Jeff wrote:

Actually, here is a better example, because it represents the situation
found in many shacks.

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding, and thereby forming a short antenna.

The short antenna, only 1/4 inch long is immediately terminated
by a 50 ohm resistance.

1. How much of the power from the coax is fed into that short antenna
despite
the claimed (by others) impedance mismatch?

2. How much of that power is radiated by that short antenna?

3. If all the power that is fed to the short antenna is radiated, does the
50 ohm resistor dissipate any of it?

4. How much of the power is dissipated in the 50 ohm resistor?

5. How much of the power is reflected back down the coax because
of the impedance mismatch of that (very) short antenna?


It depends on how you connect the 50ohm load. The ground side of the 50
ohm load must go somewhere, so logic would dictate that it was connected
to the outer of the coax. If that case the 1/4 inch is so small compared
to a wavelength that it will not present much of a mismatch to the coax
at 3.6MHz, the coax will see 50ohms in series with a very small
inductance and even smaller stray capacitance, virtually all of the
power will be dissipated in the load, a small ammopunt will be
reflected. ie you have just added a small lumped inductor in series with
the 50ohms.

To be pendatic, the short wire has a small resistance and a small impedance,
both of which are in milliohms.

Since P=I^2R and the R is very, very small, the P dissipated both by
radiation and heat are very small.


This example has nothing to do with how short whips radiate. You keep
harping back to feed lines which are not what is under discussion. We
all agree that practically matching short whips is where the problem
lies. However, there is unequivocal proof that IF power is got into a
short whip by some means or other it is ALL radiated or lost as heat in
the resistance of the element.

Read Kraus, Jasik, or do some NEC simulations yourself and you will see
and hopefully stop this ridiculous quest for an answer that is just not
there.

Jeff



--
Jim Pennino