On 3/28/17 2:36 PM, Hans Hammarquist via TowerTalk wrote:
Thank you, Rick,
I want to remember 700uF to be the answer, but how do you explain "to free
space"? Even if I am a physicist I can't really explain it. You might say that the
Earth has a charge that generate an electric field around it, a field that could deflect
electric charged particles approaching the earth. Now I am getting too theoretical. Let's
stop here.
The point is that the dipole has a capacitance, popularly called "end effect" that detune
the dipole to a lower frequency. The thicker the wire the lower the frequency. The strange part to
me is that it doesn't matter how thin the wire is, there is always an end effect that lower the
frequency from the "ideal" dipole.
because it's not "really" an end effect or capacitance. The actual
phenomenon is that X goes through zero at a frequency where the dipole
is not exactly a half wavelength long. There's a lot of ways you can
analyze around it or conceptualize it, but they're just mental models.
Another way to think about it is that the propagation velocity down the
wire is not c, but slightly slower (because the wire has inductance and
free space capacitance, and you can calculate the prop velocity as
sqrt(L/C))
Yet another way is to start with a parallel transmission line and
gradually open it up until it's a dipole.
Yet another way is to consider the dipole as two cones, and the diameter
of the cone goes to zero in the limit. (Schelkunoff)
You can also do a method of moments approach and consider the coupling
of each infinitesimal segment of the wire to every other one. (various
schemes and calculations: Hallen, Pocklington, King, etc.)
I would recommend this fine piece of work if you're interested in
"simple" models of dipole antennas:
https://www.fars.k6ya.org/docs/antenna-impedance-models.pdf
(I've actually built some of these for work to make a synthetic antenna
Z for testing)
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