[TowerTalk] wind load

[Charles Coldwell]

On Thu, Feb 18, 2010 at 2:02 PM, Gene Smar <ersmar at verizon.net> wrote:
>
>     To get these figures into engineering terms, you'd have to convert the
> wind speed into pounds-per-square-foot
> http://www.arraysolutions.com/Products/windloads.htm  and multiply by the
> EFFECTIVE area of the antenna. Then this force is applied to the top of the
> tower; let's say the tower is 100 feet tall.  In the example, the force of a
> 70 mph wind is 12.54 psf.  The total forced applied to the tower at 100 feet
> AGL would be 12.54 X 15 = 188 pounds.

Well, that is substantially less than the 450 pounds I came up with
from the "30 pounds per square foot in a 70 MPH wind" rule mentioned
upthread.

>     The trick comes in determining what the effective area (sqft) of the
> antenna really is.  Manufacturers are unclear in their literature in how
> they calculate the advertised areas of their antennas.

I suppose this is due to uncertainty in the drag coefficient since the
geometric cross-section should be unambiguous.  Rewriting the Array
Solutions formula as

F = A * 0.00256 * W * W * Cd

the effective cross-section is the product A * Cd, where A is the
geometric cross-section, and Cd is 1.2 for long cylinders and 2.0 for
flat plates.

> That calculation showed that my D40 rotatable dipole had a spec from
> Cushcraft of 1.3 sqft, whereas the "shadow area" of the elements calculated
> out to be 3.55 sqft.  Go figure.

Yeah, I can't make up that difference no matter how I play around with
drag coefficients.

-- 
Charles M. Coldwell, W1CMC
"Turn on, log in, tune out"
Winchester, Massachusetts, New England (FN42kk)

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