Charles Coldwell wrote:
> On Thu, Feb 18, 2010 at 1:34 PM, Stan Stockton <stan@aqity.com> wrote:
>> I could be wrong about this but have always thought (from an old Rohn book)
>> that 70 MPH equated to 30 pounds per SF of antenna.
>
> OK, so in my example if the maximum wind load at 70 MPH is 15 sq ft,
> and we have 30 pounds per square foot, that works out to 450 pounds of
> force. That sounds reasonable.
>
> If we believe the force grows as the wind speed squared, and that a 70
> MPH wind puts 30 pounds per square foot of force on the tower (and
> that a 0 MPH wind puts 0 pounds), then the constant of proportionality
> is 30/4900. So that means the general formula would be
>
> F = 3/490 * W * W * A
I use V^2*Area/391 for area in square feet and V in mi/hr
1/391 is the density of air in funny units..
If you want in other units (e.g. you have wind speed in meter/sec and
area in square meters), it's basically the half theair density (rho)..
about 1.2 kg/m^3 or 0.075 lb/ft^3
Drag = 1/2*rho* Cd* A * v^2, where Cd is the drag coefficient, assumed 1
here.
>
> where F is the force in pounds, W is the wind speed in MPH, and A is
> the antenna effective crosssectional area in square feet.
>
> What's the title/author of the Rohn book?
>
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