While catching up on some old Mythbusters episodes, I ran across an interesting myth about spinning bullets. Apparently when you shoot a bullet into the surface of a frozen lake, it bounces backward, bounces over the ice a little way, and keeps spinning even after it comes to rest.

This is a very curious result. At first, it kind of seems to make sense. A bullet comes out of the gun spinning at 80000 revolutions per minute, according to the show, which is pretty fast. So it makes sense that it has a sizable amount of angular momentum, which in turn means it'll take quite a bit of torque to stop it. Colliding with the ice conceivably might not be able to exert enough torque to do that, so the bullet would keep spinning.

But if that were the case, the bullet should maintain its orientation — it'd keep pointing in the same direction that it came out of the gun, because angular momentum is a vector quantity and it won't change in either magnitude or direction without an external torque. That clearly wasn't the case; the bullets the Mythbusters shot skittered across the ice, spinning in multiple directions, and landed with an orientation completely different from what they had when they were shot. How does that happen?

I'm not entirely sure what actually goes on with these spinning bullets, but I think the key lies in the direction of the torque exerted. Like angular momentum, torque is a vector quantity. In order for a torque to stop the bullet, it needs to point directly antiparallel to the angular momentum, but when a bullet collides with ice, the collision is quite complicated and there's no reason to expect that the net torque will point neatly along the angular momentum axis.

There are two main sources of torque that I can think of: friction between the edge of the bullet and the ice, and impacts between the bullet and fragments of ice that are flying around in the aftermath of the collision. The friction would certainly contribute to slowing the bullet's rotation, but the impacts would be more or less random, so they could conceivably create the right kind of torque to change the orientation of the spin without necessarily slowing it down. Plus, I can imagine that the torque provided by friction may not be all that much — after all, ice is slippery, and besides, the bullet bounces out (and is no longer touching the ice) after a very short time. If the impacts provide much more torque than the friction, it would account for the tumbling behavior you can see in some of the high-speed excerpts they put in the show, and it would also explain why the bullet keeps spinning as it bounces around the ice.

Whenever Mythbusters meet bullets — no, not literally, though this week's episode of Mythbusters does have Adam and Jamie trying to shoot cardboard cutouts of themselves — you know something wacky and interesting is about to ensue. The myth in question is that, with an unwisely aimed shot, it's possible for a bullet to bounce off three steel beams and come back to hit the shooter.

Seems straightforward enough, right? If the beams, or walls as the case may be, are lined up at right angles to each other, why shouldn't a bullet just bounce off all three and come right back to where it started?

As Adam and Jamie (re)discovered during the show, bullets don't bounce, at least not when they're moving as fast as, well, a speeding bullet. They shatter on contact with any hard enough surface, like steel, and the pieces spray out in what could be a completely different direction from what you'd naively predict.

From a physics perspective, this highlights the difference between elastic and inelastic collisions. Elastic collisions are based on the idea of a ball bouncing off a wall; it goes in at some speed and bounces right back at the same speed, at the same angle, like a laser bouncing off a mirror. The key characteristic is that the ball exits with the same amount of energy it came in with, and that's what defines an elastic collision: no kinetic energy is lost (except in experiments with subatomic particles, where "elastic collision" means something slightly different, but that's another story).

The nice thing about elastic collisions is that they're easy to analyze. If we know the orientation of the surface along which two elastically colliding bodies are going to contact each other, the laws of conservation of momentum and energy,

are enough to determine the final velocities of both objects. You just decompose the momentum (or velocity) into components parallel to and perpendicular to the collision surface; the parallel components are unaffected, and you can solve the equations for the perpendicular components. They work out to

Unfortunately, most collisions in the real world (again, this doesn't quite apply when you consider the microscopic view) are inelastic collisions, which do involve a loss of kinetic energy. The archetypal example of this is two lumps of clay, or perhaps bread dough, smacking into each other. Instead of bouncing away like rubber balls, they stick together and continue to move with one combined velocity, (try it!), which is most certainly not what the solutions for and describe. So if the answer doesn't correspond to reality, unless the math is wrong (and it's not, I checked), one or more of the underlying assumptions must be wrong.

