Bounce, part 1

When I was a kid, my dad sat my sisters and I down one day to inquire why the ceiling in our garage was peppered with blurry brown little splotches. I speculated that perhaps it was camouflage – the ceiling needed to blend into the herd. He didn’t buy it, and I accused him of not paying much attention to what the signs at the zoo had said about the ungulates when we visited two weeks before.

Juvenile creativity aside, the reason for the disfigured ceiling was that someone at school had told me about that thing with the basketball and the tennis ball. You know, the one where you hold a tennis ball directly over a basketball, drop them both straight down, and the tennis ball goes flying away crazily. You have seen that, right? It looks like this:

That experiment is why the ceiling was messed up: we had dirty tennis balls.

At that time, I didn’t know why the experiment should work. Why does the tennis ball bounce higher than it was dropped from? Not even a super-bouncer ball does that. Evidently it has something do with the basketball being so much larger than the tennis ball. So, what if I got something even bigger than a basketball? If I had a ball ten times as big as the basketball, and dropped the tennis ball on top of that, then how high would it blast up? Could I send something up to outer space this way? What if I use three balls – say a ping pong ball on top the tennis ball? What if I use four, or five, or infinity? What if I use two basketballs rather than a basketball and tennis ball? Can I still make one shoot up super-high then? What if I reverse the order, dropping the basketball on top the tennis ball? Will it bounce higher or lower than it was dropped from? Will it go higher than it would if the tennis ball wasn’t even there? Is there a way to make the basketball and tennis ball trick work even better? Does it matter that they’re ball-shaped, rather than bouncy cubes, for example?

The remarkable thing is that although I didn’t understand my experiment at the time, I could have, in principle, if I’d had someone around to explain it. That’s what I’m going to try to do now – answer the questions from the last paragraph with a series of thought experiments, rather than with forces, potential and kinetic energy, and a bevy of equations. It’ll take a few posts to do it, but the fundamental ideas are simple: that a ball bounces off a wall as fast as it came in, that you won’t get carsick if the ride is smooth enough, and that gravity pulls evenly.

Let’s begin by thinking about bouncing in the simplest context: bounce the basketball all by itself. When we do this in real life, it turns out that the basketball bounces back up, but not quite all the way. If we drop it from waist height, it might rise back to thigh-level, then bounce again, going to knee level, and so on.

Why lower than the original height? Why can’t it bounce higher? If I pump it up more, it bounces better. So if I pump it up high enough, could it bounce up higher than it began? The answer is no. You might get it to bounce better by pumping it up, but each bit of pumping would give you less and less benefit, so that you’d never quite get the ball to come all the way back up.

Suppose the ball bounced a little higher than it was dropped from. Then if it bounced a little higher on the next bounce, and a little higher the bounce after that, it would bounce to arbitrarily great heights, all by itself. Eventually it’d go to the moon. This is absurd, so the basketball can’t keep bouncing higher. But that doesn’t mean it can’t bounce higher for some particular height. Why can’t it, for example, bounce from 1 meter to 1.1 meters, then to .8 meters, then to .9 meters, then to .6 meters, etc. in some strange up-and-down pattern? Or why not bounce higher than dropped for all heights under one meter, and lower than dropped for all heights above one meter? Or higher than dropped on the first bounce, and lower than dropped on all subsequent bounces?

All of these scenarios are contrived, intuitively wrong, and even silly. Intuition can fail though, as it did for me when seeing the tennis ball bounce away so high. So it’s interesting that what intuition says about the single bouncing basketball can be derived from a simple principle.

Here is the first bit of physics: We assume that perpetual motion is impossible. By this, I mean that it is impossible to build a cuckoo clock that keeps on going forever without drawing in energy from the outside (for example, a solar-powered cuckoo clock with a battery to store energy for the night might run indefinitely, but if we block out the sun with a large wall of Coppertone bottles, the clock will wind down quickly and halt.)

