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[TowerTalk] Antenna surface area

To: <>
Subject: [TowerTalk] Antenna surface area
From: (alsopb)
Date: Sun, 10 Jun 2001 11:01:58 +0000
It appears that mother nature knows all about the calculation of wind

She knows that arrays have an effective surface area less than the sum
of the areas of the individual elements.  Another way of saying it is
the force of the wind on other "elements" in an array is less than the
"lead" element.  

1) Geese would not fly in V formations and the lead goose would not
swap with others in the flock. (Why a V not one behind another?  Is
that a lower energy use configuration or is it a simple matter of them
being to be able to better see the goose in front of them?)

2) Long distance runners (bicyclers, et al) would prefer to always be
ahead of the pack and not fall back in the pack occasionally to
conserve their energy.

The question is how much less.  My guess is that one needs a computer
type anaylysis or wind tunnel tests to determine this.  It isn't clear
to me that head on to the wind is always the worst case.  The goose
analogy indicates that the "best case" may not necessarily be what one
would guess.

73 de Brian/K3KO  

Bill Hider wrote:
> 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 <>
> To: EUGENE SMAR <>; <>
> 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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