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Re: [TowerTalk] dipole in space question

To: Towertalk List <towertalk@contesting.com>
Subject: Re: [TowerTalk] dipole in space question
From: Martin Ewing AA6E <aa6e@ewing.homedns.org>
Date: Fri, 13 Jun 2008 15:07:21 -0400
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
Brent Baum wrote:

>> Al Williams:
>> > Related to this question: Why is the radiation from a dipole over ground 
>> > have its strongest radiation from the center of the dipole.> Yes the 
>> > current is strongest at the center, but why?
>>     
>  
> Think of your typical nice, curvey sine wave that looks like an S laying on 
> its side. Now take just the top or bottom half of the S. That's your half 
> wave that's standing on the half-wave dipole and shows the instantaneous 
> current distribution at the moment of maximum radiation.
>   
>> > I have tried to find the explanation in a dozen or more antenna books but 
>> > to no avail. I can speculate but perhaps some Towertalkians can explain it 
>> > or offer their speculation?> > k7puc> 
>> > _______________________________________________> > > > 
>> > _______________________________________________> TowerTalk mailing list> 
>> > TowerTalk@contesting.com> 
>> > http://lists.contesting.com/mailman/listinfo/towertalk
>>     
Another way to think of it is to consider that we are working with the 
natural oscillation modes of a length of wire.  The "boundary condition" 
is that current must be zero at the ends - it's an open circuit.  It's 
like a rope that is held fixed at either end.  There are many possible 
resonant vibration modes.  The lowest frequency is the one where the 
center rises up and down (voltage or deflection), and the pattern is 
like half a sine wave.  That's what we call a half-wave dipole when we 
talk about an antenna.  The next mode is at twice the frequency, when 
you have a full sine wave.  But a full-wave dipole has a current "zero" 
at the center, i.e., a very high impedance that's hard to drive.  The 
third mode is 3 half waves, which has a "full" deflection in the center, 
and a reasonable feed impedance.  (So you can use an 7 MHz dipole on 21 
MHz pretty well.) And so on for higher multiples of the half-wave frequency.

As to "why", it's because those are the solutions to Maxwell's Equations 
with the appropriate boundary conditions.  (At least that's how a 
physicist might answer.)  But the situation is very similar to what you 
can do playing around with a length of clothesline in your backyard.

73,  Martin AA6E

-- 
Martin Ewing, AA6E
aa6e@ewing.homedns.org
+1-203-315-5160


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