On Wed, 29 Jun 2005 17:59:24 -0700, Frank Gilliland
wrote:

On Thu, 30 Jun 2005 00:45:25 GMT, james wrote
in :

On Wed, 29 Jun 2005 17:17:53 -0700, Frank Gilliland
wrote:

That's basically what Lance said, just in different words. So what's
the problem?
*****

I have problems when people state "current flows". That is not
actually true.


So current -doesn't- flow?


Power in the electromagnetic wave that flows down a
transmission line and currents on the center conductor and outer
conductor are dependant on the E and H fields of the wave within the
transmission line at any point on the transmission line.


Those E and H fields are created by voltage and current introduced
onto the conductors. A transmission line simply establishes a kind of
self-propogating 'symbiosis' between the fields and the current &
voltage on the conductors. Current does indeed flow on those
conductors -- if it didn't they wouldn't be necessary.

******
Please correct me if I have misunderstood your position above. I am
trying to visualize what are saying.

Then it is your assertion that in a coax a sinusoidal current flows in
the center conductor to the load, then through the load and back to
the source through the shield?


outside that I think we are primarily in agreement.


What you are missing is that different things happen on the outside of
the coax than what happens on the inside. And what Lance was pointing
out (if I may be so bold as to speak for him here) was that currents
on the outside of the coax can screw up things when you try to measure
what's happening on the inside.

*****

From what I visualize from your post, I have problems with this. Maybe
I am misunderstood you.

james

On Thu, 30 Jun 2005 11:18:37 -0500, Cecil Moore
wrote:
Sorry, I made a mistake in the equation.
....
Hows about reading it so I won't have to explain superposition to you?
Certainly no one stands any chance of understanding your explanations
given the continuing goof-ups.

On Thu, 30 Jun 2005 11:18:37 -0500, Cecil Moore
wrote:
When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w
Yowza! You, with W's help, can roll your Social Security over into
investments in the CB amplifier Market.
and must be supplied by the sources
[Hecht rolling his eyes] So, this means that Hecht's formula only
works for steady state? :-)
If both are 100W pulses, and the lasers are off before the target are
pulse illuminated -um-
1.) 100W
2.) 200W
3.) 400W
4.) no hundred W
or supplied by destructive interference from somewhere else.
Maybe two more magic lasers?
This is all explained in _Optics_, by Hecht.
Somehow, I don't think so.

How many errors can our readers count?
For N = number of words in orginal posting
errors = N!

Such is the problem of Xerox research.

On Thu, 30 Jun 2005 12:08:05 -0400, "Fred W4JLE"
wrote:
SWR=Strewn With Rumor
SWR = Strife Without Regret

Richard Clark wrote:
Certainly no one stands any chance of understanding your explanations
given the continuing goof-ups.

Richard, I cannot recall you and I ever disagreeing upon
a technical subject. Your objections are just personal
pot shots.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 30 Jun 2005 13:08:52 -0500, Cecil Moore
wrote:
Richard, I cannot recall you and I ever disagreeing upon
a technical subject.
Convenient memory.

Richard Clark wrote:

Cecil Moore wrote:
When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w

Yowza! You, with W's help, can roll your Social Security over into
investments in the CB amplifier Market.

Note the
following extremely important qualification. The extra power
must come from somewhere, either from the two sources or from
destructive interference. So says Hecht.

When you phasor add 100v to 100v, what V^2/Z0 do you get?

and must be supplied by the sources or or supplied by
destructive interference from somewhere else.

[Hecht rolling his eyes] So, this means that Hecht's formula only
works for steady state? :-)

Yep, irradiance is a quantity averaged over time. It is
steady-state by definition, an accumulated effect.

If both are 100W pulses, and the lasers are off before the target are
pulse illuminated -um-
1.) 100W
2.) 200W
3.) 400W
4.) no hundred W

You will get interference rings of 400w/unit-area and
rings of 0w/unit-area all averaging out to 200w total.
All this is covered in _Optics_. Please spare us your
ignorance and read the book.

If the sources are incapable of supplying the extra power,
For every P1+P2+2*SQRT(P1*P2), i.e. constructive interference,
there is a P1+P2-2*SQRT(P1*P2), i.e. destructive interference.
--
73, Cecil, http://www.qsl.net/w5dxp

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Earlier today, I wrote, "...HP8653..." Ooops. Belay that. It's
HP8753E.

On Thu, 30 Jun 2005 13:28:24 -0500, Cecil Moore
wrote:
You will get interference rings
[Hecht rolling his eyes] for a target that is smaller than a
wavelength?
Must be another failure of Hecht (or Hecht pupil).
of 400w/unit-area and
rings of 0w/unit-area all averaging out to 200w total.
Which, of course, cannot be found in the formula:
Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)
nor its correction:
Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta)
If the sources are incapable of supplying the extra power,
Which, in the end was a non sequitur.
For every P1+P2+2*SQRT(P1*P2), i.e. constructive interference,
Which, of course, cannot be found in the formula:
Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)
nor its correction:
Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta)
there is a P1+P2-2*SQRT(P1*P2), i.e. destructive interference.
Which, of course, cannot be found in the formula:
Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta)
nor its correction:
Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta)

When combined with the adhominem (which, of course, reveals another
inaccuracy, one of assignment):
Please spare us your ignorance and read the book.
is possibly the best advice (once the assignment is corrected), given
the continuing goof-ups.

