In reply to Jones Beene's message of Sat, 18 Feb 2012 13:24:05 -0800: Hi Jones, [snip] >-----Original Message----- >From: mix...@bigpond.com > >> 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/vortex-l@eskimo.com/msg63257.html
Quote:- "IOW - oxygen, which want to takes 2 electrons from lead's massive number of 82 (and also can have the same 54.4 eV 'hole') - can interact within these oxidation states to create an energy gap at 54.4 eV ... which, after all, is many times normal redox spreads which go from +1.7 to -1.5." At first I thought "The energy hole in Oxygen of 54.4 eV occurs when losing electrons, not when acquiring them. The same goes for the Lead, so one can't the donor and the other the acceptor.", however obviously the reverse can also happen, i.e. the highly oxidized state of Oxygen (i.e.) O+++ can also acquire electrons and return to O++, and so can the Pb+++ return to Pb, so it's conceivable that e.g. O+++ + Pb -> O++ + Pb+++. (Note that this would require an unbalanced number of electrons, which consequently would need to be absorbed by other reactions) Perhaps you would care to come up with a specific reaction (or set of reactions)? BTW you were correct about Pb and 54.4 eV. In fact it's 54.388 eV, and the actual energy hole value is 4*13.598 = 54.392 eV so they are very close indeed. However the value for Pb is probably only valid for atomic lead, not for metallic lead. Atomic lead may however have been available during combustion of Tetraethyl lead, when lead used to be added to gasoline, so it seems at least possible that cars using leaded gasoline produced more energy than they "should" have. (O++ has a value of 54.934 eV) > >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. Quote from your original email: "Coal has about 870 W-hr per pound when combusted. We are thus imagining that lead when used in this new way has about 522 W-hr per pound." This expresses energy density as energy per unit mass, which may be expressed just as well in eV/amu as in Whr/lb (since we are looking at the ratio; IOW it's ok as long as the same measure is used for both substances; the conversion factor cancels out when the ratio is taken). The volumetric density is irrelevant. (BTW I have a figure of 3720 Wh/lb for Anthracite, 3380 Wh/lb for bituminous coal, see http://www.generatorjoe.net/html/energy.html ). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
