Pi and billiard balls; a different application of π for Pi Day

      So suppose there is some sort of physical experiment, and the first time you do it, the answer comes out as 3. Then you change the experiment to make it a little more complicated, and the answer comes out as 31. Then you make it even more complicated, and the answer comes out 314. Then you ... . Of course you see where I'm going with this.

      It's March 14, or 3 14 (in the US date format), and because 314 are the first digits of π, many people use this day to share interesting and unusual appearances of π. I think you will find Pi and the Billiard Balls to be a little different!

      But first, if your statistics course did not require calculus as a prerequisite, you may be unaware that π is contained within the formula for the probability density of the Normal distribution, f(x) = $\frac{1}{\sqrt{2\pi {\sigma }^2}}{}^{\exp }$ where the 2π is necessary for the integral of the pdf to equal 1.

      A common application of π on Pi Day is the Buffon's needle problem: Given a needle of length l dropped on a plane ruled with parallel lines t units apart, what is the probability that the needle will lie across a line upon landing? There are many references to this such as Buffon, so the solution will not be repeated here.

      π appears in many places in math. One reason is that π is defined in reference to a circle. The trigonometric functions can be defined in terms of triangles within a circle. As a point traverses the circumference of the circle more than once, the trigonometric functions repeat in cycles of 2𝜋. Therefore, phenomena that repeat and can be represented by trigonometric functions are likely to have π somewhere within them. This is because π inherently relates to the periodic nature of these functions, making it indispensable in modeling cyclical behaviors.

      I recently discovered a physics problem called Pi and Billiard Balls. The original article is by G. Galperin, and I will try to summarize it. His paper is not an easy read.

      Suppose we have the first quadrant of an xy coordinate system. Suppose we have a vertical wall at x = 0, y ⪰ 0. We have ball 1 having mass m1 at initial position x1 > 0, and ball 2 having mass m2 ⪰ m1 at initial position x2 > x1. Let the ratio of the masses be a multiple of 100:   m2 / m1 = 100^N for a fixed non-negative integer N, including N = 0. Suppose ball 2 moves from right to left along the x-axis and collides with ball 1, ball 1 moves from right to left and collides with the vertical wall, and assume all collisions will be perfectly elastic. This perfect elasticity assumption implies the balls will satisfy the law of conservation of momentum, and the law of conservation of kinetic energy. We also assume the balls only move along the x-axis; there is no y movement.

      The amazing conclusion of all this is: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      For case 1, let N = 0 so m2 = m1. Now push ball 2 from right to left at initial velocity v2 until it hits ball 1. This is collision 1. Ball 1 will move from right to left at the same velocity v1' = v2, while ball 2 will now be at rest. Eventually ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3. Now ball 1 is at rest, and ball 2 is moving to the right with velocity −v1'. There have been 3 collisions (the first digit of π), and there will be no more.

      For case 2, let N = 1 so m2 / m1 = 100. This is more complicated because after collision 1, v1' will not equal v2, and ball 2 will not be at rest. The result of the two conservation equations m1v1 + m2v2 = m1v1' + m2v2' and .5m1(v1^2) + .5m2(v2^2) = .5m1(v1' ^2) + .5m2(v2' ^2), give v1' = [(m1 - m2)v1 + 2m2v2] / (m1 + m2) and v2' = [(m2 - m1)v2 + 2m1v1] / (m1 + m2). After collision 1, substituting v1 = 0 and m2 / m1 = 100 gives v1' = 200v2/101 ≈ 1.98v2 (nearly twice as fast as initial velocity v2) and v2' = 99v2/101 ≈ .98v2 (slightly less than initial velocity v2).

      Ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3.

      After collision 3, substituting primes into the two conservation equations, gives v1''' ≈ -0.96v2 and v2''' ≈ 1.96v2. Ball 1 will bounce back and forth between ball 2 and the wall many times. After each collision between the two balls, the velocity of each ball changes with ball 1 decreasing in velocity and ball 2 increasing in velocity. The relative velocity between ball 1 and ball 2 decreases with each collision, as the heavier ball will slowly transfer its momentum to the lighter ball. Eventually the relative velocity will be so small that they will effectively move together after the collision. At this point, no more collisions will occur. There will be 31 collisions (the first two digits of π).

