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

[broli]

From the equations I posted you can extract the slope of the linear portion quite easily. To cut to the chase I have attached the equation for the linear part of the energy slope.

@omnibus: To be really honest I don't agree with the fact that theory shows OU. The energy from the source and out of the resistance have the same slope in the equations.

The negative slope is indeed strange. In theory there are two ways to get a negative slope. 1 is to invert the sign of your current data and keep everything else in tact. This will be like having a capacitive system but producing inductive data. Since this is not the case in experiments we can ignore it. Another very subtle way is increasing the phase shift beyond 90°.

[blueplanet]

In reply 3553, you have described the experimental results when R=10 Ohm, and you have shown us the following graph:

http://www.overunity.com/index.php?action=dlattach;topic=8411.0;attach=45744;image

So, Vin =30 V

The current probe suggests that the amplitude of Ir = 5*0.02 = 0.1 A
So, the amplitude of Vr = Ir R = 0.1 * 10 = 1 Volts.

Also, according to the graph, the voltage Vr measured using the shunt resistance method is 1.2 Volts.

Have I missed anything?

(PS. I think I should put Vr = 0.5 V peak-to-peak and Vin = 15 V peak-to-peak. The reason I used Vin = 30 V because I was trying to match the figures to the frequency domain measurement shown in another graph.)

Like I said, Vr cannot be found in any of my theoretical calculations or in the graphs based on these theoretical calculations. Admit you're wrong and move on.

[Omnibus]

In reply 3553, you have described the experimental results when R=10 Ohm, and you have shown us the following graph:

http://www.overunity.com/index.php?action=dlattach;topic=8411.0;attach=45744;image

So, Vin =30 V

The current probe suggests that the amplitude of Ir = 5*0.02 = 0.1 A
So, the amplitude of Vr = Ir R = 0.1 * 10 = 1 Volts.

Also, according to the graph, the voltage Vr measured using the shunt resistance method is 1.2 Volts.

Have I missed anything?

(PS. I think I should put Vr = 0.5 V peak-to-peak and Vin = 15 V peak-to-peak. The reason I used Vin = 30 V because I was trying to match the figures to the frequency domain measurement shown in another graph.)

So, we're indeed done with the theoretical part. There indeed isn't Vr in my theoretical calculations. Now admit you were wrong in suggesting I used Vr in my theoretical calculations and, like I said, move on.

Here you're bringing up a question regarding my experimental results. This is a completely different topic and I'll look into that.

[Omnibus]

@broli,

Another very subtle way is increasing the phase shift beyond 90°.

Please take a look a few posts back. That's exactly what I said. This has to be explored thoroughly because neither an error in R nor an error in C can explain away the experimental results.

One last thing that may cause the observed OU to be in error is incorrectly assuming that what is measured as Vin isn't the real Vin. I mentioned that in still earlier posts (@blueplanet repeats it in his last post). I will look into this more.

As for the slope formula, you're using Vm, assuming perfectly symmetric waves. In fact, it is possible that, even theoretically, the waves to be skewed for some values of R and C, as the calculations above suggest. Therefore the theoretical Ein-t and Eout-t should reflect that. This so far can be achieved by point by point integration which, like I said already can only be provided conveniently in Excel. In Mathematica such integration is also possible but is cumbersome. To do it you have to create lists of integrals for time values (at small increments) along the entire cycle. In Excel it's somehow easier to do and it does show favorable (OU) difference in slopes. Now, @Omega_0 says it may be due to floating point error, @teslaalset appears not to agree. So we have to get to the bottom of this. But using Vm isn't the way. Again, I wish there were some analytical way (not through Vm but by having an expression of the curve from actually integrated point by point values) to demonstrate that. That'll be the ultimate proof.

[Omnibus]

@broli,

Let me make it a little clearer. Even with perfectly symmetric I and V sine waves, it appears that when doing the point by point integration (rather than the brute force Vm calculation) the Ein-t and Eout-t curves are skewed for some combinations of R and C.