On May 28, 10:16*am, Sonny Hood wrote:
* *I have an efficiency question concerning feed lines. *My present
system is RG-8X to my 75 meter inverted vee which is about 85 feet
away from the shack. *I propose to replace some 88 feet of coax with
300 ohm window ladder line that is inserted into the coax run with 4:1
baluns to match the coax on each end. *Also at the feed point of the
antenna switch from a voltage balun to a current balun (ferrite chock
type). *By my calculations with a 98 watt generator I will increase
the power to the load by about 11-20 watts and with a 985 watt
generator, 117-210 more watts will reach the load. *Figuring
theoretical total system (A) against total System (B) or by just the
difference in the 88 feet of ladder line versus coax. *What do you
think the increase will be?

I think you left out the losses in the balun. With a kilowatt even a
large (by ham standards) balun will get very warm if you leave the key
down... my guess is it's not quite as much as you're saving but
comparable. It all amounts to 10%, and that's such a tiny fraction of
a S-unit.

Now, if your antenna were an incredible mismatch to coax to begin
with, then ladder line makes good sense. Say you want to use your same
inv-V on 40M, then you'd run ladder line all the way to your tuner and
be in fat city.

Tim N3QE

Tim Shoppa wrote:

Hi,

I think you left out the losses in the balun. With a kilowatt even a
large (by ham standards) balun will get very warm if you leave the key
down... my guess is it's not quite as much as you're saving but
comparable. It all amounts to 10%, and that's such a tiny fraction of
a S-unit.

The other side probably wouldn`notice the difference.

Now, if your antenna were an incredible mismatch to coax to begin
with, then ladder line makes good sense. Say you want to use your same
inv-V on 40M, then you'd run ladder line all the way to your tuner and
be in fat city.

I would not take TV ladderline if you have a high standing wave ratio,
and this will happen if you use the dipol for multiband purpose.
I`ve tested a dipol with about 20 m length (67 ft) for 80m.
At first I got almost the same s-meter report compared with a W3DZZ trap
dipol. But after a couple of month I noticed that the received signal
was up to 20 db down compared with the trap dipol. I only used 100 watts
but it broke the TV ladder line. Here I`ve learned what has happened:

http://www.w8ji.com/vswr_reactive_power.htm

I`m planning to build a new antenna with self built ladderline for
multiband purposes for one of the traps of my W3DZZ is broken. I´ve only
bought the spacers for the new ladder line.
--
73
DJ4PB

On May 28, 1:22*pm, (Dieter Kiel) wrote:
wrote:

Hi,

I think you left out the losses in the balun. With a kilowatt even a
large (by ham standards) balun will get very warm if you leave the key
down... my guess is it's not quite as much as you're saving but
comparable. It all amounts to 10%, and that's such a tiny fraction of
a S-unit.

The other side probably wouldn`notice the difference.

Now, if your antenna were an incredible mismatch to coax to begin
with, then ladder line makes good sense. Say you want to use your same
inv-V on 40M, then you'd run ladder line all the way to your tuner and
be in fat city.

I would not take TV ladderline if you have a high standing wave ratio,
and this will happen if you use the dipol for multiband purpose.
I`ve tested a dipol with about 20 m length (67 ft) for 80m.
At first I got almost the same s-meter report compared with a W3DZZ trap
dipol. But after a couple of month I noticed that the received signal
was up to 20 db down compared with the trap dipol. I only used 100 watts
but it broke the TV ladder line. Here I`ve learned what has happened:

http://www.w8ji.com/vswr_reactive_power.htm

I`m planning to build a new antenna with self built ladderline for
multiband purposes for one of the traps of my W3DZZ is broken. I´ve only
bought the spacers for the new ladder line.

I have a 135-foot dipole fed with approx 100 feet of home-made ladder
line and am incredibly happy with its performance from 80M all the way
to 15M.

Much of the joy of the ladderline was not in buying the parts to make
it, but making it from stuff on hand. I had a sheet of 1/8"
polycarbonate that I cut into 4"x3/8" strips, then drilled and
notched, to put a spacer every foot using tie-wires. Others boil
wooden dowels in paraffin for the spacers (the method recommended by
the 1930's ARRL Handbook). Seeing the ladderline go up 85 feet to the
middle of the dipole is a pure joy!

