I think Robin may be right about the details. The explicit detail on the
website is a mention of KH(1./4) in the fourth step of the process
animnation. H(1/4) has an ionization potential of some 435 eV. Someplace I
recall an association of K3+ with H(1/4) but I have not found the reference
yet. There is mention of a regenerative cycle for the 'solid fuel', and
'conventional chemical reactions' without explicit details, an exercist to
be left for we students. An arrow of unreacted H points to the reactor, not
the regenerator. It is possible that the separation of H and hydrinos may be
nothing but a specialized membrane which blocks H but allows the smaller
hydrinos to penetrate.
All this points to a foundation for commercial development.
Mike Carrell
----- Original Message -----
From: "Robin van Spaandonk" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Saturday, April 05, 2008 7:17 PM
Subject: Re: [Vo]:Where's the beef? was: Stupid Academic stunt
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.
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