Roy Lewallen wrote:
Cecil Moore wrote:
Sorry, Roy, my theory is not elegant and/or well developed. Equations may
be possible in the future, but not right now. At the present time, the
theory is qualitative, not quantitative. . .


Somehow I expected this.

I didn't mean to imply that I don't know the equations - I do. I just
don't know the value of all the constants in the equations.

Given that a horizontal dipole 24 ft. above ground and constructed from
#16 wire will have a natural Z0 of 600 ohms:

The forward current will be an If-max value multiplied by an exponential
relating to frequency multiplied by an exponential relating to the loss
of energy due to conductor resistance and radiation.

The reflected current will be an Ir-max value multiplied by the same
exponential relating to frequency multiplied by the same exponential
relating to the loss of energy due to conductor resistance and radiation.

We know that (Vf+Vr)/(If+Ir) equals 50 ohms for a dipole whose feedpoint
impedance is 50 ohms. With a 1/2WL dipole the current equation is clear.

Here is the equation you asked for, unfortunately in ASCII:

Itot = If-max*e^-yz*e^-2az + Ir-max*e^+yz*e^-2az

same as it is for a transmission line. The I^2*R losses plus the
radiation "losses" are combined into the attenuation factor 'a'.

So I can indeed write you an equation for a wire dipole. The coil
in the mobile antenna causes another level of complication, and
that is the equation with which I am struggling at the moment.
In addition, the vertical nature of a mobile antenna means that
the Z0 is changing with length. That is a minor problem compared
to including the reflections from both ends of a loading coil in
both directions. But I have no doubt that I can solve that problem.
--
73, Cecil http://www.qsl.net/w5dxp


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