1850 Watts free energy power ? New GEGENE circuit by JL Naudin shows COP = 2.8

[TinselKoala]

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

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

The average of the square of the power is clearly 50 watts, and the square root
of that is 7.0711. . . watts. We found earlier that the equivalent heating power of
our circuit – the average power -- was 5 watts, not 7. The RMS value of power
is not the equivalent heating power and, in fact, it doesn't represent any
useful physical quantity. The RMS and average values of nearly all waveforms
are 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 audio
industry. In that context, it means the average power when reproducing a single
tone, but it's not actually the RMS value of the power.

[broli]

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.

[TinselKoala]

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.

Well, there are a couple of strangenesses in the spreadsheet. In the first place, the "RMS" values for voltage and current are using this formula: SQRT(Sum(J3:J602)/602)
But there aren't 602 values, there are only 600. This is a minor error, though. It would be better and easier to use
=SQRT(AVERAGE(J3:J602))
to calculate the RMS current and voltage values.

But the multiplication of the RMS current and the RMS voltage equals the _average_ power ..... but it's labelled "Power RMS" in column G.

The "Average Power" calculation in column F includes "negative power" values because of the way it's calculated, and these negative values bring the "average" down:
=AVERAGE(E3:E602)
where the E values are the instantaneous multiplications of the current and voltage readings, some of which are negative. The way to get a true average power here is to integrate these readings (to get the area under the power curve, in Joules) and divide by the time interval to yield an average Wattage. But the spreadsheet apparently can't  integrate or at least I can't find the integral function.

But f you take the absolute values of the instantaneous power values in column E and average those (getting rid of the negatives), you get a figure that is almost the same as the value of the Average Power (column "G" labelled "Power RMS")  calculated by multiplying the RMS voltage by the RMS current. I think the difference is just rounding error.

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.....

Where did these values come from, anyway? Is this a data dump from a DSO?

ETA: Yes, of course they are. I forgot which thread I was in for a moment, sorry.


I made a couple of edits to the spreadsheet:
https://docs.google.com/spreadsheet/ccc?key=0AmhyDBPbnYWudDlYUERKa3l5MFZuVkw1QXFNcGxxOXc
You can revert to the way it was before I got into it by displaying the "revision history" and reverting to the version before "stellanokia" edits.

[crazycut06]

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=LbAhUwHvJCE

Hi Mags, @ All,
Here's woopyjump's second attempt video!    load 7x400W halogen lamp...
Power input = 755 W
Power output = 164.5V x 22.3A = 3,668W!!!

[Magluvin]

http://www.youtube.com/watch?feature=player_detailpage&v=LbAhUwHvJCE

Hi Mags, @ All,
Here's woopyjump's second attempt video!    load 7x400W halogen lamp...
Power input = 755 W
Power output = 164.5V x 22.3A = 3,668W!!!

Thanks C

Interesting. 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. 

Mags