The Number Mysteries: A Mathematical Odyssey through Everyday Life

by Marcus du Sautoy

  When Will the Next Eclipse Be?

  If you know the time of one eclipse, you can use a mathematical equation to work out the time of another one. The calculations depend on two important numbers.

  The first is the synodic month ( S) of 29.5306 days. This is the average time it takes for the moon to go around the earth and return to the same position relative to the sun. It’s the average time between two new moons.

  The other is the draconic month ( D) of 27.2122 days. The moon’s orbit around the earth is slightly tilted with respect to the earth’s orbit around the sun. The two orbits cross each other in two places, called the nodes of the moon’s orbit, as shown in the following figure. The draconic month is the average time it takes the moon, starting from one node, to pass through the opposite node and return to the first one.

  Figure 5.1 The orbit of the moon intersects the orbit of the earth in two places, called the ascending node and the descending node.

  For every pair of whole numbers A and B you can find that will make A × S very close to B × D, you will have the date of an eclipse A × S ≈ B × D days after the last eclipse you saw. And there will be another eclipse after another A × S ≈ B × D days. The sequence of eclipses will continue for a while, but because the equation is not exact, the eclipses will eventually get less and less impressive, until the sun, moon, and earth are no longer aligned. And that will be the end of that particular cycle of eclipses.

  Here’s an example: A = 223 synodic months is very close to B = 242 draconic months, so every 223 × 29.5306 ≈ 242 × 27.2122 days after an eclipse, there will be another almost identical eclipse. That is a period of approximately 6,585 days, or about 18 years, 11 days, and 8 hours. The shift of 8 hours means that the next two such eclipses will be seen from a different location on the surface of the earth. However, the third will hit the same spot, so every three times 18 years, 11 days, and 8 hours—or approximately 19,756 full days—there will be a repeat eclipse.

  For example, the total lunar eclipse visible from North America on December 21, 2010, is a repeat of the eclipse on December 9, 1992, seen from Europe. It was last seen in America on November 18, 1956. There have been other eclipses between these dates, but they are part of other eclipse cycles that are running alongside this one. The math helps you calculate the date of the next eclipse in each cycle.

  The power of mathematics to predict happenings in the night sky relies on spotting patterns that repeat themselves. But how can we predict something new? The story of how we can use the equations of mathematics to look into the future begins with predicting the behavior of simple objects, such as a soccer ball.

  IF I DROP A FEATHER AND A SOCCER BALL, WHICH HITS THE GROUND FIRST?

  The soccer ball, of course. You don’t have to be a world-class mathematician to predict that. But what if I drop two soccer balls of the same diameter, one containing lead and the other air? For most people, their first reaction is to say that the lead ball will hit the ground first. That was certainly the belief of Aristotle, one of the greatest thinkers of all time.

  In an apocryphal experiment, the Italian scientist Galileo Galilei showed that this intuitive answer is completely wrong. He worked in Pisa, home to the world-famous Leaning Tower. Where better to chuck things over the side and have your apprentice standing at the bottom to see which lands first? Galileo proved Aristotle wrong: both balls, though of different weights, will hit the ground simultaneously.

  Galileo realized that the weight of the object didn’t come into it. What makes a feather fall more slowly than a ball is air resistance, and if you could take the air away, a feather and a ball should both fall at the same speed. One place where you could test this theory is on the airless surface of the moon. In 1971, the commander of the Apollo 15 mission to the moon, David Scott, recreated Galileo’s experiment by dropping a geological hammer and a falcon’s feather at the same time. They fell much more slowly than they would on earth because of the moon’s lower gravitational pull, but the two objects hit the ground simultaneously, just as Galileo predicted they would.

  NASA’s lunar rerun of Galileo’s experiment can be viewed at http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html or by using your smart-phone to scan this code.

  As the mission controller said later, the result was “reassuring considering both the number of viewers that witnessed the experiment and the fact that the homeward journey was based critically on the validity of the particular theory being tested.” That’s very true: space travel would be impossible to plan without having the equations of mathematics to predict a spacecraft’s flight as it is pushed and pulled by the gravity of the earth, sun, moon, and planets and the thrust of its engines.

  Once he had discovered that the weight of a falling object was irrelevant to its speed, Galileo wanted to see whether he could predict how long it would take for the object to hit the ground. Objects fell too quickly for accurate timing from something like the top of the Leaning Tower, so Galileo decided to roll balls down a slope to see how the speed varied. He discovered that if a ball rolled one unit of distance after one second, then after two seconds it would have covered four units of distance, and after three seconds it would have traveled nine units. He could then predict that after four seconds the ball should have traveled a total of 16 units of distance—in other words, the distance a body falls is proportional to the square of the time for which it has been falling. In mathematical symbols, d = ½gt 2 where d is the distance fallen and t is the time. The factor g, known as the acceleration due to gravity, told Galileo how the vertical speed of a falling object changed with each passing second. For a soccer ball dropped from the top of the Leaning Tower of Pisa, after one second its speed is g, after two seconds it is 2g, and so on. Galileo’s formula was one of the first examples of a mathematical equation being used to describe nature, of what would come to be called a law of physics.

