From: Axil Axil
➢ How many fusion reactions of 3 Mev produce that 20 megawatt burst? Hmmm… looks like single fusion reaction would be sufficient for a very short burst. You did not specify a pulse length. Chirped laser pulses reach the attosecond range of time. A Coulomb of charge resulting from an chirped pulse would work out to a single fusion reaction supplying 20 megawatts in one attosecond … ➢ Are there enough deuterium atoms in the volume to produce the energy generated in the burst of fusion? Yes, there are many times more D than needed for a very short 20 megawatt burst from a single reaction, even if the success rate is lowl. The probability of a laser focusing on an HD pair in a sea of hydrogen would be low so the average energy is way less. Of course, the reactor you are referring to may not employ a chirped laser, but the generic answer is a lesson in why one should not confuse power with energy, especially since it is almost always deceptive to use power instead of energy. Mills is the expert on this form of deception. I agree with Robin that a P-P fusion reaction is an extreme long shot, however a “failed P-P reaction” (inelastic collision) could be gainful if mass from either proton is converted into energy. The mass of the proton varies over a narrow range, leaving open the possibility that some mass, for instance gluon mass, could be lost in a certain kind of inelastic collision. <mix...@bigpond.com> wrote: About 1 in every 6500 H atoms is a D atom. If you take the energy of the H+D => He3 reaction, and divide it up among all the H atoms, you get about 800 eV / H atom. I think this would be adequate to explain the Saffire results. IOW there may not be any P-P reaction happening. Regards, Robin van Spaandonk local asymmetry = temporary success
