Abstract

In a technologically lead world, Virtual Reality is at the forefront with its endless applications within various fields. It provides an interactive interface that has the ability to immerse its users into a world they could only imagine. This investigation will be breaking down and explaining the mathematics behind 3-dimensional extensions of selected graphs including an extruded pie chart, a vector field and a 4D hypercube displayed as a 3D Tesseract. We will use virtual software including: PaintLab, Blender and Calcflow to display these graphs in a 3-dimensional Euclidean space. This will show how virtual reality can be used to visualise or explain mathematical graphs by harnessing its interactive qualities.

Contents

1 | Introduction | 3 | ||||||||

1.1 2-Dimensional Graph Origins . . . . . . . . . . | . | . | . | . | . | . | . | . | 3 | |

1.2 3D Virtual Reality Development Through Time | . | . | . | . | . | . | . | . | 5 | |

2 | Using Virtual Reality PaintLab Software | 11 | ||||||||

3 | Using Virtual Reality Blender Software | 14 |

4 Analysis of Virtual Reality Pie Chart Models 17

4.1 Case 1: Fixing the Maximum Height at 1 (hm a x = 1) . . . . . 18

4.2 Case 2: The Entire Volume Must Equal 1 (Vm a x = 1) . . . . . 25

5 Vector Field Analysis 32

5.1 2-Dimensional Vector Fields (R2 → R2) . . . . . . . . . . . . 33

5.1.1 Gradient, Divergence and Curl of A 2D Vector Field . 36

5.2 3-Dimensional Vector Fields (R3 → R3) . . . . . . . . . . . . 44

5.2.1 Gradient, Divergence and Curl of A 3D Vector Field . 44

5.2.2 Virtual reality Calcflow Software . . . . . . . . . . . . 50

5.3 Navier-Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . 53

6 | Higher Dimensions | 58 | ||||||||||||||||||||||

6.1 Hypercube . . . . . . . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | . | 60 |

Contents

6.2 Polyunfolding . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6.2.1 Conways Criterion . . . . . . . . . . . . . . . . . . . . 65

7 Conclusion 68

Chapter 1

Introduction

1.1 2-Dimensional Graph Origins

A Frenchman and philosopher, Nicole Oresme created a bar chart in a 14th century publication. These charts documented the path of a moving object by noting its height at different time intervals. He plotted its constant veloc- ity against time around 300 years before Sir Isaac Newton was accredited for the laws of motion [History of Bar Charts And Graphs , 2011]. His illustration used bars of equal width which is a feature used as a current fundamental rule when drawing bar charts today (Figure 1.1) [Stanton, 2015] [Oresme]. Following on, 20 years later a man named William Playfair was credited for inventing the bar chart, he used bars to display discrete groups of imports and exports for Scotland on a graph with labelled axis (Figure 1.2) [His- tory of Bar Charts And Graphs , 2011] [Wainer, 1996]. A major limitation for mathematicians before the technological era was their way of presenting illustrations. They were only able to use ink and paper which reduced the amount of observations and variables they could have at one time [Stanton,

2015].

Figure 1.1: Illustration of the bar charts Nicole Oresme created [History of

Bar Charts And Graphs , 2011].

Figure 1.2: Horizontal bar chart created by William Playfair showing im- ports and exports for Scotland [History of Bar Charts And Graphs , 2011].

1.2 3D Virtual Reality Development Through

Time

Virtual reality is the current source of digital disruption in the technological world. It is an immersive computer-generated interface which is mentally, emotionally and physically convincing [Woodford, 2006/2016]. It has the ability to replicate scenarios which can manipulate human emotions, such as fear or euphoria.

It is useful to provide the historic nature of virtual reality, investigating the foundations from which it began. Primarily, 360◦ wall paintings were of high interest, they provided a higher level of observer engagement than a normal portrait and therefore set the pace for further development of 3D projects [Virtual Reality Society – How Did Virtual Reality Begin? , No Date]. In the

1920s, the first flight simulator invented by Edwin Link [Sherman and Craig,

2002], provided risk-free training grounds for aspiring pilots. Displaying the important and potentially life-saving applications for virtual reality.

The production of the Sensorama in 1956 by a cinematographer Morton Heilig [Sherman and Craig, 2002], was one of the first attempts of a 3- dimensional experience. Along with a 3D image, it had sound, smell and a moving seat for added effect [Axworthy, 2016].

In 1963, computer scientist Ivan Sutherland created a virtual and augmented reality Sketchpad application, the first interactive graphics software designed before the term virtual reality was born. The images were rough and ill- defined but in 1965 he described his work as “a looking glass into a mathe- matical wonderland” [Sherman and Craig, 2002] when explaining the concept

of an interactive display that does not abide by the laws of physical reality. In 1968, a breakthrough was established at MITs Lincoln Laboratory, a com- puter was used instead of a camera to present each eye with a separate image within the head-mounted display [Sherman and Craig, 2002]. However, the equipment was very dense and had to be suspended whilst in use which meant

it was hard to manoeuvre [Axworthy, 2016] (Figure 1.3).

Figure 1.3: Ivan Sutherlands suspended head-mounted display [Sherman and Craig, 2002]

In 1974 Evans and Sutherland Computer Corp. (founded in 1968 by pro- fessors David Evans and Ivan Sutherland) reveals the first nocturnal flight simulation system, Novoview [Sherman and Craig, 2002]. It helped train pi- lots to fly during the night through replication of the night sky. In 1977 the first controllers were created in the form of gloves, called the Sayre Glove. It

transmitted finger bending information in the form of light, to a computer which interpreted the movement of the individuals hand (Figure 1.4)[Wood- ford, 2006/2016]. More than a decade later, the first virtual reality controllers were created by Jaron Lanier called the DataGlove which worked in a sim- ilar way to the Sayre Glove [Stone, 2001]. Users were able to move virtual objects which were projected within the EyePhone heads-up display [Axwor- thy, 2016] [Sherman and Craig, 2002]. NASA built on this and revealed LCD optics and head tracking, they were also researching new forms of human-

computer interaction.

Figure 1.4: The left illustration is of the DataGlove, and the right illus- tration shows how the light is transmitted when the user moves their finger [Woodford, 2006/2016]

3-Dimensional sound was created in 1987 when NASA’s Scott Fisher and Elizabeth Wenzel were able to simulate sound coming from a specific 3D location. This led to the introduction of the Convolvotron system which was able to manipulate the placement of 3-dimensional sound [Sherman and

Craig, 2002].

