Roller Coaster Design

Connor Hain

April 25, 2016

Contents

1 Introduction 2

2 Circular Loop 6

2.1 Motion in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Energy Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Air Resistance/Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3.1 High Air Resistance . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3.2 Low Air Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.4 G-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Clothoid Loop 16

3.1 Defining the Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Motion Along a General Curve . . . . . . . . . . . . . . . . . . . . . . . 20

3.3 Motion Around the Clothoid Loop . . . . . . . . . . . . . . . . . . . . . 22

3.4 General Energy Argument . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.5 G-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.6 Air Resistance/Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 Circular Helix Loop 29

4.1 Defining the Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.2 Frenet-Serret Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.3 Motion Around the Circular Helix Loop . . . . . . . . . . . . . . . . . . 32

4.4 G-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Clothoid Helix Loop 36

5.1 Motion Around the Clothoid Helix Loop . . . . . . . . . . . . . . . . . . 37

5.2 G-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6 Multi-Mass Problem 40

6.1 Two-Mass Train . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6.2 Three-Mass Train . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 N-Mass Train . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7 Conclusion 47

1

1 Introduction

Roller coasters have been the headline attractions at theme parks for many years.

Whether they are dominating the skyline or boasting a large number of inversions, they

have progressed from being relatively simplistic machines, to becoming highly complex

structures, with incredibly complicated control systems.

The term ’roller coaster’ is thought to have originated from American thrill rides,

from the late nineteenth century. These rides used hundreds of rollers that were mounted

to ramps, that allowed an object such as a sled to be ridden down. The passengers in the

sled then coasted along the rollers down the ramps, hence the term ’roller coaster’ [1].

However, despite giving these hugely popular machines their name today, these simple

American rides had no other impact on the future of roller coasters.

The first full circuit roller coaster, Promenades Aeriennes (Aerial Walk), was built

early in the nineteenth century (1817) in France (see Figure 1) [1]. Since this achieve-

ment, designers have been utilising many different methods in order to create taller,

faster and more exciting roller coasters. Today, roller coasters come in a vast array of

shapes and sizes. Ranging from a variety of track materials, to different seat configu-

rations (e.g. inverted coasters - riders are suspended underneath the track), designers

look to create the most thrilling experience whilst keeping the passengers safe.

Figure 1: Promenades Aeriennes - first full circuit roller coaster [10].

Throughout the design phase, many complications will arise due to the high number

of forces that act on the train, as it makes its way around the circuit. However, before

designers can begin to tackle these problems, the method in which the train gathers

enough speed to complete the circuit must be determined. There are two main ways in

which this is achieved:

2

I Lift Hill - The train is pulled up a hill at the beginning of the ride. At the top of

the hill, the train is released and runs freely around the track, with only the force

due to gravity accelerating it to the required speed to complete the circuit.

II Launch System - Many roller coasters now utilise a launch system, which accelerates

the train to its top speed in a matter of seconds, so that the circuit can be completed.

Once the method of accelerating the train has been determined, designers can then

begin to work on the rest of the track layout. There are many factors that influence

the layout of a roller coaster, all of which must be taken into consideration. One of the

main factors is g-force.

G-force creates a problem for designers, because humans can lose consciousness when

subjected to a g-force greater than 5g. When a roller coaster is designed, the track layout

must compensate for this and should be designed in a specific way, so that riders are

subjected to no more than this value. This puts severe restrictions on how a track can

be laid out. Early experimentations with vertical loops put riders under as much as 12g

[1]. This caused serious injuries, identifying that the shape of the vertical loop needed

adjusting.

As the train is set into motion, multiple forces act upon it. The force due to gravity

powers the roller coaster for some, if not all of its journey around the track. There will

also be a reaction force from the track acting on the train. The majority of modern

roller coaster trains are designed with special wheel configurations, so that they have

wheels on top and underneath tubular tracks [1]. This ensures that the reaction force

is never zero, indicating that the train and track always remain in contact with each

other.

Next, we have an opposing force called the contact force, or friction. This contact

force opposes the motion of the train as it moves along the track, consequently slowing

the train down by the means of energy conversion; converting kinetic energy to heat

energy. Care should be taken to ensure that the contact force is not too great, otherwise

significant damage to the track may occur.

Drag, D, is also a very important factor to consider when designing a roller coaster.

Drag is caused by a body moving through a medium (in this case, air) and pushing

the medium out of its path. By Newton’s third law, the medium then exerts equal and

opposite forces on the body, consequently slowing the body down.

Given that roller coasters often travel at very high velocities, we consider the drag

3

equation for an object with high velocity, given by [6]

D = −µ|u|u ,

where µ is the drag coefficient and u is the velocity of the object, with magnitude, u.

The drag coefficient is determined by experiment and is affected by the shape of the

object moving through the medium. It should also be noted that |D| = −µu2 , showing

that the magnitude of the drag force is proportional to the magnitude of the velocity

squared.

In order to minimise the drag coefficient, fewer and weaker collisions between the

air and the car must occur. To achieve this, the front car of the roller coaster train,

being the biggest contributor to drag, is designed in particular ways, so that the num-

ber/magnitude of the collisions that occur is reduced. The front car usually has a small

surface area and is conical in shape, so that fewer air particles collide with the train and

the angle at which the collisions occur is reduced, making the collisions weaker.

Figure 2: Standard design for the front car of a roller coaster train.

All of these forces that act on the train makes modelling a roller coasters motion

incredibly complex. It is therefore logical to begin by making some simplifications to

the system, before we attempt to perform any calculations.

Firstly, we begin by restricting the motion to the x-z plane, with the x-axis being

the horizontal and the z-axis being the vertical. At this point we will also ignore the

effects of drag and friction. Eventually, we will introduce the effects of drag into the

problem, before including the y-axis, so that motion can be studied in three dimensions.

The second simplification is that we first model the train as a point mass positioned

on top of the track. This prevents dealing with multiple forces that vary at different

points throughout a long train at specific times. Also the shape of each car, in particular

the front car, will affect the drag, which increases the complexity of the problem. Pro-

gressing from a point mass, two point masses can be connected via a light inextensible

rod, before adding further masses to the train, so that formulae governing the motion

for an n-mass train, where n ∈ N, can be determined.

4

Once these simplifications have been made, it is possible to begin considering the

motion of a point mass around sections of track; in particular the vertical loop.

The vertical loop is one of the most simplistic inversions that can be found on a

roller coaster, although the mathematics behind it isn’t. The first vertical loop to be

constructed was in Paris in 1846 (Grand Centrifugal Railway) [1]. The ride relied on

centripetal forces to keep the car and track in contact with each other. The train would

enter the loop at high speeds, ensuring that the reaction force from the track on the

train remained positive, when fully inverted at the loop’s highest point.

Early vertical loops, including the Grand Centrifugal Railway, were designed so that

they were a perfect circle. Beginning in the x-z plane, we will consider a point mass

on a circular track and attempt to calculate the magnitude of the initial velocity that

must be given to the mass, so that it passes the highest point of the track (the top of

the vertical loop).

To find a solution to this problem, we need to consider constrained motion/systems.

As the point mass is set into motion, we will consider the effects of different forces that

are acting on it. In the most simplified case, this is the force due to gravity, along with

the normal reaction force. By applying Newton’s second law of motion, we can obtain

an equation to calculate the reaction force, along with an equation governing the motion

of the point mass along the track. This can be solved using a variety of techniques to

find an equation to calculate the magnitude of the initial velocity that is required for

the point mass to reach the top of the loop. The equation governing the reaction force

is only useful for roller coasters with particular wheel configurations and will not be

investigated further. As we progress and add the drag force into the calculations, the

equation of motion will be adjusted to account for this extra force.

By studying this original problem in more depth, we will discover that this perfect

circular shape used in original vertical loop designs, meant that the g-force experienced

by the riders was far too great, often resulting in injury. These incredibly high g-forces

were caused by the sudden change in the curvature during the transition between the

horizontal and circular track. This transition caused a massive onset of centripetal ac-

celeration, resulting in the high g-force that is experienced. By finding an equation

for g-force and then calculating the turning points of this equation, we will show when

g-force takes its maximum value for the circular loop, along with calculating this max-

imum value to prove that it is too large. It is because of these high g-forces that were

experienced, that adjustments to the shape of a vertical loop were required.

The shape of the vertical loop was altered so that it delivered a smooth, comfortable

5

ride in 1975 by Werner Stengel [1]. Mathematically, this perfect vertical loop is known

as a clothoid loop, similar to the shape of an inverted teardrop and can be modelled

by a parametric plot of two integrals, known as the Fresnel integrals. A clothoid loop’s

radius of curvature decreases as the train moves from the bottom, to the top of the loop.

Along with calculating the magnitude of initial velocity that is required for a point mass

to reach the top of a clothoid loop, it will be shown that the transition from a larger

radius of curvature at the bottom of the loop, to a smaller radius of curvature at the top,

reduces the g-force experienced by the rider to a tolerable amount. For this reason, the

clothoid loop became the preferred shape of a vertical loop for roller coaster designers

and is still used today in multiple designs.

2 Circular Loop

We begin by studying the motion of a point mass around an old fashioned, circular

loop. First, consider a point mass, m, which is attached to a circular track of radius

a, in the x-z plane (see Figure 3). We will explore two different ways of calculating

the magnitude of the mass’ initial velocity, ξ0, that is required for the mass to pass the

highest point of the track; the top of the vertical loop. Initially, the only external forces

which will be considered are the force due to gravity and the reaction force. We will then

extend the problem further by introducing an extra external force, before investigating

the problems of using a vertical loop of this shape.

First, we define the variables that will be used in the problem. We set θ as the

angle between the negative z-axis and the radial unit vector r, R as the reaction force

which acts in the −r-direction with magnitude R, θ as the transverse unit vector, g

the acceleration due to gravity and i and k as the unit vectors in the x- and z-direction

respectively. Here we note that g = −gk, where g = 9.81ms−2.

2.1 Motion in Polar Coordinates

The problem in question considers a point mass being restricted to moving in a circle.

Therefore, we first look to solve the problem using polar coordinates. We define the

mass’ position vector as r, with drdt and d2r

dt2representing the velocity and acceleration

respectively. Initially, the mass is positioned at θ = 0, at time t = 0. We want to

calculate the magnitude of the initial velocity that is required, for the mass to pass the

point θ = π. We take the radius of the circle to be r = a.

We apply Newton’s second law to the mass and look for a solution by substituting

6

Figure 3: Problem setup using polar coordinates.

into this equation. From classical mechanics, we know that the velocity of a point mass

in polar coordinates is given by

dr

dt=dr

dtr + r

dθ

dtθ , (2.1)

and the acceleration is given by

d2r

dt2=

(dr

dt− r

(dθ

dt

)2)r +

(rd2θ

dt2+ 2

dr

dt

dθ

dt

)θ , (2.2)

where r(t) and θ(t) are functions of time, t.

Immediately, we can simplify (2.1) and (2.2). Since r is constant, drdt = d2r

dt2= 0.

Therefore, (2.1) reduces todr

dt= a

dθ

dtθ , (2.3)

telling us that the velocity of the mass acts in the θ-direction, tangential to the circle,

with magnitude adθdt . Similarly, from (2.2), we have

d2r

dt2= −a

(dθ

dt

)2

r + ad2θ

dt2θ . (2.4)

Applying Newton’s second law, from (2.4) we have

F = ma

(−(dθ

dt

)2

r +d2θ

dt2θ

), (2.5)

where m is the mass and F is the total force acting on the mass.

