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
