# Is the wind's force on a stationary object proportional to $v^2$?

I am on a boat docked at Cape Charles, VA, about 30 or 40 miles from the center of Hurricane Irene. This understandably got me thinking about the force of wind on the boat. Since air friction is proportional to the speed squared (except for some friction types I'm sure someone will be kind enough to remind me of), is the wind's force on the boat also proportional to the wind's speed squared? In other words, will a 70 knot wind produce almost twice the force on the stationary boat as a 50 knot wind, all other factors being equal?

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It's funny that only now, re-reading the question does the thought occur to me: "holy Batman that sounds dangerous! Get off the internet and take shelter now!" – Alan Rominger Aug 28 '11 at 3:26
It shows we have our priorities straight. – xpda Aug 28 '11 at 4:34
– Ben Crowell Apr 20 '13 at 15:46

On the most basic level I would say that both your suspicion and your reasoning are correct, although some caveats should follow.

First I should address this on the basis of simple kinematics. Yes, the energy of an air molecule is $m v^2$ but we need to formalize when and if this will translate into pressure. Imagine a pipe that spews a fluid onto a surface, and the fluid comes to a complete rest after impacting the surface. As speed changes the impulse from a molecule in the fluid increases (and impulse is the more true analog to force/pressure than energy), but so does the flow rate, $\dot{m}$. We say that area and density stays the same.

$$\dot{m} = \rho V A$$

$$P=\frac{F}{A}$$

$$F=\dot{m} \Delta V = \rho A V^2$$

$$P = \rho V^2$$

If you study thermal hydraulics, this will not be the last time you see that final expression. Not by a long shot. I should point out that it's not exactly energy per se that leads to the $V^2$ nature, I would say that it's the above math that leads to it.

The example I provide (much like a garden hose spraying someone holding a shield) is a similar system to that which you describe, which is wind in a hurricane hitting the wall of a house. There are several ways that these two systems are not identical, and science has done a lot of work to develop empirical correlations and representative models for fluid pressure losses.

We can look at the system as either a forms loss (on a human-sized scale) or as a boundary layer friction force (on the larger climate level). Equations for the change in pressure of a fluid due to friction or due to a obstacle ("forms" loss) are as follows.

$$\Delta P_{friction} = f \frac{L}{D} \frac{\rho V^2}{2}$$ $$\Delta P_{forms} = K \frac{\rho V^2}{2}$$

In the above equations, both $f$ and $K$ represent constants that we multiply the $V^2$ term by to account for all of the real-life complexity of fluid flow. Yes, these constants will change with the wind speed, so the answer is that the pressure is not perfectly proportional to $V^2$, but we wouldn't be using the above format if this wasn't the closest form that it does follow.

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Well, I think of it this way. Imagine that your boat and the wind are the only things present in a giant empty space, and let the wind blow head on onto the bow of your boat at a constant speed. Since in this space there are only the wind and your boat (and yourself on it), you can't conclude if it is the wind blowing onto your boat or that it is your boat moving through a gas of air particles. But more importantly, you can conclude that the two situations are actually equivalent. It is just a matter in which frame of reference you are viewing your situation: a frame stuck to your boat or a frame stuck to the center of mass of the wind.

You already know that if it was your boat that was moving, the frictional force that would be applied to your boat would go like F ~ v^2. And since this must be the same as a gas blowing onto your boat by the above mentioned arguments, you can conclude that the force applied to your boat by a blowing wind is proportional to it's airspeed squared. I think.

Good luck with the storm!

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