Tim N3QE

Tim Shoppa wrote:
I have a 135-foot dipole fed with approx 100 feet of home-made ladder
line and am incredibly happy with its performance from 80M all the way
to 15M.

Why those are good choices for lengths can be seen on
the following diagram:

http://www.w5dxp.com/pnts130.gif

You are relatively close to a low-impedance/current-
maximum point on all HF bands.
--
73, Cecil http://www.w5dxp.com

(Dieter Kiel) wrote in news:1ihnz3n.wxm0z7nvxaivN%
:

http://www.w8ji.com/vswr_reactive_power.htm



I assume Dieter that this is your recommendation of the article.

That article uses the term 'reactive power' in a non-conventional way,
though the term is a well known one (ie has a conventional meaning).

Conventional use is that the term 'apparent power' is applied to the
product of RMS voltage and current flowing into a two terminal load, and
the units are VoltAmps (VA), not Watts as used in the article.

Reactive power is the reactive component of apparent power, and expressed
in units of 'VoltAmpsReactive' (VAR).

'Real power' is the real component of apparent power and expressed in
units of Watts (W).

The relationship is that
apparentpower = (realpower^2 + reactive power^2)^0.5 .

This is all basic lumped component AC circuit theory, and holds at RF.

However, basic lumped component AC circuit theory is not a good way to
analyse antenna systems, expecially to determine transmission line losses
cause by highly reactive loads (which is what that article tries to do).

Owen

Owen Duffy wrote:
(Dieter Kiel) wrote in news:1ihnz3n.wxm0z7nvxaivN%
:

http://www.w8ji.com/vswr_reactive_power.htm



I assume Dieter that this is your recommendation of the article.

That article uses the term 'reactive power' in a non-conventional way,
though the term is a well known one (ie has a conventional meaning).

Conventional use is that the term 'apparent power' is applied to the
product of RMS voltage and current flowing into a two terminal load, and
the units are VoltAmps (VA), not Watts as used in the article.

Reactive power is the reactive component of apparent power, and expressed
in units of 'VoltAmpsReactive' (VAR).

'Real power' is the real component of apparent power and expressed in
units of Watts (W).

The relationship is that
apparentpower = (realpower^2 + reactive power^2)^0.5 .

This is all basic lumped component AC circuit theory, and holds at RF.

True, but we still aren't there.

It's very misleading to quote "VAR powers" in the kilowatt range,
because the only power available to melt the feedline is 100W from the
transmitter. There is no magnification of real power.

The high value of "VAR power" is a theoretical result of the large RF
currents in the system. These result from an antenna feedpoint impedance
that has a very low resistive part and is almost entirely reactive. The
large RF currents are a genuine physical phenomenon, as also are the
high voltages a quarter-wavelength back along the feedline (if the
feedline is long enough, of course)... but if there were no losses in
the feedline, these would have no further effect. In spite of the wild
values of impedance, current, voltage, VSWR etc, if there were no losses
in the feedline then all of the RF power would still reach the antenna.

In a real feedline, the effect of the high currents is to divert almost
all of the available RF power away from the antenna and into the
feedline's own resistive losses - skin-effect losses in the copper
conductor, and dielectric losses in the plastic. Both of these result in
heating and softening of the plastic, which makes the dielectric loss
even higher. This tends to divert even more of the available power into
the weak spots, where the plastic finally melts.

But there is still only 100W available to do the damage.

No argument about the final conclusion - it ain't gonna work - but I
don't care for the explanation. There's no problem with "VAR power" for
anyone who already has a firm grip on the concepts, but it is not a good
way to explain those concepts to a newcomer.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:

.... a very good posting with the exception (IMO)
of the choice of the word "magnification".

There is no magnification of real power.

There is indeed no net increase in real average power.
However, "magnification" seems to be a poor choice of
words since its definition contains such words as
"apparent", "seems", "exaggerate", "overstate", ...

