Terminal velocity

Graph of velocity versus time of a skydiver reaching a terminal velocity.

Based on air resistance, for example, the terminal speed of a skydiver in a belly-to-earth (i.e., face down) free fall position is about 55 m/s (180 ft/s).[3] This speed is the asymptotic limiting value of the speed, and the forces acting on the body balance each other more and more closely as the terminal speed is approached. In this example, a speed of 50% of terminal speed is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.

Higher speeds can be attained if the skydiver pulls in their limbs (see also freeflying). In this case, the terminal speed increases to about 90 m/s (300 ft/s),[3] which is almost the terminal speed of the peregrine falcon diving down on its prey.[4] The same terminal speed is reached for a typical .30-06 bullet dropping downwards—when it is returning to the ground having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study.[5]

Competition speed skydivers fly in a head-down position and can reach speeds of 150 m/s (490 ft/s). The current record is held by Felix Baumgartner who jumped from an altitude of 38,887 m (127,582 ft) and reached 380 m/s (1,200 ft/s), though he achieved this speed at high altitude where the density of the air is much lower than at the Earth's surface, producing a correspondingly lower drag force.[6]

The biologist J. B. S. Haldane wrote,

To the mouse and any smaller animal [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away. A rat is killed, a man is broken, a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal's length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.[7]

Using mathematical terms, terminal speed—without considering buoyancy effects—is given by

$${\displaystyle V_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}}$$

where

• ${\displaystyle V_{t}}$ represents terminal velocity,
• ${\displaystyle m}$ is the mass of the falling object,
• ${\displaystyle g}$ is the acceleration due to gravity,
• ${\displaystyle C_{d}}$ is the drag coefficient,
• ${\displaystyle \rho }$ is the density of the fluid through which the object is falling, and
• ${\displaystyle A}$ is the projected area of the object.[8]

In reality, an object approaches its terminal speed asymptotically.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass ${\displaystyle m}$ has to be reduced by the displaced fluid mass ${\displaystyle \rho V}$, with ${\displaystyle V}$ the volume of the object. So instead of ${\displaystyle m}$ use the reduced mass ${\displaystyle m_{r}=m-\rho V}$ in this and subsequent formulas.

The terminal speed of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional surface area.

Air density increases with decreasing altitude, at about 1% per 80 metres (260 ft) (see barometric formula). For objects falling through the atmosphere, for every 160 metres (520 ft) of fall, the terminal speed decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal speed.

Derivation for terminal velocity

Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

$${\displaystyle F_{\text{net}}=ma=mg-{\frac {1}{2}}\rho v^{2}AC_{d},}$$

with v(t) the velocity of the object as a function of time t.

At equilibrium, the net force is zero (F_net = 0)[9] and the velocity becomes the terminal velocity lim_t→∞ v(t) = V_t:

$${\displaystyle mg-{1 \over 2}\rho V_{t}^{2}AC_{d}=0.}$$

Solving for V_t yields

${\displaystyle V_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}.}$

(5)

Derivation of the solution for the velocity v as a function of time t

The drag equation is—assuming ρ, g and Cd to be constants: $${\displaystyle ma=m{\frac {\mathrm {d} v}{\mathrm {d} t}}=mg-{\frac {1}{2}}\rho v^{2}AC_{d}.}$$ Although this is a Riccati equation that can be solved by reduction to a second-order linear differential equation, it is easier to separate variables. A more practical form of this equation can be obtained by making the substitution α2 = ρACd/2mg. Dividing both sides by m gives $${\displaystyle {\frac {\mathrm {d} v}{\mathrm {d} t}}=g\left(1-\alpha ^{2}v^{2}\right).}$$ The equation can be re-arranged into $${\displaystyle \mathrm {d} t={\frac {\mathrm {d} v}{g(1-\alpha ^{2}v^{2})}}.}$$ Taking the integral of both sides yields $${\displaystyle \int _{0}^{t}{\mathrm {d} t'}={1 \over g}\int _{0}^{v}{\frac {\mathrm {d} v'}{1-\alpha ^{2}v^{\prime 2}}}.}$$ After integration, this becomes $${\displaystyle t-0={1 \over g}\left[{\ln(1+\alpha v') \over 2\alpha }-{\frac {\ln(1-\alpha v')}{2\alpha }}+C\right]_{v'=0}^{v'=v}={1 \over g}\left[{\ln {\frac {1+\alpha v'}{1-\alpha v'}} \over 2\alpha }+C\right]_{v'=0}^{v'=v}}$$ or in a simpler form $${\displaystyle t={1 \over 2\alpha g}\ln {\frac {1+\alpha v}{1-\alpha v}}={\frac {\mathrm {artanh} (\alpha v)}{\alpha g}},}$$ with artanh the inverse hyperbolic tangent function. Alternatively, $${\displaystyle {\frac {1}{\alpha }}\tanh(\alpha gt)=v,}$$ with tanh the hyperbolic tangent function. Assuming that g is positive (which it was defined to be), and substituting α back in, the speed v becomes $${\displaystyle v={\sqrt {\frac {2mg}{\rho AC_{d}}}}\tanh \left(t{\sqrt {\frac {g\rho AC_{d}}{2m}}}\right).}$$ Using the formula for terminal velocity $${\displaystyle V_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}}$$ the equation can be rewritten as $${\displaystyle v=V_{t}\tanh \left(t{\frac {g}{V_{t}}}\right).}$$ As time tends to infinity (t → ∞), the hyperbolic tangent tends to 1, resulting in the terminal speed $${\displaystyle V_{t}=\lim _{t\to \infty }v(t)={\sqrt {\frac {2mg}{\rho AC_{d}}}}.}$$