There’s nothing special about my example of a cuckoo clock. If you could build something else, like a train, that ran forever, you could use its motion to power a cuckoo clock. And if you could build a cuckoo clock that ran forever, you could make a million copies of it and use their combined motion to power a train. So the cuckoo clock is just a convenient example for a general idea that you can’t build any machine that goes on forever.

One way to make a cuckoo clock go is to let a basketball fall, and use this to drive the clock. What you do is tie a string to the basketball, then tie the other side of the string to the gears of the clock. As the clock ticks, the basketball pulls on the gears, forcefully turning them a little bit. The gears deliver a tiny push to the pendulum each swing to keep it going. This way, the clock can run for a long time. Eventually it wears down though, because the basketball falls all the way to the floor. Then there’s nothing pulling the gears around, so the gears can’t power the pendulum, and friction slows everything to a stop. (For more on how a pendulum clock like a cuckoo clock works, see the How Stuff Works article. The important point for this argument is that you can power the clock by carefully and slowly lowering the basketball.)

This leads into why the basketball can’t bounce higher than we dropped it from. Say we drop the basketball from 1 meter, and it bounces back up to 1.1 meters. Then we can concoct a scheme to run a cuckoo clock forever. Start by lifting the ball up to 1.1m. Tie the string to it and use it to make the cuckoo clock run. When the ball falls to just 1 meter, take off the string and drop the ball. It bounces up to 1.1 meters again. Catch it there and re-attach the string. Let the ball fall back to 1 meter, running the clock… This creates an infinite process that’s self-contained. The system is the clock, the basketball, the floor it’s bouncing off, and the Earth, which creates gravity. Nothing outside this system is supplying any energy, and yet the cuckoo clock can run forever. If we assume that such perpetual motion is impossible, then it’s also impossible for a basketball to bounce higher than it was dropped from, no matter what the height.

In the next post, we’ll look at why the basketball doesn’t bounce back up to exactly the same height, which will lay the foundation for understanding the interaction of the basketball and tennis ball.

Freezing in Warm Air

I took a bottle of soda out of the freezer today, and looked at it. Not frozen. I opened it, and was surprised to find that a minute later, there was a chunk of soda-ice floating on top. Somehow, the soda had frozen after I took it out of its cold environment and into a warm one.

This didn’t happen because of the temperature of the room. Instead, it happened because I opened the bottle. When sealed, the bottle is under higher pressure. If you put ice under pressure it will try to take up less space, and since water is more dense than ice (as you can test by observing that ice floats), high pressure melts ice. Once I released the pressure, the soda, which was below the normal freezing point, froze sitting out on my desk.

Falling Bullets and the Density of Air

Matt discusses a Mythbusters episode where they drop a bullet and shoot one horizontally from a gun, and see which falls first. I want to take this up where he left it off

They find that to within their experimental error, the two bullets hit the ground at the same time. It doesn’t matter how fast you move in the x-direction; gravity still gives you the same acceleration in the y-direction.

It turns out that with standard assumptions about air resistance, this is not true. What I hope to do here is, by assuming that bullets falling and shot horizontally fall at very nearly the same rate, put a bound on the density of air.

First we need a model for air resistance. If the bullet is a non-rotating sphere and air is non-viscous, then the bullet slows down because air smashes into it. We can imagine the bullet leaving a cylindrical trail behind it as it flies. Let’s say that as the bullet flies through its path, all the air it encounters gets brought from rest up to some set fraction of the bullet’s speed, and that does not depend on the bullet’s speed .

Then in one incremental unit of time , the bullet intersects a mass of air that is the density of air times the volume of its path over that time, and the volume of the path is given by the velocity of the bullet times is cross-sectional area. This is grams of air, where is the density of air, is the speed of the bullet, and is its radius.

The momentum imparted to the bullet is the same as that imparted to the air. The momentum imparted to the air is the mass of air encountered times the speed it’s being sped up to. This is .

The force on the bullet is the time derivative of its momentum, so . For the time being, set so the force on the bullet is just .

