The question how the mass loss associated with a fusion reaction is distributed
as heat still seems unanswered. Does Claytor have any ideas?
Requires a force with a longer range than the Coulombic.
-----Original Message-----
From: Jones Beene <jone...@pacbell.net>
To: vortex-l <vortex-l@eskimo.com>
Sent: Fri, Jul 10, 2015 10:05 am
Subject: RE: [Vo]:Re: LENR-forum: Claytor generates increased tritium with
Brillouin technique
From: Bob Cook
Ø
Ø The question how the mass loss associated with a fusion reaction is
distributed as heat still seems unanswered. Does Claytor have any ideas?
There probably is little excess heat in this type of experiment. Almost every
known neutron-free reaction which produces tritium is endothermic. Normal
deuterium fusion should produce copious neutrons plus tritium, as Claytor
acknowledges. He has recently started looking for thermal gain, according to
the video, but will likely not see it without adding lithium-6 and making
neutrons from changes in the parameters.
Of course, the hot deuteron fusion reaction itself would work to fuse to
tritium and produce thermal gain, as does the Farnworth Fusor which is not hot
or cold but somewhere in between - but this is NOT what is happening for
Claytor, or else both helium-3 and neutrons would be evident (not to mention
the super-hot neutron at 14 MeV). Obviously, this must be a unique reaction –
and as such will invoke other possibilities. The Lipinski model, mentioned here
recently provides one theory (but it too produces neutrons).
Bottom line - since Claytor has looked for neutrons and seen none (few) – even
with Li-6, he is unlikely to see excess heat and tritium together. This lack of
net thermal gain is also a function of the inefficiency of needing large pulse
discharges.
The most logical possibility for explaining this situation (if Mills is partly
correct) is that tritium formation requires a fractional species of deuterium
(dense deuterium or pychno) to be made in situ - which reaction has a lower
threshold and does not have the normal He-3 and neutron reaction pathways.
This is completely logical, since the other reaction branches (aside from
tritium) are more energetic than a tritium-only branch - and it follows that
only the least energetic branch would be seen at lower input energy.
Jones
