Mike, Bill, et al:
Hold everything! Multiplying by 0.6 or 0.67 as a shape factor is NOT
the correct number. It's 1.2  at least according to Mechanical Engineering
in Radar and Communications, one of K5IU's references. Let me go through
this again.
A long, thin cylinder's drag coefficient is 1.2 (above) or the book's
wrong! A flat plate's drag coefficient is 2.0 according to the same book.
Therefore, 1.2 is 60% of 2.0. In other words, the effective area (taking
shape into account) of a Yagi tube is 1.2 times GREATER than the projected
area (simply L X diameter), not 40% LESS. Big difference and the reason
I've begun questioning manufacturers' spec sheets.
(Jump in here anytime, Dick.)
73 de
Gene Smar AD3F
Original Message
From: Michael Tope <W4EF@dellroy.com>
To: Bill Hider <n3rr@erols.com>; EUGENE SMAR <SPELUNK.SUENO@prodigy.net>;
towertalk@contesting.com <towertalk@contesting.com>; Stu Greene
<wa2moe@doitnow.com>
To: <towertalk@contesting.com>
Date: Saturday, June 09, 2001 11:08 PM
Subject: Re: [TowerTalk] Antenna surface area
>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 windfront 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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>> Submissions: towertalk@contesting.com
>> Administrative requests: towertalkREQUEST@contesting.com
>> Problems: ownertowertalk@contesting.com
>>
>
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