Gnomonic Projections onto Various Polyhedra

By Brian Stonelake

Table of Contents Abstract ...... 3 The Gnomonic Projection ...... 4 The Polyhedra ...... 5 The Icosahedron ...... 5 The Dodecahedron ...... 5 Pentakis Dodecahedron ...... 5 Modified Pentakis Dodecahedron ...... 6 Equilateral Pentakis Dodecahedron ...... 6 Triakis Icosahedron ...... 6 Modified Triakis Icosahedron ...... 7 Equilateral Triakis Icosahedron ...... 7 Mapping a Polyhedron ...... 8 Icosahedral Great Circles ...... 8 Icosahedral Great Circle Type A ...... 8 Icosahedral Great Circle Type B ...... 9 Icosahedral Great Circle Type C ...... 9 LCD Triangles ...... 11 LCD Classification ...... 11 Values that Define the Projections ...... 13 Exact Values for Relevant Angles ...... 13 Exact Values for LCD Vertices ...... 15 Exact Values for Centers of Projection ...... 18 Icosahedron and Dodecahderon ...... 18 Center of Projection formulas ...... 18 Modifications of Dodecahedron and Icosahedron ...... 21 Distortion ...... 25 Entire LCD triangle ...... 28 Modifications of the Dodecahedron and Icosahedron ...... 31 LCD sampling ...... 32 Continuous measure ...... 33 Density Plots ...... 35 Summary of distortion ...... 39 Remaining Questions ...... 41 Calculations ...... 42

Exact Coordinates for M 2.8 Calculations ...... 43

Exact Coordinates for C2.7.8 Calculations ...... 45 Pentakis Dodecahedron Center of Projection Calculations ...... 47 Modified Pentakis Dodecahedron Center of Projection Calculations ...... 48 Equilateral Pentakis Dodecahedron Center of Projection Calculations ...... 49 Triakis Icosahedron Center of Projection Calculations ...... 50 Modified Triakis Icosahedron Center of Projection Calculations ...... 51 Equilateral Triakis Icosahedron Center of Projection Calculations ...... 52 Height of Pyramid Caps for Optimal Triakis Icosahedron ...... 53 References: ...... 54 Image Credits: ...... 54

Abstract

In this paper, I’ll attempt to build upon work done by RW Gray exploring gnomonic projections onto various polyhedra. Much of his work appears to be motivated by Buckminster Fuller, though many pioneers are recognized in the field.

While this work was motivated by the practical application of geographic mapping, this paper will focus on the mathematics involved, rather than its applications. Finding closed form representations for relevant quantities, for example, will be far more interesting to me than their decimal approximations.

After a brief introduction on the method of mapping that I will use and the polyhedra onto which I will project, I’ll discuss the base unit with which I will tile the sphere. These units, termed “LCD triangles” are referenced in Fuller’s Synergetics and expanded upon beautifully by Gray. My contributions will be to develop a logical coding convention for these units and to derive closed form notation for the coordinates of the vertices and their images, relative to a specific orientation.

I’ll also derive closed-form notation for other points necessary to explicitly define the projection, as well as functions allowing others to find relevant points for modifications of my polyhedra.

Finally, I’ll offer several quantifications of the distortion associated with these mappings. The goal here is more an exploration of the distortion that a search for an ideal polyhedron.

The Gnomonic Projection

A gnomonic projection (GP) is a bijection from the surface of a hemisphere (open) to a plane. It has the useful property that it maps great circle arcs to straight lines.

This map can be visualized by letting a sphere sit atop an image plane. If we imagine the sphere as our globe and associate the point of tangency on the sphere with the South Pole, any point in the southern hemisphere is in the domain of this map. With this picture in mind, the image of a point in the southern hemisphere is just the intersection of the image plane and a ray from the sphere’s center through the point in question. Figure 1: The Gnomonic Projection To see that great circle arcs are mapped to straight lines, one only needs to know that a great circle is, by definition, the intersection of a sphere and a plane defined by any two points on the sphere and the sphere’s center. Thus the ray that defines the GP for any point on the great circle arc lies on this plane, so the image is the intersection of two planes; a straight line (see figure 1).

Figure 1 not only shows that the GP takes all great circle arcs to straight lines, but can be used to show the converse; that any straight line is the images of a great circle arc. To prove the latter, consider an arbitrary line in the image plane and two distinct points on that line. We can define a plane using these two points and the sphere’s center, and note that the intersection of this plane and the sphere defines a great circle arc.

Because of the ambiguity in the orientation of the image plane, there are many formulations of this map. One such projection1 maps a point on the sphere with latitude and longitude of φ and λ to the x and y coordinates on the plane given by:

cosφ sin(λ − λ0 ) cosφ0 sinφ − sinφ0 cosφ cos(λ − λ0 ) x = , y = . sinφ0 sinφ + cosφ0 cosφ cos(λ − λ0 ) sinφ0 sinφ + cosφ0 cosφ cos(λ − λ0 )

In the above, φ0 and λ0 are constants representing the latitude and longitude of the point of tangency of the image plane and the sphere, which is taken in the above to be a unit sphere.

Note that when λ = λ0 x = 0 , and when φ = φ0 y = 0 .

1 From Wolfram Mathworld The Polyhedra

To map the entire globe, as opposed to a single open hemisphere, I will consider several gnomonic projections. By choosing appropriate points of tangency between the sphere and the image planes, our images can join together to form polyhedra. Points of tangency will be chosen to create the image polyhedra below.

The Icosahedron

The (regular) icosahedron is one of five platonic solids. It has twenty equilateral triangular faces with five meeting at each of its twelve vertices. This defines thirty edges and a 5 ! dihedral angle of 180 − arccos( 3 ) , or approximately 138.19 .

The Dodecahedron

Another platonic solid, the (regular) dodecahedron, is composed of twelve regular pentagons, with three meeting at each of its twenty vertices. This defines thirty edges and a −1 ! dihedral angle of arccos( 5 ) , or approximately 116.565 . A fact that will be useful in our analysis is the dual nature of the dodecahedron and the icosahedron.

Pentakis Dodecahedron

The pentakis dodecahedron is a modification of the (regular) dodecahedron, where each dodecahedral face is replaced with a (non-regular) pentagonal pyramid. Exchanging each of the twelve dodecahedral faces with a five sided pentagonal pyramid leaves it with sixty isosceles triangular faces.

As a catalan solid, the pentakis dodecahedron has as its dual an archimedean solid; the truncated icosahedron. The truncated icosahedron has a geometry often associated with a soccer ball, and can be thought of as the result of Figure 2: The Pentakis Dodecahedron truncating each vertex of the icosahedron so that a third of each end of each edge is removed. V V 1 3 The most relevant features of the pentakis C1.2.3 dodecahedron (for our purposes) are the differences in vertices. The “new vertices” at the apex of the C C pentagonal pyramids join five triangular faces, while 1.2.6 V2 2.3.8 the “old vertices” which are vertices on a V V dodecahedron, join six triangular faces. 6 8 C2.6.7 C2.7.8 These “old vertices” are further from an inscribed sphere’s center than are the new vertices. Put V7 differently, a circumscribed sphere only intersects the Pentakis Dodecahedron at its twelve “old vertices”. This tradeoff is made to preserve a 80+9 5 constant dihedral angle (a defining feature of a catalan solid), which is arccos(− 109 ) , or ! approximately 156.719 .

Figure 3: Modified Pentakis Dodecahedron Modified Pentakis Dodecahedron

What I will refer to as a modified pentakis dodecahedron is a slight alteration of the pentakis dodecahedron. Here, we increase the height of the pentagonal pyramids so that all vertices (“old” and “new”) fall on the surface of a circumscribed sphere. Doing so increases the congruent angles of the isosceles triangles.

Equilateral Pentakis Dodecahedron

Much like the modified pentakis dodecahedron, the Figure 4: A Slight Exaggeration of the equilateral pentakis dodecahedron increases the congruent Equilateral Pentakis Dodecahedron angles of the isosceles triangles to increase the height of the pentagonal pyramids. However now the purpose of increasing the height of the pentagonal pyramids is not to create a constant vertex radius, but rather to create equilateral triangular faces.

Doing so drastically increases the height of the pentagonal pyramid caps, making the radius of the “new vertices” considerably greater than that of the “old vertices”. Note that in figure 4, the height of these pyramids appears to have been increased beyond what is necessary for equilateral faces.

Triakis Icosahedron Figure 5: The Triakis Icosahedron Paralleling the construction of the pentakis dodecahedron, where we add pyramids to each face with a height that preserves a constant dihedral angle, is the triakis icosahedron. The only difference here is that we start with an icosahedron, not a dodecahedron, so our pyramid caps are triangular, not pentagonal. Because our pyramid heights were chosen to preserve a constant dihedral angle, a triakis icosahedron is also a catalan solid.

The above construction creates a polyhedron with sixty triangular faces, as each of the twenty triangular faces of the icosahedron are replaced by the three faces of a triangular pyramid. It has a dihedral −24−15 5 ! angle of arccos( 61 ) , or approximately 160.613 .

Modified Triakis Icosahedron

Paralleling the modification of the pentakis dodecahedron that produced the modified pentakis dodecahedron, the modified triakis icosahedron will increase the height of the pyramidal caps of the triakis icosahedron so that each vertex is equidistant from the center of the polyhedron. Thus, all thirty-two vertices (twelve “old” twenty “new”) fall on the surface of a circumscribing sphere.

Equilateral Triakis Icosahedron Figure 6: Equilateral Triakis Our final polyhedron is the icosahedral analogue of the Icosahedron equilateral pentakis dodecahedron. Paralleling the modification of the pentakis dodecahedron that produced the equilateral pentakis dodecahedron, the equilateral triakis icosahedron will increase the height of the pyramidal caps of the triakis icosahedron so that the triangular faces are equilateral. Doing so makes each pyramid cap a tetrahedron.

Mapping a Polyhedron

As aforementioned, in this paper I’ll be exploring projections of the entire sphere (not just an open hemisphere) so I will be using multiple gnomonic projections at once. This can be visualized by inscribing a sphere inside a polyhedron, where each face of the polyhedron defines a unique gnomonic projection. We restrict the domains of these projections so that the image of any given projection is exactly the polyhedral face that defines it.

When using these projections as geographic maps, it’s convenient to work in the plane, so we’ll eventually be unfolding (somewhat arbitrarily) these polyhedra to a plane.

Icosahedral Great Circles

As mentioned above, the (regular) icosahedron Figure 7: A Typical Icosahedral Unfolding has twenty equilateral triangular faces with five V V V V V meeting at each of its twelve vertices. A typical 1 1 1 1 1 N unfolding, with notation that I will assume for N V V V V V V the vertices is shown in figure 7. The points N 2 3 4 5 6 2 and S will later define the north and south V V V V V V poles, respectively. 7 8 9 10 11 7 S S V V Fascinating features of any polyhedron are its V12 V12 V12 12 12 rotational symmetries. We classify these rotational symmetries by the axis of rotation2, which can then used to define what is called an icosahedral great circle (IGC).

Figure 8: Order 2 Axis of Rotation Icosahedral Great Circle Type A If we set the axis of rotation to pass through the midpoint of an edge (and thus, the midpoint of the diametrically opposed edge), we create a rotational symmetry group. In figure 8, we see that this group has order two, as it must map point A about point R to either point A or B. A R B

Figure 9: Icosahedral Great Circles, Type A The plane through the V V V V V center of the icosahedron 1 1 1 1 1 orthogonal to the N N aforementioned axis V V V 2 V3 4 V5 V6 2 intersects with the icosahedron at what I will call Icosahedral Great Circle V V V V V V 7 8 9 10 11 7 Type A (IGCA). Because there are thirty edges, there are S S fifteen of this type of IGC. All fifteen are illustrated in V12 V12 V12 V12 V12 figure 9, with one boldfaced to illustrate a typical path.

