> As near as I can tell,
> all of the math involved is sound.

No it wasn't, as I thought I pointed out.

> 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.

> (ex. the power dissipated by a resistor with
> 170V peak sine wave applied to it is roughly the same as the power
> dissipated by a resistor with 120VDC applied to it). That's the
> definition they give you in 1st semester electronics.

Yes.  Note that you are talking about 170 *volts* peak sine having an RMS
value of 120 *volts*, from which you compute the *average* power.

> As I said, I made one mistake.  That was basing everything on summing
> voltage calculations to start with, rather than summing up the power.
>  That's all you needed to say.  I would have corrected myself after
> that.

Well, you are still going on about "RMS power".

> RMS power is *NOT* a flawed concept.  Perhaps you simply don't
> understand it? But you do use it in every day life.  Have you ever
> measured the power dissipated by a common light bulb?  I mean *all*
> of it.  Heat and light.  I *expect* that you'd find it was not just
> the average power, as you seem to indicate.

I certainly expect it is!

> Of course, I can't be
> *sure* since I haven't tried that.  I am, however, relying on what
> was taught to me.  On that basis, if you insist that this is wrong,
> there are mathematicians and physicists that would do a better job of
> explaining to you why RMS is not a "flawed concept" than I can.  I
> know one physicist that is quite well accomplished, who I can,
> perhaps, get to write an explanation if you wish.

OK, let's do an example.  A 0 to 12V square wave is being driven accross a 3
ohm resistor.  There are several ways to compute the average power.  First
we'll do it by finding the RMS voltage, then using P = V**2 / R to find the
power.

  RMS V = sqrt(12**2 / 2) = 8.485V

That means the resistor is dissipating the same power as with 8.485V DC
applied.  The power is:

  average power = v**2 / R = 8.485**2 / 3 = 72 / 3 = 24 Watts

Now let's find the average power by averaging the power from the two halves
of the waveform.  During the 12V phase, the power is:

  P = V**2 / R = 12**2 / 3 = 48 Watts

During the 0V phase, the power is obviously 0.  Since each phase occurs for
50% of the time, the average power is the average of 48W and 0W yielding 24
Watts.  Note that this is the *average* power dissipated by the resistor.

Now let's examine the RMS of the power signal.  The power signal is a square
wave from 0 to 48W.  The RMS of the power signal is:

  RMS of power = sqrt(48**2 / 2) = 33.94 Watts

As I said, it is possible to compute the RMS of the power waveform, but the
resulting value has no real significance.  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.


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