**user2718218**
1 week ago
1900 volts peak-to-peak, divided by two equals 950 volts.

950 volts times 0.7071 gives you 671.7 volts RMS

0.95 amps divided by two gives you 0.475 amps.

0.475 amps times 0.7071 gives you 0.3359 amps RMS.

671.7 Volts RMS x 0.3359 Amps RMS = 226 watts

The electric motor is drawing 655 watts of power.

Therefore the efficiency of the QEG is 226/655 x 100 = 34.5%.

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overunitydotcom
1 week ago (edited)
Yes, you are right, they are not overunity, at least the lamps do not shine so bright, I would expect at maximum 150 to 250 Watts all in all total , if you compare these lights to the other lights in the room..Their lights are pretty dim..

Either they can not measure correctly or they want to fool the people on and on...

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EsotericScience .
1 week ago
Your calculations for the RMS power are incorrect.

P = V x I = 1900 x 1 (as stated on video) = 1900 VA.

Divide by 2 and times .7071 = 671W.

This would give overunity, not by much but overunity nonetheless. At the very least it's close to unity depending on the exact values, but certainly much greater than 35% efficiency.

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user2718218
1 week ago
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EsotericScience .
No, my calculations are correct. P = V x I if V and I are constants. If you have a resistive load so that the current is always in phase with the voltage, and V is now V(t), some kind of arbitrary repeating waveform, then you have to take the RMS value for the V. The power can be Vrms^2/R or Irms^2 x R or Vrms x Irms, they are all the same thing.

In addition, in looking at the clip, the light bulbs look like there is about 226 watts of power dissipation, not 671 watts. This is subjective from watching a video but if you had to pick the most likely fit, 226 watts fits what is seen in the clip.

If you are in contact with "Allegedly Known as Dave" and James and HopeGirl, this serious mistake needs to be corrected.

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EsotericScience .
1 week ago (edited)
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user2718218
They are not the same. Think about it this way. Lets consider a short time interval for which there is little change in the voltage and current, say around the peak. Lets assume for this example that V=2000V, I=1A then Power dissipated for that time interval is 500 VA. If you use your method the power for the same time interval is 2000x.353 = 707 times 1 x .353 giving 250 VA which is wrong. It's the average of the squares versus the square of the average type issue.

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user2718218
1 week ago
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EsotericScience . At the peak like you are describing, the instantaneous power is 2000 watts. I don't know where you come up with "500 VA." "VA" is normally used for transformers. "RMS" means the square-root of the average of the squares.

Please do a Google search on "measuring AC power across a resistor." Click on the first link and you will be taken to a web page called "ElectronicsTutorials" and you will land in the section called "Resistors in AC Circuits." You will see that they explain exactly what I explained in a previous posting.

Dave and James

*really* got it wrong and it's a serious serious problem that needs to be corrected.

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EsotericScience .
1 week ago
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user2718218
The instantaneous power at the peak is not 2000W it is 1000Vx.5A = 500W.

You are not following my argument, the same argument as above will apply to all other intervals, if you add all the intervals to get the final power your method will give an incorrect total.

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user2718218
1 week ago
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EsotericScience .
Okay I forgot to divide by two for the voltage and the current so I can see where you get the 500 VA, or 500 watts. Can you see a problem with your original calculation of an average 671 watts already? If the peak power over the entire waveform is 500 watts, how can you say that the average power is 671 watts? That clearly is showing you that there is something amiss with your calculation. The average power must be less than 500 watts.

Please do the Google search I suggested and read the link. About two or three screens down you will see the formulas being identical to what I originally stated.

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EsotericScience .
1 week ago (edited)
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user2718218
Ok that's a fair enough point it can't be over 500W. I will have to rethink that.

Looks like they may indeed have a measurement problem, it's hard to be sure with the information we've been given so far.

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TinselKoala
1 week ago
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EsotericScience .
Do you mean it's hard to be sure if it's a measurement "problem" or a deliberate misrepresentation? Remember, they started this whole affair with the firm and solid claim that they had.. as in "we possess now"... a working prototype. Working, in this context, means "working as claimed" which you may recall includes self-running and providing excess power. Self-running.

So I put it to you this way: It is not hard at all to be sure. They are deliberately misrepresenting the measurements as valid indications of "overunity" when they know, or at least James knows, that there is no self-runner. Their present device as shown in this video is pitifully inefficient, very far from OU and in fact, even worse performing than if they simply took another identical drive motor and used their present drive motor to drive the other motor as a generator.

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