Projectile Drag/Wind Simulator

Compare ideal and drag-influenced trajectories.

Quadratic drag, on the velocity through the air	$\begin{gathered}\mathbf F=-k\,\|\mathbf v_{\mathrm{rel}}\|\,\mathbf v_{\mathrm{rel}},\qquad \mathbf v_{\mathrm{rel}}=\mathbf v-\mathbf w\\ k=\tfrac12\rho\,C_dA\end{gathered}$
Terminal speed: drag equals weight	$v_t=\sqrt{\frac{mg}{k}}$
Straight down from rest, and straight up	$\begin{gathered}v^{2}=v_t^{2}\left(1-e^{-2ky/m}\right)\\ H=\frac{m}{2k}\ln\!\left(1+\frac{v_0^{2}}{v_t^{2}}\right)\end{gathered}$

Worked example — Projectile Drag/Wind Simulator

Terminal speed	18.083 m/s
Speed through the air at launch	25 m/s
Drag at launch	18.75 N
Range in a vacuum	63.71 m
Greatest height in a vacuum	15.928 m
Range	28.283 m
Greatest height	9.643 m
Distance along at the top	16.259 m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Projectile Drag/Wind Simulator

Quadratic drag is the air resistance on anything thrown by hand: a force k|v_rel|² directed against the ball's velocity through the air, v_rel = v − w, with k = ½ρC_dA. Because |v_rel| contains both components, it ties the horizontal and vertical motions together, and the flight has no formula left. This page integrates it in steps of at most 5 ms.

What each symbol means

θ: the launch angle. Without drag its best value is 45° for every launch speed and every mass — a constant of the problem. With drag it is none of those things, which is what this lesson is about.
k/m: the drag coefficient divided by the mass. This is the only combination that appears anywhere in the equations: never k alone, never m alone. Two of the six input boxes are handles on one number.
v_rel: velocity relative to the air, v − w. Drag acts on this and never on the velocity over the ground.
R: the range. The page prints two: R₀ from the vacuum formula and R from the integration, and the gap between them is what the air took.

Where the formula comes from

Start with no air, where the two axes never speak to each other. Horizontal velocity is constant, vertical position is a parabola in time, and eliminating time leaves R = v₀² sin(2θ)/g. Maximise it: sin(2θ) is largest at 2θ = 90°. Notice what dropped out — neither v₀ nor the mass appears in that condition, so 45° is the best angle for every launch there has ever been. That is why it is remembered as a fact rather than as a result.
$R=\frac{v_0^{2}\sin2\theta}{g}\le\frac{v_0^{2}}{g}\ \Rightarrow\ \theta^{*}=45^\circ$
Add drag and the axes stop being independent. The acceleration is −g ŷ − (k/m)|v_rel|v_rel, and |v_rel| contains both components, so the horizontal equation depends on how fast the ball is climbing and the vertical one on how fast it travels downrange. The pair is non-linear and does not separate, and no elementary formula for R survives. The results table writes the range as the integral it is.
$\frac{d\mathbf v}{dt}=-g\,\hat{\mathbf y}-\frac{k}{m}\,|\mathbf v_{\mathrm{rel}}|\,\mathbf v_{\mathrm{rel}},\qquad |\mathbf v_{\mathrm{rel}}|=\sqrt{(v_x-w)^{2}+v_y^{2}}$
Divide by m and only k/m is left. A heavy ball is a light ball in thinner air. Set k = 0.03 with m = 3 and you get the trajectory of k = 0.01 with m = 1 to the last digit, so the heavy ball preset is the low-drag case under another name, and the mass box can tell you nothing the drag box could not.
$\frac{k}{m}=\frac{0.03}{3}=\frac{0.01}{1}=0.01\ \mathrm{m^{-1}}$
The best angle drops below 45°. A steep throw spends longer in the air, and the air takes speed for every second of it. In still air from ground level the best angle is therefore below 45°, and because drag grows with speed it falls further the harder you throw, which in a vacuum it never did. At the default drag, 25 m/s carries furthest at 38.939° (28.743 m, against 28.283 m at 45°) and 90 m/s at 29.34° (72.117 m, against 66.493 m). The last row of the table finds it for whatever you set. Wind moves it again: with 20 m/s behind the 25 m/s throw, the best angle is 47.093°.
$v_0=25:\ \ R(45^\circ)=28.283<R(38.939^\circ)=28.743\ \mathrm{m};\qquad v_0=90:\ \ R(45^\circ)=66.493<R(29.34^\circ)=72.117\ \mathrm{m}$

How to read what you see

The dashed curve is the same launch with the air taken away, the parabola of the previous page, so the space between the two curves is what the air cost. Set the drag to zero and the solid curve lands on the dashed one: the integrated range agrees with the vacuum formula to every digit the table prints, so a gap with drag on is the air, not the arithmetic.

