Terminal velocity

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In fluid dynamics, terminal velocity is the velocity at which the fluid resistance force (drag force) of a falling object equals the weight of the object minus the acting force due to fluid, which halts acceleration and causes speed to remain constant.

Terminal velocity with negligible buoyancy force

As the object accelerates (usually downwards movement due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will be equal to the downward force, mostly the weight $(mg)$, of the object. Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity. An object moving downward at greater than terminal velocity (for example because it was affected by a force downward or it fell from a thinner part of the atmosphere or it changed shape) will slow until it reaches the terminal velocity.

For example, the terminal velocity of a skydiver in a free-fall position with a semi-closed parachute is about 195 km/h (120 mph or 55m/s). This velocity is the asymptotic limiting value of the acceleration process, since the effective forces on the body more and more closely balance each other 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 limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 89 m/s), which is also the maximum speed of the peregrine falcon diving on its prey. Competition speed skydivers fly in the head down position reaching even higher speeds. The current world record is 614 mph (988 km/h) by Joseph Kittinger, set at high altitude where the lesser density of the atmosphere decreased drag.

An object falling on Earth will fall 9.80 meters per second faster every second (9.8 m/s²). The reason an object reaches a terminal velocity is that the drag force resisting motion is directly proportional to the square of its speed. At low speeds, the drag is much less than the gravitational force and so the object accelerates. As it accelerates, the drag increases, until it equals the weight. Drag also depends on the projected area. This is why things with a large projected area, such as parachutes, have a lower terminal velocity than small objects such as cannon balls.

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

$V\_t=\; sqrt\{frac\{2mg\}\{rho\; A\; C\_d\; \}\}$    ^{(see derivation)}

where

$V\_t$ = terminal velocity,

$m$ = mass of the falling object,

$g$ = gravitational acceleration,

$C\_d$ = drag coefficient,

$rho$ = density of the fluid the object is falling through, and

$A$ = projected area of the object.

On Earth, the terminal velocity of an object changes due to the properties of the fluid, mass and the projected area of the object.

This equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object.

Density increases with decreasing altitude, ca. 1% per 80 m (see barometric formula). Therefore, for every 160 m 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.

Velocity of a falling object after a given time

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),

$F\_\{net\}=mmathbf\{a\}=m\; g\; -\; \{1\; over\; 2\}\; rho\; v^2\; A\; C\_d$.

To make the derivation of the formula more practical a substitution is made with $k=\{1\; over\; 2\}\; rho\; A\; C\_d$.

Dividing through by $m$ gives,

$mathbf\{a\}=g\; -\; frac\{kv^2\}\{m\}$

where $a$ = the acceleration of the object.

Since

$mathbf\{a\}=frac\{dv\}\{dt\}$

the following is true,

$frac\{dv\}\{dt\}=g-frac\{kv^2\}\{m\}$.

Thus,

$dt=frac\{dv\}\{g-frac\{kv^2\}\{m\}\}$.

Taking the integral of both sides yields,

$int\_0^t\; \{dt\}=int\_0^v\; frac\{dv\}\{g-frac\{kv^2\}\{m\}\}$.

Factoring $g$ out of the denominator and making the substitution $a=sqrtfrac\{k\}\{mg\}$ (not to be confused with acceleration) leads to,

$int\_0^t\; \{dt\}=\{1\; over\; g\}int\_0^v\; frac\{dv\}\{1-a^2\; v^2\}$.

Factoring the bottom,

$int\_0^t\; \{dt\}=\{1\; over\; g\}int\_0^v\; frac\{dv\}\{(1+av)(1-av)\}$.

Partial fraction expansions tell us that,

$frac\{1\}\{(1+av)(1-av)\}=frac\{A\}\{1+av\}+frac\{B\}\{1-av\}$.

Multiplying by the denominator on the bottom gives,

$qquad\; 1=A(1-av)+B(1+av)$.

If $v=0$, then

$qquad\; A+B=1$.

Now let us suppose that $av=1$, then

$qquad\; 1=A(1-1)+B(1+1)=2B$,

and

$qquad\; B=\{1\; over\; 2\}$.

Therefore, for the sum of $A$ and $B$ to be 1,

$qquad\; A=\{1\; over\; 2\}$.

