[TowerTalk] VHF/UHF verticals wind load

[Jim Lux]

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.