7

Next we sum the forces that are acting on the mass to calculate F . This requires us

to calculate the component of each force acting in the r- and θ-direction. Resolving in

the direction of each unit vector, we find that

F = (mg cos θ −R)r − (mg sin θ)θ . (2.6)

Here we note that the magnitude of the reaction force in the θ-direction is zero, since

R and θ are orthogonal. Then, by substituting (2.6) into (2.5), we obtain

ma

(−(dθ

dt

)2

r +d2θ

dt2θ

)= (mg cos θ −R)r − (mg sin θ)θ . (2.7)

By taking the dot product of this equation with both r and θ, two scalar equations are

obtained. We find that

r · (2.7) : −ma(dθ

dt

)2

= mg cos θ −R , (2.8)

θ · (2.7) : mad2θ

dt2= −mg sin θ . (2.9)

Equation (2.8) can be rearranged so that the reaction force can be determined through-

out the mass’ motion. Equation (2.9) is a second order differential equation, which can

be solved to find an equation calculating the magnitude of the mass’ velocity.

We attempt to solve (2.9) by defining the new variable, ξ = dθdt , so that differentiating

using the chain rule gives

d2θ

dt2=

d

dt

(dθ

dt

)=

d

dθ

(dθ

dt

)dθ

dt=dξ

dθξ .

By substituting this into (2.9), we obtain a separable equation, such that

aξdξ

dθ= −g sin θ . (2.10)

Separating the variables and integrating, gives

1

2ξ2 =

g

acos θ +B , (2.11)

where B is a constant of integration. To find this constant of integration, an initial

condition is required. At t = 0, the mass is at θ = 0. Therefore, at t = 0, i = θ. Hence,

given a velocity, u, with initial magnitude ξ0, from (2.3) we have

dr

dt= u = ξ0i = a

dθ

dtθ = aξθ ,

8

at t = 0. Therefore, the initial condition is given by θ = 0 and ξ = ξ0a , at t = 0. This

condition will be used throughout Section 2. We apply this condition to equation (2.11)

and calculate B, so that after some slight simplification, we arrive at

1

2ξ2 =

g

a(cos θ − 1) +

ξ202a2

. (2.12)

To calculate ξ0 so that the mass passes the highest point of the loop (θ = π), we

introduce a final condition, such that ξ = 0, when θ = π, which will be used throughout

Section 2. Applying this condition to equation (2.12) gives an equation that determines

a critical value for the magnitude of the initial velocity that should be given to the mass,

for it to just reach the top of the loop. Should the mass be given an initial velocity with

a higher magnitude than this critical value, the mass will pass the highest point of the

loop.

After applying the final condition, equation (2.12) reduces to

ξ202a2

=2g

a,

which can be rearranged so that we have

ξ0 = 2√ag . (2.13)

Therefore, for a given radius, a, and initial velocity magnitude, ξ0, we have{ξ0 > 2

√ag mass passes θ = π

ξ0 < 2√ag mass doesn’t pass θ = π

(2.14)

We can perform a rough check of these inequalities, by calculating ξ0 for a real life

example. The tallest vertical loop in the world is the main element of the roller coaster

Full Throttle, at Six Flags Magic Mountain, California (see Figure 4) [7]. This roller

coaster utilises a launch system, allowing the vertical loop to be the tallest part of the

ride. Due to this, the top speed of the roller coaster occurs approximately at the point

when the train enters the loop, at θ = 0, to make sure the train successfully completes

the inversion. Although the loop is not a perfect circle, we will assume here that it is.

The loop has a diameter of 38.75m, therefore its radius is a = 19.375m [7]. Substitut-

ing this into (2.13), we find that ξ0 ≈ 61.68mph. This critical value of ξ0 is less than the

70mph top speed of the roller coaster [9]. Since the roller coaster successfully completes

the inversion, (2.14) holds, adding confidence that the inequalities are correct. When

calculating the initial velocity of a roller coaster, designers want to obtain a highly ac-

curate numerical value. Therefore, it should be noted that no rounding has taken place

until the final answer is found, so that the error in the calculations is minimised.

9

Figure 4: The tallest vertical loop in the world; the main element of the roller coaster

Full Throttle, at Six Flags Magic Mountain, California [8].

2.2 Energy Argument

It is possible to calculate (2.13) in an alternate way, by means of an energy argument.

From classical mechanics, we know that kinetic energy is given by T = 12m|u|

2 and

the gravitational potential energy is given by V = mgh, where h is the vertical distance

between the x-axis and the mass’ position. As the problem does not involve any resisting

Figure 5: Resolving diagram for an energy argument.

forces, no energy will be lost. Therefore, the system is a conservative system. This allows

us to write

T + V = B , (2.15)

where B is a constant.

First, using equation (2.3) we note that |u|2 = a2(dθdt

)2, giving T = 1

2ma2(dθdt

)2.

Then, resolving from Figure 5, we find that a−h = a cos θ, so that V = mga(1− cos θ).

Substituting T and V into (2.15), gives

1

2ma2

(dθ

dt

)2

+mga(1− cos θ) = B .

10

Setting ξ = dθdt and applying the initial condition defined in Section 2.1, we calculate B

and arrive at exactly equation (2.12). Equation (2.13) can then be derived from this as

before.

2.3 Air Resistance/Drag

2.3.1 High Air Resistance

We now look to add the effects of drag, D, into the problem. We begin with the case

where the drag coefficient, µ, is large (µ >> 1) and derive an equation for the critical

magnitude of the initial velocity, ξ0. An alternate equation can then be found by making

adjustments to this equation, for the case when µ is small (µ << 1).

As previously discussed in Section 1, when an object has a high velocity, the drag

force is given by

D = −µ|u|u . (2.16)

From Section 2.1, equation (2.3) tells us that u = adθdt θ, so that |u| = adθdt . Therefore,

we can rewrite D as

D = −µa2(dθ

dt

)2

θ .

Next, we apply Newton’s second law to the mass. Starting with equation (2.7), we

add the additional drag force to give

ma

(−(dθ

dt

)2

r +d2θ

dt2θ

)= (mg cos θ −R)r −

(mg sin θ − µa2

(dθ

dt

)2)θ .

Taking the dot product of this equation with θ, we obtain

mad2θ

dt2= −mg sin θ − µa2

(dθ

dt

)2

.

Then, by introducing the variable ξ = dθdt , where d2θ

dt2= ξ dξdθ , this equation becomes

maξdξ

dθ= −mg sin θ − µa2ξ2 . (2.17)

The −µa2ξ2 term makes this equation non-linear. To make this equation linear, we

define a new variable w = ξ2, so that differentiating w with respect to θ using the chain

rule givesdw

dθ=dw

dξ

dξ

dθ= 2ξ

dξ

dθ. (2.18)

Substituting w and (2.18) into (2.17) and rearranging, we then arrive at

dw

dθ+

2µaw

m= −2g

asin θ . (2.19)

11

This equation is of the form dwdθ + p(θ)w = q(θ), therefore, it can be solved by using

an integrating factor. After calculating the integrating factor and multiplying (2.19) by

this factor, we may write

d

dθ[we

2µamθ] = −2g

asin θe

2µamθ .

Integrating this equation then gives

we2µamθ = −2g

aI +A , (2.20)

where A is a constant of integration, which can be determined by the initial condition

and I is given by

I =

∫sin θe

2µamθdθ ,

which must be calculated.

To calculate I, we re-write sin θ in terms of the complex exponential. Setting α = 2µam ,

the integral becomes

I =

∫sin θeαθdθ =

∫Im(eiθ)eαθdθ ,

where i =√−1 and eiθ = cos θ + i sin θ. Adding the powers of the exponentials and

integrating gives

I = Im

(1

α+ ie(α+i)θ

).

Next, we multiply by α−iα−i and take the imaginary part, so that after some simplification,

we arrive at

I =1

α2 + 1(α sin θ − cos θ)eαθ .

Then, by substituting back in for α and substituting I into (2.20), we obtain

we2µamθ =

2gm

a(4µ2a2 +m2)(m cos θ − 2µa sin θ)e

2µamθ +A . (2.21)

Noting that the newly introduced variable was given by w = ξ2, the initial condition

introduced in Section 2.1 becomes w(0) =ξ20a2

. Using this condition, we can calculate

the constant of integration, A, so that equation (2.21) becomes

we2µamθ =

2gm

a(4µ2a2 +m2)(m cos θ − 2µa sin θ)e

2µamθ +

ξ20a2− 2gm2

a(4µ2a2 +m2). (2.22)

We now apply the final condition introduced in Section 2.1 to this equation, so that

we can determine an equation for the critical magnitude of the initial velocity. This

12

final condition requires w = 0, at θ = π. By substituting this condition into (2.22) and

rearranging, the critical magnitude equation is given by

ξ20 =2gm2a

(4µ2a2 +m2)(1 + e

2µamπ) . (2.23)

To check that this equation is consistent with the previous case where drag was neglected,

we set µ = 0, giving exactly equation (2.13) as required.

2.3.2 Low Air Resistance

Many roller coaster trains are cleverly designed to reduce the drag coefficient as much

as possible, so that the drag has minimal effect on the magnitude of the initial velocity

that is required for the train to reach the top of the vertical loop. Therefore, we now

consider the case where we have very minimal drag (µ << 1) and find an equation for

ξ0, which can be used for a train which has been made as aerodynamic as possible.

We begin by re-writing equation (2.23) as

ξ20 =2gm2a

m2(4µ2a2

m2 + 1)(1 + e

2µamπ) =

2ga

(α2 + 1)(1 + eαπ) , (2.24)

where α = 2µam . We can expand eαπ in Taylor series and (α2 + 1)−1 using a binomial

expansion, such that

eαπ ≈ 1 + απ +α2π2

2+O(α3) ,

(α2 + 1)−1 ≈ 1− α2 +O(α3) .

By noting that µ << 1 and therefore α << 1, terms of order O(α2) and higher are

negligible. Hence, using these expansions, (2.24) becomes

ξ20 ≈ 2ga(1 + (1 + απ +O(α2)))(1−O(α2)) .

Multiplying out the brackets and substituting back in for α gives

ξ0 = 2√ga(

1 +µaπ

m

) 12,

where we again use a binomial expansion, neglecting terms of order O(µ2), to give

ξ0 ≈ 2√ga(

1 +µaπ

2m

)= 2√ga+

µa32πg

12

m. (2.25)

Clearly if µ = 0, this equation for ξ0 is the same as equation (2.13), as required.

For a roller coaster train, particularly when loaded with passengers, m will be quite

large. By also noting that µ << 1, it can be seen that the extra term appearing in the

equation for ξ0 given by (2.25), will have minimal effect on the value of ξ0, showing the

benefit of a roller coaster train being aerodynamically designed.

13

2.4 G-Force

After calculating an initial velocity for a roller coaster, designers are then faced with

another obstacle known as g-force. The g-force acting on an object is defined as the ratio

between the force developed from the object accelerating by a cause other than gravity,

relative to the force due to gravity on the object [1]. Designers often use g-force to

their advantage, by creating moments of weightlessness, along with moments of extreme

heaviness for a rider. This is achieved by using specific track shapes, to continually vary

the g-force that a rider experiences, throughout the course of the roller coaster.

However, despite g-forces creating multiple thrilling sensations for a roller coaster’s

riders, they are the biggest limiting factor on the production of more extreme roller

coasters. Should riders experience greater than +5g, or less than −5g, serious injury

may occur, something designers clearly want to avoid [1]. It is for this reason why the

use of a circular loop in a roller coaster’s layout was stopped and a safer, alternative

shape was used. By deriving an equation for the g-force, we can identify the problems

of using a vertical loop of this shape and calculate where these problems occur. For

simplicity, air resistance and friction will be neglected throughout.

