# How was the Titan submersible implosion time calculated?

Many experts commented about the Titan submersible that the implosion took few milliseconds. What is that assertion based on? What physics computations lead to that time?

I had imagined an abstract situation where a 1-cubic-meter box is at 3000 m below the ocean surface. Suddenly a face of the cube collapses. How long does it take for the air inside to reach the new equilibrium, and why?

• see some calculations here linkedin.com/pulse/… Jul 1, 2023 at 18:31
• There is a youtube video (youtube channel 'Not what you think'), titled: What happens when a submarine implodes The author of that video has actually done research. In 2017 the argentinian submarine ARA San Juan was lost. Hydrophones picked up the acoustics of the implosion event. General physics: upon implosion air does not immediately disperse into separate bubbles. Due to the inertia of the fast moving implosion the air is overcompressed and then it bounces back. There is a correlation between the frequency of that bouncing and the depth. Jul 1, 2023 at 19:31
• This is a sensitive topic. But have you ever seen a railroad car tank or industrial tank buckle under external pressure? Jul 2, 2023 at 1:34
• @RC_23 MythBuster had an episode of that, and I just found there's a slow-mo video of it. Jul 2, 2023 at 6:42
• @AndrewT. That mythbusters episode is about a vastly different case. In the case of that implosion the outside pressure was 1 atmosphere of air pressure. That mains the pressure differential was at most one atmosphere, whereas a submersible at thousands of meters of depth is subject to a water pressure or hundreds of atmosphere. The question is specifically about the implosion physics under the circumstances experienced by the Titan submersible. Your link to the mythbusters video is of no help to the question asked. Jul 2, 2023 at 6:57

We could estimate a lower bound for the hull collapse duration through a scaling argument: Consider the acceleration of a suddenly detached small portion of the hull (with area $$A$$ exposed to the sea at pressure $$P$$, with negligible resistance from the air on the other side) based on the force $$F=PA$$ applied to it. The acceleration of this mass $$m=\rho h A$$ (with density $$\rho$$ and hull thickness $$h$$) is thus $$a=\frac{F}{m}=\frac{P}{\rho h}$$. A distance $$r=\frac{at^2}{2}$$ is obtained from unopposed acceleration over time $$t$$; let $$r$$ be radius of the cylindrical or spherical submersible. Then

$$t\sim\sqrt{\frac{\rho h r}{P}},$$

or about 1 ms for a vessel of 1 m radius and 0.1 m wall thickness of a 2 g/cc hull material failing at a surrounding pressure of 300 atm.

This suggests that the collapse time could be decreased essentially to zero if we had a sufficiently light or thin—but still very stiff and strong—hull. However, the water still has to accelerate itself, with the thickness $$h$$ of the relevant slug scaling with $$r$$ (i.e., most of the water motion occurs near the submersible): $$t_\text{water only}=r\sqrt{\frac{\rho_\text{water}}{P}}$$. This actually produces an estimate of around 10 ms using the above values.

With multiple processes occurring in parallel, one has to take the limiting process (in this analysis, the water movement) as dominant when calculating a minimum time. Thank you to @hft for clarifying this point.

I think you can model the collapse of the Titan as a collapsing cavitation bubble. Following the paper by Leen van Wijngaarden, "Mechanics of collapsing cavitation bubbles", Ultrasonics Sonochemistry, 29, 2016 https://doi.org/10.1016/j.ultsonch.2015.04.006

From Equation 3, the bubble wall velocity $$u$$ reaches the value $$u = \left ( \frac{2 p(\infty )}{3 \rho} \left ( \frac {R_0}{R} \right ) ^3\right ) ^ \frac{1}{2}$$ where $$R$$ is the radius of the bubble, $$R_0$$ is the initial bubble radius, $$\rho$$ the density of the liquid and $$p$$ the liquid pressure, $$p(R)$$ at the interface and $$p(\infty)$$ far away from the bubble.

$$p(\infty) = \rho \, g \, d$$ where $$g$$ is acceleration due to gravity and $$d$$ is depth. The Titanic is at 3800 m so $$p(\infty)$$ = 37.27 MPa

The internal volume of the Titan submersible is 2.4 m long with a 56" ID. This is a volume of 3.06 m$$^3$$. If we model this volume as of a sphere, $$V = \frac{4}{3} \pi R_0^3$$ where $$R_0$$ = 0.9 m.

If the final volume after collaspe is 0.35 m$$^3$$ (5 x 70 kg i. e. the volume of 5 average people) then the final radius $$R_\infty$$ is 0.44 m

The final velocity is $$u$$ = 466 m s$$^{-1}$$

The time taken to collapse is then $$t=2 s / ( u + v)$$ = 2 (0.9 m - 0.44 m) / (466 m s$$^{-1}$$ + 0 m s$$^{-1}$$) = 1.97 ms

• Could go back to another paper on this topic: Lord Rayleigh O.M. F.R.S. (1917) VIII. On the pressure developed in a liquid during the collapse of a spherical cavity , The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 34:200, 94-98, DOI: 10.1080/14786440808635681 Oct 5, 2023 at 7:55