Terminal Velocity Calculator

Terminal Velocity Guide

What's the key thing to understand about The Terminal Velocity Formula?

Terminal velocity occurs when drag force = weight. Vt = √(2mg / ρACd). Where m = mass (kg), g = 9.81 m/s², ρ = fluid density (kg/m³), A = cross-sectional area (m²), Cd = drag coefficient. For a typical skydiver (80 kg, belly-to-earth, A = 0.7 m², Cd = 1.0, sea-level air): Vt = √(2×80×9.81 / 1.225×0.7×1.0) = √(1569.6 / 0.858) = √1829.4 = 42.8 m/s ≈ 154 km/h. Head-down speed skydiving position reduces area and Cd, producing terminal velocities of 250-300 km/h.

How Drag Force Works?

As an object accelerates downward, air resistance (drag) increases with velocity squared: Fd = ½ρACdV². At low speeds, drag is small and the object accelerates. As speed increases, drag increases. Terminal velocity is reached when drag exactly equals gravitational force (weight = mg). The V² relationship explains why terminal velocity is so sensitive to area — doubling the cross-sectional area reduces terminal velocity by 1/√2 (about 29%). This is why a deployed parachute, with its large area, so dramatically reduces terminal velocity compared to free-falling without one.

What do I need to know about Terminal Velocities in Nature and Engineering?

Typical terminal velocities: raindrop (2mm diameter) approximately 7-9 m/s (25-32 km/h) — raindrops are not teardrop-shaped as commonly depicted; they are spherical or slightly flattened. Cat (compact body position): approximately 27 m/s (97 km/h) — cats have survived falls from high buildings due to their low terminal velocity and righting reflex. Skydiver belly-to-earth: 53-55 m/s (190-200 km/h). Head-down speed skydiver: 70-80 m/s (250-290 km/h). Felix Baumgartner's record-breaking jump from the stratosphere in 2012 reached speeds briefly exceeding the speed of sound, since the extremely thin air at that altitude offered far less drag than at normal skydiving altitudes.

What do I need to know about Applications in Engineering?

Car aerodynamics: reducing the drag coefficient from 0.35 to 0.30 reduces aerodynamic drag force at 70 mph by approximately 14%, directly improving fuel economy. Bicycle aerodynamics: at 40+ km/h, aerodynamic drag accounts for 80%+ of total resistance — hence the extreme aerodynamic positions in time trialling. Space reentry: spacecraft reenter at 7-8 km/s and use atmospheric drag plus heat shields to decelerate to safe parachute deployment speeds. Seed dispersal: sycamore and dandelion seeds have evolved shapes specifically to minimise their terminal velocity, letting the wind carry them much further from the parent plant before they land.

What should I know about the Terminal Velocity Formula?

Terminal velocity occurs when drag force exactly balances weight: v_term = √(2mg/(ρ×A×C_d)), where m is mass, g is 9.8 m/s², ρ is air density (≈1.225 kg/m³ at sea level), A is cross-sectional area, and C_d is the drag coefficient (shape-dependent — around 1.0 for a belly-down skydiver, 0.7 for a feet-first fall). For a typical 80kg skydiver with A≈0.7m² and C_d≈1.0 in a belly-down spread position: v_term = √(2×80×9.8/(1.225×0.7×1.0)) = √(1568/0.8575) ≈ √1829 ≈ 42.8 m/s (about 154 km/h), matching the commonly quoted skydiving terminal velocity. Reducing cross-sectional area (diving head-first, arms tucked) significantly increases terminal velocity, which is exactly how skydivers control their fall speed, reaching upward of 200 km/h in a head-down 'tracking' position.

What do I need to know about Why Heavier Objects Fall Faster in Air (But Not in a Vacuum)?

In a vacuum, all objects fall at the same acceleration regardless of mass, because both gravitational force and inertial resistance scale identically with mass — this is the famous result Galileo demonstrated and that astronauts confirmed by dropping a feather and a hammer simultaneously on the Moon. In air, however, drag force depends on cross-sectional area and shape, not mass, so two objects of the same size and shape but different mass will reach different terminal velocities: the heavier one requires more drag force to balance its greater weight, and since drag force increases with speed, it must fall faster to generate that extra drag. This is why a bowling ball and a beach ball of similar size fall very differently in air despite experiencing identical gravitational acceleration — the beach ball's much lower mass means a much lower terminal velocity is enough to balance its weight against drag.

What do I need to know about a Skydiver's Terminal Velocity?

