If one directly measures the voltage, current and associated phase angle (in
radians)
for the 3:1 mismatched line shown below (hard to do on a coax but easy on
twin line)
then, E x I x Cos Theta(rads)= 100 Watts...the applied power. This is true
for all
points on the line. Example: a 50 ohm line at least 1/2 wavelentgh long at
the
frequency of interest, terminated in a 3:1 mismatch with 100 watts applied
will have
E=77.80, I=2.06, and Theta(rad) = .9 at some point on the line.
77.80 x 2.06 x Cos(.9) = 100 watts. ((ignoring resistive losses of the
line))
73, Bob, W5AH
Original Message
And if you have an open or short circuit, VSWR is infinite, and so are
the forward and reverse power! In real life, Tx power goes somewhere 
into tuner and transmission line resistance  and VSWR is finite.
Martin
Chuck, W1HIS wrote:
...
(2) A meter between the tuner and the antenna will indicate
Forward power = 133.33 watts;
Reverse power = 33.33 watts;
% refl. power = 25 %;
VSWR = 3:1; and
the _net_ forward power is 133.33  33.33 = 100 watts.
Note that the forward power exceeds the power output
from the transmitter because the 33.33 watts of reverse
power returning to the tuner is reflected from the tuner
back toward the antenna.