It is sufficient to show that a circular loop is unsafe by showing that the magnitude

of g-force exceeds 5g, regardless of direction. Directly from the definition of g-force, we

can write an equation for the magnitude of g-force, G, that is experienced by the mass,

at any point on the loop. Recalling the definition, we have that

G =m∣∣∣d2rdt2 ∣∣∣m|g|

=

∣∣∣d2rdt2 ∣∣∣|g|

. (2.26)

Immediately it can be seen that g-force is independent of mass. This identifies that two

riders of different mass will experience the same g-force as each other.

Using equation (2.4), we can write∣∣∣d2rdt2 ∣∣∣ as

∣∣∣∣d2rdt2∣∣∣∣ = a

((dθ

dt

)4

+

(d2θ

dt2

)2) 1

2

. (2.27)

Recalling that ξ = dθdt , so that d2θ

dt2= ξ dξdθ , we can make substitutions for dθ

dt and d2θdt2

using

equations (2.10) and (2.12) respectively. Hence, by making these substitutions and then

substituting (2.27) into (2.26), we obtain

G =a

g

(g2

a2sin2 θ +

(ξ20a2− 2g

a+

2g

acos θ

)2) 1

2

. (2.28)

14

Solving dGdθ = 0 will give the turning points of equation (2.28). One of these turning

points will be the value of θ at which the magnitude of g-force takes its maximum value.

Given that we are only concerned with motion for 0 ≤ θ ≤ π, we will neglect any turning

points that are not within this interval. Furthermore, by calculating d2Gdθ2

, we can classify

the turning points, which in turn allows us to identify what part of the loop generates

this maximum value, along with calculating an exact, maximum value of G.

We begin by squaring equation (2.28), which, after simplification, gives an equation

which can be differentiated much more easily, term-by-term. Differentiating using the

chain rule, after further simplification, we arrive at

2GdG

dθ= −6 sin θ cos θ − 4ξ20

agsin θ + 8 sin θ . (2.29)

For turning points, dGdθ = 0. Setting dG

dθ = 0, equation (2.29) simplifies to

sin θ

(−6 cos θ − 4ξ20

ag+ 8

)= 0 .

To satisfy this equation, either sin θ = 0, or −6 cos θ − 4ξ20ag + 8 = 0. Using equation

(2.13), we make the substitution ξ0 = 2√ag, whereby it can be shown that the latter of

these equalities requires | cos θ| > 1. This clearly cannot be true. Therefore, the turning

points of equation (2.28), for 0 ≤ θ ≤ π, are θ = 0 and θ = π.

In order to classify these turning points, we must differentiate equation (2.29) with

respect to θ using the product rule, so that we obtain

2Gd2G

dθ2+ 2

(dG

dθ

)2

= 6 sin2 θ − 6 cos2 θ − 4ξ20ag

cos θ + 8 cos θ . (2.30)

By definition, when d2Gdθ2

> 0 we have a minimum turning point and when d2Gdθ2

< 0 we

have a maximum turning point. Noting that G > 0 by definition, we substitute θ = 0

and θ = π into equation (2.30) and use equation (2.13) to determine that{θ = 0, dGdθ = 0 2Gd2G

dθ2= 2− 4ξ20

ag < 0→ maximum

θ = π, dGdθ = 0 2Gd2Gdθ2

=4ξ20ag − 2 > 0→ minimum

Given that θ = 0 is the only maximum turning point, the magnitude of g-force is

greatest at θ = 0. This is the point of transition between the horizontal and circular

track, otherwise known as the entrance to the loop.

Since the second half of the vertical loop can be generated by reflection, the g-force

for the second half of the loop can also be generated in the same way. This implies that

for the entire vertical loop (0 ≤ θ ≤ 2π), we have another maximum turning point at

15

θ = 2π; the exit of the vertical loop. When substituted into equation (2.28), θ = 0 and

θ = 2π give an equivalent maximum value of G, denoted Gmax.

We can calculate Gmax for any circular vertical loop, by substituting θ = 0 (or

θ = 2π) into equation (2.28). We reconsider the vertical loop from the Full Throttle

roller coaster, described in Section 2.1. Substituting a = 19.375m and ξ0 = 70mph

into equation (2.28), we find that Gmax ≈ 5.15g. Since Gmax > 5g, Gmax is too large,

showing that this circular shape is not an acceptable shape for a vertical loop.

The reason for large g-forces being experienced at the entrance/exit of a circular

loop, is due to the large, sudden change in curvature at the point of transition be-

tween the horizontal and circular track. This indicates that to make the vertical loop

a more comfortable experience for riders, the current large change in curvature at the

entrance/exit of the loop must be decreased. This will reduce the g-force experienced,

so that for the entirety of the loop, the g-force remains below 5g. This can be achieved

by using properties of the clothoid curve.

3 Clothoid Loop

The clothoid curve, otherwise known as the Euler/Cornu spiral, has been utilised by

roller coaster designers to avoid exerting high g-forces on riders. It has the property

that its radius of curvature is inversely proportional to the distance from the center

of the spiral, which is equivalent to the curvature varying linearly with arc length [1].

This linear variation property means that a vertical loop modelled in the shape of this

curve will have a very minimal change in curvature at the entrance and exit of the loop.

Consequently, this should reduce the g-force experienced to a tolerable amount, making

the roller coaster safer and more enjoyable for the rider.

3.1 Defining the Curve

Before we can consider the motion of a mass around a two-dimensional clothoid loop,

we must first find a system of equations that allow us to compute the clothoid curve.

This requires some simple definitions, which can then be manipulated to generate the

curve.

We begin by considering a general curve, C, with arc length, s, in the x-z plane (see

Figure 6). A position on this curve is represented by the vector r, where r is represented

by two parametric equations, x(s) and z(s), such that r = (x(s), z(s)). We use n and t

as the unit normal and unit tangent vectors respectively.

16

Figure 6: General curve, C, with defined position and unit vectors.

We now define t and n in terms of x(s) and z(s), along with the curvature, κ. At

any point on C, the tangential unit vector is given by

t =dr

ds=dx

dsi+

dz

dsk , (3.1)

where i and k are unit vectors in the x- and z-directions respectively. Using the orthog-

onal property (t · n = 0), we then define n as

n = −dzdsi+

dx

dsk . (3.2)

Finally, the curvature is defined as [4]

κ =dt

ds· n . (3.3)

Defining θ as the angle between t and the horizontal x-axis, equation (3.3) can then

be rewritten as a relation between θ and the arc length, s. Beginning by taking the dot

product of t with i and k, we find from the definition of the dot product that i · t = cos θ

and k · t = sin θ. Therefore, from equations (3.1) and (3.2), we can deduce that

dx

ds= cos θ ,

dz

ds= sin θ , (3.4)

t = cos θi+ sin θk , (3.5)

n = − sin θi+ cos θk . (3.6)

Next we differentiate t with respect to s using the chain rule and note that dtdθ = n.

This givesdt

ds=dt

dθ

dθ

ds= n

dθ

ds. (3.7)

17

Substituting (3.7) into (3.3), we obtain

κ = n · dtds

= n · n(dθ

ds

)=dθ

ds. (3.8)

From (3.4) and (3.8), we now have a system of ordinary differential equations. This

system can be written as

d

ds

x

z

θ

=

cos θ

sin θ

κ

. (3.9)

Solving this system will produce two parametric equations, x(s) and z(s), which describe

the clothoid curve.

To begin solving (3.9), we recall the definition of the clothoid curve. Since the

clothoid’s curvature varies linearly with arc length, we can write κ(s) = αs, where α is

a constant. Taking the third equation in system (3.9) and substituting in κ = αs, we

use separation of variables to find an equation for θ, such that

θ =αs2

2+ θ0 ,

where θ0 is a constant of integration. Without loss of generality, we take θ0 = 0, so that

θ = αs2

2 .

We can now substitute θ into the first two equations in system (3.9), before separating

the variables and integrating. To integrate the right hand side, we introduce the dummy

variable s and integrate with respect to s between zero and s, to give

x =

∫ s

0cos

(αs2

2

)ds , z =

∫ s

0sin

(αs2

2

)ds . (3.10)

Then, by letting s =√π√αt, so that ds =

√π√αdt, we obtain

x =

√π√α

∫ √α√πs

0cos(π

2t2)dt , (3.11)

z =

√π√α

∫ √α√πs

0sin(π

2t2)dt . (3.12)

These two integrals are otherwise known as the Fresnel integrals [2]. A parametric plot

of these two integrals will produce the clothoid curve required.

The Fresnel integrals can also be written as two power series expansions. These

expansions converge rapidly for all s (not shown here). For a given value of α, the two

series expansions are given by

x =

√π√α

∞∑n=0

(−1)n

(√α√πs)4n+1

(π2 )2n

(2n)!(4n+ 1),

18

z =

√π√α

∞∑n=0

(−1)n

(√α√πs)4n+3

(π2 )2n+1

(2n+ 1)!(4n+ 3),

which can be found in [2]. These can be used to plot the clothoid curve on Matlab.

For a vertical loop, the entire clothoid curve is not required. It is important when

generating the curve that only the first half of a vertical loop is produced, instead of

allowing the curve to become a spiral. The second half of the loop can then be generated

by reflection. To ensure that only the first half of a vertical loop is produced, we require

a final condition.

To find this final condition, we recall the z-integral from (3.10), before the change of

variables took place. Differentiating z with respect to s and setting dzds = 0, we calculate

the first value of s (excluding s = 0), which satisfies this equation. This value represents

the arc length at the point when the clothoid curve becomes horizontal for the first

time; the top of the vertical loop. This arc length will be denoted s∗. Calculating dzds

and setting dzds = 0, we find that

dz

ds= sin

(αs2

2

)= 0 ,

which first occurs when αs2

2 = π. Hence, s∗ =√2π√α

. This value will be used as part of

the final condition in all calculations throughout this section.

Using this condition, it is now possible to compute a variety of vertical loops for

different values of α (see Figure 7). By producing a number of curves of varying size,

Figure 7: Left: positive clothoid curve. Right: variety of half clothoid loops for different

values of α.

it can be seen there is a strong relation between every clothoid curve as the value of

19

α changes. As shown in Figure 7, the end point of each curve can be connected by a

straight line passing through the origin.

We can verify this result by substituting s = s∗ =√2π√α

into (3.10), giving us two

integrals defined as

X =

√π√α

∫ √20

cos(π

2t2)dt , Z =

√π√α

∫ √20

sin(π

2t2)dt .

Then, by defining a shorthand notation for the Fresnel integrals such that

C(s) =

∫ s

0cos

(πt2

2

)dt , S(s) =

∫ s

0sin

(πt2

2

)dt ,

we obtain

Z

X=

√π√αS(√

2)√π√αC(√

2)=S(√

2)

C(√

2),

which is independent of α, thus verifying the result.

3.2 Motion Along a General Curve

Using the Fresnel integrals (given by (3.11) and (3.12)) in conjunction with the final

condition for arc length, we can now compute half of a clothoid loop of any height. To

allow us to calculate the magnitude of the initial velocity that should be given to a point

mass so that it reaches the top of a clothoid loop, we require an equation governing the

motion of a mass around the loop.

Considering the setup from Figure 6, we will apply Newton’s second law to a point

mass and look to find an equation of motion for a mass along a general curve, C. Once

calculated, equations describing a specific curve, in this case the clothoid loop, can be

applied to this equation of motion, so that we can study the motion of a mass around

the clothoid loop.

