The circuit you describe should be delivering approximately the peak
voltage, not RMS, so the formula is more like V X V / 100. At 7 volts
peak, that means you're putting out around 1/2 watt. That's a lot below
the 3 watt figure you mentioned, so you should track down why it's so
much lower. Pay special care to the circuit components in the final
transistor output (collector) circuit, and make sure the inductors are
built as described and the components are connected correctly.

Roy Lewallen, W7EL

John Sandin wrote:
. . .
Voltage is measured across my 50 ohm dummy load and rectified using a
.05 mfd ceramic disc cap and a 1N34A diode. Then I'm calculating
thusly:

Power = (Voltage x Voltage) / 50

I'm taking the author's word that this works for a ballpark estimate.
I'm measuring 7 volts across the load, maximum, which works out to 1
watt or less. I've verified that my meter is accurate.
. . .