If anyone besides me is having trouble keeping track of what Reg is
trying to say, maybe the following recent quotes will help. That is,
they'll help you understand why you're having trouble keeping track.

For what it's worth, I agree with the first quotation of 8/31. But
apparently Reg doesn't, even though he said it.

Roy Lewallen, W7EL

8/30:

Perhaps somebody might be prepared to state the power actually
radiated from feedlines in watts. At least it may create the
impression you know what you are talking about.

8/30:

Roy, do everybody a favour, by stating, numerically, how much power is
radiated from feedlines. Then somebody might have some confidence in
what you are bafflegabbing about.

8/31:

Any current which finds its way on to (3), which ought to flow in the
antenna, will result in the coax participating in the radiation
pattern of the whole antenna structure.

BUT IT IS A SILLY QUESTION TO ASK HOW MANY WATTS ARE RADIATED FROM THE
FEEDLINE.

OR, CONVERSELY, HOW MANY MICROWATTS ARE PICKED UP BY THE FEEDLINE ON
RECEIVE.

The feedline cannot be treated in isolation as if it behaves
independently of the antenna. For starters, the radiation resistances
of the feedline and antenna do not add arithmetically. They interact
with each other.

8/31:

If you erect a 10 meter 1/2-wave dipole and center-feed it with a 50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

http://www.smeter.net/feeding/feedpowr.php:

Centre-Fed Dipole - Radiation from Coaxial Feedline
Author: R.J.Edwards G4FGQ © 25th March 2003

Program Notes
.. . .
With no loss in accuracy, to simplify the model, the transmitter is
located at the dipole centre and the coaxial line is replaced by a
single conductor of the same diameter as the coaxial braid. There are 3
radiating elements. Radiation resistance, input impedance, and input
current of each element is calculated. Finally, the percent of total
power radiated by each of the three elements is calculated.

Reg Edwards wrote:
If you erect a 10 meter 1/2-wave dipole and center-feed it with a 50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

I think most RF people would consider radiant heat from the feedline
to be heat dissipation (as opposed to RF radiation).
--
73, Cecil http://www.qsl.net/w5dxp

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Reg, put up the corkscrew, your arguing with yourself. Here, we put people
in rubber rooms that exhibit these propensities...

"Reg Edwards" wrote in message
...

"Walter Maxwell" wrote
Will you please describe the details of your experiment where you
claim 90% of the power is radiated from the feedline? How did you
make
this measurement?

==================================

Walt,

If you erect a 10 meter 1/2-wave dipole and center-feed it with a 50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

SWR on the line will be about 1400 and the amount of power available
will not be very great.
----
Reg.

Roy, W7EL wrote:
"The calculations you qupte assume equal and opposite (i.e. purely
differential mode) currents in the feedline conductors. The topic under
discussion is radiation due to common mode current."

Radiation from current on the outside of a coax feedline can be
calculated too by using formulas appropriate to the actual current
distribution.

Refer to the infamous "ARRL Antenna Book" figure (Fig 24 on page 26-16
of my 19th edition), "showing various current paths at feed point of a
balanced dipole fed eith unbalanced coaxial cable."

From the text:

"The antenna current in arm 2 is thus equal to the difference between I2
and I3.

The magnitude of I3 (on coax exterior) is proportional to the relative
impedance in each current path beyond the split."

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
"The antenna current in arm 2 is thus equal to the difference between I2
and I3.
The magnitude of I3 (on coax exterior) is proportional to the relative
impedance in each current path beyond the split."

Could be Roy or Walt wrote that. :-)
--
73, Cecil, http://www.qsl.net/w5dxp

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"Walter Maxwell" wrote in message
...
On Thu, 1 Sep 2005 07:35:35 +0000 (UTC), "Reg Edwards"
wrote:


"Walter Maxwell" wrote
Will you please describe the details of your experiment where you
claim 90% of the power is radiated from the feedline? How did you
make
this measurement?

==================================

Walt,

If you erect a 10 meter 1/2-wave dipole and center-feed it with a
50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

SWR on the line will be about 1400 and the amount of power
available
will not be very great.
----
Reg.

Sorry to disagree, Reg, But if a balun at the feedline-antenna
junction is a perfect balun at 1.9 MHz, NO power will be radiated
from
the feedline with any amount of SWR, even 1400. It is true that only
0.1% of the power will be radiated, but the 99.9% will be reflected
back to the input of the line, not radiated,, making a 50-ohm
transmitter see a 1400:1 mismatch. In the steady state the
transmitter
will only deliver 0.1% of its available power into the line due to
the
mismatch.

The 1400 SWR inside the coax will not cause it to radiate, unless
the
outer conductor is full of holes.

============================
Walt,

Back along the thread you asked for details of the experiment which
demonstrated that a feedline could radiate as much as 90 percent (of
the available power). But you received no reply.

So I provided an example in which nearly all the power is radiated by
the feedline simply because the short antenna is unable to accept very
much and is inefficient. The power available at the antenna input HAS
to go somewhere if it cannot be accepted by the antenna.

Walt, my statement is correct. There is NO choke balun involved. The
unbalanced coax goes straight into the balanced antenna which
inevitable injects power into the outside of the coax. The choke is
omitted just to acheive that purpose.

I made the statement because you seemed to doubt it was possible for
90 percent of the available power ever to be radiated from an
unbalanced feedline. I may possibly have mis-interpretted you.

Admitted, there may not be very much power available at the antenna
due to high SWR and loss within the coax. But nevertheless most of the
power will be radiated by the outside of the feedline. The feedline
makes an excellent, efficient antenna wire compared with the dipole.

An approximate analysis can possibly be done by using Eznec and
placing the transmitter directly at the center of the dipole.

