Towertalk
[Top] [All Lists]

Re: [TowerTalk] Tower wind loads

To: <towertalk@contesting.com>
Subject: Re: [TowerTalk] Tower wind loads
From: "Bud Hippisley, K2KIR" <k2kir@telenet.net>
Date: Wed, 03 Sep 2003 16:20:39 -0400
List-post: <mailto:towertalk@contesting.com>
At 02:59 PM 2003-09-03, Bob Gates wrote:
> If you have a tower rated to hold 30sf of antenna at 70 mph, and 10sf at
>90mph, what correlation, if any, can you draw about the load handling ability 
>at
>80mph.  Is there a linear relationship, or does it require a whole new set of
>calcs?

The relationship is NOT linear, although in Bob's particular example, a more 
accurate answer isn't very far from the number he would get if he did assume it 
was linear.

The charts in the Rohn catalogs for decades showed the pressure on a tower and 
beam to be related to the SQUARE of the wind velocity.  Thus, the ratio of 
allowable antenna square footage at 90 mph to that at 70 mph would be 4900 / 
8100, or 60%, if the tower itself had no surface area.  So if you ignored the 
wind load on the tower, you would conclude that the allowable antenna surface 
area at 90 mph was 60% of that at 70 mph, or 18 square feet.  Clearly, the 
tower has surface area, and it's pretty likely the manufacturer has included it 
in the 90 mph calculation, which came out at 10 square feet for allowable 
antenna area, not 18.

A VERY ROUGH APPROXIMATION for finding the allowable antenna area at other wind 
speeds is to assume that the entire tower can be represented by an effective 
wind loading of "T" square feet applied at the top.  (We know that's a gross 
over-simplification, but since we don't know anything about the tower at this 
point, it's the best we can do.)  Then we surmise from the manufacturer's 70 
mph and 90 mph ratings the following equality:

(30 + T) x 70 x 70 x K = (10 + T) x 90 x 90 x K.

The equality is based on the assumption that the tower manufacturer uses the 
SAME total force from the wind loading on the tower plus antenna in his 
calculations at various different wind speeds.  K is an "arbitrary" constant 
that goes away, so we're not going to worry about its dimensions or meaning.

Dividing both sides by K and collecting all terms with T in them on the right 
hand side, we get

(147,000 - 81,000) = (8100 - 4900) x T

or:    T = 20.63 square feet (effective tower wind loading applied at the top).

Thus, at 70 mph, the total force on the tower and antenna is K x (30 + 20.63) x 
70 x 70, or K x 248,000.   Similarly, at 90 mph, we get K x (10 + 20.63) x 90 x 
90, or K x 248,000.

Now we can calculate the allowable antenna square footage (call it "F") at 80 
mph:

(F + 20.63) x 80 x 80 x K = 248,000 x K

or:  F = 18 square feet of allowable antenna area for 80 mph winds, versus the 
20 square feet Bob would calculate by assuming it was a linear curve between 
the figures for 70 and 90 mph.

Bud, K2KIR




_______________________________________________

See: http://www.mscomputer.com  for "Self Supporting Towers", "Wireless Weather 
Stations", and lot's more.  Call Toll Free, 1-800-333-9041 with any questions 
and ask for Sherman, W2FLA.

_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk

<Prev in Thread] Current Thread [Next in Thread>