Bob wrote:
On Sun, 04 Apr 2010 20:38:35 +0100, Baron
wrote:

You could get a sample and measure it !

Well, I discarded that idea because I have no idea how.

But then, on a hunch, I checked the manual that came with my MFJ-269,
and sure enough, on page 34, it tells how to measure Velocity Factor,
utilizing the distance to fault mode. It'll take a day or so to
recharge the 269's batteries, and then I'll have at it.

Bob
k5qwg

Unfortunately, it's not really simple to make measurements with
symmetrical line. You'll be exciting a common mode current which will
travel with a different velocity factor and affect the measurement. I
suggest making an approximate measurement, then doing final adjustments
of the MFJ kept as far as possible from conductive objects including
yourself. You'll have to adjust it, let go, back off and read the meter,
readjust, etc. And then it'll still be a bit off unless the length of
the MFJ meter is quite short relative to a wavelength. You'll also have
to keep the line well away from any conductors and avoid coiling it. Of
course, the same problems will exist when you install the line in
whatever system it'll be used for, unless you can get it very well balanced.

It'll be a good exercise in learning some basic measurement techniques.
Whether your results are adequately accurate depends on the application
you'll be using the line for.

I sometimes taught a class on TDR techniques, and I'd start by
connecting a foot or so of two-conductor ribbon cable -- just soldered
into and to the shell of an SMA connector -- to a high speed TDR. The
trace would show the large reflection from the open end, of course, but
a smaller reflection seemingly coming from a point about 1/4 of the way
from the end. I explained that ribbon cable isn't controlled for
impedance, so it obviously had a construction anomaly at that point, and
pinched the line, running my fingers along until the reflection from the
fingers was at the same point as the anomaly. Then I cut the line well
toward the TDR unit, discarding the portion with the anomaly. When the
audience saw the *new* reflection about 1/4 of the way from the end of
the shorter wire, I had their attention. And thus began a discussion of
differential and common mode waves.

Roy Lewallen, W7EL

On Apr 4, 4:19*pm, Bob wrote:
I'm plugging the velocity factor figure into Cecil's program for
optimum feedline lengths on a multiband dipole, IMAXMIN.EXE. Given the
approximate nature of this kind of feed, a ballpark figure is probably
okay.

Yes, given all the variables, adjusting the final length, sometimes by
a few feet (depending on wavelength) is almost always required to
achieve system resonance. Remember that this approach is designed to
eliminate the tuner and therefore eliminate tuner losses and it is
designed to be used with a 1:1 current-choke-balun. Owen's comments
are certainly valid for systems using antenna tuners and 4:1 baluns.
In fact, if one chooses a ladder-line length halfway in between my
"good" (current maximum) and "bad" (voltage maximum) lengths, one will
obtain the odd 1/8 wavelengths points that are recommended for use
with 4:1 baluns. Those points result in a ballpark impedance in the
neighborhood of Z0 +/- jZ0/4, e.g. 400+j100 ohms. For those who
understand a Smith Chart, a picture is worth a thousand words.

http://www.w5dxp.com/smith.htm
--
73, Cecil, w5dxp.com

On Mon, 5 Apr 2010 06:27:54 -0700 (PDT), Cecil Moore
wrote:

On Apr 4, 4:19*pm, Bob wrote:
I'm plugging the velocity factor figure into Cecil's program for
optimum feedline lengths on a multiband dipole, IMAXMIN.EXE. Given the
approximate nature of this kind of feed, a ballpark figure is probably
okay.

Yes, given all the variables, adjusting the final length, sometimes by
a few feet (depending on wavelength) is almost always required to
achieve system resonance. Remember that this approach is designed to
eliminate the tuner and therefore eliminate tuner losses and it is
designed to be used with a 1:1 current-choke-balun. Owen's comments
are certainly valid for systems using antenna tuners and 4:1 baluns.
In fact, if one chooses a ladder-line length halfway in between my
"good" (current maximum) and "bad" (voltage maximum) lengths, one will
obtain the odd 1/8 wavelengths points that are recommended for use
with 4:1 baluns.

The more I look at it, the odd 1/8 wavelengths is probably the way I
will go, connecting to my tuner's 4:1 balun. There will be a 130 foot
flat-top, and the 450-ohm feedline length can be somewhere between 50
to 100 feet or so. Tnx for the input!

Bob
k5qwg

Those points result in a ballpark impedance in the
neighborhood of Z0 +/- jZ0/4, e.g. 400+j100 ohms. For those who
understand a Smith Chart, a picture is worth a thousand words.

http://www.w5dxp.com/smith.htm

Roy Lewallen Inscribed thus:

Bob wrote:
On Sun, 04 Apr 2010 20:38:35 +0100, Baron
wrote:

You could get a sample and measure it !

