"Ian Jackson" wrote -
Are you sure it's as high as that, Reg? I once did a Smith Chart
plot of
the impedance at the centre of a dipole, the valued being taken from
a
table 'compiled by Wu' (LK Wu?). These only catered for a lengths up
to
a few wavelengths. As the plot progressed round and round the Smith
Chart, it seemed to be heading for something around 350 to 400 ohms.

I've just done a search on 'Wu+dipole+impedance', and one of the
results
is
http://www.fars.k6ya.org/docs/antenn...nce-models.pdf
I'll have a read of it today.
===================================

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).

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)

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.
----
Reg.