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