Experiments have shown that conservation of kinetic energy is that wrong assumption. What actually happens in an inelastic collision is that some of the kinetic energy gets transformed into some other form — heat, sound, maybe even light if the right materials are involved... but one of the most common destinations for extra energy is deformation. Real objects are not perfectly rigid; they can be squashed, squeezed, extruded, twisted, dented, bent or broken in various ways. All of that takes energy. The reason inelastic collisions are so hard to analyze is that there are so many different ways in which the energy can be used up, and exactly how much energy goes where depends on the detailed properties of the objects colliding. There's practically no way to figure out what really goes on inside a collision at the level of detail needed to calculate it from scratch.

But never fear, physics has not failed us yet ;-) The coefficient of restitution is a simple variable that describes, at the macroscopic level, how objects are affected by collisions: it's defined as the ratio of final velocity difference to initial velocity difference, in the direction perpendicular to the collision plane.

For elastic collisions, , which makes that definition equivalent to the conservation of momentum (the minus signs in the definition come from the fact that the two objects' velocities are in opposite directions). By analogy, we can write the conservation of perpendicular momentum for an inelastic collisions, , as

and the solutions for the final velocities become

The example of the two lumps of clay colliding is an extreme case, where , although that extreme case is not too far off from the bullet-on-steel impacts the Mythbusters measured.

If this were a full-blown research project, I might try to make a rough calculation of the coefficient of restitution for bullet-on-steel impacts, using the more fundamental properties of the various metals involved. But that would take way more time than I have right now (too bad, because I'm really curious to know how accurate of a model I could come up with... oh well, maybe someday). To do something quickly, let's just see what kind of a coefficient of restitution would be required for the ricochet Adam and Jamie were looking for.

I'm going to make a couple of assumptions that are not exactly correct, but I hope they're going to be close enough:

1. The steel plates (walls) don't move
2. The only force on the bullet is perpendicular to the plate

Here's why: assumption #1 means the velocity of the plate, let's call it object 2, is negligible. So the coefficient of restitution simplifies to

Assumption #2 means that the component of the bullet's velocity parallel to the plate stays the same during the collision, . Now I can use the fact that , and similarly for the outgoing angle, to write

The second assumption also means that the parallel component of the bullet's momentum is conserved,

With a little creative algebra, I can combine this with the expression for to give

The first thing to look at is the problem that Adam and Jamie kept coming up against: the collisions they were getting slowed their bullets and ball bearings down to the point where the shot would no longer be lethal once it got back to the shooter. The target velocity they were aiming for was about . Given that these bullets emerge from the gun at about , in order for a ricochet to be lethal, we need the product of the three factors

to be greater than . (By the way: a third assumption is in order here, that depends only on the materials involved in a collision and not on the angle of impact or the velocity. This is not exactly true, but again, I hope it's close enough for demonstration purposes.)

I'm going to look at a simple case: let's assume that the walls are oriented at angles, to form a rectangular area for the bullet to ricochet around in. That tells us a specific relationship between the final angle of each collision and the initial angle of the next, namely that they add up to , so the cosine of one is the sine of the other, and vice versa. This means that

Now here's something interesting I discovered: if you take the first two factors modifying the ricochet velocity,

and plug in for in terms of , you get

which is to say that under these particular conditions, regardless of the angle, a bouncing bullet's velocity drops by a factor of after two bounces! This makes the calculation considerably prettier, since the product of three square roots from above reduces to

Remember, this just has to be larger than to make a successful kill. It's easy enough to plot it for some reasonable values of to see if it's going to work:

From the Mythbusters' experiments with cinderblocks, I calculate , the yellow line in the plot, which means they might just be able to keep the bullet going fast enough... or rather, might have been, if these calculations were at all accurate ;-) In reality, it didn't quite work, and I'm not surprised because I've been making some pretty optimistic assumptions.

