What is the maximum power I can safely use on 160M and 80M on 26 ga.
copper-plated stranded steel wire? 200 watts?

What is the minimum gauge for a longwire of this material if I want it
to handle 600 watts?

Ken KC2JDY

Ken
(to reply via email
remove "zz" from address)

Your question cannot be answered unless you specify the maximum allowable
temperature of the wire and whether your antenna is in northern Alaska in
midwinter at midnight, or in New Mexico at midsummer at noon.

But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat above
ambient, when in a slight breeze.

So for a 130 foot, 160-meter dipole, allow 130 watts dissipation. This
antenna will easily handle 1 kilowatt of Tx power at a typical efficiency of
about 87 percent for copper-plated, stranded steel wire.
----
Reg, G4FGQ



"Ken" wrote in message
...
What is the maximum power I can safely use on 160M and 80M on 26 ga.
copper-plated stranded steel wire? 200 watts?

What is the minimum gauge for a longwire of this material if I want it
to handle 600 watts?

Ken KC2JDY

Ken
(to reply via email
remove "zz" from address)

But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat
above
ambient, when in a slight breeze.

But the current distribution isn't linear, and the loss is I^2R, so I would
think you'd have to know the max current, and work it that way.


--
KC6ETE Dave's Engineering Page, www.dvanhorn.org
Microcontroller Consultant, specializing in Atmel AVR

"Dave VanHorn" wrote -
Reg wrote -
But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat
above
ambient, when in a slight breeze.

But the current distribution isn't linear, and the loss is I^2R, so I
would
think you'd have to know the max current, and work it that way.

============================
Yes Dave, I know. But I did say the wire gets warm at the centre. Sorry I
forgot to say it remains cold at the ends except for the small conduction of
heat along the wire. And it does not get quite so warm when it rains or
snows.

Actually, the current distribution along a 1/2-wave dipole is exactly a
sinewaveform. This is arithmetically very convenient because the linearly
distributed wire resistance then behaves as though it is lumped at the
dipole centre with a value exactly half of the overall distributed value.

Now the end-to-end resistance of a 1/2-wave, 160-meter dipole, made with
26-gauge copper-plated wire, is about 22 ohms. Therefore the feedpoint
resistance of the antenna is the radiation resistance of 73 plus 22/2 ohms =
84 ohms.

Therefore, with Tx power equal to 1000 watts and a current at the centre
equal to 3.45 amps, the overall loss of power dissipated in the antenna wire
is equal to 3.45 squared times half of the end-to-end conductor resistance
of 11 ohms. Which (as I said in my previous message) equals 131 watts.

Which raises the wire temperature a harmless handfull of a few degrees
Celsius. Actual temperature, of course, depends on the open-air ambient
temperature at noon, in June, in the New Mexico desert.

Antenna radiating efficiency = 73/(73+11) = 84 percent, or a loss of 0.76
decibels, or 1/8th of an S-unit. Which, by no stretch of the imagination,
is likely to influence the results of a contest.

Isn't it remarkable what a little bit of ohms-law and arithmetic can do?

Mathematics - don't make me laugh! smiley

Mathematics is what English infants school teachers refer to when whining
for a pay rise. To such a level are what the western-world's standards of
education have been reduced.
----
Reg, G4FGQ

"Reg Edwards" wrote in message
...

"Dave VanHorn" wrote -
Reg wrote -
But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat
above
ambient, when in a slight breeze.

But the current distribution isn't linear, and the loss is I^2R, so I
would
think you'd have to know the max current, and work it that way.

============================
Yes Dave, I know. But I did say the wire gets warm at the centre. Sorry
I
forgot to say it remains cold at the ends except for the small conduction
of
heat along the wire. And it does not get quite so warm when it rains or
snows.

Actually, the current distribution along a 1/2-wave dipole is exactly a
sinewaveform. This is arithmetically very convenient because the linearly
distributed wire resistance then behaves as though it is lumped at the
dipole centre with a value exactly half of the overall distributed value.

Now the end-to-end resistance of a 1/2-wave, 160-meter dipole, made with
26-gauge copper-plated wire, is about 22 ohms. Therefore the feedpoint
resistance of the antenna is the radiation resistance of 73 plus 22/2 ohms
=
84 ohms.

