Partnered Output Coils - Free Energy

[synchro1]

You are shameless in how disgusting you can be.  What a horrible spectacle you are for everyone to see.  You are not even addressing your problem.  You stop harassing me with your gutter trash mouth.

If you persist in harassing me I am asking people of good will to report Synchro1 to Stefan.

I'll report you back you two faced scum bag.

[MileHigh]

I'll report you back you two faced scum bag.

The only reportable person will be you.

[TinselKoala]

Quote from conradelectro,

"Result: input of 0.55 mW through primary H1 results in an output from the partnered secondary (H2 + H3) through R2 of 1.4  mW".

Definitive proof that all my allegations about MileHigh are entirely justifiable.

@Conradelectro,

Congratulations! You may be a candidate for the annals of "World Records". 2khz is probably right on a harmonic node. (4/π)2

You are indeed an astounding hypocrite. Conrad is a "bootlicker" when his results don't please your preconceptions, and when they do, he's a candidate for "world records."


Please show how (4/π)2 relates to 2kHz in some "harmonic node" fashion.
 
4/π = 1.273239545...     
1.273239545... x 2 = 2.546479089...
Or is the "2" supposed to be an exponent?
(1.273239545...)² = 1.621138939...

None of which come anywhere near 2000 Hzor anywhere near an integer divisor or multiple of it. So explain your use of "harmonic node" and how (4/π)2 corresponds to 2kHz in such a manner.

[Pirate88179]

If Chris is now over at OUR and makes a claim of overunity without backing it up...he won't last long over there either.

Bill

[picowatt]

I did again the test of the coil with both parts of the secondary (H2 + H3) connected in parallel with R2. The resistor R1 (input) should be in the right place this time:

10 Vpp sine wave from function generator, 2 khZ

Vh = 3.04 V
Vt = 3.28 V
Vr = 3.28 - 3.04 = 0.24 V
I = 0.24 / 100 = 0.0024
Ɵ = 17°
Watt through the primary H1 (input) = 0.0024 * 0.24 * cos(17°) = 0.55 mW
V0 = 0.38
Io = 0.38 / 100 = 0.0038
Watt through R2 (output) = 0.38 * 0.0038 = 1.4 mW

Result: input of 0.55 mW through primary H1 results in an output from the partnered secondary (H2 + H3) through R2 of 1.4  mW.

As you might have noticed, it seems to be an OU result. This stems from the rather high Voltage over the primary of 3.04 Volt. This means that the primary H1 has a much higher resistance (impedance) than the 100 Ohm shunt R1.

The Voltage (RMS) over (R1 + H1) is 3.28 Volt, which means that the Voltage drop over the shunt has become very small i.e. 0.24 Volt, indicating that very little current is running from the function generator into the primary circuit.

In the "normal transformer" situation, the current through the primary circuit is about 10 times higher, due to the very low resistance (impedance) of the primary. The resistance of the 100 Ohm shunt R1 is much higher than the resistance (impedance) of the coil H1 in the "normal transformer situation" (see here http://overunity.com/15395/partnered-output-coils-free-energy/msg436051/#msg436051 ).

I do not understand the measurement, there probably is a measuring error which I can not identify at the moment.

More tests are needed, please do not get excited. I am not an expert and the chance is very high that I made a stupid error.

Would be great if someone else (may be TinselKoala) could make a similar replication. I will switch to the audio amplifier to get a higher input, but it will take time, my free time is limited in the next week.

As I recall, EMJunkie was writing about very little current through the Primary if the secondary is in a bucking configuration. And my last measurements (checked and rechecked) show exactly that. The error is not in my numbers, it must be in the measurement set up.

Greetings, Conrad

Conrad,

For input power calculations, should not "Vr" actually be "Vh", and therefore closer to 7mw in?

PW