# How deep in the ocean can a ping-pong ball go before it collapses due to pressure?

Stimulated by the calculation showing that a ping-pong ball does not pop in vacuum, I'm driven to ask how deep in the ocean a ping-pong ball can be brought before it collapses due to pressure.

This has a practical application in that boats have been refloated by the expedient of pumping ping pong balls into their hull.

• There is a telltale, that this Donald Duck episode was read as "prior art", leading to refusal of a patent application. Which patent office was involved, I do not remember. The story was told by the patents specialist of our company. Apr 19, 2011 at 9:51
• Here is a page on the patent issue: iusmentis.com/patents/priorart/donaldduck Apr 19, 2011 at 15:07
• How did they manage to get the balls into such depth? The water has much greater pressure than atmosphere down there. Did they use some special kind of pump? Apr 5, 2015 at 9:24

The collapse pressure of a ping-pong ball is probably limited by the eventual buckling of the wall, instead of a compressive failure of its walls. Assuming (unrealistically) that the ball is a perfect sphere, the critical external pressure will be given by

$$P_e - P_i = \frac{2E\,\left(\frac{h}{R}\right)^2}{\sqrt{3(1-\nu^2)}}$$

where $P_e$ is the external pressure, $P_i$ the internal pressure, $E$ the Young's modulus of the material, $h$ the shell thickness, $R$ the sphere radius and $\nu$ the Poisson's ratio.

A ping-pong ball has a diameter of 40 millimeters, a weight of 2.7 grams and is made of celluloid. As celluloid's density is $1.4\,\mathrm{g\ cm^{-3}}$, its wall thickness will be

$$h = \frac{2.7\,\mathrm{g}}{4\,\pi\,(2\,\mathrm{cm})^2\,1.4\,\mathrm{g\ cm^{-3}}} \approx 0.04\,cm$$

The Young's modulus for celluloid is approximately $1400\,\mathrm{MPa}$ and its Poisson's ratio can be estimated as 0.35. Replacing these values in the expression for the critical pressure:

$$P_e - P_i = \frac{2\cdot 1400\,\mathrm{MPa}\cdot 0.02^2}{\sqrt{3(1-0.35^2)}} = \frac{1.12\,\mathrm{MPa}}{1.62} \approx 0.7\,\mathrm{MPa}$$

Assuming a compressive strength of $50\,\mathrm{MPa}$), the ball will fail "compressively" at a pressure of

$$P_e - P_i = \frac{50\,\mathrm{MPa}\cdot 0.04\,\mathrm{cm}\cdot 2\,\pi\cdot2\,\mathrm{cm}}{\pi(2\,\mathrm{cm})^2} = 2\,\mathrm{MPa}$$

so the real failure mode will be buckling. If the internal pressure of the ball is not very different from the atmospheric pressure, the crush depth of a perfect ball will be

$$d_c \approx \frac{0.7\,\mathrm{MPa}}{0.01\,\mathrm{MPa}\ \mathrm{m^{-1}}} = 70\,\mathrm{m}$$

The real crush depth will be between a half and a quarter of this value, matching the experimental value of approximately 30 meters.

• Surprisingly deep. I must take one in my pocket next time I dive - I should see something at 18m. Apr 5, 2015 at 6:02
• Where is the first formula from? Jun 28, 2023 at 21:19

Let me assume that the answer to the previous question is right that the tensile strength is 50 MPa. I think that Georg is right that it's not correct just to equate that to the pressure. (I thought it was right at first -- dimensional analysis is often your friend!)

Here's a very heuristic argument why not. Suppose that the ball is under high pressure from the outside. Then the downward force on the upper hemisphere (or the upward force on the downward hemisphere) is $\sim PA\sim P\pi r^2$, where $r$ is the ball's radius. The part of the ball around the equator has to withstand that force. If the tensile strength is $T$, then the maximum force it can withstand is something like $T$ times the area of the ball that you'd expose if you sliced the ball in half around the equator. That is, it's $T(2\pi r)t$ where $t$ is the thickness of the material.

So the maximum pressure the ball can withstand is something like $P\sim Tt/r$. If $t/r\sim 0.01$, then $P\sim 5\times 10^5$ Pa. That's 5 atmospheres. One atmosphere is 10 meters of depth in water, so I estimate that it's of order 50 m.