With these new expanded fractions,

$int\_0^t\; \{dt\}=\{1\; over\; g\}\; int\_0^v\; frac\{1/2\}\{1+av\}dv\; +\; \{1\; over\; g\}\; int\_0^v\; frac\{1/2\}\{1-av\}dv$

We now want to evaluate both integrals on the left side. The first one is,

$int\; frac\{1/2\}\{1+av\}dv$.

Making a substitution

$qquad\; u=1+av$,

we find that,

$qquad\; \{du\; over\; a\}=dv$,

and thus,

$int\; frac\{1/2\}\{1+av\}dv=\{1\; over\; 2a\}int\; frac\{1\}\{u\}du=\{ln\; u\; over\; 2a\}+C=\{ln(1+av)\; over\; 2a\}+C$.

Now we want to evaluate the second part of the left hand side,

$int\; frac\{1/2\}\{1-av\}$.

Making a substitution

$qquad\; u=1-av$,

we find that

$qquad\; -frac\{du\}\{a\}=dv$,

and thus,

$int\; frac\{1/2\}\{1-av\}=-frac\{1\}\{2a\}\; int\; \{1\; over\; u\}\; du=-frac\{ln\; u\}\{2a\}+C=-frac\{ln(1-av)\}\{2a\}+C$.

Putting these back into our original formula, we get,

$t-0=\{1\; over\; g\}left[\{ln(1+av)\; over\; 2a\}-frac\{ln(1-av)\}\{2a\}\; right]\_\{v=0\}^\{v=v\}+C=\{1\; over\; g\}\; left[\{ln\; frac\{1+av\}\{1-av\}\; over\; 2a\}\; right]\_\{v=0\}^\{v=v\}+C$,

and thus,

$t=\{1\; over\; 2ag\}\; ln\; frac\{1+av\}\{1-av\}-0+C$.

Since $v=0$ when $t=0$, we find that $C=0$. We also have a specially defined function, the inverse hyperbolic tangent that is defined such that,

$frac\{1\}\{2\}\; ln\; frac\{1+av\}\{1-av\}=operatorname\{artanh\}(av)$.

So,

$t=frac\{operatorname\{artanh\}(av)\}\{ag\}$.

Thus,

$frac\{1\}\{a\}tanh(tag)=v$,

and thus, substituting a back in,

$v=sqrt\{frac\{mg\}\{k\}\}\; tanh\; left(sqrt\{frac\{k\}\{mg\}\}gtright)$

substituting and simplifying, assuming that g is positive (which it was defined to be), we arrive at the final equation,

$v=sqrtfrac\{2mg\}\{rho\; A\; C\_d\}\; tanh\; left(t\; sqrt\{frac\{g\; rho\; A\; C\_d\; \}\{2m\}\}right)$.

Terminal velocity in the presence of buoyancy force

When the buoyancy effects are taken into account, an object falling through a fluid under it's 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 give below:

$quad\; (2)\; qquad\; W\; =\; frac\{pi\}\{6\}\; d^3\; rho\_s\; g$

$quad\; (3)\; qquad\; F\_b\; =\; frac\{pi\}\{6\}\; d^3\; rho\; g$

$quad\; (4)\; qquad\; D\; =\; C\_d\; frac\{1\}\{2\}rho\; V^2\; A$

where

$d\; =$ diameter of the spherical object

$g\; =$ gravitational acceleration,

$rho\; =$ density of the fluid,

$rho\_s\; =$ density of the object,

$A\; =\; pi\; d^2/4\; =$ projected area of the sphere,

$C\_d\; =$ drag coefficient, and

$V\; =$ 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

$quad\; (5)\; qquad\; V\_t\; =\; sqrt\{frac\{4\; g\; d\}\{3\; C\_d\}\; left(frac\{rho\_s\; -\; rho\}\{rho\}\; right)\}$.

Terminal velocity in creeping flow

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 flow 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

$\{mathbf\; v\}$ = velocity vector field

$p$ = pressure field

$mu$ = 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

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

where the Reynolds number, $Re\; =\; frac\{rho\; d\; V\}\{mu\}$. 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 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 high viscous fluids.

References

External links

• Terminal Velocity - NASA site

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