2 One intersection of the given axis and the icosahedron is denoted R in the figures below. The fact that there are fifteen of this type is confirmed by the fact that each traverses two edges and bisects four faces on its journey around the icosahedron. Because there are thirty edges total and each is traversed exactly once, there must be 30 ÷ 2 = 15 IGCA. Alternatively, because there are twenty faces, and each can be bisected three times, we again show that there are 20 ⋅ 3 ÷ 4 = 15 IGCA.

Icosahedral Great Circle Type B Figure 10: Order 5 axis of rotation If R, a point on both the surface of the icosahedron and the axis of rotation, lies on a vertex (and thus, the vertex diametrically opposed) we create a rotational symmetry group of order five. This can be visualized by rotating point A to any of points A, B, C, D or E in figure 10. B

The plane through the center of the icosahedron orthogonal C A to the aforementioned axis intersects with the icosahedron R E at what I will call Icosahedral Great Circle Type B (IGCB). D Because there are twelve vertices, there are six of this type of IGC. All six are Figure 11: Icosahedral Great Circles, Type B illustrated in figure 11, V1 V1 V1 V1 V1 with one boldfaced to N N illustrate a typical path. V V2 V3 4 V5 V6 V2 The fact that there are six of this type is confirmed by V V V V V 7 8 9 10 11 V7 the fact that each traverses ten faces on its journey S S around the globe. Because there are twenty faces total V V and each is traversed three times, we see that there V12 V12 V12 12 12 are 20 ⋅ 3 ÷10 = 6 IGCB.

Icosahedral Great Circle Type C Figure 12: Order 3 Axis of Rotation Finally, if the axis of rotation passes through the centroid of a face (and thus, the centroid diametrically opposed), we create a rotational symmetry group of order three. To visualize this, imagine rotating point A in figure 12, about point R, to either point A, B or C. A B The plane through the center of the icosahedron R orthogonal to the aforementioned axis intersects with the icosahedron at what I will call Icosahedral Great Circle C Type C (IGCC). Because there are twenty faces, there are ten of this type of IGC. Again, I’ve illustrated all ten, but boldfaced one to depict its path around the icosahedron.

The fact that there are ten of this type is confirmed by the fact that each traverses twelve faces on its journey around the icosahedron. Because there are twenty faces total and each is traversed six times, we see that there are 20 ⋅6 ÷12 = 10 IGCC.

Figure 13: Icosahedral Great Circles, Type C Because we’ve exhausted our rotational symmetries, we’ve exhausted our IGC’s. The fact V V V V V 1 1 1 1 1 that IGCC don’t unfold to straight lines (which will N N V later be proven but for now can be visualized in 2 V V V5 V6 3 4 V2 figure 13) limits their value in some practical senses. V V V V V V 7 8 9 10 11 7 S S Although the IGC name implies that it resides on V V V12 V12 V12 12 12 the icosahedron, it’ll be helpful to also be able to think of them as great circle subsets of the sphere, or paths on other polyhedra. We can do so by projecting them back to a sphere via inverse gnomonic projections and from there, to another polyhedron via a gnomonic projection. While it may be more precise to refer to them as the inverse gnomonic projections of IGC’s when referring to them on the sphere, and some even more verbose term when they Figure 14: Icosahedral Great Circles, Types A and B are on a different polyhedron, I’ll simply use the V V V V V term IGC and let the context define their location. 1 1 1 1 1 N N V V V While both IGC type A and type B divide the 2 V3 4 V5 V6 2 icosahedral faces into congruent triangles, only V V type A divide the sphere into congruent triangles 7 8 V9 V10 V11 V7 as the icosahedral edges are not traversed by IGCB. S S While the union of ICGA and ICGB would contain V12 V12 V12 V12 V12 those edges, this union results in differently sized triangles, as illustrated in figure 14.

Thus, we will use the 120 triangles resulting from ICGA as our base units. These triangles are referred to as LCD triangles, as a result of the fact that they are the smallest unit (Least Common Denominator) with which the sphere can be tiled. In A Fuller Explanation, Amy Edmondson gives a delightful description of these triangles:

Imagine that some number of hypothetical creatures are to inhabit the surface of a large sphere and it is considered necessary that each one sit in the middle of an identical plot. The consequence of this stipulation is that no more than 120 creatures can fit on the sphere, no matter how large it is. This result is certainly counterintuitive, for assuming a large enough sphere, it seems that we should be able to accommodate as many creatures as we want. However, the unyielding laws of symmetry limit the population to the unexpectedly low number of 120.

LCD Triangles

Because LCD triangles will be our base unit, and they are derived from IGCA, it makes sense to orient the sphere so that the North and South Poles lie at an edge midpoint (the axis of rotation that defined IGCA). This explains the placement of N and S in figure 7, the original icosahedral unfolding.

However, it is not especially aesthetically pleasing (nor is it conducive to labeling LCD triangles) to have the North and South Poles at those points. Figure 15 attempts to remedy this by depicting an alternative view of the LCD triangles, so that octants of the sphere defined by the equator (in black) and two meridians (prime and 90 degree, in orange) are more visible.

Note that to do so, we’ve split several icosahedral faces Figure 15: Depiction of LCD Triangles, Arranged to Show Octants in half. Because ICG’s are relevant in contexts other than V V just on the icosahedron, doing so is not especially N 1 1 N N N troubling. V V V V4 5 V V 2 3 6 2 V V While it is clear that the degree measure of these triangles 8 V V10 V11 7 ! 9 on the icosahedron is 30 − 60 − 90 , on the sphere the S S V ! V V 12 smallest angle of these triangles is 36 , as evidenced by 12 V9 11 S the fact that the small angle of ten of these congruent S triangles meet at points on the sphere associated with icosahedral vertices. Using similar ! ! logic we see that the remaining angles still measure 60 and 90 on the sphere, leaving us ! with 6 of spherical excess in our spherical LCD triangles.

Due to the dual relationship of the icosahedron and the dodecahedron and the fact that LCD triangles pass through all vertices and centroids of the icosahedron, we can use the same LCD triangles as our base unit on the dodecahedron (and its modifications).

For example one can imagine figure 15 as our LCD triangles on an unfolded dodecahedron by picturing the points V1 through V12 as the centers of the pentagonal faces. The ten LCD triangles sharing a vertex at one of these twelve points define the face of a dodecahedron. ! Note that our LCD triangles on a dodecahedron do not measure 30 − 60 − 90 and rather ! 36 − 54 − 90 .

LCD Classification

To systematically classify the 120 LCD triangles, I label them according to their associated location on the sphere, and give each a three-digit code. The first digit of the code signifies which octant the triangle lies in. My convention is to associate 0 degrees longitude and latitude with the point on the equator furthest left in figure 15. Octants 1-4 are in the northern hemisphere, numbered starting at 0 degrees longitude and moving east. The label of an octant in the southern hemisphere is always four greater than the octant of which it is immediately south.

The second digit of the code describes which third of a Figure 16: LCD Coding for Octant 1 given octant the LCD triangle lies in. Note that each octant contains exactly one complete icosahedral face (boldfaced in figure 16), and there is exactly one IGCA from this face’s centroid to an icosahedral edge midpoint 125 that lies on the equatorial border of the octant (the V2 124 111 V3 longer of the blue lines in figure 16). We measure angle α at this face’s centroid, from this IGCA arc, counter- 123 112 clockwise to an arbitrary point in an LCD interior. We 122 113 give a second digit in the code of 1-3 if the measure of 114 ! ! ! ! 121 angle α falls in the intervals (0 ,120 ) ,(120 ,240 ) or 115 133 132 240!,360! , respectively. In figure 16, I arbitrarily ( ) 134 131 ! ! placed point P in LCD112. Note that 0 < α < 120 so the 135 second digit of the code is 1. V8 Finally, each octant third contains exactly one icosahedral vertex. We measure angle β about this vertex, counter-clockwise, so that 0! ≤ β ≤ 180! . We assign 1-5 as the third digit of the code if β is in 0!,36! , 36!,72! ( ) ( ) 72!,108! 108!,144! or 144!,180! , respectively. In figure 8, point P defines an angle ( ) ( ) ( ) ! ! so the third digit in the code is a 2. 36 < β < 72

This produces the naming conventions that will be used going forward. Figure 16 depicts the first octant using these conventions, a complete listing can be found in Figure 17 below.

Figure 17: LCD Triangle Naming Convention

Values that Define the Projections

We next tackle the task of finding exact values for the location of all LCD vertices, as well as the tangency points necessary to define our gnomonic projections. We’ll refer to these points according to their latitude and longitude, and exploit the fact that a gnomonic projection keeps these angles constant. This fact will allow us to refer to an LCD vertex without specifying whether it is on the sphere or a given polyhedron.

Exact Values for Relevant Angles Figure 18: Equatorial Cross-Section of Icosahedron We will first locate the vertices of our base units, the LCD triangles. The first step toward this end will involve an equatorial cross-section of the icosahedron, shown in figure 18.

Four of the vertices of this (non-regular) hexagon correspond with icosahedral vertices V4 , V5 , V7 and

V8 as shown. The remaining hexagonal vertices are edge midpoints on our icosahedron. They are labeled according to the icosahedral vertices that define the edge for which they are the midpoint. Thus, M 3.9 is the midpoint of the edge between icosahedral vertices

V3 and V9 . Similarly C3.8.9 corresponds with the centroid of the icosahedral face defined by vertices V3 , V8 and V9 .

Finally, r(v), r(i) and r(m) refer to three characterizations of the radius of the icosahedron. The distance from the icosahedron’s center (denoted S) to any vertex is given by r(v). Similar notation describes the distance from S to any edge midpoint as r(m) and the distance from S to any face’s centroid as r(i)3.

Assuming an icosahedron with edge length 2, the following radii values are well-known:

• r(m) = ϕ ≈ 1.618 • r(v) = ϕ 5 = ϕ 2 +1 ≈ 1.902 ϕ 2 • r(i) = ≈ 1.512 3

These could easily be derived from figure 18 and the given dihedral angle, but I will take these values as given, and use basic trig to produce the following:

3 Typically r(u) is used instead of r(v), but I find the latter more descriptive. I adopt the more standard r(i), as opposed to perhaps r(c), as the distance to the centroid, because it corresponds with the radius of an sphere inscribed in an icosahedron. Note that r(v) is the radius of a sphere circumscribed about an icosahedron. • α = 2arcsin 1 ≈ 63.43! 1 ( rv ) • α = arccos ri ≈ 37.38! 2 ( rv ) • α = arccos ri ≈ 20.91! 3 ( rm ) 90 α1 58.2825! • α 4 = α 2 +α 3 = − 2 ≈

The known dihedral gives the final angle measurements for figure 18.

• ∠V M V = arccos − 5 ≈ 138.19! 8 3.9 4 ( 3 ) 360! arccos − 5 − ( 3 ) ! • ∠V7V8 M 3.9 = ≈ 110.91 2

Note that these angles are independent of the assumed icosahedral edge length of two noted above. I will later switch to polyhedra inscribed in a unit sphere for ease of calculation.

The point P in figure 18, while not an icosahedral vertex, face centroid, nor edge midpoint, will be relevant to our analysis. To help pinpoint its location, figure 19 depicts a larger version of the bottom left quadrant of figure 18.

