The angle that drag moves

Forty-five degrees throws furthest, and the maximum is so flat that nobody notices — in vacuum. That essay ended by naming what happens when a single velocity-dependent force is added, and this one does it.

The same launches with drag 2.4 times the weight. Five launches at the same speed and at 20, 32, 45, 60, 70 degrees, drawn twice: in vacuum, where the arcs are symmetric parabolas, and with quadratic drag whose force at launch is 2.4 times the projectile's weight. Nothing about the drag figure is a parabola. Each path rises at nearly the vacuum angle, loses horizontal speed that nothing restores, and comes down far more steeply than it went up — the 32° launch leaves at 32° and arrives at 50°. The best of these angles in vacuum is 45° and in air is 32°, and the best range has fallen by 60 per cent. The asymmetry is the whole of the difference: drag removes speed in proportion to speed squared, so it takes most from the fast early part of the flight, and the descent happens at a speed the drag has already limited. — Fig. 1 Five launches at one speed, drawn in vacuum as faint arcs and with quadratic drag as solid curves. The drag force at launch is 2.4 times the projectile’s weight. The best of these angles is 45° in vacuum and 32° in air, and the best range has fallen by 60 per cent.

Why the closed form goes

In vacuum the two components of the motion are independent. The horizontal one has no force on it and the vertical one has a constant force, so each can be solved on its own and the trajectory is their combination — which is the parabola, and which is why the range has a formula.

Drag destroys the independence in a way that is worth being precise about, because it is a good example of a decomposition that survives in appearance and not in fact. The force is

\mathbf{F}_{\text{drag}} = -k\,|\mathbf{v}|\,\mathbf{v}

which can still be written component by component: the horizontal part is $-k|\mathbf{v}|\dot{x}$ and the vertical part is $-k|\mathbf{v}|\dot{y}$ . But $|\mathbf{v}| = \sqrt{\dot{x}^2+\dot{y}^2}$ contains both, so the horizontal equation now depends on the vertical speed and the vertical one on the horizontal speed. They are coupled, no rearrangement separates them, and there is no closed-form solution.

The decomposition that works is the one for a constant force: resolve it along two fixed directions and each component becomes a separate problem, solvable on its own. What makes that legitimate is that the force does not depend on the motion. The moment it does — and drag does, on both the speed and the direction — the two components stop being separate problems while continuing to look like them, which is the trap this whole subject is built on.

Every trajectory in this essay is therefore an integration. That is not a defeat; it is the ordinary situation, and the vacuum case is the unusual one. What is lost is a formula, and what remains is everything a formula would have been used for.

One number describes the drag

The equations have three constants — the launch speed, gravity and the drag coefficient — and only one combination of them matters.

Measuring lengths in $v_0^2/g$ and times in $v_0/g$ leaves a single dimensionless parameter,

B = \frac{k v_0^2}{g} = \frac{\rho C_D A v_0^2}{2mg}

which is the drag force at launch divided by the weight. Everything about the shape of the trajectory depends on that one number, and nothing else about the projectile enters at all.

Its range across ordinary objects is enormous. A shot put has $B \approx 0.008$ : the drag on it at release is under one per cent of its weight, and it is nearly a vacuum projectile. A golf ball driven at seventy metres a second has $B \approx 2.4$ — the air is pushing back more than twice as hard as gravity pulls. A shuttlecock is at about thirty, which is why it stops almost as soon as it is hit.

A resisting force that grows with speed is familiar enough on its own. What quadratic drag adds is that the force also points along the direction of travel, and that is what couples the two components: the horizontal drag depends on the vertical speed, because the vertical speed is part of the magnitude and part of the direction. Neither component can be integrated without the other.

That single parameter is the reason a table of “best angles for sports” is not a table of sports at all. It is a table of one number.

