> Wouldn't the effective area of a 38' boom at 2" diameter be 6.33SqFt.
> and 9.5SqFt. at 3" ? I assume it is the cross sectional area we are
> concerned with.
> 73s de WA6IBU, Paul
Well, without getting into a complicated explanation, it depends on which
set of standards one is using to determine what the windload will be.
The 6.33 SqFt is the projected area of the boom. I multiplied it by a 2/3
shape factor for round tubing. This would be used with the .004*v^2 formula
for wind pressure to get the wind load.
Another approach would use a shape factor of 1.2 and a different wind
pressure formula to get very close to the same wind load.
I happen to know that KLM used the first one.
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