Transmission Line = Antenna
 

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.

Reg Edwards wrote:

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)

Thanks Reg, this one is a keeper. If you made the helix long enough to
cause a phase reversal in the current, would the world come to an end?
--
73, Cecil http://www.qsl.net/w5dxp



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--
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........
"Cecil Moore" wrote in message
...
Reg Edwards wrote:

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)

Thanks Reg, this one is a keeper. If you made the helix long enough to
cause a phase reversal in the current, would the world come to an end?
--
73, Cecil

========================

I forgot to say it will also have a 'cosine' current distribution between
base and top.

The antenna 'end effect' can be accounted for if necessary.

When the helix is not close wound, the increase in velocity factor and the
reduction in Zo can be accounted for.

Axial mode radiation cannot be distinguished from normal mode but this is of
negligible consequence until helix circumference approaches 1/4 wavelength
which it never does in the present context.
----
Reg.

Reg wrote,
(snip)

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


You're right.
73,
Tom Donaly, KA6RUH

Cecil wrote,

Thanks Reg, this one is a keeper. If you made the helix long enough to
cause a phase reversal in the current, would the world come to an end?
--
73, Cecil http://www.qsl.net/w5dxp


It also doesn't mean anything. Those formulas for C and L are just
good hand-waving approximations and nothing else. Nice try, though.
73,
Tom Donaly, KA6RUH

Tdonaly wrote:

Cecil wrote,
Thanks Reg, this one is a keeper. If you made the helix long enough to
cause a phase reversal in the current, would the world come to an end?

It also doesn't mean anything.

Tom, please build a two wavelength long helical antenna, measure the
current at various places, and then come back and tell us, with a
straight face, that the current phase never goes from positive to
negative.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil wrote,

Tdonaly wrote:

Cecil wrote,
Thanks Reg, this one is a keeper. If you made the helix long enough to
cause a phase reversal in the current, would the world come to an end?

It also doesn't mean anything.

Tom, please build a two wavelength long helical antenna, measure the
current at various places, and then come back and tell us, with a
straight face, that the current phase never goes from positive to
negative.
--
73, Cecil http://www.qsl.net/w5dxp


It won't work, Cecil. I quit arguing when you quit understanding.
73,
Tom Donaly, KA6RUH

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



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On Sat, 31 Jan 2004 17:01:58 -0600, Cecil Moore
wrote:

And you would rather leave me ignorant than contribute anything
to my understanding.

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

Richard Clark wrote:

wrote:
And you would rather leave me ignorant than contribute anything
to my understanding.

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."

Richard, is one iota of scientific proof too much to ask or should
I accept it on faith like you? When I want hand-waving, I go to church.
I expect more than math-model-based religion out of this newsgroup.
Maybe I expect too much. You guys remind me of the people in one
of the Planet of the Apes movies who worshiped Tha Bomb.
--
73, Cecil http://www.qsl.net/w5dxp



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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.

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

I once had a 2m Heathkit linear amplifier for my Wilson transceiver.
I chose the wrong length of coax between the two and when I ceased
to transmit, the amplifier would oscillate. My neighbor, also a ham,
told me I was transmitting a wandering signal that keyed all the 2m
repeaters in the area. That was back before PL was popular.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 1 Feb 2004 10:10:44 -0600, "John Smith"
wrote:
If you can explain my error, please do so rather than insulting me by saying
I am hopelessly mixed up.

Hi John,

You will undoubtedly find the answer when our resident troll argues
this from the other side of the argument in a future thread. Time and
conflict finds him everywhere eventually. ;-)

73's
Richard Clark, KB7QHC

"Reg Edwards" wrote in message
...
"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.



What is the formula you used to arrive at:

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

"Richard Clark" wrote in message
...
On Sun, 1 Feb 2004 10:10:44 -0600, "John Smith"
wrote:
If you can explain my error, please do so rather than insulting me by
saying
I am hopelessly mixed up.

Hi John,

You will undoubtedly find the answer when our resident troll argues
this from the other side of the argument in a future thread. Time and
conflict finds him everywhere eventually. ;-)

73's
Richard Clark, KB7QHC


Thanks, Richard. I didn't know the meaning of the word Troll until now. I
asked for an explanation but, well, you saw his answer:

"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."

It seems to me that he could have at least let us in on his formula. If he
doesn't want to share knowledge with the group, and if he is not a troll,
why does he bother to post?

Oh, well. I guess the best thing is to ignore him.

John, KD5YI

On Sun, 1 Feb 2004 15:48:43 -0600, "John Smith"
wrote:


Thanks, Richard. I didn't know the meaning of the word Troll until now.

