Old Nick wrote (or quoted):

Throughout this site I will use the term 'RMS power'. Technically, as
far as I know, there is no such thing as 'RMS' power. This section
will help clarify and define a few terms. I will define RMS power as
the power that's calculated when using an RMS voltage or current into
a resistive load.

Then you/he are/is wrong - dead wrong. You simply do not understand the
term RMS and the reasons for its use. How's your calculus? If it's not
up to snuff, the following will do you no good - don't take offence,
just take the word of someone who knows and don't worry about it or try
to make the facts fit your misconceptions. If you _can_ handle calculus
then:

The time average (or mean) of a continuous integrable function of time,
f(t) over a time interval from t1 to t2 is DEFINED as

_ 1 t2
F = -------- Integral f(t)dt
t2-t1 t1

Note: For discrete quantities, we replace the integral with a sum and
t2-t1 with the count and get the familiar definition of average. These
definitions agree with "Reference Data for Radio Engineers", pg 982
published by IT&T. My copy is the 1957 edition but that part of
mathematics and electrical engineering is beyond even Kalifornu to
change.

Now, supose we want AVERAGE power. Given voltage, v(t), and current,
i(t), themselves continuous integrable functions of time, INSTANTANEOUS
power is DEFINED as

p(t)= v(t) i(t)

IF the load is linear and purely resistive then v(t) and i(t) look the
same except for a constant, R. Thus

v(t) = R i(t)

and two frequently useful substitutions are possible. Namely

2 2
p(t) = R i(t) or p(t) = v(t)
-----
R

What's more, if these functions, v(t) and i(t), are periodic (i.e.
repeat exactly every (say) T seconds) then the long term average is the
same as the average over a single period (cycle) and things get even
simpler. We can then say the average (or mean) power is

2
_ 1 T v(t)
P = --- Integral ---- dt
T 0 R

Since R is a constant, we can take it outside the integral to get

_ 1 1 T 2
P = - - Integral v(t) dt
R T 0

Thus AVERAGE power is the mean square voltage divided by R. Now, except
that I have natural gas heat, if I were to switch my electric heater
over to DC when the power failed, what DC VOLTAGE would I have to apply
to it to get the same heat output? Obviously I want the AVERAGE POWER
to be the same and at DC

_ 2
P = V / R

for some appropriately chosen voltage, V. For the same heat, I want

2
V 1 1 T 2
- = - - Integral v(t) dt
R R T 0

The R's cancel and we have

2 1 T 2
V = - Integral v(t) dt
T 0

or

( 1 T 2 )
V = SQRT ( - Integral v(t) dt )
( T 0 )

"root" "mean" "square"

This is the RMS voltage associated with a time varying voltage, v(t).
This is how the term arises and the reason it's useful is it relates the
AVERAGE power in a resistor load when ac applied voltage and dc applied
voltage are compared. The same argument can be applied to derive Irms
from i(t) and, for any periodic waveform into a pure resistor load

Pavg = Vrms Irms

And only the media and the hi-fi industry can manage to screw that up as
badly as they have.

Oh, sure, I can define an RMS power mathematically using the integral
but it has absolutely nothing to do with the ac voltage that is
equivalent in heating power to a dc voltage when fed into the same
resistance heater.