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

[Omnibus]

Now, back to this I-V phase shift. If the error in measuring current were the inductance then inductance cannot explain the reversal of the phase shift at 47kHz because inductance would increase with frequency and therefore the error due to it would be in one only direction. If we suppose that somehow at low frequencies there's some capacitance overwhelming the inductance and that capacitance decreases with frequency then at 47kHz the inductance and capacitance become equal and we have a condition of resonance. What good does it do, though? Resonance cannot be the cause of OU and, rightfully so, as we saw the OU effect begins to appear away from 47kHz, somewhere around 200kHz and persists to about 700kHz only to vanish again at still higher frequencies. Not an easy explanation of OU in terms of inductance, capacitance and eventual errors due to these. Evidently the core plays some hitherto unknown and unstudied role.

[lumen]

I hope you are correct, but I still have this vision that as some point in the inductance of the two windings, a standing wave is setup with the internal capacitance, and the input impedance causes an increased rejection of the input frequency. This would cause further internal heating without the increase in current into the ground plane where it would have been detected.

I am thinking of an antenna that is out of phase with the input. Nothing changed at the antenna but the reflected wave causes additional loading or rejection of the input frequency.

I do understand that the unit can be viewed as a closed system and what is put into one side must be the same as what is coming out the other. If it is not at the output then it was not at the input.

In the end it would be much easier to place another 100ohm resistor from the signal generator to the input.
The input will be measured from the input resistor and considered the actual input.

If the OU effect is still present, then all else could be eliminated, since increased input current for any unseen reason would result in reduced input voltage and the calculations would show no OU.

[Omnibus]

Like I said, I'll do this experiment on Saturday or on Sunday. If you could only tell me where exactly you want me to put that 100Ohm resistor in my schematic diagram: http://www.overunity.com/index.php?topic=8411.msg242764#msg242764 .

Now, since the reality of the effect is of the essence and nothing else, I still don't see how this

at some point in the inductance of the two windings, a standing wave is setup with the internal capacitance, and the input impedance causes an increased rejection of the input frequency. This would cause further internal heating without the increase in current into the ground plane where it would have been detected.

I am thinking of an antenna that is out of phase with the input. Nothing changed at the antenna but the reflected wave causes additional loading or rejection of the input frequency.

can explain away the effect (if such mechanism in fact does exist). The further additional internal heating you mention due to the rejection of the input frequency is triggered by an energy amount of a magnitude that can be precisely determined if voltage and current are measured correctly. The only concern would be if the current and voltage were affected, without accounting for that, by the mechanism you describe and that would lead to misrepresenting of the actually applied power. Otherwise, why should we care in terms of reality of the effect, as long as the overall balance shows OU. So, the question is, could the mechanism you describe affect the correct current measurement (overestimate the output and underestimate the input)?

[lumen]

I believe this would automatically adjust the voltage to CH2 according to any dynamic load caused by any internal conditions that may result in increased current draw from the source that would not be detected on the other side of the transformer.

If the OU effect still holds, then ........?

I don't know!

[Omnibus]

That new R is equal to the input resistor Rin used to measure the input current and isn't equal to the internal resistance of the pulse generator, correct?

Because if that new R is equal to the internal resistance of the pulse generator (50Ohm) then the OU effect is increased when using the earlier schematic (not that last one with the copper plate as common ground; I haven't tried it here).