Saturday, 7 April 2012

Introductory Mechanics: Maths of Snooker

Whether you like to think of it or not, Maths is a part of every day life and snooker exhibits a lot of mechanical properties that can explain everything about snooker.

You may have never thought exactly why a ball actually stops moving, just that it loses speed and stops. But this means that there is a deceleration (in fact this is incorrect terminology, it is better to say negative acceleration). Newton's laws of motion is where the mechanics begins to come into play, his first law is that a body remains a constant velocity or at rest unless an external force is applied.

As there is a negative acceleration when a snooker ball is hit it means that there must be another external force on the ball besides the force that was applied when it was hit, in fact there are three forces acting upon it. The weight of the ball (note that weight and mass are different!), the reaction force of the object, and the friction between the ball and the cloth. You may be thinking what about when the ball is hit, surely that is a force? And it is! But, that is a non-constant force, it is applied just once and as it is not constant it does not affect the acceleration of the ball.

To explain what these forces are I first need to explain a common misconception. Weight is not mass. Weight changes depending on the gravitational strength of a planet (this is why astronauts jump higher on the moon), Weight = Mass × Gravitational Field Strength. The reaction force is what stops an object falling through a surface, in order to remain at rest or terminal velocity all forces must be in equilibrium (equal), so if the ball is not moving off the table or through it, the reaction force is equal to its weight. Friction is proportional to the reaction force when it is about to move, it is this that allows us to work out friction easily. Friction (F) = Coefficient of Friction (μ) × Reaction Force (R); the coefficient of friction is a property between that is unique between two materials.

And we do know what the coefficient of friction is between a snooker ball and the cloth, it is 0.5 and the mass of each ball is roughly 0.15kg. The gravitational field strength of Earth is about 9.81 ms-2, this means that we can work out what the weight and the friction is between the ball and the cloth. W = 0.15kg × 9.81 ms-2, which gives W = 1.47N. Now that we have the weight we can work out what the friction between the ball and the cloth with the friction coefficient, F = 0.5 × 1.47N, which gives F = 0.736N. We now know the three constant forces acting on the snooker ball, only one actually moves the ball in anyway, and that is the friction.

So a person supplies an initial force to the ball, which gives the ball an instantaneous velocity in the direction it was hit. But this force is not constant so friction kicks in immediately and slows the ball down, but at what rate? Using Newton's second law of motion (Force = Mass × Acceleration), a = 0.736N ÷ 0.15kg so the acceleration is 4.9ms-2.

So we could get an estimate of when the ball would finish up if we knew where the ball started and what it's initial velocity was. But, as you will know if you have played, the cue ball slows considerably when it hits cushions or other balls and there is a reason for this; energy. The kinetic energy of the ball can be calculated initially from the formula ½mv2, where m is the mass of the ball and v is the initial velocity. When the ball hits a cushion or another ball it slows down as the energy of the ball is converted into heat and sound energy, the mass of the ball will not change (much) in a collision, which means the velocity has to decrease to compensate for the decrease in energy. There is no simple formula for working out the loss in energy from contact, which is why snooker players tend to avoid going into balls when not necessary because of the unpredictability of it.

That is all well and good, but what about actually playing a shot? Well it depends what sort of shot you wish to play, but I will talk about playing a pot. If you know the distance between the cue ball and the object ball, the object ball and the pocket and the angle between the two distances, you can work out a variety of things; I will illustrate this with an image.

The dotted lines are lengths and angles that can be worked out, the solid
lines are ones that would have to be measured.

But what does all of this actually mean? The angle θ created by the two balls and the pocket is the angle that has to be adhered to in order for the object ball to be potted. The angle α can be worked out first by working out N using the cosine rule and then using the sine rule with N and θ; ϕ can be worked out using SOHCAHTOA. Then by adding them: α + ϕ is the angle the cue ball has to be played at in order to hit the red ball at the correct angle to direct it towards to pocket. We can work out everything from here now, but first I need to introduce to SUVAT equations.

Where u is the initial velocity of
an object, v is the final velocity,
a is the constant acceleration, t is the
time taken to do the motion and s is
the displacement travelled.

So if the cue ball is hit with some velocity, what will the velocity of the ball be when it comes into contact with the object ball? Well we know the acceleration of the ball, it is decelerating due to friction, -4.9ms-2 (both vertically and horizontally) and the distance from the cue ball to the object ball (L × cos(α + ϕ) horizontally and L × sin(α + ϕ) vertically).

So we have u, a and s and want to find v, so we need to use the equation v2 = u2 + 2as. Inputting this information into the equation gives that (horizontally) vh = √[u2 - 9.8Lcos(α + ϕ)] and vertically vv√[u2 - 9.8Lsin(α + ϕ)]. You may notice that this means if the initial velocity of the cue ball is struck at less than √[9.8Lcos(α + ϕ)] horizontally and √[9.8Lsin(α + ϕ)] vertically it will not reach the object ball. 

How fast the object ball will travel depends on how fast the cue ball is struck, obviously. But what velocity does the object ball need to be struck at initially in order for the ball to go in?  To work this out is pretty simple, we need to know the distance the ball has to travel, the angle created from the line between then object ball and the pocket and the horizontal line from the ball (we'll call it λ) is equal to θ + α + ϕ - 180. We know that the final velocity has to be greater than 0 (v > 0) and the acceleration is -4.9ms-2 (both vertically and horizontally). So we know v, a and s and want to find u, so we need to use v2 = u2 + 2as again. Inputting the values we have we get horizontally uh > √[v2 + 9.8Mcos(λ)] and vertically uv > √[v2 + 9.8Msin(λ)]. This tells you how fast the object ball has to go, but not how fast the cue ball has to go initially.

