Richard Clark wrote:

On Tue, 11 Oct 2005 15:37:51 GMT, Cecil Moore wrote:
I dug up some calculations from sci.physics.electromag

which you recite here; then in sci.physics.electromag you can quote
their use by authorities (sic both times) in
rec.radio.amateur.antenna....

This appeal is called a circle of friendship - not evidence.

Actually, it is called an argumentum ad verecundiam, an appeal
to authority - a technical authority in this case. I don't know
Kevin G. Rhodes at Dartmouth. He merely answered my question on
s.p.e. that didn't find an answer on this newsgroup. Exactly what
did you find technically wrong with the following evidence?

****Quote****
Newsgroups: sci.physics.electromag
From: "Kevin G. Rhoads"
Date: Tue, 07 Oct 2003 12:49:14 -0400
Subject: Transmission Line Question

For 10 MHz I would expect that all other modes
would be non-propagating (i.e., evanescent) even though RG-213
is a large coax (improved RG-8 apparently). The speed of propagation
is listed as 66%, so the nominal wavelength is 3/2 times the free
space wavelength for the TEM mode. 3/2 x 30m = 45m, which implies
the decay rate in space for non-TEM modes is going to be large
as the cable diameter is .405" (jacket) which implies the
spacing from inner to outer conductors will be less than .203".
For order of magnitude estimate, assuming the lowest non-TEM mode
can be approximated using a characteristic equation that really
is only applicable in Cartesian geometries:
(1/45m)**2 = (1/.203")**2 + kz**2
Clearly, kz must be imaginary to make this work. thus an
evanescant, non-propagating wave:
kz**2 = (1/45m)**2 - (1/.203")**2
To the accuracy used to date, the first term on the right
is negligible, so the decay rate, alpha, can be estimated:
alpha**2 = - (kz)**2 = (1/2.03")**2
Or, the lowest order undesired mode should reduce intensity
by a factor of 1/e (0.37) in about 2.03"; power will reduce
by that factor squared in the same distance (.135). In
about four inches, undesired mode power is down to about
0.018, in six inches, .00248, and after a foot, 6.14x10-6

You should double check my algebra, but I think the estimate
is reasonable. To put it into other terms, since the wavelength
in the coax dielectric is 45m and the conductor to conductor
spacing is about 2", any non-TEM mode will suffer attenuation
in E-field intensity with a space-rate constant rounghly
equal to the conductor to conductor spacing. INtensity
drops by 1/e = 1/2.71828 every 2 inches. Power availalbe
drops faster, being square of intensity.

So unless almost all the power diverts into an undesireable
mode (by a factor of more than a million to one), one foot
of cable should see pure TEM at the end.
***End Quote***
--
73, Cecil http://www.qsl.net/w5dxp