominates. Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Mon, 26 Jul 2004 07:22:10 -0700: Hi, [snip] >> But fusors don't depend on Maxwellian tails. All the deuterons get accelerated to >> thousands of eV, which equates to a temperature of approx. 1E8 K. > >NO!! You are confusing kv with KeV. All Fusore depend on Maxwellian tails. The actual >average plasma temperature of the Fusor is less than 2 eV Then they are operating with way too high plasma densities. The original intent was that bare nuclei be accelerated by the full voltage of the device, shoot right through the centre, and come out the other side, from where they would be reversed and sent back again. All of which would result in the occasional head on collision, and hopefully, a fusion reaction. If the average temperature in practice is only 2 eV or less, then this has to be because most fast moving ions are being scattered before they can pass through the core, which in turn implies that the density is way too high. Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
> So there are lots of 700 keV electrons detected? How would you detect these in a Fusor? If the question is, "are secondary gammas detected" the answer is yes.
Re: "free-energy" mechanism with H2O
Robin, > As they approach, the first bond grows weaker as the second bond grows stronger:- N > P NP followed by N PNP, and 2 D has become N + He3. The force required to tear > the D apart is supplied by the growing force between the P and the other D. The net > reaction is exothermic, and the other reaction resulting in P + T works about as > well. I can't disagree except to say that the reason the two got that close may involve an 'extra' energy component which can be supplied by the stripping reaction. Now can you figure out how do we get helium4 in a three body reaction? Jones
Re: "free-energy" mechanism with H2O
Robin, > I still don't see how this can happen without supplying the 2.2 MeV binding energy > of the deuteron. The neutrino oscillation, which is elevated and instigated by the electric field, provides whatever energy is needed, but it is in the KeV range > >Whoa, Robin! You are missing the major point of post - the beauty of properly > >engineered *decay energy* over fusion energy. > I think this is a case of beauty being in the eye of the beholder. That is always a distinct possibility... for me impartiality demands some thorough explanation, and I have yet to see a better explanation, or even another thorough explanation, but admittedly this is from my POV and I may be the only observer that feels this way - with the possible exception of Fred Sparber. > >There is a HUGE advantage of not having to create fusion conditions, and not > >requiring rare metals like palladium and not having toxic ash like tritium. Sure > >D+D gives slightly more net energy per nucleon, but who really cares when your > >energy-multiple is already way in excess of a million to one?... > Conservation of mass/energy tells me that this is an imaginary reaction. Conservation is clearly present but the reactions are nuclear > I suspect you will find that all stripping reactions are in fact fusion reactions > where half of the deuteron is absorbed, thus providing the energy required to split > the deuteron, and the other half flies free. (Usually the neutron is absorbed). Another possibility, or else anoptehr candidate for not 'imaginary' just incomplete accounting... > However I won't deny the possibility that this can occur under the equivalent of CF > conditions, and yes, I agree that any resulting clean reactions would be well worth > considering. My only real point... > If you can accelerate decay, then you might note that several Pb isotopes should at > least in theory be unstable to alpha decay, though with enormously long half lives. There are better decay candidates - potassium and rubidium. > You wouldn't have reference to Oppenheimer's work would you? None of their work is online, but a google search will help and there is a lot of (misleading) information on Fusor.net > But fusors don't depend on Maxwellian tails. All the deuterons get accelerated to > thousands of eV, which equates to a temperature of approx. 1E8 K. NO!! You are confusing kv with KeV. All Fusore depend on Maxwellian tails. The actual average plasma temperature of the Fusor is less than 2 eV Jones
