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Overunity Machines Forum



Real OU-Effect to Share with everyone!!!

Started by Magnethos, February 02, 2009, 08:37:03 PM

Previous topic - Next topic

0 Members and 6 Guests are viewing this topic.

zerotensor

1)  The multimeter should be replaced with a scope.  I reckon there is indeed a lot of "bounce" in this setup.

2)  I'd recommend that a proper switch, (perhaps a SCR or isolated rotary contact), be installed before trying the experiment with a whole bunch of caps in parallel, unless you think you might enjoy a potentially lethal shock.  Remember to short-out the array while you aren't using it -- the caps can accumulate a full charge if left alone and unshorted.

good luck and for heaven's sake be careful.

MeggerMan

Energy in a capacitor = 1/2 x C(farrads) x V^2
E = 0.5 x 0.005 x 20 x 20
E = 1 joule
OK so to charge a capacitor of 5000uF to 20V requires 1 joule of energy.
So therefore 1 joule = 1 watt for 1 second.
To achieve it in a millisecond needs 1Kw for 1 mSec, quite a high power density - a scope should be able to show the real duration of charging.

So if you can charge a 5000uF cap to 20V you can discharge this at a constant voltage across a resistor of 1V at 1Amp for 1 second or 10 milliamps for 100 seconds.

Either way I think a scope shot of the cap and coil is required to check the voltage surges.
Now if this can be applied to an ultra-capacitor of say 40KWHrs capacity, now that would be something.
If I get time this evening I will try and scope this.

Regards
Rob

Koen1

Interesting setup Magnetos :)

As for the "splitting the positive" element, that reminds me of Grey's
circuits where he "split the positive" and managed to produce apparent OU...
Is this perhaps a variation on the same theme?

Also, do we or don't we need to use a bifilar coil in this?

Regards,
Koen

jogilbert

so this is basically a car coil type thing but with poor points,if the volts increase is it more power??I think not, unless there is amps then OK watts

Frederic2k1

Quoteso this is basically a car coil type thing but with poor points,if the volts increase is it more power??I think not,

With only voltage and capacitance there is no need to know the current to calculate energy.