[TowerTalk] Antenna surface area and wind force

[Tod - Minnesota]

I am sure that I must be missing something in my reasoning, but drawing from
my faint recollection of advanced calculus and/or vector calculus it seems
to me that the solution to the wind force on an antenna element will be a
double integral which integrates the wind force vector normal to an
infinitesimal surface area element over all angles from 0 to pi (0 to 180
deg) in the vertical plane and 0 to pi in the horizontal plane on the side
of the element facing the wind source.

Once the equations for the wind force on the infinitesimal surface area
element are written, it should be possible to write an equation that
describes the torque generated by that infinitesimal element. (Rotation
around the mast, tower overturning torque or similar torque of interest).

Additional integrations over all surface area infinitesimals should provide
the total torque that is of interest. "Turn the crank and get the answer",
as one of my calculus professors would often say (usually when I had no idea
what the hell he was talking about). His other pet phrase was, "The answer
is obvious!". He once said this, looked at the equations he had written on
the board, pondered them, and then left the room for five minutes. When he
came back he said, "Yes, the answer is obvious".

There may be no simple analytic solution, but I would expect that there is a
way to use numerical analysis to get good approximations to the solution.

Fearing that I have made a simplistic mistake in putting these thoughts
together, I will stand back and let the flaming begin.


Tod, KØTO




-----Original Message-----
From: owner-towertalk@contesting.com
[mailto:owner-towertalk@contesting.com]On Behalf Of EUGENE SMAR
Sent: Saturday, June 09, 2001 10:07 PM
To: towertalk@contesting.com; Stu Greene
Subject: Re: [TowerTalk] Antenna surface area


Stu:

     The area to which I refer is the area of the cylindrical shape of a
tubular element/boom as seen in shadow behind the element/boom.  This is the
area used in all wind force studies I've read over the past couple of weeks,
including those by Dick Weber, K5IU.  The area to which you refer is the
surface area of the entire element, the area of material, as it were, needed
to fabricate the element.  This area is NOT used in the computation of wind
force.

     The tapering of elements would be accounted for by measuring the length
and diameter of each taper segment as above and adding them together.

73 de
Gene Smar  AD3F
-----Original Message-----
From: Stu Greene <wa2moe@doitnow.com>
To: EUGENE SMAR <SPELUNK.SUENO@prodigy.net>; towertalk@contesting.com
<towertalk@contesting.com>
Date: Saturday, June 09, 2001 9:46 PM
Subject: Re: [TowerTalk] Antenna surface area


>At 09:06 PM 6/9/01 -0400, you wrote:
>
>>  Area = L X diameter for the exposed surface of each piece of aluminum,
>> simple as that.
>
>
>
>Shouldn't the calculation be L X (Diameter X pi) ?  Or length times
>circumference?
>
>And this doesn't reflect tapering of the elements.
>
>
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List Sponsor: Are you thinking about installing a tower this summer? Call us
for information on our fabulous Trylon Titan self-supporting towers - up to
96-feet for less than $2000! at 888-833-3104 <A HREF="http://www.ChampionRadio.com">
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