John Popelish wrote:
Go back and ponder what I wrote.
Too much has been clipped for my elaboration to have any continuity.

Sorry, I don't respond well to primrose paths. What keeps you
from simply stating your point?

It is a koan.

Sorry, I don't respond well to "nonsensical questions". What
keeps you from simply stating your point?

Is it, if your point turns out to be wrong, you want me to
take the heat?

If you have no interest in anything but butting heads with the people
who have disagreed with you, then, please stop responding to my posts.

If you are into playing games, you are responding to the
wrong person. Try W8JI or W7EL instead.
--
73, Cecil http://www.qsl.net/w5dxp

John Popelish wrote:

Cecil Moore wrote:
For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?

Somewhat less then 2 amps.

Does "somewhat" cover 24 dB of losses? :-) The point is that the
current "drops" by exactly the same amount as the voltage. That's
a characteristic of distributed networks as opposed to lumped
circuits. In a Z0 RF environment, the current has to "drop" by exactly
the same amount as the voltage to maintain the Z0 ratio. There are
really no "across" and "through" concepts as exist in DC circuitry.

I guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.

Sounds like something I might do, this afternoon.

Which, helping or imbibing? :-)
--
73, Cecil http://www.qsl.net/w5dxp

wrote in message
oups.com...

Yuri Blanarovich wrote:
Yuri wrote:
You agree that impedance along the radiator changes, being low at the
bottom, around tens of ohms, to being high at the top, around thousands
of
ohms.

)Tom replied:
)I never said that. What do you mean by reactance? The X can be very
)high but radiation resistance very low even near the open end.

I really give up. What's the point. This is a typical example of Tom's
response to technical argument or trying to go step by step. I am talking
impedance, he "knows" I mean reactance. As I said, I get better response
from a brick wall.
No wonder he duntgetit! Oh well!

Yuri,

Part of communicating is understanding the words the other person is
using. I'm only trying to understand what you intend the words you use
mean.
Please don't blame me for trying to sort out what you are saying. If
you mean "reactance", say "reactance" and not "impedance". If you mean
"through", say "through" and not "across".

Really good one Tom! Why twist and dance?
If you don't know what the impedance is please read 2005 ARRL Handbook: page
4.42 Impedance (re Inductance) and
page 22.1 Impedance (re Antennas)
If we measure current drop from one end of the the coil to the other, we see
current drop across the coil.
(You would love "through" because you can then "prove" that if it flows
through it has to be the same, right?)

When you use "across", I guessed and thought you really meant "through"
or "at each end".

When you used "impedance", I couldn't guess and figure out if you meant
the scalar impedance, vector impedance, resistance, reactance, or what.
It could mean too many things.

I meant freakantance, just could not express it :-)
Nice try! Keep on twisting. Before you didn't get it, now you don't want to
get it.

Let's back off one more giant step back to measurements. Show where W9UCW
was "cheating" in his test setup, pictures and comments at
http://www.k3bu.us/loadingcoils.htm
where he shows clearly that RF current drops significantly across (through)
the loading coil, just like it drops across (through) the resonant antenna
(piece of wire or tubing) from max at the base, to zero at the tip. I
venture to say that (most) everybody knows that impedance of such resonant
radiator is low, about few ohms, at the base to thousands of ohms at the
tip. If you don't know or get that, no point of discussing current magnitude
at the ends of the loading coil.

73 Tom


73 Yuri, K3BU

Cecil Moore wrote:
John Popelish wrote:

Cecil Moore wrote:

For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?


Somewhat less then 2 amps.


Does "somewhat" cover 24 dB of losses? :-) The point is that the
current "drops" by exactly the same amount as the voltage. That's
a characteristic of distributed networks as opposed to lumped
circuits. In a Z0 RF environment, the current has to "drop" by exactly
the same amount as the voltage to maintain the Z0 ratio. There are
really no "across" and "through" concepts as exist in DC circuitry.

I guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.


Sounds like something I might do, this afternoon.


Which, helping or imbibing? :-)

Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.
73,
Tom Donaly, KA6RUH

Cecil, W5DXP wrote:
"There is no phase information in standing wave current phase."

