What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Uwe, Wouldn't be the first time I was wrong.
If you rectify and filter an AC waveform, and the read the DC voltage, this
is Vpeak of the AC. In the example I gave, you read 2 vdc out of the detector
circuit. Multiply this times the X10 divider, and you get 20vdc which is Vpeak
across the 50 ohm load. To get the rms voltage across the load
Vpeak*.707=20*.707=14.14 Vrms. 14.14Vrms across 50 ohms=4 watts.
BTW, the reason I suggested a voltage divider (attenuator) rather than
reading directly with a VOM or scope is to reduce the likelyhood of the scope
probe or meter impedance from detuning the AC-1 causing further errors.
If you are using a scope, measure the Vpeak across the 50 ohm resistor,
multiply by .707 to get Vrms.
73 Gary N4AST