The 90’s was the time for gaming enthusiasts. W-industries started by launching the first VR system, Virtuality. It was a two player arcade system that offered higher interaction levels through a virtual multilevel shooting game. The game used a 3D head-mounted display to show an alternate re- ality to the users, they could see each other within the game in the form of avatars [Sherman and Craig, 2002].

In 1995, the CAVE was developed by students at the University of Illinois, using lightweight LED shutter glasses and wall projections to create a three- walled room that could hold multiple users, this was revolutionary as previous designs were limited to one user whereas the CAVE could hold ten. [Axwor- thy, 2016][Sherman and Craig, 2002].

Ascension Technologies Corporation introduced a product in 1996 that was able to track fourteen different parts of the body using wireless magnets, it was rightfully named MotionStar [Sherman and Craig, 2002]. Shortly after, the CyberGrasp was introduced by Virtual Technologies which enabled the virtual system to enhance the users hand movements, such as touching and closing their fingers around virtual objects.

Another breakthrough in 1999 made video tracking possible for users with a personal computer equipped with a camera input. The ARToolKit, jointly released by HITLab (Seattle) and ATR Media Integration & Communication (Japan), was relatively cheap and easy to use even though it was developed for augmented reality [Sherman and Craig, 2002].

Several years ago, in 2009, a man called Palmer Luckey crowdfunded nearly

2.5 million dollars to fund the development of the Oculus Rift in his parents

garage. Facebook acquired Oculus in a $2 billion deal in 2014, thus leading to current purchases of the Oculus Rift headsets in 2016 [Woodford, 2006/2016] [Axworthy, 2016] [Gleasure and Feller, 2016].

At present, smart phones are starting to create rival virtual reality technolo- gies along with PlayStation [Woodford, 2006/2016]. Including Google selling over 5 million cardboard VR headsets that users can combine with their smart phones to experience a virtual space in an inexpensive and portable manner. Virtual reality has seen major development in the 21st century and has large expectations for the coming years [Woodford, 2006/2016][Virtual Reality Society – How Did Virtual Reality Begin? , No Date][Axworthy, 2016]. If technological companies fail to keep up with virtual reality advances, they will fall behind leading their products and ideas to become obsolete.

There are vast applications for virtual reality technology within sectors such as art, entertainment, education, military excetera [Woodford, 2006/2016]. However, the health care and surgery sector can gain more benefits from virtual reality equipment than any other area [Satava, 1995]. As the VR technology did with helping pilots navigate, it can offer a virtual operation to lower human risk and increase training for doctors and surgeons in a near to realistic environment. VR offers a cheaper alternative when testing pro- totypes and creating new ones [How is Virtual Reality Used? , No Date]. Breakthrough VR technology such as the Oculus Rift and the HTC Vive can help within education and work environments in a cognitive way, such that the acquisition and understanding of knowledge is based on the users creations in a virtual platform [Winn, 1993]. The hope is to find a way to

present data containing several variables on a 3D virtual reality graph. This graph can be manoeuvred, shrunk, viewed in detail from every angle and even walked into. It offers a new perspective on how information can be analysed, compared and presented in fields such as data analysis for example.

Chapter 2

Using Virtual Reality PaintLab

Software

Figure 2.1: A free hand drawing of a 2D pie chart using virtual reality

PaintLab software.

Using the HTC Vive virtual reality equipment, initial attempts were made to draw a simple 3-dimensional pie chart within a virtual space using PaintLab

software. We began by constructing a 2D pie chart (Figure 2.1) which was then extruded upwards into 3-dimensions (Figure 2.2 and Figure 2.1 2.3).

Figure 2.2: A side perspective of the 3D free hand extrusion created on

PaintLab.

Figure 2.2 shows the HTC Vive hand equipment used to navigate the vir- tual space seen within the corresponding headset, the left hand was used to select artistic materials, gradients, sizes, colours etcetera and the right hand was for drawing, erasing and switching the camera angle. Both hands used in conjunction would spin, enlarge and manoeuvre the diagram. Whilst draw- ing, we discovered that the edges were difficult to connect in a 3D space and drawing completely straight lines proved to be problematic. If it was neces- sary to input accurate angles and heights onto the diagram PaintLab offered no way of doing so. All the problems combined would produce an inaccurate and unscaled extruded pie chart that is not very aesthetically pleasing. Al- though entertaining to use, this would not be a suitable method for displaying

scaled data. However, it could be used to design rough sketches.

Figure 2.3: Birds-eye perspective created on PaintLab of a free hand ex- trusion of the 2D pie chart into 3D.

Chapter 3

Using Virtual Reality Blender

Software

Blender, although not virtual, is a more complex design software that we used to create a sharper extruded pie chart. The software was initially difficult to use but once it was figured out, we were able to create several potential 3D pie chart designs as shown in Figure 3.1.

The great thing about Blender is we could export the objects onto Virtual Reality software called AM Model Viewer. This was a step forward from the rough diagrams created on PaintLab as we were now able to see a diagram in virtual space that bore more similarities to a 3D graph than the free hand drawing did, it also has the potential to be scaled as the objects were produced using nets. The design we chose from Blender to explain the two cases for potential 3D virtual reality graphs, was the extruded cylinder on the top right corner of Figure 3.1.

Figure 3.1: A 3D taurus, extrude pie chart and a birds-eye/side view of an extruded sphere designed on Blender.

Figure 3.2: A 3D extruded pie chart that was created on Blender software and printed using a 3D printer.

A 3D printer was used to develop a tangible replica of the chosen extruded pie chart in Figure 3.1, this is shown next to a 1p coin to give an idea of its dimensions (Figure 3.2).

Chapter 4

Analysis of Virtual Reality Pie

Chart Models

2D graphs offer a simple way of visually displaying and comparing data. Diagrams like this are primarily used to keep audiences engaged and pro- vide an easier way to view relatively complicated data. However, simple

2-dimensional graphs dont offer much help when trying to display more than one variable, for example pie charts only plot mutually exclusive data in the form of angle sectors that are relative to the frequency of the data being shown [Statistics: circle graphs / pie charts / pie graphs , 2017]. Each sector

is calculated using Figure 4.1.

dataf requency θ =

totalf requency

× 360◦, (4.1)

There can be no negative data values because a 2D pie chart is too simplistic

to illustrate this. Consider a 3D extruded cylinder pie chart, its segments can move to a height which represents another data variable. This would give new ways to display data using the height and volume both in relation to the angle of the sector. Thus, providing the two cases we will be investigating for pie charts.