Assumes: Air at rest apart from a steady horizontal wind blowing at the same speed at every height, uniform density all the way up, and a drag coefficient that does not vary with speed. A point mass: no lift, no spin, no Magnus force, so a sliced golf ball and a drifting shell are both outside what this model can say.

Problems solved in full

Linear or quadratic drag? 4 steps
A sphere of diameter D moving through air at speed v meets two kinds of drag: viscous drag 3πηDv (Stokes' law) and pressure drag ½ρC_d(πD²/4)v². Take η = 1.8 × 10⁻⁵ Pa·s, ρ = 1.2 kg/m³ and C_d = 0.47. Which one matters for a baseball (D = 7.4 cm) at 40 m/s, and which for an oil droplet (D = 1.5 µm) drifting at 0.10 mm/s?
1. Divide one force by the other. Most of the factors cancel, and what is left grows with both the size and the speed:
  
  $\frac{F_{\mathrm{quad}}}{F_{\mathrm{lin}}}=\frac{\tfrac12\rho C_d\,\tfrac{\pi D^{2}}{4}\,v^{2}}{3\pi\eta Dv}=\frac{\rho C_d}{24\eta}\,Dv$
2. The air's numbers make the constant:
  
  $\frac{\rho C_d}{24\eta}=\frac{1.2\times0.47}{24\times1.8\times10^{-5}}=1305.6\ \mathrm{s/m^{2}}$
3. The baseball:
  
  $1305.6\times0.074\times40=3864.4$
4. The droplet:
  
  $1305.6\times(1.5\times10^{-6})\times(1.0\times10^{-4})=1.9583\times10^{-7}$
Answer

The baseball is quadratic, its pressure drag 3864 times the viscous drag; the droplet is linear, its quadratic term 2.0 × 10⁻⁷ of the viscous one. Anything thrown by hand sits at the quadratic end, which is the law this page uses. Stokes' law is for dust, mist and Millikan's oil drops.

$\frac{F_{\mathrm{quad}}}{F_{\mathrm{lin}}}=3864\ \ (D=7.4\ \mathrm{cm}),\qquad \frac{F_{\mathrm{quad}}}{F_{\mathrm{lin}}}=2.0\times10^{-7}\ \ (D=1.5\ \mu\mathrm{m})$
Drag at launch, and the terminal speed 4 steps
A 1.0 kg ball with drag constant k = 0.030 kg/m is thrown at 25 m/s, 45° above the horizontal, in still air. Find the drag force on it at launch and compare it with the ball's weight. What is its terminal speed?

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1. In still air the velocity through the air is the velocity itself:
  
  $|\mathbf v_{\mathrm{rel}}|=|\mathbf v|=25\ \mathrm{m/s}\qquad(w=0)$
2. Quadratic drag is k times the square of that speed:
  
  $F=k\,|\mathbf v_{\mathrm{rel}}|^{2}=0.030\times25^{2}=18.75\ \mathrm{N}$
3. Compare it with the weight:
  
  $mg=1.0\times9.81=9.81\ \mathrm{N},\qquad \frac{F}{mg}=\frac{18.75}{9.81}=1.9113$
4. The terminal speed is the speed at which drag balances the weight:
  
  $k\,v_t^{2}=mg\ \Rightarrow\ v_t=\sqrt{\frac{mg}{k}}=\sqrt{\frac{1.0\times9.81}{0.030}}=18.083\ \mathrm{m/s}$
Answer

18.75 N at launch, 1.9113 times the weight; terminal speed 18.083 m/s. The ball leaves the hand faster than it could ever fall, so until it slows below 18.083 m/s the air pushes on it harder than gravity does.

