In-Reply-To: <[EMAIL PROTECTED]> On Sun, Jan 30, 2005 at 04:29:47PM +1100, Robin van Spaandonk wrote:
>When H[n=1/3 (or more)] is formed from H, a total of 108.8 eV >is liberated. Of this, 54.4 eV goes to the catalyst, leaving >54.4 eV either in the form of UV, or as kinetic energy of the >hydrino. This change is from n = 0 to n = 1/2, and n = 1/2 to n = 1/3. So each incremental change in hydrino "level" -- i.e., change of n from 1/k to 1/(k+1) -- liberates 54 ev. I thought 27.2 ev is liberated for each hydrino level increase from n = 0 to n = 1/2, or n = 1/k to n = 1/(k+1), where k > 1. But you're saying that the catalyst gets 27.2 ev per level increase and the hydrino, or a UV photon, also gets 27.2 ev. >by the time n gets to e.g. n=1/10, a drop of 2 levels, >such as would be catalyzed by O++, to n=1/12, results in an >energy release of 598 eV, ... >(For n=1/120 -> n=1/122 this is 6582 eV according to Mills). You're saying that the drop from n = 1/10 to n = 1/12 produces 598 ev, or 299 ev per level increment, and n=1/120 to n=1/122 produces 3291 ev per level increment! I thought the energy released for each increase in hydrino level was the same, 27.2 ev -- at least this is the amount that the catalyst has to absorb -- for any change of n = 1/k to n = 1/(k+1). But if the energy released per level increment increases at greater levels, then the original catalysts would only work for the first few levels. What kind of catalyst would absorb 299 ev or 3291 ev? What is the formula for energy released for an incremental hydrino level increase, at a given level?