To substitute into Newton’s second law, we must sum the total forces acting on the

mass and calculate a formula for d2rdt2

; the acceleration of the mass. First we calculated2rdt2

. Differentiating the position vector, r, with respect to t using the chain rule gives

dr

dt=dr

ds

ds

dt.

Since the velocity acts in the tangential direction, drds = t. Hence,

dr

dt= t

ds

dt. (3.13)

20

Differentiating again using the product rule, we obtain

d2r

dt2= t

d2s

d2t+dt

dt

ds

dt,

which can be re-written by use of the chain rule, such that

d2r

dt2= t

d2s

d2t+

(dt

ds

ds

dt

)ds

dt= t

d2s

d2t+dt

ds

(ds

dt

)2

. (3.14)

Using equations (3.7) and (3.8), we note that dtds = ndθds = nκ. From this, we can make

further simplifications to (3.14), so that d2rdt2

becomes

d2r

dt2= t

d2s

dt2+ nκ

(ds

dt

)2

. (3.15)

Next, we sum the forces acting on the mass, initially ignoring the effects of drag and

friction. Since the reaction force, R, always acts in the n-direction and the force due to

gravity acts in the −k-direction, the total force acting on the mass, F , is given by

F = Rn−mgk .

Hence, Newton’s second law gives

Rn−mgk = m

(td2s

dt2+ nκ

(ds

dt

)2). (3.16)

Similarly to the circular loop example in Section 2.1, we want to consider the n and

t components of this equation. Therefore, by taking the dot product of (3.16) with n

and t, we obtain

n · (3.16) : R−mgk · n = mκ

(ds

dt

)2

, (3.17)

t · (3.16) : −mgk · t = md2s

dt2. (3.18)

Recalling the definitions of t and n from (3.5) and (3.6) respectively, we substitute into

equations (3.17) and (3.18), so that after rearrangement, (3.17) gives an equation for

the reaction force, which is given by

R = mκ

(ds

dt

)2

+mg cos θ , (3.19)

whilst (3.18) gives us a second order, ordinary differential equation, such that

d2s

dt2+ g sin θ = 0 . (3.20)

By defining θ in terms of arc length for a specific curve, equation (3.20) can now be

solved to obtain an equation for the magnitude of the mass’ velocity. It should be noted

that by redefining θ in Section 2, we could also solve equation (3.20) for the circular

loop, to give us exactly equation (2.13), as before.

21

3.3 Motion Around the Clothoid Loop

We now have an equation of motion for a mass along a general curve, given by (3.20).

By applying equations used to describe the clothoid curve to this equation of motion,

it is possible to calculate an equation which determines the magnitude of the initial

velocity that is required for a point mass to reach the top of a loop, as done previously

for the circular vertical loop in Section 2.1.

Equation (3.20) is initially an equation of three variables. In Section 3.1, we found

that θ = αs2

2 for the clothoid. Substituting this into (3.20) gives

d2s

dt2+ g sin

(αs2

2

)= 0 , (3.21)

so that we now have a differential equation in terms of only two variables, s and t.

Similarly to the approach in Section 2.1, we introduce the variable ξ = dsdt , so that via

the chain rule, d2sdt2

= ξ dξds . Substituting this into equation (3.21), we obtain a separable

equation, which is given by

ξdξ

ds+ g sin

(αs2

2

)= 0 .

Introducing the dummy variable s, we separate the variables and integrate between zero

and s in anticipation of the initial condition, ξ(0) = ξ0, where ξ0 is the magnitude of

the initial velocity. This gives

1

2ξ2 = −g

∫ s

0sin

(αs2

2

)ds+

1

2ξ20 . (3.22)

The integral in this equation can be rewritten by setting s =√π√αt, so that ds =

√π√αdt,

as seen in Section 3.1. By making this change of variables, equation (3.22) becomes

1

2ξ2 = −g

√π√α

∫ √α√πs

0sin(π

2t2)dt+

1

2ξ20 . (3.23)

Recalling the shorthand notation for the Fresnel integrals, we can rewrite equation

(3.23), such that1

2ξ2 = −g

√π√αS

(√α√πs

)+

1

2ξ20 . (3.24)

To obtain an equation for the critical value of ξ0, a final condition is required. We

previously calculated in Section 3.1 that at the top of the clothoid loop, s = s∗ =√2π√α

.

Therefore, the final condition for the mass to just reach the top of the loop is given by

ξ(s∗) = 0. Applying this final condition to equation (3.24) gives

ξ20 = 2g

√π√αS(√

2) . (3.25)

22

Here we note that S(√

2) is a known value and can be calculated by the use of Maple.

Equation (3.25) can also be derived by using an energy argument. Given that |u| =|tdsdt | = ds

dt , since |t| = 1 and dsdt > 0, the kinetic energy of the system is given by

T = 12m(dsdt

)2. Then, since the system is conservative, by substituting the z-Fresnel

integral given by (3.12) into the equation for gravitational potential energy, we can

write1

2m

(ds

dt

)2

+mg

√π√αS

(√α√πs

)= B ,

where B is a constant. Reintroducing the variable ξ = dsdt and dividing by m gives

1

2ξ2 + g

√π√αS

(√α√πs

)= D ,

where D is a new constant. By applying the initial condition it can then be shown that

D = 12ξ

20 . Finally, after applying the final condition to this equation and rearranging,

we arrive at equation (3.25), as required.

For a given α, it is now possible to determine the minimum magnitude of the initial

velocity that must be given to the mass, so that it reaches the top of the vertical loop.

By specifying a height, h, for the vertical loop, we substitute s = s∗ and z = h into the

z-Fresnel integral given by (3.12), so that we can rearrange to give the required value of

α for this height. This can then be used in equation (3.25) to calculate ξ0.

It is useful to compare the required initial velocity for a clothoid loop with a circular

loop of identical height, so that it can be assessed how using a different loop shape

affects the magnitude of the initial velocity that is required.

We recall the Full Throttle example used in Section 2.1. The height of the vertical

loop is 38.75m, which occurs when s = s∗. By setting z = 38.75m, evaluating the

z-Fresnel integral at s = s∗ using Maple and rearranging, we find that α ≈ 1.07× 10−3.

Here we note that this truncated value of α is not used to calculate ξ0. Instead, the exact

value is used to minimise the error in the calculation. Substituting the exact values into

equation (3.25), gives a critical magnitude of ξ0 ≈ 61.68mph, which is identical to the

value calculated for the circular loop. To understand why these values are identical, a

general energy argument can be formulated, showing that the equation calculating the

magnitude of the initial velocity of a mass along a general curve is independent of curve

shape.

3.4 General Energy Argument

Consider a general curve with arc length s, connecting two fixed points, P and Q, in

the x-z plane (see Figure 8). P is at height z = a and Q is at height z = a+H, where

23

H is the vertical distance between P and Q and a ∈ R. Neglecting air resistance and

friction, we use simple energy definitions to find an equation to calculate the magnitude

of the initial velocity, ξ0, that is required to launch a point mass from P to Q, so that

the mass’ velocity is zero as it reaches point Q.

Figure 8: General curve between two points, P and Q.

Since all resisting forces are neglected, the system is conservative. Hence, we can

write T + V = B, where B is constant. In Section 3.3, we found that T = 12m(dsdt

)2.

Thus, noting that V = mgz, we have

1

2m

(ds

dt

)2

+mgz = B .

Defining ξ0 as the magnitude of the initial velocity, the initial condition requires thatdsdt = ξ0, at z = a. This implies that B = 1

2mξ20+mga. Next we apply the final condition,

which requires dsdt = 0, at z = a+H. After some simple rearranging, we are left with

ξ20 = 2gH , (3.26)

which is clearly independent of the path taken as required.

This result proves that when neglecting air resistance and friction, for a given height

of vertical loop, the shape of the vertical loop has no effect on the value of ξ0. Pro-

viding the system is conservative, for loop shapes which are complicated to describe

mathematically, equation (3.26) may be used as a useful shortcut to avoid complicated

calculations when determining the value of ξ0.

3.5 G-Force

Since the clothoid curve’s properties are used primarily to reduce the g-force experienced

by riders, it is vital that the g-force experienced when travelling around a clothoid loop

24

is calculated, to verify that riders are indeed safe. By recalling the definition for the

magnitude of g-force, G, we will apply equations from Section 3.3 to the definition to

acquire an equation for G, in terms of arc length, s. Air resistance and friction are

neglected throughout.

We begin by substituting equation (3.15) into the equation for the magnitude of

g-force, which is given by (2.26), so that we have

G =1

g

∣∣∣∣∣d2sdt2 t+ κ

(ds

dt

)2

n

∣∣∣∣∣ . (3.27)

A substitution can then be made for d2sdt2

using equation (3.20). Similarly, by noting thatdsdt = ξ, we can also make a substitution for

(dsdt

)2using equation (3.24). Therefore, by

substituting equations (3.20) and (3.24) into equation (3.27) and acknowledging that for

the clothoid, θ = αs2

2 and κ = αs, we obtain

G =1

g

∣∣∣∣−g sin

(αs2

2

)t+ αs

(−2g

√π√αS

(√α√πs

)+ ξ20

)n

∣∣∣∣ .This is equivalent to writing

G =1

g

(g2 sin2

(αs2

2

)+ α2s2

(−2g

√π√αS

(√α√πs

)+ ξ20

)2) 1

2

. (3.28)

Solving the equation dGds = 0 gives the turning points of equation (3.28). Further-

more, by calculating d2Gds2

, we can classify these turning points as either minimum or

maximum turning points. One of these maximum turning points is the value of s at

which the magnitude of g-force is maximum. This maximum magnitude is denoted by

Gmax. By substituting each maximum turning point into equation (3.28), the maximum

turning point which calculates Gmax can be determined, along with a numerical value

for Gmax.

For the first half of the clothoid loop, 0 ≤ s ≤ s∗. Therefore, turning points that

are not within this interval are neglected. Since the upper bound of the interval is

dependent on α, we must first specify a value of α before the turning points can be

calculated. It is useful to compare the magnitude of the g-force that is experienced from

using a clothoid loop, with a circular loop of identical height. Therefore, we recall the

vertical loop from the Full Throttle roller coaster introduced in Section 2.1, for which

we previously calculated Gmax ≈ 5.15g for the circular shape loop in Section 2.4. In

Section 3.3, it was calculated that α ≈ 1.07 × 10−3 for this vertical loop, which will

now be used in the interval, so that the turning points of equation (3.28) can be can be

calculated.

25

Given the complexity of equation (3.28), we use Maple to calculate the turning

points of the equation. By solving dGds = 0 on Maple, we find that equation (3.28) has

three turning points in the range 0 ≤ s ≤ s∗. These are given by s = 0, s = s∗ and

s ≈ 40.76m, where s ≈ 40.76m is just over halfway along the first half of the clothoid

loop. Furthermore, by calculating d2Gds2

and substituting in the turning points, we find

that s = 0 and s = s∗ are minimum turning points and s ≈ 40.76m is a maximum

turning point. Since s ≈ 40.76m is the only maximum turning point of equation (3.28),

Gmax occurs at this value of s.

Recalling that ξ0 = 70mph for the Full Throttle roller coaster, we substitute the

non-truncated values of s ≈ 40.76m, α ≈ 1.07 × 10−3 and ξ0 = 70mph into equation

(3.28), giving Gmax ≈ 2.50g. Clearly Gmax is substantially lower than 5g, showing that

the clothoid is an acceptable shape for a vertical loop.

For the clothoid loop, Gmax is less than half the value of Gmax for the circular loop.