Now let’s make some simplifying assumptions. We imagine that air resistance plays a minor role, because otherwise the Mythbusters would not have found so close to a tie as they did. This means the bullets have very nearly the same y-height at any time, and that it’s the same as the normal, physics 1a parabola falling under gravity. Let the bullet fired out of the gun have the same horizontal velocity throughout its flight. Also assume the bullet goes much faster in the horizontal than in the vertical direction. Using these assumptions, we’ll try to find the relative acceleration of the two bullets, rather than finding both of their paths and then subtracting.

For the bullet falling straight down, all motion is in the y-direction, and the upward force is . For the bullet fired out of the gun, the total force is , but only part of this is in the y-direction, so the force in the y-direction is . That can be simplified to because velocity in the y-direction is small.

Because air resistance plays a minor role, assume the bullets fall for almost the same amount of time, . Over that time, there is a difference in the force on the bullets given by . Notice that this is equivalent to assuming the dropped bullet doesn’t get slowed down at all. The shot bullet experiences so much more air resistance that we are essentially assuming only air resistance on the shot bullet matters.

The bullets are in free fall, so and the total distance between the two bullets when they fall comes from integrating their relative acceleration once for their relative velocity, and again for their relative displacement. When the mass of the ball is , we have

.

Plugging in , the falling time, for gives the distance between the bullets as they land. Dividing by their speed at that time , gives the difference in their falling times. This comes to

To check, the dimensions work out to time. The difference increases with the density of air, with the radius of the ball (for fixed mass), with the horizontal firing speed, with the height dropped, and with the factor of air resistance. It decreases with increasing gravity and mass of ball (for fixed radius). All that seems plausible enough.

We can solve this for , the density of air. Since I can cancel those.

.

Let’s make some guesses. They fired the bullet from about off the floor. It traveled some down the hall in the it had to fall, so . , and . .

That gives

.

The Mythbusters think (they said 39 milliseconds), which implies

.

And it turns out the density of air is indeed less than this – about 1 kg/m^3.

I should admit, though, that the first time I crunched through these numbers, I got that the density of air should be less than . That was plugging in for the bullet weight, thinking of a bullet about as dense as a typical rock. But the bullet I hypothetically used had radius 1cm, not diameter, and also bullets are made of metals that are pretty dense. I googled "bullet weight" and learned that google automatically interprets that to mean the weight of rugby player Tom James, nicknamed "The Bullet". (His weight was 15 st 8 lbs).

Incidentally, I think my calculation worked by accident. The aerodynamics of a bullet are probably much more complicated than I've assumed here, since the bullet spins and probably entrains air and makes eddy currents and does all kinds of ungodly things. isn't constant, but according to my model it should drop by less than in the time it takes the bullet to fall. I guessed kind of wildly on things like the height of the drop and the factor by which air is sped up when the ball passes through. Also, the experiment has error in terms of simultaneous dropping and how level the gun is. But things seem to have worked out to order of magnitude.

The Power Tower

Some time ago, Foxmaths! had a nice post about a power tower

To make this more precise, we say

.

And ask for the limit of as . A student of mine at physics camp asked me about this problem with the specific value . He was stuck. He assumed there was some limit , and then wrote

substituting an infinite tower of ’s for itself less one. This is equivalent to setting

in our recursion relation, meaning that we’re finding what the limit must be if it exists.

The problem is that by inspection there are two solutions: and . But the real answer is two. Plugging it into the calculator and going a few iterations, we see that this is correct.

I didn’t know why it was two rather than four. Neither did anyone else around. But there were a lot of people around. Pretty soon we had six or eight people all trying to find why the thing went to two rather than four.

We tried using different values, like

,

which didn’t work at all – it blew up. But use rather than for the tower and it works fine – going to about , although by straightforward algebra there still appear to be two solutions, the other being . Use and it goes to again, rather than . Use and it goes to .

