On Sat, 31 Jan 2004 18:14:18 -0600, Cecil Moore
wrote:
Richard, is one iota of scientific proof too much to ask or should
I accept it on faith like you?

Depends: is your head under water, or are you drawing a vacuum?

Cecil wrote,

Tdonaly wrote:
It won't work, Cecil. I quit arguing when you quit understanding.

And you would rather leave me ignorant than contribute anything
to my understanding. I certainly understand that. I have an
engineering degree and my IQ, according to MENSA, is in the
upper 1/2 of one percent. That you cannot post anything that
I can understand seems to be your problem, not mine. But don't
feel alone. My Southern Baptist Preacher has the identical
problem. His religion interferes with my understanding. Your
math model religion interferes with my understanding in exactly
the same way. Hint: If your math model doesn't predict reality,
it ain't worth much. The original assertion was that the current
into a coil is identical to the current out of a coil. Never mind
that the current has to travel faster than the speed of light to
make that true. Every measurement has proven that there is a
current taper through the coil. Yet, you maintain the original
assertion. There's something seriously wrong with a mind that
maintains concepts that have been proven wrong by measurements.
--
73, Cecil http://www.qsl.net/w5dxp

Boy, you sure have accused me of a lot of things I never knew I
did. I don't maintain any of that stuff. I don't even care. Furthermore,
I don't have any better idea than you do what the current distribution in
any individual coil is. I suppose it depends on its environment and its
physical description, but beyond that it means nothing to me. What I
was railing at you about was something entirely different: I simply don't
believe your theories explain what you say they do, and unless you
can give better evidence than you have, I don't see why anyone else
should believe them either. Period. That's it. The end. Kaput. Finis.
73,
Tom Donaly, KA6RUH
(P.S. You euchred me into responding this time but it won't happen again.)

Tdonaly wrote:
I simply don't
believe your theories explain what you say they do, and unless you
can give better evidence than you have, I don't see why anyone else
should believe them either.

I don't recall stating any theories except quotes from Kraus's book.
Would you be so kind as to re-state one of them for me so we can discuss
it intelligently? I don't recall you offering anything except personal
opinions and ad hominem attacks.

I have stated that there is a current taper in every real world mobile
loading coil and measurements by the very people who disagreed proved
that I was right.
--
73, Cecil http://www.qsl.net/w5dxp



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"Reg Edwards" wrote in message
...

Take a close-wound wire helix of diameter D metres and having N = 500
turns
per meter. Wire diameter approx 2mm.

All formulae available from Terman, Kraus, ARRL, etc., etc.

C = 55.5 / (Ln( 2 * H / D ) - 1) picofarads per meter.

L = Square( N * Pi * D ) / 10 microhenrys per meter.

TransLine impedance, Zo = Sqrt( L / C ) ohms.

Propagation Velocity = 1 / Sqrt( L * C ) metres per second.

Take a length of H =1.5 metres of this helix and use it as a short
vertical
antenna above a good ground.

It will resonate as a 1/4-wave vertical at 3.5 MHz.

Zo = 3243 ohms.

Velocity factor = 0.0701

Radiation Resistance = 0.176 ohms.

etc., etc.,

It's so simple you can't believe it. ;o)
----
Reg.



Somehow this doesn't seem right.

For example, RG58 has a Zo of 52.5 ohms = sqrt(L/C). It also has C=28.5
pF/ft or 93.5 pF/meter. Solving for L gives 267.8 nH.

So, are you saying that a 1 meter length of RG58 will resonate at
fo=1/sqrt(L*C) or 31.8 MHz? If that's not what you're saying, where does
your Zo=3243 come from?

Richard:

[snip]
This reminds me of a time when I challenged our Chief Bosun's mate
with: "You can lead a horse to water, but you can't make him drink."

The Chief always had the last word with logic like yours:
"You hold his head under water and suck on his ass."

73's
Richard Clark, KB7QHC
[snip]

Heh, heh... memorable. I love it... but....

Are you sure he didn't also use the "f" word?

It's my recollection that Chief Bosun's Mates do focus on the ass a lot...
but not without a few well chosen "f" word interjections.

Too many years ago to remember, I had the con on a tin can along
side a carrier at sea readying to pass a sick matloes over a bosun's chair.
I was holding the con "real steady" in heavy seas +/- 1/2 degree or so,
grinning to myself from ear to ear at how well I was doing, the Chief
Bosun's
Mate ruined my pride and reverie when he hollered down the blow tube,
"Sailor
if you don't keep this fuxxxxg tin can on fuxxxxg course I'll come down this
fuxxxxg ladder and stuff that fuxxxxg wheel up your 'ass' spoke by fuxxxxg
spoke."

