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

"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

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

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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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.

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

I'm just wondering how 99.9% of the power can be radiated
from the feedline when 80% of the power turns to heat in
the feedline. Is that what is known as gain? :-)
--
73, Cecil http://www.qsl.net/w5dxp

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

I'm actually trying to make a technical point, Reg.
Walt may be trying to make that same technical point.
The SWR is based on differential energy which doesn't
radiate from the feedline. It just seems to me that
you are trying to have your reflected power and radiate
it too. That doesn't work for radiated power any better
than it works for cake. If the SWR is 1400:1 then the
reflected power is almost equal to the forward power both
of which are associated with *differential* currents which
don't radiate.

If the SWR on the feedline is 1400:1, almost no power
is being radiated! The forward power is 100 watts
then the reflected power is 99.7 watts. Both of these
powers are based on *differential* currents and therefore
don't radiate from the feedline. There's only 0.3 watts
available to radiate.

If 99.9% of the power is being radiated by the outside
braid of the coax, then the reflected power cannot be
more than 0.1% of the power and as a result of that
fact, the SWR on the feedline must necessarily be
very low, i.e. close to 1:1. Seems to me you need to
resolve that contradiction.

You cannot radiate 99.9% your reflected power and still
have it available to the SWR measuring equipment. If the
reflected power is available to the SWR measuring equipment,
it is composed of *differential* currents and is, by definition,
not radiating. If it is radiating, then it is not available to
the SWR measuring equipment and the SWR is, therefore, low. You
simply cannot have a sky high SWR on the feedline with the
feedline radiating 99% of the power. The SWR meter cannot tell
if the antenna or coax braid is doing the radiating and will
report a very low SWR.
--
73, Cecil, http://www.qsl.net/w5dxp

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"Cecil Moore" wrote
... If 99.9% of the power is being radiated by the outside
braid of the coax, then the reflected power cannot be
more than 0.1% of the power and as a result of that
fact, the SWR on the feedline must necessarily be
very low, i.e. close to 1:1 ...
_____________

.... and therefore incapable of melting down the inner conductor of the coax,
and/or causing catastrophic failure of components in the output network of
the tx PA.

But, given enough tx power, these failure events are common when a tx tries
to supply its full output power into a very high mismatch at the end of a
run of coax. I have had to fix some of these systems after this happened to
them.

Please explain how this fits with your theory, Reg.

RF