[TowerTalk] Windload at 90 mph (was: Plumbing a tower)

[Chris Wendling]

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 at 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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