# Stereographic Projection

Stereographic Projection of a Sphere
Field: Geometry
Image Created By: Paul Nylander
Website: bugman123.com

Stereographic Projection of a Sphere

Stereographic projection maps each point on a sphere onto a plane.

# Basic Description

Figure 1
An example of a stereographic projection. Two points, P1 in the upper hemisphere and P2 in the lower hemisphere, are projected onto the x-y plane.

Stereographic projection is a map from the surface of a sphere to a plane.

A map, generally speaking, establishes a correspondence between a point in one space and a point in another space. In other words, a map is a pattern that brings us from one space to another (in this case, the two spaces are a sphere and a plane). The process for mapping to the plane, in this case, is to draw a line from the North Pole of the sphere; the line will pass through both a point on the sphere and a point on the plane. The point on the sphere is mapped to the point on the plane. The image to the left shows this process for two points.

The main image shows this process more concretely. A sphere with a spiral-like mesh is placed on a surface, and a light source is placed at the point furthest from the surface. The shadows represent the stereographic projection of the mesh onto the surface. This is a useful way to think about stereographic projection.

The following applet demonstrates how a sphere is projected onto a plane. A sphere with bands of color is stereographically projected onto a plane in the background. Rotating the sphere with the mouse will change the orientation of the colors on the sphere relative to the north pole, thereby changing the projection on the plane. The sphere and north pole remain fixed; only the colors are shifted.

ERROR: Unable to find Java Applet file: StereoProj.class.

# A More Mathematical Explanation

## Definition

A stereographic projection maps the points of a sphere onto a plane.

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## Definition

A stereographic projection maps the points of a sphere onto a plane.

Figure 2
Cross section of a sphere. Two arbitrary points on the sphere are mapped to the plane.

Specifically,

Let S be the centered at the origin; that is, the set of all points (x, y, z) that satisfy the equation $x^2 + y^2 + z^2 = 1$.
Let T be the north pole of the sphere, the point (0, 0, 1), and let T ' be the of T, the point (0, 0, -1).
Let H be the x-y plane, the horizontal plane that passes through the origin and the equator of the sphere; that is, the set of all points (x, y, z) that satisfy the equation z = 0.
A line drawn from T through any point P on the sphere S will also intersect the plane H at a unique point which we call Q. We say that Q is the stereographic projection of P, or alternatively that P is mapped to Q by stereographic projection.

Figure 2 to the right demonstrates how points are stereographically projected. Any line passing through T and P will also intersect the plane. In one sense, the figure is an example of the stereographic projection of the two-dimensional unit circle onto the x axis. The rest of this page will examine the three-dimensional case of stereographic projection of the unit sphere onto the x-y plane, although we will find that reference to the two-dimensional case is often useful, because such a cross section can be considered, by symmetry of the sphere, for the projection of any point.

Q, as the stereographic projection of P, will at times be referred to as the projected point. T could also be called the projection point, or the point from which we are projecting, and can be any point on the sphere. H, likewise, can be any plane that does not contain T and is perpendicular to the axis passing through T (so it does not have to be the x-y plane, and depending on T does not even need to be horizontal). Throughout the page, however, we will derive formulas using the conventions established earlier in this section.

Since T ' always projects to the origin on the plane, it may be called the center of the map. This phrasing will be particularly useful in discussing cartography: a map is projected from T and projected around or centered on T '. Don't confuse the center of the map with the projection point as we have defined!

Our goal in the following two sections is to derive the formulas that map P to Q; we will use the coordinates of a point on the sphere to find the coordinates of a point on the plane, and then we will find a formula to reverse the process. First we will do so given a point in rectangular coordinates; then we will do so given a point in spherical coordinates.

## Rectangular Coordinates

Using rectangular coordinates, we will refer generally to point P on the sphere as (x, y, z) and point Q on the plane as (X, Y, 0). Note that for the coordinates of Q we have used big X and big Y, which are to be distinguished from the coordinates of P. By writing X and Y in terms of x, y, and z, we will have derived a formula for projecting S onto H.

Mapping of S to H in rectangular coordinates:

$P = (x, y, z) \text{ maps to } Q = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right)$.
That is,
Eq. 1         $(X, Y, 0) = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right)$.

Derivation of Eq. 1.

Let's restate what is happening by returning to our definition of the sphere. The sphere's pole is at the point T (0, 0, 1). A line can be drawn through some point P (x, y, z) on the sphere and some point Q (X, Y, 0) on the plane. We consider the vectors drawn from T to P and from T to Q. By construction, these two vectors are colinear and parallel:
$\overrightarrow{TP} \parallel \overrightarrow{TQ}$.
Since the vectors are parallel, their cross product is the 0 vector.
$\overrightarrow{TP} \times \overrightarrow{TQ} = \overrightarrow{0}$
$\langle - x, - y, 1 - z \rangle \times \langle - X, - Y, 1 \rangle = \langle x, y, z-1 \rangle \times \langle X, Y, - 1 \rangle = \langle 0, 0, 0 \rangle$
$\langle -y - Y(z-1), X(z-1) + x, xY - yX \rangle = \langle 0, 0, 0 \rangle$
We get three equations:
$x+X(z-1) = 0 \Longrightarrow X = \frac{x}{1-z}$
$y+Y(z-1) = 0 \Longrightarrow Y = \frac{y}{1-z}$
$xY - yX = 0 \Longrightarrow \frac{xy}{1-z} - \frac{yx}{1-z} = 0$
These are the coordinates in Eq. 1.
$\blacksquare$

Since points on the sphere are mapped uniquely, or one-to-one, to points on the plane, Eq. 1 should be . In other terms, we have found a mapping from S to H and we now want to find a mapping from H back to S.

