On Thu, Feb 18, 2010 at 2:02 PM, Gene Smar <email@example.com> 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
> 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
> 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
Charles M. Coldwell, W1CMC
"Turn on, log in, tune out"
Winchester, Massachusetts, New England (FN42kk)
GPG ID: 852E052F
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