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700: Hi, [snip] >That is to say, in stripping, the energy deficit that appears to prohibit the >reaction from happening in the first place comes from the energy that should have >been left over once it happened. Of course it does, except it's not a case of before or after, but rather both at the same time. Imagine two deuterons aligned NP NP. As they approach, the first bond grows weaker as the second bond grows stronger:- N P NP followed by N PNP, and 2 D has become N + He3. The force required to tear the D apart is supplied by the growing force between the P and the other D. The net reaction is exothermic, and the other reaction resulting in P + T works about as well. Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700: Hi, [snip] >The two physicists found that, when a deuteron is fired into a target atom even >weakly, the neutron of that atom can be stripped off the proton and penetrate the >nucleus of the target. [snip] This is actually the description of a fusion reaction, where the energy to split the deuteron is derived from the formation of the new nucleus. (The splitting and the fusion in effect occurring concurrently). Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Sat, 17 Jul 2004 12:48:30 -0700: Hi, [snip] >5) The neutrons measured in a Fusor are indeed mostly fusion neutrons, but they >represent a small proportion of the total neutrons which have been produced A 2.5 MeV neutron/proton has enough energy to break apart a deuteron in a collision, which uses up 2.2 of the 2.5 MeV, resulting in much lower energy neutrons. Also collisions which don't result in a such a split will transfer considerable kinetic energy, because the deuteron is only ~twice the mass of a neutron. > >6) Most of the stripped neutrons are 'overloaded' and decay in milliseconds. > So there are lots of 700 keV electrons detected? Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Fri, 16 Jul 2004 07:49:44 -0700: Hi, [snip] >a variety of beta decay - not markedly different from the situation where the neutron >became partially disengaged and decayed. There are QM reason why the neutron cannot >itself decay within the confines of the D nucleus; which is, I suppose, what you are >getting at. But there are also QM reasons why the neutron is periodically disattached >far enough to allow decay - the double-low-probability accounting for the presumed >rarity. I still don't see how this can happen without supplying the 2.2 MeV binding energy of the deuteron. [snip] >> >Where is all this leading? [snip] a gallon of water has the energy equivalent of >> >250-300 gallons of gasoline - and if only one in a hundred neutrons is utilized >> >for its decay energy, the water still has an energy content equal to about 3 >> >gallons of gasoline. > >> More if the neutrons are allowed to fuse with other nuclei, yielding on average >> about 6-8 MeV per neutron. Considerably more than if they are allowed to decay. > >Whoa, Robin! You are missing the major point of post - the beauty of properly >engineered *decay energy* over fusion energy. I think this is a case of beauty being in the eye of the beholder. > >There is a HUGE advantage of not having to create fusion conditions, and not >requiring rare metals like palladium and not having toxic ash like tritium. Sure D+D >gives slightly more net energy per nucleon, but who really cares when your >energy-multiple is already way in excess of a million to one?... Conservation of mass/energy tells me that this is an imaginary reaction. I suspect you will find that all stripping reactions are in fact fusion reactions where half of the deuteron is absorbed, thus providing the energy required to split the deuteron, and the other half flies free. (Usually the neutron is absorbed). However I won't deny the possibility that this can occur under the equivalent of CF conditions, and yes, I agree that any resulting clean reactions would be well worth considering. > and the related problems of setting the stage for fusion are so much more demanding > than with accelerating the rate of decay, which is trivial by comparison. If you can accelerate decay, then you might note that several Pb isotopes should at least in theory be unstable to alpha decay, though with enormously long half lives. > >> > In a Farnsworth type Fusor, it has been proven beyond any doubt that a non-static >> > electric field of 10,000 volts per CM will result in a lot of free neutrons. > >> Probably as a result of fusion reactions:- > >Yes, but the proportion of