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?

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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"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.

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

"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.)