I'm doing a game that is set in a desert and there are sand storms that occur periodically. What I've done is very basic and far from realistic, which is

if (distance > 30) {
    windForce = 60000 / distance;

Beyond 30 meters you are within the eye of the storm and (supposedly) feel no wind. Given the equation above, at 30 meters the force would be 2.000 newtons and get weaker from there.

How can I know how much wind speed that is in meters/s? I know wind speed is calculated by special equipment and it depends on air density and temperature, (and probably other factors), but a rough approximation would do. And as I said, it's in a desert, so hot and dry climate.

  • $\begingroup$ Air isn't very viscous, so force shouldn't really be a function of distance (not at scales of ~30 m). $\endgroup$ – Mohammad Athar Mar 15 '18 at 13:52
  • $\begingroup$ Not even in a hurricane? I thought the closer to the eye walls the stronger the wind was, and it gradually decreased in strenght further away. Anyway, that's how I am doing it :P just wanted to know how to translate x force to wind speed in m/s $\endgroup$ – RealAnyOne Mar 15 '18 at 13:58
  • $\begingroup$ you want to use control volume analysis to get a decent estimate: jove.com/science-education/10444/… I can write up a solution later tonight (when I'm not at work), but that link should get you started $\endgroup$ – Mohammad Athar Mar 15 '18 at 14:04

The simplest model of air resistance has the force proportional to the square of the velocity. To go a distance $d$ through a fluid at speed $v$, the minimum air resistance requires you to accelerate the air to speed $v$ to move it out of the way. That requires energy:

$$ E \propto \frac 1 2 m v^2$$

with the mass, $m$, depending on $d$:

$$ m \propto \rho \cdot d$$.

The work done is:

$$ W = F\cdot d = E $$

so that:

$$ F = E/d \propto (\rho d)v^2/d \propto v^2 $$.

The hurricane model has a core of constant vorticity in the eye (basically, it rotates like a solid block). Outside the eye wall, the vorticity is zero. That require zero curl:

$$ \vec{\nabla}\times \vec{u} = 0 $$

which is solved by a circular flow with:

$$ ||u|| \propto 1/r $$.

(Note: you can watch airborne big rigs revolve around a tornado vortex, without rotating--that's zero curl irrotational flow).

So in summary:

$$ F \propto 1/r^2 $$

for $r > r_{eye}$, and:

$$ F \propto r^2 $$

for $ r < r_{eye} $.

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