Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[gyulasun]

Hi Folks,

Would like to show you some tests made a few years ago on 2 types of ferromagnetic cores by Nali2001 (a member here too).  The video is in two parts, 1st is 9 min and the 2nd is 2 min long.

www.youtube.com/watch?v=SHbQXnXK6Xc     part 1

http://www.youtube.com/watch?v=BsN2sr3U0PY   part 2

The youtube uploader (Raselli1) is not Nali2001, only collected these very useful tests when Nali2001 showed them. Raselli1 has some other videos collected from either NAli2001 or someone else, see his video channel http://www.youtube.com/user/Raselli1/videos?flow=grid&view=0

rgds, Gyula

[Grumage]

I have spent the best part of this afternoon (it's raining anyway!!) going over my results, to find all tests done on my pseudo parrametric transformer to be DUD. So I decided to recheck the tests done with the special russian transformer, these are still perplexing.
Results.

16 V P/P in (earth gounded probe) @ 20Khz
15.9 V P/P out (earth grounded probe) Into 20 W 12 V Halogen bulb.
DC input current, inc amplifier quiescent @70mA, Was 490mA
Output current determined by voltdrop across 1.09 ohm resistor in series with load.  3 V P/P (scope measured).    So 1.5 div by 0.7071 div by 1.09 = 1.946 A

Am I right? I am beginning to doubt myself!!

My reasons for persuing this avenue is because I feel it is the heart of the TK device. And that of SR193.

Hi All, Can anyone please verify my calculations, I think I am beginning to lose it!!

More test's on simple 1 to1 ratio Ferrite ring with a 0.047 mFd capacitor in parallel with output coil. Looked for maximum voltage rise against input trace by varying the applied frequency. Noted current. Then applied a small incadescent bulb to output. The input load current dropped. I did not expect this, I would have thought that any load applied to the output should be carried back to input?
BTW, I am applying full Sinus waveform, not pulsed DC from a PWM.

Comments please.

[SchubertReijiMaigo]

3V P/P = 1.5*2
Then Vpeak = 1.5
I = V/R
Then 1.5/1.09 = 1.3761...


If perfect sine then Irms = Ipeak*1/SQR2
Then 1.3761*1/SQR2 = 0.973...
Then I rms is 0.943 A


Error --> 0.973*2 = 1.946... A


What is the DC input voltage ? since we know that current is 0.490 A
The bulb is at full luminal output or not ?
In standard theory since 15.9/2*1/SQR2 = 5.62... V, the bulb will be underpowered.

[Grumage]

3V P/P = 1.5*2
Then Vpeak = 1.5
I = V/R
Then 1.5/1.09 = 1.3761...


If perfect sine then Irms = Ipeak*1/SQR2
Then 1.3761*1/SQR2 = 0.973...
Then I rms is 0.943 A


Error --> 0.973*2 = 1.946... A


What is the DC input voltage ? since we know that current is 0.490 A
The bulb is at full luminal output or not ?
In standard theory since 15.9/2*1/SQR2 = 5.62... V, the bulb will be underpowered.

Thanks for your input. Dc input to AF amp is 18 vdc. Output into primary winding is 15 V P/P. And yes you are quite right the lamp is not at full brilliance. (visual load purpose only).

[TinselKoala]

 

3 V P/P (scope measured).    So 1.5 div by 0.7071 div by 1.09 = 1.946 A

Schubert has done it correctly. Your error is indicated in red. You need to multiply here, not divide.

1.5 x 0.7071 = 1.06065        the RMS amplitude of the voltage drop across the CVR -- in this step you are "scaling" and leveling the sine wave by chopping off the peaks and filling the valleys, so the resulting "height" number must be less than the starting number, not more.

1.0606 / 1.09 = 0.973...      the RMS current, by applying Ohm's Law.

And it looks like you are supplying nearly 9 Watts input power. I can testify that filament bulbs glow more brightly when powered with HF AC, than they do with the equivalent DC power. I wonder if you are seeing darkening in the glass... my bulbs get darker after a while due to the filament metal boiling off and depositing on the inside surface, turning them black.