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Overunity Machines Forum



STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

Started by PaulLowrance, December 04, 2009, 09:13:07 AM

Previous topic - Next topic

0 Members and 16 Guests are viewing this topic.

teslaalset

@Omnibus, @broli,

I hope we can now agree on our models.
Sad but true, it can be observed that OU can not be obtained by appying full sinus periods.

However, play with the model and you can see there is still room for OU.
See below graph. It's small, but it's there.
This can be optained by switching off the input at the right moment.
(interupting the full sinus cycle)
Question is, is this feasible in practice?

Question two would be, what if we use an inductor instead of a capacitor?

Update
Seems that an inductor in series with a low value resistor has better options.
See bottom graph.

broli

Quote from: teslaalset on June 30, 2010, 02:30:29 PM
@Omnibus, @broli,

I hope we can now agree on our models.
Sad but true, it can be observed that OU can not be obtained by appying full sinus periods.

However, play with the model and you can see there is still room for OU.
See below graph. It's small but it's there.
This can be optained by switching off the input at the right moment.
(interupting the full sinus cycle)
Question is, is this feasible in practice?

The main reason wasn't uncovering OU from theory, at least not for me. I wanted to have a solid comparison between theory and omnibus's experiment, so we can find out exactly what the difference is and how it causes the apparent OU. Currently I'm leaning towards phase angle. But due to omnibus excel problem I can't analyze his data and compare it to theory. But we at least know that the current amplitude is the same.

I think it will take mainly some mathematical scrutiny to find the difference. A model can be extracted out of the data using statistical magic.

broli

Quote from: teslaalset on June 30, 2010, 02:30:29 PM
Update
Seems that an inductor in series with a low value resistor has better options.
See bottom graph.

From my humble view I would say that the slope matters not the offset. The average slope on both look about the same, the offset is due to initial conditions. An inductor stores maximum energy at maximum current and this can offset the graphs. Average slope is actual Joules / second (aka Watt) and this is what matters.

teslaalset

Quote from: broli on June 30, 2010, 04:12:27 PM
From my humble view I would say that the slope matters not the offset. The average slope on both look about the same, the offset is due to initial conditions. An inductor stores maximum energy at maximum current and this can offset the graphs. Average slope is actual Joules / second (aka Watt) and this is what matters.

@Broli,
Maybe I was not very good at explaining.
Look at the time  moment of the arrow.
Imaging one switches on at time = 0 and off at the time indicated by the arrow.
At the time of the arrow, the total input energy used is almost zero, but the total consumed energy is much higher.
So the ratio output / input energy is much greater than 1.

So, in practice, one should repeat this sequence;
1) Switch the circuit on when input sinus voltage is zero, but increasing.
2) Switch off the circuit when consumed energy is much higher than the delivered input energy.
3) Wait for the input voltage crosses 0 Volt again, so back to 1)

Omnibus

@broli and @teslaalset,

I got it. Posted it in a couple of posts back but you probably missed it. Here it is for 5Ohms. For 1Ohm you got negative slope of the input energy-time curve.