An Investigation into Virtual Reality Graphing

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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  =  θi  × 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  θi  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  =  θi        πh

×

360         i

(4.12)

Rearranging  Equation 4.12  to equal hi :

1

hi =

 θi

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  θi  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  θi  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

186.4

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  θi  πh

2

360◦        i

θp 

=      πhp

360◦

θf 

+      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   Gradient:

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   Divergence:

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   Gradient:

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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