[Towertalk] Wind pressure calculations (2/3 factor)

[Michael Keane, K1MK]

On Wed, 27 February 2002, "Richard Karlquist" wrote: 
 
> I am trying to do a simple wind 
> pressure calculation on an antenna, 
> but I am confused about how to do 
> it.  
> ....	
> Can anyone boil this down to something 
> simple? 
 
No.  
 
Part of the reason is there is no really "simple" answer is that at least two "correct" ways exist for answering the question: a pure "physics" answer and an applied "engineering" answer. Either might be appropriate depending on how one chooses to interpret the term "wind speed". 
 
The physics answer assumes that when you say 80 mph you really do mean the instantaneous wind speed that's measured at the same height as the object of interest. This is how a lot of people would interpret wind speed when they encounter it. In this case: 
 
    F =        Qs	    *	     a	       *      A 
Force = Stagnation Pressure * Drag Coefficient * Projected Area 
 
Qs(v) = 0.00256 * v^2  (Qs in lb/sq. ft.; v in mph)   
  
"a" depends on object's shape (how "streamlined" it is); a is 1.2 for a thin cylinder and 2 for a flat plate
 
"A" is the projected (shadow) area (length x width; no pi's); 1 square foot in your example.  
 
So the instantaneous pressure on a thin cylinder would be something like  19.7 lb/sq. ft at 80 mph.  
 
On the other hand, the other answer assumes that the 80 mph in your question actually refers to a specifically defined averaging (fastest mile) of a time varying wind speed having defined statistical properties that is measured at a height  of 10m. This is the precise meaning of wind speed as it is used in the building codes and the EIA spec.  
 
In that case there are corrections that depend on height (because of drag the air feels near the ground). Using the methodology given ithe UBC/EIA, for a wind speed of 80 mph, the peak pressure that's exherted on a thin cylinder located at 10m height is 24.6 lb/sq. ft.
 
That 2/3 factor you're recalling comes from an older (1976) version of the EIA spec in which several factors, such as stagnation pressure and drag for a flat plate were "mixed together". It gives the same answer although perhaps obscures some of the underlying physics. The bottom line is that if one is consistent and always uses that 2/3 factor along  with the set constants that it was intended to be used, you end up with the same sansweras the current version of the spec, i.e. 20.5 lb/sq ft.  

73, 
Mike K1MK 
k1mk@alum.mit.edu

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