The asymmetry, which is the mechanism

One launch at 45°, in air and in vacuum. A single launch at 45° with a drag force at launch 2.4 times the weight, against the same launch in vacuum. The vacuum arc is symmetric: it spends half its time going up and lands at the speed and angle it left with. The air trajectory does neither. It spends 45 per cent of its flight going up, arrives at 64° rather than 45°, and lands at 48 per cent of its launch speed. The apex is closer to the end than to the start, which is the feature that makes a real trajectory recognisable in a photograph, and the reason is that the horizontal speed is largest at the beginning and drag is quadratic in speed. — Fig. 2 One launch at 45°, in air and in vacuum. The vacuum arc spends half its time going up and lands at the speed and angle it left with. The air trajectory spends 45 per cent of its flight going up, arrives at 64°, and lands at 48 per cent of its launch speed. The apex sits closer to the end of the flight than to the start.

The trajectory in air is steeper coming down than going up, and that single fact contains the whole result.

Drag is quadratic in speed, so it takes most from the fastest part of the flight — which is the beginning. By the apex, the projectile has already lost most of the horizontal speed it will ever lose; the descent then happens at a speed drag has capped, and the horizontal distance covered on the way down is far less than the distance covered on the way up. The path is not a parabola and it is not symmetric about any vertical line.

Height, therefore, is a bad investment in air. In vacuum, hang time and horizontal speed trade off exactly, which is what makes the optimum sit halfway at 45°. In air, hang time is bought with launch speed that drag will remove before it can be spent, so the trade is no longer even and the optimum moves toward the horizontal.

The speed a fall cannot pass

The same parameter has a second meaning that is worth extracting, because it is the one that makes the numbers concrete.

A projectile falling straight down reaches a speed at which the drag equals the weight, and stops accelerating — the same balance of a driving force against a resistance that grows with speed as the terminal state of any damped motion. In the units used here that is $Bv^2 = 1$ , so the terminal speed measured in launch speeds is $1/\sqrt{B}$ — and the ballistic parameter is therefore the square of the ratio between the launch speed and the terminal speed. A shot put at $B = 0.008$ is launched at about a ninth of its terminal speed and never comes close to it. A shuttlecock at $B = 30$ is launched at five and a half times its terminal speed, which is why it decelerates so violently and then falls almost vertically.

One launch at 45°, in air and in vacuum. A single launch at 45° with a drag force at launch 12 times the weight, against the same launch in vacuum. The vacuum arc is symmetric: it spends half its time going up and lands at the speed and angle it left with. The air trajectory does neither. It spends 40 per cent of its flight going up, arrives at 75° rather than 45°, and lands at 26 per cent of its launch speed. The apex is closer to the end than to the start, which is the feature that makes a real trajectory recognisable in a photograph, and the reason is that the horizontal speed is largest at the beginning and drag is quadratic in speed. — Fig. 3 The same 45° launch at a ballistic parameter of 12, where the launch speed is three and a half times the terminal speed. The projectile is brought down to terminal speed within the first part of its flight and then falls almost vertically; the arc has stopped resembling a parabola in any respect, and the vacuum path drawn faintly behind it is eight times longer.

Reading $B$ that way explains why the optimum angle falls where it does. When the launch speed is far below terminal, the flight is over before drag has done anything and 45° survives. When it is far above, the projectile spends the first part of its flight being brought down to terminal speed, and everything after that is a slow fall — so the launch is essentially a way of buying horizontal distance in the first moment, and pointing it upward wastes it.

The optimum, over four decades

The best launch angle, against how much drag there is. The launch angle of greatest range against the ballistic parameter — the drag force at launch measured in weights — over nearly four decades, each point found by integrating the trajectory and searching. At the left the answer is 45.01°, which is the vacuum result recovered rather than assumed. It falls monotonically from there: 44.9° for shot put, 38.1° for golf ball, drag only, 28.7° for shuttlecock. The reason is asymmetry rather than loss. Drag takes most from the fastest part of the flight, so a launch spends its speed early; a lower angle keeps more of that early speed horizontal, where it buys range, and the height that a steeper launch buys is returned at a descent speed that drag has already capped. There is no formula on this chart. With quadratic drag the equations do not separate and there is no closed-form trajectory, so every point here is an integration. — Fig. 4 The launch angle of greatest range against the ballistic parameter, each point found by integrating the trajectory and searching. At the left the answer comes out at 45.0°, which is the vacuum result recovered rather than assumed. It falls monotonically: 44.9° for a shot put, 38.1° for a golf ball’s drag alone, 28.7° for a shuttlecock.

