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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

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Omnibus

Quote from: broli on June 29, 2010, 09:19:26 AM
Yes this is a good reference now. Any negative averages are real negative values.

There's something else that hasn't been discussed as far as I know. Average value of current and voltage. If dealing with ideal sine waves these should be 0 but from the inductor data you gave me the averages are non zero. I still have to think of how this should be interpreted.

Remember, there was slight offset there due to the pulse generator? That's OK, because without the offset the OU effect would've been greater.

teslaalset

Quote from: Omnibus on June 29, 2010, 09:23:44 AM
@teslaalset,
The question is what will be the analytical expression of the current which the standard theory gives, based on the given expression for V with the given R and C?

Results from my spreadsheet:

Input Voltage = 15.6 x sin(wt)
Current = 0.007880434 x sin(wt-87.12 degrees)
Voltage across the resistor = R x 0.007880434 x sin(wt-87.12 degrees)

Or to make it a bit more scientific notation:
Input voltage 15.6 sin (2*Pi*700000t)
Current 0.007880434 sin (2*Pi*700000t-87.12*2*Pi/360)
Voltage across the resistor R*0.007880434 sin (2*Pi*700000t-87.12*2*Pi/360)

broli

Quote from: Omnibus on June 29, 2010, 09:25:30 AM
Remember, there was slight offset there due to the pulse generator? That's OK, because without the offset the OU effect would've been greater.

The offset should indeed not change power input to output ratio as we have integrated instantaneous values. But I'm just curious as what the direction of the offset voltage is compared to current. Could you imagine that it's found that offset current is going in the other direction than the offset voltage.This requires you to do a controlled DC test. This should tell us what's + and - for the current probe and voltage probe. Then we can take the averaged values and see in what direction they are.

LarryC

Quote from: Omega_0 on June 29, 2010, 03:26:09 AM
Vm and Im are indeed peak values, see the diagram and the dotted line drawn from the peak. 'm' here means max, peak and max are the same thing. V and I (without subscripts) are here used for rms values. Don't confuse yourself.
That integration equation gives "average power" from Vm and Im or Vrms and Irms. Omnibus is not using any of these values in his calculation, there is no Vm, Im  or rms values, neither he uses "average power". These are not there anywhere in his data, neither he measured them (it requires true rms meter or peak meters, the old way), so the standard equations were not applied.

Part of your confusion may be due to my not showing the entire Instantaneous Power section. It is shown below, notice at the bottom average power tag, which brings you to Average Power section where they integrate Vm and Im values over a half-cycle.

Now why would anybody do that if they only have one peak value? M does mean max or peak, it is the momentary value at phase angle difference. I do agree it is pointing at peak value, but only because that is the easiest to show.

If you still don't understand, I don't care now that you finally admitted the following truth.



Quote from: Omega_0 on June 29, 2010, 03:26:09 AM
Let me repeat once again, he is using sampled data, delta-V, delta-I and delta-t, area under the plotted product of samples gives the power precisely, no equations needed.

This must be new thing for you, as this is not taught in schools or text books but experimentalists and pros are familiar with numerical integration techniques. This is my last attempt to justify the calculations, and I'm doing so only to clear the confusion of anyone else reading this thread, not necessarily to prove you wrong or prove omnibus right.

It is a new thing for me. If you had said from the beginning that you do  not want to use established scientific measurement method as taught by Universities, there would have been very little arguments.


Regards,Larry 

broli

Quote from: LarryC on June 29, 2010, 11:51:11 AM
Part of your confusion may be due to my not showing the entire Instantaneous Power section. It is shown below, notice at the bottom average power tag, which brings you to Average Power section where they integrate Vm and Im values over a half-cycle.

Now why would anybody do that if they only have one peak value? M does mean max or peak, it is the momentary value at phase angle difference. I do agree it is pointing at peak value, but only because that is the easiest to show.

If you still don't understand, I don't care now that you finally admitted the following truth.



It is a new thing for me. If you had said from the beginning that you do  not want to use established scientific measurement method as taught by Universities, there would have been very little arguments.


Regards,Larry

If I was your teacher I would have killed myself by now. How you still don't get it is beyond me. You are actually a good example of how crappy the education system has become. You can't rely on your own logic and reasoning skills, yet seek fault where there is no fault.

So really there are only 2 options left,

1) you are an idiot
2) you have some hidden agenda to disrupt progress