Jim/Steve etal:
You forget that this is just the air mass density force at a velocity.
There are modifiers beyond Cd (drag/shape) that need to be taken into account
which modify the force due to height and exposure. This other modifiers are not
insignificant. To provide a more accurate determination of the area these will
need to be taken into account.
Regards
Lonberg Design Group, Ltd.
Hank Lonberg, P.E.,S.E.
President
 Original Message 
From: "Jim Lux" <jimlux@earthlink.net>
To: towertalk@contesting.com
Sent: Thursday, October 27, 2011 7:34:35 AM
Subject: Re: [TowerTalk] Equivalent wind load to square ft
On 10/27/11 6:56 AM, Cqtestk4xs@aol.com wrote:
> A friend of mine is thinking of putting up some wind generators and has
> asked for my advice. The only information he can give me for the wind load
> is 300# at 100 MPH. I know that makes sense to some, but I'm used to square
> ft loads for ham antennas.
What they have done is calculate the total load at 100 mi/hr.
A rule of thumb is that wind load in pounds is
Area(sq ft)* windspeed (mi/hr)^2/391
So, his windload at 50 mi/hr would be 75 lbs.
However, this makes the assumption that the drag coefficient is constant
(that is, that the drag goes as the square of the airspeed), which is
probably not true for a wind generator, and certainly isn't true for
something like a round tube at airspeeds we see.
the real equation is Force = rho*Cd*A*v^2, where
rho is the density of the fluid (air)
Cd is the drag coefficient
A is the area
v is the fluid speed.
The problem is that Cd varies quite a lot with speed, for low speeds
(tens of mi/hr) and common dimensions (inches). What most people do is
either assume Cd=1 (a sort of tending to the worst case) or take the
worst case Cd (say, 1.3) and increase the "effective area" by that
amount, so you can use the equation assuming Cd=1
>
> Is there some kind of simple equivalency that I can use to figure out what
> this would mean if we were talking plain old square ft antenna load? I
> don't know if I worded this exactly right but I hope someone can help me on
> this.
What a tower designer does is take the "design wind speed" and the
"maximum force" and work backwards to a square footage rating. Here's
an example...
Say the tower is designed for 70 mi/hr winds AND can take a 100 pound
load at the top. The designer plugs in:
A = 100*391/70^2 = about 8 square feet.
Then they publish that in their spec sheet: "Can hold an 8 square foot
antenna at 70 mi/hr"
(but of course, the real design limit is some force... whether it's an
antenna in the wind or a winch cable from your evil neighbor pulling the
tower over, it doesn't matter to the tower)
So, if you wanted to convert your friend's wind generator into an
"equivalent antenna area" at 100 mi/hr, you do the same thing:
300 * 391/100^2 = 11.7 square feet.

A note about wind load ratings..
For many towers, the tower itself has more area than the antenna. This
is especially true for home TV antenna type towers which can't take very
much antenna at all.
Since most of the load on the tower is from wind pushing on the tower,
as opposed to the antenna, my example of the equivalency of winch cable
and wind load isn't quite right. Your evil neighbor will have to pull
MUCH harder to pull your tower down, than the few hundred pounds rated
antenna wind load.
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