Terminal velocity calculator
Terminal velocity in air with the time and distance needed to reach it.
Fill in the fields and the result will appear here automatically.
Find the speed limit under constant quadratic drag ½ρACd·v² and weight mg. Frontal area, Cd and density are explicit inputs; defaults are illustrative numbers, not a validated person or parachute model. For release from rest, time and distance to 95% are reported; the limit itself is approached asymptotically.
How it works
Formula and logic
Balancing weight against drag gives v = √(2 mg/(ρ·A·Cd)); the time and distance to 95 per cent come from the hyperbolic-tangent solution.
Example
With m=80 kg, A=0.7 m², Cd=1 and ρ=1.225 kg/m³, the model gives 42.776 m/s, about 154 km/h; 95% takes about 7.99 s and 217.18 m. These are stated geometry inputs, not a prediction of an actual fall.
Fields and units
- Mass — kg
- Frontal area — m²
- Drag coefficient — 1
- Air density — kg/m³
How to use
- — Use measurements or a source for the actual geometry, Reynolds number and flow regime. A table value for a similar shape is an assumption, not a guarantee. Multiplying A·Cd by four with other inputs fixed halves the speed limit.
- — Find the speed limit under constant quadratic drag ½ρACd·v² and weight mg. Frontal area, Cd and density are explicit inputs; defaults are illustrative numbers, not a validated person or parachute model. For release from rest, time and distance to 95% are reported; the limit itself is approached asymptotically.
- — Vertical release from rest, constant g=9.80665 m/s², ρ, A and Cd, and quadratic drag. Buoyancy, density changes with height, parachute deployment and speed-dependent Cd are excluded. This calculation does not determine a safe landing speed.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: quadratic drag and terminal speed
- Limitation
- Vertical release from rest, constant g=9.80665 m/s², ρ, A and Cd, and quadratic drag. Buoyancy, density changes with height, parachute deployment and speed-dependent Cd are excluded. This calculation does not determine a safe landing speed.
FAQ
Why does a heavy body fall faster than a light one?
In a vacuum it does not. In air the terminal velocity grows as the square root of mass for the same area, so a feather and a stone of equal size differ radically: for the feather drag balances weight almost at once.
Where do I get the drag coefficient?
Use measurements or a source for the actual geometry, Reynolds number and flow regime. A table value for a similar shape is an assumption, not a guarantee. Multiplying A·Cd by four with other inputs fixed halves the speed limit.
Why is terminal velocity never reached exactly?
Because the closer you get, the smaller the remaining acceleration: the speed approaches the limit as a hyperbolic tangent. The practical answer is in the 95 per cent rows.
Does terminal velocity change with altitude?
Vertical release from rest, constant g=9.80665 m/s², ρ, A and Cd, and quadratic drag. Buoyancy, density changes with height, parachute deployment and speed-dependent Cd are excluded. This calculation does not determine a safe landing speed.