"AI4QJ" wrote in
:


I put to you a scenario where there was no antenna or load inductance or
capacitance, and there was a standing wave on the transmission line.

That says to me that the standing wave is not a consequence of antenna or
load inductance or capacitance, or "a vibrational energy shift between
antenna system inductance and antenna system capacitance" as you put it.

Owen

Cecil Moore wrote:

...
I suspect what Dan is referring to is the LCLCLCLC
equivalent circuit for a transmission line. A horizontal
wire over ground is a one-wire transmission line with
Z0 = ~SQRT(L/C). A radiating antenna can be considered
to be a lossy transmission line.

A #14 horizontal wire at 30 feet calculates out to be
Z0 = 600 ohms so L/C = ~360,000. I'm pretty sure that
is the L and C that Dan is talking about - the same
L and C in which the standing wave energy is stored.

since:
2b
L = 0.00508((ln---) - 0.75) (best ascii can do)
a


whe
L = uH
a = wire dia in inches
b = wire length in inches
ln = natural logarithm
and, since #14 is .0641 inch
and 30 ft = 30*12 or 360 inches

2*360
L = 0.00508((ln(------)) - 0.75)
0.0641

L = .043568935605uH

Now, if I can just find that formula for the capacitance of a free wire
in space--I know I seen it here just a bit ago ... sorry, I'll have to
get back to you on that one ;-)

Regards,
JS

John Smith wrote:
[a bunch of kidding stuff--it's CHRISTMAS!]

There is the equation:

| |2h | b+sqrt(bsquared+asquared) |||
L = .0117|log10|---|----------------------------||| +
| | a |b+sqrt(bsquared+(4*hsquared)|||

| b |
0.0508|sqrt(bsquared+4*hsquared) - sqrt(bsquared+asquared) + - - 2h + a|
| 4 |

whe (and, sorry again, we only have ascii here)

|
(and, | is simply a bracket) or, one of these "({[" or these ")}]" but
|
then, you already knew that ...


L = uH
a = wire rad. in inches
b = wire length parallel to ground, in inches
h = wire height above ground, in inches

Now, this equation is probably a 'bit' more accurate than above--but
d*mn, still looking for that free wire (or, wire-above-ground) equation
for capacitance, for a wire in space ... :-D

Regards,
looking forward to New Years,
JS

John Smith wrote:

[more stuff his sick mind gets a kick out of]

Regards,
looking forward to New Years,
JS

Sorry, when everyone else is in bed, or his/her cups, I am still
up--just has always been like that ... and with that, a Good Night!

Regards,
JS

"AI4QJ" wrote in
:

....
not so nicely linear. The antenna is a lossy transmission line just as
Owen's example was a lossy xmission line example with a 25 ohm load at

No, my example stipulated an ideal transmission line, and by that I mean it
to be lossless amongst other things.

You and Cecil are transforming the example to suit yourselves.

Owen.

John Smith wrote:
Jay in the Mojave wrote:

...
Setting back on the side lines and watching.

Merry Christmas to all.

Jay in the Mojave

Merry Xmas Jay.

We are kindred spirits in the enjoyment of this group ...

Warm regards,
JS

Hello JS:

Yeah Ten-4 hope Christmas was a pleasant day for. I do enjoy reading the
group here. But don't have the time to look everyday.

Have a Happy and safe New Year. Stay out of jail. (humor) Maybe catch ya
on the bands some day.

Jay in the Mojave

Kreediantilas:
Not 2 many

Owen Duffy wrote:

I put to you a scenario where there was no antenna or load inductance or
capacitance, and there was a standing wave on the transmission line.

In that case, the standing wave is contained in the inductance
and capacitance in the transmission line.

That says to me that the standing wave is not a consequence of antenna or
load inductance or capacitance, or "a vibrational energy shift between
antenna system inductance and antenna system capacitance" as you put it.

When Dan said that, he was not talking about transmission lines.
He was talking about the standing wave antenna itself (without
the transmission line). You two are talking about entirely
different things.
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
"AI4QJ" wrote:
not so nicely linear. The antenna is a lossy transmission line just as
Owen's example was a lossy xmission line example with a 25 ohm load at

No, my example stipulated an ideal transmission line, and by that I mean it
to be lossless amongst other things.

What we are saying is that even if the transmission line
is lossless, the *system* is lossy because of the 25 ohm
resistor.

If there were no losses in the *system*, the waves on the
lossless transmission line would be pure standing waves.
Because of the losses in the load, the waves on the lossless
transmission line are not pure standing waves, but a mixture
of standing waves and traveling waves. In your case (#1 below)
the system is primarily a traveling wave system, closer to
flat than to an OC or SC stub because only 11% of the forward
energy is rejected by the load.

You and Cecil are transforming the example to suit yourselves.

I'm not transforming the example. You are the one who put
the lossy resistor in the system. The traveling waves are
the direct result of the installation of the resistor.

Let's look at a few different examples and assume the
measured joules/sec flowing forward toward the load is
100 joules/sec in each case.

1. Your example of 50 ohm lossless coax connected to a 25
ohm load. The forward joules/sec is 100. The reflected
joules/sec is 11.11. The joules/sec consumed by the 25
ohm load is 88.89. 89% of the forward wave is traveling
wave. 11.11% of the forward wave is used by the standing
wave. The system is primarily a traveling wave system.
The energy not delivered to the load is stored in the
standing wave in the LCLCLCLC components of the
transmission line.

2. No load on the lossless coax. The forward joules/sec
and the reflected joules/sec are equal. 100% of the energy
is standing wave energy and all of it is stored in the
LCLCLCLC components of the transmission line. It does not
move from LC to LC. It simply oscillates in place between
L and C. EZNEC confirms that the current phasor does NOT
rotate.

3. 50 ohm load on the lossless coax. The reflected joules/sec
equals zero and the system is flat. 100% of the energy is
traveling wave energy. The only energy in the transmission
line is the energy it took to fill the pipeline, the delay
between power-on and the load dissipating power. The LCLCLCLC
in this case is an energy bucket brigade.

4. 500 ohm load on the lossless coax. Of the forward 100
joules/sec, only 33 joules/sec is accepted by the load.
The other 67 joules/sec are rejected by the load and become
half of the energy in the standing wave. The system is
primarily a standing wave system. The energy not delivered
to the load is stored in the standing wave in the LCLCLCLC
components of the transmission line.
--
73, Cecil http://www.w5dxp.com

John Smith wrote:
....

Well, finished unpacking the new toy. Thanks Santa!

Bolted the neodymium magnet to the 1,000,000 R.P.S. motor (specially
constructed from the metal from crashed UFOs' recovered by the gov't.)
Shoved this rf generator into the coaxial tank to couple with the
specially constructed copper coupling constructed into the tank, and
firmly secured it. Coupled the ant to the tank with a 1 turn loop
located at a standing wave "hump" and plugged it in ... darn thing is a
little large!

Anyone have their MW radio(s) tuned to 1Mhz? :-|

Regards,
JS

Cecil Moore wrote:

What I said is the voltage and
current in a standing wave are *always* 90 degrees out of
phase and it is impossible to generate heat when the voltage
and current are 90 degrees out of phase.

So then shouldn't one expect coax to be heated uniformly along its
length at a high SWR?

73, ac6xg