Mapping of H to S in rectangular coordinates:

$Q = (X, Y, 0) \text{ maps to } P = \left(\frac{2 X}{X^2 + Y^2 + 1}, \frac{2 Y}{X^2 + Y^2 + 1}, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}\right)$.
That is,
Eq. 2         $(x, y, z) = \left(\frac{2 X}{X^2 + Y^2 + 1}, \frac{2 Y}{X^2 + Y^2 + 1}, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}\right)$.

Derivation of Eq. 2.

This inverse function is derived by substituting the coordinates from Eq. 1 back into the equation for the unit sphere and solving for z.
$x^2 + y^2 + z^2 = 1$
$(X(1-z))^2 + (Y(1-z))^2 + z^2 = 1$ (substitute)
$(X^2 + Y^2)(z^2 - 2z + 1) + z^2 - 1 = 0$ (factor and expand)
$(X^2 + Y^2 + 1)z^2 - 2(X^2 + Y^2)z + (X^2 + Y^2 - 1) = 0$ (distribute and factor)
This latter equation is is a quadratic in z, so we can solve for z using the quadratic formula, obtaining:
$z = 1, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}$
The first solution may be discarded because T = (0, 0, 1) is the pole of the circle; it is the only point for which stereographic projection is undefined. (Think about why it would not make sense to map T onto the plane. We would have to draw a line from T to T, but no line tangent to the sphere at T will ever pass through the x-y plane.) The second solution is what we expected.
An alternate route to getting this result is to note that the line drawn from T to Q intersects the sphere at two points: at T and P! We know that one result will be z1 = 1, and we must find the other. If we divide the previous quadratic by the coefficient of the second-order term, we get:
$z^2 - \frac{2(X^2 + Y^2)}{X^2 + Y^2 + 1}z + \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1} = 0$
The constant term c in a quadratic of the form x2 + bx + c = 0 must be equal to the product of the roots. As noted, z1 = 1 is one solution, so
$z_1 \cdot z_2 = c$
$1 \cdot z_2 = \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}$
$z_2 = \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}$
This second solution is, again, what we expected for the coordinate of z.

We can use this formula for z to find formulas for x and y. From Eq. 1 we know:
$X = \frac{x}{1-z}$
$x = X(1-z)$.
Substituting, we get:
\begin{align} x &= X \left(1 - \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1} \right) \\ & = X \left( \frac{X^2 + Y^2 + 1 - (X^2 + Y^2 - 1)}{X^2 + Y^2 + 1} \right)\\ & = \frac{2X}{X^2 + Y^2 + 1} \end{align}
Since x and y are interchangeable for the purpose of these formulas, the same may be repeated for y to obtain:
$y = \frac{2Y}{X^2 + Y^2 + 1}$
These are the coordinates in Eq. 2.
$\blacksquare$

## Spherical Coordinates

Figure 3
The angles used in the spherical coordinate system. θ is the azimuth, and ϕ is the zenith.

As established, a point P on the sphere has rectangular coordinates (x, y, z). We will introduce the spherical coordinate system, which describes the position of a point on a sphere by its radius, azimuth, and zenith.

These are strange terms, although they can be made familiar in relation to geography. The azimuth (θ) is longitude, the angle measurement between the positive x axis and the point's bearing in the "eastern" direction. The zenith (ϕ) is , the angle measurement between the positive z axis and the point's bearing in the "southern" direction. In addition, the following restrictions on θ and ϕ will be sufficient to describe the position of any point on the sphere:

$0^{\circ} \leq \theta < 360^{\circ}$ and $0^{\circ} \leq \phi \leq 180^{\circ}$.

So any point can be written in the following form of spherical coordinates:

$P = (1, \theta, \phi)$

The above will be our general formulation of a point P on S in spherical coordinates.

The polar coordinate system, in which the position of a point is described by its radius and angle, is the two-dimensional analog of spherical coordinates. The radius is the positive distance of the point from the origin, and the angle is measured from the positive x axis.

$Q = (R, \Theta)$

The above will be our general formulation of a point Q on H in polar coordinates.

We now know all that we need to know to reformulate Eq. 1 in terms of spherical and polar coordinates.

Mapping of S to H in spherical coordinates:

$P = (1, \theta, \phi) \text{ maps to } Q = \left( \frac{\sin \phi}{1 - \cos \phi}, \theta \right)$.
That is,
Eq. 3        $(R, \Theta) = \left( {\frac{\sin \phi}{1 - \cos \phi}}, \theta \right)$.

R, the distance of a projected point from the origin, depends entirely on $\phi$, while the azimuthal angle $\theta$ is preserved by the projection.

Derivation of Eq. 3.

Figure 4
A two-dimensional cross section of a sphere. Such a cross section could be drawn for a point at any θ.