$F=18.75\ \mathrm{N}=1.9113\,mg,\qquad v_t=18.083\ \mathrm{m/s}$
Into the wind, and with it 4 steps
The same throw is made into an 8.0 m/s headwind. Find the ball's speed through the air at launch and the drag force on it. Then repeat with the wind behind the thrower.

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1. Components over the ground, as before:
  
  $v_x=25\cos45^\circ=17.678\ \mathrm{m/s},\qquad v_y=25\sin45^\circ=17.678\ \mathrm{m/s}$
2. Into the wind the air moves at w = −8.0 m/s. Subtract the air's velocity from the ball's:
  
  $\mathbf v_{\mathrm{rel}}=(v_x-w,\ v_y)=(17.678+8.0,\ 17.678)=(25.678,\ 17.678),\qquad |\mathbf v_{\mathrm{rel}}|=31.174\ \mathrm{m/s}$
3. Drag goes with the square of that speed:
  
  $F=k\,|\mathbf v_{\mathrm{rel}}|^{2}=0.030\times31.174^{2}=29.155\ \mathrm{N}$
4. With the wind behind, w = +8.0 m/s and the subtraction goes the other way:
  
  $w=+8.0:\quad \mathbf v_{\mathrm{rel}}=(9.6777,\ 17.678),\quad |\mathbf v_{\mathrm{rel}}|=20.153\ \mathrm{m/s},\quad F=12.185\ \mathrm{N}$
Answer

Into the wind, 31.174 m/s through the air and 29.155 N of drag; with it, 20.153 m/s and 12.185 N. The ball leaves the hand at 25 m/s both times. The air sees only the difference, and the drag more than doubles between the two. Press tailwind for the second case.

$|\mathbf v_{\mathrm{rel}}|=31.174\ \mathrm{m/s},\ F=29.155\ \mathrm{N}\ \ (w=-8.0);\qquad |\mathbf v_{\mathrm{rel}}|=20.153\ \mathrm{m/s},\ F=12.185\ \mathrm{N}\ \ (w=+8.0)$
Dropped from 100 m 4 steps
The same ball (m = 1.0 kg, k = 0.030 kg/m) is dropped from rest 100 m above the ground. How fast is it moving when it lands? Compare that with its terminal speed and with the speed it would reach in a vacuum.

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1. Take y as the distance fallen and v as the speed downwards. Writing the acceleration as v dv/dy turns the equation into one in distance rather than time:
  
  $v\frac{dv}{dy}=g-\frac{k}{m}v^{2}=g\left(1-\frac{v^{2}}{v_t^{2}}\right)$
2. Separate the variables and integrate from rest:
  
  $\int_0^{v}\frac{v\,dv}{1-v^{2}/v_t^{2}}=\int_0^{y}g\,dy\ \Rightarrow\ v^{2}=v_t^{2}\left(1-e^{-2ky/m}\right)$
3. Put the numbers in, with the terminal speed of 18.083 m/s from problem 2:
  
  $\frac{2ky}{m}=\frac{2(0.030)(100)}{1.0}=6.0,\qquad v=18.083\sqrt{1-e^{-6.0}}=18.061\ \mathrm{m/s}$
4. Without air the same drop gives:
  
  $v_{\mathrm{vac}}=\sqrt{2gy}=\sqrt{2(9.81)(100)}=44.294\ \mathrm{m/s}$
Answer

18.061 m/s: 99.88% of the terminal speed, and 41% of the 44.294 m/s a vacuum would give. The page integrates the same fall without the formula and lands on the same 18.061 m/s, after 6.8066 s.

$v=18.061\ \mathrm{m/s}=0.9988\,v_t=0.41\,v_{\mathrm{vac}}$
Thrown straight up 4 steps
The same ball is thrown straight up at 25 m/s. How high does it go, and how long does that take? How fast is it moving when it gets back to the ground, and is the way down quicker or slower than the way up?

