I recall a discussion of a similar ilk, but instead of varying wind speed,
if one varies the height of tower. Ie., 10 sqr feet at 50ft vs 60ft?
Assume no other addition of winload, ie., in an ideal world where the
additional mast has no additional windload... is this also nonlinear? is
the ratio a sqr function, ie. inv proportional to the square of height?
Original Message
From: towertalkbounces@contesting.com
[mailto:towertalkbounces@contesting.com]On Behalf Of Bud Hippisley,
K2KIR
Sent: Wednesday, September 03, 2003 1:21 PM
To: towertalk@contesting.com
Subject: Re: [TowerTalk] Tower wind loads
At 02:59 PM 20030903, 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 oversimplification, 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, 18003339041 with any
questions and ask for Sherman, W2FLA.
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_______________________________________________
See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather
Stations", and lot's more. Call Toll Free, 18003339041 with any questions
and ask for Sherman, W2FLA.
_______________________________________________
TowerTalk mailing list
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http://lists.contesting.com/mailman/listinfo/towertalk