neutrons due to stripping, as opposed to real fusion has >never been anything more than guesswork. Richard Hull will tell you it is 100% fusion >- but despite his expertise, I think from Oppenheimer's old and largely neglected >work, that the correct figure is closer to 10% with fully 90% from stripping, most of >those going unmeasured as they decay in situ creating some of the AMAZING efficiency >of the device. You wouldn't have reference to Oppenheimer's work would you? > As you know, the threshold for D+D is supposed to be in the several MeV range and > yet we have it found it in the Fusor "on the tail" in the low KeV range - why? \ Primarily because you only need to get two deuterons near enough for fusion to take place which releases more energy than is required to split a deuteron, and 5-10 keV is enough to overcome Coulomb repulsion and get them close enough to increase the probability of tunnelling to within a human lifetime. ;) >well, some of that efficiency may be due to that beta decay electron coming in at >much higher energy (as the maxwellian tail of the energy distribution is nowhere near >long enough, otherwise). But fusors don't depend on Maxwellian tails. All the deuterons get accelerated to thousands of eV, which equates to a temperature of approx. 1E8 K. > >And, of course, the Fusor was mentioned by analogy, as it does not look like a device >which can be pushed above unity because of that issue you mentioned - mean free path >- which requires very low energy density (same as with BLP) > >> And why do you believe that field gradient alone is sufficient [to free a neutron] > >No. I believe that field gradient alone will NOT be enough, but there is a chance >that when combined with other synergistic field-effects, and/or acceleratd QM >effects, such as spin/isospin disruption - that neutron stripping can be accomplished >efficiently enough to allow net energy production. What is "spin/isospin disruption", and do you have an example of where it happens? > >And the main point of the previous post is that the secondary methodology, once you >have some free neutrons, should be from accelerated decay of the neutron, not fusion >and not neutron absorption (unless it something clean like boron) I wouldn't have suggested a dirty f
Re: "free-energy" mechanism with H2O
In a message dated 7/17/04 12:55:41 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: Executive summary: 1) High voltage has the demonstrated property of forcing a 'neutrino oscillation.' "There is also a one-sided emphasis on entropic (explosive, radiative, centrifugal) phenomena in physics and consequently also in current technology, to the almost total exclusion of the constructively ordering, non-entropic phenomena such as brought about by implosion and centripetal or vortex motion, Joseph Hasselberger, www.hassleberger.com" Better penetration, with less energy and more efficiency than exploding high voltage and wave energy is imploding vortex energy. Vortex energy can be made by guiding a high voltage wave down a spiral pipe with spiral groves in it to implode the energy around the neutrinos to cause the neutron to implode rather than explode to create more energy from water cold fusion more efficiently and with less energy. Josef Hassleberger has posted at www.hasslberger.com the complete plans to modify a car's exhaust and engine to allow the car to run off of water. The plans work and the US government admitted that they work but claimed that cars running off of water would effect our economy so that the government would cause problems for those who tried to mass produce the cars. A mass produced car in the US that runs off of water would produce many new automotive jobs in the US to build the cars, and reduce our dependency on foreign oil, and probably improve our economy, so that the only reason that the US government will not allow water cars to be produced in the US and USA is because US government works for China, Big Oil Companies, and the Middle East to destroy our economy and not improve it. Baron Von Volsung, www.rhfweb.com\baron, Email: www.rhfweb.com\emailform.html President Thomas D. Clark, Email: www.rhfweb.com\emailform.html, Personal Web Page: www.rhfweb.com\personal New Age Production's Inc., www.rhfweb.com\newage Star Haven Community Services, at www.rhfweb.com\sh. Radiation Health Foundation Trust at www.rhfweb.com Making a difference one person at a time Get informed. Inform others.