4.1 Case 1: Fixing the Maximum Height at

1 (hmax = 1)

Figure 4.1: An annotated extruded pie chart created in Blender software with its 2D birds-eye perspective.

As mentioned previously, 2D pie charts only use the angle to display data, in Case 1 we are considering the relationship between the volume and the

angle, assuming the height cannot exceed a boundary of 1. The variables we need to to assess this case can be seen on the annotated pink extruded sector on the pie chart in Figure 4.1:

Height (h) Radius (r) sector angle (θ)

By using the area of a chosen sector (i ) multiplied by its corresponding height

(hi ), We can find the volume of a sector (Vi ) using Equation 4.2:

V = __ ____θ____i__ × πr2

i 360

hi , (4.2)

And vice-versa if we have the volume of a sector and its angle, we can find

the respective height shown in Equation 4.3:

hi =

Vi

360 πr2

__ ____θ__

(4.3)

To keep consistency with our units of measure, we assume that the radius is always 1 – let’s say cm, which in this particular case equals the maximum value for the height of a sector. Therefore the range of values h can take after the data in question has been normalised is: 0cm 6 h 6 1cm. Normalising the data means dividing every value by its highest, thus giving a maximum value of hm ax = 1 every time, for any set of numerical data we

have. This makes it easier to display and ensures the data set is in proportion.

When we input our normalised units for the radius and height (h = r = 1 )

the Equation 4.2 changes to 4.4:

V = θi π (4.4)

i 360

Considering the maximum height is 1, the total volume the cylinder can hold would be when θ = 360◦, giving Vm a x = πcm3, with a range for V as:

0cm3 6 V 6πcm3 . Using Vm ax =π cm3 we know that if we take the sum of

all the volumes in the cylinder it cannot exceed the maximum value in the volumes range, therefore forming the summation in Equation 4.5:

V = θ1

360◦

n

πh1 +

θ2

360◦

πh2 + … +

θi

360◦

πhi

(4.5)

X __ ____θ____i____ __ πh 6 πcm3

⇒

i

360◦

i=1

The graph shown inFigure 4.2 shows the volume gradually decreasing from

πcm3→ 0 cm3 as the sector angle is decreased from 360◦ → 0◦ with a fixed

height and radius of 1cm. Here we assume there is only one extruded sector

containing the entire volume. Imagine Figure 4.1 with just the extruded pink sector while the others have no height at all, we are plotting the rela- tionship between this sectors varying angle against its volume. Thus proving the summation in 4.5 that the volume MUST be equal or fall below π.

Figure 4.2: A bar chart displaying the relationship between a decreasing angle and the volume when the height of the sector is fixed at 1cm.

__Example:__

Now that we have our main equations, we can apply this to a simple example: A group of 50 random people took a test, 38 people passed and 12 people failed.

Using this data we can calculate the angle of those who passed (θp ) and those who failed (θf ) and display them as segments on a pie chart (Figure 4.3). Substituting the given values into Equation 4.1:

38

θp = 50 × 360◦

= 273.6◦

(4.6)

12

θf = 50 × 360◦

= 86.4◦

(4.7)

Figure 4.3: A 2D pie chart displaying the data variable of those who passed who failed.

The values calculated in 4.6 and 4.7 represent the inner angle of the pie chart shown, this is equivalent to the proportion of those who passed and those who failed out of the total frequency of data, which in this case is 50.

By combining the same example combined with case 1, we can not only show how many people passed and failed but also the proportion of each group that were male. After this figure is calculated, we can extrude the sector upwards which would satisfy the 3D element of our pie chart.

Extending the example to state that out of the 38 people who passed; 21

were men, and out of the 12 people who failed; 6 were men. We can now look to extrude the respective sectors to represent this new variable.

To do this we need to normalise the figures as shown in 4.8 and 4.9 where hp and hf define the number of males that passed and failed respective to each group:

21

hp = 21

= 1cm

(4.8)

6

hf = 21

= 0.286cm(3 s.f.)

(4.9)

Now both values have been normalised, their corresponding group segments can be extruded to the heights calculated in order to represent the number of males in each group.

The sector showing who passed will contain an angle of 273.6◦ (equivalent to

the __38__

50

people who passed) and will be extruded to its normalised height of

1cm which represents the proportion of males who passed ( __21__ ). This is also the tallest sector in the pie chart as required. The sector showing who failed

38

will contain an angle of 86.4◦ (equivalent to the __12__

50

people who failed) and

will be extruded to a height of 0.286cm 1cm which represents the proportion of males who failed ( __ ____6__ ) .

12

Finally, by using the formula for the volume of a sector in 4.2 we can assess the relationship between the angle and the volume using our general example:

V = θp πh

p 360 p

273.6

=

360

19

= π

25

π(1)

(4.10)

= 2.388cm3

V = θf h

f 360 f

86.4

=

360

π(0.286)

(4.11)

= 0.217cm3

This shows that the relationship between the angle and volume is positive, meaning that if the angle increases then so will the volume, provided that the height of any sector never surpasses 1cm. Thus showing a stable 3D model we could use if we wanted to display two data variables.

4.2 Case 2: The Entire Volume Must Equal

1 (Vmax = 1)

In case 2 we are considering the relationship between the angle and the height, to do this we assume that the volume cannot exceed a boundary of

1cm3 .

Referring back to the sector volume equation in 4.2 (Vi ). We can now say Vm ax = 1cm3 and set the entire equation equal to 1 whilst also using our fixed value for the radius. Thus giving us Equation 4.12 which we will rearrange to find the upper and lower bound values that the height can take:

1cm3 = __ ____θ____i__ πh

×

360 i

(4.12)

Rearranging Equation 4.12 to equal hi :

1

hi =

__ ____θ____i__

π

360

(4.13)

To find hm in we need to set θ = 360◦ and assume the entire volume is held

within this sector:

hi =

=

1

__360__

π

360

1

cm π

(4.14)

To find hm a x we need to assume the sector angle can carry very small per- turbations of positive values between 0◦ and 1◦. If this is the case then the height will become infinite and will graphically display an asymptote as seen

in Figure 4.4. Again, this is assuming the entire volume is held within this sector.

Figure 4.4: A scatter plot displaying the relationship between the angle and height when the volume of the sector is fixed at 1cm3.

