Jim,
I can't give you "right" on all your points here:
-----Original Message-----
From: jimlux <jimlux@earthlink.net>
To: towertalk <towertalk@contesting.com>
Sent: Tue, 28 Mar 2017 17:55
Subject: Re: [TowerTalk] Fwd: Fwd: OCFD
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.
## You see a similar end effect if you use an open end (not shorted) of a coax
in a filter made with coax cables. If you increase the diameter of the tip of a
dipole, the X goes through zero at an even lower frequency. I would say that
the increase of the capacitance decrease the frequency just like it decrease
the frequency when you add capacitance to a LC circuit.
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))
## If you do the calculation you find that sqrt(L/C)=c(speed of light) if the
wire is in vacuum. The air will contribute to a very minor decrease in the
speed. The closeness of ground and its dielectric property will affect the
speed a lot more.
Yet another way is to start with a parallel transmission line and
gradually open it up until it's a dipole.
## which has to start with an open end transmission line with its end effet
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.)
## That does not dispute the ed effect I am talking about. Various methods to
calculate it is not a proof against it.
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
## reviewing it as I write.
(I've actually built some of these for work to make a synthetic antenna
Z for testing)
## What method would you suggest to explain that the resonance frequency is
reduce when you put the dipole closer to real (or ideal) ground?
73 de,
Hans - N2JFS
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