"magnify - to cause to *seem* greater" certainly describes
a feedline with 100W source power and 200W forward power.

But there is still only 100W available to do the damage.

True, but a 100W laser can do a lot of damage. :-)
And a *magnifying* glass can start a fire.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Ian White GM3SEK wrote:

... a very good posting with the exception (IMO)
of the choice of the word "magnification".

There is no magnification of real power.

There is indeed no net increase in real average power.
However, "magnification" seems to be a poor choice of
words since its definition contains such words as
"apparent", "seems", "exaggerate", "overstate", ...

"magnify - to cause to *seem* greater" certainly describes
a feedline with 100W source power and 200W forward power.

But there is still only 100W available to do the damage.

True, but a 100W laser can do a lot of damage. :-)
And a *magnifying* glass can start a fire.

I have no intention of allowing *that* particular fire to start again!


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:

Owen Duffy wrote: (Dieter Kiel) wrote in
news:1ihnz3n.wxm0z7nvxaivN% :
http://www.w8ji.com/vswr_reactive_power.htm I assume Dieter that
this is your recommendation of the article. That article uses the term
'reactive power' in a non-conventional way, though the term is a well
known one (ie has a conventional meaning). Conventional use is that the
term 'apparent power' is applied to the product of RMS voltage and
current flowing into a two terminal load, and the units are VoltAmps
(VA), not Watts as used in the article. Reactive power is the reactive
component of apparent power, and expressed in units of 'VoltAmpsReactive'
(VAR). 'Real power' is the real component of apparent power and
expressed in units of Watts (W). The relationship is that
apparentpower = (realpower^2 + reactive power^2)^0.5 . This is all
basic lumped component AC circuit theory, and holds at RF. True, but we
still aren't there.

It's very misleading to quote "VAR powers" in the kilowatt range, because
the only power available to melt the feedline is 100W from the
transmitter. There is no magnification of real power.

The high value of "VAR power" is a theoretical result of the large RF
currents in the system. These result from an antenna feedpoint impedance
that has a very low resistive part and is almost entirely reactive. The
large RF currents are a genuine physical phenomenon, as also are the high
voltages a quarter-wavelength back along the feedline (if the feedline is
long enough, of course)... but if there were no losses in the feedline,
these would have no further effect. In spite of the wild values of
impedance, current, voltage, VSWR etc, if there were no losses in the
feedline then all of the RF power would still reach the antenna.

In a real feedline, the effect of the high currents is to divert almost
all of the available RF power away from the antenna and into the
feedline's own resistive losses - skin-effect losses in the copper
conductor, and dielectric losses in the plastic. Both of these result in
heating and softening of the plastic, which makes the dielectric loss even
higher. This tends to divert even more of the available power into the
weak spots, where the plastic finally melts.

But there is still only 100W available to do the damage.

No argument about the final conclusion - it ain't gonna work - but I don't
care for the explanation. There's no problem with "VAR power" for anyone
who already has a firm grip on the concepts, but it is not a good way to
explain those concepts to a newcomer.

I think you are right, the referenced link of W8JI mixed up the
definitions. He used "kilowatts" instead of "1000 VAs". But for me the
text was very helpful. It also will give newcomers some information
about problems with short antennas. If you ask: What is responsible for
the losses of an antenna construction ? The answer would be: The losses
are proportional to the " Apparent Power VA" It`s "Scheinleistung" in
German language. I didn`t even know the term before it was mentioned
here.
I found this schematic in the wikipedia with the different terms
Real Power, Reactive Power, and Apparent Power:

http://en.wikipedia.org/wiki/Image:P...riangle_01.png

So if we use a short antenna, e.g. 20m for the 80m band or even the G5RV
we should use parts that can handle the "Apparent Power". I`ve seen
Sonny has already made up his mind using koax for the 80m Band. This
probably is a good idea, and I`m planning to build a monoband for my
favorite band and multiband antenna fed with a low loss ladder line to
cover all the bands. I`m not quite sure if a folded dipol as a monoband
antenna fed with ladder line would give better results as a coax fed
dipol. I have the equipment to match both kinds of feeding lines.