Have you ever read a book that just sticks with you for a long time? For me, it's The Black Swan: The Impact of the Highly Improbable, by Nassim Nicholas Taleb. There's a lot of great stuff in there, but one thing that I think about often is his mention of a 1978 paper by physicist M. V. Berry titled “Regular and Irregular Motion.” Berry shows how difficult it can be to predict future motion in some situations. For instance, in billiards we can calculate the result of two balls colliding. However, if you want to look at nine successive collisions, the outcome is very sensitive to the velocity of the initial ball. In fact, Berry claims that in order to correctly predict the outcome, you would have to also include the gravitational interactions between the first ball and the player who shot that ball.
OK, just to be clear—there is a gravitational interaction between all objects with mass. However, in most cases this interaction is super tiny. Suppose you have a person with a mass of 68 kilograms (about 150 pounds) holding a pool ball with a mass of 157 grams a distance of 1 meter away from their body. The gravitational force the human exerts on that ball would be around 10-9 newtons. I mean, that is so tiny that I don't even have a comparison. Even the weight of a grain of salt (its gravitational interaction with the Earth) would be about 1,000 times greater. Could such a small force really even matter? Let's find out.
I'm going to start with two colliding balls, and I'm going to make some assumptions so that we can at least get a rough answer to this question. Don't worry, it should all be fine in the end—physicists make these kinds of approximations all the time. But here are my estimations:
The balls all have a mass of 165 grams and a diameter of 57 millimeters. That seems to be fairly standard for billiards-based games.The balls move without a frictional force and without rolling. Yes, that seems silly—but really, I think this will be fine for now.Ball-on-ball collisions are completely elastic. This means that the total momentum of the balls is the same both before and after the collision. It also means that the total kinetic energy of the balls is constant. (Or, you could say momentum and kinetic energy are both conserved.) In short, this means it's a "bouncy" collision.
Let's start with a very basic collision: A cue ball moves and knocks into a second, stationary ball. Of course, it's entirely possible to find the final velocity and angle of the initially stationary ball using conservation of momentum and kinetic energy—but I like to do things in a different way. For this case, I'm going to model the collision in Python. This way, I can break the motion into tiny time steps (0.0001 seconds). During each step, I can calculate the force on each ball and use that to find the change in velocity during that short time frame.
What force is acting on the ball? That's the secret—I'm going to use springs. Yes, springs. Suppose the two balls aren’t real (because they aren’t). In my model, when they collide, the outer part of one ball overlaps with the other ball. In that case, I can calculate a spring-like force that pushes the two balls apart. The greater the overlap, the greater the repulsive spring force. Here, maybe this diagram will help:
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Using fake springs to model a collision includes something that’s super useful. Notice that the spring force pushes away from an imaginary line connecting the centers of the balls? That means that this spring model will work for "glancing" contact when the balls don't hit head on. Really, this is exactly what we want for our (partially realistic) ball collisions. If you want all the physics and Python details, I go over everything in this video.
Now that we have a ball-colliding model, we can make our first shot. I'm going to start the cue ball 20 centimeters from another stationary ball. The cue ball will have an initial velocity of 0.5 meters per second and be launched with an angle of 5 degrees away from a direct hit. A direct hit is boring.
The stationary ball is yellow, so I'm going to call it the 1 ball. (The 1 ball is yellow in pool.)
Here's what it looks like—and here is the code.
(If you want a homework assignment, you can use the Python code and check how the momentum and kinetic energy are indeed conserved. Don't worry, this won't be graded—it's just for fun.)
Now let's use our model to do some cool stuff. What happens if I launch the cue ball at different angles, instead of just 5 degrees? What effect will that have on the recoil velocity and angle of the 1 ball?
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Here is a plot of the resultant angle of the 1 ball after the collision for different initial angles of the cue ball. Notice that the data doesn't have launch angles greater than 16 degrees—this is because a larger angle would completely miss the 1 ball, at least for my starting position.
This doesn't look bad. It almost seems like a linear relationship—but it's not, it’s just close.
Now, what about the velocity of the 1 ball after the collision? Here is a plot of the speed the 1 ball has for different launch angles of the cue ball.
Obviously this is not linear. But it also seems to make sense. If the cue ball is moving with a velocity of 0.5 m/s with a zero degree launch angle (aimed right at the 1 ball), the cue ball will completely stop and the 1 ball will travel on with that 0.5 m/s velocity. That's what we expect. For larger impact angles, it's more of a glancing blow and the final velocity of the 1 ball is much smaller. This all looks fine.
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OK, now what about two collisions? I'm going to add another ball, yes—the 2 ball is blue. Here's what that looks like:
That looks pretty—but here's the real question: How difficult is this? And by difficult, I mean, what range of values for the initial angle of the cue ball will cause the 2 ball to still get hit by the 1 ball?
For the first collision, this was pretty easy to determine, because the launch angle of the cue ball would either hit or miss that 1 ball. However, for two collisions among three balls, a change in the launch angle of the cue ball will change the deflection angle of the 1 ball such that it might not hit the 2 ball.
