The secret to Overunity

[Tajerek]

Hi Tajerek. Unfortunately voltage alone does not cause a capacitor to become charged,
it takes current (flow of charge) to charge up a capacitor, which takes time (as indicated by
the RC time constant). Discharging a cap into a resonant tank circuit also does not (under normal
conditions) provide any more energy than what was stored in the capacitor. In order to achieve OU,
something very out of the ordinary will have to be going on, otherwise it should be easy to achieve OU,
but it seems it is not so easy at all to achieve OU. What do you think is very unusual about what
you are describing that will cause excess energy to be drawn into the setup you have described?

All the best...

the HV source will of course have current but you don't need significant current to push charge into a low ESR high voltage capacitor. If you have doubts try it for yourself. And the time it takes for full recharge is 5*RC time constant which is doesn't depend on current either its formula speaks for itself. It is only function of the resistance and capacitance.

As example 3kv cap of 1uF with ESR of 0.3 ohm can store 4.5 joules in 0.000000015 s. This technically means you can charge it and recharge it 333k times per second. with a frequency up to 333khz. The time can be made shorter still by increasing voltage beyond the rating of the capacitor.

I do not discharge the cap into a resonating tank circuit as you stated, but I discharge it into inductor forming together the tank circuit only during discharge then the circuit is opened to recharge the cap again.

[Void]

the HV source will of course have current but you don't need significant current to push charge into a low ESR high voltage capacitor. If you have doubts try it for yourself. And the time it takes for full recharge is 5*RC time constant which is doesn't depend on current either its formula speaks for itself. It is only function of the resistance and capacitance.

As example 3kv cap of 1uF with ESR of 0.3 ohm can store 4.5 joules in 0.000000015 s. This technically means you can charge it and recharge it 333k times per second. with a frequency up to 333khz. The time can be made shorter still by increasing voltage beyond the rating of the capacitor.

I do not discharge the cap into a resonating tank circuit as you stated, but I discharge it into inductor forming together the tank circuit only during discharge then the circuit is opened to recharge the cap again.

Sorry but that is false. Whether a capacitor has a low ESR or not it still takes a flow of charge
to charge up the capacitor. This voltage times current means an expenditure of energy over time to charge
the capacitor, again regardless of the ESR of the capacitor. Just because a capacitor may charge
more quickly, it doesn't at all mean it takes less energy to charge the capacitor than the capacitor stores. 

When the capacitor is connected across the inductor it forms a tank circuit. The energy that was stored in
the capacitor excites this tank circuit. This is a tank circuit. Whether a separate cap is used to form the
tank circuit is neither here nor there. Sorry, I don't see anything in your described setup that should
be doing anything out of the ordinary, and you haven't described anything so far that appears to be
out of the ordinary. You named this thread 'The secret to Overunity' but I see nothing out of the
ordinary with what you have described so far. Do you have a test setup that you think demonstrates anything
out of the ordinary going on?

All the best...

[Tajerek]

Sorry but that is false. Whether a capacitor has a low ESR or not it still takes a flow of charge
to charge up the capacitor.

the current when charging capacitor is not as the current across a resistive load.

The formula for finding the current while charging a capacitor is:

I=C*dV/dt

it is nothing more than the change of voltage over time. It is not related to the current of the source as much as the voltage.

take 9v source across a 3kv cap you will generate 0.0000405 joules in a time constant. Step up the same voltage to 10kv and put it across the same cap you will generate 5 joules in fraction of the time constant. You stored more energy but the power source is the same...that's the secret

[citfta]

Tajerek,

What you are proposing is very similar to the claims made by Don Smith.  In fact your circuit looks like a copy of one of Don Smith's circuits.  You should know that literally hundreds of people have tried to get at least one of his circuits to work like he claimed.  None of them to my knowledge have ever succeeded.

He made almost exactly the same claims as you.  You charge a cap with high voltage and then discharge it and somehow get extra energy.  There is no known way that discharging a cap will get any more power than was put into the cap.  That is a fact.

You need to actually build something and then prove it to yourself.  No amount of claims will convince most of us with real experience that you have discovered the secret of overunity until you can actually demonstrate it with a working circuit.  And copying Don Smith's work is not going to get you there.

Respectfully,
Carroll

[Tajerek]

There is no known way that discharging a cap will get any more power than was put into the cap.  That is a fact.

That's correct, I am not claiming to generate more than what the cap can store in one cycle, however what is in the cap is not directly related to how much you draw from the source. It is function of the voltage you put across the cap , the combined resistance R and C. Small current is only needed to get the charges across the combined resistance of the circuit, if that's small enough you literally need just voltage.

Don smith showed the devices, I am showing the formulas that prove it and therefore the ways to replicate it scientifically. Don didn't explain the theory and science behind his devices that's probably experimenters were guessing to replicate it. Also I don't think his aircore inductors were as efficient.