Dave wrote:
math please??? where do you get 4.17watts?? to me it looks like a 50 ohm
load on a 50 ohm meter so zero reflected power.

Here's the setup.

100W--tuner--75 ohm coax--Bird--1/2WL 75 ohm coax--50 ohm load

Assumptions: Losses are negligible. The tuner provides a Z0-match
so 100 watts is delivered to the load, i.e. all reflected power
is re-reflected. I think this was actually the case for the experiment.
Also assume that the impedance bump caused by the insertion of the
Bird is negligible, i.e. the same net voltage and net current exists
whether the Bird is in or out of the circuit.

The voltage reflection coefficient at the load is (50-75)/(50+75)
equals -0.2. That makes the power reflection coefficient (-0.2)^2
equals 0.04, i.e. 4% of the power incident upon the load is reflected.
If 100 watts is delivered to the load while 4% is being reflected
then the power incident upon the load is 100w/(1-0.04)=100w/0.96=
104.1667 watts. Incident power minus delivered power = reflected
power so the reflected power on the 75 ohm coax is 4.1667 watts.

The SWR on the 75 ohm coax is (1+|rho|)/(1-|rho|)= 1.2/0.8 = 1.5
THE SWR ON THE 75 OHM COAX IS 1.5:1. The Bird is in error when
it reports the SWR to be 1:1. The SWR is *NOT* 1:1 anywhere on
the load side of the system.

The Bird reports a forward power of 100w. The actual forward power
is 104.1667w. The Bird reports a reflected power of near zero. The
actual reflected power is 4.1667w. The Bird reports an SWR of 1:1.
The actual SWR is 1.5:1. The Bird is reporting false values because
it is embedded in a 75 ohm environment and is calibrated for 50 ohms.

If the ratio of voltage to current equals 50 within a 75 ohm system,
there exists an SWR of 1.5:1 and an SWR meter calibrated for 75 ohms
will verify that fact.
--
73, Cecil http://www.qsl.net/w5dxp