In the graph the points are very clustered, it is difficult to see the asymptotic

nature of the points tending to infinity on the y-axis and __1__

π

on the x-axis. To

clarify this, Figure 4.5 and Figure 4.6 show enlarged sections of the graph

from 340◦ → 360◦ and 0◦ → 10◦ respectively:

Figure 4.5: A scatter plot displaying the relationship between the angle

and height when the volume of the sector is fixed at 1cm3 from 340◦ → 360◦

showing hm in = __1__ .

π

Figure 4.6: A scatter plot displaying the relationship between the angle

and height when the volume of the sector is fixed at 1cm3 from 0◦ → 10◦

showing hm a x → ∞.

Thus giving a height range of: __1__ cm 6 h < ∞cm. This very large range of

π

values lacks precision unlike the range for the volume previously calculated

in case 1. Due to this, we can say that case 2 would amplify the smaller data values as their corresponding height would be very tall in comparison to the sectors which have a larger angle. Whereas in a 2D pie chart, small values are usually near to invisible to viewers as shown in Figure 4.7. As you can see, Group A’s value is too small to be seen on the pie chart which makes it look like it does not exist:

Figure 4.7: 2-Dimensional Pie Chart created on Microsoft Excel showing difficulties in displaying small data values.

Having a 3D pie chart that contains a sector with an infinite height would be disproportionate to its neighbouring sectors making it unaesthetic, thus defeating the purpose of the graph.

Considering the total volume of all the segments has to equal 1, we can form a summation using the equation for Vi similar to our first case giving us the

sum in 4.15:

θ1 θ2 θi

Vi = 360◦ πh1 + 360◦ πh2 + … + 360◦ πhi

n

(4.15)

⇒ X __ ____θ____i____ __ πh = 1cm3

360◦ i

i=1

This case is different in that the sum of the volumes MUST equal 1cm3 at all times. Therefore, the pie charts 2D sector variable is found as normal but its corresponding height needs to be calculated with respect to the total volume. Our answer would be one height in relation to another, shown as a multiple.

The best way to explain this would be to use the same example used earlier in case 1.

__Example:__

The example contains the same post-test results seen in Case 1:

Total frequency = 50

Passed = 38

Failed = 12

Men who passed = 21

Men who failed = 6

Recall the angles calculated previously in 4.6 and 4.7 as θp =273.6◦ and

θf =86.4◦. In this case, we need to find the volume each sector must hold

by creating an equation with two height unknowns (hp and hf ) using the summation in 4.15. We then solve this equation equal to 1 in order to have

one height in terms of the other:

1 = X __ ____θ____i____ __ πh

2

360◦ i

i

⇒ 1 = Vp + Vf

1 = θp

360◦

πhp +

θf

360◦

πhf

(4.16)

273.6◦

1 =

360◦

πhp +

86.4◦

360◦

πhf

__1____−____86____.____4__

360 πhf

π

__273____.____6__

360

= hp

From equation 4.16 we now have hp in terms of hf as required. In essence we are showing the proportion of males in one group with respect to a multiple of the males in the other group. Remember there is an inverse relationship between the height and the angle when the entire volume is held in one sector (4.6). However, when the volume is dispersed amongst two sectors (like our

example), one height will be negative when we substitute a value into the other unknown height. The negative is ignored when measuring the extrusion length for the pie chart illustration, but not within the calculations to check if the total volume is correctly equal to 1.

To explain this using our example, we set hf = 10, then by using equation

4.16 we get the value hp = −2.739065939 (full number is necessary in order

for the summation to work).

To confirm these values are correct we substitute them into the initial volume summation (4.15) and it should equal 1:

X __ ____θ____i____ __ πh

2

360◦ i

__θ____p____ __

= πhp

360◦

__θ____f____ __

+ πhf

360◦

i

273.6

=

360◦

π(−2.739065939) +

86.4

360◦

π(10)

(4.17)

= 1

Therefore confirming that our height ratio works provided the value is kept negative when performing the calculation in 4.16 but ensuring the height of the actual pie chart when drawn is always positive. To do this we take the

absolute value of the height such that |hi | and we should expect the sectors

with smaller angles have larger heights than those with large angles as proven

previously in graphs 4.5 and 4.6. Hence showing a negative relationship between the angle and the height when the volume is fixed at 1cm3.

Chapter 5

Vector Field Analysis

In this chapter the next type of graph we will be analysing are those of vector fields. As we are investigating how to illustrate graphs in a 3-dimensional space, we need to understand how vector fields work in 2D first and then extend the field into 3D space.

Primarily, a vector is defined as quantity that contains two independent properties: direction and magnitude. A vector tells us the position of a point in a space relative to another point. It is commonly illustrated using directional arrows that vary in size depending on the magnitude. Therefore, a vector field is a space that assigns a vector to each point which is seen as an array of arrows, as shown throughout this chapter. The notation of a vector field contains a function F that gives the coordinates (x, y) or (x, y, z) a 2D or 3D vector respectively denoted as F(x, y) or F(x, y, z) [Dawkins, a]. The equations contain points P, Q and R which are located on the vector field by using their corresponding position vector (x, y, z) [Schwab, 2012].

5.1 2-Dimensional Vector Fields (R2 → R2)

First, we define a vector field in a 2D real coordinate space (R2 → R2 )

using the information provided in the chapter introduction as Equation 5.1

[Dawkins, a]:

F(x, y) = P (x, y)ˆi + Q(x, y)ˆj (5.1)

Where ˆi and ˆj are Cartesian coordinates that represent unit vectors in the respective x or y direction.

We have now stated mathematically how a 2D vector field is represented using position and unit vectors. An example can be illustrated on a 2D (x, y) axis to help understand what it could represent visually.

Looking at the simple vector field F(x, y) = x + y in Figure 5.1 it is seen as a clockwise spiral rotation of arrows around the origin (0,0), this means the origin is a source point. As stated before, these arrows represent direction and magnitude of the vectors assigned to the points in the vector field [Nykamp,

No Datea]. If the equation of F was changed to F(x, y) = −x−y the direction

of arrows change and can be seen to move towards, then shy away from the

source. The origin seems to ‘repel’ the arrows away; this is called a ‘saddle point’. These points will be made clearer further along the chapter.

Figure 5.1: An illustration of a 2D vector field created on Desmos F(x, y) = x+y showing a clockwise rotational flow around the source point at the origin (0,0).

Figure 5.2: An illustration of a 2D vector field created on Desmos F(x, y) =

−x − y showing the flow towards and away from the saddle point at the origin

(0,0).

