One curve answers every slope

Throwing a ball as far as possible along level ground is a problem with a famous answer, forty-five degrees, and the angle that throws furthest derives it and explains why the maximum is so flat that nobody notices when they miss it. The answer is so well known that it is carried into situations it was never derived for: a ball thrown up a hillside, a stone thrown from the top of a cliff, a shot at a basket above the thrower’s head.

Each of those has a different answer, and at first sight each needs its own calculation. The landing point is where a parabola meets a line, and the line is a different line each time. There is a shortcut that makes all of them the same calculation, and it comes from an object already on the page.

The curve written about the hand #

Everywhere a throw can reach found that the set of points a projectile of fixed speed can reach, aimed in every direction, is bounded by a parabola — the envelope of the whole family of trajectories. Its vertex is directly above the hand at a height $v^2/2g$, it crosses level ground at the greatest range $v^2/g$, and its focus is exactly at the hand.

That last fact is the one that pays. A parabola described by its distance from its own focus has the polar equation of a conic,

$$r(\alpha) = \frac{v^2/g}{1 + \sin\alpha},$$

with $\alpha$ the elevation of the direction measured from the horizontal and $v^2/g$ the distance from the focus to the directrix. A straight line from the hand at elevation $\alpha$ meets the envelope at distance $r(\alpha)$, and since the envelope is the edge of what can be reached, $r(\alpha)$ is the farthest the throw can land along that line. Every slope problem is one evaluation of one function.

The numbers along the lines are worth reading as a set. Level ground gets the familiar $v^2/g$. A 20° hillside gets 0.745 of it and a 45° one 0.586. A slope falling away at 30° gets exactly twice the level range, because $\sin(-30°) = -\tfrac12$. And as the line turns towards straight down the denominator goes to zero and the reach grows without limit, which is correct and only sounds alarming: a stone dropped into a deep enough well keeps going.

The asymmetry is the first thing the polar form says that intuition does not. Tilting the ground up by some angle costs less than tilting it down by the same angle gains, and the two are not mirror images of one another about level ground.

Up the hill #

On a hillside the throw that feels natural lands short, and the throw that feels too steep is the best one. That is easy to see once the geometry is drawn. The hill rises to meet the ball, so a low trajectory is intercepted early; a steeper one buys height the hill has not yet reached, and loses less to the interception than it gains in time aloft. The balance point is not 45°.

The distance along the slope reached by a launch at $\theta$ can be written out exactly: the ball meets the line $y = x\tan\alpha$ after a time $2v\sin(\theta - \alpha)/(g\cos\alpha)$, and the distance along the slope is then

$$s(\theta) = \frac{2v^2}{g}\,\frac{\cos\theta\,\sin(\theta - \alpha)}{\cos^2\alpha}.$$

The product in the numerator is largest when $2\theta - \alpha = 90°$, which is to say at $\theta = 45° + \alpha/2$, and its largest value turns the whole expression back into $r(\alpha)$. The figure does not use either formula for the heavy trajectory. It searches the launches, finds the peak, and then compares.

The 45° and 70° throws landing level is the level-ground pairing of complementary angles moved round. On flat ground 30° and 60° land together because they are symmetric about 45°; on a 25° hillside the throws that land together are symmetric about 57.5°. The whole pattern of the level-ground problem survives, tilted by half the slope.

Half the slope #

“Forty-five plus half the slope” has a geometric reading that is better than the arithmetic. Forty-five degrees is halfway between the horizontal ground and the vertical. Tilt the ground by $\alpha$ and the halfway direction moves to halfway between the tilted ground and the vertical, which is $45° + \alpha/2$. The best launch bisects the angle between the line the ball must land on and the direction gravity points. On level ground that bisector happens to be 45°, and the number was always a coincidence of the ground being level.

The curves say something about errors as well. On level ground the range curve is symmetric about its peak, so a throw ten degrees too high and one ten degrees too low lose the same distance, which is the observation that makes the optimum so unimportant in practice. On a slope the symmetry goes. The uphill curve falls away faster on the low side, where the hill intercepts a flat throw, and the downhill curve faster on the high side. Level ground is the special case in which the two directions of error cost the same.

