In message
,
Cecil Moore writes
On Sep 1, 1:54*pm, Ian Jackson
wrote:
The question is, at which end of the feeder is the most absolute power
lost per unit length?

I already explained that. On HF, maximum losses depend upon where the
current/voltage-nodes/loops are located on the transmission line. One
cannot answer the above question without knowing that information and
the answer is different depending upon the conditions.

Owen has previously proved that on HF, a higher SWR can result in
lower losses than a flat line when that high SWR is near a current
node and the feedline is less than a quarter wavelength long simply
because the average standing-wave current is *lower* than the flat
line current.

Ah, I see what you mean. I was thinking in terms of an 'electrically
long' feeder (many wavelengths long).

I can see that, for a 'short' feeder (for example, if you were feeding a
halfwave dipole via a quarterwave of 300 ohm line), at the antenna
feedpoint you will have a current node (high I^2*R loss), and at the
TX/tuner end, where there is a voltage node, you will have a low I^2*R
loss. It could be that the loss of a matched relatively low-impedance
feeder of 50 or 75 ohms (being inherently higher than that of higher
impedance feeders with similar conductors etc) could be higher than the
average loss of the 300 ohm - even though the antenna will present an
SWR to the feeder of around 4 or 6 to 1.

That still leaves the question of which end of a long feeder has the
greater absolute power loss. Using 'reductio ad absurdum' and 'rule of
thumb' (two very dangerous principles!), I feel that it can only be at
the TX end. If the cable is really long, any power left to be dissipated
at the load end must be negligible compared with that available at the
TX end.
--
Ian