High Impedance Feedlines = Lower Loss - Why ?
 

I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

I guess an analogy if the above is true could be made about the 120Kv
+ power lines on tall steel towers that are about 500 feet behind my
shack. ( Lucky me ! ) They have much less loss than trying to run say
120 volts and all the current flow that would entail for the same
wattage delivered to homes, business etc ? I can imagine the size of
the conductors required to deliver the same amount of wattage at 120V
VS 120 Kv +/-.

Thanks .... Gary

Gary wrote:
I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.
--
73, Cecil http://www.qsl.net/w5dxp


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Gary:

This question is out of my league, suspect Roy will be able to answer with
no problem--probably a lot of the others too...

However, in my expermenting, I have adusted the match at an antenna fed with
300 twin lead (causing a high SWR) while watching Field Strength on a meter
which was positioned so it could only "see" a good section of the
feedline... you could watch radiation from the feedline go up with SWR--I
cringe when they say coax has even more "loss"--or perhaps this is not the
"loss" you mean...

Warmest regards,
John
--
Marbles can be used in models with excellent results! However, if forced
to keep using all of mine up... I may end up at a disadvantage... I seem
to have misplaced some!!!


"Gary" wrote in message
...
| I've read for years ( and never asked why ) that when you're operating
| into a high SWR that a high impedance feedline ( say 450 Ohm ladder
| line VS 52 Ohm coax ) provides much less loss. I think I recall
| someone in this group saying that its mostly current losses. Does the
| high impedance line have higher voltage points across its length and
| therefore less current flow for a give power level ( say 100 watts )
| than the 52 Ohm coax ?
|
| I guess an analogy if the above is true could be made about the 120Kv
| + power lines on tall steel towers that are about 500 feet behind my
| shack. ( Lucky me ! ) They have much less loss than trying to run say
| 120 volts and all the current flow that would entail for the same
| wattage delivered to homes, business etc ? I can imagine the size of
| the conductors required to deliver the same amount of wattage at 120V
| VS 120 Kv +/-.
|
| Thanks .... Gary

On Fri, 13 May 2005 23:33:40 -0500, Cecil Moore
wrote:

Gary wrote:
I've read for years ( and never asked why ) that when you're operating
into a high SWR that a high impedance feedline ( say 450 Ohm ladder
line VS 52 Ohm coax ) provides much less loss. I think I recall
someone in this group saying that its mostly current losses. Does the
high impedance line have higher voltage points across its length and
therefore less current flow for a give power level ( say 100 watts )
than the 52 Ohm coax ?

The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.

Thanks Cecil ! In your example it appears that the coax is carrying
about 3 times the current of the 450 Ohm ladder line. That explains a
lot.

73 Gary

Thanks for the reply and example John. I was referring to the loss in
power when operating into a high SWR with coax VS a high impedance
line like twinlead / ladder line / open wire. Someone in a post here
mentioned it was mainly current losses and that piqued my interest.
Cecil answered my question and gave me the formula. In his example the
coax was carrying about 3 times the current of the 450 Ohm ladder
line, which explains it.

73 Gary

On Fri, 13 May 2005 21:47:40 -0700, "John Smith"
wrote:

Gary:

This question is out of my league, suspect Roy will be able to answer with
no problem--probably a lot of the others too...

However, in my expermenting, I have adusted the match at an antenna fed with
300 twin lead (causing a high SWR) while watching Field Strength on a meter
which was positioned so it could only "see" a good section of the
feedline... you could watch radiation from the feedline go up with SWR--I
cringe when they say coax has even more "loss"--or perhaps this is not the
"loss" you mean...

Warmest regards,
John

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than the
inner conductor of coaxial lines.

Thicker wires mean lower resistance.

Lower resistance means lower loss.
----
Reg

Gary wrote:

Cecil Moore wrote:
The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
^^^^^^^
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.

Thanks Cecil ! In your example it appears that the coax is carrying
about 3 times the current of the 450 Ohm ladder line. That explains a
lot.

Yep, just noticed a typo above where "voltage" should have been
"current" above. Hope that was obvious. The ratio of the current
between 50 ohm coax and 450 ohm ladder-line is indeed 3 to 1 *FOR
EQUAL SWRs*.

Taking it to the next level of understanding, what if the SWRs are
not equal? Let's say we have 50 ohm coax and a 50 ohm load. The
system is matched and current flows only one way. Total current
for 100 watts equals SQRT(100/50) = 1.414 amps.

Now let's feed the 50 ohm load with 450 ohm ladder-line. The SWR
will be 9:1. Forward power is 278 watts and reflected power is
178 watts. Forward current is SQRT(278/450) = 0.79 amps. Reflected
current is SQRT(178/450) = 0.63 amps. Both of those currents cause
total I^2*R losses roughly equivalent to their sum. Their sum is
1.414 amps, the same as the forward current in the matched coax.

So the losses in 50 ohm coax and 450 ohm ladder-line are roughly
equivalent using similar size wire and driving a 50 ohm load.

I'll leave it as an exercise as to what happens when 50 ohm coax
vs 450 ohm ladder-line is used to drive a 450 ohm load.
--
73, Cecil http://www.qsl.net/w5dxp

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Gary wrote:
Thanks for the reply and example John. I was referring to the loss in
power when operating into a high SWR with coax VS a high impedance
line like twinlead / ladder line / open wire. Someone in a post here
mentioned it was mainly current losses and that piqued my interest.
Cecil answered my question and gave me the formula. In his example the
coax was carrying about 3 times the current of the 450 Ohm ladder
line, which explains it.

Remember, that is for the *SAME SWR*. When the SWRs are different,
as they will be for a fixed load, that loss ratio figure will vary
away from 3 to 1.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg Edwards wrote:

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than the
inner conductor of coaxial lines.

Thicker wires mean lower resistance.

