In message
,
Cecil Moore writes
On Sep 1, 10:29 am, Ian Jackson
wrote:
Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!

It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
The SWR on a feeder will always be greatest at the load end. However,
because of the feeder losses, the power will also be lowest at the load
end.

So, although the power loss per unit length (in dB) due to SWR will
steadily increase as you approach the load end, because of the feeder
loss, there will be less absolute power (in watts) to lose as you
approach the load end.

The question is, at which end of the feeder is the most absolute power
lost per unit length?
--
Ian

On Sep 1, 1:54*pm, Ian Jackson
wrote:
The question is, at which end of the feeder is the most absolute power
lost per unit length?

I already explained that. On HF, maximum losses depend upon where the
current/voltage-nodes/loops are located on the transmission line. One
cannot answer the above question without knowing that information and
the answer is different depending upon the conditions.

Owen has previously proved that on HF, a higher SWR can result in
lower losses than a flat line when that high SWR is near a current
node and the feedline is less than a quarter wavelength long simply
because the average standing-wave current is *lower* than the flat
line current.
--
73, Cecil, w5dxp.com

In message
,
Cecil Moore writes
On Sep 1, 1:54*pm, Ian Jackson
wrote:
The question is, at which end of the feeder is the most absolute power
lost per unit length?

I already explained that. On HF, maximum losses depend upon where the
current/voltage-nodes/loops are located on the transmission line. One
cannot answer the above question without knowing that information and
the answer is different depending upon the conditions.

Owen has previously proved that on HF, a higher SWR can result in
lower losses than a flat line when that high SWR is near a current
node and the feedline is less than a quarter wavelength long simply
because the average standing-wave current is *lower* than the flat
line current.

Ah, I see what you mean. I was thinking in terms of an 'electrically
long' feeder (many wavelengths long).

I can see that, for a 'short' feeder (for example, if you were feeding a
halfwave dipole via a quarterwave of 300 ohm line), at the antenna
feedpoint you will have a current node (high I^2*R loss), and at the
TX/tuner end, where there is a voltage node, you will have a low I^2*R
loss. It could be that the loss of a matched relatively low-impedance
feeder of 50 or 75 ohms (being inherently higher than that of higher
impedance feeders with similar conductors etc) could be higher than the
average loss of the 300 ohm - even though the antenna will present an
SWR to the feeder of around 4 or 6 to 1.

That still leaves the question of which end of a long feeder has the
greater absolute power loss. Using 'reductio ad absurdum' and 'rule of
thumb' (two very dangerous principles!), I feel that it can only be at
the TX end. If the cable is really long, any power left to be dissipated
at the load end must be negligible compared with that available at the
TX end.
--
Ian

Ian Jackson wrote in
news

The question is, at which end of the feeder is the most absolute power
lost per unit length?

In response to your two posts...

You might be informed of the traditional graphs (or simple underlying
formulas) of "Additional loss due to VSWR" that appear in some
publications including several ARRL publications.

The formulas on which these graphs usually depend rely on a set of
assumptions that are often, if not usually, not stated.

You could be forgiven for interpreting these things to mean that loss per
unit length (in dB) is constant along a transmission line independent of
VSWR.

The traditional RLGC model of a transmission line does give constant loss
per unit length when VSWR=1.

For most practical lines at HF, most of the loss is due to conductor loss
(the R term in RLGC) and very little due to dielectric loss (G in RLGC).
The result of that is that line loss in any incremental length is very
dependent on current, higher in regions of higher current, and lower in
regions of lower current.

Consider the loss in 1m of RG58 at 3.5MHz with a 500+j0 load. It is a
very short line electrically so current is almost uniform. VSWR=10 and
current is approximately 1/3 of that for a matched line for the same
transferred power.

Why should loss be greater than a matched line when I^2 is one tenth the
matched line case? Well, it isn't. The formulas and graphs do not apply
because the (often unstated) underlying assumptions are not met.

