Clay,
Yes, you're in the ballpark.
The formula for drag is 1/2*
rho*velocity-squared*frontal-area*coefficient-of-drag.
I calculate about 26 lbs force for 90 mph for 1 sq-ft frontal projection
area.
This assumes a coefficient of drag of 1.3 - typical for tubing or wire
profiles.
Also, the density of the atmosphere was assumed to be at STP (standard temp
and pressure at sea level.)
The largest error contributor to these calculations is most oftem the
estimate of the drag coefficient- Surface roughness, Reynolds number,
interconnections on the tower, etc., may yeild different results.
For me, to be safe, I can see why one might choose the 35 lbs force at 90
mph.
It's quite possible in certain conditions, including icing, etc. etc.
A finer analysis would require more specific information on tubing
configurations, sizes, fitting, etc. At some prudent point, a bit of
overdesign can prevent "analysis to paralysis".
P.S. If you'd like to run the numbers yourself, in the FPS (Imperial
system), use feet per second for velocity, pounds for force, rho (density) =
.00237, and use square feet for frontal projection are. The drag coefficient
is dimensionless, and for cable/wire/tubing is usually about 1.0 to 1.3.
73,
CW-AI4MI
W7CE <w7ce@curtiss.net> wrote: > One could look at the off-vertical situation
by considering that
> installing it non vertical puts a static side load on the whole thing.
> The load would be sin(theta)*weight. Say the whole thing weighs 1000
> pounds (I don't know if this is plausible.. it's just easy to calculate)
> and you're 2 degrees off vertical (about 3.5 feet in 100). The side
> load is about 35 pounds (distributed along the whole thing). That's
> pretty small compared to the wind load (90 mi/hr = about 20 lb/sq ft,
> and you know the tower has a lot more than 10 square feet of cross
> sectional area)
>
Is 20 lb/sq ft correct for 90 mi/hr winds? I've run several of the
published formulas in the past and seen other references that would
indicated that 36 lb/sq ft is a good number to use at 90 mph. I realize
that it is dependent on height above ground and other factors. Have I
miscalculated or am I in the right ballpark?
73,
Clay W7CE
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