Lower resistance means lower loss.

That is true. But the number one reason that matched line loss
for 450 ohm ladder-line is lower than matched line loss for RG-213
at HF is the effect of (characteristic impedance = load) which is
the same effect as Ohm's law.

Given RG-213 vs 450 ohm ladder-line the losses are *roughly*
equal when:

SWR(coax)/50 = SWR(ladder-line)/450

or, in general, when:

SWR1/Z01 = SWR2/Z02
--
73, Cecil http://www.qsl.net/w5dxp

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The main reason a high impedance feedline equals lower loss is that the
resistance in the feedline is the main loser. Its loss is current
squared times the total resistance. The same power can be conveyed as a
high voltage and a low current (high impedance), or as a low voltage and
a high current (low impedance).

When conveyed as a high current and a low voltage, the power extracted
by resistance is higher for a given power conveyence. Powerline voltages
are often limited by flashover voltage. That`s a reason for d-c
high-voltage power transmission. D-C is at peak value all of the time,
and needs to be insulated for no higher voltage. Maximum voltage means
minimum current for a given power. Ciurrent squared times the resistance
is lowest too.

Loss increases with line length for a particular cross section. A
rule-of-thumb for powerlines is that your transmission voltage should be
1000 volts per mile the energy is to be transported.

Power is volts x amps x cos theta. Cos theta is the power factor which
should be 1 to minimize total current (zero reactive current). Impedance
of the line is volts / amps.

Best regards, Richard Harrison, KB5WZI

Cecil said -
That is true.

===========================
Cec,

With this sort of argument you must not compare one manufacturer's
cable with another. You can't believe the sales-talk anyway

Why unnecessaily complicate the question? Why introduce standing
waves and reflections and all the other silly time wasting
distractions?

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

The exact simple mathematical relationship is -

Line attenuation = 8.69*R/2/Ro dB.

Where R is the resistance of the wire and Ro is the real component of
line impedance, all in ohms.

Make a note of it in your notebooks.

And, hopefully, that should be the end of the matter. But, knowing
you lot, it probably won't be. ;o)
----
Reg, G4FGQ















But the number one reason that matched line loss
for 450 ohm ladder-line is lower than matched line loss for RG-213
at HF is the effect of (characteristic impedance = load) which is
the same effect as Ohm's law.

Given RG-213 vs 450 ohm ladder-line the losses are *roughly*
equal when:

SWR(coax)/50 = SWR(ladder-line)/450

or, in general, when:

SWR1/Z01 = SWR2/Z02
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 14 May 2005 20:33:41 +0000 (UTC), "Reg Edwards"
wrote:

Cecil said -
That is true.

===========================
Cec,

With this sort of argument you must not compare one manufacturer's
cable with another. You can't believe the sales-talk anyway

Why unnecessaily complicate the question? Why introduce standing
waves and reflections and all the other silly time wasting
distractions?

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

The exact simple mathematical relationship is -

Line attenuation = 8.69*R/2/Ro dB.

Where R is the resistance of the wire and Ro is the real component of
line impedance, all in ohms.

Make a note of it in your notebooks.

And, hopefully, that should be the end of the matter. But, knowing
you lot, it probably won't be. ;o)
Dear Reg,

Can we also assume that your formula is for "matched line" attenuation
only, and that the attentuation for a given line will actually
increase with SWR?

Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html

"Reg Edwards" wrote
The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher. The exact simple mathematical relationship is -
Line attenuation = 8.69*R/2/Ro dB.
Where R is the resistance of the wire and Ro is the real component of
line impedance, all in ohms
____________

Is skin effect accounted for in your equation? For example at 10MHz, skin
effect confines most of the current to the outer ~21 µm of the conductor.

RF

Reg Edwards wrote:
The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

On that we can disagree. The *number one* reason for attenuation
being higher is because, in a matched feedline, the losses are
proportional to the square of the current, and the current is
inversely proportional to the characteristic impedance of the
feedline, i.e. given #20 wire, a Zo-matched 75 ohm feedline
will have Sqrt(600/75) times the I^2*R losses of a matched 75
ohm feedline. Proof:

SQRT(100w/75) = SQRT(600/75)*SQRT(100w/600)

SQRT(100w/75)/SQRT(600/75) = SQRT(100w/600)

100w/600 = 100w/600

Given that the center conductor of RG-213 is the same size wire as
a parallel feedline, a *very* large percentage of the difference
in matched line dissipation is due to the Z0. (I don't know the
size of the center wire in RG-213 but it looks like #14 or #12.)
I don't think the RG-213 center conductor is at all smaller.
--
73, Cecil http://www.qsl.net/w5dxp


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On Sat, 14 May 2005 08:57:41 +0000 (UTC), "Reg Edwards"
wrote:

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than the
inner conductor of coaxial lines.

I get balanced line from the Wireman, and the wire in his 450 ohm line
http://thewireman.com/antennap.html#balanced
is 18, 16 and 14 AWG, depending on which line you choose. Is that
appreciably bigger than coax center conductor wire?

bob
k5qwg



Thicker wires mean lower resistance.

Lower resistance means lower loss.
----
Reg

Regarding VSWR:
I would think that increase in effective current does cause greater (I*I)*R
losses (ohmic)...

.... also, an effective increase in voltage also causes increased (E*E)/R
losses in the dielectric itself...

But, what kinda of losses are we talking with both these, under 1db? (no
irate flames, please)
http://blake.prohosting.com/mailguy2/chart_1.JPG
http://blake.prohosting.com/mailguy2/chart_2.JPG

Regards,
John
--
Marbles can be used in models with excellent results! However, if forced
to keep using all of mine up... I may end up at a disadvantage... I seem
to have misplaced some already!!!