It's certainly been evident that this problem has exploded in complexity — certainly I had no idea what I was getting in to when I started writing! And we're still stuck on the rectangular box model, which only worked because of that fortunate substitution . What about arbitrary orientations, like the final cinderblock trial on the show? I don't know if there's any way to get an analytic solution, but this strikes me as an ideal opportunity to switch over to computational methods and demonstrate an evolutionary algorithm that selects the best candidates for reflected kill shots. But that's a project for another day. I'll have to leave off with this teaser graph that shows some different bullet trajectories in the rectangular box model. Color represents initial angle, opacity represents final velocity; and it's easy to tell that nothing moving fast enough to be lethal quite makes it back to the point of origin...

I guess one point of this is clear: if you're going to shoot yourself, better point the gun at your head.

This is one of those really cool things that I've often wondered about: can you really shoot a gun out of an outlaw's hand? Last week on Mythbusters, Adam and Jamie decided to test it out. Sure, it's not the kind of thing you'd think would be easy (or safe) — unless you have access to that classic Mythbusters creativity. Their first idea involved a Velcro-like gripping arm to hold the gun, and although it may not be clear just how exactly that compares to a real hand, they obtained some interesting results from comparing the different gripping positions.

Anyone who's ever tried to pry an object out of somebody's hand knows that the easiest way to do it is to twist it to apply stress on the thumb, the weakest point of the grip — not just to hit it as hard as possible. And whenever an object is twisting or rotating, the operational physical principle is torque, the rotational analogue of force. Torque can be calculated from the formula

but in most simple cases, we can identify an axis of rotation and then calculate the torque around that axis as

Here is the radius of rotation, the distance from the axis to the point at which the force is applied in the direction perpendicular to the force.

For a gun in an outlaw's hand, the easiest axis of rotation is pretty clear: it runs up and down along the handle of the gun. His fingers wrap around the handle, holding it more or less in place (as long as he's able to hold on to it), but there's not much other than friction to prevent the gun from rotating around while the handle stays in his hand. According to the formula , then, you can get a higher torque by shooting a point on the gun that's further away from this rotation axis. This is exactly why the "hostage position" on the show was the easiest one to disarm: you're shooting at a side view of the gun, which allows you to hit it near the business end of the barrel, far away from the axis of rotation. in this case can be up to the length of the gun, maybe for the type they were using on the show.

Of course, Adam and Jamie also tested two other positions, the "draw position" and the "shooting position." Neither of these gives you a side view of the gun, so if you hit it, you're shooting right in line with that axis of rotation that runs through the handle. This means that you won't be able to twist the gun around that axis. But there are other ways it can twist; specifically, the gun can rock back and forth in the outlaw's hand, and with enough force from the bullet, it can rock right out of his grip. The same rule about torque applies, namely that the further away from the axis of rotation (which now runs through the trigger hole, more or less) you hit, the more torque the bullet can apply. And that's indeed what Adam and Jamie found: in the "draw position," when you get a top view of the gun, you can again hit it up to a barrel length away from the axis, making it hard to hold on to, whereas in the "shooting position" you're facing the smallest possible target presented by the gun and are forced to hit it nearly on-axis. This tends to knock the outlaw's hand back but not really dislodge the gun.

Besides the radius from the rotation axis, the amount of torque exerted by a bullet is also influenced by the force the bullet can impart, and that in turn depends on how the bullet collides with the gun. From the high-speed footage in the show, we saw that the bullets flew apart into a lot of little pieces when they hit the gun, and the pieces spread out in all different directions. This is what we in physics call an inelastic collision (an elastic collision would have the objects involved just bouncing off each other). Inelastic collisions can be quite difficult to analyze, because not only are there many different pieces to keep track of, but those pieces could have a range of different velocities depending on how much energy is lost to heat and sound. Still, we can make some reasonable-sounding guesses.