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1. Going up, drag and gravity both point down. With v dv/dy again, the climb ends at:
  
  $v\frac{dv}{dy}=-g\left(1+\frac{v^{2}}{v_t^{2}}\right)\ \Rightarrow\ H=\frac{v_t^{2}}{2g}\ln\!\left(1+\frac{v_0^{2}}{v_t^{2}}\right)=\frac{327}{19.62}\ln\!\left(1+\frac{625}{327}\right)=17.810\ \mathrm{m}$
2. The time comes from dv/dt = −g(1 + v²/v_t²) instead:
  
  $t_{\mathrm{up}}=\frac{v_t}{g}\arctan\frac{v_0}{v_t}=\frac{18.083}{9.81}\arctan\frac{25}{18.083}=1.7412\ \mathrm{s}$
3. The way down is problem 4 from a height of 17.810 m. Since e^−2kH/m = (1 + v₀²/v_t²)⁻¹, the landing speed simplifies:
  
  $v_{\mathrm{back}}^{2}=v_t^{2}\left(1-e^{-2kH/m}\right)\ \Rightarrow\ v_{\mathrm{back}}=\frac{v_0v_t}{\sqrt{v_0^{2}+v_t^{2}}}=\frac{25\times18.083}{\sqrt{625+327}}=14.652\ \mathrm{m/s}$
4. And its time, from dv/dt = g(1 − v²/v_t²):
  
  $t_{\mathrm{down}}=\frac{v_t}{g}\operatorname{artanh}\frac{v_{\mathrm{back}}}{v_t}=\frac{18.083}{9.81}\operatorname{artanh}\frac{14.652}{18.083}=2.0789\ \mathrm{s}$
Answer

17.810 m up in 1.7412 s, and back down at 14.652 m/s after another 2.0789 s. It returns slower than it left and takes longer coming down, because on the way up drag helped gravity and on the way down it worked against it. In a vacuum the ball would rise 31.855 m and come back at 25 m/s. The page shows the total, 3.8201 s.

$H=17.810\ \mathrm{m},\quad t_{\mathrm{up}}=1.7412\ \mathrm{s},\quad v_{\mathrm{back}}=14.652\ \mathrm{m/s},\quad t_{\mathrm{down}}=2.0789\ \mathrm{s}$
Why the 63.71 m vacuum answer becomes 28.283 m 5 steps
25 m/s at 45°, a 1.0 kg ball with k = 0.030 kg/m. The vacuum formula says 63.71 m. Work out why the answer is 28.283 m, and why the height survives far better than the range.

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1. Start with the answer the air is going to spoil. Both vacuum formulas are exact, take one line, and are the only analytic ground in this problem.
  
  $R_0=\frac{v_0^{2}\sin2\theta}{g}=\frac{25^{2}\sin90^{\circ}}{9.81}=63.71\ \mathrm{m},\qquad y_{0,\max}=\frac{v_0^{2}\sin^{2}\theta}{2g}=15.928\ \mathrm{m}$
2. Now write the real equation. The drag term points against the velocity, so its horizontal part depends on the vertical speed and the other way round: the two axes stop being independent and the parabola stops existing.
  
  $\frac{d\mathbf v}{dt}=-g\,\hat{\mathbf y}-\frac{k}{m}\,|\mathbf v|\,\mathbf v$
3. Read the two losses off the panel as fractions. They differ, and the difference is the thing worth explaining.
  
  $R=28.283\ \mathrm{m}=0.4439\,R_0,\qquad y_{\max}=9.643\ \mathrm{m}=0.6054\,y_{0,\max}$
4. The terminal speed is the speed at which drag balances the weight. Comparing it with the launch speed says which force is in charge at the start.
  
  $v_t=\sqrt{\frac{mg}{k}}=\sqrt{\frac{1.0\times9.81}{0.030}}=18.083\ \mathrm{m/s}<v_0=25\ \mathrm{m/s}$
5. The drag length m/k is the distance over which the air removes a substantial fraction of the speed. Compare it with the range you were hoping for.
  
  $\ell=\frac{m}{k}=\frac{1.0}{0.030}=33.3\ \mathrm{m}\approx\tfrac12R_0$
Answer

Because drag has a length scale of 33 m, and the throw wanted to be 64 m long but only 16 m high. The range keeps 44% of its vacuum value and the peak height 61%. Two numbers say drag is not a small correction here. The terminal speed is 18.083 m/s, below the 25 m/s launch, so the ball starts out losing more to the air than to gravity; and m/k = 33.3 m is half the vacuum range, where a small correction would need it to be many times larger. There is no closed form to fall back on: drag couples the two axes, which is why the page integrates instead of substituting.

