11
$\begingroup$

I know it is possible to calculate terminal velocity using air density, mass, and drag coefficient, but is there any way to calculate the time taken until that speed is reached (assuming air density is constant for simplification purposes)?

$\endgroup$
  • 7
    $\begingroup$ The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question! $\endgroup$ – alephzero Nov 6 '18 at 17:10
  • 5
    $\begingroup$ @alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity. $\endgroup$ – Meni Rosenfeld Nov 7 '18 at 1:49
  • $\begingroup$ If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times. $\endgroup$ – Ben Nov 7 '18 at 13:58
  • 1
    $\begingroup$ @Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached. $\endgroup$ – D. Halsey Nov 7 '18 at 22:44
23
$\begingroup$

A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula $$v=\sqrt{\frac{2mg}{\rho A C_d}}\tanh{\left(t\sqrt{\frac{g\rho A C_d}{2m}}\right)}.$$ Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $\rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.

So $$v_t=\sqrt{\frac{2mg}{\rho A C_d}}$$ is the terminal velocity and $$\tau=\sqrt{\frac{2m}{g\rho A C_d}}=\frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_t\tanh{\frac{t}{\tau}}.$$ At $t=\tau$ the object is at 76% of terminal velocity. At $t=2\tau$ the object is at 96% of terminal velocity. At $t=3\tau$ it is at 99.5% of terminal velocity.

$\endgroup$
  • 7
    $\begingroup$ Note that $\tanh x \approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that. $\endgroup$ – Michael Seifert Nov 6 '18 at 20:43

Not the answer you're looking for?Browse other questions tagged or ask your own question.