Therefore, with Tx power equal to 1000 watts and a current at the centre
equal to 3.45 amps, the overall loss of power dissipated in the antenna
wire
is equal to 3.45 squared times half of the end-to-end conductor resistance
of 11 ohms. Which (as I said in my previous message) equals 131 watts.

Which raises the wire temperature a harmless handfull of a few degrees
Celsius. Actual temperature, of course, depends on the open-air ambient
temperature at noon, in June, in the New Mexico desert.

Antenna radiating efficiency = 73/(73+11) = 84 percent, or a loss of 0.76
decibels, or 1/8th of an S-unit. Which, by no stretch of the imagination,
is likely to influence the results of a contest.

Isn't it remarkable what a little bit of ohms-law and arithmetic can do?

Mathematics - don't make me laugh! smiley

Mathematics is what English infants school teachers refer to when whining
for a pay rise. To such a level are what the western-world's standards of
education have been reduced.
----
Reg, G4FGQ


Reg
I went to public schools in Texas in the 50's and 60's.
There were no standards then. School was a pathetic joke.
If I hadn't had ham radio, cars and flying as teenage passions, I never
would have learned math or physics.
73
H., NQ5H

On Tue, 21 Sep 2004 09:04:00 -0500, "Dave VanHorn"
wrote:

But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat
above
ambient, when in a slight breeze.

But the current distribution isn't linear, and the loss is I^2R, so I would
think you'd have to know the max current, and work it that way.

I don't think you should count on a perfect SWR match, either, and
should size for some amount of mismatch.

Worst case kind of thinking.

Happy trails,
Gary (net.yogi.bear)
------------------------------------------------
at the 51st percentile of ursine intelligence

Gary D. Schwartz, Needham, MA, USA
Please reply to: garyDOTschwartzATpoboxDOTcom

Reg Edwards wrote:

"Dave VanHorn" wrote -

But the current distribution isn't linear, and the loss is I^2R, so I
would
think you'd have to know the max current, and work it that way.

============================
Yes Dave, I know. But I did say the wire gets warm at the centre. Sorry I
forgot to say it remains cold at the ends except for the small conduction of
heat along the wire. And it does not get quite so warm when it rains or
snows.

This can be empirical verified by running a 1KW signal into the
antenna and feeling the ends!

(Kids! Don't try this at home!)

Irv VE6BP


--------------------
Irv Finkleman,
Grampa/Ex-Navy/Old Fart/Ham Radio VE6BP
Calgary, Alberta, Canada

There is NO sharp dividing line between rec.radio.amateur.antenna and
politics. Or any other newsgroup.

It's just a matter of keeping topics in sensible proportion.

How else would I learn about Bush, puppy-dog Blair, atmospheric polution,
the frequency of forest fires caused by 100-watt transmitters with
short-whip mobile antennas high up in trees, and the impending state of the
present excellent Californian winery's?
----
Reg, G4FGQ

zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzt!
"Irv Finkleman" wrote in message
...
Reg Edwards wrote:

"Dave VanHorn" wrote -

But the current distribution isn't linear, and the loss is I^2R, so I
would
think you'd have to know the max current, and work it that way.

============================
Yes Dave, I know. But I did say the wire gets warm at the centre.
Sorry I
forgot to say it remains cold at the ends except for the small
conduction of
heat along the wire. And it does not get quite so warm when it rains or
snows.

This can be empirical verified by running a 1KW signal into the
antenna and feeling the ends!

(Kids! Don't try this at home!)

Irv VE6BP


--------------------
Irv Finkleman,
Grampa/Ex-Navy/Old Fart/Ham Radio VE6BP
Calgary, Alberta, Canada

What effect do you expect a mismatch to have on the current or the
current distribution on the antenna?

Roy Lewallen, W7EL

Gary S. wrote:

On Tue, 21 Sep 2004 09:04:00 -0500, "Dave VanHorn"
wrote:


But to be on the safe side, allow for a dissipation of 1 watt per foot
length of dipole wire which will then get warm in the centre, somewhat
above
ambient, when in a slight breeze.

But the current distribution isn't linear, and the loss is I^2R, so I would
think you'd have to know the max current, and work it that way.


I don't think you should count on a perfect SWR match, either, and
should size for some amount of mismatch.

Worst case kind of thinking.

Happy trails,
Gary (net.yogi.bear)
------------------------------------------------
at the 51st percentile of ursine intelligence

Gary D. Schwartz, Needham, MA, USA
Please reply to: garyDOTschwartzATpoboxDOTcom