Cycloid This curve is a cycloid Construction Take a circle Make it stand on a table so that the table is tangent to the circle Mark the point where the circle and table touch as P Start rolling the circle (without sliding) The path traced by P is a cycloid If P starts from the origin of the 𝑥𝑦 plane then the cycloid’s equation is

𝑥 = 𝑟 cos!! 1 −𝑦𝑟 −

2𝑦𝑟 −

𝑦𝑟

!

where 𝑟 is the radius of the circle Animation Let us assume the circle is rolling forward at a constant rate Also, we shall assume it completes 1 revolution in 1 second 𝑓 = 1 𝐻𝑧

𝑇 = 1 𝑠 𝑡 = 0.1 𝑠 𝑡 = 0.2 𝑠

𝑡 = 0.3 𝑠 𝑡 = 0.4 𝑠

𝑡 = 0.5 𝑠 𝑡 = 0.6 𝑠

𝑡 = 0.7 𝑠 𝑡 = 0.8 𝑠

𝑡 = 0.9 𝑠 𝑡 = 1 𝑠

Figure 1

Derivation of the equation Let 𝜃 be the angle by which the circle has rotated At the start 𝜃 = 0 and at the end 𝜃 = 2𝜋 or 360° At any 𝜃 we can parameterise the 𝑥 and 𝑦 co- ordinates Let 𝑠 be the horizontal distance travelled by the circle When an angle of 𝜃 has been turned, 𝑠 = 𝑟𝜃 𝑥 = 𝑠 − 𝑟 sin 𝜃 𝑦 = 𝑟 − 𝑟𝑐𝑜𝑠 𝜃 𝑦 = 𝑟 1 − 𝑐𝑜𝑠 𝜃 ∴ cos 𝜃 = 1 −

𝑦𝑟

∴ 𝜃 = cos!! 1 − 𝑦𝑟 ∴ 𝑥 = 𝑟 𝜃 − sin 𝜃 ∴ 𝑥 = 𝑟 cos!! 1 −

𝑦𝑟 − sin cos!! 1 −

𝑦𝑟

But

sin cos!! 1 −𝑦𝑟 = 1 − cos cos!! 1 −

𝑦𝑟

!

sin cos!! 1 −𝑦𝑟 = 1 − 1 −

𝑦𝑟

!

∴ 𝑥 = 𝑟 cos!! 1 −𝑦𝑟 − 1 − 1 −

𝑦𝑟

!

And

1 − 1 −𝑦𝑟

!

= 1 − 1 − 2𝑦𝑟 +

𝑦𝑟

!

=2𝑦𝑟 −

𝑦𝑟

!

∴ 𝑥 = 𝑟 cos!! 1 −𝑦𝑟 −

2𝑦𝑟 −

𝑦𝑟

!

𝜃(𝑥, 𝑦)(0,0)

Figure 2

𝑠

Approximation Construct regular polygons where 𝑟! is the distance from each vertex to the centroid Let 𝑛 be the number of sides Now start at 𝑛 = 3 (an equilateral triangle) Place it upright on a table Mark one vertex touching the table as P Start rolling the triangle to the right (by pivoting around one vertex at a time) Trace the path followed by P The path will not be a cycloid, but an approximation Repeat the experiment, increasing 𝑛 by 1 each time As 𝑛 → ∞, the path becomes a better and better approximation of a cycloid Shape Curve 𝑛 = 3

𝑛 = 4

𝑛 = 5

𝑛 = 10 Each Approximation is produced by only one polygon If we choose a different 𝑛, we get a different approximation Also each approximation can be thought of as being composed by many smaller arcs Each approximation has 𝑛 − 1 arcs All arcs are described by two properties: The radius of the circle they are part of (𝑝) The angle of the circle they cover (𝛼) For each approximation, the centres of each arc are separated by exactly 𝑠 (Figure 2) This is the side length of the polygon that is being used for that particular approximation. This is because after each rotation, the new pivot point changes to the next vertex that touches the ground. This new vertex is exactly a distance of 𝑠 away from the previous

𝑠𝑠

p𝛼 =2𝜋5

For 𝑛 = 5 with 4 arcs

Figure 3

For each diagram, 𝛼 stays constant for all arcs and only depends on 𝑛 For 𝑛 = 3, 𝛼 = 2𝜋

3 or 120°

𝑛 = 4, 𝛼 = 𝜋2 or 90°

𝑛 = 5, 𝛼 = 2𝜋5 or 72°

𝑛 = 10, 𝛼 = 𝜋5 or 36°

Formulae for 𝛼 and 𝑝 In each case, 𝛼 is the angle of rotation for each arc 𝛼 is also the exterior angle of each polygon since the amount rotated by the polygon while making each arc is always the exterior angle

∴ 𝛼 =2𝜋𝑛

Note, the exterior angle is not the full angle outside the polygon that adds to the interior angle to give 360° It is the angle that adds to the interior to give 180°

Now we need a formula for 𝑝 If we consider the specific case of 𝑛 = 7, we can look at the frames when one rotation around a vertex is complete. We can number the arcs from 1 to 6. Let 𝑗 be the arc number. 𝑟! can be any value because the shape of the approximation doesn’t change with size. The shape is scale invariant.