  Using mathematics in this way has revolutionized the way we understand the world. Previously, people had used everyday language to describe nature, and that can be vague—you could say that something was falling, but you couldn’t say when it would land. With the language of mathematics, people could not only describe nature more precisely, but they could also predict the way it would behave in the future.

  Having worked out what happens to a ball when you drop it, Galileo’s next move was to predict what happens to it when it gets kicked.

  WHY DOES WAYNE ROONEY SOLVE A QUADRATIC EQUATION EVERY TIME HE VOLLEYS THE BALL INTO THE GOAL?

  “Beckham with the free-kick, a perfectly timed volley by Rooney . . . goal!!!”

  But how did Rooney do it? You might not think so, but Rooney has to be incredibly good at math to be able to score such a goal. Every time he gets on the end of a free kick from Beckham, he’s subconsciously working out another of the equations Galileo concocted so that he can predict where the ball will end up.

  Equations are like recipes. Take the ingredients, mix them up a certain way, and the equation spits out an outcome. To construct the equation that Rooney will solve, Galileo needs the following ingredients: the incoming ball’s horizontal speed u and vertical speed v when it left Beckham’s foot, and the effect of gravity, which is summed up in the number g and tells Rooney how the vertical speed of the soccer ball changes with each second. The value of g depends on what planet you play your soccer; on the earth, gravity increases the speed by about 9.8 meters per second per second (about 22 mph per second). Galileo’s equation then tells Rooney the height of the ball at any point relative to where the free kick was taken. For example, if the ball is a horizontal distance of x + meters from where Beckham kicked the ball, then the height above the ground will be y meters, where y is given by the equation

  The recipe is the set of mathematical instructions for what to do with all these numbers, and the outcome is the height of the ball at a certain point in its trajectory.

  For Rooney to work out how far to stand from the free kick so that he
can volley or head the ball into the net, he has to undo the equation and work backward. First, he decides that he wants to head the ball. Rooney is about 1.80 meters tall, so the ball has to be at a height y = 1.80 if he is going to head it (without jumping). He knows what u, v, and g are. Let’s choose some approximate numbers: u = 20, v = 10, and g = 10. (If you are worried about units, the speeds u and v are in meter/second, and g is in meter/second2.)

  The only thing Rooney doesn’t know is how far away from Beckham he should stand to intercept the ball correctly. But the equation does have that information encoded in it; it’s just not so visible. The equation says that Rooney should stand x + meters from Beckham, where x + is a number that makes the equation

  true. Tidying this up gives us

  x +2 – 40 x ++ 144 = 0.

  This sort of equation might look familiar—it’s one we all learned to solve in school; it’s called a quadratic equation. Think of it as a cryptic crossword clue, hiding the true value of x.

  Amazingly, the first people to start solving equations like this one were the ancient Babylonians. Their quadratic equations didn’t describe trajectories of soccer balls but appeared when they were surveying the land around the Euphrates. A quadratic equation comes about when we are trying to work out some quantity that is multiplied by itself. We call this squaring because it gives the area of a square, and it’s in the context of calculating the area of a piece of land that these quadratic equations were first formulated.

  Here’s a typical problem. If a rectangular field has an area of 55 square units and one side is six units shorter than the other, how long is the longer side? If we call the longer side x, then the problem tells us that x + × (x + – 6) = 55 or, simplifying things,

  x +2 – 6 x + – 55 = 0.

  But how do you go about unraveling this mathematical cryptic clue?

  The Babylonians came up with a neat method: they dissected the rectangle and rearranged the pieces to make a square, which is an easier shape to deal with. We can divide up the pieces of our field just as Babylonian scribes would have done thousands of years ago:

  Figure 5.2 How to solve a quadratic equation by completing a square.

  Start by cutting a small rectangle measuring 3 x (x + – 6) units off the end of the rectangle and move this around to the bottom of the rectangle. The total area hasn’t changed—just the shape. The new shape is almost a square with each side x + – 3 units long, but it’s missing a small 3 x 3 square in the corner. If we add in this small square, we increase the area of the shape by nine units. The area of this large square is therefore 55 + 9 = 64. Now we have the simple task of taking the square root of 64 to discover the length of the side, which must be 8. But the side had length x + – 3, and so x + – 3 = 8—that is, x + = 11. Although we’ve only been shuffling around imaginary parcels of land, behind what we’ve been doing lies a general method for unlocking those cryptic quadratics.

  Once algebra was created in the ninth century in Iraq, a formula could be written down that captured the Babylonian method. Algebra was developed by the director of the House of Wisdom in Baghdad, a man named Muhammad ibn-Musa al-Khwarizmi. The House of Wisdom was the top intellectual center of its day and attracted scholars from around the world to study astronomy, medicine, chemistry, zoology, geography, alchemy, astrology, and mathematics. The Muslim scholars collected and translated many ancient texts, effectively saving them for posterity—without their intervention, we may never have known about the ancient cultures of Greece, Egypt, Babylonia, and India. However, the scholars at the House of Wisdom weren’t content with translating other people’s mathematics. They wanted to create a math of their own and push the subject forward.

