On 7/27/12 8:44 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.
>
> Al
actually, the equivalent flat plate area for a cylinder can be greater
than the cross sectional area, depending on the diameter of the cylinder
and the wind speed. (that is, Cd can be >1) This is particularly true
for small diameter things like wires.. And for typical wind speeds and
diameters, there's no appreciable streamlining.
Just for some real numbers:
for 60 mi/hr wind
1" diameter is Cd=1.01 and it doesn't change much from 0.5" to 3"
about 0.8 lb drag per linear foot for the 1" tube.
So for any practical sized tubing/radome in typical amateur wind
velocities, the cross sectional area approximation is just fine.
But if you try and extrapolate to (much) faster wind speeds and thin
wires (<0.1" diameter), you can get bitten. (faster wind bites you two
ways: drag goes up as the square, and the Cd also starts going up)
A bigger deal is aeolian vibrations with this sort of size/windspeed.
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