Off a height #

The polar form was written for lines through the hand, and a flat field below a raised hand is a line that does not pass through it. The envelope still answers the question, just not in polar form: the reachable boundary meets a floor a height $h$ below the hand at a horizontal distance $\sqrt{1 + 2h}$ in units of $v^2/g$, and the trajectory touching it there leaves at $\tan\theta = 1/\sqrt{1+2h}$.

The answer falls below 45° with height, for the mirror-image reason to the hillside: a floor below the hand is a slope that runs away from the ball, and the ball has time to spend on distance rather than on height. A shot putter releasing from shoulder height at fourteen metres a second is at 0.11 on that axis and should, by this calculation, throw at 42°.

Shot putters throw at about 37°, and they are right to. Nothing in this essay has questioned the assumption that the release speed is the same in every direction, and for a human arm it is not: the body can push harder forward than upward, and a steeper throw is a slower one. That is a different kind of correction from anything here — it changes the set of throws available rather than the line they land on — and the polar form cannot see it.

The least speed that arrives #

Turn the question round. Instead of a fixed speed and the farthest landing point, fix the target and ask for the least speed that reaches it. The polar equation answers that too, because a target at distance $r$ in direction $\alpha$ is on the envelope of exactly one speed, the one for which $r = (v^2/g)/(1+\sin\alpha)$:

$$v_{\min}^2 = g\,r\,(1 + \sin\alpha),$$

and the launch that achieves it is again the bisector, $45° + \alpha/2$.

For a free throw the numbers are a hoop 3.05 metres up, a release at about 2.1 metres, and 4.19 metres of horizontal distance, which puts the target 12.8° above the horizontal. The least speed is 7.17 metres a second, at 51.4°.

What the figure adds is the reason this shot matters beyond saving effort. At the least speed the target sits exactly on the envelope, and on the envelope the two launches that reach a point — the flat one and the lobbed one — have merged into one. A merger of two solutions is a place where the quantity being solved for is stationary: the height at the target’s distance, as a function of launch angle, has a maximum there rather than a slope. So a small error in angle changes where the ball arrives only in second order. The figure’s 2° error costs a centimetre and a half at the least speed and eleven centimetres on the flatter solution at a slightly higher speed, where the curve crosses the target height with a definite slope.

This is the same event everywhere a throw can reach found at the edge of the reachable set, seen from the target rather than from the gun. It is also the event that makes the ring at twenty-two degrees bright: many ice crystals of slightly different orientation send light to nearly the same angle because the deviation is stationary at its minimum, and many throws of slightly different angle send the ball to nearly the same place because the height is stationary at the least speed. The least-effort shot and the most forgiving shot are one shot, because both are the place where two solutions become one.

The forgiveness is only in angle. At the least speed a shortfall in speed has no second solution to fall back on, and it costs first-order distance. A player is therefore trading one kind of error against the other, and the angle that minimises the total depends on which of the two a particular throwing motion controls better. The least-speed angle is where the angle error vanishes, not where the total does, and a thrower whose speed is less reliable than their aim is better served a little away from it.

A jet aimed at a window #

A stream of water from a nozzle is the same family of throws delivered continuously, and the least-speed result says how hard the water has to be pushed. Take a window ten metres above the nozzle and fifteen metres away. It is 18.0 metres off along a line 33.7° above the horizontal, so the least jet speed that reaches it is $\sqrt{g r (1 + \sin\alpha)} = 16.6$ metres a second, aimed at 61.8°.

A jet speed is a pressure. Water leaving a nozzle at speed $v$ has had $\tfrac12\rho v^2$ of pressure turned into kinetic energy, and for 16.6 metres a second that is 1.37 bar above the atmosphere. That is the floor. A real hose loses pressure along its length, a real jet breaks up and is slowed by air over eighteen metres, and fire services work at several times this nozzle pressure — but none of those losses changes the geometry, and the geometry is what fixes the angle at which the least pressure is enough.

The stationary property is visible in a hose held by hand. At the least pressure a jet swinging a few degrees up and down keeps arriving at nearly the same place, because the arrival point is at the top of its curve against angle. What it does not survive is a drop in pressure: the target is on the envelope of exactly that speed, and any less speed puts it outside. Turn the pressure up and the window can be reached two ways, a flat jet that arrives fast and a lob that comes down steeply, and firefighters choose between them for the reason everywhere a throw can reach gives an artillerist — one clears an obstacle and the other arrives first.

