Reg Edwards wrote:
Now Cecil, you can stop trying to pull my leg. ;o)

I'm actually trying to make a technical point, Reg.
Walt may be trying to make that same technical point.
The SWR is based on differential energy which doesn't
radiate from the feedline. It just seems to me that
you are trying to have your reflected power and radiate
it too. That doesn't work for radiated power any better
than it works for cake. If the SWR is 1400:1 then the
reflected power is almost equal to the forward power both
of which are associated with *differential* currents which
don't radiate.

If the SWR on the feedline is 1400:1, almost no power
is being radiated! The forward power is 100 watts
then the reflected power is 99.7 watts. Both of these
powers are based on *differential* currents and therefore
don't radiate from the feedline. There's only 0.3 watts
available to radiate.

If 99.9% of the power is being radiated by the outside
braid of the coax, then the reflected power cannot be
more than 0.1% of the power and as a result of that
fact, the SWR on the feedline must necessarily be
very low, i.e. close to 1:1. Seems to me you need to
resolve that contradiction.

You cannot radiate 99.9% your reflected power and still
have it available to the SWR measuring equipment. If the
reflected power is available to the SWR measuring equipment,
it is composed of *differential* currents and is, by definition,
not radiating. If it is radiating, then it is not available to
the SWR measuring equipment and the SWR is, therefore, low. You
simply cannot have a sky high SWR on the feedline with the
feedline radiating 99% of the power. The SWR meter cannot tell
if the antenna or coax braid is doing the radiating and will
report a very low SWR.
--
73, Cecil, http://www.qsl.net/w5dxp

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