Well, I discarded that idea because I have no idea how.

But then, on a hunch, I checked the manual that came with my MFJ-269,
and sure enough, on page 34, it tells how to measure Velocity Factor,
utilizing the distance to fault mode. It'll take a day or so to
recharge the 269's batteries, and then I'll have at it.

Bob
k5qwg

Unfortunately, it's not really simple to make measurements with
symmetrical line. You'll be exciting a common mode current which will
travel with a different velocity factor and affect the measurement.

Roy Lewallen, W7EL

Please could you elaborate on how and why a common mode current has a
different VF on a balanced line.

--
Best Regards:
Baron.

Bob wrote in
:

The more I look at it, the odd 1/8 wavelengths is probably the way I
will go, connecting to my tuner's 4:1 balun. There will be a 130 foot
flat-top, and the 450-ohm feedline length can be somewhere between 50
to 100 feet or so. Tnx for the input!

I guess then that you didn't look at the article I quoted.

Typical T match ATU's are lossier on capacitive loads than on inductive
loads.

The odd eighth wave rule of thumb is a popular one. But, alternate odd
eight waves (on a resonant load) assures the highest ATU losses for the
given SWR.

These rules of thumb, and there are plenty that are conflicting, are
usually given without explanation of why they work. We are a gullible
lot!

The same occurs with 4:1 voltage tuner baluns which anecdotal evidence
suggests assist match of a wider range of loads. There is good reason to
think that the mechanism behind this is that their own loss assists, and
it is an inefficient work-around for another problem.

Owen

On Mon, 05 Apr 2010 20:33:47 GMT, Owen Duffy wrote:

Bob wrote in
:

The more I look at it, the odd 1/8 wavelengths is probably the way I
will go, connecting to my tuner's 4:1 balun. There will be a 130 foot
flat-top, and the 450-ohm feedline length can be somewhere between 50
to 100 feet or so. Tnx for the input!

I guess then that you didn't look at the article I quoted.

Actually, I did look at the article.

It mentioned the voltage maximum problems, the current maximum
problems, and then said, "Is there a better option?" And I don't
understand the few sentences that follow that query. In other words, I
don't understand the solution -- i.e. "line lengths around 135
degrees longer than voltage maximum" :-)

Bob
k5qwg


Typical T match ATU's are lossier on capacitive loads than on inductive
loads.

The odd eighth wave rule of thumb is a popular one. But, alternate odd
eight waves (on a resonant load) assures the highest ATU losses for the
given SWR.

These rules of thumb, and there are plenty that are conflicting, are
usually given without explanation of why they work. We are a gullible
lot!

The same occurs with 4:1 voltage tuner baluns which anecdotal evidence
suggests assist match of a wider range of loads. There is good reason to
think that the mechanism behind this is that their own loss assists, and
it is an inefficient work-around for another problem.

Owen

Baron wrote:
Roy Lewallen Inscribed thus:

Bob wrote:
On Sun, 04 Apr 2010 20:38:35 +0100, Baron
wrote:

You could get a sample and measure it !
Well, I discarded that idea because I have no idea how.

But then, on a hunch, I checked the manual that came with my MFJ-269,
and sure enough, on page 34, it tells how to measure Velocity Factor,
utilizing the distance to fault mode. It'll take a day or so to
recharge the 269's batteries, and then I'll have at it.

Bob
k5qwg
Unfortunately, it's not really simple to make measurements with
symmetrical line. You'll be exciting a common mode current which will
travel with a different velocity factor and affect the measurement.

Roy Lewallen, W7EL

Please could you elaborate on how and why a common mode current has a
different VF on a balanced line.


I'll take a shot..

The VF, to a first order, depends on the dielectric constant
(permittivity) of the medium separating the conductors of the
transmission line. (more correctly, the medium containing the electric
and magnetic fields) For differential mode, it's the insulation between
the wires (for window line, a value somewhere between that of the
plastic and that of air)...

For common mode, it's more the two wires acting as one conductor against
the surroundings (e.g. earth) as the other conductor. The permittivity
of that tends to be lower than that of the medium between the wires, so
the velocity factor is "faster" for the common mode than the
differential mode.

It's not quite that simple, of course, because the field surrounds the
conductors in all directions, not just conveniently between them.

Another way to look at it is think of a balanced pair with distributed L
and C suspended above a ground plane. The C (per unit length) between
the pair is different than the C to ground, as is the L, for the common
mode vs differential mode.

