Roger wrote:

In the last 24 hours, Roy posted a revised analysis that contains
results useful here. He presented a voltage example that resulted in a
steady state with steady state voltages 4 time initial value. Under
superposition, this would equate to 4 times the initial power residing
on the transmission line under the conditions presented. This concurs
with other authors who predict power on the transmission line may exceed
the delivered power due to reflected waves.

Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from
illumination and radiation theory to describe power existing at a point
in space near a reflecting surface. If we consider space to be a
transmission media, and the reflecting surface to be a discontinuity in
the transmission media, then we have a situation very similar to an
electrical transmission line near a line discontinuity.

It is entirely reasonable to consider that the reflection ratio between
the space transmission media and the reflective surface would result in
an storage factor equaling 4 times the peak power of the initial forward
wave. By storage factor, I simply mean the ratio of forward power to
total power on the transmission media under standing wave conditions.

Under open circuit conditions, a half wavelength transmission line will
have a storage factor of 2. Roy presented an example where the storage
factor was 4. Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

Power is neither stored nor conserved, so a power "storage factor" is
meaningless. Consider a very simple example. Let's charge a capacitor
with a constant current DC source. We'll apply 1 amp to a 1 farad
capacitor for 1 second. During that time, the power begins at zero,
since the capacitor voltage is zero, then it rises linearly to one watt
as the capacitor voltage rises to one volt at the end of the one second
period. So the average power over that period was 1/2 watt, and we put
1/2 joule of energy into the capacitor. (To confirm, the energy in a
capacitor is 1/2 * C * V^2 = 1/2 joule.) Was power "put into" or stored
in the capacitor?

Now we'll connect a 0.1 amp constant current load to the capacitor, in a
direction that discharges it. We can use an ideal current source for
this. The power measured at the capacitor or source terminals begins at
0.1 watt and drops linearly to zero as the capacitor discharges. The
average is 0.05 watt. Why are we getting less power out than we put in?

"Where did the power go?" is heard over and over, and let me assure you,
anyone taking care with his mathematics and logic is going to spend a
long time looking for it. So in this capacitor problem, where did the
power go?

It takes 10 seconds to discharge the capacitor, during which the load
receives the 1/2 joule of energy stored in the capacitor. Energy was
stored. Energy was conserved. Power was neither stored nor conserved.

Roy Lewallen, W7EL