[TowerTalk] Antenna surface area

[EUGENE SMAR]

Bill:

     My comments are below.  Dick Weber K5IU's papers on the subject
includes equations for calculating force of the wind on tubular elements and
booms, all of which start with the calculation I described:  exposed tube
length X diameter.  The shape of the element/boom is taken into account by
something referred to (in the papers) as drag coefficient, which is 1.2 for
tubes used in Yagi construction.

     Most of what you question, Bill, is addressed in the very few papers
I've read so far.   If the area to be used to calculate wind force on
tapered elements is NOT the sum of the individual taper sections' shadow
areas, then I missed yet another non-obvious quirk in the aerodynamics
involved.

73 de
Gene Smar  AD3F
-----Original Message-----
From: Bill Hider <n3rr@erols.com>
To: EUGENE SMAR <SPELUNK.SUENO@prodigy.net>; towertalk@contesting.com
<towertalk@contesting.com>; Stu Greene <wa2moe@doitnow.com>
Date: Saturday, June 09, 2001 10:15 PM
Subject: Re: [TowerTalk] Antenna surface area


>The formula Gene proposed is not exactly correct, nor does he precisely
>state what to do with the taper.
>
>Regarding the formula, Gene's thinking is:  If you think of the wind as
>hitting the tube broadside (at 90 Deg to the tube), the exposed surface
area
>as seen by the wind looks like a rectangle whose length is the length of
the
>tube and whose height is the full OUTSIDE diameter of the tube.  Hence, L x
>Dia.

I agree.

>Unfortunately, it is not that simple.
>
>The tube is a cylinder as seen by the wind, hence the angle that the wind
>*hits* the tube, even if it is perpendicular to the tube, hits at 0 Deg on
>the centerline and then the angle increases to 90 Deg as the wind hits the
>tube away from the centerline (above and below the centerline of the tube
>for a horizontal element).

This is kinda like the sun's rays hitting the equator from directly overhead
at the equinoxes, while hitting the earth's surface at steeper and steeper
angles the farther away you go from the equator.

>This assumes the wind-front is wider than the
>outside diameter of the tube, which is probably a very good assumption.
So,
>the surface area exposed to the wind by the tube is not simply L x Dia.
>It's the exposed tube length times the integral from 0 to Dia of the
surface
>area of each tube (Pi x Dia), where Dia is the Outside diameter of the
tube.

The drag coefficient in the equations takes this into account.  From what I
gather, it's an empirical number, not one derived from the physics of the
problem as you're suggesting.


>Gene, this is why the manufacturer's wind area is less than what you
>calculated, and theirs is correct.

Why do you say theirs is correct, BIll?  We've been reading here about
discrepancies between specs and measurements of area taken by owners.  If
anything, the manufacturers ought to explain HOW they arrived at their
figures.  (I will say that F12's brochures define wind load as the larger of
element or boom load, but they don't say whether a shape factor or other
drag coefficient has been included.   So we can't tell if we can just plug
F12 figures into any of the windload equations in Dick's reference studies.)

 > [If anyone has trouble picturing this,
>let me know and I'll try to explain it in more detail.]
>
>But, I question the correctness of adding all of these calculations up and
>saying that's the *wind area*.  It certainly is the 1/2 of the *surface
>area* if it's done this way, but the wind cannot be simultaneously hitting
>the elements at 90 Deg and the boom at 90 Deg, so it should take that into
>account by specifying the *larger* of both calculations, but not the sum of
>both.

I agree that the larger of the two figures is what will give rise to the
maximum wind force on a Yagi.  I've been converted!  I'm advocating
collecting both boom and element areas and letting the user decide what to
do with the numbers.  I'm not advocating adding these figures to obtain the
total wind area of the Yagi.

>
>Regarding the taper:  each tube should be calculated separately for the
>*exposed* length of the tube *only*.  Remember, some tubes are *inside*
>other tubes, causing the taper.

Agreed.  That's why I said include exposed area of aluminum in the
calculation, implying that we should ignore the few inches buried inside the
next-larger diameter tube.
>
>Bill, N3RR
>
>
>----- Original Message -----
>From: Stu Greene <wa2moe@doitnow.com>
>To: EUGENE SMAR <SPELUNK.SUENO@prodigy.net>; <towertalk@contesting.com>
>Sent: Sunday, June 10, 2001 2:42 AM
>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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>


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