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Tuesday 24 September 2013

The influence of the Coriolis force on a sniper bullet

(Deutsche Version) Some time ago, I watched the film Shooter (Wikipedia). In the early stages, a shot over a distance of 1,5 km is mentioned. It is also mentioned that for such a shot, even the Coriolis force has an effect. Today, I want to calculate this.

At first a short introduction about the topic Coriolis force. The Coriolis force is one of the pseudo forces which occurs in rotating frames. The force can be derived simply by transforming the equation of motion of a free particle into a rotating frame (for more information, see e.g. H.-R. Trebin, Theoretische Physik 1, Mechanik). For my calculation, this is not necessary. Basically, the Coriolis force occurs because the earth turns away under the projectile.

Let's face the calculation. To get a maximum influence of the Coriolis force, I assume that the shot is parallel to a line of longitude. Assuming a degree of latitude of 45°, the target is about 1 km farther from the rotation axis of the earth, so the target moves quicklier perpendicular to the trajectory than the projectile. Now I want to calculate these velocities. The diameter of the earth is about 12730 km. In the 45. degree of latitude, this equals a radius of about 4500 km, leading to a perimeter of 28274 km. With a time of circulation of 86400 seconds, this leads to a velocity of 327.245 m/s. If the radius is increased about 1 km, the new perimeter is 28281 km and the new velocity 327.326 m/s. This causes a difference of 0.081 m/s.

In the film, a time of flight of 6 to 10 seconds is mentioned, this would cause a deviation of 0.5 to 0.8 meters. I personally think that the time of flight is much shorter, a projectile with 1000 m/s only needs 1.5 seconds for this distance, so the deviation is about 12 cm (which still could decide wheter it is a hit or not).

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