Then we tried fractional numbers to their own root. Take a power tower with a bunch of ’s stacked and you converge to . Same (converges to ). But converges towards . Trial and error showed that for from zero to , a power tower of converges to . Then, above , it converges to something smaller and smaller, until for very large it converges towards one. This makes some sense, because has a maximum at and converges to as .

That was as far as we got before 11PM, when I had to send them all to bed without any dessert. Until I realized: duh. Draw a picture.

The recurrence relation was

with some base we’re considering. So let’s draw as a function of That way, by going from a value on the x-axis to the value of the plotted function, we do one iteration. The relevant function is an exponential.

To do another iteration, we have to move to the correct x-value, so we can drop to the new y-value. That is, we want to go from the y-value we’re at, and make that the x-value. So plot the line , then trace over horizontally to that line. Now you can drop down to the new value of the function, conducting another iteration. Doing this over and over, we get a series of steps, and each step represents a new iteration.

A heuristic representation of the series we're talking about here. Each blue step marks a new iteration. You start at the red line, drop to the green line to iterate, then cut over horizontally to the red line to get ready to do it again.

Repeating this process, we see that where the lines intersect there are solutions to . But only one of those solutions is “stable”. If we’re close to the lower one, we’ll get closer on the next try. But if we’re close to the higher one, we’ll get further away on the next try. So the power tower converges towards the lower intersection.

As we change the base , the roots move. As increases from a small value, the roots come in towards each other because the exponential is getting steeper. When gets to , the two lines are tangent, which is an easy calculus problem. Then the roots coincide. If increases beyond that, the lines never cross and the series diverges.

We can also see why using converges to for . It's because is clearly a solution for in

.

The two roots have one bigger than and one less. And for , we converge to , because we converge to the smaller root.

I’d Tap That

Here are a few things I would tap:

Tap that and get some sugar.

Or maybe I'd bang it.

Tap the night away.

It's easy to turn on and it always puts out.

Yeah, I'd tap that.

Utilities are included in my apartment, so I don't actually have to tap that. The image is pretty suggestive, though.

Tap that? No, wait. Pat that. I meant I'd PAT that.

I swear this is a waterfall.

The Renaissance Man Uniform Gravitational Acceleration SMACKDOWN

Matt at Built On Facts posts about coriolis forces, and points out that a falling body is deflected by them one eighth as much as one tossed from the ground to the same height, and that they’re deflected in opposite directions. Here’s my attempt to explain this intuitively.

This makes me think of the competing da Vinci – Galileo laws for bodies (not their own I hope) falling freely under gravity. They stated their rules in the same basic way. I remembered these laws from watching The Mechanical Universe in high school – before taking physics from the real life version of David Goodstein three years later.

da Vinci said (or so I hear, I never met the guy) that if you fall one unit of distance in the first unit of time, you’ll fall two in the next unit, three in the one after that, then four, etc. So if you fall 5 meters in one second, in the next you’ll fall another 10 for 15 meters total.

Galileo said almost the same thing, but with odd numbers. If you fall one unit of distance in the first second, then in the second you fall three, then five more, then seven, etc. So if you again fall 5 meters in one second, in the next you’ll fall another 15, for 20 total.

Galileo was right; da Vinci wrong. But let’s not screw over our primitive-flying-device-making friend with such a cursory examination. They’re both awesome dudes, as Leonardo’s testudine counterpart would say.

Galileo was right because acceleration is constant, so the distance fallen is proportional to the square of the time. Adding Galileo’s odd numbers gives a square number. 1+3+5 = 9, for example. This is easy to see from a picture.

Each new section adds the next odd number worth of dots, and takes you to the next bigger square number when counted as a whole.

da Vinci, instead of the square numbers for total distance fallen, gave the triangular numbers. 1 + 2 + 3 = 6, which is triangular. This has its own picture.