And what about the initial velocity of the cue ball? If that changes, it will also have an effect on the deflection of the 2 ball. Let's just look at a large range of possible initial conditions and see if they result in a collision with that 2 ball. However, instead of considering the launch angle and the launch velocity, I will just treat the initial conditions in terms of the x- and y-velocity of the cue ball. (Both of those depend on the total velocity and the angle.)
It will be easier to make a plot, so here is that graph. This shows a bunch of different initial conditions for the cue ball (x- and y-velocities) and which ones result in the 2 ball getting hit. Each point on the graph is a cue ball shot that will make that 1 ball knock into the 2 ball.
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But what if I add yet another ball to the collision? Here is the 3 ball (it's red) added to the series of hits:
That animation doesn't really matter. Here's what matters: Which range of initial cue ball velocities will result in hitting the 3 ball? Here is a plot of the initial cue ball velocities (x and y) that result in that collision. Notice that I'm including the data for the 2 ball collisions from before (the blue data) so that we can make a comparison.
Think of this plot in terms of area. The area on the graph covered by the blue data (to hit the 2 ball) is much greater than the area on the graph showing the velocities required to hit the 3 ball. It’s getting much more difficult to achieve a collision that involves all four balls.
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Let's do one more. What if I add in a 4 ball to the chain of collisions?
Just to be clear, this is a comparison of the range of initial cue-ball velocities that result in the 3 ball hitting the 4 ball. Let me go over some rough ranges for the initial velocities of the cue ball.
To make the 1 ball hit the 2 ball, the x-velocity could be from close to 0 m/s to 1 m/s. (I didn't calculate the velocities greater than 1 m/s.) The y-velocities could be from about 0.02 to 0.18 m/s. That's an x-velocity range of 1 m/s and a y-velocity range of about 0.16 m/s.
In order to have the 2 ball hit the 3 ball, the x-velocity could be from 0.39 to 1 m/s with the y-velocity from 0.07 to 0.15 m/s. Notice that the x-velocity range dropped to 0.61 m/s and the y-velocity range is now 0.08 m/s.
Finally, for the 3 ball to hit the 4 ball, the x-velocity could be from 0.42 to 1 m/s and the y-velocity from 0.08 to 0.14 m/s. This gives an x-range of 0.58 m/s and y-range of 0.06 m/s.
I think you can see the trend: More collisions means a smaller range of initial values that will result in a hit on the final ball.
Now we need to test the final case: nine balls. Here is what that looks like:
OK, that works. But will that last ball still get hit if we factor in an extra gravitational force caused by the interaction between the cue ball and the player?
This is fairly easy to test. All I need to do is add some type of human. I'm going to use an approximation of a spherical human. I know, people aren’t actually spheres. But if you want to calculate the gravitational force due to a real player, you would have to do some seriously complicated calculations. Each part of the person has different mass and would be a different distance (and direction) from the ball. But if we assume the person is a sphere, then it would be the same as if all the mass was concentrated at a single point. This is a calculation we can do. And in the end, the difference in gravitational force between a real and spherical person probably wouldn’t matter too much.
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I can find the magnitude of this force with the following equation:
In this expression, G is the universal gravitational constant with a value of 6.67 x 10-11 newtons x meters2/kilogram2. This is a super small value and shows you why the gravitational force is so weak. The other variables are the masses of the two objects: mp (mass of the person) and mb (mass of the ball) and the distance between the person and ball, r.
But notice that as the ball moves away from the person, r increases and the gravitational force decreases. That would normally make this quite a bit more complicated. However, since I'm already breaking the motion into tiny time intervals, I can just recalculate the gravitational force each time the ball moves.
Let's try this out. I'm going to use a person with a mass of 68 kg (that's 150 pounds) starting with a distance of just 4 centimeters from the cue ball to give the maximum impact. But guess what? Nothing really changes. The final ball still gets hit.
In fact, I can look at the final position of the last ball both with and without this gravitational force from the human. The ball's position only changes by about 0.019 millimeters—that is super tiny. Even if the mass of the human is increased by a factor of 10, the final position only changes by 0.17 millimeters.
Why isn't this working? Let's make a rough approximation. Suppose that I have a pool ball that's just 10 centimeters from a player. The magnitude of the gravitational force on the ball will be 7.12 x 10-8 newtons. If this force continues with the same magnitude for one second (which it wouldn't, since the ball gets farther away), the ball would have a change in velocity of only 1 x 10-9 m/s. I just don't think this is going to make a noticeable difference with the trajectory of the final ball.
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There are a couple of options to consider. First, is my pool ball collision model incorrect? I don't think so—I can get a change in position of the ball with a gravitational force, but it’s just not very large.
Second, I hate to say this, but maybe M. V. Berry was wrong. His paper was published in 1978, and while it was possible to do a numerical model back then, it was not as easy as it is today. I don’t know if he did one.
There is one final option: I picked a mostly arbitrary arrangement of nine balls for this chain of collisions. It's possible that for some other arrangement, or some other initial velocity, the gravitational force from a human would have a noticeable effect.
Even though I couldn't get this to work, it's still a pretty cool problem. I guess the next step would be to find out how many pool ball collisions it takes before the gravitational force from the player actually does make that last ball miss. Yes, that will make another excellent homework problem for you.
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