The outer conductor of the coax can be represented by a thick wire
down to ground, the transmission loss inside the coax being ignored
without having the slightest effect on the experiment.

Incidentally, the length of the feedline in wavelengths and the
resistance of the bottom end ground connection will have a
considerable effect on results.

I mention Eznec because of the number of people who swear by it.

For Roy's benefit, the experiment will not tell him how many watts are
radiated from the feedline because it is impossible to separate the
three radiation resistances involved and there will be only one very
peculiar radiation pattern from their combination.

I entered this extremely long-standing discussion only to say that, in
general, the amount of "Power Radiated from a Feedline" is not worth
the effort which has been expended in waffling about it.

I use the words "Power Radiated from a Feedline" only because, to make
myself understood, I am obliged to use the same confusing language as
used by old-wives for the last 50 years. I try to be helpful.
----
Reg, G4FGQ

Reg Edwards wrote:
So I provided an example in which nearly all the power is radiated by
the feedline simply because the short antenna is unable to accept very
much and is inefficient.

But how much of that power is heat and how much is RF?
With an SWR of 1000:1, the I^2*R losses are bound to
be pretty high.
--
73, Cecil http://www.qsl.net/w5dxp


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On Thu, 1 Sep 2005 07:35:35 +0000 (UTC), "Reg Edwards"
wrote:


"Walter Maxwell" wrote
Will you please describe the details of your experiment where you
claim 90% of the power is radiated from the feedline? How did you
make
this measurement?

==================================

Walt,

If you erect a 10 meter 1/2-wave dipole and center-feed it with a 50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

SWR on the line will be about 1400 and the amount of power available
will not be very great.
----
Reg.

Sorry to disagree, Reg, But if a balun at the feedline-antenna
junction is a perfect balun at 1.9 MHz, NO power will be radiated from
the feedline with any amount of SWR, even 1400. It is true that only
0.1% of the power will be radiated, but the 99.9% will be reflected
back to the input of the line, not radiated,, making a 50-ohm
transmitter see a 1400:1 mismatch. In the steady state the transmitter
will only deliver 0.1% of its available power into the line due to the
mismatch.

The 1400 SWR inside the coax will not cause it to radiate, unless the
outer conductor is full of holes.

Walt

But how much of that power is heat and how much is RF?
With an SWR of 1000:1, the I^2*R losses are bound to
be pretty high.
=================================

Your question is not relevant. We are concerned only with the possible
high ratio of feedline radiation to radiation from the antenna.

If you are worried about transmission loss in the coax then that can
be reduced by transmitting at a higher frequency than 1.9 MHz where
the 10m dipole provides a better terminating impedance and SWR will be
lower.

With an appropriate line-length in wavelengths the coax can still be
forced into radiating better than the short 10m dipole.

But the conditions under which this occurs are NEVER those under which
the system would normally be operated.

Under normal conditions, eg., transmitting on 29 MHz, radiation from
the feedline is low enough never to cause interference to the
neighbor's television. If such an unwelcome event should occur then
it will be due to legitimate high-power radiation from the dipole and
your conscience will be clear.

Now Cecil, you can stop trying to pull my leg. ;o)
----
Reg.

On Thu, 1 Sep 2005 23:57:37 +0000 (UTC), "Reg Edwards"
wrote:


"Walter Maxwell" wrote in message
.. .
On Thu, 1 Sep 2005 07:35:35 +0000 (UTC), "Reg Edwards"
wrote:


"Walter Maxwell" wrote
Will you please describe the details of your experiment where you
claim 90% of the power is radiated from the feedline? How did you
make
this measurement?

==================================

Walt,

If you erect a 10 meter 1/2-wave dipole and center-feed it with a
50
feet length of coax, and then transmit on 1.9 MHz, 99.9 percent of
available power will be radiated from the feedline and only 0.1
percent from the antenna itself.

SWR on the line will be about 1400 and the amount of power
available
will not be very great.
----
Reg.

Sorry to disagree, Reg, But if a balun at the feedline-antenna
junction is a perfect balun at 1.9 MHz, NO power will be radiated
from
the feedline with any amount of SWR, even 1400. It is true that only
0.1% of the power will be radiated, but the 99.9% will be reflected
back to the input of the line, not radiated,, making a 50-ohm
transmitter see a 1400:1 mismatch. In the steady state the
transmitter
will only deliver 0.1% of its available power into the line due to
the
mismatch.

The 1400 SWR inside the coax will not cause it to radiate, unless
the
outer conductor is full of holes.

============================
Walt,

Back along the thread you asked for details of the experiment which
demonstrated that a feedline could radiate as much as 90 percent (of
the available power). But you received no reply.

So I provided an example in which nearly all the power is radiated by
the feedline simply because the short antenna is unable to accept very
much and is inefficient. The power available at the antenna input HAS
to go somewhere if it cannot be accepted by the antenna.

Walt, my statement is correct. There is NO choke balun involved. The
unbalanced coax goes straight into the balanced antenna which
inevitable injects power into the outside of the coax. The choke is
omitted just to acheive that purpose.

snip

I use the words "Power Radiated from a Feedline" only because, to make
myself understood, I am obliged to use the same confusing language as
used by old-wives for the last 50 years. I try to be helpful.
----
Reg, G4FGQ


Thank you for the detailed explanation, Reg, I'm sorry I began
transmission while off frequency. I jumped on too soon without
considering you were talking specifically a condition without a balun.
You are correct about the feedline radiation without the balun.
However, with the 1400 SWR in the case you described, because even
though the SWR at the input of the feedline will be substantially less
that at the load, the mismatch seen by the source will allow only an
insignificant amount of the available power to be delivered to the
feedline.

Walt