The left-hand end of that curve is a check rather than a result. Nothing in the integration knows that the answer at zero drag is forty-five degrees; the search returns 45.0°, and if it did not, everything to the right of it would be worthless.

The fall is smooth and has no feature in it. There is no drag at which something changes character — no threshold, no transition — which is worth saying because the shape of a curve is evidence about whether a mechanism is one thing or several.

Range against launch angle, at five drags. Range against launch angle for ballistic parameters of 0, 0.3, 1.2, 4, 12, each curve integrated point by point. The top curve is the vacuum case, symmetric about 45° because sin 2θ is, and every other curve is asymmetric: the peak moves left as the drag rises — to 45° at 0, 42.5° at 0.3, 40° at 1.2, 35° at 4, 32.5° at 12 — and the fall-off is much steeper on the high side than on the low. That asymmetry is the practical content. Throwing ten degrees under the optimum costs almost nothing in air; throwing ten degrees over it costs a great deal, because the extra height is bought with horizontal speed that drag then removes. Every curve is also lower than the one above it, which is the loss, but the loss is not what moves the peak. — Fig. 5 Range against launch angle at five drags. The top curve is the vacuum case, symmetric about 45° because sin 2θ is. Every other curve leans: the peak moves to 42.5°, 40°, 35° and 32.5°, and the fall-off is far steeper on the high side than on the low.

The leaning is the practically important part and it is not the same statement as the moving peak. In vacuum the curve is symmetric, so an error of ten degrees either way costs the same. In air it is not: throwing under the optimum is cheap and throwing over it is expensive, because the extra height is bought with horizontal speed that will not be returned. A thrower who does not know the optimum should aim low.

How wrong the vacuum answer is

It is worth putting a number on the error rather than describing it, because “air resistance reduces the range” is the kind of statement that sounds like a correction and is not.

Trajectories at one speed and several angles. Projectile paths launched at the same speed and five different angles. The 45° launch travels furthest, and the 20° and 70° launches land in the same place. — Fig. 6 The same five launch angles in vacuum, where the arcs are parabolas and the 45° launch goes furthest. Complementary angles land together, the paths are symmetric, and the range has a formula. None of those four statements survives the addition of one velocity-dependent force.

At a ballistic parameter of 2.4 the best achievable range is 40 per cent of the vacuum range at the same launch speed, and it is reached at 32° rather than 45°. Using the vacuum answer — launching at 45° in air — costs a further few per cent on top of the 60 already lost. So the error in the angle is small in its consequences and the error in the range is enormous, which is the opposite of the pattern an approximation usually has.

The useful habit with any approximation is to draw its error rather than assert that the error is small. Do it here and the vacuum trajectory turns out to be an approximation whose error, at a golf ball’s ballistic parameter, is larger than the answer — the vacuum range for a well-struck drive is over four hundred metres against a real one under three hundred. That is not a correction; it is a different calculation.

That is the honest summary of the vacuum model. It is not a first approximation to a real trajectory with a correction to be added; for anything light and fast it is a different problem whose answer happens to be nearby in one variable and nowhere near in another.

What is not in this model

Three things are left out, and the third is the one that matters most in practice.

The drag coefficient is constant here. In reality it depends on the Reynolds number, and for a sphere it drops abruptly — by a factor of about three — when the boundary layer becomes turbulent. A golf ball’s dimples exist to trigger that drop at a lower speed, and a projectile crossing the transition during its flight has an effective $B$ that changes underneath it.

The drag is quadratic. That is right for anything large and fast, where the resistance comes from pushing air out of the way. For something small and slow, the resistance is viscous and linear in speed instead — the force a fluid exerts by shearing rather than by being displaced — and the linear problem does have a closed-form solution, which is why textbooks solve that one.

