If you want to compute the circumference of the observable universe to within, say, the width of a human hair, you’ll need to know about 35 digits of π, though this never seems to deter a certain sort of person from memorizing the first 100, 200 or 500 digits. But it turns out there’s no need to memorize anything at all! You can recover any number of digits you like from a simple little physics experiment that I just learned about, though it was invented over ten years ago by Professor Gregory Galperin of Eastern Illinois University. His lovely little paper is here.

To see how it works, start with two identical billiards lined up in front of a wall like so:

Now push Ball 2 toward Ball 1 and count the collisions: First Ball 2 collides with Ball 1 and pushes it toward the wall. (At this point Ball 2 has transferred all its momentum to Ball 1 and stops moving). Then Ball 1 collides with the wall and bounces back toward Ball 2. Then Ball 1 collides with Ball 2 and pushes it off to a far-away place. Three collisions. That tells you that π starts with a 3.

If you want more accuracy, make Ball 2 exactly 100 times as heavy as Ball 1. This time the sequence of events is a little more complicated, but it turns out there are exactly 31 collisons. That tells you that π starts with 3.1.

Or if you prefer, make Ball 2 exactly 10,000 times as heavy as Ball 1. You’ll get exactly 314 collisions. π starts with 3.14.

You want 100 digits of accuracy? No problem. Just make Ball 2 10200 times as heavy as Ball 1 and count the collisions. There should be exactly

31,415,926,535,897,932,384,626,433,832,795,028,841,971,693,993,751,058,
209,749,445,923,078,164,062,862,089,986,280,348,253,421,170,679

of them.(Be sure not to lose count halfway through or you’ll have to start all over again!).

Granted, this is not a terribly practical way to remind yourself of the first 100 digits of π, but then there’s no reason a terribly practical person would ever care about the first 100 digits of π in the first place. Galperin’s result is way cool, and so is the argument that proves it (which, if you go for this sort of thing, you can read in his paper). I thought it was well worth sharing.

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#### 15 Responses to “Follow the Bouncing Ball”

1. 1 1 Phil

Neat! Do the balls have to be round? Or is it just about the mass?

2. 2 2 Mike H

@Phil it’s just about the perfectly elastic collisions in one dimension between particles of equal mass.

3. 3 3 Harold

It is amazing how often Pi pops up. I love the way he describes the reaction when presenting the idea at symposia “ﬁrst complete distrust and then complete acceptance, due to the obviousness of the proof” Well, obvious to some, maybe.

4. 4 4 Jonathan Kariv

Cool. Of course if I wanted say 35 digits (because yes I do want to talk about this universe sized circle and I DO need accuracy to the width of a human hair)I’d need the big ball to be bigger than earth even if the small ball had the weight of a hydrogen atom. Plus I’d probably miscount, and I’d die before I finished counting.

The idea is still gorgeous .

5. 5 5 Joel

Does this work in other base number systems?

For example, if ball 2 is 16^100 times as heavy as ball 1, will it give you pi in hexadecimal to 50 digits?

6. 6 6 Steve Landsburg

Joel: Yup! (Though of course you’ll have to record the number of collisions in hexadecimal notation.)

(At least this is almost sure to be true. The method actually would actually fail, even in decimal, if you used it to compute the first 500 digits and it turned out that digits 251 through 500 were all 9′s. It would fail in hexadecimal if you used it to compue the first 50 digits and it turned out that digits 26 through 50 were all F’s. This exception seems astronomically unlikely and there is good reason to believe it never occurs.)

7. 7 7 Joel

Steven #6: That’s not a problem; I’m in the habit of counting collisions in hexadecimal all the time. Why, just last week, I counted collisions in hexadecimal for my Aunt Gloria…

8. 8 8 Brian

Wow, what an extraordinarily lovely insight! Thanks for bringing this to our attention.

9. 9 9 Martin-2

The algebra in section 5 step 3 is kinda messed up. The first line has an ‘M’ instead of an ‘m’, and the third line has term squared for no reason. But the result is correct. Is this common in math papers?

10. 10 10 Bob Murphy

Like Joel, I was worried about the base. So what happens if someone is using binary notation? Does it not work? I.e. the first digit (when balls are equal mass) has to be a 3 right?

11. 11 11 Steve Landsburg

Martin-2: The algebra … is kinda messed up…But the result is correct. Is this common in math papers?

Extremely.

12. 12 12 Steve Landsburg

Bob Murphy: The method does fail for very small bases. It should work for at least base 5 and up, and probably (though I haven’t checked it) for base 4.

13. 13 13 Eric

Alright, what about non-integer bases :-). That way the balls can have arbitrary mass ratios?

14. 14 14 David Wallin

In my youth I heard speculation about sleep learning. So, I decided to test this by having 50 digits of pi played while I slept. My only problem was that I had no way to have those digits repeated to me in my sleep except by recording me saying them over and over into a tape recorder. Of course I learned them in the process of recording the tape. I can still do a minimum of 25 digits, which has provided me zero benefit, except for impressing the ladies……

15. 15 15 Thomas

Nice paper, but this method is severely restricted by the ratio of the weight of an atom compared to all mass in the universe (2^200 is a pretty big number). So we still need maths.