In a nutshell, the question is whether it can be faster to bounce a ball down an infinite flight of stairs than to bounce it down a ramp with the same slope.

To be more specific: this is a $2$ dimensional problem, so my ball is a disk and all the action takes place in $\mathbb{R}^2$. I imagine an infinite staircase with steps having depth $1$ and height $1$, the top corner of the top step is at the origin (and the staircase is oriented down and to the right). The ball has radius $1$, so when we toss the ball on the stairs and it will impact on an outside corner of a stair every time. Gravity acts downward with constant acceleration $g$. We compute the bounces by replacing the single corner point of the stair with an imaginary line tangent to the ball at the impact point.

Let $s(t)$ be the $x$-coordinate at time $t$ of the ball bouncing down the stairs, and let $r(t)$ be the $x$-coordinate at time $t$ of the ball rolling down the ramp.


  1. Are there initial conditions for which $s(t) > r(t)$ for all sufficiently large $t$?
  2. Does the answer depend on the value of $g$?


  1. Can we determine the distribution of the incidence angles? Is it uniform? Does the distribution govern the long-term behavior of $s$?

  2. We can replace the ball/stairs system with an equivalent one in which we track only the center of the ball. The center of the ball bounces down a "cobblestone ramp" made up of quarter circles (with radius $1$) heading down and to the right at slope $-1$.

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    $\begingroup$ On a (likely with probability 1) set of conditions, the ball will go slower, because the forces it receives from the staircase will be more in opposition to gravity than those from a ramp of similar slope. (This may be true even for balls of large radius, but is less clear.) The only time the ball might go faster is if it hits the corner of the stair in a way that the force is directed more horizontally than it would be from the ramp. This depends strongly on the ratio of ball radius to stair step dimensions. $\endgroup$ Jun 19, 2014 at 2:53
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    $\begingroup$ Now that I've seen that the ball has large enough radius, I suspect the answer is that it will be faster, because the ball will tend to impact the stair at a shallow angle and so receive more of a propelling force than a retarding force. I'm curious as to what a simulation would reveal. $\endgroup$ Jun 19, 2014 at 3:00
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    $\begingroup$ Just a tangent: When the ball has friction and can spin, a ball thrown down the stairs can actually climb up a bit. See this MO question, and especially this figure. $\endgroup$ Jun 19, 2014 at 11:43
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    $\begingroup$ I've often bounced cricket balls down stairs. If the ball catches the corner of a step just right then it can suddenly shoot forward very quickly. I don't know if this model captures enough of the physics to have the same behaviour. $\endgroup$
    – Ben Barber
    Jun 23, 2014 at 15:45
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    $\begingroup$ @BenBarber Right! and then sometimes it hits flat or almost flat and gets very little forward thrust. Then meanwhile the ball on the ramp gets the "same" proportional forward thrust each time. So this is some kind of averaging question. $\endgroup$
    – Jeff Strom
    Jun 23, 2014 at 16:05

1 Answer 1


Consider trajectories that start on the slope. We measure success by how long it takes to reach any point a given vertical distance below the starting point (which is not exactly the success measure in the question but may be a satisfactory replacement).

Let $s$ denote distance along the trajectory. Let $t(s)$ be the time at which point $s$ is reached, and $y(s)$ be the vertical displacement from the starting point (measured downwards). Also let $u$ be the initial speed. Since the collisions are elastic, conservation of energy implies that the speed of the ball when it has gone distance $s$ along its trajectory is $\sqrt{u^2+2g\, y(s)}$. From this it follows that $$t(s) = \int_{0}^s \frac{dq}{\sqrt{u^2 + 2g\, y(q)}}.$$ Now suppose you could choose any trajectory lying on or above the smooth slope. For trajectories of given length $s$, you can minimise $t(s)$ by maximising $y(q)$ as much as possible for $0\le q\le s$, which seems a bit vague value until you realise that moving down the slope makes $y(q)$ take its maximum possible value $q/\sqrt 2$ for all $q$. Since this straight line also minimises the length of the trajectory to any given vertical distance below the starting point, we can infer a theorem:

If the ball starts on the slope at speed $u$ and can be constrained by elastic collisions to take any trajectory on or above the slope, the fastest way to reach any given vertical distance below the starting point is sliding down the slope.

The staircase can be drawn so that the inner corners lie on the slope. If you have to start at such an inner corner, the above theorem shows that you can't choose any initial starting speed and direction so that the ball will reach a given vertical distance below the starting point faster than it would if you used the same initial speed down the smooth slope.

If you can start at places other than the inner corners, a bit more work needs to be done (starting with an exact definition of the problem). But it is still clear that the slope is better eventually.

A further, much harder, question is how much worse the staircase is. As Ryan commented, repeated glancing collisions seem best but I'm not sure if such trajectories are physical. Might it be that no matter how hard you try eventually the ball will encounter hard collisions that make it bounce upwards? This would be disastrous since it can bounce up to almost the same height it started from.

ADDED: In the above I treated the ball as a point mass. If it is a disk of finite radius (but still frictionless so the angular inertia is not an issue), replace what I called "the slope" by the lowest 45-degree line that the centre of the disk may touch and "the trajectory" by the path taken by the centre of the disk. Draw the staircase so that the centre of the ball is on that line when the ball is sitting on a step as far in as possible. The analysis appears to be the same.

  • $\begingroup$ Is it possible to bounce the ball down the ramp so that it approximates the slide to increasingly good accuracy? $\endgroup$
    – Jeff Strom
    Jun 24, 2014 at 16:45

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