Cecil Moore wrote:

Roy Lewallen wrote:

You seem to have convinced a few readers of the group that you know
what you're talking about. Since you're apparently not able to
express your ideas in concrete form, perhaps one of them will
volunteer to do the mundane work of developing a coherent theory to
explain it in scientific and mathematical terms.


You must have missed some of my postings. I have already expressed my
ideas in concrete form. Maybe you don't like the simplicity of

Itot = I+ + I- (I+ is forward current and I- is reflected current)


I love the simplicity of Itot = I+ + I-. I also like the simplicity of 1
+ 1 = 2, V = I/R, and e^(j*pi) = -1. But none of these is adequate to
determine the current at the top and bottom of a loading coil.

Standing wave antennas possess standing waves. Standing waves occur
when forward waves and reflected waves are superposed. The phase
rotation of these two component currents are in opposite directions.
The result is a sinusoidal function for both net voltage and net
current.

Any real-world air-core coil has a phase delay that affects the
forward current and the reflected current. Since they are phase
rotating in opposite directions, the overall effect is doubled. It's
all explained on my web page. Have you taken time to read it?

Yes, and for the life of me I can't see how to use it to find the
current at the top and bottom of a coil in a simple monopole.

But I'm kind of dense. Can one of the many other readers of this group
kindly explain how Cecil's equations and explanations are used to
actually figure out what the current will be at the top and bottom of a
loading coil? Yuri, you seem to understand it -- can you explain it to
me? Richard? Surely the necessary equations are in one of your books?

Roy Lewallen, W7EL

Roy Lewallen wrote:

Cecil,

In a simple monopole with one inductor, let L1 be the distance from
the base of an antenna to the bottom of the loading coil in meters,
L2 the length of the loading coil, L3 the distance from the top of the
loading coil to the top of the antenna. I is the base current, L the
inductance value and F the frequency. You can assume the antenna is
very thin.

Since your theory is so elegant and well developed, and you've had
such an excellent education at Texas A&M, it shouldn't be difficult at
all for you to write a couple of simple equations which give the
currents at the two ends of the coil. In the time-honored methods of
science, your equations can then be tested against modeled and
measured results to prove the validity of your theory.

Roy Lewallen, W7EL


Cecil Moore wrote:


Well, I have already posted the equations. . .

. . . Here are the equations again.

A loading coil exists in a standing-wave antenna.
The forward current through the loading coil is I+.
The reflected current through the loading coil is I-.
The net current at any point in the coil is I+ + I- (phasor addition)
The magnitude of the net current depends upon the phase of I+ and I-.

Itot = I+ + I-

There's the equation that I have already posted many, many times.
Sorry you missed it.
. . .

Helloo. . . Anybody home?

Roy Lewallen, W7EL

But I'm kind of dense. Can one of the many other readers of this group
kindly explain how Cecil's equations and explanations are used to
actually figure out what the current will be at the top and bottom of a
loading coil? Yuri, you seem to understand it -- can you explain it to
me? Richard? Surely the necessary equations are in one of your books?

Roy Lewallen, W7EL


I will leave Cecil's equation to Cecil, but I will point out that you can use
your own creation EZNEC 4+ and HELIX feature to model the quarter wave vertical
with loading coil. I did it after your pointed out HELIX feature and posted it
at the beginning of my thread, here it is again, actually using "worst" case of
10m loaded whip (not too much loading) and demonstrating significantly
different (not equal) currents at coil's ends. Did I do anything wrong?

Here it is again:

I took the time to check out the Helix feature in EZNEC 4.08 and modeled the
"worst" case - CB whip or 10 m whip with loading coil - helix half way up and
then the same helix moved up to 3/4 way up. Things will get more pronounced
when more turn, more inductance coil is used and frequencies are lower. Yes,
Virginia there is a CURRENT DROP across the loading coil, unless you have more
"appropriate" or "scientwific" term for it.