We begin by observing that the azimuth $\theta$ is the same as the polar angle $\Theta$. In other words, the longitude of any point P on the sphere is preserved when the point is mapped onto the plane.
This should be unsurprising if we visualize the projection: T, P, and Q all lie in a vertical plane. As such, we just need to find an expression for R, the distance from the origin to the projected point Q.
Figure 4 shows a cross section of the sphere. We see that R is dependent on ϕ, the zenith.
The z coordinate is labeled cos(ϕ) in this image. We see that there are two similar triangles. The larger one has legs of length R and 1; the smaller has legs of lengths sin(ϕ) and 1 - cos(ϕ), respectively. As such, we can set up the ratio:
${R \over 1} = {\sin \phi \over 1 - \cos \phi}$.
These are the coordinates in Eq. 3.
$\blacksquare$

The inverse of stereographic projection can be formulated in terms of spherical coordinates as well. In this case, we begin with Q in polar coordinates on the plane H and find the spherical coordinates of P on the sphere S.

Mapping of H to S in spherical coordinates:

$Q = (R, \Theta) \text{ maps to } P = \left(1, 2 \arctan {1 \over R}, \Theta \right)$.
That is,
Eq. 4         $(1, \phi, \theta) = \left(1, 2 \arctan {1 \over R}, \Theta \right)$.

Derivation of Eq. 4.

Using trigonometric identities, we can rearrange $R = \frac{\sin \phi}{1 - \cos \ \phi}$ in terms of $\phi$.
We will make use of the trigonometric identity:
$\frac{1 - \cos x}{\sin x} = \tan {x \over 2}$
The right-hand side of Eq. 3 is the reciprocal of this identity, so
$R = \frac{\sin \phi}{1 - \cos \phi} = \frac{1}{\tan {\phi \over 2}}$.
So:
$\tan {\phi \over 2} = {1 \over R}$
${\phi \over 2} = \arctan {1 \over R}$
$\phi = 2 \arctan {1 \over R}$.
Again, we complete our derivation by noting that the radius of the spherical point must be 1, and that $\theta = \Theta$.
$\blacksquare$

## Properties

Property 1

Points on the upper hemisphere (z > 0) of the sphere are mapped outside of the unit circle on the x-y plane (X 2 + Y 2 > 1). Points on the lower hemisphere (z < 0) of the sphere are mapped inside of the unit circle on the x-y plane (X 2 + Y 2 < 1). Points on the equator (z = 0) of the sphere are trivially mapped to the unit circle on the x-y plane (X 2 + Y 2 = 1).

The equation for a circle is R 2 = X 2 + Y 2. We will use the coordinates in Eq. 1 to write the right-hand side of this equation in terms of z. By examining that expression, we will be able to determine whether R is greater than or less than 1.
\begin{align} X^2 + Y^2 & = \frac{x^2}{(1 - z)^2} + \frac{y^2}{(1 - z)^2} = \frac{x^2 + y^2}{(1 - z)^2} \\ & = \frac{x^2 + y^2}{(1 - z)^2} + \frac{z^2}{(1 - z)^2} - \frac{z^2}{(1 - z)^2} \\ & = \frac{x^2 + y^2 + z^2}{(1 - z)^2} - \frac{z^2}{(1 - z)^2} = \frac{1 - z^2}{(1 - z)^2} \\ & = \frac{1 + z}{1 - z} \\ \end{align}
For positive z, the numerator is greater than the denominator, so R > 1 and the projection falls outside of the unit circle. For negative z, the denominator is greater than the numerator, so R < 1 and the projection falls inside of the unit circle. For z = 0, the fraction evaluates to 1, so R = 1 and the projection is on the unit circle.
This completes the proof of Property 1.
$\blacksquare$

Figure 5
A circle on the surface of the sphere is projected to a circle on the plane.

Property 2
The stereographic projection preserves circles. We distinguish between two possible cases:
• Case #1: The circle on the sphere contains the north pole T. Then the stereographic projection of the circle onto the x-y plane is a line.
• Case #2: The circle on the sphere does not contain T. Then the projection of the circle onto the x-y plane is a circle.

Here we will prove this property analytically.

Recall our definition of the unit sphere:
$S = \{ (x, y, z) \text{ } | \text{ } x^2 + y^2 + z^2 = 1 \}$.
The above is written in set notation, which is useful for describing spaces. The vertical bar means such that, so the above reads, "S is the set of all points (x, y, z) such that x2 + y2 + z2 = 1."
In general, a plane in $\mathbb{R}^3$ is defined as follows:
$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$.

Figure 6
The intersection of a sphere with a plane.

Any circle on the surface of a sphere is the intersection of a plane with the sphere. See, for example, Figure 6. So the set of all points comprising a circle on the unit sphere is given by:
Eq. 5         $W = S \bigcap V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0 \text{ and } x^2 + y^2 + z^2 = 1 \}$.
Note that we are only interested in cases in which the plane intersects the sphere at more than one point. The other planes of this form either do not intersect the unit sphere or are tangent to the unit sphere, in which cases $W$ does not contain the points of a circle.
The strategy of this proof is as follows: We want to show that the stereographic projection of W is a line on the plane when W includes T and a circle on the plane when W does not include T. Note that $0A + 0B + 1C + D = 0$ if and only if $C = -D$. Therefore, if $C = - D$,   then $T \in W$ (Case #1). If $C \neq -D$,   then $T \notin W$ (Case #2). We now have the criteria for examining the two cases of Property 2. We can distinguish between the two cases by looking at the constants C and D, which along with A and B determine which circle on the sphere we are referring to. Our goal will be to find an equation for the projection that allows us to examine these two cases.
Any point $P = (x, y, z) \in W$ is on the sphere (and the circle on the sphere) and so can be written in terms of its corresponding point on the plane, Q = (X, Y, 0). We do this by substituting Eq. 2 into Eq. 5:
$Ax + By + Cz + D = 0$
$A \frac{2X}{X^2 + Y^2 + 1} + B \frac{2Y}{X^2 + Y^2 + 1} + C \frac{X^2 + Y^2 -1}{X^2 + Y^2 + 1} + D = 0$
Multiply by $(X^2 + Y^2 + 1)$:
$2AX + 2BY + C (X^2 + Y^2 - 1) + D (X^2 + Y^2 + 1) = 0$
Distribute:
$2AX + 2BY + (C + D)X^2 + (C + D)Y^2 - C + D = 0$
Factor and rearrange to obtain:
Eq. 6         $(C + D)(X^2 + Y^2) + 2AX + 2BY - C + D = 0$