Suppose, for instance, that after the collision, all the pieces of the bullet have the same forward velocity , and the gun's forward velocity is some possibly different forward velocity . Then the total post-collision momentum is

where I've indicated that it has to be equal to the pre-collision momentum . Also, we can compare the total energy before and after the collision,

Putting these together, we get

which translates to

(I used a computer to simplify that, but you could work through it by hand if you wanted to) This places an upper limit on the momentum of the gun after the collision. And this, in turn, is useful because the force is given by

(this is a rough approximation, but it lets me avoid relying on information I don't have) We can estimate , the time over which the momentum change takes place, to be approximately the amount of time it takes the bullet to travel its own length,

and putting all this together,

So we've calculated a rough upper bound for the amount of force that you'd feel from a bullet hitting your gun, and knowing the radius from the rotation axis, we can use to calculate the maximum torque.

Why did I do that? I'd like to compare the actual impact from the bullet with the gun-shaped paddle target that Adam built to test whether he'd be able to hold on to a gun. According to Jamie, for that experiment they chose the mass and speed of the baseball bat so that it would have the same kinetic energy as a bullet. But I think they should have been comparing torque. (Then again, that calculation makes for lousy entertainment, I suppose it's too much to expect them to do it on the show ;-)

Let's start with the torque delivered by a real bullet. Obviously the numbers involved will vary from one type of bullet (and gun) to another, but Jamie quoted values of and on the show, so let's go with that. I'll have to estimate some other values, and . (if someone can identify the type of bullet and gun they used on the show so I can get better values, please let me know!) Plugging these numbers in, we get

Whoa. That's a lot of force. But still, we've just calculated an upper bound, so the real amount of force could definitely be a lot less.

Anyway, on to compare with the bat. Since the inequality from above was derived for a bullet and a real gun, we'll need to check and see how well it applies to a bat and a target... as it turns out, there really isn't much that needs to change. Everything I said about conservation of momentum and energy was completely general, so it applies equally well to both situations; we just need to come up with a new estimate for how long the collision between the bat and the target takes. Unfortunately, trying to do so analytically would be worthy of an entire post itself (probably longer than this one), and it depends on a lot of factors that we really have no information about, such as how tightly Jamie was holding the bat. I'm just going to make a guess that the collision occurs within the space of about . Then plugging in that for , again, , and , we get

which, while still a lot of force, is a whole order of magnitude less. You might object that the target-paddle-toy Adam built may have been a lot lighter than a real gun, but it turns out that that further reduces the force. So these numbers are strongly indicating that the torque from a real bullet impact on a real gun is a fair amount more than from a baseball bat hitting a paddle.

Just one problem: that's not what we saw on the show! Toward the end, when the Mythbusters tried Jamie's idea of holding a gun to see if they could withstand the shock when it was fired, Adam was able to hold on to the gun in all three positions, whereas when he tried the paddle... well, did you see the look on his face? (If you didn't: it was painful) There are a few possible explanations I can think of. First of all, Adam himself said that the "training" he got from the baseball bat experiment may have better equipped him to hold on to the gun. Also, a lot of uncertainty comes in from estimating the distance over which each collision takes place, . Those numbers could have been off by an order of magnitude or so, especially in the case of the bat, and that could mean that the forces were not so dissimilar.

I suspect, though, that the major factor might be one particular difference between a baseball bat and a bullet that I haven't really accounted for: the bat has a person on the other end, and that person can keep applying extra force throughout the collision. This essentially increases the period over which the force is applied, as well as increasing its magnitude, and that can definitely make a big difference. If you don't believe it, try opening a door by repeatedly throwing a baseball at it. (On second thought: don't try this at home) Unless the door is particularly well balanced, it doesn't budge very much, because the baseball, like a bullet, can only apply a limited impulse. Compare that with how easy it is to open a door with a push of your hand — not because you push with a larger force, but because you can sustain it even as the door starts to move away from you.

I guess this means that if you're ever in the situation of having to disarm someone without hurting them, opt for the baseball bat ;-P But seriously: as found on the show, bullets just aren't going to do the job.

