During our most recent storm, I found myself victim of the surprisingly harsh Riverside wind. With my umbrella directly out front of me, pointing into the wind, I found I barely had enough strength to hold my position, so great was the force imparted by the wind. I got to thinking about just how strong the wind was as the windspeed picked up.
In a time 
, the column of air in the figure hits and transfers its momentum to my umbrella. The force is equal to the momentum transferred to my umbrella per unit time:
How much is that? Well, the column has area 
and height 
, so the volume is 
. The change in momentum is the mass of this column times the change in velocity,
The mass of the column is its volume times the density, 
. Putting this all together,
If we allow that some of the wind retains a bit of its forward momentum, so that it's just deflected instead of dead stopped, we could add a simple coefficient 
to this. Thus,
This says that if the wind speed doubles, the force required to stay still quadruples. For a half sphere, the coefficient is about 0.2. Assuming an area of ![\unit[1]{m^2}](/images/2010/03/21/winddragIL7.png)
, the below graph shows the force in pounds for a wind speed in miles per hour.
As you can see, the force gets powerful mighty fast.
Actually, if we were to consider this scenario from a reference frame where the air was stationary, and I was moving at a speed 
, we get the expression for drag force (apart from the factor of 1/2). This was surprising to me, although it shouldn't have been. For some reason, I find it easier to derive the expression in the way I just did.