Regarding the dummy load, it might seem intuitive that loss is highest
where the VSWR is worst, but that is not correct. The correct answer is
found by calculating the loss along the line.

If you consider the case of 100m of RG58C/U with O/C termination as a
dummy load for 1296MHz...

Lets section the 100m into 10+90m.

Using TLLC (http://www.vk1od.net/calc/tl/tllc.php), the input impedance
of the last 90m is 50.00-j0.01.

Now lets look at the first 10m of our 100m, it has a termination of
50.00-j0.01 and the loss is 8.5dB, only 14% of the input power flows into
the last 90m, 86% if the input power is lost somewhere in the first 10m.

By the same method, you could find that almost 20% of input power is loss
in the first metre, even though the VSWR in that section is almost
perfect.

Owen

In message , Owen Duffy
writes
Ian Jackson wrote in
news

The question is, at which end of the feeder is the most absolute power
lost per unit length?

In response to your two posts...

You might be informed of the traditional graphs (or simple underlying
formulas) of "Additional loss due to VSWR" that appear in some
publications including several ARRL publications.

The formulas on which these graphs usually depend rely on a set of
assumptions that are often, if not usually, not stated.

You could be forgiven for interpreting these things to mean that loss per
unit length (in dB) is constant along a transmission line independent of
VSWR.

The traditional RLGC model of a transmission line does give constant loss
per unit length when VSWR=1.

For most practical lines at HF, most of the loss is due to conductor loss
(the R term in RLGC) and very little due to dielectric loss (G in RLGC).
The result of that is that line loss in any incremental length is very
dependent on current, higher in regions of higher current, and lower in
regions of lower current.

Consider the loss in 1m of RG58 at 3.5MHz with a 500+j0 load. It is a
very short line electrically so current is almost uniform. VSWR=10 and
current is approximately 1/3 of that for a matched line for the same
transferred power.

Why should loss be greater than a matched line when I^2 is one tenth the
matched line case? Well, it isn't. The formulas and graphs do not apply
because the (often unstated) underlying assumptions are not met.

Regarding the dummy load, it might seem intuitive that loss is highest
where the VSWR is worst, but that is not correct. The correct answer is
found by calculating the loss along the line.

If you consider the case of 100m of RG58C/U with O/C termination as a
dummy load for 1296MHz...

Lets section the 100m into 10+90m.

Using TLLC (http://www.vk1od.net/calc/tl/tllc.php), the input impedance
of the last 90m is 50.00-j0.01.

Now lets look at the first 10m of our 100m, it has a termination of
50.00-j0.01 and the loss is 8.5dB, only 14% of the input power flows into
the last 90m, 86% if the input power is lost somewhere in the first 10m.

By the same method, you could find that almost 20% of input power is loss
in the first metre, even though the VSWR in that section is almost
perfect.

That is indeed very interesting - and illuminating. I've always assumed
that, for electrically long feeders, most of the power got lost at the
input end - but I didn't realise that the effect was so dramatic. I must
do a few calculations for myself!

And as for electrically short feeders, I do see how the greatest power
loss is where there are current nodes etc (although, presumably,
averaged over a halfwavelength, the power loss will still be greater
towards the TX end). But I now recall that some time ago, there was a
lengthy discussion (in this NG, I'm sure) about how a matched
low-impedance feeder could actually have more loss than a mismatched
relatively high-impedance feeder.
--
Ian

Ian Jackson wrote in
news
That is indeed very interesting - and illuminating. I've always
assumed that, for electrically long feeders, most of the power got
lost at the input end - but I didn't realise that the effect was so
dramatic. I must do a few calculations for myself!

I chose a case that would be clear. When the line loss is low, it is not
so dramatic. Trying to find Rules of Thumb in all of this is fraught
with problems.