"Gary" wrote in message
...
| I've read for years ( and never asked why ) that when you're operating
| into a high SWR that a high impedance feedline ( say 450 Ohm ladder
| line VS 52 Ohm coax ) provides much less loss. I think I recall
| someone in this group saying that its mostly current losses. Does the
| high impedance line have higher voltage points across its length and
| therefore less current flow for a give power level ( say 100 watts )
| than the 52 Ohm coax ?
|
| I guess an analogy if the above is true could be made about the 120Kv
| + power lines on tall steel towers that are about 500 feet behind my
| shack. ( Lucky me ! ) They have much less loss than trying to run say
| 120 volts and all the current flow that would entail for the same
| wattage delivered to homes, business etc ? I can imagine the size of
| the conductors required to deliver the same amount of wattage at 120V
| VS 120 Kv +/-.
|
| Thanks .... Gary

On Sat, 14 May 2005 09:21:57 -0500, Cecil Moore
wrote:

Gary wrote:

Cecil Moore wrote:
The Z0 of a feedline forces the ratio of forward voltage to forward
current to be Z0. It also forces the ratio of reflected voltage to
reflected voltage to be Z0. Let's say we have 100 watts forward and
^^^^^^^
50 watts reflected on both 450 ohm feedline and 52 ohm coax. The
forward voltage on the 450 ohm feedline is SQRT(100*450). The forward
current on the 450 ohm feedline is SQRT(100/450). The forward voltage
on the 52 ohm coax is SQRT(100*52). The forward current on the 52 ohm
coax is SQRT(100/52). The same pattern holds for reflected signals.
The effect of Z0 on voltage and current is easy to see.

Thanks Cecil ! In your example it appears that the coax is carrying
about 3 times the current of the 450 Ohm ladder line. That explains a
lot.

Yep, just noticed a typo above where "voltage" should have been
"current" above. Hope that was obvious. The ratio of the current
between 50 ohm coax and 450 ohm ladder-line is indeed 3 to 1 *FOR
EQUAL SWRs*.

Taking it to the next level of understanding, what if the SWRs are
not equal? Let's say we have 50 ohm coax and a 50 ohm load. The
system is matched and current flows only one way. Total current
for 100 watts equals SQRT(100/50) = 1.414 amps.

Now let's feed the 50 ohm load with 450 ohm ladder-line. The SWR
will be 9:1. Forward power is 278 watts and reflected power is
178 watts. Forward current is SQRT(278/450) = 0.79 amps. Reflected
current is SQRT(178/450) = 0.63 amps. Both of those currents cause
total I^2*R losses roughly equivalent to their sum. Their sum is
1.414 amps, the same as the forward current in the matched coax.

So the losses in 50 ohm coax and 450 ohm ladder-line are roughly
equivalent using similar size wire and driving a 50 ohm load.

I'll leave it as an exercise as to what happens when 50 ohm coax
vs 450 ohm ladder-line is used to drive a 450 ohm load.


Thanks again for the examples Cecil, I missed the typo but your
formulas appear to be accurate and that's what I was after.

73 Gary

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance
is
higher.

The exact simple mathematical relationship is -

Line attenuation = 8.69*R/2/Ro dB.

Where R is the resistance of the wire and Ro is the real component
of
line impedance, all in ohms.

Make a note of it in your notebooks.

And, hopefully, that should be the end of the matter. But, knowing
you lot, it probably won't be. ;o)
----
Reg, G4FGQ

================================

To you all.

As predicted, I appear to have stirred up a hornet's nest.

First of all, give credit to where credit is due. The simple equation
is not due to me but to Oliver Heaviside, 1850 - 1925. May God rest
his soul. And mine!

It applies from DC to VHF where the predominent loss is due to
conductor resistance including skin effect. At higher frequencies, say
above 0.5 GHz, loss in the dielectric material begins to play an
important part.

The complete equation is -

Attenuation = R/2/Ro + G*Ro/2 Nepers

where G is the conductance of the dielectric, which is small for
materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
8.686 dB.

The Neper is the fundamental unit of transmission loss per unit length
of line, familiar to transmission line engineers. It is named after
Napier, a canny Scotsman who had something to do with the invention of
Logarithms around the 18th Century.

Attenuation is simply the basic matched loss of a particular line,
unaffected by SWR and all the other encumbrances which amateurs such
as W5DXP ;o) worry about. KISS.

Incidentally, the additional-loss versus SWR curves, published in the
ARRL books and copied by the RSGB, for many years, are based on an
incorrect mathematical analysis. But they are near enough for
practical purposes.

Not that SWR matters very much. SWR meters don't measure SWR on any
line anyway. You are all being fooled. ;o) ;o) ;o)
----
Reg, G4FGQ

I don't use the SWR meter to measure the standing wave anyway... more just
to keep the finals running kewl... lucky for me there is some kind of
relationship to replacing finals in high power amps and the dumb meter...

I can just lay a hand on the amp and start a conversation... adjusting match
for lowest heat (or least smoke)--but the meter takes a lot of guess work
out of it... I miss tubes for that very reason... you could always adjust
for least red glow on the plates... grin

Warmest regards,
John
--
If "God"--expecting an angel... if evolution--expecting an alien... just
wondering if I will be able to tell the difference!

