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?

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    $\begingroup$ 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. $\endgroup$
    – user10851
    Oct 28, 2012 at 22:55
  • $\begingroup$ Sort of similar question: physics.stackexchange.com/q/36439 $\endgroup$ Oct 29, 2012 at 11:05
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    $\begingroup$ I won't insert this as an answer, because it is more opinion and depends on other factors (what you are wearing for instance). My rule of thumb (when on mountains) is that you are on the verge of losing control at 60 mph, but can till make progress in a steady wind. You (or at least I, 60kg) become "airborne" at somewhere just over 70-75mph, not much more than that. I have had one experience in a wind measured (at a nearby weather station) to be 80+ mph and was fortunate to tell the tale - and only by anchoring with an ice-axe. $\endgroup$
    – ProfRob
    Dec 30, 2014 at 21:48

1 Answer 1


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.

  • $\begingroup$ Clarification: weight = mg = 50 kg. $\endgroup$ Oct 29, 2012 at 16:59
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    $\begingroup$ @MichaelLuciuk Kilograms are not weight. Weight is measured in newtons. $\endgroup$
    – Muphrid
    Oct 29, 2012 at 17:21
  • $\begingroup$ Oops. You're absolutely right. $\endgroup$ Oct 29, 2012 at 19:58
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    $\begingroup$ Incidentally, while I like Muphrid's calculation, the answer is not altogether consistent with observations - which suggest that even at speeds of 39–46 mph, it is still possible to walk or stand, although "Progress on foot is seriously impeded." (Beaufort $\endgroup$
    – akrasia
    Aug 15, 2014 at 21:48
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    $\begingroup$ For reference, iFly - indoor skydiving company uses a wind tunnel to create artificial winds at 100-130mph to lift off the person against gravity when person is on the belly. articles.chicagotribune.com/2014-06-26/entertainment/… $\endgroup$ Oct 12, 2016 at 23:14

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