"Reg Edwards" wrote
It is equally correct to say that the radiation resistance of a
dipole
is 2*73 = 146 ohms, since this is the value when the radiation
resistance is assumed to be uniformly distributed along its length
as
is the conductor resistance.

For calculating purposes the radiation and conductor resistances can
simply be added together.
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Hints and tips -

By the same arithmetical reasoning -

The centre feedpoint input resistance of a 1/2-wave dipole is 73 ohms
plus HALF of the conductor end-to-end HF resistance, which to be
precise sometimes matters. This is exact insofar as 73 is exact - not
an approximation.

The HF resistance of a copper wire at 20 degrees C is -

Rhf = Sqrt( F ) / 12 / d ohms per metre,

Where F is the frequency in MHz and d is the wire diameter in
millimetres.

For example, the end-to-end loss resistance of a 1/2-wave dipole at
1.9 MHz using 16-gauge copper wire is 7.07 ohms, which increases the
feedpoint resistance to 73 + 3.53 = 76.53 ohms, to give a radiating
efficiency of 95.4 percent. Some people would consider that's enough
to lose a contest!

Why not start a little notebook to record useful, simple, little
formulae such as above which don't appear in Terman et al? Or if they
do appear then you can never find the right page in the right volume.
----
Reg.