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Fred,
Given your finding that the filament resistance is about 1
ohm at 2000 K there is no way that more than 8-10 amps is being drawn, even at
full duty. That five watt figure which Naudin is claiming is actually looking
possible, if not likely, at 5% duty factor - if most of the cathode
heating is provided __by the reaction__ itself, and NOT by the meager
current. Would you agree? Perhaps we should not even call this electrode a real
cathode, until we get the wiring schematic.
That would assume that the CMOS unit seen in the image is
PNP. again, we need a schematic diagram.
Some other rough estimates (please post your
suggested corrections):
If the volume of a MAHG tube is about a half liter and the vacuum is 80 torr, then there will be rougly10e20 molecules of H2 in a MAHG tube, about 4 milligrams. If ZPE is being somehow cohered by the hydrogen in the tube (by a bare proton in one theory) and the characteristic ZPE mass/energy level for that transfer of energy is 3.4 eV (half the 6.8 eV ionization potential of virtual Ps), and the net output of the tube is 100 watts from 5 watts electrical input, then how many molecules of gas are participating in the ZPE coherence reaction per given time period, and does this reconcile with the filament temperature which is seen? 100 Watt seconds = 6.24*e20 eV = 100 joules = 24 calories. This would mean that every molecule in the tube was participating about one time per second, or in any give pulse (of 50 pulses per second), then 2% of the molecules will on average have one of the protons go "bare" for long enough to cohere the characteristic photon from ZPE (the Dirac epo field). To convert the characteristic ZPE energy of 3.4 eV into the Kelvin scale of temperature, multiply by 11,605 = ~40,000 degrees K. Bob Fickle estimates that it would take 200 joules to heat the filament to 2000 K - but that is assuming continuous temperature. If the filament were cycling between 2000 K and 700 K at the 50 Hz pulse rate than, yes the 100 joule output from the reaction (not from the input current) should seem to be sufficient to cycle the filament to that temperature for a brief period, as in the graphic chart on the MAHG site. The Mean Free Path for Electrons (or H2 molecules) is harder to determine but the number of electron collisions with hydrogen at 80 torr per linear distance in cm = ~100 per cm/sec at 80 torr, and if are 10 amps of flux, then one ampere is 6.24 × 10e18 elementary charges per second, then approximately one in every 10 collisions of a ballistic electron with an H2 gas molecule will result in a temporary dislocation of one of the protons - which becomes temporarily "bare" for an instant and is "replaced" by a positron in the molecule, while "borrowing" the virtual electron from the Ps, during this instant of time. When the temporary dislocation is over, in a matter of femptoseconds or less, half of the I.P. of the virtual Ps will remain in our 3-space (3.4 eV) - which is the ZPE energy - which as suggested takes the form of an ultraviolet photon, which is immediately down-converted to IR heat - which is the excess heat seen at the tube wall and at the cathode (through the abnormally high heating effect). Since the cool tube wall is critical, and since the
wall is sputtered, it is likely that all this OU happens within a few microns of the wall itself and none of it happens on the so-called cathode - which is just there to provide thermionic electrons - which are the "instigating" particle (being transferred by the molecule as a negative ion? or not ;-) However, this begs the question: is it possible to convert the 3.4 eV photon to electricity without letting it get first diluted down to IR low-grade heat? It also suggests a way to immediately improve the output - a very high surface area at tube wall. Actually the "sputtering" itself does this, to a degree. Maybe that is the key and that this is NOT a true anode, nor is the cathode a true cathode. More craziness to ponder...
Jones
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