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

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Subject: [Towertalk] Wind pressure calculations (2/3 factor)
From: (Red)
Date: Wed, 27 Feb 2002 11:53:42 -0600
The drag force on a cylinder perpendicular to the wind (and other forms)
is given by:

Drag = Cd X 1/2 X density of air X area X velocity squared.

(D = Cd*.5*rho*A*V^2)

Cd, coefficient of drag, is a function of a parameter called the
Reynolds number (Re).

The Re for a cylinder perpendicular to the wind is proportional to the
air density times the diameter times the velocity divided by the
viscosity of the air.

The coefficient of drag for cylinders from 0.5" to 7" dia at velocity
from 50 to 100 mph at sea level and 59°F is 1.2.

The density of air at sea level, standard pressure and temperature, is
0.00238 slug/cu ft.

Thus, the drag on a cylinder under 7" dia at 80 mph at sea level and
59°F is 16.25 pounds per square foot.

The coefficient of drag for flat plates and other sharp edged shapes
perpendicular to the wind is 1.98 (same range of Re).  The drag at 80
mph is 27 lb/sq ft.

There are added complications associated with the interference drag at
the many joints in a tower and with the question of the relative angle
of the wind to the horizontal bracing in a tower.

EIA Standard RS-222A accounts for this by specifying 30 lb/sq ft in Zone
A (80 mph) and 40 lb/sq ft in zone B (100 mph)

For more information on how this is derived, look for "Basic Fluid
Mechanics," by J. Lister Robinson: 1963 or any other text on fluid
mechanics.  It may be difficult to find parameters for small cylinders
in texts or handbooks for aeronautical engineering because they tend to
focus on aerodynamic shapes and larger Reynolds numbers.

Hope this helps.

73 de WOØW

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.  I see conflicting information
> on towertalk archives and various
> ham vendor sites.  Let me see if
> I can get a consensus on a very
> basic example.
> If I have a 1 inch diameter round
> tube 12 feet long in an 80 MPH wind,
> what is the wind pressure.
> The projected area is 1 square foot
> (ie the shadow the tubing would cast
> on a plane surface)
> The total frontal area is pi/2 times this
> or 1.59 square feet
> Some references say to multiply by 2/3
> to get "effective area" whatever that
> means.  Other references talk about 1.5,
> which is the reciprocal of 2/3 and close
> to pi/2.
> Anyway, some sort of area needs to be
> multiplied by some fudge factor and then needs to
> be multiplied by wind stagnation pressure.
> Am I correct in assuming this is 30 lbs
> per square foot for 80 MPH?  I am thinking
> this is an ideal value in free space
> (ie pushed by an airplane traveling at
> 80 MPH).
> I see references to a "UBC basic wind speed
> of 80 MPH" where 80 MPH doesn't mean 80 MPH,
> rather it refers to a table of pressure vs
> height.  I see at 70 feet this is something
> like 24 lbs/square foot.  This seems to imply
> a ground effect.
> I'm not trying to apply for a building permit,
> I'm just trying to see if an antenna will
> survive 80 MPH winds.
> Can anyone boil this down to something
> simple?
> Rick Karlquist    N6RK
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