For Carl, or any others who are not yet convinced that folding a unipole
doesn't change the ground losses: here's another way of looking at it.
Draw a picture of a vertical and underneath it some radials. Draw the end
of a piece of coax near the feedpoint. Connect a 12 ohm resistor between
the coax shield and the common point of the radials. Connect the center
conductor of the coax to the bottom of the vertical. Label all the currents
as one amp (coax center conductor, coax shield, radiator, and ground
through the ground loss resistor. The ground loss is I*I*R or 12 watts.
Draw another picture of a vertical, this one having a folding wire coming
down from the top to ground. Connect the coax center conductor to the
feedpoint, the coax shield to the bottom of the folding wire and the 12 ohm
resistor, and the other end of the 12 ohm resistor to the common point of
the radials, as before. Now label the currents as follows: coax center
conductor, one half amp. Feedpoint, one half amp. Coax shield, one half
amp into the 12 ohm resistor, and thence to ground. Folding wire, one half
amp into the 12 ohm resistor and thence to ground. Total current in the 12
ohm resistor is one amp (half an amp from the folding wire and half an amp
from the coax shield. Loss is again I*I*R=12 watts.
I think maybe the mystery arises because we assume (incorrectly) that the
coax shield current equals the current into the lossy ground system. But
the current from the folding wire is also flowing into the lossy ground
system, making the ground current exactly the same as before.
Hope this helps.
Larry, N9DX
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