-----Original Message----- From: [email protected] > An atom of Lead is about 17 times heavier than an atom of carbon, so for it to have 60% of the energy density, it would need to produce about 10 times the energy per atom that carbon produces. Since carbon produces about 4 eV / atom, when combined with oxygen to create CO2, the implication is that the lead would need to produce about 40 eV / atom. I think that's a bit of stretch, don't you?
Well, not if we allow supra-chemistry (but I was hoping for this kind of response). If we define energy release via accumulated electron orbitals as "chemical" in a quasi-Millsean way ... almost in the way you quickly mention ... > (Coincidentally, this about equal to the first shrinkage level Hydrino energy ;) Yes, this is what I was getting at - but not quite. The sum of the first three valence electrons of Pd is almost exactly 54.4 eV and lead oxides will be able to manipulate these 3 electrons in a way that is "not exactly Millsean" - as I tried to outline in this message: http://www.mail-archive.com/[email protected]/msg63257.html Robin - for the record - the density of carbon (graphite) is 2.25 gm/cm^3 and lead is 11.35 so the ratio of the two is not much more than 5:1 for those two - but the point is the same - that more than deeper orbitals needs to be involved in an "energy hole" arrangement with hydronium from the battery acid (in a new kind of reduction reaction) to see this kind of gain. Jones