$\frac{R}{R_0}=0.44,\qquad \frac{y_{\max}}{y_{0,\max}}=0.61$

Supplementary problems, answers only

A baseball of mass 0.145 kg has drag constant k = 1.3 × 10⁻³ kg/m. What is its terminal speed?

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33.08 m/s.

$v_t=\sqrt{\frac{0.145\times9.81}{1.3\times10^{-3}}}=33.08\ \mathrm{m/s}$
A shot and a ball are the same size, so k = 0.030 kg/m for both, but the shot has a mass of 3.0 kg and the ball 1.0 kg. What drag constant would the ball need to fly exactly like the shot?

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0.010 kg/m: only k/m enters the motion.

$\frac{k}{m}=\frac{0.030}{3.0}=0.010\ \mathrm{m^{-1}}\ \Rightarrow\ k=0.010\ \mathrm{kg/m}\ \ (m=1.0\ \mathrm{kg})$
The 1.0 kg ball with k = 0.030 kg/m is dropped from rest. How far has it fallen when it reaches 90% of its terminal speed?

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27.68 m.

$y=-\frac{m}{2k}\ln\!\left(1-0.90^{2}\right)=-\frac{1.0}{0.060}\ln0.19=27.68\ \mathrm{m}$
A ball with k/m = 0.010 m⁻¹ is thrown straight up at 40 m/s. How high does it go, and how fast does it come back down?

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48.37 m, and it comes back at 24.66 m/s.

$v_t^{2}=\frac{9.81}{0.010}=981,\quad H=\frac{981}{19.62}\ln\!\left(1+\frac{1600}{981}\right)=48.37\ \mathrm{m},\quad v_{\mathrm{back}}=\frac{40\times31.32}{\sqrt{1600+981}}=24.66\ \mathrm{m/s}$
A ball with k/m = 0.030 m⁻¹ falls through a steady 5.0 m/s wind. If it falls for long enough, what velocity over the ground does it settle to?

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18.76 m/s, 74.54° below the horizontal: 5.0 m/s downwind and 18.08 m/s downwards.

$\mathbf v\to(w,\,-v_t)=(5.0,\,-18.08)\ \mathrm{m/s},\qquad |\mathbf v|=18.76\ \mathrm{m/s},\quad \varphi=74.54^\circ$

Why the way down is steeper and slower

On the way up, drag and gravity both point down and stop the climb together. On the way down they pull against each other, so the ball falls more slowly than it rose and takes longer about it: at the default throw the climb lasts 1.2702 s and the fall 1.5141 s. All the while the air is taking the horizontal speed, so the second half of the flight covers less ground than the first. The top comes 16.259 m along a 28.283 m flight, past the middle, and the ball lands at 62.324° after leaving at 45°. A vacuum parabola is symmetric about its top because the equations without drag work just as well run backwards in time. Drag is the term that knows which way time runs.

It's the air you move through

Drag acts on the velocity through the air, v − w, so it does not pull a ball towards rest. It pulls it towards moving with the air. Given long enough, any ball in a steady wind ends up drifting downwind at the wind's own speed while it falls at its terminal speed. Press headwind: an 8 m/s wind into the throw strips the ball's forward speed before it lands, and it comes down at 92.328° below the horizontal, leaning back towards the thrower.

Why dimpled golf balls fly farther

Roughening a ball can reduce its drag. A golf ball's dimples trip the air next to its surface into a thin turbulent layer, which clings further round the back of the ball before it separates and so leaves a narrower low-pressure wake. A sphere's drag coefficient drops sharply once the flow is fast enough to do this by itself, the drag crisis, and the dimples bring that drop down to the speeds of a drive. This page holds k the same at every speed, so the crisis is exactly what it cannot show.

Learning path

Throwing things

References (2)

The lesson — why the coupled equations admit no elementary range formula, and how the optimum angle moves: G. W. Parker, "Projectile motion with air resistance quadratic in the speed." American Journal of Physics 45(7), 606–610, 1977.
Insight block 3 — why roughening a ball reduces its drag: P. W. Bearman and J. K. Harvey, "Golf Ball Aerodynamics." Aeronautical Quarterly 27(2), 112–122, 1976 — wind-tunnel measurements of dimpled against smooth spheres.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.