Re: "free-energy" mechanism with H2O
From: "explorecraft" > Allegedly > The Farnsworth type Fusor centers around kinetic ions > which apparently free the neutrons by collision. Not exactly. The Farnswoth Fusor is an Inertial Confinement Fusion device which beneifts from spherical convergence, but the nuetrons are not collisional (spallation), per se, according to the expert spectroscopy which ahs been done, but actual (hot) fusion-derived. IOW, it is generally assumed the all the neutrons observed are the result of the two well-known lower energy D+D fusion reactions. In fact most of these do test-out to the expected energy ~2.5 MeV. Many college kids have built Fusors from spare parts that will output 10^5 fusions per second with an input voltage of 10kv minimum, or 20kv to get a large continuours flux. Officially, the textbook "threshold" for this reaction is anywhere from low MeV up to 2.2 MeV, which is the value most often seen. Now, every energy distribution has a statistical 'tail' - and the Boltzmann tail of a Maxwellian 2.2 distro should end in 3-4 std deviations so you can figure that either the Fusor is the largest single anomaly in all of physics... or ... else, that some unknown (hopefully, 'previously' unknown) source of 'hidden' energy is being input to power this reaction above its threshold of 2.2 MeV. What are the candidate inputs? Oh heck, let's cut to the chase. This is going to long enough and complicate anyway. Executive summary: 1) High voltage has the demonstrated property of forcing a 'neutrino oscillation.' One implication of this heresy is to test the hypothesis around power lines. Are nuclear transmutations seen under power lines? Well, believe it or not, there is one fellow in Kansas who has made a lifetime commitment to find out this one fact, and here is his website: http://old.jccc.net/~rhammack/ 2) Neutrino oscillation from the 'massless' variety, into the 'electron anti-neutrino' which is suspected to have rest mass is 3.4 eV, is hypothesized to drastically increase the cross-section for nuclear interaction. The nuclear interaction with a D nucleus is sufficient to both eject the neutron (in a process called stripping - and then with the added high field gradient- to force accelerated decay of that neutron) Although the neutron is net-neutral from a distance, it has a negative near-field which extends out about 2 x 10^-15 m (2 fermi). 3) The fusor is a fairly large volume HV device which should encourage neutrino oscillation and resultant interaction with deuterium nuclei. 4) In a HV field, beta decay can be accelerated a billion-fold under certain circumstances, depending on the degree of previous neutrino interaction 5) On decay of any neutron recently ejected from the D nucleus, it will give up a beta of about 500 MeV and a proton of enough energy to bring about D+D fusion by inertial confinement technique where the 'effective' convergence voltage has been raised a few deviations form the threshold (MeV). 5) The neutrons measured in a Fusor are indeed mostly fusion neutrons, but they represent a small proportion of the total neutrons which have been produced 6) Most of the stripped neutrons are 'overloaded' and decay in milliseconds. Not too long ago, researchers completed experiments at the National Accelerator Facility in Newport News in which a high-energy electron beam interacted with deuterons to resolve details below 1 angstrom. The results indicate that contrary to most theoretical predictions - the deuteron can be adequately described as consisting of two particles loosely bound together into a pulsating dumbbell shape: they concluded that "we don't have to worry about the quarks and gluons" in describing deuteron structure at higher energies. Although the neutron binding energy in the deuteron appears to be 2.2 MeV, plasmas of a few eV and can knock neutrons out of deuterons in such a way that the neutron goes free. That much is not in doubt. In 1935 Robert Oppenheimer and Melba Phillips made a basic contribution to quantum theory, discovering what is known as the Oppenheimer-Phillips effect, and to this day the implications of it are not fully appreciated, even among high energy physicists. In fact, I have a grave suspicions the widely used deuteron plasma cross-section table that is used by physicists all over the world was constructed without correction for neutron stripping reactions. The two physicists found that, when a deuteron is fired into a target atom even weakly, the neutron of that atom can be stripped off the proton and penetrate the nucleus of the target. Before, it had been assumed that since the deuteron and target nucleus are both positively charged, each would just repel the other except in high-energy collisions. The Oppenheimer-Phillips effect suggests that electric polarization, at low energies of impinging deuterons, may act to nullify coulomb repulsion to a certain extent but that the effect is limi
RE: "free-energy" mechanism with H2O
> -Original Message- > From: Jones Beene [mailto:[EMAIL PROTECTED] > Sent: Thursday, 2004 July 15 00:59 > To: vortex > Subject: "free-energy" mechanism with H2O >...X...< > In a Farnsworth type Fusor, > it has been proven beyond any doubt that > a non-static electric field of 10,000 volts per CM > will result in a lot of free neutrons. > That works out to a gradient of only one volt per micron. The enormity of this statement begs for clarifying references. Allegedly The Farnsworth type Fusor centers around kinetic ions which apparently free the neutrons by collision. The phrase > ...a non-static electric field of 10,000 volts per CM... < suggests the release of neutrons by some mechanism of electrodynamics as opposed to collision. This would be a neat trick, if possible. There is a big difference in your suggestion that neutrons could be stimulated into decaying inside a neucleus as opposed to actually stripping them out of the nucleus. Don't get me wrong, I like either solution, but stripping them out seems a bit improbable, unless you could point to some reference reading, which is all I am asking for anyway. - The bottom line is that if we can, at will and with mere electric stresses, strip out neutrons out of a nucleus, something which could be replicated and verified, then natural decay could be observed. On the other hand: If neutron decay is stimulated inside a nucleus, then it will probably be another decade before technology exists to irrefutably prove that. cheers.