"Reg Edwards" wrote in message
...
| The number one reason for attenuation being higher is because the
| conductor diameter is smaller and, as a consequence, its resistance
| is
| higher.
|
| The exact simple mathematical relationship is -
|
| Line attenuation = 8.69*R/2/Ro dB.
|
| Where R is the resistance of the wire and Ro is the real component
| of
| line impedance, all in ohms.
|
| Make a note of it in your notebooks.
|
| And, hopefully, that should be the end of the matter. But, knowing
| you lot, it probably won't be. ;o)
| ----
| Reg, G4FGQ
|
| ================================
|
| To you all.
|
| As predicted, I appear to have stirred up a hornet's nest.
|
| First of all, give credit to where credit is due. The simple equation
| is not due to me but to Oliver Heaviside, 1850 - 1925. May God rest
| his soul. And mine!
|
| It applies from DC to VHF where the predominent loss is due to
| conductor resistance including skin effect. At higher frequencies, say
| above 0.5 GHz, loss in the dielectric material begins to play an
| important part.
|
| The complete equation is -
|
| Attenuation = R/2/Ro + G*Ro/2 Nepers
|
| where G is the conductance of the dielectric, which is small for
| materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
| 8.686 dB.
|
| The Neper is the fundamental unit of transmission loss per unit length
| of line, familiar to transmission line engineers. It is named after
| Napier, a canny Scotsman who had something to do with the invention of
| Logarithms around the 18th Century.
|
| Attenuation is simply the basic matched loss of a particular line,
| unaffected by SWR and all the other encumbrances which amateurs such
| as W5DXP ;o) worry about. KISS.
|
| Incidentally, the additional-loss versus SWR curves, published in the
| ARRL books and copied by the RSGB, for many years, are based on an
| incorrect mathematical analysis. But they are near enough for
| practical purposes.
|
| Not that SWR matters very much. SWR meters don't measure SWR on any
| line anyway. You are all being fooled. ;o) ;o) ;o)
| ----
| Reg, G4FGQ
|
|

John,

All the so-called SWR meter tells you is whether or not the
transmitter is being loaded with 50 ohms.

This may be a useful thing to know. But it is NOT SWR. Where is the
line on which the SWR is supposed to be measured?

By the way, I think I am receiving all your emails. But you do not
appear to be receiving any of mine. Don't think I do not wish to
speak to you.

Could you check that you can receive other people's emails?
----
Reg, G4FGQ

Reg Edwards wrote:
The complete equation is -

Attenuation = R/2/Ro + G*Ro/2 Nepers

where G is the conductance of the dielectric, which is small for
materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
8.686 dB.

Reg, I didn't disagree with your equation. I disagreed with this
statement of yours:

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

That is simply not a true statement. #13 RG-213 wire is actually
***LARGER*** than #18 ladder-line wire yet the coax still has the
higher matched-line loss. If your statement were true, #13 RG-213
would have lower losses than #18 ladder-line but it doesn't.

The number one reason that coax has higher matched line losses
than ladder-line is NOT primarily due to wire size. It is primarily
due to the differences in characteristic impedance, as I said earlier,
and as proved by your equation above.
--
73, Cecil http://www.qsl.net/w5dxp


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Reg Edwards wrote:
All the so-called SWR meter tells you is whether or not the
transmitter is being loaded with 50 ohms.

This may be a useful thing to know. But it is NOT SWR. Where is the
line on which the SWR is supposed to be measured?

If the reflected power on the line between the transmitter and
the meter equals zero:

SWR = [SQRT(Pf)+SQRT(Pr)]/[SQRT(Pf)-SQRT(Pr)]

SWR = [SQRT(Pf)+0]/[SQRT(Pf)-0] = 1:1
--
73, Cecil http://www.qsl.net/w5dxp


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Reg:

You kinda relate SWR meters to "time", huh? Well, I am used to being fooled
by "non-existant" "things"-"values" which we plug into formulas and somehow
get workable results back!!!! The "Unified Antenna Theory" will be a nice
thing--when it finally gets here... so, a SWR meter which doesn't work as
"mentally modeled"--hey, perfectly understandable!!!

So, not in anyway in disagreement with you, just poking a bit of fun at the
way things work...

Like I say, I miss the old PA tubes--dull almost dark red glow from the
plates--good to excellent... brighter almost cherry red--danger will
robinson!!!!... orange'ish-red--time to buy new finals!!!!! grin

I am at a complete loss on the email... may have picked up some malicious
email from a "fan" of mine... have worn myself out attempting to find the
problem--you can count on me NOT giving up until it is fixed... I keep a
seperate computer for use here--this demonstrates why...

Warmest regards,
John
--
If "God"--expecting an angel... if evolution--expecting an alien... just
wondering if I will be able to tell the difference!

"Reg Edwards" wrote in message
...
| John,
|
| All the so-called SWR meter tells you is whether or not the
| transmitter is being loaded with 50 ohms.
|
| This may be a useful thing to know. But it is NOT SWR. Where is the
| line on which the SWR is supposed to be measured?
|
| By the way, I think I am receiving all your emails. But you do not
| appear to be receiving any of mine. Don't think I do not wish to
| speak to you.
|
| Could you check that you can receive other people's emails?
| ----
| Reg, G4FGQ
|
|

Cecil Moore wrote:
Reg Edwards wrote:

The complete equation is -

Attenuation = R/2/Ro + G*Ro/2 Nepers

where G is the conductance of the dielectric, which is small for
materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
8.686 dB.


Reg, I didn't disagree with your equation. I disagreed with this
statement of yours:

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.

As an illustrated example: Assume a parallel feedline made from #24
wire and having a characteristic impedance of 600 ohms.

What size would the wire in 50 ohm coax have to be to equal the HF
matched line loss of the #24 600 ohm line? (The wire in the coax has
to be 12 times as conductive as the wire in the parallel feedline in
order to offset the effect of Z0.)

A rough estimate indicates that the #24 600 ohm line has approximately
the same matched line loss as RG-213 with its #13 wire.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:
Reg Edwards wrote:

You will get all sorts of technical reasons for lower loss. But
essentially -

The wires in high impedance balanced pair lines are thicker than
the
inner conductor of coaxial lines.

Thicker wires mean lower resistance.

Lower resistance means lower loss.

That is true. But the number one reason that matched line loss
for 450 ohm ladder-line is lower than matched line loss for RG-213
at HF is the effect of (characteristic impedance =3D load) which is
the same effect as Ohm's law.

