Terminal Velocity Calculator (drag)

Understanding the terminal velocity calculator — concept and principle

Terminal velocity is the steady speed a falling object reaches when air resistance pushing up becomes exactly as large as gravity pulling down. From that moment the net force is zero, so the object stops accelerating and falls at a constant speed. It is used for skydiving, parachutes and drones.

Drag grows with speed until it equals the 735.5 N weight — at 41.4 m/s.

How it is calculated

The calculator uses v_t = √(2mg ÷ (ρ C_d A)), and also converts the result to km/h by multiplying by 3.6.

m is the mass in kg; a heavier object needs more drag to balance its weight, so it falls faster.
g = 9.80665 m/s², the standard acceleration of gravity.
ρ (rho) is the air density in kg/m³; 1.225 kg/m³ is the sea-level value of the standard atmosphere at 15 °C.
C_d is the dimensionless drag coefficient, which describes how streamlined the shape is.
A is the frontal area in m², the area the object presents to the oncoming air.

The principle

Aerodynamic drag on a body moving fast through air follows the drag equation F_d = ½ρv²C_dA. It grows with the square of speed, so as a falling object speeds up, drag rises quickly until it matches the weight mg. Setting ½ρv²C_dA = mg and solving for v gives the formula above. The model assumes constant C_d, still air and constant density; real air thins with altitude, so a high-altitude jumper falls faster at first.

Worked example

Inputs: m = 75 kg, C_d = 1.0, A = 0.7 m², ρ = 1.225 kg/m³.
Weight: 2mg = 2 × 75 × 9.80665 = 1,470.9975 N.
Denominator: ρC_dA = 1.225 × 1.0 × 0.7 = 0.8575 kg/m.
Ratio: 1,470.9975 ÷ 0.8575 = 1,715.45 m²/s².
Square root: v_t = √1,715.45 ≈ 41.418 m/s, and 41.418 × 3.6 ≈ 149.10 km/h.

Each point is the calculator re-run with only “Drag coefficient Cd” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Drag coefficient Cd	Terminal velocity (m/s)
0.25	82.83595788
0.5	58.57386755
0.75	47.82536258
1	41.41797894
1.25	37.04536656
1.5	33.81763819
2	29.28693377

How to read the result

The answer is the speed at which the fall stops accelerating, not a speed reached instantly; the object approaches it gradually. Because v_t depends on √(m ÷ C_dA), doubling the mass raises it only by √2 (about 41%), while halving the effective area C_dA does the same. A belly-down skydiver is commonly quoted at roughly 50–55 m/s (about 180–200 km/h); a head-down position reduces the area and gives much higher speeds. An open parachute raises C_dA enough to cut the speed to a few metres per second.

Common mistakes

Entering weight in newtons or pounds instead of mass in kilograms.
Using total surface area instead of frontal (projected) area.

Keep in mind: The result assumes constant drag coefficient, density and still air. It is an educational estimate, not a design value for parachutes, drones or other safety-critical equipment.

Frequently asked questions

Does a heavier object really fall faster?

In a vacuum no, all objects fall with the same acceleration. In air yes: for the same shape and size, a heavier object needs a higher speed before drag can balance its weight.

How long does it take to reach terminal velocity?

Speed approaches v_t asymptotically. A rough time scale is v_t ÷ g, about 4 s here.

Why is the default air density 1.225 kg/m³?

It is the standard-atmosphere value at sea level and 15 °C; at 3,000 m it drops to about 0.91 kg/m³, which raises v_t by about 16%.

Sources: NASA Glenn Research Center, Beginner's Guide to Aeronautics — The Drag Equation and Terminal Velocity · ISO 2533:1975, Standard Atmosphere · Halliday, Resnick and Walker, Fundamentals of Physics, Wiley

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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