Figure 19: SW Quadrant of Equatorial Cross-Section Figure 20: Icosahedral Face Containing Point P

The most important addition in figure 19 is the inclusion of the edge lengths between V8 , P,

C3.8.9 and M 3.9 . These lengths are derived in figure 20, which show the icosahedral face on

which point P lies. The take away from these images is the measures of α 5 below: 1 ⎛ tan 30⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ! α = arctan = arctan 2 = arctan ≈ 10.81 • 5 ⎜ ⎟ ⎜ 2ϕ ⎟ ⎜ 2 ⎟ ⎝ 2r(i) ⎠ ⎝ 3 ⎠ ⎝ 2ϕ ⎠ α α = 1 −α • 5 2 3

Exact Values for LCD Vertices

Using the angles above, we can begin labeling the coordinates of the LCD vertices. I’ll only do so for those in the first octant, relying on symmetry for the rest. Notation for octant 1 LCD vertices is shown in figure 21, with values supplied in table 23 and justifications below.

N: We choose the point N as the North Pole, and calculate the latitude and longitude of the remaining vertices relative to this point.

M 7.8 : Because M 7.8 lies on an IGCA one fourth of the way around the globe, we can ! associate this point with 0 latitude and longitude, orienting our sphere to have IGCA as the four main meridians and the equator.

M 3.9 : Traveling east on the equator, the first intersection with an IGCA that passes through ! ! the North Pole is at M 3.9 . Thus, this point has latitude 0 and longitude 90 .

V8 : As a point on the equator, it’s clear that V8 has Figure 21: Octant 1 LCD Vertices ! latitude 0 . Using figure 18, we see that the angle N α1 C1. 2. 3 between M 7.8 and V8 is half of α1 , thus its longitude is 2 .

V2 M2.3 V3 V2 : Lying between N and M 7.8 on an IGCA, V2 has longitude

0. Using symmetry, we see that the angle between V2 and N C2. 3 . 8 is equal to the angle between M and V . Thus V has 7.8 8 2 M2.8 M3 .9 longitude 90! - α1 . 2 C2. 7. 8 M3 .8 C3 . 8 . 9

V : Lying on an IGCA that passes through N and intersects 3 M7.8 the equator one quarter of the equator’s journey around the V8 ! globe, we see that V3 has a longitude of 90 . Using symmetry, we see that the angle between V3 and M 3.9 is equal to the angle between M 7.8

α1 and V8 . Because M 3.9 lies on the equator, we see that V3 has longitude 2 .

C3.8.9 : The point depicted α 3 degrees short of M 3.9 in the equatorial cross-section of figure ! ! 18 is C3.8.9 . Thus it has latitude 0 and longitude 90 -α 3 .

! C1.2.3 : Exploiting the symmetry between the equatorial IGCA and the 90 meridian IGCA, we ! ! see that C1.2.3 has longitude 90 and latitude 90 -α 3 .

C2.7.8 : Using symmetry between the equatorial IGCA and the prime meridian IGCA, we see ! the C2.7.8 has longitude 0 and latitude α 3 .

M 2.3 : Because this point lies on an IGCB that passes through N and point P on the equator, its longitude is equal to that of point P. Using figure 19, we see that the angle between M 3.9

α1 and point P is α 3 +α 5 which as noted above is equal to 2 . Thus, the longitude of M 2.3 is 90! α1 . To determine this point’s latitude, we take advantage of the same IGCB and the − 2 ! fact that M 2.3 is three fifths of the way from the equator to N, to calculate a latitude of 54 .

M 3.8 : Similar to M 2.3 , the location of M 3.8 on the IGCB between points P and N yield a

! α1 longitude of 90 - 2 . Because it falls one fifth of the way from the equator to N, its latitude is ! 18 .

M 2.8 : By virtue of residing on an IGCC that passes through N, we can calculate a latitude of ! 30 for M 2.8 by noting that it lies one third of Figure 22: Midpoint Formulas the way from the equator to N. For its longitude, there is no doubt a clever argument that that can be made, but instead

I’ll use the midpoint formula with points V2 and V8 . As shown in the calculations appendix, we are able to evaluate the longitude to be α 3 .

C2.3.8 is not a midpoint of any known points. However, we see that the midpoint of it and is . Manipulating the midpoint formula to solve for the location of is quite C2.7.8 M 2.8 C2.3.8 cumbersome. Instead, I note that if a GP using M 2.8 as the point of tangency projects C2.7.8 to

(x, y) it would have to project C2.3.8 to (−x,−y) . The inverse GP of this point yields the 1 ! closed form for the location of C2.3.8 as latitude of arcsin( 3) and longitude 45 , as seen in ! the calculations appendix. It’s worth noting that its longitude of 45 is probably “obvious” using some reasoning that I’ve missed.

Figure 23: Location in Degrees of 1st Octant LCD Vertices

LCD Vertex Exact Approximate Exact Approximate Longitude λ Longitude Latitude φ Latitude N Undefined Undefined 0 0

α1 V2 0 0 90 - 2 58.2825

α1 V3 90 90 2 31.7175

α1 V8 2 31.7175 0 0

α1 M 2.3 90 - 2 58.2825 54 54

M 2.8 α 3 20.9052 30 30

α1 M 3.8 90 - 2 58.2825 18 18

M 3.9 90 90 0 0

M 7.8 0 0 0 0

C1.2.3 90 90 90-α 3 69.0948 C 45 45 1 35.2644 2.3.8 arcsin( 3)

C2.7.8 0 0 α 3 20.9052

C3.8.9 90-α 3 69.0948 0 0

With the above exact coordinates for the LCD vertices, all we need to calculate projections is the point of tangency between the inscribed sphere and the plane onto which we are projecting.

Exact Values for Centers of Projection

Describing our gnomonic projection’s center using the term “point of tangency” is intuitively wise, but a bit of a misnomer as we begin to consider more exotic polyhedra. Thus, for the sake of precision, I will use the term “center of projection” going forward. In this section I will find a closed-form expressions for one of the centers of projection for each of the polyhedra.

Icosahedron and Dodecahderon

In the cases of the dodecahedron and the icosahedron, the centers of projection are the points at which an inscribed sphere is tangent to the polyhedra, which are simply the centroids of each face. Those points are LCD vertices; Cx.x.x and Vx respectively. My analysis will later focus on LCD triangle 135, so the relevant centers of projection for the polyhedra are C2.7.8 and V8 , respectively. ! For the rest of our polyhedra, it will be advantageous to first derive a center of projection formula.

Center of Projection formulas

Modifications of Dodecahedron Figure 24: Pentagonal Pyramid Cap To find a center of projection for each of the three modifications of the dodecahedra that contain pyramid caps of varying heights, we first define a V ' new vertex for the top of the pyramid cap. While 8 both the center of the pentagonal base and the pyramid’s apex are the image of the same point on V8 the sphere (V8 for the pyramid relevant to LCD triangle 135 which I will be analyzing), for ease of C7.8.12 M7.8 notation I’ll refer to the center of the pentagonal C2.7.8 base as V8 and the apex of the pyramid as V8′ .

Because line segment V8 M 7.8 is an apothem of the regular pentagonal face, its length in terms pentagon S 3 s ϕ side length s is ! = s . Thus, assuming a unit 2 tan 36 4 5 side length4 and letting h represent the height of the pyramid cap, we see that angle

ϕ 3 x = arctan 2 . ( 4h 5 )

4 an assumption inconsequential once we take trigonometric ratios, but otherwise crucial ! We now exploit the fact that P’, the center of projection relevant for LCD triangle 135, onto any of these three modifications of the dodecahedron, lies on the line V8′M 7.8 . Thus, the angle measured at the sphere’s center between line segments S V8′ and S P′ is the compliment of x (see figure 25).

α1 α1 Finally, using the fact that the longitude of V8′ is 2 , the longitude of P′ is 2 − (90 − x)

3 1 ϕ = arcsin 2 + arctan 2 − 90 . Figure 25: Center of Projection on Pyramid Cap ( ϕ +1 ) ( 4h 5 )

To find a simplified closed-form expression for

3 V ' 1 ϕ 8 arcsin + arctan 2 − 90 , we let ( ϕ 2 +1 ) ( 4h 5 )

3 1 ϕ P' s = arcsin , t = arctan 2 . ( ϕ 2 +1 ) ( 4h 5 ) C7.8.12 M7.8 cos(s + t − 90) = sin(s + t) = sin scost + cosssint C2.7.8 1 4h2 5 ϕ ϕ 3 = + ϕ 2 +1 4h2 5 +ϕ 3 ϕ 2 +1 4h2 5 +ϕ 3 4h2 5 + ϕ 5 = S (ϕ + 2)(2ϕ +1+ 4h2 5 )

⎛ 2 5 ⎞ 3 1 ϕ 4h 5 + ϕ Thus, 2 ⎜ ⎟ . arcsin 2 + arctan 4h 5 − 90 = arccos ( ϕ +1 ) ( ) ⎜ 2 ⎟ ⎝ (ϕ + 2)(2ϕ +1+ 4h 5 ) ⎠

This gives a general formula producing Figure 26: Center of Projection on Modified Dodecahedron formula the location of the relevant center of ⎛ ⎞ projection for any such modification of the 2 5 ⎜ 4h 5 + ϕ ⎟ dodecahedron, which I will call the Center of f (h) = arccos ⎜ 2 ⎟ Projection for Modified Dodecahedron ⎝ (ϕ + 2)(2ϕ +1+ 4h 5 ) ⎠ function, or CPMD for short.

Because I switch between different sized polyhedra when it suits my calculations, it’s important to note that the CPMD function takes as its inputs the height of the pyramid cap, assuming a unit length edge. The output is independent of the polyhedra’s size, but the input is not. Further, the domain of this function should technically be restricted to [0, 0.425325] to avoid issues with trigonometric sign changes ignored in the simplification. However, values outside of this interval are handled easily below.

Once we have values for h, we can find closed-form expressions for the longitudes of these centers of projection for our three modifications of the dodecahedron. However, before we move forward, I’ll analogously derive a function that yields the center of projection if it is an icosahedron, rather than a dodecahedron, being modified. !

Modifications of the Icosahedron Figure 27: Triangular Pyramid Cap To find a center of projection for each of the three modifications of the icosahedron that contain pyramid caps of C2.7.8' varying heights, we mirror the strategy used above in the dodecahedron cases.

First, using the fact that the apothem of a unit edge equilateral C2.7.8 V 3 8 triangle is 6 , we can calculate angle x shown in figure 25 as M7.8 ! V7 3 arctan( 6h ) and thus conclude that the difference in latitude 90 arctan 3 from that of C2.7.8 is − ( 6h ) .

Finally, using the fact that the latitude of C2.7.8 is S α = arccos ri = arccos ϕ , we deduce that the latitude of the 3 ( rm ) ( 3 ) center of projection of one of our three modifications of the icosahedron is α − 90 − arctan 3 = arccos ϕ + arctan 3 − 90 . Figure 28: Center of Projection on 3 ( ( 6h )) ( 3 ) ( 6h ) Pyramid Cap

C ' 2.7.8 We simplify this expression is a similar fashion to the CPMD s = arccos ϕ t = arctan 3 formula by letting ( 3 ) ( 6h ) and noting that

P cos(s + t − 90) = sin(s + t) = sin scost + cosssint −1 ϕ 6h ϕ 3 V8 = 3 2 + 3 2 M 3+36h 3+36h V 7.8 2 7 = 6h+ϕ 3 3ϕ 1+12h2

Which yields our Center of Projection for Modified Icosahedron 6h+ϕ 2 3 formula (CPMI) as g(h) = arccos 2 . Again, note that h S ( 3ϕ 1+12h ) refers to the height of the Figure 29: Center of Projection on pyramid cap, assuming a unit length edge, and should Modified Icosahedron Formula

3 technically be restricted to ⎡0, 2 ⎤ to avoid issues with 2 ⎣ 6ϕ ⎦ 6h+ϕ 3 g(h) = arccos 2 sign changes that were ignored in the simplification. ( 3ϕ 1+12h ) However, again, we can (and will) easily handle heights outside of this range by simply accounting for the sign change manually.

