[TowerTalk] Wind Surface Area estimates

[Larry]

If you access to Leeson's "Physical Design of Yagi Antennas" (or something 
close
to that) it covers this topic and how to "fix" it.

73, Larry  W6NWS
----- Original Message ----- 
From: "Gary Slagel" <gdslagel at yahoo.com>
To: "Jim Lux" <jimlux at earthlink.net>; "TowerTalk" <towertalk at contesting.com>
Sent: Friday, March 09, 2012 1:32 PM
Subject: Re: [TowerTalk] Wind Surface Area estimates


Shoot... using the .707 figure for 45 deg I'm back right at the worst case 
scenario. Where does the .707 come from?

I was using a simplified version of windload calcs from the force 12 web 
site that goes something like '..wind load is either the total element wind 
load OR the boom windload, whichever is the larger resistance to the 
wind....'.

Thanks,

Gary Slagel
KT0A




________________________________
From: Jim Lux <jimlux at earthlink.net>
To: towertalk at contesting.com
Sent: Friday, March 9, 2012 10:24 AM
Subject: Re: [TowerTalk] Wind Surface Area estimates

On 3/9/12 6:50 AM, Gary Slagel wrote:
> Hi guys... more questions!
>
> I'm thinking looking at putting an A4S and and some kind of 40M yagi on 
> the same mast. I plan to rotate the 40M yagi 90 deg from the A4S to 
> minimize interaction. I'm thinking somewhere in the future I may even add 
> another A4S somewhere down the tower a ways.
>
> The A4S has an advertised windload of 5.5 sq ft. The 40M yagis around the 
> same 5.5 sq ft. But, I'm thinking the 40M yagi is 5.5 sq ft looking at it 
> from the front. Looking at the side of the 40M yagi (2" boom and 22.6 ft) 
> I'd guess more like 2.5 sq ft. Same with the A4S from the side.... a 2" 
> 18' boom should be around 2 sq ft. So... total windload with the yagi's 
> arranged 90 deg from each other should be 5.5 + 2 = 7.5 sq ft from one 
> side and 5.5 + 2.5 = 8 sq ft from the other side.
>

A tube looking end on has a drag much more than the cross sectional area
of the tube.

Also consider what happens when the wind is at a 45 degree angle..

But.. let's for a minute assume that the drag end on is zero, and cross
ways to the tube it's 1. And when the wind comes at 45 degrees, the drag
is 0.707 (the projected cross sectional area)

You have two tubes at right angles, one (Tube A) is oriented N-S and the
other (Tube B) is E-W.

When the wind is from the north, Tube A has no drag and Tube B has a
drag of 1, for a total load of 1.

When the wind is from the east, Tube A has a drag of 1 and Tube B has a
drag of 0, for a total load of 1

But, when the wind is from the north east, tube A has a drag of 0.7 and
tube B has a drag of 0.7, for a total drag of 1.4.


So, your calculation is off by a bit (without looking at the end on drag
of a tube).


(I don't recall Cd for flow along a tube.. let's say it's Cd=0.01, since
this is skin friction, the area is the wetted surface area
(diameter*pi*length). A 2" diameter tube that is 20 feet long will be
about 10 square feet. That works out to 0.1 square feet "equivalent
flat plate area". however, the Cd could easily be 0.05 or 0.1 so you
could be as big as 1 square foot)

I have a handbook that gives Cd for tubes at various angles and Reynolds
numbers, but I can't seem to lay my hands on it right now)


There's also the practical detail that a lot of the drag will come from
"interference drag" where the crossing members are.

Maybe there's some charts showing the drag polar diagram of an antenna
out there. (the Yagi design book might have some?)



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