On Tue, 11 Oct 2005 09:11:19 +0100, Ian White G/GM3SEK
wrote:

The subsequent conversion to VSWR is a mathematical relationship only.

Hi Ian,

This seems to be a particularly notable difference - to which
absolutely NO ONE has ever deviated from in ANY determination of SWR!

That is to say, this "mathematical" distinction that some rely on to
differentiate their arguments has not got one scintilla of difference
over any other method.

The only way to claim you "directly" measure SWR is to find some way
to place two probes of a meter along the line such that one probe goes
into the trough and the other into the peak and the meter reads SWR
directly. Unfortunately for rhetoric's sake, this STILL renders the
determination in terms of a mathematical relationship. It cannot be
escaped.

Why this keeps on being revisited must be to allow the new lurkers to
observe my correction.

73's
Richard Clark, KB7QHC

Reg Edwards wrote:
It measures the MAGNITUDE of the reflection coefficient. It discards
the information which is contained in the phase angle of the
reflection coefficient. As a consequence the only use which can be
made of the magnitude is to calculate the SWR on an imaginary 50-ohm
line.

Reg, I dug up some calculations from sci.physics.electromag
from about a year ago that indicate one foot of 50 ohm coax
on each side of the Bird is enough to make the line real,
i.e. not imaginary, and that's a conservative estimate.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
.. .
Reg Edwards wrote:
It measures the MAGNITUDE of the reflection coefficient. It discards
the information which is contained in the phase angle of the
reflection coefficient. As a consequence the only use which can be
made of the magnitude is to calculate the SWR on an imaginary 50-ohm
line.

Reg, I dug up some calculations from sci.physics.electromag
from about a year ago that indicate one foot of 50 ohm coax
on each side of the Bird is enough to make the line real,
i.e. not imaginary, and that's a conservative estimate.
--
73, Cecil http://www.qsl.net/w5dxp

i like imaginary!

Reg Edwards wrote:
Forward and especially reflected power are even more imaginary than
the SWR on a non-existent 50-ohm line.

My old Heathkit HM-15 SWR meter samples the peak forward and
reflected currents. A pot is used to set the meter to full
scale using a voltage proportional to the peak forward current.

When the voltage proportional to the peak reflected current is
then switched into the meter circuit, it reads SWR from the
precalibrated scale which is linear with |rho|, i.e. at half-scale,
SWR=3 and |rho|=(3-1)/(3+1)=0.5
--
73, Cecil http://www.qsl.net/w5dxp

Ian White G/GM3SEK wrote:
All agreed. Along with the math that Cecil has retrieved and quoted
again, everything points towards the distance in question being a
function of coax diameter only; and not wavelength.

Please forgive my previous senior moment.
It was ~2% of a wavelength at 10 MHz for RG-213.
It appears that one foot of coax on each side of
a Bird wattmeter is enough to establish Z0 at
50 ohms which forces Vfor/Ifor=Vref/Iref=50,
the necessary Bird boundary conditions.
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 11 Oct 2005 15:40:55 -0000, "Dave" wrote:
i like imaginary!
fantasy is more appropriate.

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.

Richard Clark wrote:
That is to say, this "mathematical" distinction that some rely on to
differentiate their arguments has not got one scintilla of difference
over any other method.

I must have a dozen equations for SWR, all mathematically consistent
with each other. A lot of the math is performed by simply calibrating
a meter face. For instance, given a linear meter reading for |rho|
with full-scale equal to 1.0, SWR values can be applied to the
meter face with 3.0 at half-scale. (Ever notice how many SWR meters
have 3.0 at half scale?)
--
73, Cecil http://www.qsl.net/w5dxp

On Tue, 11 Oct 2005 16:37:42 GMT, Cecil Moore wrote:
Ever notice how many SWR meters have 3.0 at half scale?
Amateur grade - 1 in 4;
Professional grade - none.

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