The numbers are used by tower manufacturers to tell us Hams how much
antenna area a particular tower model can support at a given wind speed
In the example you cited, that vendor is telling us that his tower can
support a maximum antenna area of 15 square feet (usually within a foot or
two of the tower top) at a wind speed of 70 mph. More than that, either in
antenna area or wind speed, and the forces on the tower are greater than
what the tower was designed to withstand. Something will break.
To get these figures into engineering terms, you'd have to convert the
wind speed into pounds-per-square-foot
http://www.arraysolutions.com/Products/windloads.htm and multiply by the
EFFECTIVE area of the antenna. Then this force is applied to the top of the
tower; let's say the tower is 100 feet tall. In the example, the force of a
70 mph wind is 12.54 psf. The total forced applied to the tower at 100 feet
AGL would be 12.54 X 15 = 188 pounds. Think of this as someone pulling
sideways with a force of 188 pounds at the top of the tower, trying to
When you are deciding what tower to purchase, you have to take into
consideration all the area of all the antennas you intend to mount on the
tower and its mast. If you plan on having only one antenna, say a triband
Yagi, on the example tower, then it can be no more than 15 sqft in effective
area and be mounted no higher than one or two feet above your tower top. If
you install a larger antenna, or install that 15 sqft antenna ten feet above
the tower top, then all bets are off. In winds of 70 mph the tower will
likely bend over at some point along its length. The structural steel
members that make up the tower will be subjected to forces they were not
designed to handle safely.
The trick comes in determining what the effective area (sqft) of the
antenna really is. Manufacturers are unclear in their literature in how
they calculate the advertised areas of their antennas. This is important
because one way of looking at antenna area is how big is the shadow of each
antenna element? A twenty-foot long element that is one inch in diameter
would cast a shadow of 240 X 1 = 240 sqin or almost two square feet.
However, because the element is actually a cylinder, and not a flat plate,
the actual area of this element that is presented to the wind (this is the
effective area) is different from 240 sqin.
As an example of this ambiguity in antenna areas, see a posting I
placed here in 2001.
That calculation showed that my D40 rotatable dipole had a spec from
Cushcraft of 1.3 sqft, whereas the "shadow area" of the elements calculated
out to be 3.55 sqft. Go figure.
I'm not intending to open up a debate of which is the right way to
calculate the effective area of antenna elements. It would be like trying
to ask the proper way to determine true north. I'm simply saying that you
ought to be conservative in how you interpret any manufacturer's antenna
area specs. Otherwise, you might accidentally exceed the ratings of your
tower and see it bent over or down on the ground (or your roof) after a
Gene Smar AD3F
----- Original Message -----
From: "Charles Coldwell" <firstname.lastname@example.org>
Sent: Thursday, February 18, 2010 1:16 PM
Subject: [TowerTalk] wind load
> I've been looking at various tower specifications, and often see
> something like this
> Maximum Wind Load
> 70 MPH 15 sq. ft.
> I don't really understand how to interpret this. I think a wind load
> is a (static) force, and therefore should be measured in either
> newtons or pounds. IIUC, it should be proportional to the square of
> the wind speed and that the constant of proportionality should itself
> be proportional to the cross-sectional area to the wind. So 15 sq ft
> is an area, and 70 MPH is a wind speed, but I'm still missing some
> factors in order to calculate a force.
> Can anyone shed some light?
> Charles M. Coldwell, W1CMC
> "Turn on, log in, tune out"
> Winchester, Massachusetts, New England (FN42kk)
> GPG ID: 852E052F
> GPG FPR: 77E5 2B51 4907 F08A 7E92 DE80 AFA9 9A8F 852E 052F
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