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In the recent Spielberg/Jackson Tintin movie, there is a scene where Red Rackham and Captain Haddock's ships are fighting, and cannons are fired. The cannonball is shown at one point to go through a wave, and inflict serious damage on the other ship. I know that bullets stop in water; do cannonballs, with their greater weight, continue with enough force to inflict damage?

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migrated from Jul 2 '12 at 15:46

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I dont have the time to work out a complete answer, but generally, yes, the momentum of a cannonball would likely be sufficient to both pass through a wave and damage a ship. Of course, it depends on the relative sizes, but, think of it as the wave serving as a shield; the momentum would be reduced, but the ball likely wouldn't stop. –  AdamRedwine Jul 2 '12 at 17:34
It also depends on the length of water the cannonball goes through and on the projectile's shape: bullets are far more aerodynamically shaped than cannonballs. –  Emilio Pisanty Jul 3 '12 at 9:53
However, keep in mind that ship cannons were originally used to break the other ship's masts or damage its hull below the waterline in order to sink it. –  Emilio Pisanty Jul 3 '12 at 9:54

2 Answers 2

up vote 8 down vote accepted

What distance can a cannonball traverse thru water without losing too much kinetic energy? For a back-of-the-envelope calculation we start from the observation that this distance scales with the ratio of the kinetic energy of the cannonball and the drag force exerted on the cannonball.

Let's denote the ball's radius by $R$, its speed by $v$, and its mass density by $\rho_{ball}$. The kinetic energy $E_k$ equals $\frac 1 2 M v^2 = \frac{2 \pi}{3} \rho_{ball} R^3 v^2$.

The drag force $F_d$ is given by $\frac 1 2 C_d \rho_{water} v^2 A = \frac {\pi}{2} C_d \rho_{water} v^2 R^2$. Here, $C_d$ denotes the drag coefficient for a sphere.

The maximum distance $L _{max}$ that can be traversed by a cannonball $L_{max} = E_k/F_d$ is therefore $\frac 4 3 \frac {R}{C_d} \frac {\rho_{ball}}{\rho_{water}}$. For typical values ( $\frac{\rho_{ball}}{\rho_{water}} < 8$ and $C_d > 0.1$, see here), we find $L_{max} < 100 R$.

In other words, a cannonball loses much of its kinetic energy when it traverses a layer of water larger than about fifty times its diameter.

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Thank you so much! –  NiceOrc May 13 '13 at 11:49

The accepted answer is not complete. A projectile may travel very long distances underwater if it is supercavitating.

A supercavitating object is a high speed submerged object that is designed to initiate a cavitation bubble at the nose which (either naturally or augmented with internally generated gas) extends past the aft end of the object, substantially reducing the skin friction drag that would be present if the sides of the object were in contact with the liquid in which the object is submerged.

A round nosed subsonic pistol bullet will travel much further than a supersonic pointed nose rifle bullet underwater because of this effect.

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protected by Qmechanic Dec 12 '14 at 22:07

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