[TowerTalk] Wind Loading Question

[k6ll@juno.com]

I think the point that some people may be missing is that
"projected area" of a piece of tubing is equal to diameter
times length.

For example, a 1" O.D. piece of tubing, 120 inches long,
has a projected area of 1 x 120 = 120 square inches, or
.833 square feet.

To get the "effective area," which factors in the dynamics of
wind flow over a cylindrical body, this figure is multiplied
by .67

The problem lies in the fact that towers usually use Projected
Area as the specification, and what the antenna manufacturers
use is uncertain. If the antenna spec says "windload," or
"wind surface area," I think it's a pretty safe bet that the
antenna manufacturer has already multiplied by .67, so you need
to multiply by 1.5 to match it up with the tower spec. If the
antenna spec says "surface area," it's totally ambiguous, but I
would multiply by 1.5 unless you can find out for certain to the
contrary from the manufacturer.

There aren't that many antenna manufacturers out there. You would
think that we could get a definitive statement from them,
especially since their are potential liability and consumer
misinformation issues associated with multiplying the projected
area by .67.


Dave Hachadorian, K6LL
Yuma, AZ
K6LL@juno.com























___________________________________________________________________
Get the Internet just the way you want it.
Free software, free e-mail, and free Internet access for a month!
Try Juno Web: http://dl.www.juno.com/dynoget/tagj.

--
FAQ on WWW:               http://www.contesting.com/towertalkfaq.html
Submissions:              towertalk@contesting.com
Administrative requests:  towertalk-REQUEST@contesting.com
Problems:                 owner-towertalk@contesting.com
Search:                   http://www.contesting.com/km9p/search.htm