Jim Kelley wrote:
The purpose of pointing out the trigonometric relationship between the
sum of sines and the product of sine and cosine was to illustrate that,
contrary to your assertion, there isn't a difference in the waves.

Jim, please go ask the head of your math department if
there's any difference in those equations. That you don't
see any difference is just extreme ignorance on your part.

Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the
trig identities until you alleviate your ignorance.

You are extremely wrong and ignorant of mathematics.
You will realize that fact when you are unable to
prove your assertions even to yourself.

Hint: cos(kx+wt) = cos(kx)*cos(wt) - sin(kx)*sin(wt)
You are obviously missing half of the terms when you
say there "isn't a difference in the waves". (FYI,
anyone who knows anything about mathematics is laughing
at you.)
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
As a general rule, the voltage can't be computed because of the spacing
between the wire and whatever reference point you're measuring voltage
to. The voltage depends on the path taken between the two points
(conceptually, how you arrange the voltmeter leads), so there's no
single answer. That's why EZNEC computes only current.

This is also one of the several problems with treating an antenna like a
transmission line. The similarity between the two is just enough to
tempt people to take the analogy farther than it holds.

Just because the voltage is hard to measure doesn't mean it
isn't proportional to the E-field.
--
73, Cecil http://www.w5dxp.com

AI4QJ wrote:
"Cecil, Can you help me understand that the end of a dipole is an open
circuit?
Nothing falls out because the current changes direction ala AC because
the end of a dipole equates with the time that requires the current
changes direction. So where is the open circuit? I also cannot see why
radiation reduces reflection if the circuit can be completed with
iron!"

OK Cecil, I'm waiting for your answer on this one! ;-)

The end of a dipole is like an open circuit transmission line.
The traveling wave hits the open-circuit and is reflected.
The forward current and reflected current are equal in magnitude
at the end point and 180 degrees out of phase. Therefore, the
current is zero. That's logical.

The forward voltage and reflected voltage are in phase. Therefore,
the voltage is at a maximum at the ends of the dipole. That's known.
That voltage can be calculated to a certain accuracy.

At the feedpoint of a dipole, the forward wave energy is about
20% higher than the reflected wave energy. That 20% difference
between forward wave and reflected wave is the amount of power
lost to losses and radiation.

A #14 horizontal wire 30 feet in the air has a characteristic
impedance of 600 ohms. That's all you need to know to perform
the voltage and current calculations for a particular power
input to the antenna.
--
73, Cecil http://www.w5dxp.com

On Mon, 7 Jan 2008 21:03:35 -0500, "AI4QJ" wrote:

Well, you said the only way to measure SWR on a receive antenna is by
measuring current (sort of as an aside). I wanted to know the significance
of that. Now I know that the reason you said that was to support your SWR
calculation for a receive antenna using EZNEC, which is based on current
(method of moments). You were not saying it in terms of making physical
measurements but only to support the accuracy of your statement (i.e.
there's no better way to determine SWR on a receiving antenna than to
measure current, and, EZNEC is based on current). I know this was not your
main point, it was just an aside, but I don't agree with it; however, that
in itself does not diminish your main point.

Hi Dan,

It is merely a response to the "framing" of a specific expectation
within the context of Cecil's citation. The citation demanded an
externality, I supplied a stimulus external to the antenna.

However, this does not mean that there is no other indicator of
Standing Waves on a Traveling Wave antenna, and it does not make this
exotic testing the only proof of Standing Waves on a Traveling Wave
antenna. There is, after all, the concept of reciprocity.

If you look at the reciprocal actions offered by exciting a Traveling
Wave antenna with a source directly attached to it (and Cecil's last
example proves this), you find Standing Waves. This may confound the
SWR meter, but then that meter doesn't indicate what is on the wires,
it indicates what is impressed upon the finals.

Now, if you want to discard EZNEC (which for some odd reason you seem
to approach method of moments with a sneer), conventional methods
would still bear out the same results. Lord knows I've sat at the
bench doing it the conventional way for thousands of measurements.
I've probably made more physical measurements in a day, than anyone
here has in a lifetime.

Others, don't bore us with indignities about all your SWR meter
readings in reply to that last statement. :-)

So now to the shoe you dropped:
I know this was not your
main point, it was just an aside, but I don't agree with it

What was my main point, and how is yours conflict with it? Is yours a
philosophical triviality so common to these threads, or does it come
with physical measurements experience?

73's
Richard Clark, KB7QHC

My daughter's surgery went well.
I will be away from my computer for a week.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
My daughter's surgery went well.
I will be away from my computer for a week.

I cannot express the relief I feel that all has gone well for your
daughterj--and yourself ...

We will be here when you get back, take your time and take care OM ...

Warmest regards,
JS

Cecil Moore wrote:

That you don't
see any difference is just extreme ignorance on your part.

Set cos(kx-wt) = cos(kx)*cos(wt) and wrestle with the
trig identities until you alleviate your ignorance.

