Curving bullets

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This week the Mythbusters tackled the question of whether you can make a bullet follow a curved flight path, as in the movie Wanted. The characters in the movie are able to do this using some fancy flick of the wrist as they fire the gun, but is it really possible? Apparently a lot of people were wondering.

The short, simple answer is no. It’s an obvious application of Newton’s first law of motion: objects moving in a straight line will continue moving in a straight line at constant speed, unless subject to an external force. But there are only a couple of external forces that can act on a bullet: air resistance and gravity. Gravity certainly isn’t going to make the bullet curve sideways as we see in the movie, and all air resistance will do is slow it down, not change its direction.

Then again, Kari, Grant, and Tory hit on an important point: bullets are highly symmetric and are typically ejected from the gun barrel with high spin. All this is optimized for motion in a straight line toward whatever you’re aiming the gun at. What happens if you use asymmetric, oddly shaped bullets?

In order to follow a curved trajectory, the bullet would need to experience a force perpendicular to its motion. Air resistance is normally antiparallel to the motion, but in principle, say by shaping one side of the bullet into a “ramp” for the oncoming air, you could redirect some of that force in a perpendicular direction to make the bullet curve. The thing is, this effect is pretty small, and any force coming from air resistance would only be able to make a microscopic nudge over the time it takes a bullet to traverse, say, a gun range. Specifically, drag force can be calculated as

$$F_D=\frac{1}{2}\rho v^2 A C_D$$

where \(\rho\) is the density of the fluid, \(v\) is the bullet’s velocity, \(A\) is the cross-sectional area of the bullet, and \(C_D\) is a unitless constant on the order of 1. If we could somehow direct all of this drag force into curving the bullet, then the centripetal force would be equal to the drag force,

$$\frac{1}{2}\rho v^2 A C_D = \frac{mv^2}{r}$$

which means that the bullet would have a radius of curvature of

$$r = \frac{2m}{\rho A C_D}$$

The radius of curvature is a standard way to quantify curvature in physics; basically, it’s what you would get if you took a part of a curved path, extended it into a circle, and measured the radius of that circle. Putting it some reasonable numbers — a bullet mass of about \(\unit{5}{\gram}\), an area of about \(\unit{1}{\centi\meter\squared}\), and the density of air as \(\unit{1.3}{\kilo\gram\per\meter\cubed}\) — this comes out to about \(\unit{77}{\meter}\). So the best curve that would be ideally possible is a circle of radius \(\unit{77}{\meter}\), or \(\unit{252}{\foot}\). That’s pretty big, but you could probably still notice it over the span of a gun range, if you had a laser pointer as the Mythbusters did. Of course, the best curve you could really get in practice would be far, far straighter, since you couldn’t possibly hope to harness all the drag force to curve the bullet.