5.1.1 Gradient, Divergence and Curl of A 2D Vector

Field

__2-D____ ____ ____Gradie____n____t:__

The gradient is defined as a rate of change of a given function. In other words, it is a vector that represents the direction of the slope at a selected point within the function. If the gradient is calculated to be zero then it means there is a ‘local maximum’ or ‘local minimum’ as there is no particular point where the function increases (Figure 5.3). The gradient is found using Equation 5.2 [Azad, No Datea].

grad(f ) = ∇f

∂ ∂

= f + f

∂x ∂y

(5.2)

Where (∇) pronounced “nabla” or “del” represents the derivative of the

scalar field in the case of the gradient.

__Example:__

The gradient of the scalar field f = (x2 + y2) can be found using Equation

5.2.

grad(f ) = ∇f

∂ ∂

= f + f

∂x ∂y

(5.3)

= 2x + 2y

The gradient we calculated in 5.3 is a vector that represents the direction we need to travel in so that our function reaches its local maximum or mini- mum point the fastest way possible. This also confirms that once there, the

gradient will be zero as there is no further distance to climb.

Figure 5.3: Illustration on the right showing the gradient for F(x, y) = x2 −

y2 accompanied with its parabolic surface for reference on the left [Kramer,

2014]. The arrows point in the direction of greatest rate of change. There are no arrows in the center representing a gradient value of zero i.e. it is a local point.

__2-D____ ____ ____Di____v____ergence:__

The divergence uses partial differentiation on a vector field to calculate a scalar quantity. This quantity defines the changing magnitude at a chosen

point and can be found using Equation 5.4 below [Pfeffer, 1986]:

F =

P (x, y)

Q(x, y)

(5.4)

= ∇·F

∂ ∂

= P + Q

∂x ∂y

Example:

Given the following vector field, find its divergence:

F=

P (x2 + y)

Q(y2 + x)

Using Equation 5.4 the divergence can be written in terms of Cartesian

coordinates to denote the vector in terms of its respective planes:

F = ∇·F

∂ ∂

= P + Q

∂x ∂y

(5.5)

= (2x)ˆi + (2y)ˆj

If we were to substitute an arbitrary point for (x, y) into the divergence found in 5.5 we would find the change in density of the fluid at that point within the vector field.

Vector field divergence points can be seen in figure 5.4, these are the result of substituting values into (x, y). The three common calculated results for divergence and their respective illustrated field shape are as follows:

∇ ·F < 0 ‘sink’

∇ ·F > 0 ‘source’

∇ ·F = 0 ‘No divergence’

Figure 5.4: Types of divergence created on Microsoft Publisher.

Using the result from the example, we can state that (2x)ˆi + (2y)ˆj < 0 is true

at the arbitrary point P (−1, −1) therefore, there is a sink at P as shown in

Figure 5.5.

Figure 5.5: An illustration produced on GeoGebra [Muro, 2015] represent- ing the vector field in our divergence example. The red dot represents the

point P (−1, −1). As you can see the arrows around the point are exactly like

the ‘sink’ illustration in figure 5.4. Thus proving there is a sink at P (−1, −1).

__2-D____ ____ ____Curl:__

The curl shows the rotational direction of fluid in a given field at a selected point. In a 2D space it is presented as a scalar because no matter what point the curl is calculated at, the direction remains. The same can also be defined as the ‘vorticity’ of F [Lugt, 1979] and is found by using the vector cross

product of ∇ and F, as shown in (equation 5.6) [Knill, 2011] [MIT, 2010]:

curlF =

P (x, y)

Q(x, y)

(5.6)

= ∇ × F

∂ ∂

= ∂x Q − ∂y P

As seen, the resulting equation is the same as the kˆ

component in the 3-

dimensional curl, shown further in the chapter when we look at 3D vector fields.

__Example:__

Find the curl of the following vector field:

F =

P (x2 + y)

Q(y2 + x)

To do this, we use the equation in 5.6:

curlF =

P (x2 + y)

Q(y2 + x)

= ∇ × F

∂ ∂

(5.7)

= ∂x Q − ∂y P

= 1 − 1

= 0

If the curlF = 0 as shown in 5.7 then the vector field is irrotational i.e. there is no turning effect or ‘vorticity’ at any point within the vector field as illustrated in Figure 5.6

Figure 5.6: An illustration produced on GeoGebra [Muro, 2015] represent- ing the vector field in our curl example. As you can see the arrows are pointing in irregular directions proving that if the red point P was placed anywhere on the field, it would not have a rotation. Therefore, the vector field has no curl.

5.2 3-Dimensional Vector Fields (R3 → R3)

The 2-dimensional vector field equation (5.1) can now be extended into a

3D coordinate space (R3 → R3) where the field is extended to include a

z-plane as shown in equation 5.8 through the addition of a new point R

with position vector (x, y, z) and Cartesian coordinate kˆ [Dawkins, a]:

F = P (x, y, z)ˆi + Q(x, y, z)ˆj + R(x, y, z)kˆ

(5.8)

5.2.1 Gradient, Divergence and Curl of A 3D Vector

Field

__3-D____ ____ ____Gradie____n____t:__

A vector field is created using the gradient operator ∇ on a scalar field

f (x, y, z). Similar to the 2D vector field, the 3D gradient at any point is

defined using partial differentiation to give Equation 5.9 in terms of Carte- sian coordinates (as you can see this equation is the same as 5.2 but with an extension into the z plane):

grad(f ) = ∇f

∂

= f +

∂x

∂ ∂

f + f

∂y ∂z

(5.9)

__Example:__

The gradient of scalar field f = (−x2 + z) can be found using Equation 5.9.

grad(f ) = ∇f

∂

= f +

∂x

∂ ∂

f + f

∂y ∂z

(5.10)

=(-2x)+(0)+(1)

Similar to the concept of the previous 2D example, this gradient vector points us in the direction of largest change in order to quickly approach the local maximum/minimum. Figure 5.7 shows a 3D surface plot of the scalar field, the darker shades of orange represent a steeper area. If arrows were to point us towards the local maximum/minimum they would point us towards the lighter and darker shades of orange.

Figure 5.7: An illustration produced on WolframAlpha [WolframAlpha, No

Date] showing the gradient of the scalar field in the example above.

