[TowerTalk] Antenna surface area

[Michael Tope]

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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> 
> 
> 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
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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">
www.ChampionRadio.com</A>

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