Figure 7
Stereographic projection of circles on a sphere. Those passing through the north pole are projected to lines; those not passing through the north pole are projected to circles.

© The Mathematical Association of America

Eq. 6 is written in terms of X and Y (the coordinates of the projected point $Q \in H$) and A, B, C, and D (the coefficients that determine the plane V). In other words, by examining Eq. 6, we can see what patterns emerge in the map as a result of the equation for the plane V.
As previously mentioned, when $C = -D$,   $T \in W$ (refer to Eq. 5 to confirm this). Since our goal in this proof is to distinguish between the projections of circles on the sphere that do and do not contain T, we just need to look at Eq. 6 with respect to C and D:
• Case #1: When $C = - D$, it is true that $T \in W$, and the coefficients of $X^2$ and $Y^2$ in Eq. 6 are 0, so Eq. 6 is the formula for a line in H. Therefore, the stereographic projection of circles on the sphere which contain T, the projection point, are lines.
• Case #2: When $C \neq - D$, it is true that $T \notin W$, and the coefficients of $X^2$ and $Y^2$ in Eq. 6 are the same and nonzero, so Eq. 6 is the formula for a circle in H. Therefore, the stereographic projection of circles on the sphere which do not contain T, the projection point, are circles.
This completes the proof of Property 2.
$\blacksquare$
Figure 7 is an illustration of this result. The circles which pass through the north pole are projected to lines on the plane. The circles which do not pass through the north pole are projected to circles on the plane.

In each of these cases, it is possible to manipulate Eq. 6, the equation of the projected points, further. Keep in mind that A, B, C, and D are coefficients that determine the plane V that intersects the sphere S. Therefore, the coefficients A, B, C, and D merely describe the circle W that is being projected. Based on the equation of that circle, we can determine what the projection looks like. We will do this for both Case #1 and Case #2.

In the former case where $C = -D$, we can simplify:
$(C + D)(X^2 + Y^2) + 2AX + 2BY -C + D = 0$
$0(X^2 + Y^2) + 2AX + 2BY -C -C = 0$
$2AX + 2BY = 2C$
$AX + BY = C$ (line in standard form)
$Y = - \frac{A}{B} X + \frac{C}{B}$ (line in slope-intercept form)
Note the impressive simplicity of these results!
So when $C = -D$, the projection is a line with slope
$- \frac{A}{B}$
and y-intercept
$\frac{C}{B} = -\frac{D}{B}$.

In the latter case where $C \neq - D$, we can divide Eq. 6 through by $(C + D)$:
$X^2 + Y^2 + \frac{2A}{C + D}X + \frac{2B}{C + D}Y + \frac{-C + D}{C + D} = 0$
$X^2 + \frac{2A}{C + D}X + Y^2 + \frac{2B}{C + D}Y = \frac{C - D}{C + D}$
By completing the square, this becomes
$X^2 + \frac{2A}{C + D}X + \left( \frac{A}{C + D} \right)^2 + Y^2 + \frac{2B}{C + D}Y + \left( \frac{B}{C + D} \right)^2 = \frac{C - D}{C + D} + \left( \frac{A}{C + D} \right)^2 + \left( \frac{B}{C + D} \right)^2$
$\left( X + \frac{A}{C + D} \right)^2 + \left( Y + \frac{B}{C + D} \right)^2 = \frac{A^2 + B^2 + C^2 - D^2}{(C + D)^2}$
Therefore, the stereographic projection of $W = S \bigcap V$ onto H is a circle centered at
$\left( - \frac{A}{C + D}, -\frac{B}{C + D}\right)$
$\frac{\sqrt{A^2 + B^2 + C^2 - D^2}}{C + D}$.

Property 3
Maps onto planes other than the x-y are scaled in proportion to their distance from T.

In defining stereographic projection, we elected to use the convention that we were projecting onto the x-y plane, the set of all points (x, y, z) such that z = 0, which we called H:

$H = \{ (x, y, z) \text{ } | \text{ } z = 0 \}$.

Observant readers might have noticed that, although stereographic projection was defined as projection onto the x-y plane, some figures used so far project onto other planes that do not intersect the unit sphere at its equator. In particular, Figure 5 and Figure 7 in the proof of Property 2 have not yet been rigorously justified. Are these still examples of stereographic projection, or should we expect them to behave differently because they project onto a different plane? The proof of this property will justify such representations and show that the relationships are still valid.

Figure 8
Cross section of a sphere in which a point is stereographically projected onto three different planes.