Rough dimensions: 1m mast (5 mm copper wire/tubing), 20 cm long coil/helix with
5 cm diameter turns, 5mm wire diameter, 10 turns, spacing 2 cm followed by 1 m
whip
Resonated at 27.05 MHz
With base current 1 A, at the end of mast/start of coil the current is 0.87457
A
at the end of coil/start of whip the current is 0.66884 A - a decent drop of
0.20573 A or 20.5 % - not an "EQUAL" (you DC coil believer types!!!)

Then I moved the same coil up 50 cm, so the mast was 1.5 m, same coil, followed
by .5 m of whip.
Again with base current of 1 A, the bottom of the coil had current this time
was 0.65479 A, while top of the coil 0.37127 with larger drop of 0.28352 A or
28.3 % - even bigger not "EQUAL" with resonant frequency moving up to 28.7 MHz,
which corresponds to REALITY measured, experienced and finally properly (close
enough) modeled. Even M0RON (with apologies if there is call like that issued
:-) can see the nice current drop across the coil displayed in the VIEW.

Thank you Roy (now you believe it?), Cecil, Richard. Now the unbelievers can
even model this case themselves and SEE it properly. So ON4UN, K3BU, W9UCW,
W5DXP, KB5WZI were and are right. W8JI, G3SEK et al are sooooo wrong :-) Some
still persist, some are converted and many will be enlightened.

Now if Roy can incorporate elegant way of modeling real life coil/inductance by
inputing Inductance L and its physical size and have it calculate things
without modeling turns, that would be a winner and a segment saver.


Not sure if you saw it, but it looks to me that even EZNEC 4+ is on "our" side.

73 Yuri, K3BU.us

Roy Lewallen wrote:
I love the simplicity of Itot = I+ + I-. I also like the simplicity of 1
+ 1 = 2, V = I/R, and e^(j*pi) = -1. But none of these is adequate to
determine the current at the top and bottom of a loading coil.

We don't have to determine the current at the top and bottom of a
loading coil. All we have to determine is that the current at the
top and bottom of a loading coil cannot be equal in magnitude. It's
a simple true/false logic question, not a question of degree of
accuracy.

EZNEC says the currents are not equal when the helix feature is
used to simulate a real-world coil.

Your and Tom's measurements proved that the currents are not equal
except for Tom's one special toroidal case where he probably located
the current maximum point within the inductor.

Yes, and for the life of me I can't see how to use it to find the
current at the top and bottom of a coil in a simple monopole.

You don't have any equations either but that doesn't stop you from
making assertions about those currents. Why are you attempting to
hold me to a higher standard than you hold yourself or Tom? I'm
not asserting anything new. I learned the qualitative basics of this
subject at Texas A&M almost half a century ago. I'm amazed that those
basics don't seem to be taught any more.

But I'm kind of dense. Can one of the many other readers of this group
kindly explain how Cecil's equations and explanations are used to
actually figure out what the current will be at the top and bottom of a
loading coil? Yuri, you seem to understand it -- can you explain it to
me? Richard? Surely the necessary equations are in one of your books?

It is not necessary to figure out what the current will be. That is
just diverting the issue. All that is necessary for the present
argument is to show that, in the 90 electrical degrees of a mobile
antenna, that the two component phasor currents at each end of the
coil don't add up to the same value. I have done that.

Since the bugcatcher coil shifts the phase of both the forward current
and reflected current by the same amount in opposite directions, it
is impossible for the current to have the same magnitude at both
ends of the coil when the loading coil is installed at the 45
degree (center-loaded) point in an electrical 1/4WL antenna.
--
73, Cecil http://www.qsl.net/w5dxp


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Yuri Blanarovich wrote:
I will leave Cecil's equation to Cecil, ...

The equation, Itot=Ifor+Iref, doesn't look complex, but it is. :-)

Not sure if you saw it, but it looks to me that even EZNEC 4+ is on "our" side.

