On May 9, 1:56*pm, wrote:
Tom,

OK, I tried what you suggested. I put my loading coil midway up a 20ft
vertical wire in the EZNEC model. I reduced the number of turns to
lift the resonant frequency to 5.6MHz. EZNEC predicted that the
magnitude of the current at the top of the coil would be 77% of the
magnitude at the bottom.

Then I removed the coil in the model, replaced it with a straight wire
containing an EZNEC lumped load, and adjusted that load for antenna
resonance at 5.6MHz again. I needed +j1630.

Given the dimensions of the coil, the Corum calculator predicted a
lumped circuit equivalent reactance of *+j1573, and it predicted a
current fall-off across the coil of 78%.

Hi Steve,

OK, I'm wondering now exactly what "Corum calulator" you are using
that predices "a current fall-off across the coil of 78%." The
inductance calculator on the HamWaves website that I thought we were
talking about doesn't seem to say anything about "current fall-off" in
coils, though perhaps I'm missing it.

Cheers,
Tom

Tom Ring wrote:

And denigrating slide rules is silly. Most of the world that surrounds
you was calculated with a slide rule's resolution. When used properly
they give answers that are as accurate as is needed for engineering.

I was one of the last classes in school to use a slide rule - they went
to calculators the next year.

I have to say that using a slide rule changed my outlook on math in all
it's forms. Took a absolute idiot at math to the dilettante I am today.
8^)

I use calculators all the time now, but I still have a slide rule that I
use in the garage....

- 73 de Mike N3LI -

Hi Tom,

I should have been more explicit.

I took the "Axial Propagation Factor" (4.372 rad/m) figure which was
given by the HamWaves calculator and multiplied it by the coil length
(155mm) to find the effective electrical length of the coil (38.83
degrees). Then I took cos(38.83)=0.779 as the fall-off in current
across the coil.

73,
Steve G3TXQ


On May 11, 2:46*am, K7ITM wrote:

Hi Steve,

OK, I'm wondering now exactly what "Corum calulator" you are using
that predices "a current fall-off across the coil of 78%." *The
inductance calculator on the HamWaves website that I thought we were
talking about doesn't seem to say anything about "current fall-off" in
coils, though perhaps I'm missing it.

Cheers,
Tom

wrote:
I took the "Axial Propagation Factor" (4.372 rad/m) figure which was
given by the HamWaves calculator and multiplied it by the coil length
(155mm) to find the effective electrical length of the coil (38.83
degrees). Then I took cos(38.83)=0.779 as the fall-off in current
across the coil.

Interesting. W8JI's coil through which he measured a 3 nS
delay was 100t, 2" dia, 10" long, #18 wire.

http://www.w8ji.com/inductor_current_time_delay.htm

Converting everything to metric and entering the data into
the HamWaves calculator at 4 MHz, yields a calculated delay
of 21.5 nS through the W8JI coil and a VF of ~0.04 at 4 MHz.

So which are we to believe? W8JI's measurements or ON4AA's
calculator. There's a 7x difference between 3 nS and 21 nS.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On May 11, 2:26*am, wrote:
Hi Tom,

I should have been more explicit.

I took the "Axial Propagation Factor" (4.372 rad/m) figure which was
given by the HamWaves calculator and multiplied it by the coil length
(155mm) to find the effective electrical length of the coil (38.83
degrees). Then I took cos(38.83)=0.779 as the fall-off in current
across the coil.

73,
Steve G3TXQ

Hi Steve,

OK, so I suppose you are assuming that the current distribution will
follow a cosine along electrical degrees of your antenna, with a
maximum at the base/feedpoint. If that's the case, then would you not
account for the bottom 10 feet of wire, about 20.5 electrical
degrees? If I do that and assume 1 amp at the feedpoint, I should see
about .9367 amps at 20.5 degrees and 0.5101 amps at (20.5+38.83)
electrical degrees. 0.5101/.9367 would then be the ratio of currents
between the ends of the coil, and that's 0.5446, only a 45.54 percent
fall-off.