Re: "free-energy" mechanism with H2O
From: "Robin van Spaandonk" > How should D decay? a variety of beta decay - not markedly different from the situation where the neutron became partially disengaged and decayed. There are QM reason why the neutron cannot itself decay within the confines of the D nucleus; which is, I suppose, what you are getting at. But there are also QM reasons why the neutron is periodically disattached far enough to allow decay - the double-low-probability accounting for the presumed rarity. BUT...consider that 'setting the stage' for the latter may hasten the former... or vice versa. > >Where is all this leading? [snip] a gallon of water has the energy equivalent of > >250-300 gallons of gasoline - and if only one in a hundred neutrons is utilized for > >its decay energy, the water still has an energy content equal to about 3 gallons of > >gasoline. > More if the neutrons are allowed to fuse with other nuclei, yielding on average > about 6-8 MeV per neutron. Considerably more than if they are allowed to decay. Whoa, Robin! You are missing the major point of post - the beauty of properly engineered *decay energy* over fusion energy. There is a HUGE advantage of not having to create fusion conditions, and not requiring rare metals like palladium and not having toxic ash like tritium. Sure D+D gives slightly more net energy per nucleon, but who really cares when your energy-multiple is already way in excess of a million to one?... and the related problems of setting the stage for fusion are so much more demanding than with accelerating the rate of decay, which is trivial by comparison. > > In a Farnsworth type Fusor, it has been proven beyond any doubt that a non-static > > electric field of 10,000 volts per CM will result in a lot of free neutrons. > Probably as a result of fusion reactions:- Yes, but the proportion of neutrons due to stripping, as opposed to real fusion has never been anything more than guesswork. Richard Hull will tell you it is 100% fusion - but despite his expertise, I think from Oppenheimer's old and largely neglected work, that the correct figure is closer to 10% with fully 90% from stripping, most of those going unmeasured as they decay in situ creating some of the AMAZING efficiency of the device. As you know, the threshold for D+D is supposed to be in the several MeV range and yet we have it found it in the Fusor "on the tail" in the low KeV range - why? well, some of that efficiency may be due to that beta decay electron coming in at much higher energy (as the maxwellian tail of the energy distribution is nowhere near long enough, otherwise). And, of course, the Fusor was mentioned by analogy, as it does not look like a device which can be pushed above unity because of that issue you mentioned - mean free path - which requires very low energy density (same as with BLP) > And why do you believe that field gradient alone is sufficient [to free a neutron] No. I believe that field gradient alone will NOT be enough, but there is a chance that when combined with other synergistic field-effects, and/or acceleratd QM effects, such as spin/isospin disruption - that neutron stripping can be accomplished efficiently enough to allow net energy production. And the main point of the previous post is that the secondary methodology, once you have some free neutrons, should be from accelerated decay of the neutron, not fusion and not neutron absorption (unless it something clean like boron) as these create enormous secondary problems and residual radioactivity which cannot be allowed in a transportation-fuel setting. Even the tiny amount of tritium from CF would likely not be permitted in automobiles, as tritium is hygroscopic and extraordinarily toxic if ingested in water vapor. > Personally, I think that hydrino formation using O++ formed by the spark stands a > better chance. Yes, I agree with that for sure. And... this might be the way to proceed IF you accept all that Mills has claimed, and that he has been up-front in his disclosures. However... and please don't misinterpret this - as I am convinced that the redundant ground state of hydrogen is real - but not necessarily most of Mills' theory, nor necessarily the positive energy balance claimed by Mills, nor even the validity of some of the techniques Mills' has claimed. The Arie de Guess *theory* of hydrino formation, for instance, makes more sense to me, even though he has not yet publically demonstrated a working device either. It may be impossible, in either case, due to low energy density and low cross-section and negative scaling factor. Lately, more and more doubts are creeping in about Mills results, not to mention candor and honesty. When the guy can't tell you the density of HHCs, which should be well known by now, one can only assume that his silence is indicative of having either been caught in a lie or at best needlessly trying to protect something as a trade secre
