Ian White GM3SEK wrote:
Pipe the signal back from the far end into the shack, feed
it into a circulator, and add it to the outgoing signal. Cecil will
explain what happens to the power :-)

Dr. Best, VE9SRB, in his 2001 QEX articles explained what
would happen. Based on his idea that 75w + 8.33w = 133.33w,
one could route the unused Rhombic power back to the source,
recycle it, and cause voltage superposition to multiply the
power up to a factor of 4. :-) To anyone who thinks I am kidding
about Dr. Best posting the above equation, it can probably
be verified by Google circa May 2001 on this newsgroup.

Dr. Best proved his assertions with the following power equation:

Ptot = P1 + P2 + 2[SQRT(P1*P2)]cos(A)

where A is the angle between V1 and V2. If we make the angle
between V1 and V2 equal to zero, we can take the P1 power
from the source and the P2 power routed back from the load
and increase our total power output by a factor equal to
2[SQRT(P1*P2)]. Who says there is no such thing as a free
lunch? :-)

Discussed by me in May 2001 was the fact that his term,
2[SQRT(P1*P2)] is constructive interference energy which must
necessarily be exactly balanced by 2[SQRT(P1*P2)] watts of
destructive interference energy or else the conservation of
energy principle is violated. At the time, Dr. Best did not
understand where the necessary destructive interference energy
was coming from. It comes from the Z0-match between the feedline
and the source and it works exactly like the thin-film layer on
non-reflective glass. Dr. Best's Ptot equation above is true for
A = 0 if and only if Ptot is being supplied with destructive
interference energy where A is probably equal to 180 degrees.
--
73, Cecil http://www.w5dxp.com