If you are thinking of E²/R, you should note this is division, not
multiplication.

Thanks Richard. What I am thinking is: assuming a perfect 50 ohm load, scale
the voltage divider rectifier combo to give 10.0 VDC at full scale watts, 1500
in my case. I now have a relative power indicator as suggested by Reg. I can
take care of the squaring and division by putting a Log scale(watts) on a 0-10V
analog meter movement. But, what I intend to do is run the voltage divider
signal to a multiplier whose output will be V**2/10. If V is 10. volts for
1500 watts then the multipier output will be 10.0 volts. let 10.0 volts equal
1500 watts on the meter scale. Checking another point, assume 500 watts, then
the output of the divider will be 5.76 volts, and the output of the multiplier
will be 3.33 volts, which is 33% of 10 volts. The wattmeter will read 33% of
1500 or 500 watts. The scale will be linear.
I have not seen it done this way before, although I am sure it has. For this
method to accurate your load must be near 50 ohms. The AD633 is a cheap,
easy-to-use multiplier chip. Using a +/- 15 VDC supply, it looks like it will
have the dynamic range I need.
73 Gary N4AST