According to da Vinci, each new row is how much you fall in one additional second.

da Vinci’s fub may have been in misunderstanding the relationship between speed and distance. If da Vinci’s rule had been giving the speed at the end on each second, rather than the incremental distance fallen, he’d have been right. If you’re going 10m/s after one second, you go 20m/s after two, and 30 m/s after three, etc. The problem is that you can’t find the distance traveled in a second by taking the speed at the end of that second and multiplying by time. If you do that, you get only an approximation to the correct integral, like this:

Don't worry about the numerical details. I stole this from the internet somewhere. da Vinci's law overestimates distance fallen every second by assuming your speed at the end of the second was you speed for the entire second.

It’s possible that da Vinci was actually right on about the kinematics, but that he made a mathematical error in reporting his result. I wanted to follow up on this, so I checked online to see precisely what Leonardo said. I did not succeed. Fritjof Capra’s book quotes da Vinci:

The natural motion of heavy things at each degree of its descent acquires a degree of velocity. And for this reason, such motion, as it acquires power, is represented by the figure of a pyramid.

But when I search online texts of Da Vinci’s notebooks, I can’t find this passage. I can’t find the relevant passages in my Dover copy of Richter’s translation, either. In fact, I can’t find this passage anywhere else on the entire internet, except one book that doesn’t cite the source. So I’m not sure what to make of this. da Vinci’s writings on falling bodies must be somewhere, if we know about them. But as of now I’m still uncertain. Based on the preface to my translation of the notebooks, it looks like they decided to omit some of Leonardo’s physics, since that is obviously unimportant and uninteresting to readers of his notebooks.

Let’s assume Leo had the right idea, but brain farted on the integration thing. Considering how clever Da Vinci was, his mistake is very surprising, because his law is not only empirically wrong, it is logically impossible.

To see what I mean, let’s carry out Da Vinci’s argument a little further. According to his rule, in four units of time you fall 1+2+3+4 = 10 units of distance. But the choice of how long a unit of time is was arbitrary. So let’s do it again, but consider the unit of time to be twice as long as it was previously. We’ll call these “shmunits” of time. In one shmunit of time, you have to fall three units of distance to be consistent with the first calculation. Then you fall six units of distance in the second shmunit of time, because the second has you falling twice as far as the first. After two shmunits of time, you fall a total of nine units of distance. But we already said that with the same law you fall ten units of distance! Surely if Leonardo had considered his law carefully he’d have seen this error, right?

Unless it’s not an error. What if Leonardo actually meant that you have to take the limit as your unit of time becomes infinitely short? In that case, Leonardo’s law

can simply be reduced to the correct law

.

Could this really have been what Leonardo had in mind? I think it’s possible, but not likely. The Greeks explored the basic ideas here. They approximated using the method of exhaustion, and Archimedes is said to have been doing what amounted to integral calculus. If Leonardo was aware of this research, he might have stated such a law accurately. But it seems far-fetched.

Now I Know SCIENCE!

This is supposed to be a blog. It isn’t. It’s just the standard WordPress template with posts coming in irregular bursts.

Part of being a blog is that you’re supposed to be integrated into the blogosphere. This is because if you take a bunch of meaningless things and endlessly interconnect them, you get what’s known in science as an emergent phenomenon, in this case endless confusion. So I really ought to hook up to the blogosphere and get in on that.

On the other hand, it turns out there are these people who want blogs to be useful. So what they do is sift, organize, centralize, and bring similar bloggers together. GrrlScientist (real name, I think. You wouldn’t put a fake name in the internet, right?) is one such person. She (or he, I suppose) runs Scientia Pro Publica, a blog carnival whose goal it is to make science understandable to everyone by writing in Latin. She’s hosted the 13^th edition of Scientia today on her blog, including about 20 essays in popular science. They’re heavy on the life sciences, discussing such topics as parasites that eat weaverfish tongues and why I still don’t understand evolution. (Seriously, I don’t.)

There’s some meta-discussion of science, a post about the suckiness of iPhone camera, and they graciously included my post about retroreflectors on the moon as the only physics/astronomy representative.

