How this calculator works
At terminal speed, gravity minus buoyancy balances the Stokes drag on a sphere. A denser particle settles, while a lighter particle rises; downward velocity is positive here. Stokes drag assumes a rigid isolated sphere, a Newtonian fluid, negligible wall effects and a particle Reynolds number much smaller than one. A numerical result does not establish those assumptions, so the output includes a Reynolds check. It does not model the time needed to reach terminal speed.
Formula
F_drag = 3 pi mu d abs(v); v_terminal = g(rho_particle-rho_medium)d^2/(18 mu); Re_particle = rho_medium abs(v)d/mu.
Worked example
With g = 10 m/s2, density difference 1000 kg/m3, diameter 0.001 m and viscosity 1 Pa s, terminal speed is 0.000555556 m/s downward. Halving diameter reduces it to 0.000138889 m/s.
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Frequently asked questions
Can the terminal velocity be negative?
Yes. A particle less dense than its medium rises, which is negative under the downward-positive convention.
Why include Reynolds number?
The Stokes relation becomes inaccurate when inertial effects are no longer negligible.
Does this work for bubbles?
A mobile bubble interface has different drag. This model assumes a rigid no-slip sphere.