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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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

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:
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
|
|