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

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World!
120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via
Encryption =----

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


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

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
|
|

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


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---

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
|
|