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 movies.stackexchange.com Jul 2 '12 at 15:46
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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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