As alluded to when H was defined, it is possible to project onto any plane perpendicular to the axis that passes through T, the projection point (so long that the plane does not contain T). In other terms, we can define H more broadly:
$H = \{ (x, y, z) \text{ } | \text{ } z = k, \text{ where } k \in \mathbb{R} \text{ and } k \neq 1 \}$.
The line from T to P will still intersect H at a unique point Q, so the mapping holds and is one-to-one.
This property is easy to demonstrate using basic trigonometry. Consider the two-dimensional cross section of a sphere in Figure 8 to the right. As was true in the proof of Eq. 3, by symmetry we are justified in taking such a cross section since the distance of a point from the origin is independent of the azimuthal angle.
$\angle ATQ_1$ is shared by $\bigtriangleup ATQ_1$, $\bigtriangleup BTQ_2$, and $\bigtriangleup CTQ_3$, and the angles formed with the vertical axis by the planes H1, H2, and H3 are, by construction, right and therefore congruent. Therefore $\bigtriangleup ATQ_1$, $\bigtriangleup BTQ_2$, and $\bigtriangleup CTQ_3$ are similar.
By triangle similarity,
$\frac{ \overline{AT} }{ \overline{AQ_1} } : \frac{ \overline{BT} }{ \overline{BQ_2} } : \frac{ \overline{CT} }{ \overline{CQ_3} } : \frac{1-k}{R}$.
In other words, 1 - k, the distance between T and the plane H, is proportional to R, the distance of the projected point Q from the origin.
This completes the proof.
$\blacksquare$

Note that this does effect some properties of stereographic projection. While circles are still projected to circles as Property 2 demonstrates, it is no longer true that the equator of the unit sphere maps to the unit circle, as Property 1 shows for projections onto the x-y plane! For planes with $k \neq 0$, the radius of the circle resulting from the projection of the equator would have a radius proportional to the distance of the plane from the pole T. More broadly, the map still "looks" the same if we shift the plane up or down. It is just scaled accordingly.

If we were to project directly onto another plane, it would be necessary to reformulate each of the equations for the coordinates of stereographic projection given earlier. The maps would have to be, instead, from P = (x, y, z) to Q = (X, Y, k), where k is the z-coordinate of all points on the plane H, rather than to Q = (X, Y, 0) as we've done on this page. We will not derive such equations here since, under the conventions established initially, it is possible and even easy to scale any resulting projection (we just would need to multiply each coordinate by a common factor), so there is not much practical, mathematical utility to projecting onto any plane other than the x-y plane. The importance of Property 3 is that it justifies visual representations as such.

Property 4

The shortest distance between T ' (the antipode of T) and any other point on the sphere is shown as a straight line (shortest distance) on the map.

It is easy to see that T ' is mapped to (0, 0, 0), the origin of the x-y plane. Let us refer to the origin as O.
The shortest distance between any two points on a sphere is the arc length along a great circle passing through both of the points. Let W be a great circle that passes through T ' = (0, 0, -1) and some other point P (and let Q be the stereographic projection of P). All great circles that pass through a point also pass through the point's antipode; therefore, W also contains T, the antipode of T '. By Property 2, then, W is mapped to a line on the plane (because it is a circle which contains the projection point). Therefore, the curve representing the shortest distance between the T ' and any other point is represented on the map as a line, which is the shortest distance between points on the two-dimensional plane.
In other words, any stereographic map is centered on some point T ' (the point opposite the projection point). The projection of the arc from T ' to any P (the shortest distance between the two points on the surface of the sphere) is the line through O and Q (the shortest distance between the two points on the plane).

This property has particular relevance to cartography. The meridians will pass through the center of a ; as such, the shortest distance on the map will be lines passing through the origin, representing the meridians.

Since, in principle, we can center a stereographic map on any T ', this means that we can choose a point such that all straight lines passing through the point's projection will represent the shortest distance on the sphere. This has clear advantages for navigation: a stereographic map centered on an airport, for example, would show the shortest distances to any other point on earth as a straight line passing through the center of the map.

This only works, of course, for the shortest distances from the origin of the map; any other lines on the map would not represent the shortest distance, but there are circumstances (like the map centered on an airport) in which this technicality is not really a disadvantage.

Figure 9
The inverse of the stereographic projection. A rectangular coordinate system is projected back onto a sphere, demonstrating that the projection is conformal.

Property 5

Stereographic projection is conformal; that is, any angles formed on the surface of the sphere by the intersection of two paths are the preserved in the corresponding projection onto the plane. This means that the projection preserves shape locally, since all angles are the same on the sphere and on the plane.

Stereographic projection does not, however, preserve the area of regions on the sphere (it is not equiareal). Intuitively, this makes sense because the sphere has finite surface area, but can be projected onto the entire x-y plane, which extends without bound.

In fact, there are no ideal map projections; all map projections from a sphere to a plane distort either angles or areas (or, in some cases, both!). Figure 9 to the left shows this property of stereographic projection. In general, the area of each enclosed region on the rectangular grid is not equal to that of its projection on the sphere. However, the angles do seem to be preserved locally; each path on both the plane and the sphere seems to intersect at a right angle, even though on the sphere the paths are curved. Are angles actually preserved, as this property claims?

Here we will prove Property 5.

We begin by distinguishing the necessary conditions for a conformal map projection from the sphere to the plane.

Figure 10
Spherical rectangle.

Figure 11
Polar rectangle.