New Problem: Discreet Weighing

On a segment from the PBS show of my childhood, “Square One”, a wrestler is challenged to weigh a dog. He captures the dog and puts him on a scale, but the dog keeps jumping off before the wrestler can get his measurement. The solution, whose cleverness so overwhelmed me I repeated it to my parents during dinner several times, was to weigh himself, pick up the dog, weigh himself and the dog together, and subtract his weight to get the dog’s weight.

I tried to emulate the technique by weighing such things as my Fisher Price Wild West Saloon, but pretty much everything I could pick up and carry around with me at that age weighed less than a pound, and the effort failed. Also, I just now realized for the first time it wasn’t actually a wrestler in that show, but a grown up pretending to be one. Damn.

Here’s a variation on that old theme. You’re at the gym when you notice a cute guy/girl bouncing on an exercise ball (pilates ball, big round blue thing, whatever it’s called). You want to figure out how much they weigh, because your bunk bed has a certain tolerance you can’t exceed. It would be impolitic to ask directly what they weigh, and you’re unlikely to pick them up and carry them to the scale in the locker room. Still, by observing them for a short while, and a bit of experimentation, you can figure it out. How?

The Root of the Issue

I … regard it as mysterious that an object moving in an inverse square force law traces out a conic section. There are lots of ways to prove it, of course. Newton did it using Euclidean geometry. My homework problems above give two other ways. The one using the Runge-Lenz vector is pretty… but I’m still looking for the truly beautiful way, where you leave the room saying: “Inverse square force law… conic sections… of course! Now the connection is obvious!” – John Baez, Mysteries of The Gravitational 2-Body Problem

I’m not ready to going to give the beautiful proof Baez has been searching for throughout a long career in mathematical physics. I’d like to, but I’m not brilliant enough to be that easy to understand. Instead, I’m going to talk about some middle school math.

I’ve found that in most circumstances, if I’m willing to sit and be confused for long enough, I can normally understand what people are trying to tell me about math. That’s nice. It doesn’t work that way for everyone, and I’m glad I was lucky enough to be able to enjoy it, to the extent I can. But more than knowing the answer to a difficult problem, we humans like to know where the answer comes from.

Suppose, for example, I tell you that the solution to

is

.

It’s easy to verify – just plug the purported solution back into the original equation and you’ll see that it works. But if you have a soul you’ll wonder where in Hell I got that from. (Which is the sort of thinking that, ironically, will cost you your soul due to blasphemy.) Actually where I got it from was my seventh grade math teacher. But I should have gotten it by drawing a picture.

We start with . It looks like this:

We'll represent the values on each side of the equation by the area of a figure. This is how we start.

Next is . Where is in that picture? is the side of the square, so if we add on to that with a rectangle of width , we’ll have a region , which is the left hand side of the original equation.

The total area is now x^2 + ax. I know the picture says bx where it should say ax. This is the internet. You get what you pay for.

But that is not really fair to the other side of the square, who didn’t get any region added on. Let’s be symmetric about this, and add to both sides.

That’s the left hand side of the equation. The right hand side is just .

Here's a picture of the entire equation: x^2 + ax = c.

The thing on the left, though, is so damn close to being tractable. It’s almost a square. Once we’ve drawn this picture, it’s natural to “complete the square” by adding in that little region that’s .

We add a little region to both sides, completing the square.

Now let’s do some morphing and stretch that right hand side out into whatever shape we want. Also a square, perhaps?

The equation discusses area, not shape, so we can make the shape on the right into any shape we want, as long as we give it the same area. We'll merge the red and green of the previous image to make purple, and form a square to give each side of the equation the same shape.

Great. Now we have two squares that are the same size. Therefore, the lengths of their edges are the same. And the length of the edge on the right hand side is just the square root of the area . The length of the edge of the square on the left is , so equating them gives

or

.

Cool.

Now let and . Then .