Modifications of Dodecahedron and Icosahedron

Using the CPMD and CPMI functions with the respective pyramid heights, we can finally calculate the centers of projections for our remaining polyhedra.

Pentakis Dodecahedron

For the pentakis dodecahedron, the pyramid height, assuming edge length of one, is known to be 1 1 .25153. 19 5 (65 + 22 5 ) ≈

Using h 1 1 65 22 5 to evaluate the CPMD function, and noticing that = 19 5 ( + ) 2 2 2 , we find the longitude of the center of projection for the relevant 4h 5 = ( 19 ) (22 +13 5 ) face of the pentakis dodecahedron as:

⎛ ⎞ 2 5 1 1 ⎜ 19 22 +13 5 + ϕ ⎟ f 19 5 (65 + 22 5 ) = arccos ( ) ⎜ 1+ 5 1+ 5 2 2 ⎟ ⎜ ( 2 + 2) 2( 2 ) +1+ ( 19 ) 22 +13 5 ⎟ ⎝ ( ( )) ⎠ ⎛ ⎞ 2 22 +13 5 +19 ϕ 5 = arccos⎜ ⎟ ⎜ 5 ⎟ 2 1223+ 575 5 ⎝ ( ) ⎠ ! ≈ 11.64072

Modified Pentakis Dodecahedron

For the modified pentakis dodecahedron, because the pyramid’s apex is raised to lie on the surface of a circumscribing sphere, the pyramid height is the difference between the radii of an inscribed and a circumscribed sphere about a dodecahedron. Those values, for a unit 2 length edge, are known to be ϕ and 3ϕ . Thus the height of a pyramid cap on the 2 3−ϕ 2

2 modified pentakis dodecahedron is 3ϕ ϕ . 2 − 2 3−ϕ

3ϕ ϕ 2 2 26+14 5−(7+3 5 ) 30−6 5 Using − as our value for h in the CPMD, we see that 4h 5 = , 2 2 3−ϕ 2 and evaluate the CPMD function as: ⎛ ⎞ 2 26+14 5−(7+3 5 ) 30−6 5 5 ⎛ 3ϕ ϕ ⎞ ⎜ 2 + ϕ ⎟ f ⎜ − ⎟ = arccos⎜ ⎟ ⎝ 2 2 3−ϕ ⎠ ⎜ ⎛ 26+14 5−(7+3 5 ) 30−6 5 ⎞ ⎟ (ϕ + 2) 2ϕ +1+ 2 ⎝⎜ ⎝ ⎠ ⎠⎟ ⎛ ⎞ 52 + 28 5 − (14 + 6 5 ) 30 − 6 5 + 22 +10 5 = arccos⎜ ⎟ ⎜ ⎟ ⎝⎜ 230 +110 5 − (50 + 22 5 ) 30 − 6 5 ⎠⎟ ! ≈ 9.02699

Equilateral Pentakis Dodecahedron

For the equilateral pentakis dodecahedron, using the fact that the pyramid caps are regular 5 − 5 pentagonal pyramids, we can calculate the height of each one as . This height will 10 cause the pyramidal caps to protrude out of the sphere that was circumscribing the dodecahedron. This fact is perhaps noteworthy, but not problematic or even relevant to the calculations.

However, this height is outside of the domain of our CPMD function, so to evaluate the longitude we make the adjustment below (or simply use an unsimplified version of the CPMD formula).

Using 5− 5 and noticing that 2 5−1 , we evaluate the h = 10 ≈ 0.5257 4h 5 = 4( 2 ) = 4(ϕ −1) CPMD function as:

⎛ 5 − 5 ⎞ ⎛ 4(ϕ −1) + ϕ 5 ⎞ f ⎜ ⎟ = arccos⎜ ⎟ 10 ⎝ ⎠ ⎝⎜ (ϕ + 2)(2ϕ +1+ 4(ϕ −1)) ⎠⎟

⎛ 4(ϕ −1) + ϕ 5 ⎞ = arccos⎜ ⎟ ⎝ 15ϕ ⎠ ! ≈ 5.65989

However, because the distance from V8′ to P′ is greater than the distance from V8′ to M 7.8 , the true solution of longitude is: ⎛ 5 ⎞ 4(ϕ −1) + ϕ ! − arccos⎜ ⎟ ≈ −5.65989 . ⎝ 15ϕ ⎠

Triakis Icosahedron

A unit length edge triakis icosahedron is known to have a pyramid height of 1 h = ≈ 0.057 3(5ϕ + 2) which we can input into out CPMI formula to yield:

⎛ 6 1 +ϕ 2 3 ⎞ 1 ( 3(5ϕ+2) ) g 3 5ϕ+2 = arccos⎜ 2 ⎟ ( ( ) ) 1 ⎜ 3ϕ 1+12 ⎟ ⎝ ( 3(5ϕ+2) ) ⎠ = arccos ϕ+2 ≈ 9.69372! ( 4ϕ+7 )

Modified Triakis Icosahedron

The defining feature of our modified triakis icosahedron is that its triangular pyramid caps are chosen to have heights that make all of its vertices fall on the surface of a sphere circumscribing the modified triakis icosahedron. As such, the height of the pyramids is the difference between the distance from the center of an icosahedron to a vertex and the distance from the center of an icosahedron to any of its face’s centroids. Assuming a unit edge length, this height is:

2 2 4 2 1 ϕ 3ϕ 5 −ϕ ϕ +4 −ϕ h = 2 ϕ 5 − 2 3 = 2 3 = 2 3 . ≈ 0.195

We use the above pyramid height and our CPMI formula below:

⎛ 4 4 2 ⎞ ⎛ 6 ϕ + −ϕ +ϕ 2 3 ⎞ ϕ 4 +4 −ϕ 2 ⎜ ⎝⎜ 2 3 ⎠⎟ ⎟ g 2 3 = arccos 2 ( ) ⎛ ϕ4+4−ϕ2 ⎞ ⎜ 3ϕ 1+12 ⎟ ⎜ 2 3 ⎟ ⎝ ⎝ ⎠ ⎠ −1 ϕ+2 ! = arccos ϕ 6 9 2 2 3 6 ≈ 13.17408 ( ϕ+ − ϕ ϕ+ )

However, because the distance from C2.′ 7.8 to P′ is greater than the distance from C2.′ 7.8 to

M 7.8 (or equivalently, because the pyramid height is outside of the domain of the CPMI function) the solution is actually:

−1 ϕ+2 ! 2 . − arccos ϕ 6 9 2 3 6 ≈ −13.17408 ( ϕ+ − ϕ ϕ+ )

Equilateral Triakis Icosahedron

Finally, we analyze the equilateral triakis icosahedron to find its relevant center of projection. The defining feature of this polyhedron is that the faces of its pyramid caps are equilateral triangles, and this the pyramid caps themselves are regular tetrahedra. We exploit this fact to see that (assuming a unit length edge) our pyramid caps have heights 6 h = 3 .

We use the above pyramid height and our CPMI formula below:

⎛ 6 6 +ϕ 2 3 ⎞ 6 ( 3 ) g 3 = arccos 2 ⎜ 6 ⎟ ( ) 3ϕ 1+12 ⎝ ( 3 ) ⎠

= arccos 4 6+3 3+ 15 ≈ 49.62362! ( 9+9 5 )

But again, because the distance from C2.′ 7.8 to P′ is greater than the distance from C2.′ 7.8 to

M 7.8 , the solution for the longitude is actually:

− arccos 4 6+3 3+ 15 ≈ −49.62362! ( 9+9 5 )

Our centers of projections relevant to LCD triangle 135 are summarized in the table below.

Figure 30: Centers of Projection

Polyhedron Approximate Approximate Exact Longitude Exact Latitudeφ λ Longitude Latitude

Dodecahedron α1 2 31.7175 0 0 Pentakis ⎛ ⎞ Dodecahedron 2 22 +13 5 +19 ϕ 5 arccos⎜ ⎟ 11.6407 0 0 ⎜ 5 1223+ 575 5 ⎟ ⎝ 2 ( ) ⎠

Modified ⎛ ⎞ 52 + 28 5 − (14 + 6 5 ) 30 − 6 5 + 22 +10 5 arccos⎜ ⎟ Pentakis ⎜ ⎟ 9.0270 0 0 Dodecahedron ⎝⎜ 230 +110 5 − (50 + 22 5 ) 30 − 6 5 ⎠⎟ Equilateral ⎛ 4(ϕ −1) + ϕ 5 ⎞ Pentakis − arccos⎜ ⎟ -5.6599 0 0 Dodecahedron ⎝ 15ϕ ⎠ Icosahedron 0 0 α 3 20.9052

Triakis ϕ+2 0 0 arccos 9.6937 Icosahedron ( 4ϕ+7 ) Modified −1 ϕ+2 Triakis 0 0 − arccos ϕ 2 -13.1741 ( 6ϕ+9−2ϕ 3ϕ+6 ) Icosahedron Equilateral Triakis 0 0 − arccos 4 6+3 3+ 15 -49.6236 ( 9+9 5 ) Icosahedron

Distortion

Now that we have exact values for the vertices of our LCD triangles and the centers of projections, we can explicitly apply our gnomonic projections. We will do so to define different measures of the distortion inherent in the gnomonic projection.

Going forward, I’ll assume a unit sphere. This choice is for the simplicity of the image LCD vertices below. However, because the distortion calculations are measured relative to the pre-image, this decision is inconsequential to the results.

We must use caution in specifying planar values for images of points in different polyhedral faces, as we want to preserve freedom in how we unfold our polyhedra. Fortunately, we can derive all of our analysis from the projection of a single LCD triangle (which never cross polyhedral edges) and symmetry.

For the sake of specificity, I chose (somewhat arbitrarily) LCD triangle 135 for my analysis. The charts below show exact and approximate coordinates for the images of these vertices using the relevant centers of projection for each polyhedron.

I will take this opportunity to introduce a new polyhedron, which is a modification of the triakis icosahedron with pyramid cap heights between those of the triakis icosahedron and the modified triakis icosahedron. This genesis of this “optimal triakis icosahedron” will be explained later.