In fact, it seems to me that the idea of cos(38.83 degrees) = .779
would imply a fall-off of 22.1%... and that tells me that perhaps I'm
still not understanding your model very well. Maybe you are NOT
assuming the current along the electrical degrees of the antenna, up
from the feedpoint, will have a cosine distribution. At this point, I
have to say that I'm just not at all sure what your model really is.
Perhaps you are making different assumptions about the current
distribution...

Also, if you still have the model around, try adding a top hat to the
upper wire. For simplicity, you can just use a simple "T" structure,
where the top horizontal wire is, say, five feet long total. With
such a configuration, what's the current distribution along the
radiating element going to be?

Of course, what I'm suggesting here is that one must be careful to
test ones models at corner cases before putting too much faith in
them, and even then, one must always be wary of cases where the model
may go awry.

Cheers,
Tom

K7ITM wrote:
OK, so I suppose you are assuming that the current distribution will
follow a cosine along electrical degrees of your antenna, with a
maximum at the base/feedpoint.

This is a good assumption for horizontal 1/2WL thin-wire
dipoles as presented by Kraus. It doesn't seem to be valid
for loaded vertical antennas where there is an instantaneous
phase shift at the impedance discontinuities. There is a
definite change in the slope of the current profile at
such boundaries.

And there is the nagging current bulge in the loading coil
causing a rise in current in adjacent turns. Normally a
current maximum would indicate a purely resistive impedance
but that doesn't seem to be the case inside a loading coil.

Years ago, I gave up on the current cosine argument for
loaded mobile antenna current in favor of loading the
coil with its characteristic impedance and using traveling
wave current to measure the electrical length of the coil.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Mon, 11 May 2009 02:26:04 -0700 (PDT), wrote:

I should have been more explicit.

I took the "Axial Propagation Factor" (4.372 rad/m) figure which was
given by the HamWaves calculator and multiplied it by the coil length
(155mm) to find the effective electrical length of the coil (38.83
degrees). Then I took cos(38.83)=0.779 as the fall-off in current
across the coil.

Hi Steve,

I don't often drop into this side-thread as the topic had drifted into
a stagnated intellectual backwater.

On this and one prior posting by you:
On Sat, 9 May 2009 13:56:31 -0700 (PDT), wrote:
OK, I tried what you suggested. I put my loading coil midway up a 20ft
vertical wire in the EZNEC model. I reduced the number of turns to
lift the resonant frequency to 5.6MHz.

I note how little Corrum really has to offer when you had to take the
same:
effective electrical length of the coil (38.83 degrees)
and change it (to the same effective electrical length? I think not.)
to fit the same available wire, at the same specific frequency - only
at a different height along the available wire.

By my quick read on the stale crisis of current "fall-off" and proving
Corum by EZNEC; it seems quite apparent that EZNEC (the authority) is
driving the coil requirements which are then force fitted by Corum's
inappropriate application.

After all, Corum says nothing of:
1. Application;
2. Base loading;
3. Mid or Top loading;
4. Stinger selection;
and yet all solutions seem to derive from their math with the elegance
of an ad-hoc "missing degrees" provision (that is quickly discarded as
shown above when current becomes the focus).

Corum DOES say that the formula is only applicable for certain
constraints which I note are NEVER observed in the application nor the
breach. All of the commentary proceeds through equation (32) when
every argument is an instance of equation (31).

How much are you willing to accept of that paper (which is another way
of asking how much you are willing to discard)?

I will ask one ace-buster question that I expect no one will answer:
Show me the computation for M (= tau · a)
which would be appropriate for the NON-quarterwave resonance of the
coil in question at 3.85 MHz.

For extra credit:
1. What is the wave number, k for 3.85 MHz?
2. What is the phase velocity for the original (not changed) coil?
3. What is tau for the original (not changed) coil at 3.85 MHz?

Yes, this is intimidating to ask; but seeing there are so many
authorities on Corum; and that these considerations would have been
done by the authors themselves; then their solutions must reside
somewhere in notes or as marginalia for quick reporting (or could be
summoned up through running through the same math as before).

