Re: How would I determine the force on an area, generated by wind?

>How would I determine the force on an area, generated by wind?

>Example:
>Flying a kite
>Find the tension in the string?
>Wind=25 MPH
>area of kite= 300 in^2

Let's sneak up on this problem one bit at a time.  First of all, let's talk 
about drag.  When air moves over a body, it pushes against it and this is 
called "drag".  There are several components of drag.  There is "pressure 
drag", which is the force caused by simlpy resisting the airflow.  There is 
"skin friction drag", which is the drag produced on a body as the air 
"sticks" slightly to the body's skin as it moves over it.  Lastly, there is 
"induced drag", which is the drag caused by lift.

We'll skip the second two and just concentrate on pressure drag for now.  
To experience pressure drag directly, hold your hand slightly outside of a 
car window and feel the pressure of the oncoming wind.  The pressure drag 
can be calculated by this formula (as long as the velocity isn't anywhere 
close to the speed of sound):

   Drag = Drag Coeficient * Dynamic Pressure * Area

Here's what these terms mean:

   Drag - a force, in pounds
   Drag Coeficient - a special coeficient that is different for 
      different shapes
   Dynamic pressure - a term for this: 1/2 * density * (velocity)^2
   Area - area, in square feet, that is perpendicular to the flow

The dynamic pressure pops up again and again when you're doing aerodynamic 
stuff.  The density of the atmosphere at sea level is about 0.00238 (a 
number I just looked up in a book) and the velocity is in feet per second 
(which gets squared).

The drag coeficient for a flat plate that is approximately square (that is, 
its length is about the same as its width) is 1.16 (another number I just 
looked up it's about 1.12 for a circular plate and about 0.63 for a 
circular cylinder and so on).

So, let's put this all together for your example.  If a flat plate with an 
area of 300 square inches (about 2 square feet) in a wind of 25 mph (about 
37 feet per second) at sea level pressure and temperature will experience 
this much drag:

Drag Coeficient: 1.16
Dynamic pressure: 1/2 * 0.00238 * 37^2 = 1.63 
Area: 2.08

Drag = 1.16 * 1.63 * 2.08 = 3.93 lb

So, we have a drag of about four pounds.  This number isn't exact, since we 
haven't paid attention to skin drag (which probably isn't much anyway).

Notice some additional things:  the drag is proportional to the SQUARE of 
the velocity.  Double the velocity and you get FOUR TIMES the drag.  Let's 
go back to the case where you hold your hand out in a 60 mph airflow.  Your 
hand is much smaller, about 1/5 - 1/7 of a square foot, but the velocity is 
a lot higher.

Run the numbers, and you get about two pounds of force.  Down at the store, 
pick up a two-pound can of coffee and see if it doesn't feel about the same 
as the force on your hand out the window.  (Of course, you want to be very 
careful when holding your hand out the window of a moving car and you 
certainly don't want to do it while you're the driver).

Also, notice that the area counts proportionately.  If you double the area, 
you double the drag.  If your kite was a four by eight foot sheet of 
plywood, the drag would be about 16 times as much (because its area is 32 
sqaure feet instead of 2 square feet and 32 is 16 times 2) or about 64 
pounds.

Think about this:  A sheet of plywood in a hurricane (where the winds are 
100 mph, four times what it is for your kite) would have a drag that is 16 
times more for size than your kite AND 16 times more for wind speed.  This 
works out to around 4 * 16 * 16 or about 1000 pounds!  This is why 
hurricanes can lift houses up and turn them over.

NOW, if you're going to apply this to a kite, you will have to include 
induced drag, which is drag caused by lift.  Induced drag is much more 
complex, however, not something we can go into here.  Since the amount of 
lift the kite generates is small (just enough to support the weight of the 
kite and string), we can assume that the induced drag will also be small 
and won't change the above number very much.

So, if your kite string is perpendicular to the airflow (and it usually 
isn't, you'll have to correct for that) and if the airflow isn't near the 
speed of sound (and it had better not be) and if your kite is more or less 
square (and it usually is), then we can say that the force on the string 
for this kite would be about four or five pounds.

Getting a really accurate number for things like this takes very large 
equations solved by very large computers.