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1. [Antennaware] Cylindrical Model for Triangular Tower (score: 1): Author: btippett@alum.mit.edu (Bill Tippett); Date: Tue, 21 May 2002 14:09:54 +0100; 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; /archives//html/Antennaware/2002-05/msg00001.html (7,055 bytes)
2. [Antennaware] Cylindrical Model for Triangular Tower (score: 1): Author: k2av@contesting.com (Guy Olinger, K2AV); Date: Tue, 21 May 2002 10:47:59 -0400; 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 i; /archives//html/Antennaware/2002-05/msg00002.html (8,117 bytes)
3. [Antennaware] Cylindrical Model for Triangular Tower (score: 1): Author: john.w1fv@telocity.com (John Kaufmann); Date: Tue, 21 May 2002 09:10:56 -0700 (PDT); cylindrical * F^2)/2] Rohn 45, D = 1.25" 11.7446" valid? This formula originally appeared in the IEEE Transactions on Broadcasting. I posted a note about it on this reflector back in 1998. See: http:; /archives//html/Antennaware/2002-05/msg00004.html (7,674 bytes)

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