We define: A spherical rectangle is a region on the sphere bounded by two and two (for example, see the white region of Figure 10). By construction, an angle formed by the intersection of a meridian and a parallel is a right angle, so the angles of a spherical rectangle are all right angles. A polar rectangle is a region on the polar coordinate plane bounded between the arcs of two circles and two polar angles (for example, see the yellow region of Figure 11).
 Please note that the above is a nonstandard definition of spherical rectangle. In spherical geometry, the term spherical rectangle typically refers to the region on a sphere bounded by 4 great circles, such that the angles at which the segments intersect are all equal. Such angles are not, in fact, 90°, and such a figure is not stereographically projected to a polar rectangle! It will not be of use to us in this proof.
Any spherical rectangle is mapped by stereographic projection to a polar rectangle. Assure yourself of this fact by looking to Figure 12 below. Note that the polar grid is divided up into polar rectangles and the sphere is divided up into spherical rectangles.
This should makes sense as well in light of what we have proven already. By Property 2, meridians are mapped to lines that pass through the origin; by Eq. 3, stereographic projection preserves azimuth, so the meridians are mapped to lines on the polar grid. Also by Property 2, circles are preserved. In this case, because the circles on the sphere are parallels, they are mapped to circles on the plane centered at the origin.

Figure 12
The inverse of the stereographic projection; a polar coordinate system projected back onto a sphere.

What relevance do spherical and polar rectangles have for this proof? Consider any line segment in the plane. We can draw a rectangle around it such that it represents the diagonal of the rectangle. We can even make the rectangles smaller and smaller, sharing smaller portions of the segment as their diagonals. We can do the same on the surface of a sphere; given a path D, we can form a spherical rectangle by drawing in two meridians and two parallels. By choosing meridians that are very close together (ie. by choosing meridians with very similar azimuths) and by choosing parallels that are very close together, we can make the resulting spherical rectangle very small, so the local direction of a path on the sphere at any point along the path can be represented by the diagonal of an infinitesimally small rectangle.
We then stereographically project the spherical rectangle onto the plane. As mentioned before, a spherical rectangle stereographically projects to a polar rectangle. If the along any meridian (which we will call $k_m$) is equal to the scale factor along the parallel (which we will call $k_p$), then the angle of the diagonal will be preserved by the projection. In other words, we need to show that stereographic projection simply maps a spherical rectangle to a proportionally larger or small rectangle on the plane. This will show that the direction of any line segment is preserved. Figure 13 demonstrates this process for a planar rectangle.

Figure 13
A rectangle and its projection (the larger rectangle) when scale factors km and kp are equal. If the scale factors were unequal, then the diagonals of the two rectangles would not have the same direction, or angle.

How is this fact useful for proving that the stereographic projection is conformal? As mentioned, we can represent the local direction of any path using a very small spherical rectangle. Naturally, this means that we can represent the local direction of any two paths on the sphere using very small spherical rectangles; we can construct such spherical rectangles, notably, around the point of intersection. The direction of both of the paths would be preserved if both rectangles are projected. Therefore, the difference between the angles of the two paths, or the angle of intersection, would be preserved from the sphere to the plane.

Figure 14
When the scaling factor along the meridian is equal to the scaling factor along the parallel, and the direction of diagonals are preserved.

In other words, our proof will consist of demonstrating that $k_p = k_m$. This is sufficient to show that the angle of the diagonal will be preserved by the projection, and the angle of intersection between any two paths on the sphere will be preserved.

Figure 15
Cross section of stereographic projection of a point. The shortest distance from the vertical axis of any point with zenith $\phi$ will be $\sin \phi$, so a parallel at $\phi$ has radius $\sin \phi$.

First we will look at the scale factor of the parallel. Parallels lie on planes parallel to the x-y plane; they are circles on the surface of the sphere, with centers on the vertical axis. Each point on a given parallel has the same zenith, $\phi$. (See Spherical Coordinates to review the spherical coordinate system.) The radius of a parallel is the shortest distance to the vertical axis, on which is the parallel's center. Figure 15 shows that this distance is $\sin \phi$. Therefore, an arc on a parallel, between points with azimuths $\theta _1$ and $\theta _2$ has arc length:
$|\sin \phi (\theta _2 - \theta _1)|$.
Any parallel is stereographically mapped to a circle centered at the origin on the x-y plane. Since all points on a parallel have the same zenith, the radius of the projected circle can be obtained from Eq. 3:
$\frac{\sin \phi}{1 - \cos \phi}$.
Since stereographic projection preserves the azimuth of all points, the arc on the parallel between $\theta _1$ and $\theta _2$ is projected to an arc on the plane between the same polar angles. This projected arc has length:
$\left| \frac{\sin \phi}{1 - \cos \phi} (\theta _2 - \theta _1) \right|$.
Since we've found the lengths of a generalized region on the parallel and its projection, we can find the scale factor for that region:
$k_p = \frac{\frac{\sin \phi}{1 - \cos \phi} (\theta _2 - \theta _1)}{\sin \phi (\theta _2 - \theta _1)} = \frac{1}{1 - \cos \phi}$.
For a given $\phi$, this scale factor holds regardless of the arc length. It would even hold for very small arc lengths, and so could be used to represent the instantaneous scale factor along the parallel.

Figure 16
Stereographic projection of a region of a meridian. The region and its projection are highlighted in red on the sphere and on the plane.

