I am assuming the reflection angle of 0j terminations would all be the same
(you said 180), this includes both resistive loads and an antenna at
resonance. The method I am using relies on that.

I am confused about one thing in the smith chart program. Cables
identified as stubs create a circle that goes through infinity and do not
exhibit a constant SWR, while simple cables create a constant SWR circle
with the center and 50 Ohm 0j.

I believe the setup I am testing exhibits the constant SWR characteristic.

Dan

Dan, if the load is resistive, and less than 50 ohms (assuming the center of
the Smith Chart is considered to be 50 ohms), then the angle of the
reflection coefficient is 180 degrees. If the load is resistive, and
greater than 50 ohms, then the reflection coefficient angle is 0 degrees.
As you approach the center of the Smith Chart the angle becomes less and
less relevant.

Shunt stubs - either open or shorted - always pass through "zero" (And
infinity - which you will not see since it is in parallel with the load
impedance) and the selected impedance on the Smith Chart. This is based on
the effect that an open stub, at quarter wave multiples, will exhibit
successive open and short circuits; as will a shorted stub.

Your method will exhibit a constant reflection coefficient circle (and
VSWR), with the angle varying from 180 degrees through zero degrees and then
back, through negative angles, to 180 degrees.

Frank