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I have quickly processed the data from Naudin's new test #5 measurements.And I must say I'm a bit confused myself now :p. I calculated the power output using two different methods, one is using the mean the is using the root mean square. Both of these results should have given the same result or atleast very close to eachother, however in these calculations the average is lower than the RMS. Does anyone have a clue what might have caused this?Here's the spreadsheet: http://goo.gl/1Zk1G

The average of the square of the power is clearly 50 watts, and the square rootof that is 7.0711. . . watts. We found earlier that the equivalent heating power ofour circuit – the average power -- was 5 watts, not 7. The RMS value of poweris not the equivalent heating power and, in fact, it doesn’t represent anyuseful physical quantity. The RMS and average values of nearly all waveformsare different. A notable exception is a steady DC waveform (of constant value),for which the average, RMS, and peak values are all the same.It should be noted that the term “RMS power” is (mis)used in the consumer audioindustry. In that context, it means the average power when reproducing a singletone, but it’s not actually the RMS value of the power.

http://www.eznec.com/Amateur/RMS_Power.pdf

Thanks for the pdf however that's not the issue here though. In the pdf he talks about the RMS OF the resulting power waveform which is indeed a pointless thing to calculate. I added this to the previous spreadsheet as an example, but as he states it has no real usage.However calculating the power with RMS current and voltage should give the same result as taking the average (not RMS) of the power waveform as he also states in the pdf.

Or, maybe 1 receiver coil on both sides of the transmitter coil? I dont know if one will take away from the other.Mags

http://www.youtube.com/watch?feature=player_detailpage&v=LbAhUwHvJCEHi Mags, @ All,Here's woopyjump's second attempt video! load 7x400W halogen lamp...Power input = 755 WPower output = 164.5V x 22.3A = 3,668W!!!

Power output = 164.5V x 22.3A = 3,668W!!!

Thanks CInteresting. less than nominal change to the input and the large load coil when adding 100w. AND, it seems so far that his readings and calculations are, what they are.Woopy? Have you delved into finding what freq the heater is driving its coil? Most meters will specify what freq range they can read accurately.quad fi coil. Instead of having 100v lets say at the input across the coil and 50v pot difference between adjacent windings like a bifi, the quad will have 25v between adjacent windings with an input of 100v. Both bifi and quad with the same number of turns.I wonder if the quad acts differently in a transformer?Woopy? How does the quadfi work on its own, without the other bifi coil on the heater?Nice work, as always.

I don't know if this is right or wrong; I don't know how the original data were gathered or how the scope was set. There have been power computation errors introduced by incorrect use of AC coupling in some other cases.....

Could Woopy or somebody try placing a DIODE (not a bridge) in SERIES with one of the wires comingfrom the Bifilar Coil. Perhaps a minimum of a 5 or 6 Amp Diode should do the job just to try it.