3-D Divergence:

The 3D divergence is similar to the 2D divergence. It takes the partial derivatives in every direction of the vector field F and adds them together using the ‘dot’ product, which gives a scalar value. The equation is as follows (5.11) [Azad, No Dateb]:

divF = ∇·F

∂ ∂ ∂

= P + Q + R

∂x ∂y ∂z

(5.11)

__Example:__ Using the vector field below, find its divergence using equation

(5.11):

F=

P (y3 − 3)

Q(x3 − 6)

R(z3 − 9)

divF = ∇·F

∂ ∂ ∂

= P + Q + R

∂x ∂y ∂z

(5.12)

=(0) + (0) + (3z2 )

=(3z2)

Here, we can see there is only a divergence in the z direction which represents the magnitude at that point. A positive, negative or zero number indicates whether that particular point is respectively a source, sink or a neutrally stable point in the field. In our example, the point on the vector field will be a source as it is always positive because z2 will always give a number above zero.

__3-D____ ____ ____Curl:__

Now we know that the divergence calculates the density of the fluid. The curl in a 3D space is presented as a vector, it represents the direction in which the flow is rotating around a point in the vector field [Dawkins, b].

A good way to imagine the curl is to use the ‘Eulers Right Hand Rule’ seen in Figure 5.8. Using your right hand point your thumb in the direction of the vector field and curl your fingers round like a ‘thumbs up’. The direction in which your fingers ‘curl’ around your hand defines the vector fields curl.

Figure 5.8: Eulers Right Hand Rule method to help visualise the curl of a vector field [Right-hand Rule, 2017].

To calculate the curl of a vector field:

F=

P (x, y, z)

Q(x, y, z)

R(x, y, z)

We need to find the determinant of the matrix containing the partial deriva- tives of the vector field points as shown in Equation 5.13 to give a vector

in the form of Cartesian coordinates [Nykamp, No Dateb], we do this using

the vector cross product:

curlF =

P (x, y, z)

Q(x, y, z)

R(x, y, z)

= ∇ × F

ˆi ˆ ˆ

k

j

__∂__ __∂__ __∂__

=

∂x ∂y ∂z

P Q R

(5.13)

∂ ∂

= ˆi R Q

−

∂y ∂z

∂ ∂

− ˆj ∂x R − ∂z P

∂ ∂

+ kˆ __ ____ __ Q P

−

∂x ∂y

__∂__

__∂__

∂y R − ∂z Q

__∂__

=

__ ____∂__

∂x R − ∂z P

__ ____∂__ __∂__

∂x Q − ∂y P

__Example:__

Find the curl of the following 3-Dimensional vector field:

F =

P (y3 − 3)

Q(x3 − 6)

.

R(z3 − 9)

Uisng equation 5.13

C urlF =

P (y3 − 3)

Q(x3 − 6)

R(z3 − 9)

= ∇ × F

ˆi ˆ ˆ

k

j

__∂__ __∂__ __∂__

=

∂x ∂y ∂z

y3 − 3 x3 − 6 z3 − 9

∂ ∂

= ˆi R Q

−

∂y ∂z

∂ ∂

− ˆj ∂x R − ∂z P

∂ ∂

+ kˆ __ ____ __ Q P

−

∂x ∂y

(5.14)

__∂__

__∂__

∂y R − ∂z Q

__∂__

=

__ ____∂__

∂x R − ∂z P

__ ____∂__ __∂__

∂x Q − ∂y P

0

=

0

3×2 − 3y2

Here we can see the x and y components of the vector are zero – showing no curl. The z component of the vector can be ignored because the direction of the curl is only affected by the x and y components. When you narrow the vector field down to one point and pin a ball there (allowing it to spin), the rotation arises only if there is a force in the x or y direction, whereas the z component has no effect on rotation. Therefore, we can say the curl for example 5.14 is zero and irrotational.

5.2.2 Virtual reality Calcflow Software

In order to visualise the vector field of a 3D function, we used virtual reality software called Calcflow to see if a ‘hands on’ virtual approach would help explain a vector field and its flow to users.

Using the vector field from example 5.14 (shown again below), the points P, Q and R were input into the vector field generator using their correspond- ing position vectors. We did this with the controllers ‘pointer’ tool by se- lecting letters and values on a mathematical keyboard. Thus, the following

vector field was created in figure 5.9:

F =

P (y3 − 3)

Q(x3 − 6)

R(z3 − 9)

Calcflow gives us the 3D graph shown in Figure 5.9 and 5.10. The virtual software enabled graph expansion and reduction, rotation and interaction through changing the position of the white sphere (representing a point on the field) to show and track (using the ‘pen’ tool) its trajectory within the vector field – if there was any. It would help discover whether the vector field contained any sources, sinks or neutrally stable points purely by eye, before mathematically calculating anything. This advanced piece of software was designed specifically to help users understand vector mathematics with options to view Stokes Theorem, the curl of a vector field and many more.

Figure 5.9: Calcflow software displaying a relatively 2D angle showing predominantly the x and y axis of the 3D vector field in example 5.14.

The illustration also confirms the positive divergence we calculated in ex- ample 5.12 as the field has a source point at its origin. This is confirmed by placing the sphere on the origin and noticing that there is no variation of trajectory, the flow simply enters the field from one end and leaves through the other.

Figure 5.10: All three axis of the vector field are within view. However, the sphere has been moved to a vector field point with a stronger velocity. Thus showing how the trajectory changes when moved to a point containing a stronger magnitude and direction.

5.3 Navier-Stokes Theorem

Navier-Stokes theorem defines how a surface integral over the surface S is related to the boundary curves line integral [Broda, 2004].

Before explaining the formula we need to know that in order for Stokes theorem to work the following conditions need to be satisfied [Mathonline, No Date] [Itoh et al., 2007]:

1. The surface is ‘piecewise smooth’. Therefore, it has continuous derivatives i.e. a gradual slope, no vertexes (Figure 5.11).

2. The surface is a simple closed boundary i.e. there are no intersections

(Figure 5.12).

Stokes Theorem, represented by Equation 5.15, states that the integral of a field (F) around a closed piecewise smooth boundary (where C =∂S) is equal to the surface integral of the curlF in the 3-dimensional Euclidean space. As we know from Subsection 6.2.1 the 3D curl is calculated using the cross

product (∇ × F). Therefore, the formula for Stokes theorem is as follows

[Broda, 2004] [Nykamp, No Datec]:

Z Z

∇ × F·dS =

S

I

F·dr (5.15)

C

Figure 5.11: An illustration created in Microsoft Publisher to show how a surface can be split in order to satisfy the definition of ‘piecewise smooth’

(last image) in Stokes first condition.

