[TowerTalk] Antenna surface area

Bill Hider n3rr@erols.com
Sun, 10 Jun 2001 03:11:48 +0100


No Stu, it wouldn't be precise.

It would be the integral, as I stated.

Your formula  would result in a surface area exposed to the wind
much larger than even Gene's calculation (especially, if his is reduced for
1/2 of the tube facing the wind), when in fact, it's actually less.
Take a look at what you said:  0.5 x Pi x D.  Well Pi x 0.5 = approx. 1.55.
If you multiply 1.55 x the diameter times the length, that's *much* more
than Gene's original calculation!!!

It is the Integral over the complete half of the diameter.  Not a simple
calculation - it requires a computer.

As far as taking 1/2 of each of the elements and the boom, it all depends on
your definition of *wind area*.  Just define it, and you're fine.
So, your definition is fine with me, as is mine.  Whatever definition the
mfgr uses, it is Ok with me, as long as we all know what they use.
If you use the larger of the two, that will be larger than the sum of 1/2 of
each, so it's a bit more conservative.  That's why I use it.

Bill


----- Original Message -----
From: Stu Greene <wa2moe@doitnow.com>
To: Bill Hider <n3rr@erols.com>; <towertalk@contesting.com>
Sent: Sunday, June 10, 2001 3:50 AM
Subject: Re: [TowerTalk] Antenna surface area


> At 02:13 AM 6/10/01 +0100, you wrote:
> >The formula Gene proposed is not exactly correct, nor does he precisely
> >state what to do with the taper.
> Following your reasoning, would A = 1/2 [L X (pi X D)]  be close
> enough?  That does not account for taper, but it is half the circumference
> of the element times its length.
>
> You added
>
> >  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
>
> Bill, irrespective of the wind direction which can change momentarily,
> surface area is surface area, so that an accurate measurement should be
the
> sum of half the boom area plus half the element area.
>
> I wonder if antenna manufacturers, when calculating wind area and
> advertising it, do it your way or mine or perhaps something a bit more
> arcane.  I'm sure that someone will straighten out the error in my logic.
>
>                                 73   Stu
>
>
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