To visualise how G varies as the mass travels around the two loop shapes, we plot G

against s using equations (2.28) and (3.28) (see Figure 9), noting that s = aθ for the

circular loop.

Figure 9: Graph to show how the magnitude of g-force, G, varies with arc length, s,

for a circular loop (red) and a clothoid loop (blue) of identical height (38.75m), for

0 ≤ s ≤√2π√α

.

The main problem with the circular loop is the large sudden change in curvature at

26

the point of transition between the horizontal and circular track. It can be seen from

Figure 9 that the clothoid loop has removed this problem, because G steadily increases

from zero, as opposed to the circular loop, where G decreases from its maximum value.

It is also clear that Gmax for the circular loop is approximately double Gmax for the

clothoid loop, adding confidence that the calculations of Gmax in Sections 2.4 and 3.5

are correct. It should also be noted that s∗ for the circular loop is smaller than s∗ for

the clothoid loop. Therefore, in Figure 9, the g-force for the circular loop is plotted for

more than the first half of the loop.

3.6 Air Resistance/Drag

Now that we have verified that the use of the clothoid curve to model a vertical loop

reduces the g-force experienced by riders to a tolerable amount, we look to add an extra

external force into the problem; the drag force.

The drag force that we add into the problem is given by equation (2.16), such that

D = −µ|u|u. Since the velocity, u, points in the tangential direction, we can rewrite

D in terms of t. We can then find an equation of motion, so that an equation which

determines the magnitude of the initial velocity that is required for a point mass to

reach the top of a clothoid loop, with drag included, can be found.

From equation (3.13), we know that u = tdsdt . Substituting this into (2.16), we have

D = −µ∣∣∣∣dsdt∣∣∣∣ dsdt t = −µ

(ds

dt

)2

t ,

since |t| = 1 and dsdt > 0.

Now that we have a relation between D and t, we can substitute into Newton’s

second law, giving us equation (3.16) with an additional term on the left hand side, such

that

Rn−mgk − µ(ds

dt

)2

t = m

(td2s

dt2+ nκ

(ds

dt

)2). (3.29)

In Section 3.1, we found that θ = αs2

2 for the clothoid. Hence, by taking the dot product

of equation (3.29) with t and substituting in for θ, we obtain

−mg sin

(αs2

2

)− µ

(ds

dt

)2

= md2s

dt2. (3.30)

To solve this equation, we will make use of a substitution. We again introduce the

variable ξ = dsdt , so that via the chain rule, d2s

dt2= ξ dξds . Substituting this variable into

equation (3.30) gives

mξdξ

ds= −µξ2 −mg sin

(αs2

2

). (3.31)

27

By making a further substitution of the form w = ξ2, so that via the chain rule, dwds =

2ξ dξds , we can rearrange equation (3.31) to give

dw

ds+

2µw

m= −2g sin

(αs2

2

). (3.32)

This equation is now of the correct form so that it can be solved by using an integrating

factor. Letting β = 2µm , we calculate the integrating factor and multiply equation (3.32)

by this factor. Then we may write

d

ds

[weβs

]= −2g sin

(αs2

2

)eβs . (3.33)

We now integrate this equation, introducing the limits, zero and s, in anticipation of

the initial condition, w(0) = ξ20 , where ξ0 is the magnitude of the initial velocity, so that

we have

weβs = −2gI + ξ20 , (3.34)

where I is given by

I =

∫ s

0sin

(αs2

2

)eβsds ,

which must be determined. We note that the introduction of the limits has required the

use of a dummy variable, s.

To calculate I, we first note that

sin

(αs2

2

)= Im

(ei(αs2

2

))= Re

(−iei

(αs2

2

)).

Therefore, I can be written as

I = Re

[−i∫ s

0ei(αs2

2

)+βs

ds

].

Next, we complete the square for the power of the exponential in the integrand, giving

I = Re

[−iei

β2

2α

∫ s

0ei(α2 (s+ β

iα)2)ds

].

Anticipating the appearance of the Fresnel integrals, we write√π√2t =

(s+ β

iα

) √α√2,

so that ds =√π√αdt. Substituting this into I, gives

I =

√π√αRe

[−iei

β2

2α

∫ √α√πs

0ei(

π2t2)dt

].

It follows that

I =

√π√αRe

[−iei

β2

2α

{∫ √α√πs

0cos(π

2t2)dt+ i

∫ √α√πs

0sin(π

2t2)dt

}],

28

where the two integrals appearing in I are the Fresnel integrals defined in Section 3.1,

given by (3.11) and (3.12) respectively. Thus, by recalling the shorthand notation for

these integrals defined in Section 3.1, we may write

I =

√π√αRe

[−i(

cos

(β2

2α

)+ i sin

(β2

2α

)){C

(√α√πs

)+ iS

(√α√πs

)}].

Taking the real part of this and substituting into equation (3.34), we obtain an equation

for w, which is given by

w = −2ge−βs√π√α

[cos

(β2

2α

)S

(√α√πs

)+ sin

(β2

2α

)C

(√α√πs

)]+ ξ20 . (3.35)

We can now apply the final condition to this equation, to obtain an equation which

determines the critical value of ξ0. This final condition is given by ξ(s∗) = 0. Hence,

in terms of the new variable, w, the final condition becomes w(s∗) = 0. Applying this

condition to equation (3.35), we find that the equation for the critical magnitude of the

initial velocity is given by

ξ20 = 2g

√π√αe−β

√2π√α

[cos

(β2

2α

)S(√

2) + sin

(β2

2α

)C(√

2)

]. (3.36)

By setting µ = 0, so that β = 0, we arrive back at equation (3.25), which is the initial

velocity with air resistance neglected, adding confidence that this result is correct.

4 Circular Helix Loop

So far, all calculations have been restricted to two dimensions. Although useful, these

calculations do not resemble the complete characteristics of a vertical loop. In two

dimensions, a vertical loop self intersects, whereas in three dimensions, the track does

not intersect. Therefore, to obtain more realistic values for the initial velocity and g-

force, we must consider the circular loop and the clothoid loop in three dimensions. We

will begin by considering the three-dimensional circular loop, also known as the circular

helix loop.

4.1 Defining the Loop

We can describe a two-dimensional circular loop in the x-z plane using two parametric

equations, which are functions of a variable, t. For a loop of radius a which is centered

at (0, a), these two equations are given by x(t) = a sin(t) and z(t) = −a cos(t) + a. In

addition to these equations, we can introduce a third parametric equation, which adds

29

a linear variance in the y-direction, so that the loop becomes three-dimensional. This

linear variance can be written as y(t) = bt, where b represents the pitch of the helix.

Restricting 0 ≤ t ≤ 2π, the circular helix loop is therefore described parametrically by

x(t) = a sin t , y(t) = bt , z(t) = −a cos t+ a . (4.1)

For a real life roller coaster, the distance between the entrance and exit of a vertical

loop is usually between one and two meters. This distance is kept small so that the

loop doesn’t lose its vertical aspect. For this reason, we can therefore assume that old

fashioned circular vertical loops also had this property.

Given that the distance between the entrance and exit of the circular helix loop, d,

is given by d = |y(2π) − y(0)|, we have that b must be in the interval 12π ≤ b ≤ 1

π , to

ensure that the entrance and exit of the vertical loop are between one and two meters

apart. The value of b that is chosen from this interval is largely dependent on the overall

size of the loop. We recall the vertical loop element of the Full Throttle roller coaster,

introduced in Section 2.1, which has a radius of a = 19.375m. To model this loop as a

circular helix loop, we will use b = 1π , giving a distance of 2m between the entrance and

exit of the loop.

4.2 Frenet-Serret Formulas

To help us understand motion in three dimensions, we will derive and make use of the

Frenet-Serret formulas. Consider a general curve, C, in three-dimensional space with

arc length, s (see Figure 10). We define r(s) as the position vector of a point on the

curve, t as the tangential unit vector pointing in the direction of motion and n as the

unit normal vector. Since t and n are orthogonal unit vectors, they satisfy t · n = 0.

Then, we introduce a new unit vector known as the binormal unit vector, b, which is

defined as b = t× n [5].

The Frenet-Serret formulas are a set of relations between the three unit vectors,

defining how each unit vector varies with arc length. Defining κ as the curvature, we

know from (3.7) and (3.8) thatdt

ds= κn . (4.2)

Next we differentiate b with respect to s, so that we obtain

db

ds=

d

ds(t× n) =

dt

ds× n+ t× dn

ds. (4.3)

Noting that n× n = 0, we substitute (4.2) into (4.3) to give

db

ds= t× dn

ds.

30

Figure 10: General curve, C, in three dimensional space, with the position vector for a

point on C given by r(s) and the tangential, normal and binormal unit vectors given by

t, n and b respectively.

From the definition of a vector product, we know from this equation that dbds is orthogonal

to t. We also know that b · b = 1. Therefore,

db

ds· b = 0 =

db

ds· t .

This implies that dbds must be in the direction of n. Thus, defining the magnitude of db

ds

as −τ , we havedb

ds= −τ n . (4.4)

The variable τ is known as the torsion of a curve. Torsion is a measure of the rate at

which a curve is twisting out of its osculating plane at a point [3].

Next, we differentiate the unit vector n, with respect to s. Since b = t× n, we know

from use of the right hand rule that n = b× t and t = n× b. Hence, by differentiating

n with respect to s and using (4.2) and (4.4), we obtain

dn

ds=

d

ds(b× t) =

db

ds× t+ b× dt

ds= −τ n× t+ b× κn = −κt+ τ b . (4.5)

The combination of equations (4.2), (4.4) and (4.5) is known as the Frenet-Serret Triad:dtds = κn

dbds = −τ ndnds = −κt+ τ b

(4.6)

31

These equations can be found in [5] and [3]. By defining a position vector for a point

on a three-dimensional curve, we can use the Frenet-Serret Triad and other geometry

definitions to calculate t, n and b, along with κ and τ , which can then be used to derive

the governing equations of motion along the curve.

4.3 Motion Around the Circular Helix Loop

Using the definition of a circular helix loop, we now define a position vector for a point on

the curve. Using the Frenet-Serret Triad and other definitions, we can use this position

vector to calculate t, n and b, along with κ and τ , which will allow us to study the

motion of a point mass around the loop.

We begin by defining r(t) as the position vector of a point along the curve. Using

(4.1), we can write

r(t) = a sin ti+ btj + (−a cos t+ a)k . (4.7)

Next we parameterize this position vector, so that it is given in terms of arc length, s.

Arc length is given by [3]

s =

∫ ∣∣∣∣drdt∣∣∣∣ dt =

∫ √(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt . (4.8)

Therefore, substituting (4.7) into (4.8) gives

s =

∫ √(a cos t)2 + b2 + (a sin t)2dt =

∫ √a2 + b2dt = t

√a2 + b2 .

Hence, by defining λ =√a2 + b2, (4.7) can be written as

r(s) = a sin( sλ

)i+

bs

λj + (−a cos

( sλ

)+ a)k .

Now that the position vector is in terms of arc length, we can use equation (3.1) to

calculate t. Differentiating r with respect to s, we find that

t =a

λcos( sλ

)i+

b

λj +

a

λsin( sλ

)k . (4.9)

We now use the first equation of the Frenet-Serret Triad, also given by (4.2), to calculate

n and κ. Differentiating t with respect to s, we obtain

κ =a

λ2, n = − sin

( sλ

)i+ cos

( sλ

)k . (4.10)

Now that we have defined t and n, we can apply them to the definition of the binormal

unit vector, so that we can obtain b. Calculating t× n, we have

b =b

λcos( sλ

)i− a

λj +

b

λsin( sλ

)k . (4.11)

32

Finally, we use the second equation of the Frenet-Serret Triad, also given by (4.4), along

with equation (4.10), to calculate the torsion. By differentiating b with respect to s, we

find that the torsion is given by

τ = − b

λ2.

