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Stokes Settling Velocity — Reference

Terminal settling velocity of discrete particles, and the surface-overflow-rate basis for sedimentation basins, forebays, and grit chambers.

Settling Velocity by Regime

Stokes (Re < 1):   vs = g(ρs − ρ)d² / (18μ)
Newton (Re > 1000):   vs ≈ 1.74·√( g(Ss − 1)d )
RegimeParticle ReApplies to
Stokes (laminar)< 1Silt, clay, fine sand
Transition1–1000Medium–coarse sand
Newton (turbulent)> 1000Gravel, large grit
Sizing a sediment basin, not one particle? HydroComplete runs a seven-bin PSD through the basin and reports trap efficiency.

Particle Re = ρ·vs·d/μ. Ss = ρs/ρ (quartz sand ≈ 2.65). Water at 20°C: μ = 1.00×10−3 Pa·s, ρ = 998 kg/m³.

Typical Settling Velocities (quartz, 20°C water)

ParticleDiametervs (approx.)
Coarse sand1.0 mm~100 mm/s
Medium sand0.5 mm~50 mm/s
Fine sand0.1 mm~6–8 mm/s
Very fine sand0.05 mm~2 mm/s
Silt0.01 mm~0.08 mm/s
Clay0.001 mm~0.0008 mm/s

Fine silt/clay settle far too slowly for practical basins — they require flocculation or filtration. The d² dependence is why removing the fine fraction is so hard.

Surface overflow rate governs removal.
vo = Q / As
In ideal Type I settling, every particle with vs ≥ vo is fully captured — removal depends on surface area, not depth. Size sedimentation basins and forebays by overflow rate, then check detention time and short-circuiting.

Sources: Camp, T.R. (1946), "Sedimentation and the Design of Settling Tanks," Trans. ASCE. Metcalf & Eddy, Wastewater Engineering. Reynolds & Richards, Unit Operations and Processes in Environmental Engineering.

Sizing a basin or forebay? Open the settling velocity calculator → · Detention time? Detention time. · Want the whole set? Print all 38 cards →

Worked example: sizing a basin for a target particle

Sediment basin, Q = 2.0 cfs (0.0566 m³/s), quartz particles, water at 20°C

Given ρs = 2,650 kg/m³, ρ = 998 kg/m³, μ = 1.00×10⁻³ Pa·s
Find surface area needed to capture 0.05 mm and 0.01 mm particles
0.05 mm: vs = 9.81 × (2,650 − 998) × (5×10⁻⁵)² / (18 × 10⁻³) = 2.25 mm/s
Check Stokes: Re = 998 × 0.00225 × 5×10⁻⁵ / 10⁻³ = 0.11 < 1, so the Stokes form is valid
As = Q/vs = 0.0566 / 0.00225 = 25.2 m² ≈ 271 ft²
0.01 mm: vs = 0.090 mm/s, As = 629 m² ≈ 6,770 ft², 25 times the area for a particle one-fifth the size
Same 0.05 mm particle in 5°C water (μ = 1.52×10⁻³): vs = 1.48 mm/s, 34% slower

The d² term drives most basin decisions. A basin sized for fine sand is reasonable. One sized for fine silt is usually not, and clay will not settle in any practical footprint. Below roughly 0.01–0.02 mm the realistic options are flocculants, filtration, or accepting the turbidity. Design for the cold-weather viscosity if the basin has to perform in winter.

Common mistakes

When this card doesn’t apply

Stokes’ law assumes isolated, spherical, rigid particles in still water. Platy clay minerals, organic flocs and aggregates depart from it, often by an order of magnitude. Regulatory sediment-basin sizing, such as state erosion-control manuals with fixed surface areas per acre or per cfs, may override the overflow-rate calculation. Use the governing manual where one applies.

Related cheat sheets and tools

Use vs with the settling velocity tool and check detention time for the basin. Protect the basin inlet/outlet with riprap. For full RUSLE sediment load and BMP/forebay sizing across a watershed, see HydroComplete. Fluid properties for these calculations are in the water viscosity & density table.

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