Robin van Spaandonk
Tue, 10 Jun 2008 15:37:50 -0700
In reply to Jones Beene's message of Tue, 10 Jun 2008 06:19:12 -0700 (PDT): Hi, [snip] >Where are the gammas then? I see two possibilities (and I suspect Arata of adhering to the first). 1) The cluster of four deuterium atoms creates 2 alphas concurrently (or 1 Be8* that subsequently fissions). This means two fast alphas, with equal energy, and opposite momentum, and presumably no gammas. 2) The energy is carried away by fast electrons that were Mills "shrunken" electrons in a previous life. :) These electrons (two of them?) then lose most of their energy ionizing surrounding atoms. Though it seems probable that at least some of the energy would show up as bremsstrahlung x-rays in this case. (Has anyone looked for it?) [snip] >This is called "Mills-lite" instead of Millsean, since >the redundant shrunken ground-state can be (and >probably is: temporary, and not permanent). This is >also in keeping with Mills experiments, where lots of >UV is seen there, but where water-bath calorimetry can >find only a COP of less than 2 when the ion energy >suggests it should be 40-100. There could be a simpler explanation for Mills' results. Most of the energy they had to add to the system didn't result in shrinkage and excess energy production. In short, the efficiency of the whole was low. However when the amount of excess energy *per consumed atom* is calculated, the result is way beyond chemistry. e.g. It takes 1 kW to create enough catalyst ions and H atoms to liberate an extra 300 W. Most of the ions simply recombine again without catalyzing a shrinkage reaction. Most of the H recombines to H2 without undergoing shrinkage. In short at any one time, only a small percentage of the overall atomic population undergoes shrinkage. That means you need a high power input to maintain the overall population, of which only a tiny fraction is undergoing the desired reaction. As near as I can tell, this has been Mills' main problem since day one. It's also why I suggested to him that he try heating KH. When heated, it breaks up into one K atom, and one H atom, i.e. one catalyst atom and one Hydrino candidate - a perfect match. I seems however that he has progressed beyond that to NaH. (BTW very good insulation of the experiment should also solve the problem, since that would ensure that less energy leaked away, and thus was "reused". There is probably also an optimal temperature at which the reaction operates, which would vary depending on which catalyst was used). [snip] Regards, Robin van Spaandonk The shrub is a plant.