Table 31a: LDC135 Projection onto Dodecahedron

Dodecahedron x -coordinate y -coordinate x - coordinate y - coordinate approximation approximation

V8 0 0 0 0 −1 M 7.8 −ϕ 0 -0.618034 0 C −1 -0.618034 0.449028 2.7.8 −ϕ 25 −11 5

2

Table 31b: LDC135 Projection onto Pentakis Dodecahedron

Pentakis x -coordinate y -coordinate x - coordinate y - coordinate Dodecahedron approximation approximation 6 V8 ϕ −11 0 0.365488 0

19 M −1 0 -0.206011 0 7.8 3ϕ C 1 -0.206011 0.389987 2.7.8 − 1 81− 35 5

3ϕ 3 2

Table 31c: LDC135 Projection onto Modified Pentakis Dodecahedron

Modified x -coordinate y -coordinate x - coordinate y - coordinate Pentakis approximation approximation Dodecahedron V −4 0 0.418114 0 8 −ϕ + 4 +ϕ 6 M 7.8 1+ϕ 0 -0.158867 0 ϕ − 6 C 6 -0.158867 0.386756 2.7.8 1+ϕ 1 2 4 ϕ − 1− − 1+ 6 3ϕ 7 3 ϕ 4

Table 31d: LDC135 Projection onto Equilateral Pentakis Dodecahedron

Equilateral x -coordinate y -coordinate x - coordinate y - coordinate Pentakis approximation approximation Dodecahedron V 2 0 0.763932 0 8 ϕ 2 M 1 0 0.0991064 0 7.8 ϕ 5 −1

C2.7.8 1 0.0991064 0.383837 5 1 15 103− 45 5 ϕ −1 ( ) 11 2

Table 31e: LDC135 Projection onto Icosahedron

Icosahedron x -coordinate y -coordinate x - coordinate y - coordinate approximation approximation

V8 3 −1 0.661585 -0.381966 2 ϕ 2 ϕ M 0 −1 0 -0.381966 7.8 ϕ 2

C2.7.8 0 0 0 0

Table 31f: LDC135 Projection onto Triakis Icosahedron

Triakis x -coordinate y -coordinate x - coordinate y - coordinate Icosahedron approximation approximation

V8 1 ⎛ 1 ⎞ 0.626986 -0.17082 2 ⎛ 1 ⎞ 1− ⎜ 4 −1⎟ 5 ⎝⎜ 2ϕ 7 ⎠⎟ 5 ⎝ ϕ ⎠

M 7.8 0 1 ⎛ 1 ⎞ 0 -0.17082 −1 5 ⎝⎜ ϕ 4 ⎠⎟

C2.7.8 0 2 ⎛ 1 ⎞ 0 0.198213 1+ 11⎝⎜ ϕ 5 ⎠⎟

Table 31g: LDC135 Projection onto Modified Triakis Icosahedron

Modified x -coordinate y -coordinate x - coordinate y - coordinate Triakis Approximation approximation Icosahedron 4 V8 3(7 +ϕ) 3(3+ 4ϕ) 2 ϕ −1 1 0.634739 0.234071 − 2 ϕ − 4 + 5 5 5 ϕ 4 M 0 ϕ 4 −1 1 0 0.234071 7.8 ϕ 2 − 4 + 5 ϕ 4

C 0 2 2 0 0.676522 2.7.8 −ϕ + 3(1+ϕ )

Table 31h: LDC135 Projection onto Equilateral Triakis Icosahedron

Equilateral x -coordinate y -coordinate x - coordinate y - coordinate Triakis Approximation approximation Icosahedron 4 V8 6 3 ϕ 3− 2 2 0.954042 1.17598 ( ) 3+ 4 2 + 5 M 0 4 0 1.17598 7.8 ϕ (3− 2 2 )

C2.7.8 0 2 2 0 2.82843

Table 31i: LDC135 Projection onto Optimal Triakis Icosahedron

Equilateral x -coordinate y -coordinate x - coordinate y - coordinate Triakis Approximation approximation Icosahedron −1 V8 ϕ 0 0.618034 0

M 7.8 0 0 0 0 −2 C2.7.8 0 ϕ 0 0.381966

Entire LCD triangle

Our first measure of distortion will be to analyze the area and angles of the right triangular5 projection of an LCD triangle, and compare that with the spherical LCD triangle. I’ll use the above projections of LCD triangle 135, which by symmetry will suffice for any other LCD triangle, given a gnomonic projection.

Each LCD triangle has exactly one “pair” LCD triangle on its face, which shares an edge and combines to make a larger triangle. The measure of the congruent angles at the vertices not shared between these pair triangles is reported in table 31 below (the rationale for this choice will follow).

I first calculate my basis of comparison, the spherical surface area of the LCD triangle. I found it most instructive to use the familiar spherical excess formula, after calculating spherical angles using the spherical law of cosines. I was able to confirm my calculations using L’Huilier’s theorem.

Using the above I found that the spherical surface area of an LCD triangle is 4 arctan ( 2 −1)( 6 − 5 )(−3− 2 5 + 6(5 + 2 5 ))(−2 − 5 + 10 + 4 5 ) or approximately .10472. It then (somewhat embarrassingly) occurred to me that the 120 LCD triangles have equal surface area, so the above surface area6 is just π 30 .

Using the coordinates from table 31 and Heron’s formula for the area of a triangle, I calculated the area of the image triangles. Here I define “LCD distortion” as the percentage increase of the area of the image triangle over the surface area of the spherical triangle. Table 32 below summarizes these results.

5 We know the image of this spherical triangle is a triangle as the gnomonic projection maps the great circle boundaries of the spherical triangle to straight lines in the plane. Because two of these GC’s are perpendicular, the image is a right triangle. 6 ⎛ π ⎞ This implies that tan = 2 −1 6 − 5 −3− 2 5 + 6 5 + 2 5 −2 − 5 + 10 + 4 5 . ⎝⎜ 120⎠⎟ ( )( )( ( ))( ) Table 32: LCD 135 Distortion

Approx. “Outer” Exact Exact Planar Planar Angle Spherical Area of LCD Polyhedron Area of Measure in Area of Image of Distortion LCD135 Image LCD135 LCD135 Image Triangle

65 − 29 5 ! Dodecahedron π 30 0.138757 33% 54 8 Pentakis π 30 1 323,309 −143,289 5 ! Dodecahedron 0.111439 6% 55.69064 342 2 Modified 1 2175 − 607 5 − 6 113,910 − 42,654 5 ! Pentakis π 30 0.111576 7% 56.16567 Dodecahedron 6 2 Equilateral 1 Pentakis π 30 20,734 206 ! + 0.127592 22% 60 Dodecahedron 675 3 5

3 Icosahedron π 30 0.126351 21% 30! 2ϕ 4

Triakis 61 61 π 30 0.115689 10% 30.48032! Icosahedron 5 339,506 −151,542 5 Optimal 1 Triakis ! π 30 3 0.118034 13% 31.71747 Icosahedron 2ϕ Modified 1 3,123−1,251 5 − 2 4,386,990 +1,955,766 5 ! Triakis π 30 0.140421 34% 34.87885 Icosahedron 2 10 Equilateral 108 6(511+ 336 2 +165 5 +128 10 ) Triakis ! π 30 4 0.788253 653% 60 Icosahedron (3+ 4 2 + 5 )

All things being equal, we prefer a polyhedron with more faces in order to limit distortion. The rationale is that each face defines a center of projection, and more centers of projection generally mean less distortion. This is evident in the 20-faced Icosahedron’s distortion being less than that of the 12-faced Dodecahedron.

It’s also noteworthy that the icosahedron’s image LCD triangles measure 30-60-90, while the dodecahedron’s measure 36-54-90. The former may seem advantageous, as equilateral triangles are ideal for the practical application of mapping. However, the layout of the LCD triangles in the icosahedron prevent the pairing of two 30-60-90 triangles to create an equilateral triangle, as it is the short leg that they share (see figure 33). Thus pairs of co- planar LCD triangles actually create a 30-30-120 triangle7.

7 The reported “outer” angle from table 32 gives the measure of the congruent angles in the single triangular image formed from the two paired LCD triangles. This explains the extreme results associated with the equilateral triakis icosahedron, as a large modification of the triakis icosahedron is necessary to create equilateral faces.

Figure 33: Centers of Projection for Icosahedron and All Modifications of Icosahedron

As shown in figure 34, the dodecahedral LCD images pair to create a triangle that measures 54-54-72, closer to the equilateral ideal.

Figure 34: Centers of Projection for Dodecahedron and All Modifications of Dodecahedron

Increasing the total amount of faces, generally speaking, decreases distortion. Thus, when we add pyramid caps to these shapes, the playing field is leveled, as all modifications of either shape has 60 faces, each face having exactly two “paired” LCD triangles. However, due to the differences in the LCD triangle layouts, it’s most instructive to group modifications based on whether they were derived from a dodecahedron or an icosahedron

Modifications of the Dodecahedron and Icosahedron

As indicated in table 31, our polyhedron with the least distortion is the Pentakis Dodecahedron, with approximately 6.4161%. However, we can actually do slightly better if we increase the pentakis dodecahedron’s pyramid heights by about 4.5%. Doing so places ! the center of projection at a longitude of approximately 10.8123 , between those of the ! modified pentakis dodecahedron and the pentakis dodecahedron measuring 9.0270 and ! 11.6407 , respectively. This reduces the distortion to a minimal value of approximately 6.3803%.

While this may seem noteworthy, measuring distortion in this way is not consistent with the more practical goal of minimizing the maximal point of distortion. As such, I only report approximate values above, and don’t bother to give this new polyhedron a name.

Similarly, minimal area for an icosahedral modification is not attained from the triakis icosahedron, and instead found by raising its pyramidal caps by approximately 24.6%. ! Doing so moves the center of projection from a latitude of approximately 9.6937 ! southward to a latitude of approximately 7.0371 . This reduces distortion from approximately 10.4752% to 10.1086%.

Again, I’ll mention this only in passing and fight the urge to find closed-form representations quantifying these two unnamed polyhedra. LCD sampling Our second measure of distortion will be to measure the area of the projections of arbitrarily small regions of an LCD triangle, and compare that with the spherical surface area of the pre-images. The logic here is that a map is only as good as its “worst” areas, so we measure distortion on the far reaches of the triangle to see how bad a given map can get.

I’ll again use LCD triangle 135, which by symmetry will suffice for all of the analysis. I used ! squares of length 0.02 centered on the vertices of LCD 135 for this analysis. When necessary, I extended the image of the GP’s to include any part of an arbitrarily small square’s image that extended across a polyhedral edge.

The use of a square as my sampling unit allows me to easily analyze the change in shape, in addition to the change in area. I’ll refer to the former as “shape distortion” which I will define as the percentage increase in the length of the largest edge to the length of the shortest edge, in the image quadrilateral.

Table 15: LCD Sampling, Area and Shape Distortion

Sampling Sampling Sampling Shape Shape Shape Polyhedron Distortion Distortion Distortion Distortion Distortion Distortion V8 M 7.8 C2.7.8 V8 M 7.8 C2.7.8 Dodecahedron 0% 62% 99% 0% 18% 5% Pentakis Dodecahedron 21% 6% 31% 6% 2% 12% Modified Pentakis 27% 4% 27% 8% 1% 13% Dodecahedron Equilateral Pentakis 99% 1% 24% 26% 0% 14% Dodecahedron Icosahedron 99% 23% 0% 7% 7% 7% Triakis Icosahedron 70% 4% 6% 15% 1% 9% Modified Triakis Icosahedron 76% 8% 76% 13% 3% 29% Equilateral Triakis 498% 268% 2600% 45% 54% 221% Icosahedron Optimal Triakis Icosahedron 62% 0% 23% 18% 0% 15%

Note the lack of distortion8 for the dodecahedron, icosahedron and optimal triakis

icosahedron at vertices V8 , C2.7.8 and M 7.8 , respectively. These results are not surprising, as these vertices correspond with the centers of projection for the polyhedra. I will delay any further analysis of table 35, as it will be redundant with that of the continuous measure of distortion below.

8 rounds to zero, but because this measure is discrete, not exactly zero Continuous measure

Unlike the above, my final measure of distortion is a Figure 36: Image Distance from continuous measure. The logic here is to take advantage of Center of Projection the fact that distortion at a given point is simply a y = tan(x) function of the distance that point is from the center of x projection. Thus, I’ll define a function that gives the distance that the image of a point is on the plane from the origin as a function of the (arclength) distance that point is from the center of projection. The derivative of this 1 function will be my continuous distortion function.

