I disagree with Tom this time.
If I have a vertical with a feedpoint resistance of 10 ohms, and one
with a feedpoint resistance of 100ohms, the ratio of the currents will
be square root 10. An inductor with a series resistance of 1 ohm will
dissipate 10 times the amount of power in the 10 ohm case. So I squared
R losses are reduced by using a higher feedpoint impedance.
To see matching components catch fire, just look at 500KHz 1KW marine
tx's running into 30 foot whips - we had tank coils wound with Litz wire
(243 strands of 38 gauge) with unloaded Q's of 400 to 500, and still had
problems. As far as the Q problem goes, most of the amateur ATU's,
transmatches - call them what you will - commercially available suffer
from excess Q when matching large impedance ratios, and in mnay cases,
this is evidenced by tank coil heating. There was a review in QST
recently of ATU's that evidenced this.
I guess on radiation resistance that there is an interesting dichotomy
of views, with various published textbooks taking different stands.
Taking the definition
" The total power radiated as EM energy divided by the square of
effective current causing that radiation taken at the current maximum
of the radiator."
In a folded unipole, where is the current maximum, and what is the
effective current? I guess we all accept that the feed impedance has
gone up, and the feed amps gone down. Does this not mean that it doesn't
radiate as well as a non folded unipole if the radiation resistance is
the same as the non folded unipole?
Plus the fact that it indubitably works!
73
Peter G3RZP
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