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-   -   Current in antenna loading coils controversy (https://www.radiobanter.com/antenna/670-current-antenna-loading-coils-controversy.html)

Yuri Blanarovich November 9th 03 11:55 PM

On his web site, Yuri quoted W9UCW as measuring the currents at the ends
of a toroid mounted at the base of the antenna as being 100 mA at the
bottom and 79 at the top. You must, then, believe these measurements to
be in error.

Roy Lewallen, W7EL



Not!
Here what Yuri has on his web site, quote by W9UCW:

"Because of the constant claim that this must be due to the fact that the coil
is so big compared to a wavelength, I measured the in and out current on a
TOROIDAL loading coil used on a 20m mobile antenna. It was a 78" base mast
(including spring and mount) with a 38" top whip (including 12" of alum. tubing
for adjustment).
Below --100ma & Above --79ma
When I moved the coil to the top of the mast and made a horizontal "X" top hat
to resonate it back on the same freq, I got
Below --100ma & Above --47ma

So, It happens even in a totally shielded loading coil with miniscule power
going thru it! Kirchoff has no laws about current being the same on both ends
of inductors. His current law is about one POINT in a circuit and his voltage
law is about a closed loop."

He described exactly how it was done, definitely not at the base.

Yuri

Yuri Blanarovich November 10th 03 12:16 AM


That's a pretty good theory, Yuri. I'd like to know where you got
this "Cosine law" you keep talking about.


From ancient Greeks I presume, it was way before my time.
ON4UN has it shown in his book and pictures.
Current in the radiator Ir = Imax x cos Alpha
where Imax is the maximum current (like on the bottom of quarter wave radiator)
Alpha is the angle (distance) in degrees like from 0 to 90 in quarter wave
radiator.

I can't seem to find
mention of any such _law_ anywhere but on this newsgroup.


It is one of the first things in trigonometry books, look into your high school
library.

Does
that mean I should throw away my method of moments software
because I don't need it any more?


No, keep it. It is good for number of things, like arguing against reality. If
you threw it away, what would you use as an argument?

And what is a current
drop? I've heard of voltage drops and cough drops but never
current drops.


It is right next to voltage drops. If it drops from one of the component to the
other, that is called current drop across the component, like loading coil in
the antenna, or coil in PI network, or in RF chokes, etc.

Finally, how do you measure the "current in
the radiator (in degrees)?" Why not use amperes like everyone
else?


We use amperes like everyone else. My ammeters are calibrated in A. Radiator
length can be expressed in electrical degrees, if Roy claims that it can't be
expressed in electrical length.
Like quarter wave vertical would be 90 deg., half wave dipole would be 180 deg,
Full wave quad would be 360 deg or 2 x 180 deg (same thing :-)

I won't believe your theory, Yuri, until you and Cecil take the
time to present it in terms of field theory. Since you guys have taken
EM classes in college you should have no trouble doing this, right?


You can worship what you wish, itsa free country. Or you can measure things and
then you can choose whom to believe and where to send your contribution.

73,
Tom Donaly, KA6RUH


Yuri, K3BUm

Jer November 10th 03 12:22 AM

I think I'm learning something, but I'm getting a headache trying to follow
you guys!

de jerry...

"Roy Lewallen" wrote in message
...
Can I conclude from this that if I were to make a coil with more or less
inductance, then I would see a current difference between the ends of
the coil?

So tell you what. If you'll pull out your equations and calculate the
expected current difference, I'll replace the coil with one of 100 ohms
reactance and remeasure. How much current difference (magnitude andd
phase, of course) between the ends of a 100 ohm inductor at the base of
that same antenna?

Roy Lewallen, W7EL

Cecil Moore wrote:
Yuri Blanarovich wrote:

Judging by description, I would guess that there wasn't much

difference.


