Dave wrote:
just enough for any fringe effects to become negligible... no more than a
couple diameters of the coax as a rough guess.

Apparently, the 2% of a wavelength that I was remembering was
at 10 MHz. 1'/(98.4*0.66) rounded to 2%. All I was interested
in at the time was proving to Reg that one foot of coax forces
Vfor/Ifor = Vref/Iref = Z0, the boundary conditions assumed
by Bird Wattmeter designers. Since Kevin was not familiar with
PL-239's, I erred on the side of caution with the 2% estimate.

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

Cecil wrote:
It addresses it adequately but doesn't answer any particulars.
Given PL-239 connectors and RG-213 coax, I wonder what the
answer would be for 10 MHz?

I'm not familiar with the connectors in question. Assuming
they are properly attached, they should not introduce much
mode diversion. 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.

HTH
Kevin
****End Quote****
--
73, Cecil http://www.qsl.net/w5dxp