ENERGY AMPLIFICATION

[br549]

Hi Br549,

Very good, thanks for the efforts. 

So the measurements with the 1 Ohm resistor gives output power of about 3mW (0.055² / 1= 0.003)  If I assume that the input voltage was about 3.9 Vrms again at the now measured input current of 6.373mA then the input power was about 24.8mW,  (COP=3/24.8 i.e. under 1).

You may also wish to consider to tune out the output coils inductive reactance with a similar amount of series capacitive reactance.  And how the input current draw changes when the load changes at the output, that would be also a good info.


rgds,  Gyula

Gyulasun:  So if I use the same one OHM resistor and insert in series with the input, and measure the voltage across it (6.5 mv) and use the same calculation used on the output, (0.006.5² / 1= 0.00004225) than
the input current seems to be 0.042 ma, making the COP = .003/0.00004225 = 71 which doesn't seem right to me.

[Farmhand]

@farmhand,a working perpetual motion device doesnt get energy from nowhere,it sucks in ambient heat on one side and spits it out on the other,chek out the theory page.

Profitis, can you show me one ? I'm as keen as anyone to see one, no joke,

Forest, it's a measurement error. Br549 simply made an error, anybody can make errors, that includes every human on the planet.

Cheers

[Farmhand]

This post below should be deleted. That is the stuff that annoys me, a declaration of OU when there is none and without any checking just assume and declare it as OU and proven.
It's too much speculation, and should not be ignored. That is the post that caused me to reply in this thread. No other.

Cheers

Posted by Madddann

@br549
Congratulations on making an overunity device! Also thank you for sharing it openly!
The Flynn paralell path solid state setup is on my "to build" list for quite a long time and I was about to give it a go in about two months. I thought noone was on this project, but I'm glad I was mistaken.
I knew all along this setup has great potential and you just prooved it... kudos to you!

[gyulasun]

Gyulasun:  So if I use the same one OHM resistor and insert in series with the input, and measure the voltage across it (6.5 mv) and use the same calculation used on the output, (0.006.5² / 1= 0.00004225) than
the input current seems to be 0.042 ma, making the COP = .003/0.00004225 = 71 which doesn't seem right to me.

Br549,

No, I did not say your input power calculation was wrong. It was correct. (What you say as an example for the "input power" (0.006.5² / 1= 0.00004225) just gives the power dissipated in the 1 Ohm resistor due to the 6.37mA input current taken by the input coils, right?  (the 6.37mA may have been 6.5mA now that you measured 6.5mV voltage drop across the 1 Ohm).  The 6.5mV across it is not the input voltage to your transformer's input coils.

However, the 1 Ohm resistor directly across the output coil terminates the output, it is the load, just like your ammeter was in your video: with its 1 Ohm inner resistance the ammeter served also as a load of 1 Ohm.  And in the video you measured 1.5V rms voltage across your ammeter terminals and later you measured 0.055V rms across an outside 1 Ohm resistor. How can you explain this?  The 1 Ohm inner resistance of the ammeter and an outside 1 Ohm resistor should give the same load to the output coils, hence the rms voltages must be the same across both (within a few percent).

Gyula

[br549]

I'm not sure yet if I have a measurement error or not, or if there is, which way the error goes.
The input and output frequency and waveforms are very simular, and I am using identical meters and scope channels to measure both. Even if the meters and scope was not measuring exactly dead nuts, the ratio between the input and the output should be accurate. (apples to apples), since the input conditions and the output conditions are the same??