Am 19.07.19 um 19:32 schrieb Jones Beene:
Which is to say: until we get mass spec readings for significant amounts of helium after a long run, fusion remains just a fall-back assumption based on old electrolysis results - and possibly unjustified for anything else.

Surprisingly, even with a reactor operating at 3 kw for extended periods - there is no report of tritium or transmutation of any kind.


In nuclear transformation (LENR) D* adds like a double proton and H* adds like a neutron. That's what we see (exactly measure) from the gamma radiation signature of complex reactions.

Tritium is not possible with most D-D LENR. Only 3-He will be seen. (If you would use neutron rich isotopes then the range of results would be much larger. But this is for large state labs.)

Transmutations always happen as side reaction with about 10E-8 of the reaction power equivalent. We at Ecalox measure LENR gamma radiation since about 2 years now. (> 10000 spectra stored now)


However, Your answer does not answer Robins question about your model’s accurate energy calculations.

Separately there are tables of numerous measured isomeric energy states, which could be compared directly with the detailed calculations of your physical 6-D modeling.   Are you or anyone else doing such calculations in way of validating you models of nuclei?

Bob Cook

Currently I have other priorities than extending my model. But, as a strongly handicapped person I would welcome any help. Such a person would of course learn much more than has been written so far!

There is good reason to believe that a nice piece of software that models the EM collapse of two ring currents in SO(4) could give some deep insight into the nature of physical constants.

Currently I follow a refined approach to model the D-D fusion process that seems to give the same resonance 1000eV as the magnetic moment model and is in good agreement with Mills measurements.


But for this you will have to wait as there are to many open tasks now.

And to remind everybody: A model is only as exact as experiments tell you. Thus I give you "the most likely value" you can see in a simple experiment. An exact value makes no sense as nobody has an exact measurement...

Jürg


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Jürg Wyttenbach
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