Why does an artillery shell rotate when fired?

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An artillery shell rotates when fired due to its shape and the aerodynamic forces acting on it, which help orient the shell nose-first upon impact. As the shell follows a parabolic trajectory, air resistance combined with its longitudinal spin influences its orientation. The design of the shell, particularly its pointed forward end, contributes to this rotation, similar to how an airplane's elevators affect its nose position. While rifling grooves are used to prevent tumbling in bullets and shells, most mortars, which are not rifled, still manage to hit nose-first. Understanding these dynamics is crucial for optimizing range and accuracy in artillery fire.
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An artillery shell is fired at, let's say, a 45 degree angle. The shell will rise to a maximun height and then fall back to earth, but the shell will also rotate considerably so that it strikes the ground with the nose forward.

What accounts for the rotation? It would seem that the center of gravity would tend to orient the shell somewhat vertically even with the aerodynamic forces acting on the body. Also, the shell is spinning about its longitudinal axis and this would prevent any rotation.
 
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Frank Peters said:
An artillery shell is fired at, let's say, a 45 degree angle. The shell will rise to a maximun height and then fall back to earth, but the shell will also rotate considerably so that it strikes the ground with the nose forward.

What accounts for the rotation? It would seem that the center of gravity would tend to orient the shell somewhat vertically even with the aerodynamic forces acting on the body. Also, the shell is spinning about its longitudinal axis and this would prevent any rotation.
The rotation you're asking about (not the rotation about the longitudinal axis of the projectile) is due to the shape of the shell, being pointed on the forward end. The natural tendency of the shell would be to maintain the same angle with respect to the horizontal, but as the shell rises, its center of mass follows a roughly parabolic arc, so I would imagine that air resistance plus its longitudinal spin force the shell to be oriented tangent to the curve it's following.
One way to think about this is with the elevators on an airplane's rear wing. If both elevators are in the lowered position, the plane's nose drops. I think something like this happens as the shell starts to drop. The rear of the shell is hanging down, relative to motion through the air, which exerts an upward force on the rear of the shell, and causing the front of the shell to drop.
 
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It is drag from aerodynamic forces. A spherical cannonball will behave differently.

Somettimes bullets or artillery shells tumble through the air. Trying to prevent that is the purpose of the spin put on be rifling grooves.
 
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anorlunda said:
It is drag from aerodynamic forces. A spherical cannonball will behave differently.

Somettimes bullets or artillery shells tumble through the air. Trying to prevent that is the purpose of the spin put on be rifling grooves.
Most mortars are not rifled, but still hit nose first, the rifling makes a difference in range - at least for the US 120mm mortar. The only bullets I know that tumble are ones that strike something first (this was common with the 5.56 M16 round. If it hit a twig it would tumble.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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