Cecil,

There are an infinite number of different functions one could write to
describe an infinite number of different possible wave shapes. And
none of them would necessarily be mathematically equivalent to
another. But when you write the equation for the superposition of
traveling waves and claim that resultant standing wave is a different
kind of electromagnetic wave, that is a misguided point of view. The
equation for a standing wave is simply a different way of writing the
sum of two traveling waves. Being that there is only one kind to
chose from, there cannot be a different kind of electromagnetic wave.
Is it impossible for you to acknowledge this simple point in a
gentlemanly fashion?

You are extremely wrong and ignorant of mathematics.

Any 'extreme wrongness' notwithstanding, what I don't know about
mathematics could fill a book. On the other hand, you almost
inevitably end up lying and turning technical discussions into
personal attacks. Ask anyone.

ac6xg

On 7 Jan, 22:19, Cecil Moore wrote:
AI4QJ wrote:
"Cecil, Can you help me understand that the end of a dipole is an open
circuit?
Nothing falls out because the current changes direction ala AC because
the end of a dipole equates with the time that requires *the current
changes direction. So where is the open circuit? I also cannot see why
radiation reduces reflection if the circuit can be completed with
iron!"

OK Cecil, I'm waiting for your answer on this one! ;-)

The end of a dipole is like an open circuit transmission line.
The traveling wave hits the open-circuit and is reflected.
snip.
--
STOP STOP STOOOOOOOOOP RIGHT THERE
Thank you Lord for this opportinity you have given meto bring Ham
radio back to reality after Terman went to heaven
Gentleman this is a relativistic statement where Einstein tried to get
a unification theory
\before he died. He failed.You all know your houses are wired for AC,
usually a ring circuit.
Yes 50/60 is frequency just like 10 metrrs. Now go homw and cut the
wire in the kitchen to make it an open circuit Now ham r5adio says the
cuyrrent will turn around where you cutt it and it will go back. Well
science has not seen a trace of this new frequency of 100, 110 0r 220
line double its frequency to twice what it was before. On top of that
the lights went out on the circuit before and after the wire was cut
which means the current stopped when it was cut.
Same for a antenna which usually has a higher frequency. No the
current does not turn back or fall of the end. Right from the get go
the current knows that it must travel forward regards until the time
of half a period comes about. It cannot turn around and collide with
current that is obeying the rules. When it first started its travels
up the radiator the current travelled on or above the skin that
encloses the conductor knowing it can only referse at a specific
time. When the current6 reaches the top it will turn around if the
time alloted has also exspired. If this hasn't happenned it sees that
the inside of the antenna has nothing but decaying particles on its
inside which is really just a resistor. The current flows down the
center of this resistance. True it is not radiating at this point as
it is on the wrongside
of this decaying skin. Fortunately forward time expires just at the
time the current reached the bottom end of the radiator at the same
time half a period expired. So the current retraces its path up the
center and down the outside of the antenna where it can again radiate.
It reaches its starting point at exactly the same time as two haf
periods have elavsed which is one cycle. Yes the cuurent traveled a
full circuit which it must do as we know current flow today. No open
circuits and no current flow each and every way because it was forced
somehow to change direction and ofcourse all the lights went out.

All of this started because physicsts could not reconcile the laws of
electricity or Maxwell with those of Newton and ofcourse Gauss. This
thinking continues to this day even tho Einstein died without proving
things. Was this thread really worth it just to carry on the legacy of
Terman and Einstein regardless of the facts? Now look at the URL i
have supplied on another thread and note how a diamagnetic field is
somehow different from what you have been taught.
Note especially that the lines of force do not follow the same
direction that you expected.
If a paricle gets carried by these special lines of force the particle
becomes ejected AWAY from the field. Now.Is all this "blabbering" or
is it worth pursueing in the absence of Einstein? Can't we get back to
classical physics that the masters of which Newton, Maxwell and Gauss
stood on the shoulders of. Their science and our science is the same.
Let us not kick the legs from under them by inventing new sciences so
that our difficulties in understanding can be resolved.
Read the URL PLEASE .
I have found for you a description which also has picture as well as
words and mull over the implications of what I place before you.
Art Unwin KB9MZ...xg (uk)




73, Cecil *http://www.w5dxp.com

"AI4QJ" wrote in message
...

The VSWR meter on the ham rig is merely looking at the balance of forward
and reflected "power" and it is calibrated to read it out as VSWR (or
SWR). It may as well say "ISWR"; it is all the same thing. But it is not
measured by sensing either voltage ot current going into the antenna...it
measures the delta power.


how do you measure 'power'?? you don't. and no swr meter in the world
measures 'power'. they all take samples of voltage and/or current and drive
a simple meter circuit that just happens to be calibrated in units of watts
because thats what most users of cheap meters want to see. they could just
as easily be calibrated in volts or amps referenced to 50 ohms

On Tue, 08 Jan 2008 23:20:44 GMT, "Dave" wrote:

they could just
as easily be calibrated in volts or amps referenced to 50 ohms

Which is power.