"Antonio Vernucci" wrote in
:

I realized my error!
....
ARRL is always correct!

The ARRL information on "extra loss due to VSWR" is may be incomplete in
that it may not the assumptions that underly the formula used for the
graphs.

The very concept that SWR necessarily increases loss from the matched line
loss figure is flawed.

Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to
calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms
(both VSWR=10).

Now refer to the ARRL... does it explain the difference?

Owen

Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete in
that it may not the assumptions that underly the formula used for the
graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete
in that it may not the assumptions that underly the formula used for
the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.
--
Ian.

I'd never thought of that. I suppose it applies to any situation where the
feeder is electrically short, and the majority of the current is less than it
would be when matched. I presume that the moral is that formulas only really
work when the feeder is electrically long enough for you to be concerned about
what the losses might be.
--

Conversely, for a very short line closed on 5 ohm (instead of 500 ohm), the
extra loss caused by SWR would be higher than that shown on the ARRL graph
(apart from the fact that, when attenuation is so low, the extra attenuation is
generally not of much interest, nor it can be read on the ARRL chart).

Evidently the ARRL chart shows some average between the two cases. On the other
hand they probably had no better way to synthetically illustrate a concept
without giving too many details.

73

Tony I0JX

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

Owen's explicit method should get it right in all cases. If you select
say 0.125 wavelengths of RG213 in Owen's online calculator, the
predicted loss with a 100 ohm load resistance is *less* than the matched
loss. If you change the load to 25 ohms, the predicted loss is *greater*
than the matched loss.

Both of these results make perfect physical sense because the largest
part of the loss is proportional to the square of the current, which
will be greater with the lower-resistance load. The two different
resistances correctly give different results, yet they both have a VSWR
of 2 (based on the 50-ohm system impedance). This shows that VSWR does
not contain sufficient information to give an explicit single-valued
result.

Owen's program will accept a VSWR input, but it correctly posts a bold
red warning that the result is an estimate. If you let the program
select the worst-case load impedance for the supplied value of VSWR,
you're back on track and it can calculate an explicit result.


Although we're debating fractions of a milliBel here, the debate has
shown how often the terms "VSWR" and "load impedance" are used
interchangeably - which they aren't. It isn't a big mistake here, but it
can be in other applications. For example, a solid-state PA designed
for a 50 ohm load will respond very differently to load *impedances* of
100 or 25 ohms, yet the load *VSWR* is the same in both cases.



--

73 from Ian GM3SEK

Ian White GM3SEK wrote in
:

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

I think the ARRL graph is based on a well known, but apparently not well
understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

On terminology, I prefer to not use the term 'extra loss due to VSWR'
because the name implies to many, that it is always positive. IMHO a
better way to speak of the loss is as line loss under mismatched
conditions... and those conditions are more specific than just a VSWR
figure.

In a lot of cases, the approximation is sufficiently accurate... but you
lose visibility of the error when you assume that the approximation is
ALWAYS sufficiently accurate, a Rule of Thumb or ROT for short.

Owen

Owen Duffy wrote in
:

....
I think the ARRL graph is based on a well known, but apparently not
well understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

The formula is developed by integrating I^2 over an electrical half wave
of line on one side or the other from the observation point, according
to PS. (I don't know the origin of the formula, I am not suggesting that
PS invented it, not that he didn't.)

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave. A requirement for this is that line loss
must be relatively low, that the exponential real term in the
transmission line equations is close to zero.

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

For a line that is many half waves, the formula is fine so long as VSWR
is approximately constant (now a very low loss line). If the line is
longer than many half waves, but not an exact integral number of half
waves, then the error in the partial section will be somewhat diminished
relatively by the loss in the complete half wave sections.

If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

So, in summary, the formula is good for low loss half wave lines, or
even longish random length low loss lines, but not good for short random
length lines or very long lines.

So, why is the formula so popular?

Could it be that it underpins one of the popular myths of ham radio,
that VSWR necessarily increases line loss?

Modern computation tools are better than the 70 year old graphical
method. Publication of the formula without qualification with the
underlying assumptions treats the reader as a dummy.

Owen

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

Ian Jackson wrote:
I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

I suspect the ARRL additional loss due to SWR charts
are based on 1/2WL increments of feedlines.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein