Drag and Terminal Velocity

Beginner

Gravity keeps pulling, yet a falling skydiver stops speeding up. Let's see why, one step at a time.

Step 1: fluids push back. Hold your hand out of a moving car's window and the air shoves it backwards. Air, water and every other fluid push against anything moving through them. This push is called drag, or air resistance. It grows quickly with speed (twice as fast means about four times the push) and with the area facing the flow.

Step 2: a falling body stops speeding up. Watch the skydiver. Just after the jump the speed is low, so the air resistance (red arrow, up) is small and the weight (green arrow, down) wins: the skydiver speeds up and the blue speed bar climbs. But more speed means more air resistance. After about ten seconds the red arrow is as long as the green one. The forces cancel, so the speed stops changing: the skydiver falls steadily at about 200 km/h, the terminal velocity.

Step 3: the parachute. The open canopy has a huge area, so the air resistance suddenly becomes much bigger than the weight and the skydiver slows down. As the speed drops, so does the push, until it balances the weight again at a gentle 18 km/h.

Step 4: shape and stickiness. A smooth, pointed shape lets the fluid close in behind it, so it feels little drag: think of fish and racing cyclists. Thick liquids such as honey resist even slow motion; this inner stickiness is called viscosity.

Check yourself: a raindrop falls at a steady speed. Is the air resistance on it smaller than, equal to or bigger than its weight? (Answer: equal. A steady speed means the forces balance.)

Intermediate

What drag is and what it depends on

A falling skydiver stops speeding up, although the weight stays the same. Drag is the force a fluid exerts on a body moving through it. It points against the body's velocity relative to the fluid: a cyclist riding at 10 m/s into a 5 m/s headwind meets the air at 15 m/s.

For balls, cars and skydivers the drag follows the quadratic law
$F_d=\frac{1}{2}\rho C_{d}A{v}^2$
Here $\rho$ is the density of the fluid (1.2 kg/m³ for air near the ground), $A$ the frontal area facing the flow and $v$ the speed. The drag coefficient $C_d$ is a pure number set by the shape: about 0.05 for a teardrop, 0.3 for a modern car, 0.47 for a sphere and 1.5 for a round parachute.

Why the square? Each second the body shoves aside a fluid mass of about $\rho Av$ and gives it a speed comparable to $v$. The momentum handed over per second, which is the force, is then about $\rho A{v}^2$: twice the speed, four times the drag.

Viscosity and the world of the small and slow

Viscosity, $\eta$, is a fluid's internal friction, the resistance of its layers to sliding past one another, measured in pascal seconds (Pa·s). At 20 °C air has $1.8\times 10^{-5}\mathrm{Pa}\cdot \mathrm{s}$, water $1.0\times 10^{-3}\mathrm{Pa}\cdot \mathrm{s}$, olive oil about 0.08 Pa·s and glycerol 1.4 Pa·s; warm liquids are runnier.

Tiny or slow objects, such as dust, fog droplets or a bead sinking in honey, feel a drag proportional to $v$ instead. For a sphere of radius $r$ this is Stokes' law,
$F_d=6\pi \eta rv\text{(Stokes, }\mathrm{Re}\ll 1\text{)}$
The density does not appear: the fluid is not flung aside but sheared, layer against layer.

Which law applies is decided by the Reynolds number, built from the fluid's density and viscosity, the speed and the size $L$ of the body:
$\mathrm{Re}=\frac{\rho vL}{\eta }$
It compares the fluid's inertia with its viscosity. Below about 1, Stokes' law holds; above about 1000, the quadratic law holds, with a $C_d$ set mainly by the shape. A fog droplet has $\mathrm{Re}\approx 0.02$; the skydiver, with $L\approx 0.5\mathrm{m}$, about 2 million.

Falling with drag: terminal velocity

Take down as positive. A falling body feels its weight $mg$ downward and the drag upward, so Newton's second law reads
$ma=mg-F_d$
At the start $v=0$, there is no drag and $a=g$. As the body speeds up, the drag grows and the acceleration shrinks, until the drag equals the weight. From then on $a=0$ and the body falls at a constant speed, the terminal velocity. Setting $\frac{1}{2}\rho C_{d}A{v}_t^2=mg$ gives
$v_t=\sqrt{\frac{2mg}{\rho C_{d}A}}$

For a small sphere in the Stokes regime, the weight minus the buoyant force (see Buoyancy and Archimedes' Principle) is balanced by $6\pi \eta rv$, which gives
$v_t=\frac{2r^2({\rho }_s-{\rho }_f)g}{9\eta }\text{(Stokes sphere)}$
where ${\rho }_s$ and ${\rho }_f$ are the densities of the sphere and of the fluid. In air, buoyancy is only about 0.1 % of a skydiver's weight, and we ignore it. A body of the same shape and material but twice as large falls $\sqrt{2}\approx 1.41$ times faster with quadratic drag ($m\propto r^3$, $A\propto r^2$), which helps small animals survive falls, but four times faster in the Stokes regime ($v_t\propto r^2$).

Worked example: the skydiver in the animation

The jumper has $m=80\mathrm{kg}$ and, belly-down, $C_{d}A=0.43{\mathrm{m}}^2$, in air with $\rho =1.2kg/{\mathrm{m}}^3$. The weight is $mg=80\cdot 9.81=785\mathrm{N}$, the green line among the readouts, and the terminal velocity is
$v_t=\sqrt{\frac{2\cdot 785}{1.2\cdot 0.43}}\approx \sqrt{3040}\approx 55.2m/s$
about 199 km/h.

Since $mg=\frac{1}{2}\rho C_{d}A{v}_t^2$, the drag at any speed is $F_d=mg(v/v_t{)}^2$. When the $v$ readout shows 30 m/s, at $t\approx 3.4\mathrm{s}$, $F_d=\frac{1}{2}\cdot 1.2\cdot 0.43\cdot 30^2\approx 232\mathrm{N}$ and $a=(785-232)/80\approx 6.9m/{\mathrm{s}}^2$. The approach is slow: at 13 s, when the canopy opens, the readouts show $v\approx 54.1m/s$, $F_d\approx 755\mathrm{N}$ and $a\approx 0.4m/{\mathrm{s}}^2$. The $t$ readout counts real seconds; the animation runs at double speed.

The canopy has $C_{d}A=52.5{\mathrm{m}}^2$, 122 times more, so $v_t$ falls by a factor $\sqrt{122}\approx 11$, to $55.2/11.05\approx 5.0m/s$ (18 km/h), as the parachute line under the formulas says. While the canopy inflates, the drag peaks near 3340 N, 4.3 times the weight, at $t\approx 14\mathrm{s}$, and the $a$ readout dips to about $-32m/{\mathrm{s}}^2$, a deceleration of 3.3 g. By about 19 s the readouts settle at $v=5.0m/s$, $F_d=785\mathrm{N}$ and $a=0$.

Worked example: fog droplets and raindrops

Why do clouds float while rain falls? A fog droplet of radius $r=10\mu \mathrm{m}$ falls through air ($\eta =1.8\times 10^{-5}\mathrm{Pa}\cdot \mathrm{s}$), with ${\rho }_s=1000kg/{\mathrm{m}}^3$ and ${\rho }_f=1.2kg/{\mathrm{m}}^3$. The Stokes formula gives
$v_t=\frac{2\cdot (10^{-5}{)}^2\cdot 998.8\cdot 9.81}{9\cdot 1.8\times 10^{-5}}\approx 0.012m/s$
just 1.2 cm/s. Check the regime: $\mathrm{Re}=1.2\cdot 0.012\cdot 2\times 10^{-5}/(1.8\times 10^{-5})\approx 0.016$, far below 1, so Stokes' law was the right choice. An updraft of a few centimetres per second keeps such droplets aloft.

A raindrop of radius 1 mm is 100 times larger. Stokes' law would give $10^4$ times more, 121 m/s, at a Reynolds number of about 16 000, far outside its range. With the quadratic law instead, $C_d\approx 0.5$, $m={\rho }_s\cdot \frac{4}{3}\pi r^3=4.19\times 10^{-6}\mathrm{kg}$ and $A=\pi r^2=3.14\times 10^{-6}{\mathrm{m}}^2$:
$v_t=\sqrt{\frac{2\cdot 4.19\times 10^{-6}\cdot 9.81}{1.2\cdot 0.5\cdot 3.14\times 10^{-6}}}\approx 6.6m/s$
close to the 6.5 m/s measured for drops 2 mm across, at $\mathrm{Re}\approx 900$. Find the regime first, then choose the law.

Reading the force arrows

The two arrows share one scale in pixels per newton. The green $mg$ arrow never changes; the red $F_d$ arrow goes through four phases.

• Speeding up: red shorter than green, so the net force points down and $a$ shrinks from $g$ towards zero.
• Terminal velocity: both arrows equally long, zero net force, a steady speed near 55 m/s.
• Opening: red about 4.3 times longer than green, so the net force points up and $a$ turns negative while the jumper still falls.
• Balanced again: red back to the length of green, at 5.0 m/s, with the grey air streaks barely creeping past.

Common mistakes

• Balanced forces do not mean rest: at terminal velocity the net force is zero, yet the jumper still falls at 55 m/s.
• An opening parachute does not pull the jumper up, whatever videos from a still-falling camera flyer suggest: the velocity stays downward, dropping from 54 to 5 m/s, while the acceleration points up, about $-32m/{\mathrm{s}}^2$ at its peak.
• Everyday drag is not proportional to speed: at half the terminal speed, 27.6 m/s, the drag on the jumper is a quarter of the weight, 196 N, not half.
• Heavier bodies do fall faster in air, but only as $\sqrt{m}$: 40 kg of extra gear with the same $C_{d}A$ raises $v_t$ by a factor $\sqrt{1.5}\approx 1.22$, to 67.5 m/s.
• Viscosity is not density: olive oil floats on water, yet it is about 80 times more viscous.

Where you meet it

Drag sets the energy bill of every vehicle. Moving steadily through still air takes the power
$P=F_{d}v=\frac{1}{2}\rho C_{d}A{v}^3$
For a car with $C_{d}A=0.66{\mathrm{m}}^2$ at 90 km/h (25 m/s) this is $\frac{1}{2}\cdot 1.2\cdot 0.66\cdot 25^3\approx 6.2\mathrm{kW}$; at 130 km/h it is 18.6 kW, three times as much, and each kilometre costs about twice the energy.

Cyclists crouch to cut $C_{d}A$, golf-ball dimples reduce drag and parachutes exploit it, while centrifuges speed up the slow Stokes settling of blood cells. Bernoulli's Principle explains the pressure $\frac{1}{2}\rho v^2$ behind the drag law, Damping and Resonance treats drag that slows an oscillation, and Work, Energy and Power follows where the energy goes.

Advanced

Viscosity: momentum flowing between layers

Viscosity is a fluid's resistance to having its layers slide over one another. Every drag force in this lesson comes from it, directly or indirectly, so we begin with the simplest flow that measures it.

Legend

$d$	Width of the gap between the two plates.	$\mathrm{m}$
$A$	Area of the upper plate.	${\mathrm{m}}^2$
$F$	Force that pulls the upper plate along at a steady speed.	$\mathrm{N}$
$v_0$	Speed of the upper plate; the lower plate is at rest.	$m/s$
$v(y)$	Speed of the fluid at height $y$ above the lower plate: zero at the bottom, $v_0$ at the top.	$m/s$
$\eta$	Viscosity of the fluid: the stress needed per unit velocity gradient.	$\mathrm{Pa}\mathrm{s}$

Figure 1 shows that flow. A fluid fills the gap of width $d$ between two large parallel plates. The lower plate is held still. The upper plate, of area $A$, is pulled by a force $F$ and slides at the steady speed $v_0$. Each blue arrow gives the fluid's speed $v(y)$ at the height $y$ above the lower plate.

Two facts fix the lengths of the arrows. First, fluid in contact with a solid surface moves with that surface. This is the no-slip condition, and it holds for every ordinary fluid. So the arrow at the bottom has length zero and the one at the top has length $v_0$. Second, neighbouring layers pull on each other with a force per unit area proportional to how quickly the speed changes with height, $\eta dv/dy$. This is what defines the viscosity $\eta$. In steady flow no layer speeds up, so the pull from the faster layer above it must balance the drag from the slower layer below: the stress, and with it $dv/dy$, is the same at every height.

That is why the arrows grow in equal steps and the dashed line through their tips is straight. The speed rises by $v_0$ across the gap $d$, so $dv/dy=v_0/d$ everywhere, and the force needed per unit area of the plate is
$\frac{F}{A}=\eta \frac{v_0}{d}$
Doubling the plate's speed, or halving the gap, doubles the force. Since $F/A$ is a stress, $\eta$ is measured in Pa·s.

The same law reads more deeply as a flow of momentum. Each layer pulls the slower layer below it forward and is held back by it in return. Momentum along the flow therefore passes from fast fluid to slow, at a rate set by $\eta$. In Figure 1 it enters at the moving plate, crosses every layer downwards and leaves through the still plate; $F$ is the rate, in momentum per second, at which the upper plate feeds it in.

This picture explains how viscosity depends on the state of the fluid. In a gas the momentum is carried by molecules hopping between layers: a molecule arriving from a faster layer brings extra forward momentum into a slower one. Kinetic theory gives $\eta \approx \frac{1}{3}\rho \bar{v}\lambda$. Here $\rho$ is the gas density, $\bar{v}$ the mean molecular speed and $\lambda$ the mean free path, the average distance a molecule travels between collisions. Because $\lambda \propto 1/\rho$, the product $\rho \lambda$ stays the same when the gas is compressed, so gas viscosity is independent of pressure. It rises with temperature, because $\bar{v}$ does.

Liquids behave the opposite way: their molecules are held by their neighbours, heat helps them break free, and the viscosity falls steeply on heating. Water has 1.79 mPa·s at 0 °C but only 0.28 mPa·s at 100 °C.

The viscous force also completes the equation of motion of a fluid. For an ideal fluid, Newton's second law per unit volume is Euler's equation: density times acceleration equals the pressure force $-\nabla p$ plus the weight $\rho \vec{g}$, where $p$ is the pressure. In an incompressible fluid the net viscous force per unit volume is $\eta {\nabla }^2\vec{v}$. The Laplacian ${\nabla }^2$ measures how far the velocity at a point differs from the average of its surroundings, so this force pulls each parcel towards the motion of its neighbours. Adding it gives the Navier–Stokes equation:
$\rho \left(\frac{\partial \vec{v}}{\partial t}+(\vec{v}\cdot \nabla )\vec{v}\right)=-\nabla p+\rho \vec{g}+\eta {\nabla }^2\vec{v}$
The bracket is the acceleration of a fluid parcel: the change of velocity at a fixed point plus the change that comes from moving to a place where the flow is different. The equation is solved with the no-slip condition of Figure 1: at every solid surface the fluid moves with the surface, so $\vec{v}=0$ on a body at rest.

Dimensional analysis and the Reynolds number

The flow around a real body is far too complicated to calculate from the Navier–Stokes equation. Dimensional analysis shows how much can be said without solving it, and it produces the one number that classifies every flow.

For a body of a given shape, the drag $F$ can depend only on the body's size $L$, its speed $v$ through the fluid and the fluid's density $\rho$ and viscosity $\eta$. These five quantities are built from three units, the kilogram, the metre and the second, so they can be combined into only $5-3=2$ independent dimensionless numbers. A convenient pair is $F/(\rho v^2{L}^2)$ and the Reynolds number
$\mathrm{Re}=\frac{\rho vL}{\eta }$
A physical law cannot depend on the choice of units, so it must be a relation between these two numbers: the first is a function of the second. Replacing $L^2$ by the body's frontal area $A$ and adding the conventional factor $\frac{1}{2}$ gives the drag law
$F_d=\frac{1}{2}\rho v^{2}A{C}_d(\mathrm{Re})$
The drag coefficient $C_d$ is a pure number. All the hard physics now hides in one function of one variable, which has to be measured only once for each shape.

Physically, $\mathrm{Re}$ compares inertia with viscosity. The velocity changes by about $v$ over distances of about $L$, so the inertial term $\rho (\vec{v}\cdot \nabla )\vec{v}$ is about $\rho v^2/L$ and the viscous term $\eta {\nabla }^2\vec{v}$ is about $\eta v/L^2$. Their ratio is $\mathrm{Re}$. At large $\mathrm{Re}$ inertia dominates; at small $\mathrm{Re}$ viscosity does.

It follows that flows past geometrically similar bodies at equal $\mathrm{Re}$ are scaled copies of each other, which is what makes model tests work. With the kinematic viscosity $\nu =\eta /\rho$, the Reynolds number is $\mathrm{Re}=vL/\nu$; $\nu$ is $1.0\times 10^{-6}{\mathrm{m}}^2/s$ for water and $1.5\times 10^{-5}{\mathrm{m}}^2/s$ for air. A 1 : 10 model of a car has one tenth of the length, so in water it matches the real car in air when its speed is $10{\nu }_{\mathrm{water}}/{\nu }_{\mathrm{air}}\approx 0.67$ of the car's speed.

The range of $\mathrm{Re}$ in nature is enormous: about $10^{-4}$ for a swimming bacterium, 0.016 for a fog droplet, 900 for a raindrop and $2\times 10^6$ for the skydiver.

Stokes flow and the drag curve

The function $C_d(\mathrm{Re})$ has been measured for many shapes. The sphere is the standard example, and its curve shows the two very different regimes of drag.

Legend

$C_d$	Drag coefficient, defined by $F_d=\frac{1}{2}\rho v^{2}A{C}_d$.
$Re$	Reynolds number of the flow, $\rho vL/\eta$, with $L$ the sphere's diameter.
$R{e}_c$	Reynolds number of the drag crisis, about $3\times 10^5$.

Figure 2 plots the drag coefficient $C_d$ of a smooth sphere against the Reynolds number $\mathrm{Re}$, based on the sphere's diameter. Both axes are logarithmic: each factor of 10 takes the same distance, so a power law $C_d\propto {\mathrm{Re}}^n$ appears as a straight line of slope $n$. The dashed line is $C_d=24/\mathrm{Re}$, of slope $-1$.

Follow the measured curve from left to right. For $\mathrm{Re}\ll 1$ it runs along the dashed line. Near $\mathrm{Re}\approx 1$ it bends above it, and between $\mathrm{Re}\approx 10^3$ and $2\times 10^5$ it flattens into a plateau at 0.4 to 0.5. At ${\mathrm{Re}}_c\approx 3\times 10^5$ it plunges to about 0.1: this is the drag crisis. Beyond it the curve recovers part of the way, to roughly 0.2 at $\mathrm{Re}\approx 10^7$. The plateau and the plunge come from the shape of the wake, the subject of the next section.

The straight part is Stokes flow. At $\mathrm{Re}\ll 1$ the inertial term is negligible, so the density $\rho$ drops out of the problem. The drag on a sphere of radius $r$ can then depend only on $\eta$, $v$ and $r$, and the only product of these with the unit of force is $\eta rv$. Solving the slow-flow equations fixes the numerical factor:
$F_d=6\pi \eta rv$
Dividing by $\frac{1}{2}\rho v^2\pi r^2$ gives $C_d=12\eta /(\rho vr)=24\eta /(\rho v\cdot 2r)=24/\mathrm{Re}$, which is the dashed line.

A small sphere sinking through a liquid reaches its terminal velocity when this drag balances its weight minus the buoyancy, $\frac{4}{3}\pi r^3({\rho }_s-{\rho }_f)g$, where ${\rho }_s$ is the density of the sphere and ${\rho }_f$ that of the fluid. Setting the two equal gives
$v_t=\frac{2r^2({\rho }_s-{\rho }_f)g}{9\eta }$
proportional to $r^2$: a sphere twice as wide sinks four times as fast. Timing a sinking ball and solving for $\eta$ is a standard way to measure the viscosity of oils.

Separation, the wake and the drag crisis

Why does $C_d$ stop falling and level off, and why does it suddenly drop near ${\mathrm{Re}}_c$? The answer lies in a thin layer of fluid next to the body's surface.

Legend

$v$	Speed of the stream far upstream.	$m/s$
$r$	Radius of the sphere.	$\mathrm{m}$
$\delta$	Thickness of the boundary layer, across which the speed rises from zero at the surface to that of the stream. Drawn far too thick.	$\mathrm{m}$
$S_1$	Point where the laminar boundary layer separates, about 80° from the front.
$S_2$	Point where the turbulent boundary layer separates, farther back, about 120° from the front.
$w_1$	Width of the wake behind the sphere with a laminar boundary layer.	$\mathrm{m}$
$w_2$	Width of the wake with a turbulent boundary layer.	$\mathrm{m}$

Figure 3 shows the flow past a sphere of radius $r$ in a stream of speed $v$. At the surface the fluid is at rest (no slip). The thin layer in which its speed climbs to that of the stream is the boundary layer, of thickness $\delta$; it is drawn far thicker than it really is. Over the front half of the sphere the pressure falls, which pushes the layer along. Behind the widest point the pressure rises again, and the slow fluid near the wall does not have the momentum to move against it.

In the upper picture the boundary layer is laminar, meaning smooth and orderly. It runs out of momentum early and separates at $S_1$, about 80° round from the front: it leaves the surface along the tangent as a free shear layer. Behind it lies a broad wake of slow, swirling, low-pressure fluid, of width $w_1$. In the lower picture the boundary layer is turbulent. Its eddies keep stirring fast fluid from outside down towards the wall, so it clings to the surface longer and separates farther back, at $S_2$, about 120° from the front. The wake, of width $w_2$, is much narrower.

Now connect this to Figure 2. Above $\mathrm{Re}\approx 10^3$ the flow is of the upper kind: it cannot follow the rear of the body. The front feels nearly the full stagnation pressure $\frac{1}{2}\rho v^2$ while the wide wake behind sits at low pressure. This difference, the pressure drag, does not depend on $\eta$, so $C_d$ levels off: that is the plateau. Near ${\mathrm{Re}}_c$ the boundary layer becomes turbulent before it can separate. The flow switches to the lower kind, the wake narrows, the pressure behind recovers partly, and $C_d$ drops to about 0.1: that is the drag crisis.

Golf-ball dimples make the boundary layer turbulent much earlier, triggering the crisis near $\mathrm{Re}\approx 5\times 10^4$. A drive covers $\mathrm{Re}\approx 10^5$ to $2\times 10^5$, so over the whole flight the dimples roughly halve the drag. Bodies with sharp edges, such as a flat plate, a spread-eagled jumper or a canopy, behave differently: the flow separates at their edges at any high $\mathrm{Re}$, so their $C_d$ stays nearly constant. That is why the animation can use one value of $C_{d}A$ for each phase of the jump. The $v^2$ law summarises a complicated flow over a limited range of $\mathrm{Re}$; it is not a law of nature like $F=ma$.

The boundary layer also resolves an old puzzle. An ideal fluid, with $\eta =0$, flowing steadily past a body closes up perfectly behind it and exerts no drag at all: d'Alembert's paradox of 1752. One might expect real flows to approach this as $\eta \to 0$, but they do not.

In the Navier–Stokes equation $\eta$ multiplies the term with the highest derivatives. Dropping that term lowers the order of the equation, and the remaining equation cannot satisfy the no-slip condition. However small $\eta$ is, the change of velocity therefore crowds into a boundary layer, the $\delta$ of Figure 3, where the gradients are steep enough for viscosity to matter. Its thickness shrinks only as $L/\sqrt{\mathrm{Re}}$; for the skydiver it is a fraction of a millimetre. The layer still separates, the wake survives and the drag stays finite. Zero viscosity is a singular limit, as Prandtl's boundary-layer theory of 1904 showed.

Integrating the fall: the hyperbolic tangent

The skydiver of the animation (mass 80 kg, belly-down) falls at Reynolds numbers near $2\times 10^6$, where $C_d$ is constant and the drag grows as $v^2$. This section solves the motion exactly.

Write $k=\frac{1}{2}\rho C_{d}A$, so that the drag is $k{v}^2$, and take down as positive. The weight $mg$ acts down and the drag acts up, so Newton's second law reads $mdv/dt=mg-k{v}^2$. The jumper stops accelerating when the two forces balance, at the terminal velocity $v_t$ with $v_t^2=mg/k$. Dividing by $m$ and using this definition gives $dv/dt=g(1-v^2/v_t^2)$. The speed and the time can now be put on opposite sides and integrated from the jump, when $t=0$ and $v=0$:
${\int }_0^v\frac{du}{1-u^2/v_t^2}=v_t\mathrm{artanh}(v/v_t)=gt$
Solving for $v$ gives
$v(t)=v_t\tanh \left(\frac{gt}{v_t}\right)$
Since the integral of $\tanh$ is $\ln \cosh$, one more integration gives the distance fallen, $x(t)=(v_t^2/g)\ln \cosh (gt/v_t)$.

The natural time scale is $\tau =v_t/g$, the time free fall would need to reach $v_t$. For the animation's jumper, with $v_t=55.2m/s$, it is $\tau =5.6\mathrm{s}$. For $t\ll \tau$, $\tanh s\approx s$, so $v\approx gt$: the first seconds are almost free fall. For $t\gg \tau$, $\tanh s\approx 1-2e^{-2s}$ and $\ln \cosh s\approx s-\ln 2$, so
$v\approx v_t(1-2e^{-2t/\tau }),x\approx v_{t}t-(v_t^2/g)\ln 2$
The jumper ends up a fixed 215 m behind an imaginary companion who had fallen at $v_t$ from the start.

Compare linear drag, $mdv/dt=mg-bv$, which gives $v=v_t(1-e^{-t/\tau })$ with again $\tau =v_t/g$, here equal to $m/b$. The start is the same, with slope $g$, but the final approach is slower: the gap to $v_t$ shrinks at the rate $1/\tau$ instead of $2/\tau$. At terminal velocity nothing speeds up, yet gravity is still doing work at the rate $mg{v}_t=785\cdot 55.2\approx 43\mathrm{kW}$. All of it goes into stirring the air, whose eddies finally turn it into heat.

Reading the graphs: approach and opening shock

The formulas become concrete on the animation's graphs. Slopes and areas on them show how fast the jumper speeds up and slows down, how far the jumper falls and how hard the canopy pulls.

Legend

$v$	Speed of the jumper.	$m/s$
$t$	Time since the jump.	$\mathrm{s}$
$v_t$	Terminal speed of the belly-down jumper, 55.2 m/s.	$m/s$
$v_t^{\prime }$	Terminal speed under the open canopy, 5.0 m/s.	$m/s$
$gt$	Speed that free fall would reach: the tangent to $v(t)$ at $t=0$, of slope $g$.	$m/s$
$\tau$	Time scale $v_t/g=5.6\mathrm{s}$: the tangent $gt$ meets the line $v_t$ at $t=\tau$.	$\mathrm{s}$
$t_1$	Moment the canopy starts to open, 13 s after the jump.	$\mathrm{s}$
$x$	Distance fallen before $t_1$: the shaded area under $v(t)$.	$\mathrm{m}$

Figure 4 is the speed graph of the animation's expert view, drawn from the same model. The blue curve is the jumper's speed $v(t)$. It leaves the origin along the green tangent $gt$, of slope $g$, which meets the upper dashed line $v_t$ at $t=\tau =5.6\mathrm{s}$, the first dashed vertical. Look at the lower dot there: the curve has reached only $\tanh 1\approx 0.76$ of $v_t$, 42 m/s, because drag has been growing all the way. It reaches 96 % of $v_t$ after $2\tau$, and 54.1 m/s by $t_1=13\mathrm{s}$, the second dashed vertical, when the canopy starts to open.

The animation also draws a second graph, not shown here: the drag $F_d(t)=mg{\tanh }^2(gt/v_t)$ in red. It starts flat, because drag grows as $v^2$, and creeps up towards the dashed $mg$ line.

At $t_1$ the canopy's diameter starts to grow steadily, for 4 s. Its area grows as the square of the diameter, so the extra $C_{d}A$ grows as the square of the time since opening. At first $C_{d}A$ rises faster than $v^2$ falls, so the drag shoots up, to a peak of about 3340 N, or $4.3mg$, at $t\approx 14.0\mathrm{s}$ and $v\approx 37m/s$. This is the opening shock. In Figure 4, after $t_1$ the blue curve turns steeply down onto the lower dashed line, $v_t^{\prime }=5.0m/s$, and reaches it by about 18 s. The approach is quick because the new time scale is only $v_t^{\prime }/g=0.51\mathrm{s}$.

The areas carry meaning too. The area under $v(t)$ is the height lost: 505 m before the opening (the shaded area $x$ in Figure 4), 92 m during the inflation and 25 m more at the final speed until 22 s, 622 m in all. On the drag graph, Newton's second law $mdv/dt=mg-F_d$ shows that the net area between $F_d(t)$ and the $mg$ line after the opening is the momentum removed: $m\Delta v=80\cdot (54.1-5.0)\approx 3930\mathrm{N}\cdot \mathrm{s}$.

Worked problem: a canopy that opens at once

Why must a canopy open gradually? Suppose the full canopy, with $C_{d}A=52.5{\mathrm{m}}^2$, appeared instantly at 13 s, when the speed is $v_0=54.1m/s$. With this $C_{d}A$ the terminal speed is $v_t^{\prime }=4.99m/s$, which means the drag equals $mg$ at $v_t^{\prime }$. Drag grows as $v^2$, so at $v_0$ it would jump to
$F_d=mg{\left(\frac{v_0}{v_t^{\prime }}\right)}^2=785\cdot {\left(\frac{54.1}{4.99}\right)}^2\approx 9.2\times 10^4\mathrm{N}$
This is 117 times the weight. Subtracting the weight leaves a net force of 116 times the weight, a deceleration of about 116 g.

How quickly would the speed drop? Now $v>v_t^{\prime }$, so the drag exceeds the weight and the same equation as before, $dv/dt=-g\left(v^2/(v_t^{\prime }{)}^2-1\right)$, makes the speed fall. Above the terminal speed its solution uses $\coth$ in place of $\tanh$. With $t$ counted from the opening,
$v(t)=v_t^{\prime }\coth \left(\frac{gt}{v_t^{\prime }}+u_0\right),\coth u_0=\frac{v_0}{v_t^{\prime }}$
Here $\coth u_0=10.8$, so $u_0=0.093$. The speed is down to $2v_t^{\prime }\approx 10m/s$ when the bracket reaches $\mathrm{arcoth}2=0.549$. With $v_t^{\prime }/g=0.509\mathrm{s}$ that takes $(0.549-0.093)\cdot 0.509\approx 0.23\mathrm{s}$. The integral of $\coth$ is $\ln \sinh$, so the distance covered is $((v_t^{\prime }{)}^2/g)\ln (\sinh 0.549/\sinh 0.093)\approx 2.54\cdot 1.83\approx 4.6\mathrm{m}$.

An energy balance gives the average force. Between 54.1 and 10 m/s the jumper loses $\frac{1}{2}\cdot 80\cdot (54.1^2-10^2)\approx 113\mathrm{kJ}$ of kinetic energy. The drag must absorb this, plus the 3.6 kJ of potential energy released on the way, within 4.6 m: an average force of 32 times the weight. The animation's canopy spreads the opening over 4 s and 92 m. There the drag absorbs the 116 kJ of kinetic energy lost plus 72 kJ of potential energy, an average of $2.6mg$ with a peak of $4.3mg$. Real parachutes use sliders or reefing lines to slow the opening for exactly this reason.

Limits of validity

The model keeps $C_{d}A$ fixed in each phase and the air density $\rho$ fixed throughout. Both are approximations. Head-down, a jumper roughly halves $C_{d}A$, and since $v_t\propto 1/\sqrt{C_{d}A}$, the terminal speed rises by a factor $\sqrt{2}$, to about 78 m/s. Since $v_t\propto 1/\sqrt{\rho }$ as well, it also changes with height: at 4000 m, where the air density is about 0.82 kg/m³ instead of 1.2 kg/m³, the belly-down jumper falls at $55.2\cdot \sqrt{1.2/0.82}\approx 67m/s$. Terminal velocity therefore depends on posture and on height.

Air behaves as incompressible only while the Mach number, the ratio of $v$ to the speed of sound, stays below about 0.3; the skydiver is at 0.16. At 39 km the air is some 250 times thinner, so $v_t$ is about 16 times higher on paper. Record jumps from there have passed the speed of sound, where shock waves raise $C_d$.

Stokes' law needs $\mathrm{Re}$ well below 1: its first correction, a factor $1+3\mathrm{Re}/16$, already adds about 10 % at $\mathrm{Re}=0.5$. At the other end of the size scale, particles smaller than about 0.1 μm are comparable with the 68 nm mean free path of air molecules. The air then no longer acts as a continuous fluid: it slips at their surface, and Brownian motion moves them more than settling does. Finally, the model ignores unsteady flow, so the animation's smooth growth of $C_{d}A$ only approximates the violent real opening shock.

Where this leads

Bernoulli's Principle describes the ideal flow outside the boundary layer and explains why drag is measured in units of $\frac{1}{2}\rho v^2$. In Damping and Resonance a linear drag $-bv$ becomes the damping term of an oscillator. Kinetic Theory of Gases derives a gas's viscosity from molecular collisions. Electric Current and Resistance has the same structure as Stokes settling: electrons drift at a terminal velocity at which collisions balance the electric force, so the current is proportional to the field. Conduction, Convection and Radiation uses boundary layers to explain how wind carries heat away, and the paths of Projectile Motion become shorter and lopsided once drag is added.