Yes, I guess alternatively you could calculate the SWR at the
half-power points from

(E/Eo)^2=(1-((1-SWR)/(1+SWR))^2)

With E/Eo = 1/sqrt(2)

I get that this is about SWR = 5.8, and I understand that the MFJ-259B
is less and less accurate as the load moves away from 50 ohms
resistive, so it might be better to use the 88% power, 2:1 SWR
bandwidth points.

- - - - - -

I've thought of an additional question regarding this method. The
ground has an effect on antenna feedpoint impedance. This effect
contains both modification to the radiation resistance and the loss
resistance components of the feedpoint resistance.

Is the effect on the radiation resistance different for lossy earth and
for perfect earth? Obviously the loss increases with proximity to
ground, but can the effect of ground reflections on radiation
resistance be accurately modeled by putting the antenna over perfect
ground?

73,
Dan
N3OX