Given RG-213 vs 450 ohm ladder-line the losses are *roughly*
equal when:

SWR(coax)/50 =3D SWR(ladder-line)/450

or, in general, when:

SWR1/Z01 =3D SWR2/Z02

Wunnerful. But out here in the realities of practical (God forbid)
applications of the various types of backyard feedlines there's a
persistent rumor going back decades to the effect that decent open-wire
feedlines have significantly lower dielectric losses than "ham-level"
coax under all VSWR condx. So there are conductor *and* dielectric I=B2R
losses to consider in this discussion yes?

--
73, Cecil http://www.qsl.net/w5dxp

w3rv


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Brian Kelly wrote:
So there are conductor *and* dielectric I²R
losses to consider in this discussion yes?

Dielectric losses are usually considered to be
negligible at HF.
--
73, Cecil http://www.qsl.net/w5dxp

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Reg:

I am not after your scalp, trust me...

However, as I ready for bed, I was thinking--on the age of my coax...
although some may be as new as 3 years old... most is greater than 5, and I
bet the run to my 1/2 vertical is 20 years or better....

Sometime in the past, I remember reviewing data on loss in coax going up
with age.... not that it would amount to an important loss... but still, it
must be a measurable amount...

Oh, and strange how this all keeps touching on the matter I am constantly
holding at hand... but that "skin effect"... seems like copper becomes an
"impedance" at high freqs.... those little electrons in the wire just can't
keep pumping the charge fast enough... seems like that old rf there is
considering the ether itself (dielectric in coax) as a better choice of
travel than the copper atoms...

Warmest regards,
John
--
If "God"--expecting an angel... if evolution--expecting an alien... just
wondering if I will be able to tell the difference!

"Reg Edwards" wrote in message
...
| The number one reason for attenuation being higher is because the
| conductor diameter is smaller and, as a consequence, its resistance
| is
| higher.
|
| The exact simple mathematical relationship is -
|
| Line attenuation = 8.69*R/2/Ro dB.
|
| Where R is the resistance of the wire and Ro is the real component
| of
| line impedance, all in ohms.
|
| Make a note of it in your notebooks.
|
| And, hopefully, that should be the end of the matter. But, knowing
| you lot, it probably won't be. ;o)
| ----
| Reg, G4FGQ
|
| ================================
|
| To you all.
|
| As predicted, I appear to have stirred up a hornet's nest.
|
| First of all, give credit to where credit is due. The simple equation
| is not due to me but to Oliver Heaviside, 1850 - 1925. May God rest
| his soul. And mine!
|
| It applies from DC to VHF where the predominent loss is due to
| conductor resistance including skin effect. At higher frequencies, say
| above 0.5 GHz, loss in the dielectric material begins to play an
| important part.
|
| The complete equation is -
|
| Attenuation = R/2/Ro + G*Ro/2 Nepers
|
| where G is the conductance of the dielectric, which is small for
| materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
| 8.686 dB.
|
| The Neper is the fundamental unit of transmission loss per unit length
| of line, familiar to transmission line engineers. It is named after
| Napier, a canny Scotsman who had something to do with the invention of
| Logarithms around the 18th Century.
|
| Attenuation is simply the basic matched loss of a particular line,
| unaffected by SWR and all the other encumbrances which amateurs such
| as W5DXP ;o) worry about. KISS.
|
| Incidentally, the additional-loss versus SWR curves, published in the
| ARRL books and copied by the RSGB, for many years, are based on an
| incorrect mathematical analysis. But they are near enough for
| practical purposes.
|
| Not that SWR matters very much. SWR meters don't measure SWR on any
| line anyway. You are all being fooled. ;o) ;o) ;o)
| ----
| Reg, G4FGQ
|
|

I have never heard of ageing effects in copper or polyethylene - or
ancient ebonite spacers even.

The very first coaxial carrier communications cable was laid in Great
Britain by the Post Office, around 1937, between the cities of
Manchester and Leeds. There were 4 coaxial tubes inside a lead
sheath. Outer conductors = 0.375". Inner conductors = 0.1", which
later became the standard. Mostly air spaced. Inner conductors were
supported by ebonite disks or similar material, spaced at about 1.5".
Polyethylene was still waiting to be invented. Working frequencies
from 60 kHz to about 2 MHz. Repeater spacing about 5 miles.

Around 1960 I had the opportunity to test sections of this cable. As
far as I could judge it was in perfect working order. Bear in mind it
is possible to detect small changes in attenuation only by looping
back on very long lengths. It cannot be done in the lab.

I imagine coax, with temperature expansion and contraction, very
slowly 'breathes' through the ends and draws in humid atmospheric
pollution. Perhaps after 50 years it may have some minute detectable
effect on attenuation and appearance. Attenuation is the last
parameter to fail. Far more serious things have to happen to a
transmission line before loss becomes noticeable.

For example, a coax line can be almost flattened with a hammer over a
length of several feet which will make a shocking mess of impedance.
Yet, provided the inner and outer conductors are not in contact with
each other, additional loss will be undetectable.

=========================

Nothing happens to metallic copper with frequency. But copper
conductors also have internal inductance in addition to conductance.
Inductive reactance increases with frequency. The increase in
inductive reactance begins at the centre of the conductor and drives
the current outwards towards the surface or perimeter. At
sufficiently high frequencies the current is forced to flow only on
the conductor's skin.

The conductance of copper remains the same. But the cross-section of
the conductor allowed to the current is very much reduced and so the
effective resistance per unit length increases together with the
inductive reactance.

It's an interesting fact that at frequencies where skin effect is
fully operative, conductor inductive reactance and resistance become
equal to each other. Measure one and you also know the other.
----
Reg, G4FGQ

I should have mentioned, the Manchester-Leeds Number 1 Coaxial Cable
had an impedance of 75 ohms. The impedance at which, for a given
price of copper, in those far-off days, had the lowest attenuation per
mile. 75 ohms has stuck as the Standard..

