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> > As near as I can tell,
> > all of the math involved is sound.
>
> No it wasn't, as I thought I pointed out.

I'm just going to ignore that.  All mathematical principles were
observed.  I didn't divide one side of the equation and multiply the
other, for example.

> > RMS power *IS* a meaningful concept.  It is used to determine the
> > D.C. power *EQUIVALENT* to the changing waveform power (AC+DC)
> > dissipated in a load.
>
> What you are describing is average power.  The RMS concept is used
> for AC voltage and current to find the equivalent DC voltage and
> current, but this concept does not apply to power.  The result of
> those RMS voltage and current, (or DC voltage and current)
> calculations is *average* power.

I'm beginning to get the feeling that we have a semantic seperation
on this one.  I would view average power as the average voltage
squared over the resistance.  I.e.

4V square wave. 2ohm resistor.

(4V *.5)^2/2 = 2W

where I would see RMS power as AC + DC power.

2W(from the above) + 2V(peak)^2/2ohms
2W + 2W = 4W

I suspect you would call this the average power.  You are saying that
that would be average power, I suspect, because in the case of a
square wave, you happen to get lucky and the full voltage power/2
happens to be the RMS power.

As it happens, the formula I gave also works for this value.

> If you measure the heat coming
> from the resistor, you will get 24 Watts.  If you measure the
> electrical power delivered to the resistor, you will get 24 Watts.
> The 33.94 Watts figure has no physical relevance to this example.
> It is merely the result of shuffling some numbers around in a
> meanigless way.

As I mentioned in the first post, I didn't believe that the equations
I came up with could possibly work due to the second law of
thermodynamics.  Your answer to that was

"Nonsense perpetuated.  Going deeper into this gibberish is a waste
of time."

Well, anyways, I think that the primary stumbling block here has been
a semantic difference.  I view average power as just that: the
average, or DC component.  The sum of the powers divided by the
period of the waveform.  That is what average means.  I see RMS power
as the RMS voltage^2 divided by the load.

What I've said about RMS power being greater than average power holds
true as long as you stick to my definitions (which I hold to be the
correct ones) of RMS and average power.

I think that about clears things up.  Oh, and yes, everyone else I've
ever talked to about this has the same definitions of average and RMS
power as I do.  I guess I shouldn't assume that those are universal.

The rest is moot.

- --Brendan

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