Modeling the London Eye and a hypothetical ferris wheel on the opposite side of the River Thames through the field of trigonometry

Since a very young age, wherever I travelled, going on a Ferris wheel was an unavoidable activity to do. Seeing a place from above was beautiful and I could spot out historical buildings from angles that I had never seen before. A few months ago, while I was visiting the magic city of London, I had to embark the London eye and observe the city from higher up. It is Europe's tallest cantilevered observation wheel, and is the most popular paid tourist attraction in the United Kingdom with over 3 million visitors annually (O’Ceallaigh). As I was on the wheel I stared at the beautiful other side of the river where the houses of the Parliament are and suddenly an idea came to my mind. I wondered if I could match a cosine wave to the rotation of the wheel.

Without adding much thought to it while I was in London in June, I now have decided to model the London eye as well as a hypothetical Ferris Wheel. I am going to London in December for my best friend’s birthday. If I was to be on the London eye and she was to be on a hypothetical Ferris wheel in front of me on the other side of the river, my aim is to investigate the height and time at which we could wave at each other directly across. The periods and speeds of the Ferris wheels will be different therefore I will have to investigate when both models intersect.

Trigonometric functions have periodic behaviors allowing us to model things such as sound waves and the
spinning of a Ferris wheel. Thus, they will be useful for this investigation, modelling the London Eye.

Oscillation is the repetitive or periodic variation, typically in time of some measure about a central value (Dourmashkin). A Ferris wheel moves at regular oscillations at fixed periods of time. Indeed, on a circle particularly the unit circle, sin \(\theta\) is equal to the \(\theta\) −coordinate of your point on the circle and cos \(\theta\) is equal to the \(\theta\) coordinate. Equally, each Ferris wheel has a fixed diameter and height which grants us the elements used in a sinusoidal function. With a sinusoidal wave, we can plot the wheel’s position (height) as a function of time.

To begin the investigation, I first had to find the measurements of the London eye, such as the height, the diameter and extra information. With some research I found the following (The London Eye) -

• Height - 135m
• Diameter - 120m
• Number of capsules on the wheel - 32
• Time to complete one turn - 30 minutes

The motion of the London Eye can be described by a periodic function which is a function that repeats its values in regular periods or intervals. Knowing the dimensions of the wheel, I could now see how they can fit into the sinusoidal function - 

\(y=a \,cos (b(x-c))+d\)

Where a is the amplitude, b is the horizontal stretch, c is the horizontal translation, d is the vertical translation and x is time further referred to as variable t. To use this equation, these formulas are needed -

\(Amplitude: \frac{Y_{max}-Y_{min}}{2}\)

\(Period: \frac{2\pi}{b}\)

To model the London eye with this equation, I had to first find the amplitude. While I was researching, I did not find any information on the initial height of the London Eye or its minimum height. The initial height (It), is the lowest point on the wheel itself as one embarks the capsule. I had to calculate this as there are stairs that one must go up leading them to a platform where they enter the capsulesto begin their ride. Therefore, I figured that if the overall height of the wheel is 135 meters and the diameter is 120 meters then the minimum height also known as the initial height must be 15 meters.

\(It =135-120=15\)

This then allowed me to figure out the first component of our equation, amplitude, a.

\(a=\frac{135-15}{2}\)

\(=60\)

I started by graphing a simple cosine wave on DESMOS this is where I graphed all my functions. My functions are graphed in radians as this is calculus. I graphed the following function -

\(y=60 \,cos\,x\)

I chose to model the Ferris wheels on a cosine wave as when the vertical movement of a Ferris wheel is plotted on a time axis which is horizontal, the cosine wave gets its shape. The cosine wave oscillates back and forth from the equilibrium position. Therefore, it can be visualised as the horizontal component of a circular motion. The wave starts at its minimum as the people embark the ride whereas a sine wave would start on the principal axis. So, I decided not to model both Ferris wheels on a sine wave as it would have been illogical and visually complicated to understand. I have chosen to use the domain of {0 ≤ x ≤ 60} on the graphs of the cosine waves which displays two full periods of the Ferris wheel.

I can now move on to the horizontal stretch b of the function. To do this I will use the formula, &' \(\frac{2\pi}{b}\). The letter b is one period, I found out before that the London Eye takes 30 minutes to complete one turn. Hence this is the period.

\(\frac{2\pi}{b}=30\)

\(30b=2\pi\)

\(b=\frac{\pi}{15}\)

My sinusoidal function now looked like this -

\(y=60\,cos\,(\frac{\pi}{15}x)\)

Moreover, I needed to find the vertical translation, d. The Ferris wheel is above the ground, so a positive value is expected. I needed to find how high the center of the wheel is above the ground. To do this I added the initial height and the radius of the wheel together.

