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@contesting.com [towertalk-bounces@contesting.com] On
> Behalf Of Mario [marionow@gmail.com]
> Sent: Friday, July 27, 2012 11:14 AM
> To: towertalk@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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