As illustrated in figure 36, the function referenced above is simply f (x) = tan x , so our continuous distortion function is f ′(x) = sec2 x . An intuitive description of this distortion measure is the percentage increase in the speed of the shadow on polyhedron, relative to a point on the sphere.

To allow comparison to the LCD sampling analysis, I’ll first evaluate this distortion function at the vertices of LCD135. These values are shown in table 37 below.

Table 37: Continuous Distortion Measure

Continuous Continuous Continuous Polyhedron Distortion V8 Distortion M 7.8 Distortion C2.7.8 Dodecahedron 0% 38% 58% Pentakis Dodecahedron 13% 4% 19% Modified Pentakis Dodecahedron 17% 3% 17% Equilateral Pentakis Dodecahedron 58% 1% 16% Icosahedron 58% 15% 0% Triakis Icosahedron 42% 3% 4% Modified Triakis Icosahedron 46% 5% 46% Equilateral Triakis Icosahedron 229% 138% 800% Optimal Triakis Icosahedron 38% 0% 15%

Again, we notice the lack of distortion9 for the dodecahedron, icosahedron and optimal triakis icosahedron at their centers of projection. In terms of minimizing the maximal distortion, the modified pentakis dodecahedron appears to perform best. As we will see below, not only does it best the rest of our polyhedra, it is optimal in a sense that we will later make concrete.

9 now exactly zero, thanks to the continuous nature of this measure One important distinction that can be overlooked by only examining the distortion at the vertices is whether there is a point of zero distortion in the image of the LCD triangle, or if the center of projection lies outside of the image. Figure 38 shows the centers of projection relevant to LCD triangle 135. The contour maps and density plots in the next section give a more holistic view of the continuous measure of distortion.

Figure 38: Centers of Projection for LCD135

Density Plots

Dodecahedron We first examine distortion of the dodecahedron. Note that the SE corner of the contour plot, which corresponds with LCD vertex V8 , has a height of zero. This indicates that there is 0 distortion at this point. This result is expected, as V8 is also the center of projection of the Dodecahedron. Distortion increases the further west or north we travel, with a maximal distortion value of 58% in the NW corner, vertex C2.7.8 .

Figure 39: Dodecahedron Distortion Plot Figure 40: Dodecahedron Distortion Contour Map

Pentakis Dodecahedron We next turn our attention to the pentakis dodecahedron. Relative to the dodecahedron, the point of tangency has moved due west from V8 toward M 7.8 . This is evident in the point ! of zero distortion occurring at a longitude of roughly 11.64 (latitude 0). Note that this causes the distortion to be more evenly spread across the triangle, reducing the maximal distortion to just 19% in the NW corner at C2.7.8 .

Figure 41: Pentakis Dodecahedron Distortion Plot Figure 42: Pentakis Dodecahedron Distortion Contour Map

Modified Pentakis Dodecahedron In terms of distortion, the modified pentakis dodecahedron is almost identical to the ! pentakis dodecahedron, as their centers of projection differ by only about 3 of longitude. The small westward shift of the center of projection better allocates the distortion in that the maximal value is 17% at both then NW and SE corners, vertices C2.7.8 and V8 respectively. The equivalence of these distortions is not just an approximation, as the distances from the center of projection to the points V8 and C2.7.8 are identical. It is not obvious to me why this is true, but its truth makes the distortion here optimal (in a sense that I will elaborate more on later).

Figure 43: Modified Pentakis Dodecahedron Figure 44: Modified Pentakis Dodecahedron Distortion Plot Distortion Contour Map

Equilateral Pentakis Dodecahedron ! The center of projection for the equilateral pentakis dodecahedron is roughly 5.66 west of M 7.8 . As a result, all points in (and on the boundary of) the triangle have positive distortion. That value is minimized at only 1% at M 7.8 , and grows to 16% and 58% at C2.7.8 and V8 , respectively.

Figure 45: Equilateral Pentakis Dodecahedron Distortion Plot Figure 46: Equilateral Pentakis Dodecahedron Distortion Contour Map

Icosahedron

For the icosahedron, our center of projection is in the NW corner at vertex C2.7.8 . As such, we have 0 distortion at this point, and increase as we move south and east. The distortion is maximal in the SE corner at V8 , measuring 58%. Note that because the center of projection and point of maximal distortion are reversed from the dodecahedron case, the distortion (a function of distance from the center of projection) at the maximal point is identical to the dodecahedral case. However, because the measure of the LCD triangle’s angle at C2.7.8 is greater than that of V8 , using C2.7.8 as the center of projection reduces the total distortion of the triangle compared to the dodecahedal case, as we saw in table 32.

Figure 47: Icosahedron Distortion Contour Map Figure 48: Icosahedron Distortion Contour Map

Triakis Icosahedron ! The center of projection of the triakis icosahedron is roughly 11.2 due south of that of the icosahedron. The center of projection is still furthest from V8 where distortion is maximized at 42%.

Figure 49: Triakis Icosahedron Distortion Plot Figure 50: Triakis Icosahedron Distortion Contour Map

Modified Triakis Icosahedron Much like the equilateral pentakis dodecahedron, the center of projection for the modified triakis icosahedron lies outside of the image triangle. Thus the distortion is positive at all ! points in the triangle. The center of projection lies roughly 13 south of vertex M 7.8 , so distortion there is minimized at 5%.

Similar to the case of the modified pentakis dodecahedron, the point of tangency is equidistant from its furthest vertices. However this equitable allocation of distortion is not “optimal” as the point of tangency lies outside of the LCD triangle.

Figure 51: Modified Triakis Icosahedron Distortion Figure 52: Modified Triakis Icosahedron Contour Map Distortion Contour Map

Equilateral Triakis Icosahedron The equilateral triakis icosahedron has far more distortion than any of its predecessors. This is the result of the center of projection being nowhere near the LCD triangle. In fact, ! the center of projection is located nearly 50 south of vertex M 7.8 . This large distance results in large distortion at this vertex, and even more at the other two vertices, which are even further from the center of projection. The distortion at the vertices are approximately

138% 229% and 800%, at vertices M 7.8 , C2.7.8 and V8 , respectively.

Figure 53: Equilateral Triakis Icosahedron Distortion Plot Figure 54: Equilateral Triakis Icosahedron Distortion Contour Map

Optimal Triakis Icosahedron The center of projection for the optimal triakis icosahedron is in the SW corner of the LCD triangle. This placement minimizes the distance to the SE corner, reducing distortion there to only 38%. To avoiding increasing distortion to all points in the paired LCD triangle, we cannot move the center of projection off of the western edge of LCD 135. As such, the optimal triakis icosahedron minimizes maximal distortion for any icosahedral modification.

Figure 55: Optimal Triakis Icosahedron Distortion Plot Figure 56: Optimal Triakis Icosahedron Distortion Contour Plot Plot

Summary of distortion In allocating distortion, one might assume that placing the center of projection somewhere in the middle of an LCD triangle to minimize the maximal distance to any vertex would be optimal. However, we must take caution in recognizing that the center of projection is defined for the entire face of the polyhedron, and all of our polyhdera have multiple LCD triangles on them.

Referring back to figure 34, we have the centers of projection for the dodecahedron and its modifications, as well as the LCD triangles in the image of these projections. Note that there are ten LCD triangles in the image of the dodecahedron, but only two (shown in black) in the image of the dodecahedral modifications.

There is not much analysis for the center of projection of the ten LCD triangles. Anywhere other than the dodecahedron’s center of projection would reduce distortion in some triangles at the cost of the others, for a net loss.

However, picking the best center of projection knowing that the polyhedral face only contains two LCD triangles (135 and 511) is more interesting. Optimal from the perspective of a map being only as good as its worst part, is to minimize the maximal distance to a vertex. As we saw in the above analysis the distance from the modified pentakis dodecahedron’s center to its NW and SE vertices are equal. Using symmetry, we see that these lengths are equidistant to the SW corner of LCD 511. Thus, in terms of minimizing maximal distortion, the modified pentakis dodecahedron is optimal.

Similar to above, there is not much analysis for the center of projection of the six LCD triangles in each face of the icosahedron that we saw back in figure 33. Anywhere other than the icosahedron’s center of projection would reduce distortion in some triangles at the cost of the others, for a net loss.

However, again, picking the best center of projection knowing that the polyhedral face only contains two LCD triangles (135 and 411) is more interesting. The modified triakis icosahedron is equidistant from the NW and SE corners of LCD 135 and by symmetry equidistant from the SW corner of LCD 411. However, unlike in the modified dodecahedral analysis, we can reduce the distance to all three vertices. To do so and maintain equidistance to the SW corner of 411 and the SE corner of 135, we shift the center of projection north.

We continue to improve distortion to all points in both triangles until we hit the origin. If we continue north past the origin, we decrease the distance to C2.7.8 , increasing distance to all other vertices of LCD triangles 411 and 135. Thus, to minimize the maximal distortion, we’d chose a modification of the triakis icosahedron with pyramid heights chosen so that the center of projection was at the SW and SE corners of LCD 135 and 411, respectively. This height can be found using our CPMI formula and solving for an image of 0. Setting

2 3 arccos 6h+ϕ 3 = 0 we solve for h = ≈ 0.110264 (see appendix for calculations.) This ( 3ϕ 1+12h2 ) 6ϕ 2 defines what I’ve been referring to as the optimal triakis icosahedron.

Remaining Questions

One aspect of this project that I found especially enjoyable was the incorporation of the golden ratio into the closed-form representations of various quantities. It is quite enjoyable to work algebraically with this quantity. However, for the sake of time, I frequently opted against expressing quantities in terms of phi, even if they seemed to naturally fit that mold. I imagine software could easily make these conversions, perhaps yielding more aesthetically pleasing quantities throughout the paper.

A natural extension of this work would be to examine similar questions under projections other than the gnomonic. Further we could make the work more applied by considering the projection of an oblate spheroid, or some other shape more closely resembling that of the earth. One could also weigh distortion of landmasses more heavily than distortion of large bodies of water producing differently optimal results and an optimal orientation.

Finally, the polyhedra chosen for this analysis was somewhat arbitrary. A more thorough examination of various polyhedra may be fruitful.

Calculations Proof of “kink” in IGCC

First we note the existence of an IGCC passing through points M 2.7 and M 2.3 below. The centroids through which the axis of rotation that define this IGCC passes, are C1.5.6 and

C8.9.12 .

The claim is that the image of this IGCC passes through the point P2 pictured below, and not P1 , the midpoint of the line segment connecting the images of M 2.7 and M 2.3 .

We first note that if we assume a unit length edge of the image triangles, the distance from

V2 to P1 is ¼. One way to see this is noting that our image triangles are equilateral, so image triangle M 2.7V2P1 measures 30-60-90. The fact that image V2 M 2.7 has length ½ implies that image V2P1 has length ¼. We use this to determine the coordinates of the image of P1 , which we will later apply inverse gnomonic projection to.

The projections of a GP (centered at C2.7.8 ) of V2 and V8 , and the resulting coordinates for the image of P1 are summarized below:

x -coordinate y -coordinate x - coordinate y - coordinate approximation approximation −2 V2 0 2φ 0 0.763932 −2 −2 V8 3φ −φ 0.661585 -0.381966 3 5 P1 2 0.165396 0.477458 4φ 2 4φ

To locate P2 , I first project M 2.7 and M 2.3 using C2.7.8 as the center of projection. It’s worth noting that because M 2.3 is not on the same face as C2.7.8 it is not in the domain of that projection. However, the points M 2.7 and M 2.3 define a unique great circle on the sphere.

The intersection of this great circle and the one defined by V2V8 is the point P2 , so the fact that we later restrict the domain of this projection is not relevant.