Figure 5.12: Illustration created in Microsoft Publisher to show how a boundary can be split in order to satisfy the definition of a ‘piecewise smooth closed boundary’ in Stokes second condition. The lower image is an alternate piecewise smooth closed boundary.

Example:

Use Stokes theorem to evaluate the surface integral of the curlF where

F = (z2)ˆi − (6xy)ˆj + (x2 y2 )kˆ and the surface S is a section of z = 9 − x2 − y2

on the plane z = 0 as shown below in Figure 5.13:

Figure 5.13: Illustration created in Microsoft Publisher showing the section

of z = 9 − x2 − y2 on the plane z = 0 (drawing is brief and therefore not to

scale).

The closed piecewise smooth boundary C of surface S is oriented using the Right Hand Rule (met in Chapter 6.2.1, Figure 5.8). We need to point our thumb in the direction of the norm (z-axis) and then integrate C in the direction our fingers curl around our hand (shown as red arrows in figure

5.13) [Nykamp, No Datec].

First, using Stokes Theorem we can separate the right hand side of the equa-

tion as follows:

Z Z

∇ × F·dS =

S

I

F·dr

C

(5.16)

Z 2π

=

0

F(r(t))·r0(t)dt

To find the equation for the parameter of the plane we set z = 0 and find the radius (r) of the boundary C:

We know z = 9 − x2 − y2 so setting z = 0 we solve:

0 = 9 − x2 − y2

x2 + y2 =9

Thus, r = √9 = ±3

Using the radius we can set the parameter of C as r(t) where x = 3cos(t)

and y = 3sin(t):

r(t) = (3cos(t))ˆi + (3sin(t))ˆj (5.17) From this we can find the derivative r0(t):

r0(t) = (−3sin(t))ˆi + (3cos(t))ˆj (5.18)

Where in both equations (5.17 and 5.18) t has the range 0 6 t 6 2π.

Now we can find F evaluated on the curve by substituting our known values

for x, y and z into the equation of F(r(t)) where z = 0:

F(r(t)) = (0)2ˆi − 6(3cos(t))(3sin(t))ˆj + (3cos(t))2 (3sin(t))2kˆ

= −54cos(t)sin(t)ˆj + 81cos2 (t)sin2 (t)kˆ

(5.19)

Substituting what we found in 5.18 and 5.19 into equation 5.18:

Z Z

∇ × F·dS =

S

Z 2π

0

− 54cos(t)sin(t)ˆj + 81cos2(t)sin2(t)kˆ

· − 3sin(t)ˆi + 3cos(t)ˆj

dt

Z 2π

=

0

− 162cos2(t)(sin(t)) dt

= −54cos3(t) 2π

0

= [−54cos3(2π)] − [−54cos3(0)]

= 0

(5.20)

A value of zero means that Stokes Theorem has shown the vector field F as conservative [Lo, 2008]. This is to be expected as the area we are integrating is a simple closed loop showing that the vector flow into the field equals the flow out proving there is an independence of paths.

Chapter 6

Higher Dimensions

After the third dimension there may be a fourth, it is said to exist in an- other universe and therefore cannot be seen because it is a Euclidean space containing time, space and gravity [Rucker, 2014].

We have seen a 2D and 3D graph with (x, y) axis and (x, y, z) axis respec- tively, so it is only right to assume that a 4D graph would contain four perpendicular axis, lets say (x, y, z, w). However, this axis could only be

projected as a 4D space onto 3D space (R4 → R3) meaning that all four

points can’t be independent as they would if projected from R4 → R4 . As

an example, Figure 6.1 displays a 2D, 3D and 4D mapping of a square,

cube and hypercube respectively:

Figure 6.1: An illustration of a 2D, 3D and 4D square, cube and hypercube on an axis relative to their dimension [Resta, 2006].

6.1 Hypercube

The most known 4D shape (mentioned in figure 6.1) is the Hypercube – also predominantly known as a Tesseract (‘3D cross’) when unfolded [Ram´ırez and Aguila, 2002]. It is a cube projected into 4 dimensions, in fact, it is a cube within a cube as seen below in Figure 6.2:

Figure 6.2: An illustration of a Hypercube [Weisstein, 2001].

As you know: a 2D square contains 4 vertexes, 4 edges and 1 face; a 3D cube has 8 vertexes, 12 edges, 6 faces and is itself 1 cube. The 4D Hypercube contains: 14 vertexes, 32 edges 24 faces and 8 cubes [Weisstein, 2001].

Using a 3D virtual reality Euclidean space, an attempt was made at drawing the Hypercube using the PaintLab software we met earlier on in Chapter 3. Figures 6.3, 6.4 and 6.5 are evidence to show how difficult freehand draw- ing is in virtual reality, the scales are not accurate at all and the Hypercube looks more like a ‘Hypercuboid’. Difficulties aside, this is an example of a

4D shape in a 3D space. It looks relatively simple here, but remember, it moves in a way that looks like the smaller cube comes out of the larger one, then proceeds to enlarge and engulf the larger cube which then becomes the

smaller one – confusing right! This happens over and over again in a 4D space.

Figure 6.3: Front angle of the 3D Hypercube drawn in PaintLab.

Figure 6.4: Side angle of the 3D Hypercube drawn in PaintLab.

Figure 6.5: Bottom angle of the 3D Hypercube drawn in PaintLab.

6.2 Polyunfolding

As previously mentioned, another way to display the 4D Hypercube is to transform it into 3D and unfold it like a net. It will be in the form of a

3D cross formally called the ‘Tesseract’ which is composed of eight cubes as shown in Figure 6.7. As you can see the Tesseract is a similar shape as the net of a 3D cube (Figure 6.6). Therefore, we can say that the 3D Tesseract is a net of the 4D Hypercube.

A standard 3D cube can be unfolded in eleven different ways, whereas the

Tesseract can be unfolded in 261 ways! [Constant, 2016]

Figure 6.6: Net of a cube. [Ram´ırez and Aguila, 2002].

Figure 6.7: The Hypercube unfolded into a 3D cross containing 8 cubes, known as the Tesseract. It is also shown on a 3D axis on the right [Ram´ırez and Aguila, 2002].

Now, unfolding the cross further in figure 6.7 we create a polyomino ‘tile’

that satisfies something called Conways Criterion Figure 6.8.

