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

[gravityblock]

One thing appears to be clear now--the adamant insistence by almost every professor of electrical engineering I spoke with, let alone others, that measuring current across shunts (by measuring the voltage ac the shunt and dividing it by the resistance of the shunt) is the way to go, turned out not to be the case and it has to be understood why (even at frequencies as low as 1Hz where inductance is out of the question).

@Omnibus:  Outstanding work!

Intrinsic Inductance: Sometimes called internal inductance; this inductance is the result of changes in the magnetic field produced from the current in the wire itself. It is not the result of magnetic field changes entering the wire from the surroundings.  Essentially, the wire itself opposes changes to the current through it. Classical theory claims this relationship is linearly proportional to wire length and independent of wire thickness.  Furthermore, simple experimentation teaches that intrinsic inductance, unlike the classically derived equation, is a function of wire thickness.

According to the classical understanding of inductance, if we construct two circular loops of wire, both with the same loop shape, but with different wire gauge, then both should have the same inductance. But this is not the case as demonstrated by the following experimental data which is found in the paper New Induction (ni.pdf).

48 inch        Area (sq. in)      26 AWG wire                     22 AWG wire
perimeter                            (Measured)                       (Measured)
shapes
----------------------------------------------------------------------------------------------
Circle          183                   2253nH                            2055nH
Square        144                   2144nH                           1950nH

Since the thickness of wire does affect the intrinsic inductance, then the classical model for intrinsic inductance is incorrect.  As you can see from the above, a thinner wire will have a higher inductance and a higher resistance than a thicker wire of the same length. 

We need to test different gauge wires of the same length to see if there is a relationship to the OU effect between wire length, wire diameter, and resistance along with the frequency, and if there is, what would be the best ratio, etc.  Then we will need to test different gauge wires with varying lengths of equal resistances along with the frequency.  I think by performing these tests, then the OU effect can be understood and exploited to its fullest.  I hope this can be helpful in some kind of way.  Maybe a hollow copper wire is what we're looking for.  LOL.

Thanks,

GB

[LarryC]

You are using average V and I to calculate P and are not integrating the product of the  momentary V and I values, correct? If so, you're making a major mistake and until it is corrected any further discussion of your results is senseless.

Really, major mistake???, so Rms values are not used in Power calculations, well only in Omni world.

From Electricity 1-7: The Rms values of an a-c voltage or current is the value that will cause the same amount of heat to be produced in a circuit containing only resistance that would be caused by a d-c voltage or current of the same value.

Check http://en.wikipedia.org/wiki/Root_mean_square

I can collect momentary values in my scope and download to Excel, but it would be such a waste of time when Rms values can be used.

When you used a resistor only a-c circuit you still showed OU. How strange, the following is my test using a resistor only a-c circuit.

The first pic shows that the resistance across several carbon resistors is 49.5. Need to add the 1 Ohm amperage resistor, so 50.5 is total resistance.

The second pic shows that the Rms voltage is 6.59.

The third pic shows that Rms amperage is .130.

So with my test circuit, if P=I^2*R= .8534W and P=V*I = .8567W.

No OU with a correct test setup.


Regards, Larry

[IceStorm]

Really, major mistake???, so Rms values are not used in Power calculations, well only in Omni world.

From Electricity 1-7: The Rms values of an a-c voltage or current is the value that will cause the same amount of heat to be produced in a circuit containing only resistance that would be caused by a d-c voltage or current of the same value.

Check http://en.wikipedia.org/wiki/Root_mean_square

I can collect momentary values in my scope and download to Excel, but it would be such a waste of time when Rms values can be used.

When you used a resistor only a-c circuit you still showed OU. How strange, the following is my test using a resistor only a-c circuit.

The first pic shows that the resistance across several carbon resistors is 49.5. Need to add the 1 Ohm amperage resistor, so 50.5 is total resistance.

The second pic shows that the Rms voltage is 6.59.

The third pic shows that Rms amperage is .130.

So with my test circuit, if P=I^2*R= .8534W and P=V*I = .8567W.

No OU with a correct test setup.


Regards, Larry

LarryC , Omni don't want to learn anything , he will say you are wrong even if you show to him 10000000 empirical proof wrote on million books saying you are right.Best thing we can do is letting him realize that by himself.

[Omnibus]

LarryC , Omni don't want to learn anything , he will say you are wrong even if you show to him 10000000 empirical proof wrote on million books saying you are right.Best thing we can do is letting him realize that by himself.

Not at all. As I showed @LarryC is wrong on this issue. If he shows correct measurements and analysis of these measurements I will be more than happy to comment on them and if they are serious I'll take them into consideration. So far arguments are wanting. I will post in a moment results with the current probe and some problems that have to be addressed (real problems, not problems due to confusion as in the recent attempts at debunking).

[Omnibus]

@All,

Here are some data taken with the current probe. As you can see these data are very similar qualitatively to the data I posted earlier when using the shunt. Understandably, these data are the more rigorous, however. We're moving closer and closer to the truth.

Now, from the discussions I had these days with experts all boiled down to the following far-fetched objection (which in some way was implied in @IceStorm's posts, although he probably didn't even realize it because he was emphasizing on some obviously confused proposition). One of my friends came up with the idea that Ohmic resistance may be frequency dependent. He couldn't cite references to back up that supposition but nevertheless he insisted that that may somehow be the case. The suggestion to solve that problem was to do what our own @Omega_0 was insisting on from the get go.

Instead, I carried out parallel measurements of the current using the resistor on the one hand and the probe on the other. First, I used a precision 1Ohm resistor (0.05% tolerance) which was unfortunately wound and, while the coincidence from 1Hz to up to 100kHz went very well (almost perfect), I was starting to see a phase shift and a slight change of amplitude from 100kHz on. Therefore, I resorted to 1Ohm, 10Ohm and 100Ohm metal-oxide RadioShack resistors (the 1Ohm were in fact 10 10Ohm resistors in parallel; I will order on Monday a precision un-wound resistor of 0.05% tolerance). The coincidence between what the probe showed as current and the current measured with the resistor was just perfect. What more can I say? No frequency dependence of the active resistance. Period. Curiously, the 0.001Ohm shunt I got the other day was showing parasitic inductance and that may be the reason for the funny results I reported couple of days back. So, we're all set--no frequency dependence of the active resistance (of course, no one of the experts ever questions what some here are trying to do, namely, that it is only the real part of the impedance, the active resistance that is, that is responsible for the losses in our case, where the core losses, which would contribute even more to the OU, are ignored).