we don't know how the current is switcher, and it seems to be not a Delta but a V wiring... there are triac, unknown circuitry.
it is funny to see many skeptics so prompt to assume there is no complexity in the device, just to assume the wattmeter is broke. the wattmeter is much more reliable than our imagination. provided there is practically no HF or DC voltage, even if the is HF and DC current, the wattmeter will do the job. since the transparency case is solved, tha last question is the emissivity curve, and the IR cam algorithm to compute temperature... you can add questions about the convection, but unless you assume the convection of the blank was much more huge than imagined from the radiation estimated, the convection have a non determinant impact on the final result, and mostly it can be a cause of undersestimation... the games about the wires heat loss is laughable compared to the main problem... it is like questioning the weigh of the poo of an elephant in the kitchen, and ignore the impact of the elephant on the floor resistance. 2014-10-29 2:41 GMT+01:00 Jones Beene <jone...@pacbell.net>: > This guy makes an interesting point that is not clear to me. > > Terry, Dave, Bob et al - what do you EEs who have looked at the input > power think about this approach? > > “The currents in the three C1 wires are all equal and they are measured > by the true RMS > > instrument PCE-830. The three heating resistors are also equal and > therefore they will > > all be heated by equal currents, I2. The authors of the report have > assumed that I2 is half > > of the current in the C1 wires. That turns out to be not true. Instead the > full current I1 is alternating between the two wires in the C2 wire > pairs, so the voltage drop will be the same as for a single wire. For > calculation of the resistance Re in the wire system, see paragraph E1in > the spreadsheet and reference. > > *From:* Brad Lowe > > *http://lenr.fysik.org/eCat/COP=1_or_3.pdf* > <http://lenr.fysik.org/eCat/COP=1_or_3.pdf> > > Sent from my iPhone >