The distance between Manchester (then the centre of the cotton
industry) and Leeds (then the centre of the woolen industry), by road,
over the beautiful Lancashire and Yorkshire moors, is about 40 English
miles. By correct choice of impedance the conscientious engineers of
that age could have saved as much as £5,000 per mile in the price of
copper, to be formed in the manufactories into copper tapes for outer
coaxial conductors, and drawing copper wire from 3-ton billet-form
down to exact precision-size wire through water-cooled diamond dies.
It was and still is a precision manufacturing industry.

More savings occur in the distance between repeater stations. If
attenuation performance requirements can be met with one fewer
repeater station, the cost of a whole building, power supplies and
transmission equipment can be saved.

Although communications have shifted to digital, cables still matter.
But eventually optical fibers will take over the long distance
communications.

Radio Amateurs, with a little money to burn, never become involved
with such mundane matters. They are more interested in what they
imagine the SWR meter tells them. But if that keeps them happy then
so be it. I am an amateur myself. I have a call sign which sounds
very nice in morse code. Why should I disillusion them?
----
Reg, G4FGQ

On 15 May 2005 17:59:50 -0700, "Brian Kelly" wrote:

So there are conductor *and* dielectric I²R
losses to consider in this discussion yes?


No.

Cecil Moore wrote:

Reg Edwards wrote:

The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher.


On that we can disagree. The *number one* reason for attenuation
being higher is because, in a matched feedline, the losses are
proportional to the square of the current, and the current is
inversely proportional to the characteristic impedance of the
feedline, i.e. given #20 wire, a Zo-matched 75 ohm feedline
will have Sqrt(600/75) times the I^2*R losses of a matched 75
ohm feedline. Proof:

SQRT(100w/75) = SQRT(600/75)*SQRT(100w/600)

SQRT(100w/75)/SQRT(600/75) = SQRT(100w/600)

100w/600 = 100w/600

Given that the center conductor of RG-213 is the same size wire as
a parallel feedline, a *very* large percentage of the difference
in matched line dissipation is due to the Z0. (I don't know the
size of the center wire in RG-213 but it looks like #14 or #12.)
I don't think the RG-213 center conductor is at all smaller.

Resistivity is the 'R' in I^2R, as Reg indicated.

ac6xg

Wes Stewart wrote:
On 15 May 2005 17:59:50 -0700, "Brian Kelly" wrote:

So there are conductor *and* dielectric I=B2R
losses to consider in this discussion yes?
=20
=20
No.

.. . that's unambiguous enough . .

Whilst on the romantic subject of coaxial attenuation -

Attenuation is the number-one characteristic of all transmission
lines. From power frequencies and upwards. Yet, quantitativly, it is
the smallest parameter per mile and the most difficult to measure
accurately. An innocent observer might think it is hardly worth
bothering about.

It is inextricably mixed up with system economics. An exact knowledge
even of the temperature coefficient of attenuation is vital to
communications system design.

Around the 1950's I was involved with measurement of attenuation (and
other characteristics) of the first oceanic submarine telephone
cables. A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels. The temperature
coefficient of attenuation is half of the resistance temperature
coefficient of copper which is 0.4 percent per degree C. Which was
well known to Oliver Heaviside around 1875.

Thus, a change in temperature on the ocean bottom of 0.1 degree
results in a change of 80 dB in the signal level at the far end.
Unless corrected in the repeaters (of which there were about 100) this
is enough to shift signals between the thermal noise level of an
amplifier and its overloaded cross-modulation level.

One of the cable factories was located in Southampton Docks. As cable
came off the production machinery at about 1 mile per hour, it was
coiled in giant circular concrete tanks below ground level, the same
size as an 8000-ton cable-laying ship's hold. Attenuation and other
measurements were made in the tanks by automatic testing equipment.
The cable was then loaded onto a cable ship waiting for it in the
nearby dock.

I designed a special attenuation and phase-shift tester for research
purposes. It did not incorporate an SWR meter. It did incorporate a
phase-locked-loop but it was not until several years later that I came
by chance upon a learned paper by Gruen and discovered how a PLL
really works. The equipment was all tubes. A whole mobile rack of it!
The final output meter was a moving-coil instrument with a scale
calibrated in 0.001 decibels. There were also home-brewed 0.001 dB
stepped attenuators which I had to calibrate myself.

To determine attenuation temperature coefficients in was necessary to
bring tons of ice by lorry from Billingsgate fish market in central
London. One of the concrete tanks was flooded with sea water and the
ice dumped in. It took 24 hours for the temperature to stabilise. I
spent much of the waiting time in a pub in Southampton Town. I never
knew who organised and paid for delivery of the ice which must have
been the most intricate and illegal part of the whole operation.

The data accumulated was rushed to the boffins who immediately began
designing even higher frequency oceanic systems. I was rewarded with a
pat on the back and told to keep my mouth shut. Politics were
involved somewhere.

My tester should have eventually been installed in the Science Museum,
Kensington, London. But long after the job was finished it was stolen
by some unfeeling person and cannibalised for the spare parts. There
was a BC221, straight-line frequency variable tuning capacitor built
into it. I would have liked that for myself.

Hope you enjoyed the story. From what I remember it's mostly true.
Southampton makes a change from Manchester and Leeds. The Queen Mary
was berthed not far from the Cable Ship Monarch.
----
Reg, G4FGQ

Reg Edwards wrote:

A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels. . .

Just to get a little context here. . .

