# What does the wind speed have to be to blow away a person?

With the approaching hurricane, I am curious about what would happen if I go outside, in particular whether the wind gusts might be fast enough to blow me away. How fast would the wind have to be to blow away a person?

-
In Europe, we have access to the internet too. –  NiftyKitty95 Oct 28 '12 at 21:43
When it comes to hurricanes and wind (leaving rain and storm surge aside), the danger is more from small, hard objects hitting you than you being picked up by the wind. High-speed horizontal raindrops, though not actually dangerous, also make being outside uncomfortable enough that you probably won't want to do it. –  Chris White Oct 28 '12 at 22:55
Thanks, hadn't though of that –  Ana Oct 29 '12 at 1:59
Hi Ana - we can't address whether or not any particular thing is safe, but since the safety aspect was easily removed from your question, I made an edit that did so. Please review it and make sure it reflects what you wanted to ask, and make any changes necessary. –  David Z Oct 29 '12 at 4:00
Sort of similar question: physics.stackexchange.com/questions/36439/… –  Mitchell Porter Oct 29 '12 at 11:05

Let's do math before we look for information. First, what is the force that keeps you anchored to the ground? This is the force of static friction, which is $F_s = \mu m g$. What is this force opposing? The force of drag from the wind pushing on you. For the velocities involved (a high Reynolds number regime), the drag is quadratic in velocity, $F_d = \frac{1}{2} \rho v^2 C_d A$, where $\rho$ is the density of atmosphere, $v$ is the velocity, $C_d$ is a dimensionless drag coefficient, and $A$ is your body's cross-sectional area. So let's set the forces equal and solve for the velocity:

$$v^2 = \frac{2\mu m g}{\rho C_d A}$$

We'll be very ballpark about this. The density of air is $\rho \approx 1.2 \text{ kg/m}^3$. I'll say your mass is $50 \text{ kg}$. Per this paper, we'll say $C_d A \approx 0.84 \text{ m}^2$. Per this thread, we'll say $\mu = 0.4$.

Putting all these numbers in gives us $v \approx 20 \text{ m/s}$, or about 45 mph. But, this is just enough to make your body move (compared to standing still on the ground). It would take at least a 70 mph wind to overcome the force of gravity, and even then, that's assuming the wind keeps pushing on you with your body turned to face it (or away from it), not sideways. Hard thing to guarantee given how the body is likely to tumble or spin.

It's hard to be exact about this sort of thing, but let's just say this: going out in this kind of storm is a bad idea. The numbers aren't clear-cut enough to say you're safe, so better safe than sorry.

-
Thanks so much!!! –  Ana Oct 29 '12 at 10:22
Clarification: weight = mg = 50 kg. –  Michael Luciuk Oct 29 '12 at 16:59
@MichaelLuciuk Kilograms are not weight. Weight is measured in newtons. –  Muphrid Oct 29 '12 at 17:21
Oops. You're absolutely right. –  Michael Luciuk Oct 29 '12 at 19:58