C = 160 * pi^2 ~ 1579.

This is exactly 4 times the radiation resistance of a short dipole with
linear current distribution (i.e., one without a top hat), since the
average current is twice the amount for the same radiated power.

Of course, this assumes an infinitely thin wire. Any real wire will have
a higher radiation resistance than this.

Roy Lewallen, W7EL

Reg Edwards wrote:
I am not trolling.

What I want to know is the radiation resistance, referred to the base,
of a short vertical wire above a perfect ground, the current in the
wire being assumed uniformly distributed.

The radiation resistance at the base is in the form of -

C * Square( Length / Lambda )

where Length is the physical length or height of the wire and Lambda
is the free-space wavelength.

What is the value of the constant C ?

Thank you.
----
Reg.

Roy Lewallen wrote:

C = 160 * pi^2 ~ 1579.

This is exactly 4 times the radiation resistance of a short dipole with
linear current distribution (i.e., one without a top hat), since the
average current is twice the amount for the same radiated power.

Since the resistance is inversely proportional to the current,
shouldn't you have divided by 4 instead of multiplying by 4?
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:
C = 160 * pi^2 ~ 1579.

This is exactly 4 times the radiation resistance of a short dipole with
linear current distribution (i.e., one without a top hat), since the
average current is twice the amount for the same radiated power.

Roy's formula above is correct. It is approximatey 1580 times the
square of effective height over the wavelength.

http://www.w8ji.com/radiat1.gif

http://www.w8ji.com/radiation_resistance.htm


Cecil's answer is not correct, but I'm sure you figured that out on
your own.

73 Tom

wrote:
Cecil's answer is not correct, but I'm sure you figured that out on
your own.

Silly me, I was assuming the length of the monopole was
assumed to be 1/2 the length of the dipole.
--
73, Cecil http://www.qsl.net/w5dxp

"Roy Lewallen" wrote

C = 160 * pi^2 ~ 1579.

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

Thank you Roy.
----
Reg