And there is no lift. A spinning ball generates a force perpendicular to its motion, and for a golf ball driven with backspin that force is comparable with its weight. It is why the real optimum launch angle for a driver is between ten and fifteen degrees rather than the 38° this calculation gives: the lift holds the ball up, so hang time no longer has to be bought with launch angle at all.

What sets the launch speed in the first place is a collision: a struck body taking momentum from a heavier one, with a restitution below one. Every number in this essay is conditional on that speed, and in a real throw or strike the speed and the angle are not independent choices at all. A thrower who changes the angle changes the speed too — the shoulder is stronger in some directions than others — which moves the optimum again, by an amount no aerodynamics can supply.

That last point generalises. The optimum computed here is over launch angle at fixed launch speed, and a human being cannot hold speed fixed while changing angle: the shoulder is stronger horizontally than vertically, and the measured optimum for a thrown ball is a few degrees below even the drag calculation for that reason.

The crisis a dimple triggers

The remark that the drag coefficient is not constant deserves its numbers, because the effect it names is larger than anything else in this essay and it is the reason a golf ball has dimples.

For a smooth sphere the drag coefficient sits near 0.5 over a wide range of speeds, and then does something abrupt: at a Reynolds number around $3\times10^5$ it falls, by a factor of four or five, over a narrow band. Going faster makes the drag force smaller in absolute terms, not merely relatively — which is a startling thing for a resistance to do.

The mechanism is a change in the flow rather than in the sphere. Below the transition the boundary layer on the sphere’s surface is smooth, and it separates from the surface early — a little before the widest point — leaving a broad turbulent wake behind. Above it the boundary layer itself becomes turbulent, which makes it stick to the surface further round before separating, so the wake is narrower and the pressure behind the ball is higher.

A dimpled ball triggers that transition deliberately and early. The dimples trip the boundary layer into turbulence at a Reynolds number around $4\times10^4$ — nearly an order of magnitude below the smooth sphere’s — which is comfortably below the speeds a driven ball travels at. So the ball is on the low-drag side of the transition for its whole flight, where a smooth ball of the same size and speed would be on the high-drag side.

The gain is not marginal. A dimpled ball travels roughly twice as far as a smooth one struck identically, and about half of that is the drag reduction, with the rest coming from the lift the dimples also help generate. It is a rare case of surface roughness reducing resistance, and it is the standing counterexample to the intuition that a smoother object always moves more easily.

The tables that needed a machine

The essay’s remark that artillery tables are computed rather than derived is worth expanding, because the consequences reached further than ballistics.

A gun’s firing table gives, for each combination of charge, elevation and shell type, the range and the time of flight. Every entry is a numerical integration of exactly the equations in this essay, with a drag law measured in a wind tunnel and corrections for air density, wind and the Earth’s rotation. There is no formula and there never has been.

Before machines, the integrations were done by hand, by teams of people whose job title was computer. A single trajectory took a skilled person a day or two; a full table needed thousands, and a new gun or a new shell meant the whole exercise again. During the Second World War the backlog at the American ballistics laboratory ran to months, with mechanical differential analysers helping and not enough of them.

The response was to build a machine, and the machine was ENIAC — commissioned explicitly to compute firing tables, finished just after the war it was built for, and general-purpose almost by accident. Its first substantial calculations were not ballistics at all.

That sequence is worth recording as a corrective to a common view of how computing arrived. The first large electronic computer was not built to do arithmetic that had never been possible; it was built to do arithmetic that was entirely possible and unbearably slow, on a problem whose equations were fully understood and had no closed-form solution. The absence of a formula, in a problem of the shape this essay describes, is what paid for it.

The angle a body can actually produce

The last caution in this essay — that a thrower cannot hold the speed fixed while changing the angle — is worth putting numbers on, because it moves the answer further than the drag does for some events.

The optimum computed here is over launch angle at a fixed launch speed, and that is the right question for a gun. It is the wrong question for a person. The muscles that throw are stronger in some directions than others, and the release speed a shot putter achieves falls measurably as the release angle is raised — by something like a metre per second over ten degrees.

