Terminal velocity

For other uses, see Terminal velocity (disambiguation).

Terminal velocity occurs when the downward force of gravity (F_g) equals the air resistance/force of drag (F_d) and buoyancy, such that velocity becomes asymptotic to a constant value with net acceleration = 0

The downward force of gravity (F_g) equals the restraining force of drag (F_d). The net force on the object then, is zero, and the result is that the velocity of the object remains constant.

Terminal velocity is the highest velocity attainable by an object as it falls through air. It occurs once the sum of the drag force (F_d) and buoyancy equals the downward force of gravity (F_G) acting on the object. Since the net force on the object is zero, the object has zero acceleration.

In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving.

As the speed of an object increases, the drag force acting on the object, resultant of the substance (e.g., air or water) it is passing through, increases. At some speed, the drag or force of resistance will equal the gravitational pull on the object (buoyancy is considered below). At this point the object ceases to accelerate and continues falling at a constant speed called terminal velocity (also called settling velocity). An object moving downward with greater than terminal velocity (for example because it was thrown downwards or it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity. Drag depends on the projected area, and this is why objects with a large projected area relative to mass, such as parachutes, have a lower terminal velocity than objects with a small projected area relative to mass, such as bullets.

Examples

Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s). This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity 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 his or her limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s), which is almost the terminal velocity of the peregrine falcon diving down on its prey. The same terminal velocity is reached for a typical .30-06 bullet dropping downwards—when it is returning to earth having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study.

Competition speed skydivers fly in a head-down position and can reach speeds of 530 km/h (330 mph); the current record is held by Felix Baumgartner who jumped from a height of 128,100 feet (39,000 m) and reached 1,342 km/h (834 mph).

Physics

Mathematically, terminal velocity—without considering buoyancy effects—is given by

$V_t= \sqrt{\frac{2mg}{\rho A C_d }}$

where

• $V_t$ is terminal velocity,
• $m$ is the mass of the falling object,
• $g$ is the acceleration due to gravity,
• $C_d$ is the drag coefficient,
• $\rho$ is the density of the fluid through which the object is falling, and
• $A$ is the projected area of the object.

In reality, an object approaches its terminal velocity 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 $m$ has to be reduced by the displaced fluid mass $\rho\mathcal{V}$, with $\mathcal{V}$ the volume of the object. So instead of $m$ use the reduced mass $m_r=m-\rho\mathcal{V}$ in this and subsequent formulas.

On Earth, the terminal velocity 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, ca. 1% per 80 metres (260 ft) (see barometric formula). For objects falling through the atmosphere, for every 160 metres (520 ft) of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.

Derivation for terminal velocity

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

$F_{net} = m a = m g - {1 \over 2} \rho v^2 A C_\mathrm{d}$

At equilibrium, the net force is zero (F = 0);

$m g - {1 \over 2} \rho v^2 A C_\mathrm{d} = 0$

Solving for v yields

$v = \sqrt\frac{2mg}{\rho A C_\mathrm{d}}$

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

The drag equation is $m a = m \frac{\mathrm{d}v}{\mathrm{d}t} = m g - \frac{1}{2} \rho v^2 A C_\mathrm{d}$ 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 k = ​1⁄2ρACd. Dividing both sides by m gives $\frac{\mathrm{d}v}{\mathrm{d}t}=g-\frac{kv^2}{m}$ The equation can be re-arranged into $dt = \frac{\mathrm{d}v}{g - \frac{kv^2}{m}}$ Taking the integral of both sides yields $\int_0^t {\mathrm{d}t^\prime} = \int_0^v \frac{\mathrm{d}v^\prime}{g-\frac{kv^{\prime 2}}{m}} = {1 \over g}\int_0^v \frac{\mathrm{d}v^\prime}{1-\alpha^2 v^{\prime 2}}$ where α = ( ​k⁄mg )​1⁄2. After integration, this becomes $t - 0 = {1 \over g}\left[{\ln(1 + \alpha v^\prime) \over 2\alpha} - \frac{\ln(1 - \alpha v^\prime)}{2\alpha} + C \right]_{v^\prime=0}^{v^\prime=v}={1 \over g} \left[{\ln \frac{1 + \alpha v^\prime}{1 - \alpha v^\prime} \over 2\alpha} + C \right]_{v^\prime=0}^{v^\prime=v}$ or in simpler a form $t = {1 \over 2\alpha g} \ln \frac{1 + \alpha v}{1 - \alpha v}$ The inverse hyperbolic tangent is defined as: $\frac{1}{2} \ln \frac{1+\alpha v}{1-\alpha v}=\mathrm{arctanh}(\alpha v)$ So the solution of the integral is $t = \frac{\mathrm{arctanh}(\alpha v)}{\alpha g}$ or alternatively, $\frac{1}{\alpha}\tanh(\alpha g t) = v$ with tanh the hyperbolic tangent function. Assuming that g is positive (which it was defined to be), and substituting α back in, the velocity v becomes $v=\sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{k}{mg}} g t\right)$ Next, after k = ​1⁄2ρACd has been substituted, the velocity v is in the desired form: $v = \sqrt\frac{2mg}{\rho A C_d} \tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right)$ As time tends to infinity ( t → ∞ ), the hyperbolic tangent tends to 1, resulting in the terminal velocity $v = \sqrt\frac{2mg}{\rho A C_d}$

