Richard Harrison wrote:
We know that a monopole has half the resistance of a dipole. Example: 73
ohms and 36.5 ohms. 790 / 2 = 395. That`s not a resistance, it is only
the value of a constant which must be multiplied by (L/lambda) squared
to give the radiation resistance of a very short monopole.

Does it matter that for a vertical that is 1/2 of the length
of the dipole, (L/lamda)^2 is different by a factor of 4?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil, W5DXP wrote:
"Does it matter that for a vertical that the length of a dipole
(L/lambda)squared is different by a factor of 4?"

It doesn`t make a ratio different than two to one in the ratio of
resistances of the 1/2-wave dipole to the 1/4-wave monopole. We are not
comparing a monopole that is the the length of a dipole with the dipole.
We are comparing a monopole that is 1//2 the length of a dipole to the
dipole when we make the resistance ratio.

The small dipole is working against a perfect ground in Reg`s
specification. It would see its reflection in that perfect ground, so
its equivalent length is doubled. Kraus` dipole is presumed to be in
free space.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"Does it matter that for a vertical that the length of a dipole
(L/lambda)squared is different by a factor of 4?"

It doesn`t make a ratio different than two to one in the ratio of
resistances of the 1/2-wave dipole to the 1/4-wave monopole. We are not
comparing a monopole that is the the length of a dipole with the dipole.
We are comparing a monopole that is 1//2 the length of a dipole to the
dipole when we make the resistance ratio.

Richard, Balanis doesn't say that the 'L' in the monopole
formula is 1/2 the 'L' in the dipole formula. Does Kraus?
--
73, Cecil http://www.qsl.net/w5dxp