Well, actually you have been conned by the bait and switch.

I asked for an explanation but, well, you saw his answer:

"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."

That is the troll. Reggie likes to rail against software - except his
own. He likes to rail against citing authorities - until he gushes on
about Lord Plushbottom. He absolute fulminates against passivity in
learning - and then offers software without divulging the concrete
fundamentals behind them. Then there are the side topics where he
regales how he is the only Brit to have lost WWII to the Americans.

It seems to me that he could have at least let us in on his formula. If he
doesn't want to share knowledge with the group, and if he is not a troll,
why does he bother to post?

About the only consistent advice you could expect are his reports of
the metaphysical clarity of view in looking through the bottom of a
wine bottle.

Oh, well. I guess the best thing is to ignore him.

And miss all this comedy? Residency comes from time and experience!
When Reggie showed up in this group, he was preceeded by one of his
countrymen from the UK amateur newsgroup begging our indulgence for
his advanced years and eccentric habits. Somehow there was the
breathlessness of "he's yours now" in that plea. ;-)

Anyway, taken over the average of his arguing both sides of an issue,
you will come away with something and unlike others who mince about,
Reggie can actually write code and offer results from first
principles.

73's
Richard Clark, KB7QHC

John Smith wrote,


Thanks, Richard. I didn't know the meaning of the word Troll until now. I
asked for an explanation but, well, you saw his answer:

"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."

It seems to me that he could have at least let us in on his formula. If he
doesn't want to share knowledge with the group, and if he is not a troll,
why does he bother to post?

Oh, well. I guess the best thing is to ignore him.

John, KD5YI


Actually, some of Reg's stuff is close enough to be pretty good. He doesn't
like to give us any symbolic derivations, though. Maybe he's ashamed of them.
Anyway, like everyone else on this newsgroup, you can take him or leave him.
73,
Tom Donaly, KA6RUH
(PS You've just learned a valuable lesson about grouchy old Brits.)

"Tdonaly" wrote in message
...
John Smith wrote,


Thanks, Richard. I didn't know the meaning of the word Troll until now. I
asked for an explanation but, well, you saw his answer:

"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."

It seems to me that he could have at least let us in on his formula. If
he
doesn't want to share knowledge with the group, and if he is not a troll,
why does he bother to post?

Oh, well. I guess the best thing is to ignore him.

John, KD5YI


Actually, some of Reg's stuff is close enough to be pretty good. He
doesn't
like to give us any symbolic derivations, though. Maybe he's ashamed of
them.
Anyway, like everyone else on this newsgroup, you can take him or leave
him.
73,
Tom Donaly, KA6RUH
(PS You've just learned a valuable lesson about grouchy old Brits.)



Thanks, Tom, and you, too, Richard. I appreciate the heads-up.

John, KD5YI

Sorry John, I didn't realise you are a novice.

To answer your question about how to calculate the 1/4-wave resonant
frequency of a vertical antenna, use the following formula -

Fres = 75 / Height * Velocity Factor

As given in the simple example in my first posting, Height = 1.5 metres and
Velocity Factor = 0.0701

And so the resonant frequency calculates to Fres = 3.5 MHz.

If you are unfamiliar with "velocity factor" then I suggest you do a Google
on it.
---
Reg

Tdonaly wrote:
"---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."

Kraus` story of his discovery of the axial mode helical antenna starts
on page 222 of "Antennas". Kraus was a new Ohio State University faculty
member in 1946 when a famous scientist told him a helix wouldn`t work as
an an antenna as he had already tried it with no success.

The expert`s statement challenged Kraus to try for himself, and the rest
is history.

The first coil he wound was one wavelength in circumference and had 7
turns.. Kraus was pleased with the sharp end-fire beam of
circularly-polarized energy from the open end of the coil. Kraus found
the helical beam antenna had a resistive input and a wide frequency
bandwidth.

At low frequencies, outgoing and reflected waves along the helix were
almost equal in magnitude but as frequency increased, current
distribution changed dramatically in the helix.

At a circumference of one wavelength, there were 3 regions. Near the
antenna input, current decayed exponentially. Near the open end of the
helix, there was a standing wave for a short distance. In the middle
region, there was a relatively uniform current (small VSWR) which
extended over most of the helix. Kraus gives the plots of the currents
along the helix in Figure 8 on page 224.

Certainly, Tdonaly is correct. Different coils have different current
distributions, but Kraus` contribution interests me.

Best regards, Richard Harrison, KB5WZI