Redoing the optimisation with that dependence included lowers the answer substantially. A shot put has a ballistic parameter under a hundredth, so drag moves its optimum from 45° by only a fraction of a degree; the release height above the ground takes off another degree or two; and the speed-angle trade takes off most of the remaining ten. Measured release angles for elite throwers cluster around 30 to 34 degrees, and the calculation that reproduces them has almost no aerodynamics in it at all.

The lesson generalises past sport. An optimisation is over the variables that can actually be varied independently, and a variable that drags another one along with it is not one of them. Computing the best angle at fixed speed and then reporting it as the best angle is a category error, and it is common enough in applied optimisation to be worth naming: the constraint set is part of the problem, and a correct answer to the wrong constraint set is not an approximation to the right one.

Where the same shape appears

In the ballistics that named the parameter. Artillery tables are computed rather than derived, and have been since the eighteenth century — the first substantial use of numerical integration in engineering, and one of the first jobs given to a digital computer. The ballistic coefficient in those tables is $B$ under another name.

In every falling object. A projectile launched straight up in air comes down more slowly than it went up, for the same reason and by the same arithmetic. The terminal speed is where drag equals weight, which in this essay’s units is at $B v^2 = 1$ ; a shuttlecock’s is about seven metres a second, which is why it hangs.

In the orbital limit at the other end. Everything here assumes flat ground and constant gravity. Launch fast enough and the ground curves away as fast as the projectile falls, at which point the range formula stops meaning anything and the trajectory is an orbit — a different conic with different rules, reached continuously from this one.

And the uniform field is itself an approximation. Under a central force a projectile’s path is an ellipse with its far focus at the centre of the Earth, and the parabola everybody draws is what that ellipse looks like when only a few kilometres of it are drawn. For a golf ball the difference is far below the drag correction; for a shell fired forty kilometres it is not, which is why long-range gunnery tables carry a curvature term as well as a drag one.

And in the sensitivity of the answer to the model. The calculation is stable — the equations are not chaotic and the integration converges — but its conclusions move a long way when a term is added. An error that doubles on a schedule is one way to be wrong about a trajectory; a missing force is a much more ordinary one, and it does not announce itself with divergence.

Underneath every number here is one optimisation. Range as a function of launch angle is a smooth hill, and finding its top is the whole of the question. In vacuum the hill is symmetric and its top sits at 45°; drag tilts it, lowering the high-angle side more than the low, and only the tilt moves the answer. That is why the optimum falls rather than rises, and why it falls further the harder the ball is hit.

What the pictures cannot show

Nothing here computes a drag force. The coefficient is an input, taken from measurement, and the essay is about what a stated resistance does to a trajectory rather than about where that resistance comes from. A calculation of the flow around the projectile is a different subject with a different apparatus.

The ground is flat and the projectile is a point. Launching from a height above the landing ground lowers the optimum by a further degree or two — which is most of what is left in the shot-put case — and a real projectile with a size has a lever arm and a torque.

The integration is fixed-step. Every trajectory here is a fourth-order Runge–Kutta with a step small enough that halving it changes the range in the fifth decimal, and the landing point is found by interpolating the last step rather than by stopping at whichever sample fell below the ground. That second detail matters more than the first: stopping at a sample makes the computed range depend on the step size in a way that looks like physics.

And the wind is zero. A headwind changes the answer more than the entire effect this essay describes, and it changes it in a direction that depends on the projectile’s speed relative to the air rather than to the ground.

The ladder from here

Later rungs on this anchor: the Magnus force and the trajectory of a spinning ball, which is what a golf ball needs; launching onto sloping ground, where the optimum shifts by about half the slope angle; the envelope of all trajectories at one speed, which is itself a parabola and bounds everywhere that can be reached; and the transition to orbit, where the ground curves away as fast as the projectile falls and the range formula stops meaning anything.

The neighbouring ladders are the vacuum result that this essay perturbs, a resisting force that takes what it needs, which is the other way a force can depend on the motion, and the fourth power in a pipe, where the resistance is viscous and the arithmetic is exact.