The projections of M 2.7 and M 2.3 under C2.7.8 are shown in the table below. The intersection of the lines defined by the images of M 2.7 and M 2.3 and those of V2 and V8

(shown above) is the image of P2 . That image is also shown in the table below:

x -coordinate y -coordinate x - coordinate y - coordinate approximation approximation 3 1 M 2.7 − 2 -0.330792 0.190983 2φ 2 2φ 3 5 M 2.3 * 0.866025 1.11803 2 2

3 1 −5 P2 1+φ 0.126351 0.545085 2φ 4 2 ( )

*Projected by extending the domain of the GP centered at C2.7.8 .

We see that the images of P1 and P2 differ, thus proving the existence of the aforementioned kink. To better analyze the kink, we calculate the coordinates of these points on the sphere by applying the inverse gnomonic projections. The table below summarizes these values:

Approximate Approximate Exact Longitude Exact Latitudeφ λ Longitude Latitude ⎛ 3 4 +φ 2 ⎞ P 2 arcsin ! ! 1 arccot(2 +φ ) ⎜ ⎟ 12.2183 45.7718 ⎝ 2 29 ⎠ 1 2 ⎛ −4 ⎞ ⎛ φ ⎞ ! ! P2 arctan⎜ (1−φ )⎟ arcsin 9.69372 49.0919 ⎝ 5 ⎠ ⎝⎜ 2 3 ⎠⎟

We now calculate angle and verify10 that is a right angle using the P2P1M 2.7 P1P2 M 2.7 spherical law of cosines and the arc lengths shown below: ⎛ 5 19 + 7 5 ⎞ PP = arccos⎜ ( ) ⎟ ≈ 0.0651552 1 2 ⎜ 174 ⎟ ⎝ ⎠ ⎛ ⎞ 1 5 19 + 7 5 P M = arccos⎜ ( ) ⎟ ≈ 0.527262 1 2.7 ⎜ 2 58 ⎟ ⎝ ⎠ π P M = ≈ 0.523599 2 2.7 6 ⎛ 81− 35 5 ⎞ ! ∠P2P1M 2.7 = arccos⎜ ⎟ ≈ 83.5658 ⎝ 218 ⎠

∠PP M = arccos 0 = 90! 1 2 2.7 ( )

10 a result of the fact that the IGCA containing V2 and V8 passes through C1.5.6 and C8.9.12 , which makes it perpendicular to the IGCA containing M 2.7 and M 2.8 Exact Coordinates for M 2.8 Calculations

Figure 22: Midpoint Formulas To calculate the longitude of M 2.8 we apply the midpoint formula on points V and V 2 8 . α φ = 90 − 1 λ = 0 1 2 1

α φ = 0 λ = 1 2 2 2

⎛ cosφ2 sin λ1 − λ2 ⎞ λm = λ1 + arctan⎜ ⎟ ⎝ cosφ1 + cosφ2 cos λ1 − λ2 ⎠

α1 ⎛ cos0sin 0 − 2 ⎞ = 0 + arctan ⎜ α1 α1 ⎟ ⎝ cos(90 − 2 ) + cos0cos 0 − 2 ⎠ ⎛ sin α1 ⎞ = arctan 2 ⎜ α1 α1 ⎟ ⎝ cos(90 − 2 ) + cos 2 ⎠ ⎛ sin α1 ⎞ = arctan 2 ⎜ α1 α1 ⎟ ⎝ sin 2 + cos 2 ⎠ ⎛ ⎛ ⎛ 1 ⎞ ⎞ ⎞ sin arcsin ⎜ ⎜ ⎜ 2 ⎟ ⎟ ⎟ ⎜ ⎝ ⎝ ϕ +1⎠ ⎠ ⎟ = arctan⎜ ⎟ ⎛ ⎛ 1 ⎞ ⎞ ⎛ ⎛ 1 ⎞ ⎞ ⎜ sin arcsin + cos arcsin ⎟ ⎜ ⎜ ⎜ 2 ⎟ ⎟ ⎜ ⎜ 2 ⎟ ⎟ ⎟ ⎝ ⎝ ⎝ ϕ +1⎠ ⎠ ⎝ ⎝ ϕ +1⎠ ⎠ ⎠ ⎛ 1 ⎞ ⎜ ϕ 2 +1 ⎟ = arctan⎜ ⎟ 1 ϕ ⎜ + ⎟ ⎝⎜ ϕ 2 +1 ϕ 2 +1 ⎠⎟ ⎛ 1 ⎞ = arctan ⎝⎜ ϕ +1⎠⎟ ⎛ ϕ ⎞ = arccos⎜ ⎟ ⎝ 3 ⎠

= α 3

Exact Coordinates for C2.7.8 Calculations

To calculate the location of the gnomonic projection of C2.3.8 , we first project C2.7.8 using a projection centered at M 2.8 . We have λ0 = α 3 and φ1 = 30 and project λ = 0 and φ = α 3 using the gnomonic projection formulas:

cosφ sin(λ − λ0 ) cosα 3 sin(−α 3 ) x = = sinφ0 sinφ + cosφ0 cosφ cos(λ − λ0 ) sin 30sinα 3 + cos30cosα 3 cosα 3 ϕ − 3 3ϕ −2 3ϕ −2 3ϕ −2 3ϕ −2 3ϕ = 3 = = = = 1 3 3 ϕ ϕ 3 3 3 3 3 1 2 6 1 6 2 2 3ϕ + 2 3 3 + ϕ + ( + ϕ) ( +ϕ) ϕ −1 = 3ϕ

cosφ0 sinφ − sinφ0 cosφ cos(λ − λ0 ) cos30sinα 3 − sin 30cosα 3 cos(−α 3 ) y = = sinφ0 sinφ + cosφ0 cosφ cos(λ − λ0 ) sin 30sinα 3 + cos30cosα 3 cosα 3

3 3 1 ϕ ϕ 3 −1 2 3ϕ − 2 3 3 3 3 − 3ϕ 3 3 − 3(1+ 2ϕ) 2 3(1−ϕ) − 3ϕ = = = = = 1 3 3 ϕ ϕ 3 3 3 3 3 1 2 6 1 3 2 2 3ϕ + 2 3 3 + ϕ + ( + ϕ) ( +ϕ) ϕ −1 = 3 3ϕ

Now, because our center of projection is the midpoint of C2.3.8 and C2.7.8 , we reverse the ⎛ 1 1 ⎞ signs of the x and y coordinates above to get the projection of C , namely , . 2.3.8 ⎝⎜ 3ϕ 3ϕ 3 ⎠⎟ Finally, we map these back to the sphere via the following inverse gnomonic projection formulas:

2 2 4 2 2 2 1 1 1 1 ϕ +1 2+3ϕ+1 3ϕ 1 ρ = x + y = + 3 = 2 + 6 = 6 = 6 = 6 = 2 ( 3ϕ ) ( 3ϕ ) 3ϕ 3ϕ 3ϕ 3ϕ 3ϕ ϕ

−1 1 3 ⎛ 3 ⎞ ⎛ ysin(arctan ρ)cosφ1 ⎞ ϕ 1 3ϕ 3ϕ 2 φ = arcsin cos arctan ρ sinφ + = arcsin + ⎜ ( ) 1 ⎟ ⎜ 1 ⎟ 3 2 2 ⎝ ρ ⎠ ⎝ ϕ ⎠

⎛ ϕ 3 +1 ⎞ ⎛ 1+ 2ϕ +1 ⎞ ⎛ 2ϕ 2 ⎞ = arcsin = arcsin = arcsin ⎜ 2 3 2 ⎟ ⎜ 2 3 1 ⎟ ⎜ 2 3 2 ⎟ ⎝ ϕ ⎠ ⎝ ( +ϕ)⎠ ⎝ ϕ ⎠ ⎛ 1 ⎞ = arcsin⎜ ⎟ ⎝ 1 3 ⎠

⎛ 1 1 ⎞ ⎛ xsin arctan ρ ⎞ 3φ 3φ arctan ( ) = α + arctan λ = λ0 + ⎜ ⎟ 3 ⎜ 1 3 φ 1 1 1 ⎟ ⎝ ρ cosφ cos arctan ρ − ysinφ sin arctan ρ ⎠ 2 2 − 3 2 1 ( ) 1 ( ) ⎝ ϕ 3 3φ 3φ ⎠ ⎛ 2ϕ 2 ⎞ ⎛ 1+ϕ ⎞ ⎛ 1+ϕ ⎞ = α + arctan = α + arctan = α + arctan 3 ⎝⎜ 3ϕ 3 −1⎠⎟ 3 ⎝⎜ 1+ 3ϕ ⎠⎟ 3 ⎝⎜ 1+ϕ 4 − 2⎠⎟

⎛ ϕ +1 ⎞ ⎛ 1 ⎞ = α 3 + arctan⎜ 2 ⎟ = α 3 + arctan⎜ 2 ⎟ ϕ +1 ϕ −1 ϕ +1 ϕ −1 ϕ +1 ⎝ ( )( )( )⎠ ⎝ ( )( )⎠ ⎛ ϕ ⎞ ! = α 3 + arctan⎜ 2 ⎟ ⎝ ϕ +1⎠

ri ϕ ϕ Letting t = α = arccos = arccos and s = arctan 2 we draw the triangles in figure 3 ( rm ) ( 3 ) ( ϕ +1 ) 57 for reference. Figure 57: Reference Triangles 2 2 x2 = ( 3(ϕ 2 +1)) − (ϕ (ϕ 2 +1)) 2 2 = 3(ϕ + 2) −ϕ 2 (ϕ + 2) 2 = (3−ϕ 2 )(ϕ + 2) 2 = (2 −ϕ)(ϕ + 2) = (4 −ϕ 2 )(ϕ + 2) = (3−ϕ)(ϕ + 2) = −ϕ 2 +ϕ + 6 = 5 , so x = 5

ϕ 2 +1 ϕ ϕ 5 cos(s + t) = cosscost − sin ssint = − 6ϕ 3 6ϕ 3(ϕ 2 +1) 2 2 ϕ 2 +1 5 (ϕ +1) − 5 5ϕ + 5 − 5 = − = = 3 2 3 2 (ϕ 2 +1) 3 2 (ϕ 2 +1) 3 2 (ϕ + 2) ⎛ 1+ 5 ⎞ 5 + 5 − 5 ⎝⎜ 2 ⎠⎟ 15 + 3 5 1 = = = ⎛ 1+ 5 ⎞ 3 2 5 + 5 2 3 2 + 2 ( ) ⎝⎜ 2 ⎠⎟ So s + t = arccos 1 = 45! . ( 2 )

Pentakis Dodecahedron Center of Projection Calculations

Using h 1 1 65 22 5 to evaluate the CPMD function, and noticing that = 19 5 ( + ) 2 2 2 , we find the longitude of the center of projection for the relevant 4h 5 = ( 19 ) (22 +13 5 ) face of the Pentakis Dodecahedron as:

⎛ ⎞ 2 5 1 1 ⎜ 19 22 +13 5 + ϕ ⎟ f 19 5 (65 + 22 5 ) = arccos ( ) ⎜ 1+ 5 1+ 5 2 2 ⎟ ⎜ 2 + 2 2 2 +1+ ( 19 ) 22 +13 5 ⎟ ⎝ ( )( ( ) ( )) ⎠ ⎛ ⎞ 2 22 +13 5 + ϕ 5 = arccos⎜ 19 ⎟ ⎜ 1 2 2 2 2 ⎟ 2 5 + 5 2 ⋅19 +19 5 + 2 ⋅22 + 2 ⋅13 5 ⎝ 2⋅19 ( )( ) ⎠ ⎛ ⎞ 2 22 +13 5 + ϕ 5 = arccos⎜ 19 ⎟ ⎜ 1 ⎟ 2 5 + 5 810 + 413 5 ⎝ 2⋅19 ( )( ) ⎠ ⎛ ⎞ 2 22 +13 5 + ϕ 5 = arccos⎜ 19 ⎟ ⎜ 1 1 ⎟ 19 2 6115 + 2875 5 ⎝ ( ) ⎠ ⎛ ⎞ 2 22 +13 5 +19 ϕ 5 = arccos⎜ ⎟ ⎜ 5 ⎟ 2 1223+ 575 5 ⎝ ( ) ⎠ ! ≈ 11.64072

Modified Pentakis Dodecahedron Center of Projection Calculations

3ϕ ϕ 2 2 26+14 5−(7+3 5 ) 30−6 5 Using − as our value for h in the CPMD, we see that 4h 5 = , 2 2 3−ϕ 2 and evaluate the CPMD function as:

⎛ ⎞ 2 26+14 5−(7+3 5 ) 30−6 5 5 ⎛ 3ϕ ϕ ⎞ ⎜ 2 + ϕ ⎟ f ⎜ − ⎟ = arccos⎜ ⎟ ⎝ 2 2 3−ϕ ⎠ ⎜ ⎛ 26+14 5−(7+3 5 ) 30−6 5 ⎞ ⎟ (ϕ + 2) 2ϕ +1+ 2 ⎝⎜ ⎝ ⎠ ⎠⎟ ⎛ ⎞ 26+14 5−(7+3 5 ) 30−6 5 ⎜ + 11+ 5 ⎟ = arccos⎜ 2 2 ⎟ 26+14 5−(7+3 5 ) 30−6 5 ⎜ 5+ 5 ⎛ 2 + 5 + ⎞ ⎟ ⎜ ( 2 )⎝ 2 ⎠ ⎟ ⎝ ⎠ ⎛ ⎞ 52 + 28 5 − (14 + 6 5 ) 30 − 6 5 + 22 +10 5 = arccos⎜ ⎟ ⎜ ⎟ ⎜ 230 +110 5 − 50 + 22 5 30 − 6 5 ⎟ ⎝ ( ) ⎠ ! ≈ 9.02699

Equilateral Pentakis Dodecahedron Center of Projection Calculations

Using h 5− 5 and noticing that 4h2 5 4 5−1 4 1 , we evaluate the CPMD = 10 = ( 2 ) = (ϕ − ) function as:

⎛ 5 − 5 ⎞ ⎛ 4(ϕ −1) + ϕ 5 ⎞ f ⎜ ⎟ = arccos⎜ ⎟ 10 ⎝ ⎠ ⎝⎜ (ϕ + 2)(2ϕ +1+ 4(ϕ −1)) ⎠⎟

⎛ 4(ϕ −1) + ϕ 5 ⎞ = arccos ⎜ 2 ⎟ ⎝ 6ϕ + 9ϕ − 6 ⎠

⎛ 4(ϕ −1) + ϕ 5 ⎞ = arccos⎜ ⎟ ⎝ 15ϕ ⎠ ! ≈ 5.65989

However, because the distance from V8 ' to P' is greater than the distance from V8 'to M 7.8 , ⎛ 5 ⎞ 4(ϕ −1) + ϕ ! the solution is actually − arccos⎜ ⎟ ≈ −5.65989 . ⎝ 15ϕ ⎠

Triakis Icosahedron Center of Projection Calculations

A unit length edge triakis icosahedron is known to have a pyramid height of h 1 , = 3(5ϕ+2) which we can input into out CPMI formula to yield:

⎛ 6 1 +ϕ 2 3 ⎞ 1 ( 3(5ϕ+2) ) g 3 5ϕ+2 = arccos⎜ 2 ⎟ ( ( ) ) 1 ⎜ 3ϕ 1+12 ⎟ ⎝ ( 3(5ϕ+2) ) ⎠

Simplifying the numerator and denominator of the arc cosine argument separately for aesthetic value, we see:

⎛ ⎞ 2 2 6 ⎜ 1 ⎟ + ϕ 3 3ϕ 1+12 1 ⎝ 3(5ϕ+2)⎠ ( 3(5ϕ+2) ) ϕ 2 3 3 5ϕ+2 6 ( ) 12 = + = 3ϕ 1+ 2 3(5ϕ+2) 3(5ϕ+2) 3(5ϕ+2) 2 2 6+3ϕ (5ϕ+2) (5ϕ+2) +4 = = 3ϕ 2 3(5ϕ+2) (5ϕ+2) 2 = 6+3(ϕ+1)(5ϕ+2) = 3ϕ 25ϕ +20ϕ+8 3(5ϕ+2) 5ϕ+2 2 = 15ϕ +21ϕ+12 3(5ϕ+2) = 36ϕ+27 3(5ϕ+2) 9(4ϕ+3) = 3(5ϕ+2)

⎛ 9(4ϕ+3) ⎞ 1 3(5ϕ+2) so g = arccos⎜ 2 ⎟ ( 3(5ϕ+2) ) ⎜ 3ϕ 25ϕ +20ϕ+8 ⎟ ⎝ 5ϕ+2 ⎠

9(4ϕ+3) = arccos 2 ( 3ϕ 3 25ϕ +20ϕ+8 ) 3(4ϕ+3) = arccos 2 ( ϕ 75ϕ +60ϕ+24 ) = arccos 3(4ϕ+3) ( ϕ 135ϕ+99 ) = arccos 4ϕ+3 ( ϕ 15ϕ+11 ) ϕ 3+2ϕ 2 = arccos 2 ( ϕ 4ϕ +11ϕ+7 ) = arccos ϕ(ϕ+2) ( (4ϕ+7)(ϕ+1) ) ϕ(ϕ+2) = arccos 2 ( ϕ (4ϕ+7) ) = arccos ϕ+2 ≈ 9.69372! ( 4ϕ+7 )

Modified Triakis Icosahedron Center of Projection Calculations

The height of the triangular pyramid caps of the modified triakis icosahedron are:

2 2 4 2 1 ϕ 3ϕ 5 −ϕ ϕ +4 −ϕ h = 2 ϕ 5 − 2 3 = 2 3 = 2 3 .

We use the above pyramid height and our CPMI formula below:

⎛ 4 4 2 ⎞ ⎛ 6 ϕ + −ϕ +ϕ 2 3 ⎞ ϕ 4 +4 −ϕ 2 ⎜ ⎝⎜ 2 3 ⎠⎟ ⎟ g 2 3 = arccos 2 ( ) ⎛ ϕ4+4−ϕ2 ⎞ ⎜ 3ϕ 1+12 ⎟ ⎜ 2 3 ⎟ ⎝ ⎝ ⎠ ⎠ I’ll simplify the numerator and denominator of the arccos argument separately:

2 ⎛ ϕ4+4−ϕ2 ⎞ 2 ⎛ 4 4 2 ⎞ 6 ⎜ ⎟ + ϕ 3 3ϕ 1+12 ϕ + −ϕ ⎝ 2 3 ⎠ ⎝⎜ 2 3 ⎠⎟

3 ϕ 4 +4 −3ϕ 2 3ϕ 2 2 + = 3ϕ 1+ ϕ 4 +4 −ϕ 2 = 3 3 ( )

3 ϕ 4 +4 = = 3ϕ 1+ϕ 4 +4−2ϕ 2 ϕ 4 +4 +ϕ 4 3 3 3ϕ+6 = = 3ϕ 2ϕ 4 +5−2ϕ 2 ϕ 4 +4 3 = 3 ϕ + 2

4 2 so g ϕ +4 −ϕ arccos⎛ 3 ϕ+2 ⎞ = ⎜ 4 2 4 ⎟ ( 2 3 ) ⎝ 3ϕ 2ϕ +5−2ϕ ϕ +4 ⎠ = arccos ϕ −1 ϕ+2 ( 2ϕ 4 +5−2ϕ 2 ϕ 4 +4 ) −1 ϕ+2 ! = arccos ϕ 6 9 2 2 3 6 ≈ 13.17408 ( ϕ+ − ϕ ϕ+ )

However, because the distance from C '2.7.8 to P' is greater than the distance from C '2.7.8 to

M 7.8 , the solution actually:

−1 ϕ+2 ! 2 . − arccos ϕ 6 9 2 3 6 ≈ −13.17408 ( ϕ+ − ϕ ϕ+ )

Equilateral Triakis Icosahedron Center of Projection Calculations

The height of the triangular pyramid caps of the equilateral triakis icosahedron are: 6 h = 3 .

We use the above pyramid height and our CPMI formula below:

⎛ 6 6 +ϕ 2 3 ⎞ 6 ( 3 ) g 3 = arccos 2 ⎜ 6 ⎟ ( ) 3ϕ 1+12 ⎝ ( 3 ) ⎠ 2 = arccos 2 6+ϕ 3 ( 3ϕ 1+8 ) = arccos 2 6+(1+ϕ) 3 ( 9ϕ ) = arccos 4 6+3 3+ 15 ≈ 49.62362! ( 9+9 5 )

But again, because the distance from C '2.7.8 to P' is greater than the distance from C '2.7.8 to

M 7.8 , the solution is actually:

− arccos 4 6+3 3+ 15 ≈ −49.62362! ( 9+9 5 )

Height of Pyramid Caps for Optimal Triakis Icosahedron

2 0 = arccos 6h+ϕ 3 ( 3ϕ 1+12h2 ) 6h+ϕ 2 3 1 = 2 3ϕ 1+12h 3ϕ 1+12h2 = 6h +ϕ 2 3

9ϕ 2 (1+12h2 ) = 36h2 +12ϕ 2 3h + 3ϕ 4

3ϕ 2 + 36ϕ 2h2 = 12h2 + 4ϕ 2 3h +ϕ 4

(36ϕ 2 −12)h2 − 4ϕ 2 3h + 3ϕ 2 −ϕ 4 = 0

4ϕ 2 3 ± 48ϕ 4 − 4(36ϕ 2 −12)(3ϕ 2 −ϕ 4 ) h = 72ϕ 2 − 24 ϕ 2 3 ± 3ϕ 4 − (9ϕ 2 − 3)(3ϕ 2 −ϕ 4 ) = 18ϕ 2 − 6 ϕ 2 3 ± 3ϕ 4 + 9ϕ 6 − 30ϕ 4 + 9ϕ 2 = 18ϕ 2 − 6 ϕ 2 3 ± 3ϕ ϕ 4 − 3ϕ 2 +1 = 18ϕ 2 − 6 ϕ 2 3 ± 3ϕ 3ϕ + 2 − 3ϕ − 3+1 = 18ϕ 2 − 6 ϕ 2 3 = 18ϕ 2 − 6 ϕ 2 3 = 6(3ϕ 2 −1) ϕ 2 3 = 6(3ϕ + 2) ϕ 2 3 = 6ϕ 4 3 = 2 ≈ 0.110264 6ϕ

References:

http://www.rwgrayprojects.com/ http://www.movable-type.co.uk/scripts/latlong.html http://mathworld.wolfram.com/GnomonicProjection.html http://www.synearth.net/afullerex/127.htm http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/

Image Credits:

Figure 1: en.wikipedia.org/wiki/Gnomonic_projection#mediaviewer/File:Gnomonic.png Figure 2: apollonius.math.nthu.edu.tw/d1/dg-07-exe/943209/abcd/ Figures 3 through 6: mathworld.wolfram.com/Cumulation.html

All other images created from scratch.