Yikes... this thread is still running.
As I explained before there is a "common" usage which has unfortunately been
adopted (and printed in the almighty and "always correct" text), which uses
the words "RMS power" to actually mean the *average power*. An earlier
poster explained that this term was also adoppted (still non mathetically
correct) in audio circles to indicate a specific test.
This is not a correct usage because the term "RMS" means that it actually is
a Root Mean Square value. [[square it, take the mean, then take the square
root]] We just don't do this with power waveforms.
Sidebar:
I did extensive calculations of this type as a result of an article in QST
earlier this year where it was proposed to use a VOM/DVM to measure line
voltage and current(via a small series dropping resistor) to arrive at power
draw of common ham equipment...OOPS!
I was, however, a bit puzzeled, Roy, why you went to the trouble of
calculating the the RMS value of a power earlier. I suspect it was just to
show that the value is indeed different (didn't check your math).
I will differ with Roy on one issue. The RMS value of voltage and current
have, for many years, also been referred to as the "effective" values. This
was to relate it to the DC heating effect (of resistance) we are all
familiar with. It is, indeed just as "effective" as the same DC value, in
producing power. This is another terminology issue I suspect some of you
may wish to squabbling about, but is is not a 'what is correct technically'
issue. It is cleat that probably all of you understand the math, but this
is simply an nomenclature issue.
I feel sorry for Bill because he seems to understand the math:
understand about deriving the RMS power from the instantaneous power
in the same way that RMS voltage or current is derived,
....yet he said:
I just don't
accept the definition. It's been taught the other way all my life, even
though you say it's incorrect. I see your point, I just don't accept
it.
I don't understand when you say "I just don't accept the definition."
Does this mean that you do not accept thst "RMS Power" implies (to some of
us) that you have done the Root Mean Square calculation on the power
waveform? Which deffinition is giving you grief?
I am not not trying to prolong the pain (or this thread), it is just that I
was born with a bone in my head that makes it hard for me to give up
explaining some basic concept like this. (yep, it can be a curse) You're so
close.
--
Steve N, K,9;d, c. i My email has no u's.
"Roy Lewallen" wrote in message
...
Bill Turner wrote:
On Tue, 19 Oct 2004 00:02:50 -0700, Roy Lewallen wrote:
And now the real question: How much DC voltage would you have to apply
to the resistor to get exactly the same power dissipation?
100 volts. The dissipation would be 100 watts.
__________________________________________________ _______
Well. The answer you gave is exactly the answer I would have given, but
you say my answer is wrong.
Where have I said that's wrong? Of course it's not.
I understand about deriving the RMS power from the instantaneous power
in the same way that RMS voltage or current is derived, I just don't
accept the definition. It's been taught the other way all my life, even
though you say it's incorrect. I see your point, I just don't accept
it.
I have no idea who taught that to you, since the definition of RMS is in
its very name (the square Root of the Mean of the Square of the
function). It's your choice to ignore the accepted definition. I can
only hope you don't teach your mistaken idea to others, who will then
someday say the same thing.
I will QRT for now, but thanks for taking the time to explain. I mean
that sincerely and I do respect your point of view.
You're welcome. The only reason I've taken the time for these postings
is in the hope that it will help people understand and learn. Even if it
hasn't worked for you, I hope some other readers have benefitted.
Roy Lewallen, W7EL
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