"Richard Harrison" wrote in message
...
I wrote:
"RMS is the effective value, not the average value of an a-c ampere.

Steve Nosko responded:
"I will differ here. The RMS value is more appropriately described as
the power producing value of ANY wave form.."

Richard,

I'm not sure what your point of contention is here for most of these
comments. Perhaps I was not clear in my lone objection. All I was
correcting was your reference to "a-c ampere". My intent was to point out
that pulses to one side of zero also have an RMS value. If you do the
"square root ...average...squares..instantaneous
values..quantity..throughout one period." thing, you will get it. So
therefore, tube's current pulses have some RMS value; and that's what'll
heat the tube and cause the efficiency stuff. That's where I was going. If
we can define or approximate the shape, then we could calculate an RMS. I
wonder if a "true RMS" DVM can handel this?

A periodic function is not necessarily a sine wave but when I look up
alternating current in my dictionary, I find the diagram of a sine wave
and it is tagged: "alternating current".

Sure. I'm ok with that. A sine is indeed one example of a periodic
function.

By fourier series, any complex waveform may be resolved into a
fundamental plus a finite number of terms involving its harmonics.

Also ok, but not sure how it plays into the RMS discussion. See below.

My electronics dictionary says:
"The rms value of a periodic quantity is equal to the square root of the
average of the squares of the instantaneous values of the quantity taken
throughout one period. If that quantity is a sinewave, its rms amplitude
is 0.707 of its peak value."

No issue here. (from me, that is) The "average" turns out to be an
integration and I had to drag out the son's calc book to remember how to do
it to see some of the answers myself.

My dictionary also says:
"Effective current---That value of alternating current which will give
the same heating effect as the corresponding value of direct current.
For sine-wave alternating currents, the effective value is 0.707 times
the peak value."

Still none.


The average value for one alternation of a sine wave is 2/pi or 0.637
times the maximum value.

Yea. untill I went through the trouble of doing the calculation, I had
never known the source of the 0.637 number. Now I know. Pretty cool to
actually do it! The integration for this is pretty simple. Took me a few
looks to figure out where the pi came from. It's the old indefinite to
definite integral thingy.


I make lots of mistakes, but fortunately, I don`t think I made any of
consequence in this thread. I probably didn`t go into enough detail, but
I was just an engineer, not a professor.

I'm not a "professor", just an Enginer and "adjunct faculty" for two
community colleges for some years. (that's a fancy name for part time
teacher)
73, Steve