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[antennaware] Simplified Towers

To: <>
Subject: [antennaware] Simplified Towers
From: (Eric Gustafson)
Date: Mon, 19 Oct 1998 18:32:51 -0700 (MST)

Hi John,

Thanks for the reply.

I'll compare that result with the results of a formula I just got
out of the Antenna Engineering Handbook 3rd ed.  Basically, they
have e fudge factor to use depending on the shape of the tower
cross section polygon.  Their claim is that the equivalent
conductor radius for a triangular cross section is about 0.42
times the radius of an outscribed circle.  That makes Rohn 45
look like a conductor with a diameter of about 6.5 inches.

So, for Rohn 45, we have:

D = 1.25


F = 18

So ((D*F^2)/2)^0.33 = 5.77 inch radius.  Or 11.5 inch diameter.

Now I can't wait to grid dip 70 feet or so of Rohn 45!  Both of
these confirm my initial findings that using either the face
width or the circle outscribed resulted in conductors with too
much diameter.

Thanks a lot for the very useful reference.  I can get to it at
the UA library.

73, Eric  N7CL

To: <>
>Date: Mon, 19 Oct 1998 21:14:17 -0400
>From: John Kaufmann <>
>At 11:15 AM 10/19/98 -0700, Eric Gustafson wrote:
>>Does anyone have or know of any single conductor diameter
>>equivalences to use in models of system involving towers?
>If you have access to the June 1998 issue of IEEE Transactions
>on Broadcasting, check out the article: "A Method for Modeling
>Array Elements When Using NEC and MININEC" by J.L. Smith.  The
>author discusses methods for modeling towers and in particular,
>describes a single-wire representation for a triangular cross
>section tower.  His result is that the radius of the equivalent
>wire equals the cube root of the quantity (D*F^2)/2 where D is
>the diameter of each tower leg and F is the width of the tower
>face with all dimensions in the same units.  Some references are
>cited to justify this result.
>73, John W1FV

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