\(d=15 +60= 75\)

The value of c is inexistant since there is no horizontal translation, we are just looking at the height. The vertical translation allowed me to create the following equation -

\(y= 60\,cos\, (\frac{\pi}{15}x)+75\)

After creating this graph, I noticed that something was incorrect. While the x is the time (t), when t = 0 the capsule on the wheel should be at the minimum height of 15 meters and not at the maximum height of 135 meters high. When an individual gets on a Ferris wheel, they begin their ride at the bottom. To correct this, I added a transformation to the function which had to be a vertical reflection -

\(y=-60\,cos\,(\frac{\pi}{15}x)+75\)

With the information collected from the calculations above, this cosine function then permitted me to model the height of a given capsule on the London Eye Ferris wheel, in meters above the ground.

This cosine wave displays the relationship between the time taken for one full rotation in minutes and the height of a capsule above the ground, in meters on the London eye. I will only be on the London Eye for one rotation; therefore, one oscillation is shown.

\(y=-60\,cos\;(\frac{\pi}{15}x)+75\)

As seen from the graph above, as an individual enters one of the capsules, the wheel begins at 15 meters which is the minimum height. After 15 minutes on the ride, the capsule reaches the maximum height of 135 meters and then slowly comes back down to the minimum height after a full period of 30 minutes which is relevant to the domain. After conducting the steps above, my original, cosine wave was transformed into a cosine function that models the oscillation of the London Eye. This was done through trigonometry and transformations. I believe that the most challenging part was to find which element of the trigonometric equation matched with each diameter of a Ferris wheel.

I wanted to complete one last calculation that will be interesting to find out in relation to this investigation. I wanted to know the speed of the London Eye. The London Eye never stops apart from certain exceptions. It slows down when people disembark and embark however it never stops. Therefore, to calculate the speed of the wheel, I used the tangential velocity (Vt) equation which consists of dividing the circumference (C) of the wheel by the period. I will be rounding the final value to 3 significant figures as it is the IB standard.

C = 2 × π × 60

C = 376.9911184

• To calculate the speed of the wheel in kilometers per hour, the unit of the Ferris wheel’s circumference had to be changed from meters to kilometers by dividing the distance by 1000.

376.9911184 ÷ 1000 = 0.37699... km

• To convert the unit of time from minutes to hours the value of 30 had to be divided by 60.

30 ÷ 60 = 0.5

\(V_t = \frac{0.3769911184}{0.5}\)

\(V_t=0.754\,km/ houe\)​​​​​​​

I will now be modelling the hypothetical Ferris wheel. Let’s imagine a Christmas market was to settle in right in front of the London Eye on the other side of the Thames River. This Christmas market so happens to have a Ferris wheel that is built for the month of December however it is far smaller than the London eye since it is temporary. Hence, I will be using the same sinusoidal function to model this Ferris wheel. Here are its hypothetical dimensions.

• Height - 55m
• Diameter - 45m
• Number of carts on the wheel - 27
• Time to complete one turn - 12 minutes
• Initial height: 10m (55 − 45)
• One ride has three rotations on the wheel.

As seen from the graph above, as an individual enters one of the capsules, the wheel begins at 10 meters which is the minimum height. After 6 minutes on the ride, the capsule reaches the maximum height of 55 meters and then slowly comes back down to the minimum height after a full period of 12 minutes. The graph shows 3 rotations of 12 minutes which is the equivalent of 1 ride on the hypothetical Ferris wheel. This is therefore 36 minutes, representing 3 cosine waves which is relevant to the domain. I will now be calculating the speed of the hypothetical Ferris wheel.

C = × π × 22.5

= 141.3716694

_{\(V_t = \frac{0.1413716694}{0.2}\)} 

V_{t = 0.707 km / hour}

The graph above displays the two different cosine functions, the blue cosine wave represents the London Eye Ferris wheel, and the red cosine wave is the hypothetical Ferris wheel. On the graph, I decided to show one full rotation of the London Eye which is one ride as I will be on the London eye during this time. The hypothetical Ferris wheel is not graphed for a full ride of three rotations as anyways I would not be able to see my friend on the third rotation since I would have ended my ride on the London eye. This is due to the

difference in time of both rides, the London Eye being 30 minutes long which is relevant to my domain whereas the hypothetical wheel is 36 minutes long.

The heights and periods of both Ferris wheels are different however we can observe that they do intersect three times at different heights and times. I will now be finding out when and at what height the wheels intersect so that I am able to wave at my friend from the London Eye while she is on the hypothetical Ferris wheel directly across.

1. Firstly, the cosine functions will be intercepting when they are equal -

Y₁ = Y₂

_{\(-\,60\cos\,(\frac{\pi}{15}x)+75=-\,22.5\,cos\,(frac{\pi}{6}x)+32.5\)}

_{2. Next, I will be making an estimation from Figure 1.6 of the height and time of each intersection. I did my estimates using 2 decimal points as this is suitable for estimates from a graph.}

3. As estimations from the human eye are not accurate due to human error, by graphing both equations on my graphic display calculator, I will be able to calculate the height and time of each intersect. I will enter both my equations in my calculator as y = x.