Aaaaaaaaaaaw PHOOEY! I don’t know what the square root of two is. I am a Pythagorean and lived thousands of years ago. I believe that everything is ratios of integers and that you can’t eat beans because they resemble a fetus! But wait, I do have a little trick to get good rational approximations of root two. The trick is to make a table with two columns, and use a rule to generate new entries as you go down. It looks like this:

1	1
2	3
5	7
12	17
29	41

The rule is, to put a new entry in the left hand column, add the two entries in the row above it. To put a new entry in the right hand column, again go to the row above, but this time add twice the left hand column to the right hand column. Just try; you’ll see.

The miraculous part comes when you divide the right hand column by the left. Let’s do this again.

X	Y	Y/X
1	1	1
2	3	1.5
5	7	1.4
12	17	1.417
29	41	1.4138

It’s getting suspiciously close to root two. It’s not hard to verify that the table isn’t going to make a mistake; it will get closer and closer to root two. First we’ll write down formulas explaining the rule. Call the entry in the left column and row and right column . Then the rule is

.

Because we think the ratio is going towards , let . Then algebra tells you (on simplification)

from which we see that the ratio converges to root two, oscillating above and below it each iteration.

The method is a general one for extracting square roots. Simply let and you’ll extract the square root of . It’s similar to an even-simpler method for finding rational approximations to the golden ratio – take successive terms of the Fibonacci sequence, which converges to the exponential of the golden ratio plus a constant. But I don’t see where it comes from.

I suppose we could write the iteration equations as a matrix, and talk about having as an eigenvector, etc. But the only reason I think to do that is that linear algebra and calculus are about the only math I know, so I try to use them for everything. If we start by assuming we’re going to find some rule that lets us make a table to find rational approximations of root two, it’s not hard to believe someone would come upon the rule. But who would get this idea of the table (or some similar equivalent idea) in the first place? I don’t know. But I’m happy people told me about it.

And this is only one of many ways to extract a square root. We could guess-and-check at will, trying first 1.5, then 1.4, then 1.45, then 1.42, then 1.41, then 1.415, etc. That’s a very intuitive rule, although a poor one from an efficiency standpoint. Or we could use some other rule to generate a new guess. For example, first take a guess at root two. Then average your guess with . This generates a new, better guess. Repeating the process creates a sequence of rationals that converges to root two, just like the original method did. This new guess, though, has a wonderfully-intuitive explanation, and is almost obvious. It’s Newton’s method. I’ve seen one or two other methods that were more complicated – enough that I didn’t care to work through them in great detail when, after all, I can tell Wolfram Alpha to do whatever I want. (Well, not quite. Just now I told it to make me a sandwich, and it replied that it was assuming that “sandwich” was a city in Massachusetts).

So sometimes things make a lot of sense. Sometimes they only make a medium amount. Bit by bit, as I learn, the sphere of things that makes sense gradually expands. On the other hand, the sphere of things I heard about but don’t make sense expands even faster, and on top of that, the sphere of things I’ve heard about but understand so poorly I can’t even claim the results don’t make sense to me, since I don’t know what the results are saying, outstrips them all.

Reflections on the Moon

Q: Why don’t elephants play tennis?
A: They prefer squash.

Two players compete in a fast-paced game at the gym. I exercise across from them, watching as they smash a blue rubber ball in turns. The game is in a small indoor court, and the ball moves very quickly. Its motion, followed from a distance, appears nearly rectilinear. All of the walls, the floor, and the ceiling are in play, and the ball’s wild meanderings, jots of wild color, mesmerize me as I stretch.

The game is squash (Wikipedia), which I had heard of but not seen played before. Watching the squash players, I was at first surprised by their ability to predict the seemingly-chaotic motion of the ball. However, a geometric property of the court aids them.