Now we must find the scale factor along the meridian. Each point on a meridian has the same azimuth ($\theta$). For this reason, we will consider in Figure 16 a cross section of the sphere to get an idea of what it means to project a region of a segment of a meridian.
A region of a meridian is the region bounded between two zeniths, which we will for now call $\phi _1$ and $\phi _2$ (we will assume that $\phi _1 > \phi _2$, but doing so will have no bearing on our proof since they are interchangeable). Since a meridian is a great circle on the sphere, it has the same radius as the sphere (in this case, radius 1). Therefore, the length of a general region on a meridian is the arc length:
$1(\phi _1 - \phi _2)$.
Now, using Eq. 3, we can project this region onto the plane and obtain a line segment with length:
$\frac{\sin \phi _2}{1 - \cos \phi _2} - \frac{\sin \phi _1}{1 - \cos \phi _1}$.
So the scale factor of the generalized region of a meridian is:
$k_m = \frac{\frac{\sin \phi _2}{1 - \cos \phi _2} - \frac{\sin \phi _1}{1 - \cos \phi _1}}{\phi _1 - \phi _2}$.
Here we encounter a problem: it is necessary to use two zenith measurements to characterize an arc along a meridian, but the scale factor $k_p$ determined earlier was a function of a single zenith measurement. To determine whether $k_p = k_m$, we must formulate the scale factor of the meridian likewise in terms of a single zenith.
Recall that we are considering localized preservation of angles. Stereographic projection distorts the lengths of line segments, especially as they are projected further and further away from the origin. It is not the scale factor for some arbitrary region bounded between two zeniths that we are looking for; we need to find the instantaneous scale factor along the meridian as a function of $\phi$. How might we do this? We will alter the above equation for $k_m$. We will choose two points on the meridian that are very close together (imagine if $P_1$ were brought closer and closer to $P_2$ in Figure 16). In other words, we will take the limit as $\phi _1$ goes to $\phi _2$:
$k_m = \lim _{\phi _1 \rightarrow \phi _2} \left( \frac{\frac{\sin \phi _2}{1 - \cos \phi _2} - \frac{\sin \phi _1}{1 - \cos \phi _1}}{\phi _1 - \phi _2} \right)$.
Both the numerator and the denominator of the expression inside the limit go to 0 as $\phi _1$ goes to $\phi _2$. We can still try to evaluate the limit using l'Hôpital's rule and taking the derivative of the numerator and the denominator. We will, for this purpose, fix $\phi _2$ as a constant and take the derivative with respect to $\phi _1$.
The derivative of the denominator $\phi _1 - \phi _2$ is 1, since $\phi _2$, as a constant, goes to 0. We then differentiate the numerator using the quotient rule (the first term, containing $\phi _2$, is a constant and so goes to 0):
\begin{align} \frac{d}{d \phi _1} \left( \frac{\sin \phi _2}{1 - \cos \phi _2} - \frac{\sin \phi _1}{1 - \cos \phi _1} \right) &= 0 - \frac{\cos \phi _1 (1 - \cos \phi _1) - \sin \phi _1 (\sin \phi _1)}{(1 - \cos \phi _1) ^2} \\ &= - \frac{\cos \phi _1 - \cos ^2 \phi _1 - \sin ^2 \phi _1}{(1 - \cos \phi _1) ^2} \\ &= \frac{1 - \cos \phi _1}{(1 - \cos \phi _1) ^2} \\ &= \frac{1}{1 - \cos \phi _1} \end{align}.
Differentiating the numerator has (as desired) eliminated $\phi _2$. Henceforth we will simply replace $\phi _1$ with $\phi$. This is justified because we could choose to fix $\phi _2$ at any value, and still could let $\phi _1$ approach it.
By l'Hôpital's rule, then, we obtain the solution:
$k_m = \frac{1}{1 - \cos \phi}$.
We have confirmed that $k_m = k_p$. Therefore, stereographic projection preserves the direction of any path on the sphere, which can be represented as the diagonal of a spherical rectangle. Furthermore, by the same token, stereographic projection locally preserves the directions of all paths at a given point, and so preserves the angles of intersection between all paths on the sphere, so the stereographic projection is, by definition, conformal.
This completes the proof of Property 5.
$\blacksquare$

## Complex Analysis

An important mathematical application of stereographic projections in complex analysis is the Riemann sphere.

The complex plane can be thought of as the x-y plane, in which the x axis is the real axis and the y axis is the imaginary axis. It becomes useful to extend the complex plane by introducing a symbol for infinity, $\infty$. We call this set the extended complex plane, which is often denoted $\mathbb{C} \bigcup \infty$ (the union of $\mathbb{C}$ with $\infty$). We define the following rules:

$a + \infty = \infty + a = \infty$
$b \cdot \infty = \infty \cdot b = \infty$
$\frac{a}{0} = \infty$ for $a \neq 0$
$\frac{b}{\infty} = 0$ for $b \neq \infty$

As you can see, we have defined some interesting and usually unallowed operations involving 0 and infinity! A problem, however, remains: there is no single point on the complex plane that can represent infinity. This is where stereographic projection comes in. The Riemann sphere is merely the result of projecting the extended complex plane onto a unit sphere (inverse stereographic projection, Eq. 2). Each point (x, y, z) on the Riemann sphere corresponds with a point on the complex number plane:

$z = \frac{x + iy}{1 - z}$.

