I'm not sure who the "we" is in your statement, but it certainly doesn't
include people who understand the concepts and mathematics involved. The
problem with your approach is that there *is* such as thing as RMS power
(as I described in an earlier posting), and it is *not* the amount that
gives the same heating effect as the same amount of DC power. So while
you may think of RMS power in this way, your thinking is wrong.

I can calculate my RMS speed in going from point A to point B. But it
isn't equal to the distance I went divided by the time it took to get
there (unless, of course, I went the whole way at a constant speed) --
for exactly the same reason that the RMS power isn't the equivalent
heating power. You can think of the RMS speed as being the distance
divided by the time, but it isn't, and your thinking doesn't make it so.
And that also doesn't make the RMS speed a useful value.

Roy Lewallen, W7EL

Paul Burridge wrote:

I guess you're right, but when we speak of RMS power (sic) what we
actually envisage is the amount of AC power dissipated in a resistive
load that gives rise to the same heating effect as the equivalent
amount of DC power. Sorry if that's not well put, but you no doubt get
my drift...
So it may be a misnomer, but that doesn't make it useless.