𝑝 is the radius of each arc For each arc, 𝑝 takes on the value of 1 diagonal of the polygon. Since each polygon has multiple diagonals, the diagonal it is depends on 𝑗 𝑝 is different for each 𝑗 We need a formula for 𝑝 in terms of 𝑗 For each 𝑗, the diagonal that forms the arc’s radius has its vertices separated by a certain number of sides of the polygon Let 𝑘 be the number of sides separating the two vertices of a diagonal. So 2 adjacent vertices will have 𝑘 = 1 and if 𝑘 = 0, the vertices must be the same. Look at Figure 5 to understand what 𝑘 is. 𝑘 also varies with 𝑗 Looking back at Figure 4, When

𝛼Figure 4

𝑗 = 1 𝑗 = 2

𝑗 = 3𝑗 = 4

𝑗 = 5 𝑗 = 6

Figure 5

𝑝 𝑝

𝑝 𝑝

𝑝

𝑝

𝑗 = 1, 𝑘 = 1 𝑗 = 2, 𝑘 = 2 𝑗 = 3, 𝑘 = 3 𝑗 = 4, 𝑘 = 3 𝑗 = 5, 𝑘 = 2 𝑗 = 6, 𝑘 = 1 This is only true when 𝑛 = 7 [Note, the number of sides of separation can be counted by going around the polygon in 2 different ways. For example in Figure 5, the diagonal with a separation of 𝑘 = 3 in one direction has a separation of 𝑘 = 4 in the other. However, we will always take the 𝑘 value that is the smaller of the 2.] If we conducted the same exercise on an octagon 𝑛 = 8, counting the 𝑘 values (sides between vertices of the diagonal) for each 𝑗 (arc), we would get the following 𝑗 and 𝑘 values When 𝑗 = 1, 𝑘 = 1 𝑗 = 2, 𝑘 = 2 𝑗 = 3, 𝑘 = 3 𝑗 = 4, 𝑘 = 4 𝑗 = 5, 𝑘 = 3 𝑗 = 6, 𝑘 = 2 𝑗 = 7, 𝑘 = 1 For every, polygon, the maximum value 𝑗 takes on is 𝑛 − 1 The relationship between 𝑘 and 𝑗 is

𝑘 =𝑛2 −

𝑛2 − 𝑗

This is because the value of 𝑘 depends on the distance of 𝑗 from !! on the number line. For example, when 𝑛 = 7, 3.5 can be thought of

as a turning point. The moment 𝑗 crosses 3.5 (half of 𝑛) we need to start counting backwards. The distance between 𝑗 and !! is 𝑛2 − 𝑗 ,

but 𝑘 is not equal to this. It is the midpoint minus this distance and !! is the midpoint.

In order to create a general formula for 𝑝 (arc radius), we need a general formula for the length of the diagonals of a polygon The first step is to identify what 𝑝 depends on Firstly, the diagonal length depends on the size of the polygon This is determined by 𝑟! which is the distance from each vertex of the polygon to the centroid Then it depends on the number of sides of the polygon Lastly, to specify which diagonal we want the length of, we need to know how many sides separate the two ends of the diagonal. This is 𝑘 To summarise, 𝑝 depends on the following variables 𝑛, 𝑟!, 𝑘 Let us now consider a general polygon with any random diagonal (Figure 4) If we connect the two vertices of the diagonal to the centroid, we get an isosceles triangle

𝑘 = 2𝑘 = 3

𝑘 = 1

Diagonals with 3 different 𝑘 values for 𝑛 = 7

Figure 6

The angle at the centre is !!"!

. We can construct a right triangle by drawing a line that bisects this angle and extending it till it touches the diagonal.

sin𝜋𝑘𝑛 =

𝑝2𝑟!

∴ 𝑝 = 2𝑟! sin𝜋𝑘𝑛

We can now replace the 𝑘 in the equation with the expression for 𝑘 we got earlier in terms of 𝑗. This will give us a formula for 𝑝 in terms of 𝑗 (arc number) 𝑘 =

𝑛2 −

𝑛2 − 𝑗

∴ 𝑝 = 2𝑟! sin𝜋𝑛𝑛2 −

𝑛2 − 𝑗

But 𝜋𝑛𝑛2 −

𝑛2 − 𝑗 =

𝜋2 −

𝜋𝑛𝑛2 − 𝑗

And sin

𝜋2 − 𝜃 = cos 𝜃

∴ sin

𝜋2 −

𝜋𝑛𝑛2 − 𝑗 = cos

𝜋𝑛𝑛2 − 𝑗

∴ 𝑝 = 2𝑟! cos

𝜋𝑛𝑛2 − 𝑗

∴ 𝑝 = 2𝑟! cos𝜋2 −

𝜋𝑗𝑛

But cos 𝜃 = cos 𝜃 Because cos −𝜃 = cos 𝜃

∴ 𝑝 = 2𝑟! cos𝜋2 −

𝜋𝑗𝑛

∴ 𝑝 = 2𝑟! sin𝜋𝑗𝑛

So, we have succeeded in approximating a cycloid. We can construct all the arcs formed by polygon with 𝑛 sides from arc 1 to arc 𝑛 − 1. We have just have to use our 2 equations for 𝛼 and 𝑝

𝑝 = 2𝑟! sin𝜋𝑗𝑛

𝛼 =2𝜋𝑛

Figure 7