Kraus and Terman agree with Cecil.

Kraus writes on page 239 of his 1950 edition of "Antennas":
"It is generally assumed that current distribution of an infinitesimally
thin antenna (l/a=infinity) is sinusoidal, and that the phase is
constant over a 1/2-wavelength interval, changing abruptly by
180-degrees between intervals."

This agrees with Terman who writes on page 94 of his 1955 edition of
"Electronic and Radio Engineering":
"When the load impedance does not equal the characteristic impedance (as
at the open-circuit at the standing-wave antenna tip), the phase
relations are complicated by the presence of the reflected wave. The
phase of the resulting voltage (or current) then oscillates about the
phase of the voltage (or current) of the incident wave , as illustrated
in Fig. 4-5.The phase shift under these conditions tends to be
concentrated in regions where the voltage (or current) goes through a
minimum;----."

Pity the fool who argues with Kraus or Terman.

Best regards, Richard Harrison, KB5WZI

John Popelish wrote:
Roy Lewallen wrote:
John Popelish wrote:

A point of clarification to John's posting:

When a standing wave exists on a transmission line, the phase of the
voltage or current is fixed (other than periodic phase reversals) with
position only if the end of the line is open or short circuited.
Otherwise, the phase of voltage and current will change with position.

Is that because the result is not a pure standing wave (superposition of
two equal and oppositely traveling waves), but a superposition of a pair
of traveling oppositely traveling waves of different amplitudes?

Yes, but I wouldn't put it quite that way. I prefer to say that this is
simply a special case of the more general result you get when you sum
forward and reverse waves. Nothing magical or abrupt happens when the
two traveling waves are equal in amplitude -- if they're slightly
different, you get a little phase shift of the total current with
position along the wire, the current minima aren't quite zero, and the
spatial shape of the amplitude of the total current -- that is, the
shape of the standing wave -- isn't quite sinusoidal. Making the
amplitudes more and more different smoothly transitions the nature of
the total current until in the special case of the reverse traveling
wave being zero you have the distribution of a pure traveling wave.

Roy Lewallen, W7EL

Cecil Moore wrote:

Tom Ring wrote:

wrote:

Answer the question Cecil, how can we have charge movement over a small
length of conductor (in terms of the wavelength) in two directions at
the same time, or a drift velocity in two directions at once?

Cecil and Co. are not interested in real physics, math, or
engineering. They have made up their own. As I said to Roy, you may
as well give up.


Tom, I learned this stuff at Texas A&M in the 50's and it was
decades old already, having been developed before I was born.
Are you also willing to deny the existence of simultaneous
forward and reflected EM waves?

Nope, but that's not what you're arguing about.

tom
K0TAR

Tom Donaly wrote:
Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.

What do you ship out to Libertarians?
--
73, Cecil http://www.qsl.net/w5dxp

Yuri Blanarovich wrote:
Let's back off one more giant step back to measurements. Show where W9UCW
was "cheating" in his test setup, pictures and comments at
http://www.k3bu.us/loadingcoils.htm
where he shows clearly that RF current drops significantly across (through)
the loading coil, just like it drops across (through) the resonant antenna
(piece of wire or tubing) from max at the base, to zero at the tip.

Displacement currents in the inductor and the very high reactance of
the very short antenna above the coil explain current difference.

Current cannot vanish Yuri. It has to have an alternative path.

I can change current difference at each end of the inductor all over
the place depending on the design of the loading coil, with NO change
in the loading coil position or antenna lengths.

It is the theory you have, that the current is tied to the "missing
degrees", that is wrong.

Without displacement currents there is no current difference at each
end of the coil, it is not caused by "missing degrees. It is caused by
the capacitance above the coil being very low and the capacitance of
the coil to the outside world being much larger. The current is not all
gone in the first few turns either. It is a series reactance/shunt
reactance problem.

73 Tom

Cecil Moore wrote:
Tom Donaly wrote:

Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.


What do you ship out to Libertarians?

Libertarians don't believe in handouts.
73,
Tom Donaly, KA6RUH