Years ago when I was a little bored, I determined that the ratio of the
light output from a common two cell flashlight to the entire light
output of the Sun is a mere 280 dB (10^28). So if you attenuate the Sun
by 280 dB you get the light of a flashlight beam. Well now, if you took
that flashlight beam and attenuated it again by the same amount, then
did that again, and again, 14 times altogether, you still wouldn't quite
have totaled 4000 dB. It's a staggering number, incomprehensible except
by some pretty abstract thinking. It's real, though. I remember reading
a paper long ago about transatlantic cables, and those are the numbers
they work with.

Roy Lewallen, W7EL

Reg Edwards wrote:
Hope you enjoyed the story.

That's a really enjoyable story, Reg. Thanks for sharing.
During that time I was involved in smuggling operations -
smuggling girls into my Texas A&M dorm. :-) Today, there
are girls all over the Texas A&M dorms.
--
73, Cecil http://www.qsl.net/w5dxp


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Reg Edwards wrote:
Whilst on the romantic subject of coaxial attenuation -

A follow on from Reg's story, is that (in submarine the cables with
which I was familiar), there was another grade of cable also used, a
repair cable that had about two thirds the loss of the regular cable.

It was used when repair was necessary (eg repeater failure, cable damage
from anchors , earthquakes etc). It is not possible to effect a repair
without cutting the original cable in order to get the cable to the surface.

The technique used on CS Monarch and the like, was to use a special
grapnell that caught the cable (often after many days of steaming back
and forth across the suspected cable position) and hauled it up until a
pre-determine tension was reached which activated a cutter in the
grapnell, which separated each end off on a separate hauling line. One
was buoyed off, and the CS steamed toward the other cable end until
sound cable was retrieved. It was sealed and buoyed off. They then
steamed back to the other buoy and retreived the other end, again
steaming until sound cable was found. Calculations were then done of
depth, position, losses to find how much more cable was to be removed so
that when the repair cable was inserted, the S/N into each of the
affected repeaters was sufficient to allow normal operation (these were
linear FDM or carrier telephone cables). (In some cases, so much cable
was affected that a mix of original cable and repair cable was used.)

This operation could take several days in good weather, worse in bad seas.

Owen

PS: 4000 dB sounds a lot, but when it is stated as 40dB between
repeaters it sounds more manageable.

Really, to all you guys:

There is sense, and there is non-sense here...

Never doubted you ALL had the the sense, just pleased you can enjoy a bit of
non-sense...

And, yes, the first time I found out I had to increase effective radiated
power by a factor of 4 to achieve a factor of 2 on someones S-Meter--I was
disapointed--not sure I have fully recoved from the meaning of that to this
very day--frankly, I expected more... I expect if I consulted a
psychiatrist on all this--he would, most likely, chalk it up to "penis
envy"... and that is why I have not... grin

Warmest regards,
John

"Roy Lewallen" wrote in message
...
Reg Edwards wrote:

A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels. . .

Just to get a little context here. . .

Years ago when I was a little bored, I determined that the ratio of the
light output from a common two cell flashlight to the entire light output
of the Sun is a mere 280 dB (10^28). So if you attenuate the Sun by 280 dB
you get the light of a flashlight beam. Well now, if you took that
flashlight beam and attenuated it again by the same amount, then did that
again, and again, 14 times altogether, you still wouldn't quite have
totaled 4000 dB. It's a staggering number, incomprehensible except by some
pretty abstract thinking. It's real, though. I remember reading a paper
long ago about transatlantic cables, and those are the numbers they work
with.

Roy Lewallen, W7EL

Reg, G4FGQ wrote:
"A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels."

That sounds reasonable as it is only 2 dB per mile. A mile is 52.8
increments of 100 feet, so that would produce about 0.038 dB/100 feet.

The lowest loss 75-ohm cable I found listed in the ARRL Antenna Book
table is 7/8-inch Hard-line at about 0.1 dB/100 feet at 5 MHz. For an
intercontinental link, you would strive for better as Reg indicated.

One problem of cable is that it has constant loss. Every cebtimeter of
length takes the same percentage loss of the remaining energy. Hence,
dB/ 100 feet.

Not so with radio in free-space. Getting rid of the wires ends their
attenuation. Loss is then due to decreased signal in a square unit of
the wavefront caused by expansion or thinning of the signal. The
"unattenuated" signal decline is 6 dB every time distance from the
source doubles, be it one mile or 1000 miles. The signal power level at
a point is 1/4 the power it had for the same area at 1/2 the distance
from the source.

We could`not communicate by wire with our space probes due to too much
loss even were the wires a practical alternative.

To cross an ocean, cable solves the problem of repeater placement. The
signal must be regenerated before it falls into noise. The repeaters are
"simply" integrated into the cable at proper intervals. The first
transatlantic cable message was sent by Queen Victoria to the American
President.

Ashore and on distant offshore platforms, I`ve puzzled why microwave was
not used instead of cable. Privacy may be one reason, but encryption,
route switching and other techniques could make theft of information
from thin air more difficult than other theft. There are always
beneficiaries of the status quo who make change difficult to impose.

In the early 1950`s, Houston`s Transcontinental Gas Pipeline Company
(Ken Lay was an officer of "Transco" before moving to Enron) built a
private microwave system from its heasdquarters to New Jersey along its
pipeline. I recall looking the new system over. It was supplied by
Philco Corporation and used Pulse Code Modulation, if I recall. The
microwave system was sold to a communications common carrier (now
Sprint) after a few years but it is still in service, I believe. Transco
(now Williams Pipeline Company) is one of many subscribers to the
service I believe.

Microwave repeaters located at about 20-mile intervals can provide
low-noise and high-reliability communications when properly designed.

In the 1950`s, I marveled as I commuted to work on a stretch of road
which ran between Lisbon and O`Porto, watching the Portuguese Post,
Telephone, and Telegraph Company laying coaxial cable alongside.

Cable is more vulnerable to damage, harder to repair, and surely costs
more than microwave. It was none of my business. I was a foreigner in
their country.