Figure 6.8: A colour-coded polyunfolding of the Tesseract [Langerman and

Winslow].

To help us, some faces are colour coded so we can see their positioning on the 2D tile once the Tesseract has been unfolded into a net. The light grey lines represent ‘mountain folds’ – where the fold is pointing upwards, and the darker grey lines represent the ‘valley folds’ – where the folds peak is inside [Langerman and Winslow].

6.2.1 Conways Criterion

Certain types of polycubes (3D polygon with identical length edges) have vertex and edge unfoldings that tile the plane satisfying Conways Criterion. It is a condition for tiling a plane with tiles that have 180◦ rotational sym- metry [Rhoads, 2005]. In other words, for a shape to ‘tile’ it must be able to fit identical shapes around it like a puzzle with no gaps, as shown in Figure

6.9:

Figure 6.9: A tile plane created using the Tesseract unfold in figure

6.8[Langerman and Winslow].

In order for Conways Criterion to be met, a closed topological disk (or poly- omino) with 6 consecutive points A, B, C, D, E and F must have the following three requirements [Langerman and Winslow] [Rhoads, 2005] [Guttmann]:

1. The boundary from A → B is identical to the boundary from E → D

by translation.

2. BC, C D, EF and F A boundaries are identical when rotated 180◦ around their center points (centrosymmetric).

3. Some of the 6 points may coincide but at least one must be distinct.

Where a polyomino represents an orthogonal polygon with the same length edges.

__Example:__

Figure 6.10: A heptomino meeting the requirements of Conways Criterion

[Rhoads, 2005].

In figure 6.10 above there is a heptomino which is a polyomino made up of seven squares. It satisfies the three requirements of Conways Criterion which enables it to be tiled onto a plane [Rhoads, 2005]. Using the labels on the diagram, in order for us to tile the shape, we need to make a duplicate, turn it 180◦ and line edge A up with D as shown in Figure 6.11 below:

Figure 6.11: An illustration created in Paint using figure 6.10 to demon- strate a 180◦ rotation into a 2-strip tiling.

This 2-strip piece can now be infinitely translated to create a plane tiling similar to our original Tesseract net [Rhoads, 2005].

Chapter 7

Conclusion

Virtual reality has changed the way we can view and study graphs, provid- ing unbelievable methods to convey data beyond the previous potential of technological thresholds. However, virtual technology is dependent upon a computers operating system, highlighting the importance of acquiring the sufficient technology required to power the virtual software.

During my investigation, the HTC Vive used was connected to an up-to-date computer; however the process was not entirely smooth. In the course of the second scheduled session, visuals in the headset would turn red, resulting in a system crash and a loss of any existing progress. It can be deduced that this inconvenience was due to the involvement of another HTC Vive which was simultaneously connected to the internet. To recover any losses from insufficient technology sources, more time was allocated into restarting and recreating previous graphs, therefore reducing time allotted to retrieve relevant screen captures for the investigation. Whilst the Vive was running

with no interference, free-hand drawing in a virtual space proved to be very dicult and chaotic, as shown in the investigation where PaintLab was used. The challenges arose once two lines, which connected in a 2D perspective, were rotated and viewed in a 3D perspective. The lines were seen to distort upon rotation, disconnecting in the process and proving 2D efforts to be inaccurate. We are able to determine that freely drawing graphs in a virtual reality space is not helpful when it is essential for data to be presented with accuracy.

An example of this would be conveyed in the pie chart cases we investigated. Out of the two cases assessed, case 1 was easier to produce as it had specic upper and lower boundary values for the volume of a sector, providing the height was xed. Ensuring the pie chat was in proportion (through normali- sation), leading to greater aesthetics when it is virtually viewed. Whereas, case 2 had an innite boundary range for its height when the volume was xed. A pie chart would become disproportionate if its data set held small values, giving it an abnormal appearance in virtual space. However, it could be ad- vantageous to have a data set with very small values, increasing the ability for viewers to identify these values. The smaller values would be highlighted due to their representation as a very thin, tall sector. Overall, the use of case 1 or 2 when designing a 3D extruded pie chart is dependent upon the values in the data set. It would be beneficial to use case 2 where there are small values which require attention. On the other hand, case 1 would be sucient for the majority of data sets as it would be more compact, smarter and aesthetically pleasing for the user.

When further evaluating the HTC Vive, we can conclude that it was helpful when using Calcow software for vector eld analysis. The 3D div, grad and curl took some time to mentally visualise with respect to the vector eld. Calcow simplified understanding through interaction as it created an advanced visual learning platform. It also had several vector eld options including the ability to interact with Stokes Theorem and analyse the curl of a vector eld, these would have been included in the investigation given more allocated time with the HTC Vive.

Another constraint in this investigation can be seen in Chapter 7, it is un- derstandable that we could only view the 4D diagrams in a 3D space. The barrier to illustrate a 4D shape was due to the virtual technology only hav- ing a 3-dimensional interface; the fourth dimension cannot be seen and might not even exist to some people. Attempts were made to draw the 4D cube in a 3D space but the result was cluttered. Lines were not straight and the ratio did not meet that of a correct Hypercube. The 3D Tesseract was briey analysed using Conways Criterion and unfolding techniques. There is still a vast amount of research and computing to be done before we could display information in a 4D space, it is questionable whether it is even a realistic possibility.

Given more time, potential cases for a 3D bar chart could be mathematically discussed and accompanied with 3D designs on Blender. Non-zero cases for Stokes Theorem would have been assessed with Calcow illustrations. Fur- thermore, a 4D extension of vector elds could have been investigated along with determining whether data could be displayed in a 4-dimensional Eu-

clidean space, and if so, mathematically analysing the number of variables it has the potential to display.

In the future, virtual reality technology is expected to become common amongst schools, manufacturers, hospitals, inside households etcetera [Lari- jani, 1993]. It has an adaptive quality which means it can be injected into many sectors such as gaming, virtual travel (allowing simulation travel to highly expensive destinations) and simulating surgical procedures for trainee surgeons. Virtual reality technology has reinvented the methods used to educate and test because it provides a realistic, risk-free and relatively inex- pensive platform for almost anything. Therefore, using virtually interactive graphs to conduct detailed analytical investigations of data which would hold signicant value in elds such as data analysis and stochastic forecasting. Con- tinued development of virtual reality software, together with the evolution of user-friendly devices in current markets, could provide a pathway for self- constructed virtual reality systems in homes around the globe.

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