This completes the calculation of each component of the Frenet-Serret Triad. A useful

check to add confidence that these calculations are correct is to substitute into the final

equation of the Triad, also given by (4.5), and check that this equation holds.

Ignoring air resistance and friction, we now apply Newton’s second law to the mass

and resolve in the direction of the three unit vectors. This will give us a governing equa-

tion of motion, which can be solved to find an equation which calculates the magnitude

of the initial velocity that is required for the mass to successfully reach the top of a

circular helix loop. Given that air resistance and friction are neglected and the loop is

circular, by use of the energy argument in Section 3.4, we expect the equation for this

magnitude to be identical to the two-dimensional equation, given by (2.13).

We begin by summing the forces acting on the mass and use equation (3.15), so that

Newton’s second law gives

R−mgk = m

(td2s

d2t+ nκ

(ds

dt

)2). (4.12)

Now that we are working in three-dimensional space, we must take into consideration

that the reaction force, R, will also have a component in the b-direction. We write

R = Rnn+Rbb , (4.13)

where Rn and Rb represent the magnitude of the reaction force in the n- and b-direction

respectively. Hence, by substituting (4.13) into equation (4.12) and taking the dot

product of this equation with t, n and b, we obtain

t · (4.12) : −mgaλ

sin( sλ

)= m

d2s

dt2, (4.14)

n · (4.12) : Rn −mg cos( sλ

)= mκ

(ds

dt

)2

, (4.15)

b · (4.12) : Rb −mgb

λsin( sλ

)= 0 . (4.16)

Equations (4.15) and (4.16) can be rearranged to give an equation to calculate the

magnitude of the reaction force in the n- and b-direction respectively. Equation (4.14)

is a second order, ordinary differential equation. This can be integrated so that we can

33

find an equation for the magnitude of the mass’ velocity at any point on the circular

helix loop.

To integrate equation (4.14), we introduce the variable ξ = dsdt , so that d2s

dt2= ξ dξds .

Substituting this into (4.14), gives a separable equation. By separating the variables,

integrating by substitution and determining the constant of integration by use of the

initial condition, ξ(0) = ξ0, where ξ0 is the magnitude of the initial velocity, we obtain

1

2ξ2 = ag cos

( sλ

)+

1

2ξ20 − ag . (4.17)

Now that we are working in three dimensions, we require a new final condition. At

the top of the circular helix loop, s = λπ. Therefore, the new final condition is given by

ξ(λπ) = 0. Applying this condition to equation (4.17), we find that

ξ0 = 2√ag , (4.18)

which is exactly equation (2.13), as expected.

4.4 G-Force

In Section 2.4, we proved that the magnitude of g-force experienced was too large, as a

point mass travelled around a two-dimensional circular loop. Although the circular helix

loop is also circular in shape, it is useful to see how adding variance in the y-direction to

the problem affects the g-force experienced, compared to the two-dimensional problem

in Section 2.4.

Ignoring air resistance and friction, we substitute equations from Section 4.3 into

equation (2.26). This gives us an equation for the magnitude of g-force, G, for which we

will calculate the turning points and classify whether these are minimum or maximum

turning points. The maximum magnitude of g-force, Gmax, can then be calculated, so

that we can determine whether the circular helix is an acceptable shape for a vertical

loop.

Recalling the equation for the magnitude of g-force given by (2.26), we substitute

(4.14) and (4.17) into (2.26) and recall from (4.10) that κ = aλ2

for the circular helix

loop, so that

G =1

g

∣∣∣−agλ

sin( sλ

)t+

a

λ2

(2ag

(cos( sλ

)− 1)

+ ξ20

)n∣∣∣ .

This is the same as writing

G =1

g

(a2g2

λ2sin2

( sλ

)+a2

λ4

(2ag

(cos( sλ

)− 1)

+ ξ20

)2) 12

. (4.19)

34

Next we calculate the turning points of this equation. Following the method from

Section 2.4, we square equation (4.19) and differentiate with respect to s to give

2GdG

ds= sin

( sλ

)cos( sλ

)(2a2

λ3− 8a4

λ5

)+ sin

( sλ

)(8a4

λ5− 4a3ξ20

gλ5

). (4.20)

The turning points are given by dGds = 0. This occurs when s = nλπ, where n ∈ N, or

when

cos( sλ

)=

8a4

λ5− 4a3ξ20

gλ5= 0 . (4.21)

Using equation (4.18), we make the substitution ξ0 = 2√ag and find that equation (4.21)

cannot be satisfied. Therefore, the only turning points are given by s = nλπ. Given

that 0 ≤ s ≤ λπ for the first half of the circular helix loop, we neglect turning points

that are not within this interval. Hence, the turning points of equation (4.19) that we

are interested in are s = 0 and s = λπ.

By differentiating equation (4.20), we can classify these turning points. We define

two constants:

β =2a2

λ3− 8a4

λ5, γ =

8a4

λ5− 4a3ξ20

gλ5,

and then differentiate (4.20) with respect to s, to give

2

(dG

ds

)2

+ 2Gd2G

ds2=β

λcos2

( sλ

)− β

λsin2

( sλ

)+γ

λcos( sλ

).

By substituting the two turning points into this equation, we find that{s = 0, dGds = 0 2Gd2G

ds2= β

λ + γλ < 0→ maximum

s = λπ, dGds = 0 2Gd2Gds2

= −βλ −

γλ > 0→ minimum

Since s = 0 is the only maximum turning point, substituting s = 0 into equation (4.19)

will give the maximum magnitude of g-force that is experienced. Recalling the Full

Throttle example in Section 2.1, we take a = 19.375m, ξ0 = 70mph and recall from

Section 4.3 that λ =√a2 + b2, where b = 1

π . Substituting these values and s = 0 into

equation (4.19), we find that Gmax ≈ 5.15g. Because Gmax > 5g, the circular helix loop

is not an acceptable shape for a vertical loop.

By comparing this maximum value of Gmax with the value calculated for the two-

dimensional circular loop in Section 2.4, we see that the three-dimensional circular loop

produces approximately the same value of Gmax. However, since G is dependent upon

the pitch, b, this only holds for certain values of b.

When b = 0, the vertical loop becomes two-dimensional and the maximum value

of g-force will be equal to the value calculated in Section 2.4. For b 6= 0, the vertical

35

loop becomes three-dimensional and we find that as b increases, the g-force experienced

decreases. Hence, by increasing b, it would be possible to reduce the g-force so that

Gmax < 5g. However, as previously discussed in Section 4.1, when b becomes larger

than a certain value (typically b > 1π for a circular helix loop), the loop loses its vertical

aspect and therefore b cannot be increased. Because of this restriction on b, we can

conclude that Gmax always has approximately the same value for the two-dimensional

and three-dimensional circular loop. By applying properties of the clothoid curve, we

can produce a three-dimensional clothoid helix loop, in an attempt to reduce this value

to a tolerable amount.

5 Clothoid Helix Loop

In two dimensions, the first half of a clothoid loop is defined parametrically by the

Fresnel integrals for 0 ≤ s ≤ s∗, with the second half being generated by reflection, as

seen in Section 3.1. After redefining these equations, we can introduce a third parametric

equation, which adds a linear variance in the y-direction, so that the loop becomes three-

dimensional. This three-dimensional loop is known as the clothoid helix loop.

We begin by redefining the two Fresnel integrals in terms of the variables t and ζ,

such that

x(t) =

√π√α

∫ √α√πt

0cos(π

2ζ2)dζ , z(t) =

√π√α

∫ √α√πt

0sin(π

2ζ2)dζ .

The linear variance in the y-direction can be written as y(t) = bt, where b is the pitch

of the helix. The combination of these three parametric equations, in addition with

the condition 0 ≤ t ≤ t∗, defines the first half of the clothoid helix loop. In shorthand

notation, these equations are

x(t) =

√π√αC

(√α√πt

), y(t) = bt , z(t) =

√π√αS

(√α√πt

). (5.1)

For the circular helix in Section 4.1, we set b = 1π , giving a distance of 2m between the

entrance and exit of the loop. So that we are consistent with the distance between the

entrance and exit of the two loop shapes, we will choose b so that the clothoid helix

loop also has a 2m separation between its entrance and exit. Since the second half of a

clothoid helix loop is generated by reflection, t∗ = 2√2π√α

for the full clothoid helix loop.

Therefore, we set b =√α√2π

.

36

5.1 Motion Around the Clothoid Helix Loop

We now consider the motion of a point mass around the clothoid helix loop defined

by (5.1). We look to calculate an equation to determine the magnitude of the initial

velocity that is required for the mass to just reach the top of the vertical loop. We

will achieve this by calculating all components of the Frenet-Serret Triad and applying

Newton’s second law to the mass, with air resistance and friction neglected, so that

an equation of motion can be found. When solving this equation, we expect to find

a critical magnitude equation that is identical to the equation for the two-dimensional

clothoid loop given by (3.25), because of the energy argument in Section 3.4.

We begin by using (5.1) to define the position vector of a point on the loop, r(t),

such that

r(t) =

√π√αC

(√α√πt

)i+ btj +

√π√αS

(√α√πt

)k . (5.2)

Next, we parameterize (5.2), so that r is in terms of arc length, s. Using (4.8), we have

s =

∫ (cos2

(αt2

2

)+ b2 + sin2

(αt2

2

)) 12

dt =

∫(1 + b2)

12dt ,

giving t = sλ , where λ =

√(1 + b2). Therefore, (5.2) becomes

r(t) =

√π√αC

(√α√π

s

λ

)i+ btj +

√π√αS

(√α√π

s

λ

)k . (5.3)

Now that the position vector is in terms of s, we can use the Frenet-Serret Triad

and other geometry definitions to calculate the unit vectors t, n, and b. Along with

these unit vectors, we can also calculate the torsion and curvature, given by τ and κ

respectively.

We start by using equation (3.1) to calculate t. Differentiating (5.3) with respect to

s, we find thatdr

ds= t =

1

λcos

(αs2

2λ2

)i+

b

λj +

1

λsin

(αs2

2λ2

)k . (5.4)

Using this equation for t and the first equation from the Frenet-Serret Triad, we can

then calculate n and κ. Differentiating (5.4) with respect to s, we obtain

κ =αs

λ3, n = − sin

(αs2

2λ2

)i+ cos

(αs2

2λ2

)k . (5.5)

Next we use the definition of the binormal unit vector to find b. Using (5.4) and (5.5)

to calculate t× n, we find that b is given by

b =b

λcos

(αs2

2λ2

)i− 1

λj +

b

λsin

(αs2

2λ2

)k . (5.6)

37

Finally, we use the second equation of the Frenet-Serret Triad and equation (5.5) to

calculate the torsion. By differentiating (5.6) with respect to s, we calculate that

τ = −bαsλ3

.

This completes the calculation of all components of the Frenet-Serret Triad.

