Jones, isn't "stored nuclear energy " a new concept? If you fission/split an atom, are you releasing its stored energy? And hence, aren't all the modes of storing nuke energy known?
On Tuesday, June 6, 2017, Jones Beene <jone...@pacbell.net> wrote: > No, your conclusion is both wrong and short-sighted - and apparently you > forgot to actually look at the data and go on memory. > > The data in the last Table shows that in run 2, 5.5 MJ of input was > unaccounted for and could have been stored. You clearly missed that, but it > is the tip of the iceberg. > > The data says nothing about ongoing nuclear changes which could have > reduced the apparent gain in those runs with apparent gain - therefore in > all seven runs, there could have been both exotherm and endotherm taking > place in the same electrode, such that stored nuclear changes were > absorbing some of the gain which was occurring with a delay from prior > stored changes. > > I assume you had read this report years ago but please try reread the > papers again before making unjustified conclusions. But you main error is > the assumption that the entire electrode is either in an endothermic phase > or exothermic, when it is much more complicated and both phases can take > place simultaneously, with only the net effect being recorded. > > Jones > > On 6/6/2017 8:02 AM, Jed Rothwell wrote: > > Jones Beene wrote: > >> Any calorimeter that can measure a positive exothermic reaction of X >> watts can measure an endothermic reaction of -X watts equally well. >> >> Energy storage is ruled out. >> >> >> Not really ruled out. Let's be exact: energy storage by the conventional >> chemical redox reaction is and always has been ruled out - OK - we can go >> that far. >> > > You are missing the point. Energy storage is ruled out because *the data > shows that no energy was stored*. The balance was zero. There was no > endothermic phase. There would have to be such a phase if energy was > stored. It would have to show up in the calorimeter data. The negative > signal would be stronger than the positive signal that followed during the > exothermic phase, because the endothermic phase would be shorter. > > - Jed > > >