Terminal speed in a creeping flow

Creeping flow past a sphere: streamlines, drag force F_d and force by gravity F_g

For very slow motion of the fluid, the inertia forces of the fluid are negligible (assumption of massless fluid) in comparison to other forces. Such flows are called creeping or Stokes flows and the condition to be satisfied for the flows to be creeping flows is the Reynolds number, ${\displaystyle Re\ll 1}$. The equation of motion for creeping flow (simplified Navier–Stokes equation) is given by

$${\displaystyle {\mathbf {\nabla } }p=\mu \nabla ^{2}{\mathbf {v} }}$$

where

• ${\displaystyle \mathbf {v} }$ is the fluid velocity vector field,
• ${\displaystyle p}$ is the fluid pressure field,
• ${\displaystyle \mu }$ is the liquid/fluid viscosity.

The analytical solution for the creeping flow around a sphere was first given by Stokes in 1851.[10] From Stokes' solution, the drag force acting on the sphere of diameter ${\displaystyle d}$ can be obtained as

${\displaystyle D=3\pi \mu dV\qquad {\text{or}}\qquad C_{d}={\frac {24}{Re}}}$

(6)

where the Reynolds number, ${\displaystyle Re={\frac {\rho d}{\mu }}V}$. The expression for the drag force given by equation (6) is called Stokes' law.

When the value of ${\displaystyle C_{d}}$ is substituted in the equation (5), we obtain the expression for terminal speed of a spherical object moving under creeping flow conditions:[11]

$${\displaystyle V_{t}={\frac {gd^{2}}{18\mu }}\left(\rho _{s}-\rho \right),}$$

where ${\displaystyle \rho _{s}}$ is the density of the object.

Settling velocity W_s of a sand grain (diameter d, density 2650 kg/m³) in water at 20 °C, computed with the formula of Soulsby (1997).

When the buoyancy effects are taken into account, an object falling through a fluid under its own weight can reach a terminal velocity (settling velocity) if the net force acting on the object becomes zero. When the terminal velocity is reached the weight of the object is exactly balanced by the upward buoyancy force and drag force. That is

${\displaystyle W=F_{b}+D}$

(1)

where

• ${\displaystyle W}$ is the weight of the object,
• ${\displaystyle F_{b}}$ is the buoyancy force acting on the object, and
• ${\displaystyle D}$ is the drag force acting on the object.

If the falling object is spherical in shape, the expression for the three forces are given below:

${\displaystyle W={\frac {\pi }{6}}d^{3}\rho _{s}g,}$

(2)

${\displaystyle F_{b}={\frac {\pi }{6}}d^{3}\rho g,}$

(3)

${\displaystyle D=C_{d}{\frac {1}{2}}\rho V^{2}A,}$

(4)

where

• ${\displaystyle d}$ is the diameter of the spherical object,
• ${\displaystyle g}$ is the gravitational acceleration,
• ${\displaystyle \rho }$ is the density of the fluid,
• ${\displaystyle \rho _{s}}$ is the density of the object,
• ${\displaystyle A={\frac {1}{4}}\pi d^{2}}$ is the projected area of the sphere,
• ${\displaystyle C_{d}}$ is the drag coefficient, and
• ${\displaystyle V}$ is the characteristic velocity (taken as terminal velocity, ${\displaystyle V_{t}}$).

Substitution of equations (2–4) in equation (1) and solving for terminal velocity, ${\displaystyle V_{t}}$ to yield the following expression

${\displaystyle V_{t}={\sqrt {{\frac {4gd}{3C_{d}}}\left({\frac {\rho _{s}-\rho }{\rho }}\right)}}.}$

(5)

In equation (1), it is assumed that the object is denser than the fluid. If not, the sign of the drag force should be made negative since the object will be moving upwards, against gravity. Examples are bubbles formed at the bottom of a champagne glass and helium balloons. The terminal velocity in such cases will have a negative value, corresponding to the rate of rising up.

• Stokes's law
• Terminal ballistics

• Terminal Velocity - NASA site
• Onboard video of Space Shuttle Solid Rocket Boosters rapidly decelerating to terminal velocity on entry to the thicker atmosphere, from 2,900 miles per hour (Mach 3.8) at 5:15 in the video, to 220  mph at 6:45 when the parachutes are deployed 90 seconds later—NASA video and sound, @ io9.com.
• Terminal settling velocity of a sphere at all realistic Reynolds Numbers, by Heywood Tables approach.