We now apply Newton’s second law to the mass and resolve in the direction of the

three unit vectors. Ignoring friction and air resistance and recalling the equation for the

reaction force, R, in three dimensions given by (4.13), Newton’s second law gives

Rnn+Rbb−mgk = m

(td2s

d2t+ nκ

(ds

dt

)2). (5.7)

Taking the dot product of this equation with t, n and b, we have

t · (5.7) : −mgλ

sin

(αs2

2λ2

)= m

d2s

dt2, (5.8)

n · (5.7) : Rn −mg cos

(αs2

2λ2

)= κ

(ds

dt

)2

, (5.9)

b · (5.7) : Rb −mg cos

(αs2

2λ2

)= 0 . (5.10)

By introducing the variable ξ = dsdt , so that via the chain rule, d2s

dt2= ξ dξds , we substi-

tute ξ into equation (5.8) and attempt to find an equation to determine the magnitude

of the initial velocity that is required for a mass to just reach the top of the loop.

Substituting ξ into (5.8), we have

ξdξ

ds= − g

λsin

(αs2

2λ2

).

Introducing the dummy variable s, we separate the variables and integrate this equation

between zero and s in anticipation of the initial condition, ξ(0) = ξ0, where ξ0 is the

magnitude of the initial velocity. This gives

1

2ξ2 = − g

λ

∫ s

0sin

(αs2

2λ2

)ds+

1

2ξ20 . (5.11)

Then, in anticipation of the appearance of a Fresnel integral, we write s = λ√π√αt, so that

ds = λ√π√αdt. Therefore, equation (5.11) becomes

1

2ξ2 = −g

√π√α

∫ 1λ

√α√πs

0sin(π

2t2)dt+

1

2ξ20 , (5.12)

38

To obtain an equation for the critical value of ξ0, a new final condition is now

required. The arc length at the top of the vertical loop, s∗, is given by the first value of

s, excluding s = 0, where dzds = 0. Therefore, given that

z =

∫ s

0sin

(αs2

2λ2

)ds ,

we havedz

ds= sin

(αs2

2λ2

)= 0 ,

giving s∗ = λ√2π√α

. Thus, the final condition is given by ξ = 0, at s = λ√2π√α

. Applying

this final condition to equation (5.12), we find that the equation determining the critical

value of ξ0 is given by

ξ20 = 2g

√π√αS(√

2) ,

which is exactly (3.25), as expected.

5.2 G-Force

In Section 3.5, we discovered that using a clothoid loop instead of a circular loop reduces

the magnitude of g-force experienced when travelling around the loop, to a tolerable

amount. Using equations from Section 5.1 where air resistance and friction were ne-

glected, we can verify that the three-dimensional clothoid helix loop also reduces the

magnitude of g-force, along with comparing how the g-force varies compared to the

two-dimensional clothoid loop.

The magnitude of g-force, G, is given by equation (3.27), such that

G =1

g

∣∣∣∣∣d2sdt2 t+ κ

(ds

dt

)2

n

∣∣∣∣∣ .Given that ξ = ds

dt , we substitute equations (5.8) and (5.12) into G, whilst noting that

for a clothoid helix loop, κ = αsλ3

, where λ =√

1 + b2, so that we obtain

G =1

g

∣∣∣∣ gλ sin

(αs2

2λ2

)t+

αs

λ3

(−2g

√π√αS

(1

λ

√α√πs

)+ ξ20

)n

∣∣∣∣ .This is equivalent to writing

G =1

g

(g2

λ2sin2

(αs2

2λ2

)+α2s2

λ6

(−2g

√π√αS

(1

λ

√α√πs

)+ ξ20

)2) 1

2

. (5.13)

By differentiating equation (5.13) with respect to s and setting dGds = 0, we can

calculate the turning points. These can then be classified by taking the second derivative

of G with respect to s and substituting in the turning points.

39

Due to the complexity of equation (5.13), we differentiate using Maple. So that

we can compare the three-dimensional clothoid loop with the two-dimensional clothoid

loop, we will use the same size vertical loop from the roller coaster Full Throttle for

the three-dimensional clothoid loop, as we did for the two-dimensional clothoid loop in

Section 3.3. Noting that α is not affected by the problem becoming three-dimensional,

we take α ≈ 1.07× 10−3 and ξ0 = 70mph, with b =√α√2π

. Differentiating equation (5.13)

once with respect to s and setting dGds = 0, we find that the turning points of (5.13) are

given by s = 0, s = s∗ and s ≈ 40.76m. Differentiating again, we classify s = 0 and

s = s∗ as minimum turning points and s ≈ 40.76m as a maximum turning point. Given

that we only have one maximum, we substitute the non-truncated value of s ≈ 40.76m

into equation (5.13), giving Gmax ≈ 2.50g, which is sufficiently small enough to make

the vertical loop safe.

Comparing the truncated value of s where the g-force is maximum with the truncated

value of s found in Section 3.5 for the two-dimensional clothoid loop, it can be seen that

the maximum g-force occurs at approximately the same point on the three-dimensional

loop as it does on the two-dimensional loop. Similarly, the truncated values of Gmax

are approximately the same in two and three dimensions. This shows that increasing

the dimensions of the problem has minimal effect on the g-force and the point where it

takes its maximum value. As previously discussed in Section 4.4, since equation (5.13)

is dependent upon b, it is possible to reduce Gmax further by increasing b. However,

this leads to the vertical loop losing its vertical aspect and because Gmax is already

sufficiently small enough, increasing b is not required.

6 Multi-Mass Problem

Although some roller coaster trains are made up of one carriage, the majority of roller

coasters use a train made up of five to seven carriages. As an extension to the problems

we have seen so far, we can consider a multi-mass system, where multiple point masses

are joined together, forming a train.

Beginning with two connected masses on a two-dimensional circular loop, an equation

of motion can be derived for the two masses. Then, by adding further masses to the

train, a formula calculating an equation of motion for n masses, where n ∈ N, can be

determined. By analysing this equation, the effects of adding extra masses to the train

can be discovered. For simplicity, air resistance and friction will be neglected.

40

6.1 Two-Mass Train

Consider two point masses, m1 and m2, where m1 = m2, that are connected by a light,

inextensible rod, on a two-dimensional circular loop of radius a (see Figure 11). The

tension in the rod is given by T , with magnitude T . We define θ as the angle between

the negative z-axis and m2, with β being the constant angular separation of m1 and

m2. The transverse unit vector is defined as θi and the radial unit vector is defined as

ri, where i = 1, 2, defining which mass the unit vectors correspond to. Similarly, the

position vector of each mass is given by ri and the reaction force for each mass is given

by Ri, with magnitude Ri. It should be noted that 0 < β < π. This condition implies

that when θ = 0, the first mass in the train has not reached the top of the loop.

Figure 11: Problem setup for a two-mass train on a circular loop.

We apply Newton’s second law to each mass individually, starting with m1. The total

angle at any point between the negative z-axis and m1 is given by θ + β. Recalling the

equation for the acceleration of a mass in polar coordinates given by (2.4) and defining

ψ = θ + β, we apply Newton’s second law to m1, which gives

−ma(dψ

dt

)2

r1 +mad2ψ

dt2θ1 = −mg sinψθ1 −R1r1 − T x , (6.1)

where x is a unit vector pointing in the direction of the tension. Given that r1 = ar1

and r2 = ar2, we can write x as

x =r2 − r1|r2 − r1|

=a(r2 − r1)|r2 − r1|

.

By calculating |r2 − r1|2 and using the definition of the dot product, it can be shown

41

that

x =a(r2 − r1)

2 sin β2

.

Next, we apply Newton’s second law to m2, noting that the angle between this mass

and the negative z-axis is θ. This gives

−ma(dθ

dt

)2

r2 +mad2θ

dt2θ2 = −mg sin θθ2 −R2r2 + T x . (6.2)

We wish to solve in the direction of motion for each mass. Therefore, we calculate

(6.1) · θ1 and (6.2) · θ2. So that we can calculate x · θi, we convert θi and ri into

Cartesian coordinates, such that

ri = sin(φ)i+ cos(φ)j , (6.3)

θi = − cos(φ)i+ sin(φ)j , (6.4)

where φ is the total angle between the negative z-axis and the mass in question. Calcu-

lating the dot products, we find that

(6.1) · θ1 : mad2ψ

dt2= −mg sinψ − Ta sinβ

2 sin β2

, (6.5)

(6.2) · θ2 : mad2θ

dt2= −mg sin θ +

Ta sinβ

2 sin β2

. (6.6)

Since β is constant, d2ψdt2

= d2θdt2

. Therefore, by adding equations (6.5) and (6.6) and using

a trigonometric identity, we have

d2θ

dt2+g2a

sin

(θ +

β

2

)= 0 , (6.7)

where g2 = g cos β2 .

Clearly if β = 0, we obtain the equation of motion for a single mass train given by

(2.9), adding confidence that this equation is correct. Equation (6.7) also tells us that

adding an extra mass to the train reduces the effect of gravity that it experiences, where

g2 is this new gravitational constant, which is dependent on β. In addition, we note that

the angle β2 is the midpoint of the arc between m1 and m2, indicating that the train

behaves as though one mass is positioned at the midpoint between the two masses on

the circle.

42

6.2 Three-Mass Train

We now add a third mass, m3, to the train and attempt to find an equation of motion

in a similar way to the two-mass problem in the previous section. We attach m3 to m2

by a light, inextensible rod, so that θ is now the angle between the negative z-axis and

m3 and β is the constant angular separation between each mass, where 0 < 2β < π

(see Figure 12). The radial unit vector of each mass is denoted ri and the transverse

unit vector is denoted θi, where i = 1, 2, 3. Similarly, the position vector of each mass is

given by ri and the reaction force is given by Ri, with magnitude Ri. The magnitudes of

the tension in the rod connecting m1 and m2 are denoted T12 and T21, where T12 = T21.

Likewise, the magnitudes of the tension in the rod connecting m2 and m3 are denoted

T23 and T32, where T23 = T32.

Figure 12: Problem setup for a three-mass train on a circular loop.

We apply the same approach as seen in Section 6.1, starting with m1. Given that

the total angle at any point between the negative z-axis and m1 is given by θ + 2β, we

define η = θ + 2β, so that Newton’s second law gives

−ma(dη

dt

)2

r1 +mad2η

dt2θ1 = −mg sin ηθ1 −R1r1 − T12x , (6.8)

where x is a unit vector pointing in the direction of T21. Similarly, by defining ψ = θ+β

and applying Newton’s second law to m2, we obtain

−ma(dψ

dt

)2

r2 +mad2ψ

dt2θ2 = −mg sinψθ2 −R2r2 + T21x− T23y , (6.9)

43

where y is a unit vector pointing in the direction of T32. Finally, applying Newton’s

second law to m3, we have

−ma(dθ

dt

)2

r3 +mad2θ

dt2θ3 = −mg sin θθ3 −R3r3 + T32y . (6.10)

Following the method discussed in Section 6.1, it can be shown that x and y are

given by

x =a(r2 − r1)

2 sin β2

, y =a(r3 − r2)

2 sin β2

.

Substituting x and y into (6.8), (6.9) and (6.10) and recalling the definitions of ri and

θi in Cartesian coordinates, which are given by (6.3) and (6.4) respectively, we take the

dot product of equations (6.8), (6.9) and (6.10) with their corresponding transverse unit

vector. This gives

(6.8) · θ1 : mad2η

dt2= −mg sin η − T12a sinβ

2 sin β2

, (6.11)

(6.9) · θ2 : mad2ψ

dt2= −mg sinψ +

T21a sinβ

2 sin β2

− T23a sinβ

2 sin β2

, (6.12)

(6.10) · θ3 : mad2θ

dt2= −mg sin θ +

T32a sinβ

2 sin β2

. (6.13)

Since β is a constant, d2ηdt2

= d2ψdt2

= d2θdt2

. Therefore, by adding equations (6.11), (6.12)

and (6.13) and applying a trigonometric identity, we obtain

d2θ

dt2+g3a

(sin(θ + β)) = 0 , (6.14)

where g3 = g3(1 + 2 cos 2β).