With their season premiere this week, the Mythbusters are testing a classic physics story, so of course I had to comment on it. The myth in question is that if you fire a bullet from a gun held horizontally, it will hit the ground at the exact same time as a bullet dropped without any horizontal motion at all.

Of course, in the mind of any physicist, this is no myth at all — the laws of physics that tell us this should happen are so well established that they're almost beyond question. Specifically, it's the linear independence of orthogonal vectors, which means that components of motion that are perpendicular to each other, like gravity (vertical) and constant velocity (horizontal), don't get in each other's way. You can split the motion of the bullet into two perpendicular components and analyze each one separately. This is, in fact, one of the first things students learn in an introductory physics class: analyzing the motion of a fallen or thrown object. The equations and work for both the fallen bullet and the dropped bullet, just with in the latter case. But that velocity doesn't make any difference at all to the vertical motion, since doesn't appear in the equation for at all.

Unfortunately, (practically) nothing in real-world physics is really that simple. In reality, we have air resistance to deal with. Air resistance slows down a speeding bullet, and so you'd think that it would hold the fired bullet back, so that the dropped bullet would hit the ground first. But wait! The dropped bullet is also moving through the air, and it experiences air resistance as well. So does that mean that they would hit the ground at the same time after all?

The drag force on an object moving through a fluid can be calculated as

where is the density of the fluid, is the object's velocity, is its cross-sectional area, and is the drag coefficient, which is typically on the order of 0.1 for streamlined shapes or 1 for rounder or blockier shapes. Wikipedia reports for some bullets so let's go with that. The first thing to notice is that this formula depends on velocity squared — it's non-linear. That means that when drag force is involved, the horizontal and vertical components of velocity may not be independent anymore.

Drag force always acts antiparallel to the overall velocity of a moving object. So if the bullet is moving straight horizontally, as it starts out, the drag force will also be horizontal, but as the bullet's trajectory tilts downward, the drag force will angle upward to remain antiparallel. The horizontal and vertical components of the drag force are

and including that effect modifies the equations of motion to

Clearly, this is no longer a simple set of equations. You can't solve it exactly, but you can simulate it on a computer.

I set up a simulation of the track defined by these equations, using the Open Source Physics toolkit to do the calculation and rendering. In the show, Adam and Jamie said that they found the two bullets (fired and dropped) hitting the floor within about 36 milliseconds of each other. What the simulation can tell us is whether air resistance accounts for that difference. So does it? Well, here's the simulated result:

This graph is zoomed in closely on the point where both bullets would be hitting the floor (that would be the 0 on the vertical axis). The green track represents the dropped bullet and the red track represents the fired bullet. According to this output, the dropped bullet in the simulation hits the ground after falling for and the fired bullet hits the ground after failling for slightly under — a difference of only , about a quarter of what the Mythbusters observed.

So, does this mean that a dropped bullet and a fired bullet actually don't hit the ground at the same time? Actually, we really can't tell based on the Mythbusters' single experiment and one simple simulation. is a pretty small time compared to the total flight time of the bullets, and if we wanted to be able to confidently compare the simulation's result to reality, we'd need to have a much more accurate model of the forces acting on the bullet in flight. And even if our simple model is reasonably accurate, there are only a few tens of milliseconds left to be accounted for by other factors. That's not very much. Everyday devices like the gun and the drop rig used in the experiment just aren't designed to activate with that level of precision, so it's easy to imagine that some sort of mechanical randomness could introduce a fraction of a second's delay one way or the other, enough to account for the discrepancy between the experiment and the simulation. It would have been great if the Mythbusters took the time to repeat their shot to try to eliminate statistical errors from the data, but the next 20 trials probably aren't such great television.

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

where is the density of the fluid, is the bullet's velocity, is the cross-sectional area of the bullet, and 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,

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

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 , an area of about , and the density of air as — this comes out to about . So the best curve that would be ideally possible is a circle of radius , or . 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.