Bill,
Gene specifically said that he was calculating projected area without any shape
factors. The projected surface area at right angles to a cylindrical tube is
simply
diameter x length. The effective area is computed by including the drag
coefficient
of the object. For a cylinder, the drag coefficient is around 0.6. Thus the
effective
area of an antenna with round members is about 40% smaller than the projected
area (a round tube of length, L and diameter D, is more aerodynamic than a flat
plate length L and width, D).
73 de Mike, W4EF....................
----- Original Message -----
From: "Bill Hider" <n3rr@erols.com>
To: "EUGENE SMAR" <SPELUNK.SUENO@prodigy.net>; <towertalk@contesting.com>; "Stu
Greene" <wa2moe@doitnow.com>
Sent: Saturday, June 09, 2001 6:13 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. 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 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.
> Gene, this is why the manufacturer's wind area is less than what you
> calculated, and theirs is correct. [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.
>
> 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.
>
> 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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> >
>
>
> List Sponsor: Are you thinking about installing a tower this summer? Call us
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List Sponsor: Are you thinking about installing a tower this summer? Call us
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