Why the focus is the right place to stand #

A parabola written about its focus is a conic section written the way orbits are written. A body moving under an inverse-square force traces $r = p/(1 + e\cos\phi)$ about the attracting centre, and the orbit that does not come back to itself makes that the special property of the inverse square. The envelope here is the $e = 1$ member of that family, about the hand instead of about a sun, and the analogy is not an accident of notation: the trajectory of a thrown stone is itself an ellipse with a focus at the centre of the Earth, flattened into a parabola by the smallness of a garden compared with a planet.

The foci of the individual trajectories carry a second fact, and the hero figure draws it. Every trajectory at a given speed is a parabola with a vertical axis, and every one of them has the same directrix — the horizontal line at height $v^2/2g$, where the ball would have used up all its kinetic energy climbing, which is an energy written as a height. Parabolas sharing a directrix and passing through a common point have foci at a common distance from it, so every trajectory’s focus lies on a circle of radius $v^2/2g$ about the hand, at an angle $2\theta - 90°$ that turns twice as fast as the launch angle.

The farthest throw along a line is then the one whose focus lies on that line: set $2\theta - 90° = \alpha$ and the bisector rule comes back. The figure checks it from the vertex of each best trajectory rather than from the rule. It is a neat way of carrying the whole problem in the head — swing the focus round its circle until it points at the target — and it is the same move that makes a paraboloidal mirror send every ray from its focus out parallel to its axis, which the surface a spin decides uses for a telescope made of rotating mercury.

Where this stops being right #

There is no air. Everything above is the vacuum envelope, and the angle that drag moves shows the level-ground optimum falling from 45° towards the thirties once drag is comparable with the weight. On a slope drag lowers the best angle in the same way and the bisector rule becomes a starting estimate; for a basketball over four metres the drag is a few per cent of the weight and the correction is small, and for a golf ball on a hillside it is not.

The ground is a plane. A real hillside curves, and a line that is the right description of a slope over a few metres is the wrong one over a few hundred. The polar form answers any straight line through the hand exactly; a curved hillside needs the envelope intersected with the curve, which is still easy and no longer one line.

The speed is the same in every direction. That is the assumption the shot put section found wrong, and it is the deepest of the three. An arm, a leg, a catapult and a moving vehicle all produce different speeds in different directions, and the envelope drawn here belongs to a thrower who does not exist.

And the target is a point. A basket is 45 centimetres across, the ball is 24 centimetres, and the question a player actually faces is how to put the ball through a window of a certain width rather than onto a point — which is an argument about the width of the stationary region rather than about the point at its centre.

What the drawings cannot show #

Every figure here is drawn in the vertical plane of the throw. The reachable set is really that region rotated about the vertical, a paraboloid with its focus at the hand, and a hillside that is not aligned with the direction of the throw cuts it obliquely. The rule for a sidelong slope is the same rule applied to the elevation of the line actually thrown along, which is shallower than the hill’s steepest slope, and none of the figures shows that geometry.

Nor can a picture of trajectories show the thrower. The least-speed figure draws a ball that leaves at one speed and one angle with no account of how either is produced or how reliably; the stationary point it marks is a property of the flight and says nothing about whether a particular throwing motion can hit it. The trade between angle errors and speed errors is real, and its terms are set by muscles rather than by parabolas.

Still open: what happens when the throw is not the same in every direction #

The polar form rests on one assumption this essay never relaxed: that every direction is available at one speed, so that the set of possible launches is a circle. A thrower on a moving platform adds the platform’s velocity to every throw and shifts that circle. A shot putter who pushes harder forward than up squashes it. A long jumper, whose horizontal speed comes from a run-up and whose vertical speed comes from one leg, turns it into something with corners. None of those is a circle, and for none of them is the best launch 45° or anything this page computes.

The question that replaces the one answered here is therefore about shape rather than about lines: given the set of launch velocities a body can actually produce, where on it is the best throw? The answer turns out to be as short as the polar equation, once range is drawn as a map over launch velocities instead of over angles — and it contains every number above as the case of a circle.

The habit worth carrying away is the one the focus made possible. When a family of problems each seems to need its own calculation, look for the object they are all intersections with. Four questions about hills, cliffs and baskets were four lines drawn across one parabola, and the parabola was already known.