I remember sometime being told to use a one foot diameter conductor
for a tower, but that was a while ago, and there are some older email
caches that got lost in a HD crash.
That's difficult to use if you are trying to model a tower plus beams
etc as a vertical with the NEC problems with junctions of dissimilar
wires.
Guy.
----- Original Message -----
From: "Bill Tippett" <btippett@alum.mit.edu>
To: <antennaware@contesting.com>
Sent: Tuesday, May 21, 2002 9:09 AM
Subject: [Antennaware] Cylindrical Model for Triangular Tower
> Somewhere I got the following formula for a cylindrical
> equivalent to triangular tower like Rohn 45.
>
> Equivalent Cylindrical Diameter = 2 * CUBEROOT [(D * F^2)/2]
>
> where D = tube diameter and F = Face width. For Rohn 45, D = 1.25"
> and F = 18", so cylindrical diameter equivalent = 11.7446"
>
> Does anyone know where this came from or if it is valid?
>
> 73, Bill W4ZV
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