In reply to Mike Carrell's message of Sat, 5 Apr 2008 12:02:25 -0400: Hi, [snip] >Jones, you are a clever and sophisticated observer and can do better than >that if you want to be objective. The voluminous journal papers and >experimental reports are hardly 'vaporware'. They require study. The new >reactor configuration embodies solutions to vexing problems with the earlier >exploratory lab work. Producing H and catalyst [presumptively K3+] in a
K3+ is not a catalyst AFAIK. It is the consequence of the catalytic reaction, the product if you will. Only after it has captured free electrons does it once again become a catalyst. Mills would however mention this product as important, because he sees it as an indication that Hydrino catalysis reactions are taking place. The only reasonable alternative would be the presence of ionizing radiation. This is of course always present to some extent in K due to the decay of K-40. OTOH, such ionizing radiation should also yield a few more highly ionized atoms of K, e.g. K4+, K5+ and their spectral lines should also show up. If these lines are absent, or extremely weak, while those of K3+ are strong, then the catalysis reaction is strongly indicated. Regards, Robin van Spaandonk The shrub is a plant.