I modeled an eight-sided coil in EZNEC 2.0 before I knew EZNEC+
had a helical feature. The result is at:

http://www.qsl.net/w5dxp/octcoil.gif

EZNEC reports that the current at the top of the coil is 0.5326
amps while the current at the bottom of the coil is 0.9956 amps.
I plotted the current in all 40 EZNEC 2.0 coil segments - looks
close to a cosine function to me with the coil occupying
roughly 50 degrees of the antenna.
--
73, Cecil http://www.qsl.net/w5dxp


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H. Adam Stevens, NQ5H wrote:
Now where was that chapter in E&M about "superposition of currents"?
Can't seem to find it.

From: "Transmission Lines and Networks", Walter C. Johnson, page 244:
"10.4 The Principle of Superposition. If a number of electrical sources
exist in a linear network, the resulting CURRENT or voltage in any part
of the system is equal to the sum of the CURRENTS or voltages that would
be caused by each source acting separately."

Emphasis mine. Now you know where it is. Here's another interesting
quote from page 150:

"P = |E+|^2/Z0 - |E-|^2/Z0 We can regard the first term in this
expression as the power associated with the forward-traveling wave,
and the second term as the reflected power."

Roy, why do you bother?

Because he knows I might learn something from him or he might learn
something from me. How about you?
--
73, Cecil http://www.qsl.net/w5dxp


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On Wed, 27 Oct 2004 23:44:33 -0700, Roy Lewallen
wrote:
can you explain it to me? Richard?

Hi Roy,

This Richard could with his books, but I doubt you meant me (that's
why the FCC assigns call signs).

So, given my toe-hold of ambiguity....

The explanation was offered by my posting quoting Poe:

"The error of our progenitors was quite analogous with that of the
wiseacre who fancies he must necessarily see an object the more
distinctly, the more closely he holds it to his eyes. They blinded
themselves, too, with the impalpable, titillating Scotch snuff of
detail; and thus the boasted facts of the Hog-ites were by no
means always facts -- a point of little importance but for the
assumption that they always were. The vital taint, however, in
Baconianism -- its most lamentable fount of error -- lay in its
tendency to throw power and consideration into the hands of merely
perceptive men -- of those inter-Tritonic minnows, the
microscopical savans -- the diggers and pedlers of minute facts,
for the most part in physical science -- facts all of which they
retailed at the same price upon the highway; their value
depending, it was supposed, simply upon the fact of their fact,
without reference to their applicability or inapplicability in the
development of those ultimate and only legitimate facts, called
Law."

Of course, this was encumbered by more commentary from Poe, and as
this group has such difficulty in reading English, then perhaps the
nuances were lost to the greater appreciation of what was being said.

I will hit the hi-lights:

The term "wiseacre" hasn't lost its currency over the course of 165
years, but I suppose many english (lower case) readers probably have
lost track of the meaning of "minute" to presume it is only a unit of
time.

"... the wiseacre who ... must necessarily see an object the more
distinctly, the more closely he holds it to his eyes. ... the
diggers and pedlers of minute facts, ... their value depending ...
simply upon the fact of their fact, without reference to their
applicability ..."

To translate this Classic Comix edition of Poe into techno-jargon:

"Current Drop" accounts for less than 1dB in the far field.

To boil it down further for those who parse only Anglo-Saxon

BFD

73's
Richard Clark, KB7QHC

Roy Lewallen wrote in message ...
Roy Lewallen wrote:
it shouldn't be difficult at
all for you to write a couple of simple equations which give the
currents at the two ends of the coil.

Itot = I+ + I-

There's the equation that I have already posted many, many times.
Sorry you missed it.

Helloo. . . Anybody home?

Roy, you asked for a couple of simple equations which give the currents
at the two ends of the coil. Here they are again:

Itot(bottom) = I+(bottom) + I-(bottom)

Itot(top) = I+(top) + I-(top)

Exactly what equations do you use when you make your assertions about the
current at each end of a coil? Please be specific.
--
73, Cecil, W5DXP

Roy Lewallen wrote:

Perhaps it's your background as a manager that's in evidence here. I was
one of the people that "concept" people like you "let" provide the
equations. I have an apropos Doonesbury cartoon on my office wall:

Pointy-haired boss, pointing to simple graph labeled "Sales": "Sales are
dropping like a rock."