Now that's Chief Bosun's Mate talk! :-)

--
Peter K1PO
Indialantic By-the-Sea, FL

Reg: I think youv'e lost em now! --Peter K1PO

"John Smith" wrote in message
...

"Reg Edwards" wrote in message
...

Take a close-wound wire helix of diameter D metres and having N = 500
turns
per meter. Wire diameter approx 2mm.

All formulae available from Terman, Kraus, ARRL, etc., etc.

C = 55.5 / (Ln( 2 * H / D ) - 1) picofarads per meter.

L = Square( N * Pi * D ) / 10 microhenrys per meter.

TransLine impedance, Zo = Sqrt( L / C ) ohms.

Propagation Velocity = 1 / Sqrt( L * C ) metres per second.

Take a length of H =1.5 metres of this helix and use it as a short
vertical
antenna above a good ground.

It will resonate as a 1/4-wave vertical at 3.5 MHz.

Zo = 3243 ohms.

Velocity factor = 0.0701

Radiation Resistance = 0.176 ohms.

etc., etc.,

It's so simple you can't believe it. ;o)
----
Reg.



Somehow this doesn't seem right.

For example, RG58 has a Zo of 52.5 ohms = sqrt(L/C). It also has C=28.5
pF/ft or 93.5 pF/meter. Solving for L gives 267.8 nH.

So, are you saying that a 1 meter length of RG58 will resonate at
fo=1/sqrt(L*C) or 31.8 MHz? If that's not what you're saying, where does
your Zo=3243 come from?

On Sun, 01 Feb 2004 05:02:56 GMT, "Peter O. Brackett"
wrote:
It's my recollection that Chief Bosun's Mates do focus on the ass a lot...
but not without a few well chosen "f" word interjections.

Hi Peter,

It would put this thread to shame to offer that when I was in Nixon's
Canoe Club, we could use that word as an adjective; adverb; verb
(transitive or intransitive); noun; relative pronoun; in a
prepositional phrase; as an indirect/direct object; as a (get this)
connective; subjective complement; often as a nominative absolute
phrase; as a gerund; and quite often to split an infinitive. That
hardly allowed for the full scope of usage and there was never any
doubt about meaning, application, intent, or subject that so attends
"current." in these rather drab and limp wrist tea parties called
debate.

73's
Richard Clark, KB7QHC, ET1

I'm afraid you are hopelessly mixed up.

What on Earth has it got to do with RG58?

As I have already said, using the well known formulae -

Calculate C pF/m
Calculate L uH/m
Insert L and C in Zo = Sqrt(L/C)

and Bingo! Zo = 3243 ohms.

I'm afraid you are hopelessly mixed up.

What on Earth has it got to do with RG58?

As I have already said, using the well known formulae -

Calculate C pF/m
Calculate L uH/m
Insert L and C in Zo = Sqrt(L/C)

and Bingo! Zo = 3243 ohms.


What it has to do with RG58 is as follows:

C = 55.5 / (Ln( 2 * H / D ) - 1) picofarads per meter.

RG58 has 93.5 pF/meter. (from ARRL Antenna Book)

L = Square( N * Pi * D ) / 10 microhenrys per meter.

RG58 has 258 nH/meter. (from Zo=sqrt(L/C) and Zo from ARRL Antenna Book)

TransLine impedance, Zo = Sqrt( L / C ) ohms.

RG58 is sqrt(L/C) = 52.5 Ohms.

Propagation Velocity = 1 / Sqrt( L * C ) metres per second.

Vp = 1/sqrt(L*C) = 203.7 m/s

Take a length of H =1.5 metres of this helix and use it as a short
vertical
antenna above a good ground.

It will resonate as a 1/4-wave vertical at 3.5 MHz.

Using a 1 meter length of RG58, it will resonate at f = 1/(2*pi*sqrt(L*C)) =
32.4 MHz.

Zo = 3243 ohms.

Zo = 52.5 Ohms.

Velocity factor = 0.0701

Velocity factor = 203.6/300 = .68


In other words, Reg, I don't see why I can't apply the same equations to
determine the resonant frequency of a piece of RG58. Are you saying you can
use these equations and I can't? After all, you are implying you can analyze
your distributed coil/transmission line using non-distributed, lumped
components.

If you can explain my error, please do so rather than insulting me by saying
I am hopelessly mixed up.

"John Smith" wrote -
Using a 1 meter length of RG58, it will resonate at f = 1/(2*pi*sqrt(L*C))
=
32.4 MHz.

============================
Obviously not. You're miles away from the right ball park.

And you are using the wrong formula for the resonant frequency of a
transnmission line anyway.