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Old May 14th 05, 09:44 PM
W9DMK
 
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On Sat, 14 May 2005 20:33:41 +0000 (UTC), "Reg Edwards"
wrote:

Cecil said -
That is true.


===========================
Cec,

With this sort of argument you must not compare one manufacturer's
cable with another. You can't believe the sales-talk anyway

Why unnecessaily complicate the question? Why introduce standing
waves and reflections and all the other silly time wasting
distractions?

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

The exact simple mathematical relationship is -

Line attenuation = 8.69*R/2/Ro dB.

Where R is the resistance of the wire and Ro is the real component of
line impedance, all in ohms.

Make a note of it in your notebooks.

And, hopefully, that should be the end of the matter. But, knowing
you lot, it probably won't be. ;o)

Dear Reg,

Can we also assume that your formula is for "matched line" attenuation
only, and that the attentuation for a given line will actually
increase with SWR?

Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html

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