Pointy-haired boss, pointing to a graph labeled "Future", with single
upward line: "Our plan is to invent some sort of doohickey that everyone
wants to buy."

Pointy-haired boss, to Dilbert: "The visionary leadership work is done.
How long will your part take?"

At review time, the boss judged me on whether the "doohickey" worked
according to specifications that he and some marketing people came up
with in "concept" meetings. All I had to do was to understand the
science, come up with the equations, develop new technology as required,
create the device, and make it work -- all on a schedule and within a
budget which were also dictated by the "visionaries". Just grunt work,
not worthy of the visionary people who were doing more important things.

I don't believe for a minute that the boss really understood how the
device worked.

I like your Dilbert cartoon, Roy. On the other hand, imagine the boss's
disappointment at seeing the same old doohickey reappear each time with
only a couple new doodads hanging from it. Is it time for a new boss,
or time for a new engineer?

73, Jim AC6XG

Cecil,

I decided to take a look at the question you asked below, and I came up
with a really simple modeling experiment.

Set up a simple quarter-wave vertical in EZNEC, resonant at 7 MHz. Run
the source and current functions and save the data. Now change the
frequency to 3.5 MHz and repeat the source and current functions. Do not
scale the antenna or change anything else. I believe most people would
now view this antenna as one-eighth wave at the new frequency, or
perhaps representative of a whip above a loading coil (at 3.5 MHz).

This experiment demonstrates what happens to the "remaining eight feet"
when confronted with the conflict between the "need" for 90 degrees and
the availability of only 45 degrees.

My computer did not blow up, and I suspect yours will survive as well.

Any number of permutations can be tried. Change the length instead of
the frequency, scale up, scale down, and so on. The current always
starts at 1.0, and it always goes to 0.0 at the tip. The reactance and
driving voltage can be awesome, but the current remained unfazed (or is
that unphased?).

This is not a revelation. Antenna books point out that the current in a
short antenna decreases in a straight line, not a sine curve, from the
feed point to the tip. (E.g. Kraus, 2nd Ed. page 216)

Since your traveling wave model seems to be based on a 90 degree
requirement, you may want to consider incorporating this additional
information before submitting your new model for publication.

73,
Gene
W4SZ

(The "eight feet" is taken from your message. In this experiment the
whip length is quite a bit larger, of course. Rescale the entire
experiment if you like.)


Cecil Moore wrote:


Here's an unanswered question: If the loading coil occupies zero
degrees, how can the remaining eight feet of the antenna occupy
the entire 90 electrical degrees? Wouldn't the coil have to
change the frequency for that to happen?

Gene Fuller wrote:

Antenna books point out that the current in a
short antenna decreases in a straight line, not a sine curve, from the
feed point to the tip. (E.g. Kraus, 2nd Ed. page 216)

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG



Looks like it, back in those days it was simpler to draw the straight line
approximating end of sine wave curve than bother to draw it precisly. Close
enough for unbelievers :-)

Yuri, K3BU.us

Tom Donaly wrote:
We're all marching in lockstep right back to the middle ages.

Walter Johnson didn't live in the middle ages, Tom. I would be
careful about disagreeing with him.
--
73, Cecil http://www.qsl.net/w5dxp


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Gene Fuller wrote:
This experiment demonstrates what happens to the "remaining eight feet"
when confronted with the conflict between the "need" for 90 degrees and
the availability of only 45 degrees.