Again, if β = 0, we obtain the equation of motion for a single mass train given

by (2.9). The three-mass train also experiences reduced gravity, where g3 is the new

gravitational constant, which is also dependent on β. Since the angular separation of

the masses must be within the interval 0 < 2β < π, we have g3 > g2. Thus, the addition

of a third mass has further reduced the effects of gravity. To generalise the behaviour

of g as the number of masses in the train increases, we can calculate a formula for gn,

where n ∈ N is the number of masses in the train.

6.3 N-Mass Train

Applying ideas and methods from Sections 6.1 and 6.2, we can easily derive equations of

motion for a four, five and six-mass train, which can be used in conjunction with (2.9),

44

(6.7) and (6.14), to derive a general formula calculating the equation of motion for an

n-mass train, where n ∈ N. This formula is given by

d2θ

dt2+gna

(sin

(θ +

(n− 1)

2β

))= 0 . (6.15)

Here gn represents the reduced gravitational constant, θ is the angle between the negative

z-axis and mn and β is the constant angular separation of the masses, where 0 <

(n− 1)β < π. We find that gn is dependent upon whether n is odd or even, such that

n is odd : gn =g

n(1 + (n− 1) cos 2β) , (6.16)

n is even : gn =2g

n

12n−1∑i=0

cos

(β

2+ iβ

). (6.17)

For a roller coaster, a typical value of n is given by 1 ≤ n ≤ 7. Using (6.16) and

(6.17), we can calculate g1 up to g7 and plot how these vary with β. This allows us to

analyse how adding extra masses to the train varies the effects of gravity (see Figure 13

and Figure 14).

Figure 13: Graph showing how gn varies with β, for 0 < β < π6 and 1 ≤ n ≤ 7, where

n ∈ N.

From Figure 13, we can see that g3 > g4 for 0 < β < π6 , meaning the addition of

an extra mass to the three mass train has not further reduced the gravity experienced.

Therefore, given that we found that g3 > g2, it is not possible to draw up any conclusion

about the behaviour of gn, for general n.

45

Figure 14: Left: graph showing how gn varies with β, where n = 2k, 1 ≤ 2k ≤ 7, and

k ∈ N, for 0 < β < π6 . Right: graph showing how gn varies with β, where n = 2k − 1,

1 ≤ 2k − 1 ≤ 7, and k ∈ N, for 0 < β < π6 .

However, if we consider separate cases for n odd and n even, we can see from Figure

14 that increasing n, decreases the value of gn. From Figure 13 and Figure 14, it is

also clear that decreasing the value of β, increases the value of gn. Therefore, given the

restriction 0 < (n − 1)β < π, if extra masses are continually added to the train, the

angular separation between them must subsequently decrease, so that the effect that

each action has on the value of gn should eventually cancel out.

We can prove this result by letting n become large and subsequently decreasing β.

Let β(n − 1) = φ (φ constant), so that as n → ∞, β → 0. When n is odd, clearly as

β → 0, (1 + (n − 1) cos 2β) → n, so that gn → g. For the case when n is even, we let

n = 2k, where k ∈ N, so that

limβ→0

k−1∑i=0

cos

(β

2+ iβ

)= k =

n

2.

Hence, gn → g. Therefore, by adding more masses to the train and subsequently reduc-

ing the angle between them, the reduction in g continually decreases, until there is no

reduction whatsoever.

Using this result, we note that as n→∞ and β → 0, equation (6.15) becomes

d2θ

dt2+g

a

(sin

(θ +

φ

2

))= 0 ,

where the angle φ2 is the midpoint of the arc between the first and last mass of the train.

This implies that the train will behave as though a single mass is positioned at this

46

midpoint. It should be noted that this midpoint signifies the center of mass for all of

the combined point masses.

Using the formulas for gn, we can compute the reduced gravitational constant for

a real life roller coaster. We recall the roller coaster Full Throttle that was introduced

in Section 2.1. This roller coaster consists of three cars per train. Assuming that the

vertical loop is circular, we use equation (6.16) to calculate g3. Taking into consideration

the size of the loop and that β would be small for a real life roller coaster, we approximate

the angular separation of the cars as β = π16 , so that g3 ≈ 9.31ms−2. Since g = 9.81ms−2,

it is clear from this calculation that for a realistic setup, the reduction in the effects of

gravity are minimal.

7 Conclusion

From the calculations we have seen, it is evident that the mathematics behind the design

of a roller coaster is incredibly complex. After using trial and error in an attempt to

design new and exciting track layouts, multiple problems arose. Eventually, designers

applied rigorous mathematical techniques to the design process, allowing them to un-

derstand what was causing these particular problems with their designs, before seeking

solutions to the problems. The vertical loop is an excellent example of this process.

The main aim of this project was to replicate the calculations that designers would

have made when designing roller coaster vertical loops, for two different loop shapes; the

circular loop and the clothoid loop. These calculations included finding the magnitude

of the initial velocity that is required for a mass to reach the top of a vertical loop, along

with the g-force experienced as the mass travelled around the loop, so that we could

identify any problems/benefits of using either loop shape.

We began by considering the two-dimensional circular loop (Section 2). The first

calculation that was performed was the derivation of an equation linking the height of

the vertical loop, to the magnitude of the initial velocity, ξ0, that the train must be given

to successfully reach the top of the loop. To simplify this calculation, we used a point

mass instead of a full length train. This simplification was used throughout Sections

2-5. We also began by neglecting drag and friction.

Using polar coordinates, we found an equation of motion for the point mass on the

loop. This equation was then solved and after applying an initial and final condition, we

obtained an equation for the critical value of ξ0. This equation was also found using an

energy argument. As an extension to this calculation, the drag force was included and

47

a similar method was used to find an alternate equation for ξ0. Given the form of the

solution, further adjustments using Taylor/binomial expansions were made, allowing an

equation for ξ0 to be determined where the drag coefficient was small.

After building many roller coasters with circular vertical loops, riders complained

about the uncomfortable experiences they were having travelling around these loops.

To demonstrate this problem, we mathematically defined the g-force directly from its

definition and used equations from Sections 2.1, to find an equation for the magnitude of

g-force, G, that is experienced as the mass travelled around the loop. Using values from

the roller coaster Full Throttle, it was calculated that this circular shape loop produced

a maximum magnitude of g-force which was too high (Gmax ≈ 5.15g), which occurred

as the mass entered and exited the loop. This showed why the circular loop is not an

acceptable shape for a vertical loop.

Next we considered the two-dimensional clothoid loop, which required us to define

the loop parametrically. Starting with the basic definition of a clothoid curve, differential

geometry was used to derive two integrals known as the Fresnel integrals. Parametrically

plotting these integrals produced the clothoid curve. Properties of this curve were then

used to create half of a clothoid loop.

By using techniques similar to those seen in Section 2.1, an equation determining the

critical value of ξ0, with air resistance and friction neglected, was found. By studying

the Full Throttle roller coaster, it was shown that the critical value of ξ0 was identical

to the value for the circular loop. The reasoning behind this result was then confirmed

by an energy argument in Section 3.4. However, the most important calculation was

proving that this loop shape does indeed reduce the g-force experienced. For the Full

Throttle example, we found that the maximum magnitude of g-force experienced was

Gmax ≈ 2.50g, which is slightly less than half of the value for a circular loop of the same

height. Thus, it was proven that the clothoid loop was an acceptable loop shape to be

used.

Next we studied both loop shapes in three dimensions. To understand motion in

three-dimensional space, a binormal unit vector was introduced and the Frenet-Serret

Triad was derived, which was used to help derive an equation for ξ0. Both loop shapes

were found to produce almost identical values of Gmax to their two-dimensional counter-

parts. The equations for ξ0 were also identical to the two-dimensional examples, which

was verified by the energy argument in Section 3.4.

Finally, as an extension to the single mass problem, the behaviour of a train of point

masses travelling around a two-dimensional circular loop, with air resistance and friction

48

neglected, was investigated. It was discovered that a two-mass train reduced the effects

of gravity on the train, which was dependent upon the angular separation of the masses.

As more masses were added, we found that the reduction of gravity was dependant on

whether the number of cars in the train, n, was even or odd. In both cases, for constant

β and 0 < (n − 1)β < π, the reduction in gravity became bigger as the n increased. A

formula to calculate the equation of motion for an n-mass train was then determined,

which included two formulae for the reduced gravitational constant, gn. Whilst studying

the behaviour of gn as n→∞, so that β → 0, it was seen that gn → g.

Vertical loops are one of a variety of inversions that are found on a roller coaster. To

further the calculations we have seen, we could mathematically define different inversions

and apply these definitions to the equation of motion along a general curve. Then,

for each inversion, the g-force and critical value of ξ0 could be calculated by following

similar methods that have been seen throughout this project for the vertical loop. These

calculations would also require knowledge of other fundamental concepts of roller coaster

design, such as banking the track, which is discussed in [1], pages 47-50. An interesting

example of an inversion that could be studied is the tilt-incline loop; a vertical loop that

is tilted at an angle (see Figure 15).

Figure 15: Tilt-incline loop; an inversion found on The Swarm, at Thorpe Park [11].

Inversions, particularly vertical loops, are included in track layouts for hundreds of

roller coasters which are operational today. These inversions are used to maximise the

thrill and experience of riding a roller coaster. As designers continue to invent taller

and faster roller coasters, along with new and exciting inversions, mathematics is vitally

important to prove that the passengers will be safe when riding these roller coasters. It is

these new inventions that encourage people to visit theme parks every year, highlighting

the important role that mathematics plays in the industry and the future construction

of all roller coasters.

49

References

[1] Weisenberg, N. (2013) Coasters 101- An Engineer’s Guide to Roller Coaster Design.

3rd Edition. pp 11-54

[2] Abramowitz, M. and Stegun, I.A. (1972) Handbook of Mathematical Functions with

Formulas, Graphs and Mathematical Tables. 10th Edition. New York: Dover Publi-

cations. pp 300-301

[3] Widder, D.V. (1961) Advanced Calculus. 2nd Edition. Englewood Cliffs, N.J:

Prentice-Hall. pp 89, 97, 102

[4] Chisholm, J.S.R. and Morris, R.M. (1966) Mathematical Methods in Physics. 2nd

Edition. Amsterdam: North-Holland Publishing Company. pp 376, 386

[5] Christie, D.E. (1964) Vector Mechanics. 2nd Edition. New York: McGraw-Hill Book

Company Inc. pp 214

[6] Blyth, M. (2013) Lectures in Mathematical Problem Solving, Mechanics and Mod-

elling. University of East Anglia. High Speed Air Resistance, Chapter 5.3.

[7] Guinness World Records (2014) 8 Roller Coaster Records for National Roller Coaster

Day [Illus.]. Available at: http://www.guinnessworldrecords.com/news/2014/8/

8-roller-coaster-records-for-national-roller-coaster-day-59634

[8] Niles, R. (2013) Ride Review: Full Throttle at Six Flags Magic Mountain [Illus.].

Available at: http://www.themeparkinsider.com/flume/201306/3534/

[9] Marden, D. (2016) Roller Coaster Database. Available at: http://rcdb.com/10948.

htm

[10] PicassoMio (2016) Gallery, Open Listings [Illus.]. Available at: http://www.

picassomio.com/promenades-aeriennes-jardin-baujon-poster-1590161.html

[11] Total Thorpe Park (2016) Features, Dictionary [Illus.]. Available at: http://www.

totalthorpepark.co.uk/features/dictionary.shtml

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