And as for electrically short feeders, I do see how the greatest power
loss is where there are current nodes etc (although, presumably,
averaged over a halfwavelength, the power loss will still be greater
towards the TX end). But I now recall that some time ago, there was a
lengthy discussion (in this NG, I'm sure) about how a matched
low-impedance feeder could actually have more loss than a mismatched
relatively high-impedance feeder.

Well, if you do some calcs yourself, what you learn will stick in your
mind. Just be sure that your study is reasonably comprehensive so you
don't come away with some new, but flawed, Rules of Thumb (RoT).

You could persue the dummy load example a little further. How long does
the line need to be for VSWR1.1, and if the continuous dissipation
limit of RG58 is say 6W/m, what is the continuous power rating?

Owen

On Sep 1, 3:21*pm, Ian Jackson
wrote:
Ah, I see what you mean. I was thinking in terms of an 'electrically
long' feeder (many wavelengths long).

Even then, the same principle applies, at least on HF. If the standing
wave current maximum point is located near the load and the standing
wave current minimum is located near the source, the maximum
transmission line losses due to SWR will be near the load. If one
reverses those conditions, the results will reverse. As Owen suggests,
plot a graph of I^2 along the line - pick your particular load point
and particular source point and almost any outcome is possible.
--
73, Cecil, w5dxp.com

On 9/1/2010 11:35 AM, Cecil Moore wrote:

...
When I vary the length of my ladder-line to obtain system resonance, I
have to be satisfied with purely resistive impedances between 35 ohms
and 85 ohms. I cannot achieve a perfect 50 ohms on all HF bands with
my matching method. That would require the addition of an actual N:1
transformer which is certainly possible but probably not worth the
effort. Since 35-85 ohms is perfectly acceptable to my SC-500 amp, I
don't need a high-power tuner. And since the SC-500 is spec'ed to
handle an SWR of 6:1, I doubt that an 35-85 ohm load makes it
"unhappy". If my SC-500 has ever been "unhappy", I failed to observe
that "unhappiness" but maybe I am just oblivious to such?
--
73, Cecil, w5dxp.com

No, I got you. Soon as you stepped though the logic, the match became
quite clear as a "third leg" ... glad you have a "happy amp" ... but if
the amp goes overboard, escalates its' state of happiness and goes into
the "gay zone", DUMP IT at the nearest swap! stoopid grin

Regards,
JS

On 9/1/2010 11:54 AM, Ian Jackson wrote:
The SWR on a feeder will always be greatest at the load end. However,
because of the feeder losses, the power will also be lowest at the load
end.

So, although the power loss per unit length (in dB) due to SWR will
steadily increase as you approach the load end, because of the feeder
loss, there will be less absolute power (in watts) to lose as you
approach the load end.

The question is, at which end of the feeder is the most absolute power
lost per unit length?

At the transmitter end. Suppose you have a 100 watt transmitter and a
cable with 3 dB loss per 100 feet.

50 watts is lost in the first 100 feet
25 watts is lost in the next 100 feet
12.5 watts is lost in the next 100 feet

and so forth. If you were to divide this up more finely and plot it,
you'd find an exponential decay of watts loss per unit length versus
distance from the transmitter.

Roy Lewallen, W7EL

Ian Jackson wrote in
news
The question is, at which end of the feeder is the most absolute power
lost per unit length?

You might find Fig 10 at
http://vk1od.net/transmissionline/VSWR/displacement.htm interesting. It
is a plot of the power along a length of mismatched line.

The plot doesn't answer your question directly, but the slope of the line
in Fig 10 is the answer to your question for the scenario modelled. The
slope is steepest around the current maximum nearest the source.

Fig 9 is interesting as it shows the loss for a mismatched line can dip
below the loss line for VSWR=1.

The thing that becomes apparent from Fig 9 is that the common practice of
doubling loss for double the line length works for the VSWR=1 case, but
not for the general case. As a mismatched line is lengthed, the
*increased* loss for each *extra* metre of length *approaches* the
matched line loss and the line length approaches infinity.

Owen