Bob wrote in
:

....
It mentioned the voltage maximum problems, the current maximum
problems, and then said, "Is there a better option?" And I don't
understand the few sentences that follow that query. In other words, I
don't understand the solution -- i.e. "line lengths around 135
degrees longer than voltage maximum" :-)

The location of voltage maxima depends on the load on the line. If you
were to plot the impedance at various lengths of line, it is highest (and
purely resistive) when fed at a voltage maximum. As the line is lengthed,
that impedance becomes capacitive, and lower, eventually becoming lowest
(purely resistive again) at the current maximum (90° longer than the
point of voltage maximum). Increasing the length further, impedance
becomes inductive and increases eventually becoming highest at the next
voltage maximum. At a point of about 135° longer than the voltage
maximum, the impedance presented to the T match is in the region where it
is most efficient. Alternatively, you could state this as 45° shorter
than a voltage maximum.

This is not your odd eighth wave (from a resonant load) rule, because
that also encourages the capacitive region where losses are higher.

Owen

Baron wrote:

Please could you elaborate on how and why a common mode current has a
different VF on a balanced line.

Sure.

First, a balanced line, whether it's twinlead or coax, doesn't have any
common mode current, by definition -- the lack of common mode is what
makes it balanced. We're talking about a physically symmetrical line.

Whenever you have a two conductor line, you effectively have two
transmission lines, differential mode and common mode. Although you
actually have only one current on each conductor, by taking advantage of
the principle of superposition you can mathematically separate the two
currents into two *sets* or components of currents, analyze their
effects separately to gain a better understanding, and simply add the
results if you want to know the overall solution. The sum of the common
mode and differential currents are the actual conductor currents, and
the sum of the common mode and differential responses is the actual
response.

The differential or transmission line mode waves (voltage and current)
are the components which are equal and opposite on the two conductors,
so the field is strongest between the two conductors, fringing outward
in the case of ladder line. The presence of the dielectric material in a
major portion of the field slows down the waves, lowering the velocity
factor. In the case of coax, the field is entirely within the dielectric
so we can easily calculate the velocity factor if we know the dielectric
constant of the material. In the case of ladder line, we don't know what
fraction of the field is in the air and what's in the dielectric without
a very advanced computer program, so we have to measure the velocity
factor. The fraction and therefore velocity factor changes, by the way,
with frequency, a phenomenon known as dispersion.

The common or antenna mode waves are the components that are equal and
in the same direction or polarity on the two conductors. The field is
the same as it would be if the two conductors were connected together to
make a single conductor. One conductor of the common mode transmission
line is the two conductors of the ladder line, and the other is the
Earth and/or surrounding conductors. These two common mode transmission
line conductors are usually much farther apart than the ladder line
conductors, so the common mode characteristic impedance is higher than
the differential mode impedance. The velocity factor is usually higher,
too, because the field is between the two common mode conductors -- the
ladder line and the Earth --, and almost none of it is in the line
dielectric. So its velocity factor is nearly 1. In my TDR demonstration,
the common mode open end reflection occurred before the larger
differential mode reflection because of the higher velocity factor, so
it looked like a differential mode reflection from a point short of the
end. (And I helped reinforce this mistake in order to get the audience's
attention.)

Any two conductor line supports both modes and behave the same, but coax
is a little easier to understand because the differential and common
mode currents are actually physically separate -- so no mathematical
hocus-pocus is necessary. The differential currents and waves are
entirely inside the cable, and the common mode currents and waves are
outside. The velocity factor inside (differential mode) is determined by
the dielectric material, and the velocity factor of the outside (common
mode) is nearly 1.

Roy Lewallen, W7EL

Bob wrote in
:

....

But then, on a hunch, I checked the manual that came with my MFJ-269,
and sure enough, on page 34, it tells how to measure Velocity Factor,
utilizing the distance to fault mode. It'll take a day or so to
recharge the 269's batteries, and then I'll have at it.

As Roy has explained, you need to stop common mode current from
significantly altering your measurement.

I have had sucess with placing a balun of a string of ferrite cores over
the line. It is easy to observe the effectiveness using a VNA sweep, a
bit tricker with the MFJ269.

I have also found that stretching the line out straight causes the worst
common mode problems, but if you coil it, you have to keep adjacent turns
much further apart than the line's conductor separation.

All this has to be done with the line suspended in the air, well clear of
other dielectrics or conductors. (Hint: fishing line can be your friend!)

Before these analysers, we measured the resonant frequency of a line
section using a GDO. By very loosely coupling the GDO, and reading the
GDO frequency from a calibrated receiver, good results could be obtained.

Owen