Best regards, Richard Harrison, KB5WZI

Cable doesn't fade from atmospherics.

--

73
Hank WD5JFR
"Richard Harrison" wrote in message
...
Reg, G4FGQ wrote:
"A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels."

That sounds reasonable as it is only 2 dB per mile. A mile is 52.8
increments of 100 feet, so that would produce about 0.038 dB/100 feet.

The lowest loss 75-ohm cable I found listed in the ARRL Antenna Book
table is 7/8-inch Hard-line at about 0.1 dB/100 feet at 5 MHz. For an
intercontinental link, you would strive for better as Reg indicated.

One problem of cable is that it has constant loss. Every cebtimeter of
length takes the same percentage loss of the remaining energy. Hence,
dB/ 100 feet.

Not so with radio in free-space. Getting rid of the wires ends their
attenuation. Loss is then due to decreased signal in a square unit of
the wavefront caused by expansion or thinning of the signal. The
"unattenuated" signal decline is 6 dB every time distance from the
source doubles, be it one mile or 1000 miles. The signal power level at
a point is 1/4 the power it had for the same area at 1/2 the distance
from the source.

We could`not communicate by wire with our space probes due to too much
loss even were the wires a practical alternative.

To cross an ocean, cable solves the problem of repeater placement. The
signal must be regenerated before it falls into noise. The repeaters are
"simply" integrated into the cable at proper intervals. The first
transatlantic cable message was sent by Queen Victoria to the American
President.

Ashore and on distant offshore platforms, I`ve puzzled why microwave was
not used instead of cable. Privacy may be one reason, but encryption,
route switching and other techniques could make theft of information
from thin air more difficult than other theft. There are always
beneficiaries of the status quo who make change difficult to impose.

In the early 1950`s, Houston`s Transcontinental Gas Pipeline Company
(Ken Lay was an officer of "Transco" before moving to Enron) built a
private microwave system from its heasdquarters to New Jersey along its
pipeline. I recall looking the new system over. It was supplied by
Philco Corporation and used Pulse Code Modulation, if I recall. The
microwave system was sold to a communications common carrier (now
Sprint) after a few years but it is still in service, I believe. Transco
(now Williams Pipeline Company) is one of many subscribers to the
service I believe.

Microwave repeaters located at about 20-mile intervals can provide
low-noise and high-reliability communications when properly designed.

In the 1950`s, I marveled as I commuted to work on a stretch of road
which ran between Lisbon and O`Porto, watching the Portuguese Post,
Telephone, and Telegraph Company laying coaxial cable alongside.

Cable is more vulnerable to damage, harder to repair, and surely costs
more than microwave. It was none of my business. I was a foreigner in
their country.

Best regards, Richard Harrison, KB5WZI

Rich H. wrote:
"Getting rid of the wires ends their attenuation."
--and--
"Loss is then due to decreased signal in a square unit of
the wavefront caused by expansion or thinning of the signal."

I missed this on my first read of your post--I think there is some
measureable amount of attentuation by the
(a)ether (is the (a)ether a superconductor?)... it is just a component of
the "loss" noted in the second clip, above...

.... not wishing to make a bit point of it... just pointing it out...

Warmest regards,
John

"Richard Harrison" wrote in message
...
Reg, G4FGQ wrote:
"A transatlantic coaxial cable, 2000 miles long, has an overall
attenuation at 5 MHz of around 4000 decibels."

That sounds reasonable as it is only 2 dB per mile. A mile is 52.8
increments of 100 feet, so that would produce about 0.038 dB/100 feet.

The lowest loss 75-ohm cable I found listed in the ARRL Antenna Book
table is 7/8-inch Hard-line at about 0.1 dB/100 feet at 5 MHz. For an
intercontinental link, you would strive for better as Reg indicated.

One problem of cable is that it has constant loss. Every cebtimeter of
length takes the same percentage loss of the remaining energy. Hence,
dB/ 100 feet.

Not so with radio in free-space. Getting rid of the wires ends their
attenuation. Loss is then due to decreased signal in a square unit of
the wavefront caused by expansion or thinning of the signal. The
"unattenuated" signal decline is 6 dB every time distance from the
source doubles, be it one mile or 1000 miles. The signal power level at
a point is 1/4 the power it had for the same area at 1/2 the distance
from the source.

We could`not communicate by wire with our space probes due to too much
loss even were the wires a practical alternative.

To cross an ocean, cable solves the problem of repeater placement. The
signal must be regenerated before it falls into noise. The repeaters are
"simply" integrated into the cable at proper intervals. The first
transatlantic cable message was sent by Queen Victoria to the American
President.

Ashore and on distant offshore platforms, I`ve puzzled why microwave was
not used instead of cable. Privacy may be one reason, but encryption,
route switching and other techniques could make theft of information
from thin air more difficult than other theft. There are always
beneficiaries of the status quo who make change difficult to impose.

In the early 1950`s, Houston`s Transcontinental Gas Pipeline Company
(Ken Lay was an officer of "Transco" before moving to Enron) built a
private microwave system from its heasdquarters to New Jersey along its
pipeline. I recall looking the new system over. It was supplied by
Philco Corporation and used Pulse Code Modulation, if I recall. The
microwave system was sold to a communications common carrier (now
Sprint) after a few years but it is still in service, I believe. Transco
(now Williams Pipeline Company) is one of many subscribers to the
service I believe.

Microwave repeaters located at about 20-mile intervals can provide
low-noise and high-reliability communications when properly designed.

In the 1950`s, I marveled as I commuted to work on a stretch of road
which ran between Lisbon and O`Porto, watching the Portuguese Post,
Telephone, and Telegraph Company laying coaxial cable alongside.

Cable is more vulnerable to damage, harder to repair, and surely costs
more than microwave. It was none of my business. I was a foreigner in
their country.

Best regards, Richard Harrison, KB5WZI