Re: "free-energy" mechanism with H2O
In reply to Jones Beene's message of Wed, 14 Jul 2004 10:59:15 -0700: Hi, [snip] >two added neutrons (tritium) are rapidly unstable. The yield on D decay is over 1.4 >MeV. As with the neutron, there still exists a considerable measure of uncertainty as >to the precise value of deuterium half-life, but it is not unfair to suggest that >like many other forms of beta decay, the deuterium decay rate, which is difficult to >distinguish from deuterium "stripping" may be influenced by strong proximate electric >fields. 2*H -> D + 1.4 MeV. How should D decay? > >Where is all this leading? > >In a gallon of water, there is about a gram of 'potentially' free neutrons. At any >given time most of these are somewhat firmly attached to hydrogen in a deuterium >nucleus, which exists in one part in 3-6,000 in water, depending on its source. By >the way, just the gram of neutrons in that gallon of water have the energy equivalent >of 250-300 gallons of gasoline - and if only one in a hundred neutrons is utilized >for its decay energy, the water still has an energy content equal to about 3 gallons >of gasoline. More if the neutrons are allowed to fuse with other nuclei, yielding on average about 6-8 MeV per neutron. Considerably more than if they are allowed to decay. > >If both the firmness of that p-n attachment in the D nucleus, and the resultant >half-life of the free n, can be modulated by proximate electric fields, and also by >*spin/isospin* coupling (which sounds exotic, but might end up being a mundane >variable) then many possibilities emerge. In a Farnsworth type Fusor, it has been >proven beyond any doubt that a non-static electric field of 10,000 volts per CM will >result in a lot of free neutrons. Probably as a result of fusion reactions:- D + D -> T + neutron > That works out to a gradient of only one volt per micron. As long as the mean free path in the plasma is large enough to ensure that the D get accelerated to 5-10 keV. >Normally this will also be close to the static field strength needed to change the >decay rate of those neutrons which are freed. Can the two mechanisms ever accomplish >this simultaneously? And how does one create a non-static electric field of this >gradient at interatomic distances efficiently? And why do you believe that field gradient alone is sufficient? [snip] >BTW this post is a continuation of an extended effort to get a handle on any possible >mechanisms which can be utilized to explain the numerous reports of water being used >either as a stand-alone fuel or as an active combustion booster. > >It is admittedly beyond speculative in one sense - grasping at straws even. But I am >convinced from the past few years of experiments and looking at the work of others, >that 'water-fuel' can be and has already been accomplished, hit-or-miss fashion, yet >the variables are so poorly understood that reproducibility (and scientific disdain) >are more pronounced than even in other forms of LENR. Personally, I think that hydrino formation using O++ formed by the spark stands a better chance. > >Those few who have thought about the subject of water-fuel in the context of the >natural deuterium content of water may have noticed my agenda in this speculation, >which is twofold. > >First there is the problem of efficiency, which boils down to how does one >efficiently spread a parasitic electric field over a large mass of molecules, when >only one in 4000 (or less) is active. The partial answer to that may involve using a >very low natural pH, which can be made even lower by compression at the instant of >ignition. If field gradient alone were sufficient, then just about any ion (pair) would do the trick. The difference in voltage between them is on the order of volts, and they can easily be separated by distances on the order of Angstroms, which leads to voltage gradients on the order of billions of volts / meter. [snip] Regards, Robin van Spaandonk Hot fusion is sort of like Heaven, It's the reward you get long after everyone's dead
Re: "free-energy" mechanism with H2O
