Roger wrote:
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)".

Visualize visible interference rings between two equal
waves each of P magnitude, where the darkest of the rings
is completely black. If it is not obvious, the power in
the brightest of the rings would have to be 4P for the
energy to average out to the 2P in the source waves. That's
all there is to "black" destructive interference vs "bright"
constructive interference. (4P+0P)/2 = 2P = average power
It's a no-brainer for most folks.

On the source side of a Z0-match, the reflections are "black".
On the load side of a Z0-match with reflections, the forward
wave is "bright". What could be simpler?
--
73, Cecil http://www.w5dxp.com