  Intellectual curiosity was actively encouraged in the early centuries of the Islamic empire. The Koran taught that worldly knowledge brought people closer to holy knowledge. In fact, Islam required mathematical skills, because devout Muslims needed to have the times of prayer calculated and needed to know the direction of Mecca to pray toward.

  The algebra of al-Khwarizmi revolutionized mathematics. Algebra is a language that explains the patterns that lie behind the behavior of numbers, and its grammar underlies the way numbers interact. A bit like a code for running a program, it will work with whatever numbers you feed into the program. Although the ancient Babylonians had devised a cunning method to solve particular quadratic equations, it was al-Khwarizmi’s algebraic formulation that ultimately led to a formula that could be used to solve any quadratic equation.

  Whenever you have a quadratic equation ax +2 + bx ++ c = 0, where a, b, and c are numbers, then the geometric juggling can be translated into a formula with x + on one side and a recipe combining the numbers a, b, and c on the other:

  It is this formula that allows Rooney to undo the equation controlling the flight of the ball and work out how far away to stand. We left him knowing that he had to stand x + meters from the position of the free kick, where

  x +2 – 40 x ++ 144 = 0

  Using algebra, he can work out that he should stand 36 meters away from Beckham to intercept the ball with his head.

  How did he do this? Well, in the quadratic equation controlling Beckham’s free kick, a = 1, b = –40, and c = 144. So the formula for undoing this equation tells us that the distance Rooney should stand from Beckham is

  Interestingly, because –32 is also the square root of 1,024, we get another solution: x + = 4 meters. This is the point where the ball is heading upward on its trajectory; Rooney will wait until the ball starts coming down again. Because there is always a negative as well as a positive square root, we always get two solutions out of this formula. To indicate this, sometimes the equation comes with a +– instead of a + in front of the square root symbol.

  Of course, Rooney is using a much more intuitive approach—one that doesn’t require him to do mental math for 90 minutes. But it does show how the human brain is almost programmed by evolution to be good at making predictions.

  WHY DOES A BOOMERANG COME BACK?

  Strange things happen to objects when they spin. When you kick a soccer ball off-center, it bends in the air, and when you toss a tennis racket into the air, it always spins before you catch it. A spinning gyroscope seems to defy gravity by leaning horizontally. But the classic example of the strange behavior of a spinning object is the way a boomerang comes back.

  The dynamics of spinning objects are very complicated and have foxed generations of scientists. But we now know that the reason a boomerang comes back has to do with two different factors. The first relates to the lift of an airplane’s wing, and the second is called the gyroscopic effect. Mathematical equations help to explain and ultimately predict how the geometry of a wing generates a force that pushes up, counteracting the force of gravity that is pulling the airplane down. An airplane’s wings are shaped so that the air flows faster over the top of the wing than underneath it. The air on top gets squashed and pushed faster over the wing. It’s the same principle as water flowing through a pipe: where the pipe narrows, the water flows faster.

  A second equation, called the Bernoulli equation, shows that the higher speed of the air over the top of the wing leads to a lower pressure above the wing, and the lower speed of the air below the wing leads to a higher pressure. This difference in pressures above and below the wing creates force, which lifts the airplane.

  If you look at a classic boomerang closely, you can see that each arm is shaped like the wing of an airplane. This is what makes the boomerang turn. To throw a boomerang with the hope that it will return, you need to launch it from a vertical position in such a way that (thinking of it like an airplane) its right wing is at the top and its left wing is at the bottom. The same force that lifts an airplane’s wing now pushes the boomerang to the left.

  But there’s something slightly more subtle happening here. If the boomerang were simply behaving like an airplane, then the force would just send it to the left, not make it come back. It comes back because when it’s thr
own, it’s given spin, which, thanks to the gyroscopic effect, causes the force pushing it to the left to constantly change direction, with the result that the boomerang is pushed around the arc of a circle.

  When I throw the boomerang, the top part is spinning forward and the bottom part is spinning backward. The upper part is like an airplane’s wing, traveling faster through the air. On an airplane flying horizontally, that faster movement would create more lift. But on a boomerang being launched vertically, it causes the boomerang to tilt, its top leaning into the eventual arc of its flight.

  Now the gyroscopic effect comes into play. When you place a spinning gyroscope on a stand in a vertical position, it behaves like a top. But if you tilt it so that its axis of rotation is at an angle to the vertical, something called precession happens: the axis of rotation itself begins to rotate. This is what happens with the spinning boomerang. Its axis of rotation is an imaginary line running through its center, and as this axis rotates, so the boomerang is pushed around a circle.

  Figure 5.3 The forces acting on a boomerang. F is the force created by the lift, V is the speed at which the center of the boomerang moves, R is the radius of the path of the boomerang, and W is the rate of precession.

  Anyone who has thrown a boomerang will know that getting it to come back isn’t easy. You need to throw the boomerang in such a way that V, the speed at which it leaves your hand, and S, the spin rate you give the boomerang when you launch it, satisfy the formula