hi swr low loss??
 

i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?


even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?


boy i am mixed up

"ml" wrote in message
...
i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?


even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?


boy i am mixed up

High SWR does sort of mean loss. It is just that with ladder line the loss
is very low compaired to coax. If you have almost no loss and multiply it
by 10 due to high swr you still have very small loss. If you start out with
coax and have a much larger loss to start with and multiply it by the same
number , you will have much larger loss.

On Sun, 09 Jan 2005 22:07:00 GMT, ml wrote:
this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?
even stranger was the same equation as above but run for coax,

Hi OM,

Line loss is only specified into a matched load (equal to the Zo of
the line). Into other loads, you have to calculate/measure it for
yourself (SWR multiplies line loss).

Loss is in the resistance of the copper wire. The copper wire of the
center conductor of coax is the limiting factor. Except for the
largest ($$$) grade of coax, that copper center conductor is usually
smaller than the average (¢¢¢) grade of twin lead's conductors.

Given the same SWR, you probably can now appreciate why one is less
lossy than the other. If your twin lead still has too much loss,
replace it with larger conductor (still ¢¢¢ at most bulk wire
outlets).

73's
Richard Clark, KB7QHC

In article ,
Richard Clark wrote:

Loss is in the resistance of the copper wire. The copper wire of the
center conductor of coax is the limiting factor. Except for the
largest ($$$) grade of coax, that copper center conductor is usually
smaller than the average (¢¢¢) grade of twin lead's conductors.

Given the same SWR, you probably can now appreciate why one is less
lossy than the other. If your twin lead still has too much loss,
replace it with larger conductor (still ¢¢¢ at most bulk wire
outlets).

It's not just the wire gauge. Twinlead and ladder line and open-wire
line have a higher characteristic impedance than coax. Assuming a
matched source and load, the ladder line will carry any given amount
of power with a higher voltage across the line, and a lower current
through the line, than a coax will.

Losses are proportional to (I^2*R), and so you can decrease the losses
by either decreasing the resistance (larger wire gauge), or using a
higher-impedance matched feedline (lower currents).

--
Dave Platt AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

ml wrote:

i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?

even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?

boy i am mixed up

The losses at HF are primarily proportional to current. Current is
inversely proportional to the characteristic impedance of the feedline,
i.e. I=V/Z0

So I=V/450 is nine times lower than I=V/50. Even with an "SWR around
5" on the ladder-line and 1:1 on the coax, V/450 is probably lower
than V/50. In fact, for the current to be the same on 450 ohm ladder-
line and 50 ohm coax (for the same power) The RMS voltage on the
ladder-line would need to be nine times the RMS voltage on the
coax. So you see ladder-line can tolerate a relatively high SWR
before its RMS current equals the RMS coax current.
--
73, Cecil http://www.qsl.net/w5dxp


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In article , ml
wrote:

i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?


even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?


boy i am mixed up

hmm thanks to all that replyed, i guess i do understand what you
explaned, as it was straight fwd, but i am still feeling strange about
it, i'll run the numbers again, and if i don't 'see the light' perhaps
i can post it more specifically here again

guess the part that bothers me is having such a high swr in the ladder
and such a low loss, i followed mathmatically what youall said, that
adds up, but justt seems well i guess i just thought mentally that say a
5 or 6 swr had to mean some loss greaater than 4watts "100ft run
don't know why i thought that guess it depends "" on all the factors
you guys mentioned

cool thing we have modeling sofware


m

Dear ml

Here is the part that no on this thread has come right out and talked
about, and it is the answer to what is bothering you.

If you have an IDEAL transmission line - that is, one with no
resistivity in the conductors and no dielectric loss in the
insulation, then the SWR can be sky high, but there will be NO LOSSES.

SWR is nothing more than an unfortunate summation of voltages and
currents that cause the voltages and currents to look more like a
roller coaster than a straight line. Voltages and currents do not
necessarily multiply to give power. Sometimes they multiply to zero -
when the angle between them is 90 degrees. And that's what you have
when there are no losses.

OK?

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

"ml" wrote in message
...
In article , ml
wrote:

i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?


even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?


boy i am mixed up

hmm thanks to all that replyed, i guess i do understand what you
explaned, as it was straight fwd, but i am still feeling strange about
it, i'll run the numbers again, and if i don't 'see the light' perhaps
i can post it more specifically here again

guess the part that bothers me is having such a high swr in the ladder
and such a low loss, i followed mathmatically what youall said, that
adds up, but justt seems well i guess i just thought mentally that say a
5 or 6 swr had to mean some loss greaater than 4watts "100ft run
don't know why i thought that guess it depends "" on all the factors
you guys mentioned

cool thing we have modeling sofware


m

m

Do you have access to a ARRL habdbook?? I see a very informative graph
on page 16-14 of the 1990 ARRL handbook that clearly indicates how
transmission line loss and VSWR are related.

Jerry

In article , ml
wrote:

In article , ml
wrote:

i was reading some technical spec's on a ladder line feed.

the numbers showed it had sorta hi swr around 5 yet the power loss was
really low for it's run, like about 2watts

this confuses me, i always thought that high swr automatically ment
'loss' so what is the piece i am missing?


even stranger was the same equation as above but run for coax, which had
lower swr but higher power lost, which i then attrib to something like
the resistane of the coax?


boy i am mixed up

hmm thanks to all that replyed, i guess i do understand what you
explaned, as it was straight fwd, but i am still feeling strange about
it, i'll run the numbers again, and if i don't 'see the light' perhaps
i can post it more specifically here again

guess the part that bothers me is having such a high swr in the ladder
and such a low loss, i followed mathmatically what youall said, that
adds up, but justt seems well i guess i just thought mentally that say a
5 or 6 swr had to mean some loss greaater than 4watts "100ft run
don't know why i thought that guess it depends "" on all the factors
you guys mentioned

cool thing we have modeling sofware


m

the posts are starting to add up, espically the latest posts, very
greatful for all the help i can see the math now, but need some more
quality time to think/study and try to see what is going on to fully
digest

man i am ****sed i didn't get to learn this stuff in school or anyplace
elese it's all super neetOhh

thanks all