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
