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

Mike W4EF@dellroy.com
Wed, 27 Feb 2002 09:23:08 -0800


Richard,

It depends on which specification you use. Correctly applied
you should get the same answer either way. I think the newer
specs carry the 2/3 drag coefficient for round members in the
wind pressure calculation. Flat plates are then multiplier by
a factor greater than 1 to give the equivalent area. In the old
system, I think it was the other way around.

I use .004*v^2 for the wind pressure, and then multiply
the projected area (shadow area) of the round member by
the drag coefficient of 0.67 to get the effective surface area.
Multiply this by the wind pressure to get the force. At 80
MPH this method yields a wind pressure of 25 psf.

Somebody will probably chime in a tell me that this isn't
consistent with the latest rev. of EIA-222, but I think it
gives a useable answer.

73 de Mike, W4EF........................................................

----- Original Message -----
From: "Richard Karlquist" <richard@karlquist.com>
To: <towertalk@contesting.com>
Sent: Wednesday, February 27, 2002 8:32 AM
Subject: [Towertalk] Wind pressure calculations (2/3 factor)


> 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
> richard@karlquist.com
> www.n6rk.com
> www.karlquist.com
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