The feedpoint of the radiator alone is 35-j185. The impedance of the
loading
toroid is 0.6+j193. Assuming perfect predictability, that gives the

antenna
system a feedpoint impedance of 35.6+j8, i.e. it is *longer* than

resonant.
That moves the current maximum point inside the toroid making the

current
in and out even closer to equal. If a coil is installed at a current
maximum
point or a current minimum point, the current in and out will be the

same.
If a coil is installed at a place where the slope of the current

envelope
is positive, the current will actually increase through the coil.





Yuri Blanarovich November 10th 03 12:23 AM


So, as a successful and award-winning engineer, what do you calculate as
being the ratio of currents across my inductor, and how did you
calculate it?

Roy Lewallen, W7EL


Judging by the picture, I wouldn't be able to include all the variables present
in the setup.

I will first measure it in same way as W9UCW did, about 2/3 up the mast and no
wires hanging.

Yuri

Yuri Blanarovich November 10th 03 12:32 AM


Roy Lewallen wrote:

Yuri, the inductor I put at the base of the antenna "replaced" something
like 20 - 45 degrees,


Nope, it didn't, Roy. Your 33' vertical was already equivalent to a 50'
vertical apparently due to extraneous loading. I calculate that your
coil replaced 18 degrees of wire with a current maximum point located
inside the coil.
--
73, Cecil http://www.qsl.net/w5dxp


In that case,
If the feedpoint current was at 0 deg of the radiator length, and coil replaces
18 deg of wire, the cos 18 deg = 0.951 which should make difference, drop in
the coil current 5% (or half, 2.5 deg?)
Providing current maximum is exactly at the bottom end of the coil.

Yuri, K3BU.us

Yuri Blanarovich November 10th 03 12:37 AM


I think I'm learning something, but I'm getting a headache trying to follow
you guys!

de jerry...



It not our fault, it is them :-)

Yuri

Yuri Blanarovich November 10th 03 12:41 AM

2.5 deg?)

make that
2.5%?)

Roy Lewallen November 10th 03 01:10 AM

Well, what would it be if my setup was perfect?

Roy Lewallen, W7EL

Yuri Blanarovich wrote:
So, as a successful and award-winning engineer, what do you calculate as
being the ratio of currents across my inductor, and how did you
calculate it?

Roy Lewallen, W7EL



Judging by the picture, I wouldn't be able to include all the variables present
in the setup.

I will first measure it in same way as W9UCW did, about 2/3 up the mast and no
wires hanging.

Yuri



Roy Lewallen November 10th 03 01:50 AM

I apologize. I read and was referring to the same quote, and interpreted
it to mean that the first measurement was made with the coil at the base
of the antenna. So where was it -- 78" from the bottom?

Roy Lewallen, W7EL

Yuri Blanarovich wrote:
On his web site, Yuri quoted W9UCW as measuring the currents at the ends
of a toroid mounted at the base of the antenna as being 100 mA at the
bottom and 79 at the top. You must, then, believe these measurements to
be in error.

Roy Lewallen, W7EL




Not!
Here what Yuri has on his web site, quote by W9UCW:

"Because of the constant claim that this must be due to the fact that the coil
is so big compared to a wavelength, I measured the in and out current on a
TOROIDAL loading coil used on a 20m mobile antenna. It was a 78" base mast
(including spring and mount) with a 38" top whip (including 12" of alum. tubing
for adjustment).
Below --100ma & Above --79ma
When I moved the coil to the top of the mast and made a horizontal "X" top hat
to resonate it back on the same freq, I got
Below --100ma & Above --47ma

So, It happens even in a totally shielded loading coil with miniscule power
going thru it! Kirchoff has no laws about current being the same on both ends
of inductors. His current law is about one POINT in a circuit and his voltage
law is about a closed loop."

He described exactly how it was done, definitely not at the base.

Yuri



Roy Lewallen November 10th 03 01:53 AM

I'm sorry, I didn't catch the step where you got from cos(18 degrees) =
0.951 to 2.5%.

Roy Lewallen, W7EL

Yuri Blanarovich wrote:
2.5 deg?)



make that
2.5%?)




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