Once again, you misunderstand what I am saying which is: For an unloaded
vertical antenna to have a purely resistive feedpoint impedance, i.e.
resonant, it must be at least 1/4WL long fed against a counterpoise. A
45 degree unloaded antenna does NOT have a purely resistive feedpoint
impedance, i.e. it is NOT resonant. We are discussing only resonant antennas
here. A properly loaded short mobile antenna does indeed have a purely resistive
feedpoint impedance, i.e. at resonance, it must exhibit something in the ballpark
of 90 degrees of antenna. Hint: There is 90 degrees between the purely resistive
current maximum point and the open end of the antenna. The coil MUST occupy some
of that 90 degrees.

I'm not saying the coil occupies every degree not occupied by the
wires but it does NOT occupy zero degrees. The argument is whether it
occupies zero degrees or not.

Please stop misunderstanding what I am saying. :-)

My computer did not blow up, and I suspect yours will survive as well.

Since you misunderstood, the rest of your posting is proceeding under
false premises.
--
73, Cecil http://www.qsl.net/w5dxp


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"Tom Donaly" wrote in message
m...
snip
Roy, why do you bother?
73
H.



Hi H.,
this is interesting in a sick sort of way. You've got Cecil,
a cloud cuckoo land philosopher espousing his inchoate, hand waving
"theory," Yuri, the consummate empiric, who thinks he can understand
nature by experiment alone, and Richard Harrison, who understands the
natural world almost solely through the agency of written authority; and
all of these erstwhile fellows together can't get it right.
We're all marching in lockstep right back to the middle ages.
73,
Tom Donaly, KA6RUH
Not me, baby. I'm going kicking and screaming.
73, H.

BTW
I took graduate E&M out of J D Jackson's "Classical Electrodynamics" in
1980.
The second semester was taught by one of the two profs I presently work for
at U Texas Physics.
My main boss is an old friend of Dave Jackson and one of my colleagues is a
student of Kraus.

"Jim Kelley" wrote in message
...
Gene Fuller wrote:

Antenna books point out that the current in a
short antenna decreases in a straight line, not a sine curve, from the
feed point to the tip. (E.g. Kraus, 2nd Ed. page 216)

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG


yup

Jim Kelley wrote:
Gene Fuller wrote:

Antenna books point out that the current in a short antenna decreases
in a straight line, not a sine curve, from the feed point to the tip.
(E.g. Kraus, 2nd Ed. page 216)


Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG


Actually, no one really knows whether it's really a sine curve
or not because no one has ever been able to solve the integral
equation that would give an exact answer. The sine approximation
works o.k. because the far field is relatively insensitive to
changes in the shape of the current curve back at the antenna.
The best thing to do is to approximate the curve with a
moment method program on a computer. That's what the moment
method does best.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Actually, no one really knows whether it's really a sine curve
or not because no one has ever been able to solve the integral
equation that would give an exact answer.

It would only be a sine curve if the reflected current was equal
to the forward current, i.e. the antenna was lossless (no I^2*R losses
and no radiation). We know that the reflected current is roughly about
90% of the value of the forward current at the feedpoint of a dipole.
So the total current distribution approximates a cosine wave. In the
textbooks you will find general assumption statements like Kraus':

"IT IS GENERALLY ASSUMED that the current distribution of an infinitesimally
thin antenna is sinusoidal, and that the phase is constant over a 1/2WL
interval, changing abruptly by 180 degrees between intervals."

For all real-world current waves, there is an attenuation factor. The
reflected current arriving back at the feedpoint is always less than
the forward current. That's why the feedpoint impedance,
(Vfor+Vref)/(Ifor+Iref) is low but never zero for a dipole. The net
current cosine function is a ballpark assumption, not actual reality.

This is interesting because (Vfor+Vref)/(Ifor+Iref) is 75 ohms for a
1/2WL dipole but is about 12 ohms for a 75m bugcatcher. That means
the reflected waves are closer in magnitude to the forward waves
in the 75m bugcatcher than they are for a 1/2WL dipole. That makes
sense since the tip of the antenna is closer to the feedpoint for
the 75m bugcatcher than for the 1/2WL dipole.
--
73, Cecil http://www.qsl.net/w5dxp


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