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References: [ +subject:/^(?:^\s*(re|sv|fwd|fw)[\[\]\d]*[:>-]+\s*)*\[TowerTalk\]\s+VHF\/UHF\s+verticals\s+wind\s+load\s*$/: 4 ]

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1. [TowerTalk] VHF/UHF verticals wind load (score: 1): Author: "Mario" <marionow@gmail.com>; Date: Fri, 27 Jul 2012 11:14:21 -0400; I'm going to install in the near future a dual band VHF/UHF Diamond X510HDN, this antenna has over 17ft? Does anyone know what would be the wind load? I've checked the manufacturer website and search; /archives//html/Towertalk/2012-07/msg00352.html (7,242 bytes)
2. Re: [TowerTalk] VHF/UHF verticals wind load (score: 1): Author: Al Kozakiewicz <akozak@hourglass.com>; Date: Fri, 27 Jul 2012 11:44:15 -0400; If you know the diameter of the element you can make a first order approximation. Multiply the diameter (in inches) by the length (in inches) for square inches, then divide by 144 for square feet. Th; /archives//html/Towertalk/2012-07/msg00353.html (8,122 bytes)
3. Re: [TowerTalk] VHF/UHF verticals wind load (score: 1): Author: Jim Lux <jimlux@earthlink.net>; Date: Fri, 27 Jul 2012 09:01:11 -0700; actually, the equivalent flat plate area for a cylinder can be greater than the cross sectional area, depending on the diameter of the cylinder and the wind speed. (that is, Cd can be >1) This is par; /archives//html/Towertalk/2012-07/msg00354.html (8,348 bytes)
4. Re: [TowerTalk] VHF/UHF verticals wind load (score: 1): Author: K8RI <K8RI-on-TowerTalk@tm.net>; Date: Fri, 27 Jul 2012 14:30:45 -0400; Don't forget this is an asymmetrical load so in a high wind there is a lot of leverage. 73 Roger (K8RI) _______________________________________________ _______________________________________________; /archives//html/Towertalk/2012-07/msg00357.html (9,255 bytes)

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