[TowerTalk] VHF/UHF verticals wind load

K8RI K8RI-on-TowerTalk at tm.net
Fri Jul 27 11:30:45 PDT 2012


On 7/27/2012 11:44 AM, Al Kozakiewicz wrote:
> 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.  That's actually worse than the real wind load, as it assumes no streamlining benefit from tubular elements.
>
> A 1 inch diameter element 17 feet long would calculate out to 1.4 sq feet of wind load using the above. The real wind load is certainly less than that.
>

Don't forget this is an asymmetrical load so in a high wind there is a 
lot of leverage.

73

Roger (K8RI)



> Al
> AB2ZY
>
> ________________________________________
> From: towertalk-bounces at contesting.com [towertalk-bounces at contesting.com] On Behalf Of Mario [marionow at gmail.com]
> Sent: Friday, July 27, 2012 11:14 AM
> To: towertalk at contesting.com
> Subject: [TowerTalk] VHF/UHF verticals wind load
>
> 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 searched Google but found nothing
> about that.
>
>
>
> Thanks
>
> Mario
>
> KC8P
>
>
>
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