Elliptic integrals would be preferable to watching the Seattle Mariners blow
another ninth-inning lead :-)
Thanks, Jim - this is going to get me close enough!
73, Ward N0AX
----- Original Message -----
From: "jimlux" <firstname.lastname@example.org>
To: "Ward Silver" <email@example.com>
Cc: "Towertalk Reflector" <firstname.lastname@example.org>
Sent: Tuesday, July 29, 2008 10:58 AM
Subject: Re: [TowerTalk] Feedpoint Impedance
> Ward Silver wrote:
>>>>> Radiation resistance should be fixed..
>>>> Oh, right...duh.
>>>> The current equation will do - but the radiator (I may have confused
>>>> things by referring to it as a "dipole") may be anywhere between
>>>> one-half and one wavelength long, so the current won't be a simple
>>>> cosine function. It *will* be zero at the ends :-)
>>> I think Orfanidis's book has the equation you're looking for.. actually
>>> several approximations, of varying fidelity..
>>> I'm pretty sure there's some standard assumptions of the current
>>> distribution shape (ranging from uniform for very short, to triangular
>>> to sinusoidal to something else), possibly broken up into segments.
>>> What sort of accuracy are you looking for?
>> 20% ought to do it. There will be significant variation depending on
>> proximity of ground, type of ground, and so forth. At this point, I'm
>> just doing a feasibility study.
>> 73, Ward N0AX
> Here you go...
> Current on a thin wire of length 2*h
> k is propagation constant (i.e. pi gives you a half wavelength dipole)
> z is the distance from the center
> I(0) is the current at the center
> I(z) = I(0) * sin( k *(h-abs(z)))/sin(k*h)
> (Eq 21.4.2 from Orfanidis's book)
> Seems to match (by eye) what I got from a series of NEC models..
> I haven't looked into the off center fed aspect...This is just the current
> distribution on a center fed wire.
> Later pages in Chapter 21 of the book talk about King's three term
> approximation (which works up to about 1.25 lambda). Figure 21.6.1 shows
> a comparison between sinusoidal, King and numerical integration.
> If you want real pain, he goes on to work out an exact kernel for the
> solution of Pocklington's equation using, why yes, elliptic integrals.
> (I'd just use his canned Matlab routines....)
> Jim, W6RMK
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