The squash ball moves very quickly (faster than a major-leaguer’s fastball), so over the short distances of the court, we can approximate its motion as roughly a straight line. The court is a rectangular prism, and this shape has the property that if a player smashes the ball at a corner, the ball will pop back out right at them, parallel to the way it came in. In two dimensions this is shown in the following diagram:

The squash ball follows the dark blue path, bouncing off the corners at points A and B. The light blue line shows the continuation of the paths, with a hypothetical meeting point D if the incoming and outgoing lines are not parallel.

The ball comes into the corner bouncing at point , making an angle with the wall. By hypothesis, it bounces off with the same angle, then comes into the next wall with an angle at point . It bounces off with this angle as well and returns to the court. We’re trying to show that the incoming and outgoing paths are parallel.

To do this, I’ve drawn in light blue the continuation of the incoming and outgoing paths. If they’re parallel, they never meet, and the angle drawn as should be zero. Notice that and are the small angles of a right triangle , so they add to a right angle. Angle is opposite , and so equal to it. That means angle is and similarly angle is . Those two angles, along with form the triangle . However, they add to a straight angle by themselves, and so we must have , showing that the ball pops out parallel to the way it came in, allowing the players to predict its motion easily.

In three dimensions, this is just the same, except that you have to work through the argument over again. A student of mine pointed out a different argument to come to the same result. Set up and coordinate system at the corner along the intersections of the planes. Then one wall works by flipping the -coordinate of the incoming ball’s velocity vector, and the other two flip the ball’s and -coordinates, so that after bouncing off all three, the velocity vector is reversed.

In squash, this result is far from perfect because gravity affects the ball’s motion, and its rotation, along with friction from the walls, may affect its angle of reflection. Energy is lost in each reflection as well, and the ball will slide some against the wall, so all in all it’s a rough approximation.

For light the approximation is much better as long as the wavelength of light is much smaller than the size of the mirrors. The setup with three orthogonal plane mirrors like this is a called a retroreflector because it reflects light back the way it came. If you look into one from any angle, you will see your own pupil at the corner, because at the corner the incoming and outgoing rays are not only parallel, but on top of one another, so light must start and end at the same place after reflecting there. All the light you see ends at your pupil, so that’s what you see in the corner. If you have three hand mirrors, it’s an easy experiment to try.

One interesting application of this idea is shown here:

A retroreflector on the moon's surface. The Apollo missions left this array of hundreds of reflectors intended for extremely accurate measurements of the Eart-Moon distance.

This is a retroreflector array on the moon. When an Earth-based laser sends a pulse of light at the moon, the retroreflectors send the pulse back to Earth. If you could measure the trip time very precisely, you can multiply by the speed of light to find the distance to the moon. The APOLLO project (not the lunar orbiters, but the ground-based Apache Point Observatory Lunar Laser-ranging Operation) is trying to do this to an accuracy of one millimeter.

I’ve sometimes heard silly things like “the Campbell’s soup cans thrown out by Americans in a single month could stretch to the Moon and back three times.” I say this is silly because

• Why would you want to do that?
• I totally just made the statistic up because it is meaningless and nonmemorable. Things like per-capita consumption, percentages of usable land being turned into dumps, and statistics about ecological impact actually mean something.
• No they can’t. They would fall down if they tried.

Regardless, if you know the distance to the moon with one millimeter accuracy, then your estimate of how many Campbell’s soup cans away it is is limited by how accurately you know the size of a Campbell’s soup can until you measure the can to an accuracy of single atom (and Campbell’s soup cans vary from one to another by a lot more than that, and don’t stack perfectly regularly, and shift around, and get hotter and colder, etc).

There are a few good reasons you’d want to know the position to the moon so precisely. Perhaps the most striking is as an extremely tight test of the equivalence principle of general relativity. The Earth and Moon have different densities, and so might conceivably fall towards the sun at different accelerations, even when they’re the same distance away. Modern cosmology and theoretical physics frequently explore theories of modified gravity in attempts to explain the acceleration of the universe’s expansion or create quantum theories of gravity. If the equivalence principle doesn’t hold, watching the acceleration of the moon very closely could be the first place we’d get a hint of it.