73's
Richard Clark, KB7QHC

On Sun, 10 May 2009 14:52:06 -0700, "Tom Donaly"
wrote:

The presence of anything at all near the coil should lower its resonant
frequency. Even the measuring apparatus should have an effect. I think
it would require very careful planning and implementation to find an
exact resonant frequency. You'd have to ask Richard Clark how to do it
if you wanted high accuracy. I'm unwilling to find fault with either
Corum or EZNEC at this point. Making accurate models can be just
as hard as making valid experiments, and I wish you luck working with
your models. I do urge you to make a coil to test your results against,
though. It isn't difficult, and with a little help from some of your
fellow experimenters, you should get results that are very close to
being meaningful.
73,
Tom Donaly, KA6RUH

Hi Tom,

Thanx for the flowers. In point of fact, inductor and capacitor
standards are shielded. They are three terminal devices. To remove
the effects of the shield you drive it at the same potential (which is
to say the shield is floating with respect to everything/one around
it).

For the practicality of things, Reggie was never very far off the mark
and you following him as an exemplar is suitable to other's inventions
of proximities that have no defining moment in their references.

I can well guess the remainder of your method as it was well defined
in most Ham manuals (derived from conventional EE methods) when I read
up on it 40 odd years ago. If Cecil every gets over this intellectual
pebble in the path, please quote it so that I can see how much strain
that ascent took.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
I note how little Corrum really has to offer when you had to take the
same:
effective electrical length of the coil (38.83 degrees)
and change it (to the same effective electrical length? I think not.)
to fit the same available wire, at the same specific frequency - only
at a different height along the available wire.

Richard, I explained that phenomenon in a posting last
week which you obviously didn't read. Please go back and
read my posting of 5-9-09 at 1:08pm to this thread.

It is also explained on my web page at:

http://www.w5dxp.com/shrtstub.htm
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

Tom,

Firstly, I'm guilty of a "sloppy" choice of words. Whenever I've been
using the phrase "drop off in current" I've meant the current at the
top of the coil as a percentage of the current at the bottom. So when
I've quoted 70% the current will have reduced by 30%. Apologies!

Secondly, you're testing the limits of my understanding with the
overall current distribution from base section, through the coil, to
the top section. However I think the point is that you can't simply
"add electrical degrees" through the various sections when the
characteristic impedances of the sections are so disparate. That was
Cecil's point in the very first posting. We also know that, as
expected, summing the "degrees" for the three sections gets nowhere
near a total of 90 degrees, so clearly you can't assume a cosine
distribution that is contiguous across all three sections.

I'll investigate what happens with a "top hat".

73,
Steve G3TXQ



On May 11, 6:56*pm, K7ITM wrote:

Hi Steve,

OK, so I suppose you are assuming that the current distribution will
follow a cosine along electrical degrees of your antenna, with a
maximum at the base/feedpoint. *If that's the case, then would you not
account for the bottom 10 feet of wire, about 20.5 electrical
degrees? *If I do that and assume 1 amp at the feedpoint, I should see
about .9367 amps at 20.5 degrees and 0.5101 amps at (20.5+38.83)
electrical degrees. *0.5101/.9367 would then be the ratio of currents
between the ends of the coil, and that's 0.5446, only a 45.54 percent
fall-off.

In fact, it seems to me that the idea of cos(38.83 degrees) = .779
would imply a fall-off of 22.1%... and that tells me that perhaps I'm
still not understanding your model very well. *Maybe you are NOT
assuming the current along the electrical degrees of the antenna, up
from the feedpoint, will have a cosine distribution. *At this point, I
have to say that I'm just not at all sure what your model really is.
Perhaps you are making different assumptions about the current
distribution...

Also, if you still have the model around, try adding a top hat to the
upper wire. *For simplicity, you can just use a simple "T" structure,
where the top horizontal wire is, say, five feet long total. *With
such a configuration, what's the current distribution along the
radiating element going to be?

Of course, what I'm suggesting here is that one must be careful to
test ones models at corner cases before putting too much faith in
them, and even then, one must always be wary of cases where the model
may go awry.

Cheers,
Tom