Terminal velocity in 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 flows and the condition to be satisfied for the flows to be creeping flows is the Reynolds number, $Re \ll 1$. The equation of motion for creeping flow (simplified Navier-Stokes equation) is given by

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

where

• ${\mathbf v}$ is the fluid velocity vector field
• $p$ is the fluid pressure field
• $\mu$ is the fluid viscosity

The analytical solution for the creeping flow around a sphere was first given by Stokes in 1851. From Stokes' solution, the drag force acting on the sphere can be obtained as

$\quad (6) \qquad D = 3\pi \mu d V \qquad \qquad \text{or} \qquad \qquad C_d = \frac{24}{Re}$

where the Reynolds number, $Re = \frac{1}{\mu} \rho d V$. The expression for the drag force given by equation (6) is called Stokes' law.

When the value of $C_d$ is substituted in the equation (5), we obtain the expression for terminal velocity of a spherical object moving under creeping flow conditions:

$V_t = \frac{g d^2}{18 \mu} \left(\rho_s - \rho \right)$

Applications

The creeping flow results can be applied in order to study the settling of sediment particles near the ocean bottom and the fall of moisture drops in the atmosphere. The principle is also applied in the falling sphere viscometer, an experimental device used to measure the viscosity of highly viscous (sticky) fluids, for example oil, parrafin, tar etc.

Finding the terminal velocity when the drag coefficient is not known

In principle one doesn't know beforehand whether to apply the creeping flow solution, or what coefficient of drag to use, because the coefficient depends on the speed. What one can do in this situation is to calculate the product of the coefficient of drag and the square of the Reynolds number:

$C_d \mathrm{Re}^2 = \frac{mgD^2}{A\rho\nu^2}$

where ν is the kinematic viscosity, equal to μ/ρ. This product is a function of Reynolds number, and one can consult a graph of C_d versus Re to find where along the curve the product attains the correct value (a qualitative example of such a graph for spheres is found at this NASA site: [1]) From this one knows the coefficient of drag and one can then use the formula given above to find the terminal velocity.

For a spherical object, the above-mentioned product can be simplified:

$C_d \mathrm{Re}^2 = \frac{4mg}{\pi\rho\nu^2}$

We can see from this that the regime and the drag coefficient depend only on the sphere's weight and the fluid properties. There are three regimes: creeping flow, intermediate-Reynolds number Newton's Law (almost constant drag coefficient), and a high-Reynolds number regime. In the latter regime the boundary layer is everywhere turbulent (see Golf ball#Aerodynamics). These regimes are given in the following table. The weight range for each regime is given for water and air at 1 atm pressure and 25 °C. Note though that the weight (given in terms of mass in standard gravity) is the weight in the fluid, which is less than the mass times the local gravity because of buoyancy.

Regime	Range of Reynolds number	Range of CdRe2	Range of weight in water	Range of weight in air
Creeping flow	Quite accurate up to 0.3	Up to 7.2	Up to 0.00058 mgf (5.7 nN)	Up to 0.00017 mgf (1.7 nN)
Cd between 0.4 and 0.5	1000 to 200000	500000 to 2×1010	40 mgf (0.39 mN) to 1.6 kgf (16 N)	11 mgf (0.11 mN) to 470 gf (4.6 N)
Cd between 0.1 and 0.2	Over 400000	Over 1.6×1010	Over 1.3 kgf (13 N)	Over 375 gf (3.68 N)

Between the first two regimes there is a smooth transition. But notice that there is overlap between the ranges of C_dRe² for the last two regimes. Spheres in this weight range have two stable terminal velocities. A rough surface, such as of a dimpled golf ball, allows transition to the lower drag coefficient at a lower Reynolds number.

Terminal velocity in the presence of buoyancy force

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

$\quad (1) \qquad W = F_b + D$

where

• $W$ = weight of the object,
• $F_b$ = buoyancy force acting on the object, and
• $D$ = drag force acting on the object.

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

$\begin{align} \quad &(2) \qquad& W &= \frac{\pi}{6} d^3 \rho_s g \\ \quad &(3) \qquad& F_b &= \frac{\pi}{6} d^3 \rho g \\ \quad &(4) \qquad& D &= C_d \frac{1}{2} \rho V^2 A \end{align}$

where

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

Substitution of equations (2–4) in equation (1) and solving for terminal velocity, $V_t$ to yield the following expression

$\quad (5) \qquad V_t = \sqrt{\frac{4 g d}{3 C_d} \left( \frac{\rho_s - \rho}{\rho} \right)}$

See also

• Stokes' law
• Free fall
• Terminal ballistics

External links

• 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.