# Projectiles, spin, momentum, and range

[SOLVED] Projectiles, spin, momentum, and range

Greetings. I am simply seeking information regarding the matters of the subject line.

Here is the situation:

A spherical projectile (6mm pellet) is fired out of a device at say 400 feet per second. It is fired out level to the ground (as if you are aiming a gun at a target - no real trajectory above horizontal).

To complicate things, a backspin is placed on the projectile inside of the barrel... the backspin causing a change in pressure above the pellet as it spins, thus to some degree counteracting gravity.

So if I were to speculate on the different ranges (distances the pellet would travel before hitting the ground) of a pellet fired at 250 feet per second vs. one fired at 500 fps, what all would come into play?

Thanks for any input.

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Gold Member
Originally posted by MattK
the backspin causing a change in pressure above the pellet as it spins, thus to some degree counteracting gravity.
The spinning causes no pressure effects -- if it were to, it would affect the top and bottom equally. Bullets are spun by a rifle barrel not to "counteract gravity," but to make them aerodynamically stable. The spinning bullet takes advantage of gyroscopic rigidity in space, and resists its direction being changed by moving air.
So if I were to speculate on the different ranges (distances the pellet would travel before hitting the ground) of a pellet fired at 250 feet per second vs. one fired at 500 fps, what all would come into play?
If you neglect air resistance (which is non-linear -- it's quadratic in the velocity), the problem is simple. The bullet falls vertically at the same acceleration (9.8 m/s2) as anything else. You can calculate how long the bullet's in the air just as a function of how high above the ground it is fired. You can calculate the distance covered by multiplying its muzzle velocity by that time.

- Warren

Originally posted by chroot
The spinning causes no pressure effects -- if it were to, it would affect the top and bottom equally. Bullets are spun by a rifle barrel not to "counteract gravity," but to make them aerodynamically stable. The spinning bullet takes advantage of gyroscopic rigidity in space, and resists its direction being changed by moving air.
Its not rifling I am talking about. The pellets are sperical, and they have actual backspin on them. The backspin allows the air to move faster over the top surface of the BB, and slower over the bottom surface.. thus producing lift as seen in an airplane wing.
If you neglect air resistance (which is non-linear -- it's quadratic in the velocity), the problem is simple. The bullet falls vertically at the same acceleration (9.8 m/s2) as anything else. You can calculate how long the bullet's in the air just as a function of how high above the ground it is fired. You can calculate the distance covered by multiplying its muzzle velocity by that time.
But air resistance is in fact what I am interested in. :) Thanks.

Homework Helper
Gold Member
The lift effect of spinning has a lot to do with the exact shape of the ball and turbulent flows around it -- this is part of the reason baseballs have seams IIRC. I think it's one of those really complicated fluid mechanics problems... I certainly don't know of any simple formulae or approximation for the effect.

Air resistance can be roughly given by the drag force

F = C*rho*v^2*A/2

where A is the area of the projectile (pi*r^2 here), rho is the density of the fluid (air), and C is the drag coefficient -- probably about 0.5.

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