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Towers and Wind Load


Wind Load

Subject: Re: [TowerTalk] Tower wind loads
Date: Wed, 03 Sep 2003 16:20:39 -0400

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

2003 by CCO