At 10:59 am 14-07-04 -0700, you wrote: >For whatever reason, the subject of "accelerated radioactive decay" may be the most >neglected concept in free energy research. At the same time, accelerated decay may >offer an immediate solution to the problem of finding an acceptable transportation >fuel, or fuel additive, for the future. Here are some further off-the-wall thoughts >on the prospect that this effort should involve a closer look at both accelerated >decay and water. > >First, I think one reason for the neglect of this subject matter by some scientists >relates to religious fundamentalism. Some religious zealots have co-opted the >'accelerated decay' field for their own agenda (yes, there are a few who do care >about science): which agenda is to justify (unnecessarily) a literal doctrine of >'creationism' - their interpretation of ancient text which should not require >scientific justification to begin with, but nevertheless... >http://www.answersingenesis.org/docs2001/0321acc_beta_decay.asp > >Although it has been mentioned before in less detail, a possible but yet unproven >mechanism to explain energy anomalies with water, including many with other forms of >LENR, involves the fundamental particle, the *neutron*. The neutron is one of the few >candidate particles which will decay leaving a large amount of energy with a small >gamma 'footprint,' so as to be nearly undetectable. > >Lets look only at 'accelerated' decay of the neutron for a moment, and forget fusion >(hot, cold or warm) for the time being. I won't recite here the litany of >reproducible experimental validations for the concept of accelerated decay, except to >say that billion-to1 rate changes have been documented, as in the cite above, not to >mention the Barker patents. And now the concept is being grudgingly accepted by >mainstream physics. > >Free neutrons are unstable with a half life of *about* 636 seconds. Unlike the >disinformation which you will find in many university-level textbooks, I have >emphasized the operative word: "about" because one can find cited 'authoritative' >ranges for the half-life of neutrons in respectable journal literature that go all >the way from 600 to 1000 seconds, a gigantic range for such a fundamental and >important particle... and that doesn't include the 'anomalies,' which are many! >Anecdotal stories from researchers who have tried to determine the neutron half-life, >and given up in frustration, may sound like science fiction. > >At any rate the energy yield is also wildly variable averaging close to 1.3 MeV of >which a fourth to a half is usually carried away by the electron, but the variation >is so great that there is no precise footprint. This is about 500,000 times more >energetic than combustion - actually if you look at from an equal mass standpoint, a >neutron decay gives about four million times more heat, pound-for-pound than burning >hydrogen in air. Although there still exists this considerable measure of uncertainty >as to the precise value of the neutron half-life and why the decay energy is so >variable, but it is not unfair to suggest that- like with many other forms of beta >decay, the rate can be massively influenced by strong proximate electric fields. It >is also probable that the existence of electric fields is the reason why this very >fundamental value, the neutron half-life, cannot be stated accurately and has not >been measured by any laboratory with any degree of certainty. The mass-s! pectrometer, for instance, is an instrument which uses intense electric fields which probably changes the half-life dramatically. > >It is also not well-appreciated that deuterium is unstable, although the half life is >normally deemed to be so long that it can be, and usually is, ignored for most >purposes. Why shouldn't it be unstable? - the added neutron in D serves no real >purpose (except as an energy 'dump' following the 'big bang') and probably interferes >with atomic charge balance, and we know that two added neutrons (tritium) are rapidly >unstable. The yield on D decay is over 1.4 MeV. As with the neutron, there still >exists a considerable measure of uncertainty as to the precise value of deuterium >half-life, but it is not unfair to suggest that like many other forms of beta decay, >the deuterium decay rate, which is difficult to distinguish from deuterium >"stripping" may be influenced by strong proximate electric fields. > >Where is all this leading? > >In a gallon of water, there is about a gram of 'potentially' free neutrons. At any >given time most of these are somewhat firmly attached to hydrogen in a deuterium >nucleus, which exists in one part in 3-6,000 in water, depending on its source. By >the way, just the gram of neutrons in that gallon of water have the energy equivalent >of 250-300 gallons of gasoline - and if only one in a hundred neutrons is utilized >for its decay energy, the water still has an energy con
