Consider a bouncy ball, like a basketball or golf ball what-have-you. When the ball bounces, it will compress vertically. The mass of air inside the ball remains constant, but we expect the pressure to change throughout the volume i.e. the pressure distribution is no longer uniform, and the shear strain on the surface of the ball is expected to become higher at the edges than on the the top of the ball. We model the "bounce" as an isothermal process at first.

My question is, will the ball's volume change when it bounces? Will the volume of the ball increase momentarily, or will the increased strain at the edges counteract the decreased strain on top? Perhaps I've made some incorrect assumptions as well - how might one model the process of a ball bouncing using say, thermodynamics, and stress and strain within the ball?

The process is quite transient of course, so the behavior might not be outright simple. But surely there is a way to model the bounce.

Thanks in advance.



I've found that the subject of dynamic deforming solids is contact mechanics which distinguishes between adhesive and non-adhesive contact. This question assumes the non-adhesive case. This is a "sphere in contact with a plane" problem it would appear.

[1] https://en.wikipedia.org/wiki/Contact_mechanics

[2] http://citeseerx.ist.psu.edu/viewdoc/download?doi= (Note: PDF)

  • $\begingroup$ If you're assuming the gas inside undergoes an isothermal process, and if you assume the air inside is an ideal gas, then the ideal gas law says that increasing pressure decreases volume; $V = NkT/P$. Of course, these assumptions only hold to some degree of accuracy that needs to be evaluated in more detail, but that's a simple first order analysis. $\endgroup$ – joshphysics Jul 19 '17 at 18:55

Let's consider the 2D case and ask, which shape includes the larges area if the circumference is fixed. As you probably know, the shape is given by a circle. If we consider the 3D case and ask, which shape includes the larges volume if the surface area is fixed, we obtain a sphere.

  • For a basketball I think it is save to assume that the surface area will not change significantly, if it bounces of the floor. Therefore, due to the argument given above, the volume will shrink.
  • For a golf ball this is not immediately clear, because is not filled by a gas. Therefore, the argument with the fixed surface area does not hold. However, since the Poisson ratio of all materials I know (be aware, I am not a material scientist!) is smaller than one, the volume would shrink as well.
| cite | improve this answer | |
  • $\begingroup$ Comment on answer v1: I don't see how it's safe to assume that the surface area will not change significantly. It seems totally reasonable to me that then the basketball flattens, its surface area increases in such a way that the volume is preserved. This answer would be a lot more convincing if that point were addressed more explicitly. $\endgroup$ – joshphysics Jul 19 '17 at 21:50
  • $\begingroup$ Solids are much less compressible than gases. Hence, if the ball bounces, the gas will be compressed significantly compared to the solid. As you wrote in your comment, the formula which describes this process is given by the ideal gas law. However, I would expect an adiabatic process, $\delta Q \approx 0$, because the duration of a single bounce is "short" so that heat can't be exchanged. But the formula just obscures the simple picture: ... $\endgroup$ – Semoi Jul 20 '17 at 10:45
  • $\begingroup$ ... If you have an empty basketball it will be "scrambled". If we pump it up, the surface area will not change, but the volume will increase. If we keep going, both the volume and the surface area will not change significantly, because they are correlated. However, the basketball will explodes eventually. The explosion point is given by the "fracture point" of the surface area. Hence, once the surface area changes significantly, the ball explodes. This also applies to the bouncing ball: However, as for the empty ball, the volume doesn't have to be constant. $\endgroup$ – Semoi Jul 20 '17 at 10:54
  • $\begingroup$ Elastic solids and fibrous materials might be more susceptible than a typical material, but I see your point. I disagree that the volume inside the ball will increase when the ball is pumped up. The surface is closed, with a fixed volume, and filling the ball with gas should not increase the volume apart from elastic expansion of the ball's material. As well, if the volume did not change, we would not expect a ball bounced extremely hard to pop. But I cannot imagine that a ball fired at a wall with enough force would break at the seams. $\endgroup$ – Sam Gallagher Jul 22 '17 at 1:58
  • $\begingroup$ I have to admit, I never saw an empty basketball. However, I reckon they are similar to soccer balls. Soccer balls without (!) air are "scrambled": Their shape is not spherical. Hence, if we agree that the surface area in this state does not differ significantly from the final state (as you pointed out yourself, elasticity $> 0$), I'm sure we do not disagree. $\endgroup$ – Semoi Jul 22 '17 at 4:43

This paper addresses exactly that question


A free version is available on ResearchGate


| cite | improve this answer | |
  • $\begingroup$ Please mention what is written in the paper, instead of adding to just refer to it. It would be very helpful if you could instead say what is written $\endgroup$ – SK Dash Mar 16 at 1:28
  • $\begingroup$ Sigh, downvoting relevant answers. I guess I should have known no good deed goes unpunished. Read the paper, it's free. Section 2 addresses exactly the topic of what pressure forces and shear forces arise from the isothermal compression of a ball. $\endgroup$ – Gaëtan Landry Mar 17 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.