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Overunity Machines Forum



Mysterious Resonant Circuit

Started by EMdevices, July 24, 2008, 10:04:51 PM

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0 Members and 5 Guests are viewing this topic.

duff

EM,

So as the impedance of the battery increases, the COP increases.

Have you calculated or measured the input impedance of the oscillator?


Do you think the COP you are seeing is frequency dependant?


-Duff



innovation_station

you all seen this long ago im sure...  i sure have   just never bothered to build it

however im sure it works along the same lines.....

almost the same unit em.... ;D

http://youtube.com/watch?v=gTAqGKt64WM

the joule thief

so what if you took the output of one of thease units  and used it as your 9v supply.....  or in place of it ...  what will be your return  or your COP


ist

http://youtube.com/watch?v=bRaVI4Solg0

even better vid.... ;D ;D ;D

To understand the action of the local condenser E in fig.2 let a single discharge be first considered. the discharge has 2 paths offered~~ one to the condenser E the other through the part L of the working circuit C. The part L  however  by virtue of its self induction  offers a strong opposition to such a sudden discharge  wile the condenser on the other hand offers no such opposition ......TESLA..

THE !STORE IS UP AND RUNNING ...  WE ARE TAKEING ORDERS ..  NOW ..   ISTEAM.CA   AND WE CAN AND WILL BUILD CUSTOM COILS ...  OF   LARGER  OUTPUT ...

CAN YOU SAY GOOD BYE TO YESTERDAY?!?!?!?!

xee

@ EMdevices
It is the internal resistor temperature that matters, but I doubt it gets to 800 C so it is probably not the resistor. The only other thing I can think of that you might be doing wrong is your calculations. They seem to be correct. But if 17 volts is your peak to peak voltage instead of your peak voltage as you stated, then the power out would only be 0.24 watts.

Vpp = 17
Vp = 17/2 = 8.5
Vrms = Vp / SQR(2) = 6.01   for perfect sine wave
watts = (Vrms)^2 / R = 6.01^2 / 150 = 0.24

Is your peak to peak voltage really 34 volts?  If so, I can't think of any reason why you are not OU.


EMdevices

QuoteIs your peak to peak voltage really 34 volts?

yes xee,  it is and you can see it in the scope shots I posted.

Look, even if we get picky and say the waveform might not be perfectly sinusoidal (and shape matters), so let's assume an even lower voltage, let's say 15 volts peak amplitude, then

Pout = 0.5 * (15 volts)^2 / 147 = 0.765 watts
Pin = 0.54 watts
COP = 0.765 / 0.54 = 1.40

so we are still way into OU teritory, with 40% more energy output.

@all, I'll report more after I try to close the loop.  I will stop by Radioshack and try and buy a rectifyer bridge.  Hopefully it doesn't eat up to much power.



EM

eldarion

Quote from: EMdevices on July 28, 2008, 10:35:36 AM
@all, I'll report more after I try to close the loop.  I will stop by Radioshack and try and buy a rectifyer bridge.  Hopefully it doesn't eat up to much power.

You don't want to do that--a standard bridge rectifier is designed to run at 60Hz and will chew up all your output power at higher frequencies.  Buy some 1N4148 diodes instead and make a bridge rectifier from them.

Just my $0.02--I want to see this work (it has a lot of  similarity to Bob Boyce's original toroid experiment; well 1/3 of it anyway... :D

EDIT: Could you do one small test for me?  I can't help but notice that your setup is completely isolated from Earth ground--if you connect the negative lead of your circuit to Earth ground does the effect remain, diminish, or completely disappear?  I have been wondering for some time if the fact that my bench is grounded could be destroying the very effects I built it to look for... :-\
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