Reg Edwards wrote:
===================================

The characteristic impedance of an infinitely long wire is Zo.

If we cut the line and measure between the two ends we obtain an input
impedance of twice Zo. Which is the answer to our problem.

Zo is a function of wavelength, conductor diameter and conductor
resistance R where R includes the uniformly distributed radiation
resistance. On a high Zo line the radiation resistance is small
compared with Zo and the only effect of the radiation resistance is to
give Zo a small negative angle. Which when estimating Zo can be
ignored. (It is conductor resistance which at HF gives Zo of ALL
lines a very small negative angle).

This assumption is correct only when the transmission line conductors
are closely spaced. That isn't at all true for the halves of a dipole.

In the problem posed, the current is also uniformly distributed along
the low-loss line and radiation resistance is not the value we are
familiar with and what we might do with it.

And so we get approximately -

Rin = 120 * ( Ln( Wavelength / 2 / d ) - 1 )

At a wavelength of 2 metres and a conductor diameter of 10mm the input
resistance = 433 ohms.

I cannot guarantee the above formula to be correct. But is it low
enough for you? ;o)

I can't see how it can possibly be correct. Unless I'm mistaken, you've
completely ignored the effect of radiation in calculating the radiation
resistance. It sure makes the calculation a lot simpler, though!

Mr Wu calculates radiation resistance which is not the same as input
impedance unless correctly referenced. It is usual in technical papers
to calculate Radres at one end of the antenna. Or it may be the
distributed value. I havn't the time to find and study the full text.
From past experience, with me, it usually ends up as a wild goose
chase.

It depends on the author. Kraus uses feedpoint resistance and radiation
resistance interchangeably when loss is assumed to be zero. It's
traditional in AM broadcasting to give radiation resistance referred to
a current maximum. The conclusion is that radiation resistance can be
referred to any point along an antenna you wish, which means that it's
essential to state what point you're using as a reference.

Roy Lewallen, W7EL

Roy, if you don't like my simple approximate formula, can YOU produce
a better one without plagiarising?
----
Reg.

Reg Edwards wrote:
Roy, if you don't like my simple approximate formula, can YOU produce
a better one without plagiarising?
----
Reg.

Sure. 42. It might not be better, but it's just as good. Formulas can be
made very simple if you simply ignore any inconvenient facts. Like
radiation from an antenna.

Roy Lewallen, W7EL

"Roy Lewallen" wrote
Reg Edwards wrote:
Roy, if you don't like my simple approximate formula, can YOU
produce
a better one without plagiarising?
----
Reg.
======================================
Sure. 42. It might not be better, but it's just as good. Formulas
can be
made very simple if you simply ignore any inconvenient facts. Like
radiation from an antenna.

Roy Lewallen, W7EL
=====================================
Don't be silly. I didn't ignore radiation resistance. I said it was
small enough in comparison with Zo, as an approximation, to forget
about.

And remember Lord Kelvin.
----
Reg.

Reg Edwards wrote:


And remember Lord Kelvin.
----
Reg.


"To measure is to know."

also

"X-rays will prove to be a hoax."

http://zapatopi.net/kelvin/quotes/

ac6xg

| ----------------------------------------------------------
| "Jim Kelley"
| wrote in message ...
|
| [...]
|
| "X-rays will prove to be a hoax."
|
| http://zapatopi.net/kelvin/quotes/
|
| ac6xg
| ----------------------------------------------------------

A
I hope you will excuse me the next example.

Let

f = Sin[x]

I choose:

x = 2*k*pi

and k goes to infinity one by one: 0, 1, 2, ...

Definitely then I found correctly f(oo) = 0.


Cecil chooses:

x = 2*k*pi + pi/2

and k goes to infinity one by one,
as before.

Definitely he founds correctly f(oo) = +1


Lord Kelvin chose:

x = 2*k*pi - pi/2

and k went to infinity one by one,
as above.

Definitely he founded correctly f(oo) = -1


All of us
we are correct in all steps,
but the value

f(oo)

does not exist as a single one.
In fact f(oo) takes every value between -1 and +1.

f(oo) definitely depends
on the way in which each one of us
went to infinity.

IMHO:
this is the kind of behavior of Zinp.

B
But in addition to that there is one more to say:

Zinp is a result of
the order in which we consider the limits
for the wire radius and the length to wavelength ratio.

If
a is the wire radius and
L/wl is the ratio of length to wavelength
then
I can imagine five cases:

1
First the a is going to zero,
a formula is produced for Zinp,
then the L/wl is going to infinity
and a number may or may not be the result for Zinp.

2
First the L/wl is going to infinity
a formula is produced for Zinp,
then the a is going to zero
and a number may or may not be the result for Zinp.

3
Simultaneously,
both
the L/wl is going to infinity
and
the a is going to zero,
and a number may or may not be the result for Zinp.

4
We keep a constant value for L/wl,
then a is going to zero
and a number may or may not be the result for Zinp.

5
We keep a constant value for a,
then L/wl is going to infinity
and a number may or may not be the result for Zinp.

[ On the occasion I have to confess that the movie at
[ http://antennas.ee.duth.gr/ftp/visua...s/fu010100.zip
[ 850 KB
[ belongs to the last case.

For a possible conclusion
let me mention a remarkable note from a Mathematical book:

"The biggest source of erroneous conclusions
have to do with the order we consider the limits"

(and which order we tend then to forget ... )

Sincerely,

pezSV7BAXdag

On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote:

Reg Edwards wrote:


And remember Lord Kelvin.
----
Reg.


"To measure is to know."

also

"X-rays will prove to be a hoax."

http://zapatopi.net/kelvin/quotes/

ac6xg

Terrific reference, Jim, I've added it to my 'favorites'.

Walt

On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote:
Reg Edwards wrote:
And remember Lord Kelvin.
"X-rays will prove to be a hoax."
http://zapatopi.net/kelvin/quotes/

Hi Jim,

Certainly a trove of complexity. Keeping to the tenor of your choice:
"I have not had a moment's peace or happiness in respect to
electromagnetic theory since November 28, 1846. All this time I
have been liable to fits of ether dipsomania, kept away at
intervals only by rigorous abstention from thought on the
subject."

I sippose that California varietals were not available as an
alternative to ether. Oh how we can celebrate the modern march of
progress by viewing the problems of electromagnetic theory through the
bottom of a wine glass. Such libation allows one to simultaneously to
offer solutions and to observe that "rigorous abstention from
thought." Lord Kelvinator would be proud.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote:

Reg Edwards wrote:

And remember Lord Kelvin.

"X-rays will prove to be a hoax."
http://zapatopi.net/kelvin/quotes/


Hi Jim,

Certainly a trove of complexity. Keeping to the tenor of your choice:
"I have not had a moment's peace or happiness in respect to
electromagnetic theory since November 28, 1846. All this time I
have been liable to fits of ether dipsomania, kept away at
intervals only by rigorous abstention from thought on the
subject."

Or as is my case, intervals of rigorous abstention from participating in
discussions on rec.radio.amateur.antenna.

ac6xg

On Mon, 19 Sep 2005 16:43:59 -0700, Jim Kelley
wrote:

"X-rays will prove to be a hoax."
http://zapatopi.net/kelvin/quotes/

"Radio has no future."

Hi Jim,

Almost any lesson can be drawn from the hazard of quote choices:

"The wireless telegraphy is one of the most wonderful inventions
the world has ever seen."

One has to allow that he was either a world class loon, or had the
ability to change his mind.

No doubt the latter characteristic prevails, but this is hardly the
forum for its celebration.

Thanx for the link to such a trove.

73's
Richard Clark, KB7QHC