A skydiver in the standard belly-to-earth, arms-and-legs-spread position reaches a terminal velocity of roughly 195 km/h (54 m/s), limited by the large frontal area catching air resistance. Tucking into a head-down 'tracking' or dive position dramatically cuts the drag coefficient and frontal area, letting experienced skydivers reach 300–320+ km/h before opening their parachute. Opening the parachute suddenly multiplies the effective drag area many times over, reaching a new, much lower terminal velocity (typically around 18–25 km/h) within seconds — which is the whole point of the canopy.

What should I know about the Drag Coefficient and Reynolds Number?

The terminal velocity formula relies on a drag coefficient Cd, which is often treated as a fixed property of an object's shape, but Cd actually depends on the Reynolds number Re — a dimensionless quantity describing whether the flow around the object is smooth (laminar) or chaotic (turbulent). At low Reynolds numbers, typical of very small or very slow-moving objects in viscous fluids, flow stays laminar and drag is dominated by viscosity, following Stokes' law where drag is proportional to velocity rather than velocity squared. At the higher Reynolds numbers typical of falling skydivers, vehicles, or sports balls, flow becomes turbulent and drag follows the quadratic form F = ½ρv²CdA used in the standard terminal velocity formula, with Cd roughly constant over a wide speed range for a given shape. In between these regimes, Cd can actually change non-monotonically with Reynolds number — famously, a smooth sphere's drag coefficient drops sharply at a critical Reynolds number where the boundary layer transitions from laminar to turbulent separation, which is exactly why golf balls are dimpled: the dimples deliberately induce earlier turbulent separation, reducing drag and letting the ball fly farther than a smooth sphere of the same size and speed would.

What should I know about Terminal Velocity in Liquids (Stokes' Law)?

For very small or slow-moving objects, the standard quadratic-drag formula above doesn't apply — the relevant equation becomes Stokes' Law, valid when flow around the object stays laminar (Reynolds number below about 1). In this regime, drag force is proportional to velocity rather than velocity squared: Fd = 6πμrv, where μ is the fluid's dynamic viscosity (Pa·s) and r is the sphere's radius. Setting drag equal to the net weight (true weight minus buoyancy) gives the Stokes' Law terminal velocity: v = 2r²(ρp − ρf)g / (9μ), where ρp is the particle's density and ρf is the fluid's density. Worked example: a glass bead of radius 0.5mm (0.0005m) and density 2,500 kg/m³ falls through glycerin (density 1,260 kg/m³, viscosity 1.5 Pa·s at room temperature). v = 2 × (0.0005)² × (2500−1260) × 9.81 / (9 × 1.5) = 2 × 2.5×10⁻⁷ × 1240 × 9.81 / 13.5 ≈ 6.08×10⁻³ / 13.5 ≈ 4.5×10⁻⁴ m/s, or about 0.45 mm/s. Checking the Reynolds number confirms Stokes' law was valid here: Re = ρf v d/μ = 1260 × 4.5×10⁻⁴ × 0.001/1.5 ≈ 3.8×10⁻⁴, far below the threshold of 1 where laminar flow holds. This is the same physics behind sedimentation analysis in geology and biology (settling rates of silt particles, or centrifuging blood cells), where particle size is inferred from how fast it settles through a known fluid.

What should I know about Drag Coefficients for Different Shapes?

The drag coefficient Cd depends almost entirely on an object's shape, and reference values for common shapes let you estimate terminal velocity for objects beyond the usual skydiver example:

Notice how dramatically shape alone changes Cd even for objects of similar size — a teardrop shape has roughly a twelfth the drag coefficient of a flat plate, which is why vehicle and aircraft designers invest heavily in streamlining even when every other dimension stays the same.

What should I know about Buoyancy and Terminal Velocity in a Fluid?

The terminal velocity formula used for skydivers in air quietly ignores one force that becomes important in denser fluids: buoyancy. Strictly, an object falling through any fluid reaches terminal velocity when drag plus buoyant force balances weight, not when drag alone balances weight. For a body of volume V and density ρp falling through a fluid of density ρf, the buoyancy-corrected terminal velocity (in the high-Reynolds, quadratic-drag regime) is v = √(2V(ρp − ρf)g / (ρf A Cd)). In air, buoyancy is almost always negligible because air's density (1.225 kg/m³) is tiny compared with the density of falling objects, so the simplified formula v=√(2mg/(ρACd)) works well. In water or other liquids, buoyancy can matter enormously: an object whose density is less than the fluid's (ρp < ρf) has a negative net weight and never reaches a downward terminal velocity at all — it instead accelerates upward until it reaches an upward terminal velocity where drag balances the buoyant excess. This is exactly why a cork released underwater shoots upward rather than sinking, and why hot air balloons, submarines adjusting ballast, and fish with swim bladders all exploit the same buoyancy-versus-weight balance that the basic air-only terminal velocity formula leaves out.