[TowerTalk] Fwd: Fwd: Fwd: OCFD

[Hans Hammarquist]

 Jim,

I can't give you "right" on all your points here:

 

 

-----Original Message-----
From: jimlux <jimlux at earthlink.net>
To: towertalk <towertalk at 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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