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