To work this out we need to use the conservation of momentum (Momentum = Mass × Velocity), the conservation of momentum states that momentum before a collision = momentum after the collision. Both balls have a mass of 0.15kg. The initial momentum is 0.15 × u, after the collision the balls move at right angles to each other with the same speed, this means that:

Where λ = θ + α + ϕ - 180, M is the distance between the
object ball and the pocket and v is the velocity you want the
object ball to finish with.

If you want the minimum velocity the cue ball has to be struck at for the object ball to go into the pocket you simply let v = 0 which gives you:

As you can see, this is a pretty tough calculation and not something you want to have to do before every shot! But the human brain is a remarkable thing and it is able to not do calculations of this magnitude by using past experiences to predict about where to hit the cue ball to hit the object ball in the correct way with the correct pace.

Some of the maths here would not work for an angle on the different side, or behind the ball, etc. You would need to remodel what I have done slightly, but it would just mean creating relationships between triangles slightly differently. I did skim over some mathematical steps in this post because it would have required a lot more text and potentially disengaged readers, if you have any problems with any of the maths here please leave a comment and I will explain.

Wednesday, 4 April 2012

Solving Trigonometric Equations

If you have done maths up to any level past the age of 14 it is likely that you have encountered trigonometry in some shape or form. You may remember SOHCAHTOA, meaning sin(x) = opposite/adjacent in a right angled triangle; and this is the simplest form of trigonometry you will have encountered.

It is likely that you have also encountered some basic trigonometric equations. A basic equation would be cos(x) = 1, and you may know that this means that x = 0; but this is not the only value of x that solves the equation. If you have ever gone far enough as to draw the graphs then you may have noticed that there are multiple solutions to every trigonometric equation. This is because trigonometric functions are periodic, so they repeat values of y for different values of x.

To illustrate this consider cos(x) = a, let cos(α) = a, this means that α is a solution to the equation. But it isn't the only one, I will show this on the graph of y = cos(x); note that the graph is in radians and I will be explaining in terms of radians, if you do not understand please leave a comment.

Each red dot represents a solution to an equation
involving cos(x). Notice how they all lay on a
horizontal line.
So we know that the first solution is for x = α, but what are the next ones? You have to look at when the graph repeats itself. The cosine graph has a period of 2π and it is symmetrical at x = 0, 2π, ..., 2nπ; this means that cos(x) = cos(-x). This implies that x = ±α at least, and as the function repeats every 2π you can add that on to x again and again to get the same result. This means that if cos(x) = cos(α) then x = 2πn ± α. This gives us a general solution to cosine function!

Now to do with sin. Suppose that we have a solution to some equation involving sin where sin(x) = sin(α). Again you have to look at the sine graph for when it repeats itself and where it has lines of symmetry. In fact the sine graph is the same as the cosine graph, just it has been translated to the right by π/2, this means it still repeats every 2π but the line of symmetries are at x = π/2, 3π/2, ...(2n-1)π/2. The first of these mean that when it is an even multiple of π you can find another solution by adding α that is: x = 2πn + α. The second of these mean that whenever it is an odd multiple of π subtracting α gives another solution, that is: x = (2n+1)π + α. To find a range of solutions for sin(x) = sin(α) then x = 2πn + α or 2πn + π + α.  This gives us a general solution to sine function!

And the easiest one of all is finding solutions to tan(x) = tan(α). The tan graph repeats every π and has no lines of symmetry. This simply means that if x = α, it also equals π + α and 2π + α, etc. To find a range of solutions for tan(x) = tan(α) then x = πn + α. This gives a general solution to the tangent function!
That is fine for when you have simple trigonometric equations, but what if you have one that contains more than one trigonometric function? Well there are trigonometric identities that help you to rearrange equations into a form that is easier to solve. I will simply list them in this post, but I will prove them in a later post. For UK readers, next to the identity I will state in which module you need to utilise them in.

- sinӨ/cosӨ = tanӨ; this is required for Core 2 and beyond.
- cos2Ө + sin2Ө = 1; this is required for Core 2 and beyond.
- cosӨ/sinӨ = cotӨ; this is required for Core 3 and beyond.
- tan2Ө + 1 = sec2Ө ; this is required for Core 3 and beyond.
- cot2Ө + 1 = cosec2Ө; this is required for Core 3 and beyond.
- sin(A + B) = sinAcosB + cosAsinB; this is required for Core 4.
- cos(A + B) = cosAcosB - sinAsinB; this is required for Core 4.
- sin(A - B) = sinAcosB - cosAsinB; this is required for Core 4.
- cos(A - B) = cosAcosB + sinAsinB; this is required for Core 4.
- tan(A ± B) = (tanA ± tanB) / (1 ∓ tanAtanB); this is required for Core 4.
- sin2Ө = 2sinӨcosӨ; this is required for Core 4.
- cos2Ө = 1 - 2sin2Ө; this is required for Core 4.
- tan2Ө = 2tanӨ / (1 - 2tan2Ө); this is required for Core 4.
- sin3Ө = 3sinӨ - 4sin3Ө; this is required for Core 4.
- cos3Ө = 4cos3Ө - 3cosӨ; this is required for Core 4.
- tan3Ө = (3tanӨ - tan3Ө) / (1 - 3tan2Ө); this is required for Core 4.