Sounds a bit like having a desired result in mind – in advance, but finding that it did not calculate correctly with 2 protons, more were “found” <g>
How many proton masses are required to make the calculation work? At any rate, and given that monatomic hydrogen is an atomic BEC – why not use spillover hydrogen as the active agent instead of protons, or else Miley’s rationalization of IRH (inverted Rydberg hydrogen)…? From: fznidar...@aol.com The mass of several protons acting in concert. In the formula for frequency K/M the M is n times M. I wonder if Rossi's sparking mechanism excites the second harmonic. -----Original Message----- From: Jones Beene < Please define “heavy proton” Thanks From: fznidar...@aol.com It odd. Fixing the wave number at 50nm the frequency finds itself at 5 x 10^^12 hertz. Five terahertz. The system needs heavy protons. A lot of light protons, acting together, does the trick. Heavy loading is required. Fixing the frequency and letting with wave number find itself the active domain length is 17.5 meters. The system only requires two protons acting together. I don't like the long wire. Perhaps both resonances could be hit with a cathode designed like a loop radio antenna. It would act like an inductor and externally add proton mass. -----Original Message----- From: fznidarsic <fznidar...@aol.com> To: vortex-l <vortex-l@eskimo.com> Sent: Wed, Feb 13, 2013 7:47 pm Subject: [Vo]:17.5 meters I have said for a long time that 50 nm was the correct domain for the thermal cold fusion reaction. This reaction occurs at high loading. Using the same math I have computed another resonance. It is for a thin palladium or nickel wire 17.5 meters in length. This resonance is for light loading. I don't know what to make of it. Frank Znidarsic