      For case 3, let N = 2 and m2 / m1 = 100², there are 314 collisions. And so on.

      Not surprisingly there is a circle lurking under all of this, due to the conservation of kinetic energy equation .5m1(v1^2) + .5m2(v2^2) = constant, and there is a trigonometric function and a calculation involving an angle and π. Galperin proves the conclusion in general: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      In addition to the Galperin paper, a good explanation is here, but this is not an easy read.

      You are welcome to try this experiment yourself, if you can create the condition of perfect elasticity. Happy Pi Day.

Happy Pi Approximation Day

      Many people know March 14 is celebrated as Pi Day because 3, 1, and 4 are the first three significant digits of π (using the month, day date format). I just learned that July 22 is celebrated as Pi Approximation Day (using the day/month date format) because 22/7 is a common approximation of π .

      π Is defined as the ratio of a circle’s circumference to its diameter. π Is an irrational number (it cannot be expressed as the ratio of two integers), and it has an infinite number of non-repeating digits. Approximations of π date back to ancient civilizations and continue today as people compete to calculate π to billions of decimal places on supercomputers.

      The 22/7 approximation only matches π to the second digit after the decimal place, 3.14, and 22/7 is greater than π, a fact known by Archimedes. The error in the approximation is only about 0.04%, which is close enough for most of us.

      People also compete in the number of decimal places they can recite by memory such as using mnemonic techniques with words, where the length of each word represents a digit of π . There are many creative π mnemonics , but I am content to remember the 15 word "How I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."

      A quite impressive feat is Apu from the Simpsons who claimed to be able to recite 40,000 digits of π, and proved it by correctly stating that the 40,000th digit is 1. Apu

      The R computer language carries 15 or 16 digits. That seems like enough. A NASA engineer says he can’t think of a practical application that would require more than 15 digits of π . NASA

      π appears in many areas of math besides geometry and trigonometry. It is hidden away in statistics where the probability density function formula of the normal curve has a √(2π) term in the denominator to get the integral equal to 1, and elsewhere in other branches of math.

      Happy Pi Approximation Day.

      Here is some R code:

library(dplyr)
library(tidytext)

# compare 15 digits of 22/7 to pi
print(22/7, digits=15)
print(pi, digits=15)

# count word length & number of words in this mnemonic
textfile <- c("How I need a drink",
"alcoholic of course",
"after the heavy lectures involving quantum mechanics.")

df<-data.frame(line=1:length(textfile), text=textfile)
df_words <- df %>% unnest_tokens(word, text) %>% mutate(word_length = nchar(word))
df_words
n <- nrow(df_words)
cat("Number of words: ", n)

These drinking glasses are too short!

These drinking glasses are too short!

Some of my reinsurance and math teacher friends may remember that when I am out of town and having an adult beverage with friends, I have been known to stare at the drinking glass and say something like, "I don't mean to be rude, but the glasses are certainly short here. They are much shorter than what we have back home. In fact, they are so short, that I think that the circumference of the top of the glass is larger than the height."

Then there is generally a long pause as the group considers this. The reinsurance group may require some reminder of what circumference means.

Either group (unless they have heard this before, or unless they can guess that this is a setup) will likely disagree with me. I will reply that I am pretty sure about this, and I am willing to bet a dollar.

How do you measure this in a bar or restaurant? I use a paper or cloth napkin to measure the circumference from one end of the napkin to somewhere in the middle of the napkin, and then I use that length to compare to the height.

I have done this enough times so that I am nearly always right. Try it with your own drinking glasses. The only time it consistently fails is with champagne glasses.

