I see what you are saying and I see that there is validity to it. I do not
mean to sound arrogant, but in 35 years of discussing this type of thing,
this is the first time that I have doubted myself. I was totally against
the idea that you have presented, but I always try to stay aware of the fact
that RF is different and is a surprise, once in a while--if not more
However, I think that I see a small flaw in your thesis, but I am not sure
without giving it considerable more thought and possibly doing some
Dielectric loses are not something that I have studied with any greart
ferver, but I ^ R losses operate the same for AC as they do for DC--don't
they. If you have a given amount of current flowing through a piece of
insulated copper wire where there is a given amount of I ^ R loss which will
cause the voltage at the output end of the wire to be diminished from that
at the source end. If then you place another identical wire in parallel
with it, the amount of lost voltage due to I ^ R losses will be 1/2 of the
single wire. If you put 10 volts into one end of the wire and measure 9.5
at the output of the wire then place another wire identical in parallel with
it, the output voltage will rise to 9.725.
Therefore, I cannot understand why the losses of the cable will stay the
same for two wires as it is for one. The total I ^ R losses for two wires
have to be cut in half if the same current is available to the two wires as
was to the single wire even if the wires are inside an insulated shielded
In the case of the dielectric loss, I can make NO comment on at this time.
This will require more study before I have any confidence in anything that I
would say about it.
Billy Dean Ward
>From: "Gary Breed" <firstname.lastname@example.org>
>To: "Billy Ward" <email@example.com>
>Subject: Re: [antennaware] Twin-Coax-Cable as single feed line to 1/2
>Date: Wed, 20 Dec 2000 18:22:59 -0500
>Here's example to show that losses do not drop by half with parallel
>feedlines. Let's look at your system:
>Let's say you are using two RG59 type cables with 2 dB loss per 100 ft.
>at your operating frequency. We can also assume that your lines are 50
>feet long, with 1 dB loss in each cable when perfectly matched to its
>First, we'll look at one feedline. Assume that your transmitter delivers
>watts into 72 ohms. Then assume that you have a perfect, lossless circuit
>at the feedpoint that matches the 36 ohm antenna to the 72 ohm line. The 1
>feedline loss means that 80 watts will be delivered to the antenna.
>Next, we can represent the two parallel lines by saying that the parallel
>connection is equivalent to a perfect, lossless circuit that transforms the
>impedance and splits the power into two 72 ohm lines, each one to carry 50
>watts. Each line has 1 dB loss, so 40 watts is delivered to the antenna by
>each line, which totals 80 watts, the same as with a single line.
>There are advantages to parallel feedlines: convenient impedance matching,
>reducing mis-match losses, and increasing power handling capability, but
>reduced feedline loss.
>In a matched line, losses are the sum of I^2 R losses and dielectric
>both of which increase with frequency. This is where the manufacturer's
>vs. frequency curves come from. I^2 R behavior alone is only approximated
>very short lines or at very low frequencies where skin effect is absent and
>dielectric losses are minimal. The bottom line is that RF is definitely
>not DC. That's what makes it fun!
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