On Thu, 30 Jun 2005 07:36:21 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:

Cecil Moore wrote:
But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a
50 ohm SWR meter in a 75 ohm coaxial line.

If that were true then the mere existence of standing waves could
render any measurements worthless. Regardless, I did the experiment a
long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm
line -- it gives you almost the same measurement (in fact, I didn't
see -any- difference at all).

Please run it again in the following configuration:

Xmtr--1/4WL 75 ohm line--SWR meter--1/4WL 75 ohm line--50 ohm load

The SWR meter will read 2.25:1 when the actual SWR is 1.5:1

Xmtr--1/2WL 75 ohm line--SWR meter--1/2WL 75 ohm line--50 ohm load

The SWR meter will read 1:1 when the actual SWR is 1.5:1


I'm not going to argue this -- either you can play with theory and
speculate about the results, or you can do the experiment yourself,
observe the empirical evidence, and -then- use theory to explain the
results. When you get around to doing the latter give me a holler in
rrcb since I'm done cross posting on this topic.

And BTW, the best location for the directional coupler is at the
feedpoint of the antenna. Barring that, the next best place is at the
transmitter. Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.





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Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.

The results above obey the laws of physics. What laws do your
results obey?
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.

The results above obey the laws of physics. What laws do your
results obey?

You guys are just itchin' for a visit from the coax length police. :D

On Thu, 30 Jun 2005 17:28:45 GMT, james wrote
in :

On Wed, 29 Jun 2005 17:59:24 -0700, Frank Gilliland
wrote:

On Thu, 30 Jun 2005 00:45:25 GMT, james wrote
in :

On Wed, 29 Jun 2005 17:17:53 -0700, Frank Gilliland
wrote:

That's basically what Lance said, just in different words. So what's
the problem?
*****

I have problems when people state "current flows". That is not
actually true.


So current -doesn't- flow?


Power in the electromagnetic wave that flows down a
transmission line and currents on the center conductor and outer
conductor are dependant on the E and H fields of the wave within the
transmission line at any point on the transmission line.


Those E and H fields are created by voltage and current introduced
onto the conductors. A transmission line simply establishes a kind of
self-propogating 'symbiosis' between the fields and the current &
voltage on the conductors. Current does indeed flow on those
conductors -- if it didn't they wouldn't be necessary.

******
Please correct me if I have misunderstood your position above. I am
trying to visualize what are saying.

Then it is your assertion that in a coax a sinusoidal current flows in
the center conductor to the load, then through the load and back to
the source through the shield?


I'm not going to educate you about transmission lines in a newsgroup.
It would take several large posts and I simply don't have the time to
write that much. If you want to understand how coax works then dig out
your EM textbook and review the basics.







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On Thu, 30 Jun 2005 16:55:13 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the
coupler floating with the coax or you will end up with results like
what you describe above.

The results above obey the laws of physics. What laws do your
results obey?


The law of "been there, done that, and what I -thought- would happen
is not what -really- happened".






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"Tom Ring" wrote in message
. ..
Walter Maxwell wrote:
Reading the mail appearing in this thread is more fun than watching Saturday
Night Live!

Walt, W2DU

Some of the people involved appear to be listening from inside Faraday Cages!

tom
K0TAR

Faraday used the cages after the monkeys were through with them. The monkeys
left them so fouled up that EM waves couldn't penetrate the walls.

Walt, W2DU

But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a
50 ohm SWR meter in a 75 ohm coaxial line.

If that were true then the mere existence of standing waves could
render any measurements worthless. Regardless, I did the experiment a
long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm
line -- it gives you almost the same measurement (in fact, I didn't
see -any- difference at all). Any small error you might see is, as I
said before, insignificant, especially considering the reason you are
measuring SWR in the first place. The objective is simply to get the
reading as low as practially possible. If you feel the need to quibble
about a couple tenths of a point on a ratio then maybe you're spending
a little too much time playing with the calculator instead of the
antenna.


I'm running RG-6 out to my 2 meter antenna. I put my cheap RS HF meter inline
to see what I'd read. I got my expected 1.5:1. Wattage read 1/2 of what the
radio is rated for. It gets out and I'm not worried about it.

On Thu, 30 Jun 2005 14:59:31 -0700, Frank Gilliland
wrote:

I'm not going to educate you about transmission lines in a newsgroup.
It would take several large posts and I simply don't have the time to
write that much. If you want to understand how coax works then dig out
your EM textbook and review the basics.
*****

I am not seeking your education just seeking better understanding of
your position.

But yes I think this has gone well to far.

james