Recently it occurred to me that there must be a website with a wide variety of glasses and their measurements, and I found Dimensions.com, https://www.dimensions.com . Dimensions.com is a database of drawings with standard measurements. Measurements are based on industry standards and averages and may differ among manufacturers and regions. Here is a sample of glasses with their images and measurements which are used here with permission, plus my calculations in the last three columns of the table. Volumes are in ounces, heights and diameters are in centimeters. The source is https://www.dimensions.com/collection/drinking-glasses and https://www.dimensions.com/collection/wine-glasses.

Here is some R code to do the calculations and to draw the above graph. Note that pi (lower case) is an inbuilt R constant whose value is approximately 3.141593. (Yes, I am well aware that π is an infinite, non-repeating decimal, and I believe R carries 16 decimal digits, but that is beyond the scope of this article.)

df <- data.frame(glass = c("Kalina10", "Pokal22", "Chardonnay", "XL Oversized","Cordial", "Shooter", "Champagne"),
volume = c(10, 22, 12.3, 25.36, 1.5, 2, 9),
height = c(11, 17.75, 19.8, 22.9, 15.9, 10.5, 23.5),
diameter = c(8, 9.5, 7.9, 10.8, 5.1, 4.13, 6.35))
df$circumference <- round(pi * df$diameter, 1)
df$larger <- ifelse(df$circumference > df$height, "Circumference", "Height")
df$c_to_h <- round(df$circumference / df$height, 1)
df

library(ggplot2)
ggplot(df, aes(x=factor(glass, level = c("Kalina10", "Pokal22", "Chardonnay", "XL Oversize","Cordial", "Shooter", "Champagne")), y=c_to_h, fill = glass, color="black")) +
geom_col(width = 1, position = position_dodge(1)) +
geom_hline(yintercept=1) +
ggtitle("Ratio of Circumference to Height by Glass") + xlab("Glass") + ylab("Ratio") +
theme(plot.title = element_text(face="bold", size=12)) +
theme(axis.text.x = element_text(face="bold", size=12)) +
theme(axis.text.y = element_text(size=12, face="bold")) +
theme(legend.position="none") +
scale_fill_manual("glass", values=c("red", "yellow", "blue", "green", "grey", "brown", "violet"))

I think the reason this is a good bet is that the mind can not easily compare a circular length to a linear length (I don't know if that is scientifically accurate), plus perhaps we look at the diameter but we forget we are comparing the height not to the diameter, but rather to π times the diameter.

Feel free to make this bet with your friends or your students. How about sharing 10% of your winnings with me as a commission?

Incidentally, a beverage can is approximately a right circular cylinder. (But not exactly; look at the top and bottom to see why.) Calculus students can derive that the cylinder with the largest volume for a given surface area (the surface area can be thought of as the rectangular area of the paper label around the entire can), has height equal to diameter. A typical 12 ounce soda can does not have height equal to diameter, but its circumference is greater than its height. A fun supermarket experiment is to examine different shaped cans (a soup can, a tuna fish can, etc.) to determine which meet the largest volume criterion.

To my reinsurance friends: I learned the circumference greater than height trick from Paul Hawksworth of M&G.

Mr. Finch and pi

March 14, Pi Day, is coming up. Is there any more to be said about π that hasn’t been said?



In the January 3, 2013 episode of the TV show “Person of Interest” (“you are being watched ... ”), computer genius Mr. Finch says that since π is an infinite non-repeating decimal, “contained within this string of decimals is every single other number. Your birth date, combination to your locker, your social security number ...”

Really? I didn’t think that is necessarily so. Mr. Finch doesn’t explain why.

There is a site that lets you search for a specific string of digits within the first 200 million digits of π, http://www.angio.net/pi/piquery, but of course just because your number doesn’t appear, doesn’t prove anything. 200 million is a long way from infinity.

Are the digits in $\pi$ truly random? There is something called a normal number, which is a real number whose infinite sequence of digits is distributed uniformly. See http://en.wikipedia.org/wiki/Normal_number which says it is believed that π is normal, but this has not been proven.

So we think Mr. Fitch was right, but we’re not 100% sure.