The Riemann gives us a way of concretely representing the whole extended complex plane: The north pole of the sphere, $T = (0, 0, 1)$, is defined as the point at infinity, so on the Riemann sphere, every point on the extended complex plane is represented.

This may seem strange. What justifies doing this? Earlier we alluded to the fact that in stereographic projection T cannot be projected onto the plane. The coordinates of the projected point Q become much larger as z approaches 1 (observe the limit of Eq. 1 as z goes to 1, or visualize the projection of points with z close to 1). If we could project T, it would be infinitely far from the origin, since a line from T to T would be tangent to the sphere at T, and therefore parallel to the plane H. Property 2 implies this by showing that circles passing through T are projected to lines, which extend without bound in two directions on the plane. This makes sense, because as we project points closer and closer to T, they project further and further away on the plane.

While we cannot perform addition and multiplication on the Riemann sphere, it is useful because it allows the point at infinity to be a concrete point.

# Why It's Interesting

Stereographic projections are used in cartography and photography.

## Cartography

Cartographers have always struggled to create maps that are as accurate and usable as possible. Map projections are projections of earth's surface, represented usually as an idealized three-dimensional sphere, onto a two-dimensional plot. (The earth is actually closer to an oblate spheroid, and is of course contoured when we consider landmasses like mountains and valleys.) An ideal map projection would be both and . Each type of map projection has its merit, but unfortunately, an ideal map projection is impossible. We just have to choose one that best suits our needs.

As one might have guessed, the stereographic projection is one of those methods of mapping! The process of stereographic projection has been known for at least 2000 years; Hipparchus and Ptolemy both used stereographic projection to map the stars. It was not until the 16th century, however, that Gualterious Lud used stereographic projection to map the earth.

The advantage to stereographic maps, as is demonstrated by Property 5 in the More Mathematical Explanation, is that they are conformal; they preserve angles locally. Conformal maps tend to be useful for navigation, although the relative sizes of landmasses are often distorted. In this section, we will look at some stereographic representations of the earth. We will discuss what it means for the stereographic projection to be azimuthal, and what some other azimuthal projections are. We will also consider some other popular "competing" maps, like the Mercator and the Gall-Peters projections.

### Types of Stereographic Projections

Polar

Figure 17
Polar stereographic projection. Source

The two stereographic maps in Figure 17 are polar stereographic projections. This means that the center of the map is one of the poles, in these cases the North Pole. The points on the sphere of the earth are projected from the South Pole. In the terms established in the definition, the South Pole is T, and the North Pole is T '.

Both of these stereographic projections are projected from the South Pole and are centered around the North Pole. But how, then, do they look so different?! They are technically the same map projection, since they are projected from the same point. They appear different because the second is more "zoomed in." The cartographer chose to show only the Northern Hemisphere in the second projection. Such a decision is inevitable with stereographic projection, because the true map of the sphere is projected onto the whole x-y plane, and it's impossible to show it all on a finitely bounded plane (like a piece of paper, or a computer screen).

In the first map in Figure 17, there is a red circle that denotes the equator, outside of which (in the Southern Hemisphere) the map becomes rather distorted. While the United States, Greenland, and Russia may appear fairly "normal" on these maps, one would probably not rely on it too much, for instance, around Chile or Antarctica. Does this mean that a stereographic map cannot be used in those regions? Certainly not! If one wanted to navigate the Southern Hemisphere, then one could create another polar stereographic map projection from the North Pole, centered around the South Pole. Such a map would appear "normal" in the Southern Hemisphere and distorted in the Northern Hemisphere. Nevertheless, all stereographic projections are still conformal at all points, but one hemisphere will appear distorted because the distortions of area become too great.

Equatorial

Figure 18
Equatorial stereographic projection. Source

If someone told you to picture a map of the world, you probably would not think of anything like either of the maps in Figure 17 because we really don't think of the world from a "polar" stereographic perspective. Can a stereographic projection still accommodate our needs for a map that looks more familiar? Above we discussed that it was possible and legitimate to shift the projection point from the South Pole to the North Pole in order to focus on a different hemisphere. We can likewise choose to focus on the Eastern or Western Hemisphere, rather than the Northern or Southern. Such a stereographic projection is equatorial, and examples are given in Figure 18.

Once again, the second map here is simply the first map "zoomed in"; the cartographer decides to show just the part of the world which seems, to our eyes at least, to be represented "normally." An equatorial stereographic projection is constructed by projecting from some point on the equator rather than from one of the poles. This map is centered at 0°N 0°E (where the equator and prime meridian meet, slightly off the coast of western Africa). The projection is from 0°N 180°E, the antipode of 0°N 0°E.

Figure 19
Equatorial stereographic projection of a sphere with meridians drawn in.

The advantages of such a variation of the stereographic map are obvious: landmasses appear "normal" in the places where most people live and travel. The second map of the two above resembles other common map projections rather closely. A more direct comparison can be made later, when we introduce some of the alternatives to the stereographic projection.

One might still look at Figure 18 and doubt that it is truly stereographic. The first of the two equatorial maps looks very little like anything we are familiar with: the various continents and landmasses seemed to be jumbled around in a massive ocean! To say the least, it seems to bear little resemblance to the polar projections of Figure 17, and does not really look like other common map projections. However, this should be what we expect in a equatorial map projection that "shows too much." Consider Figure 19 to the right, however. This more abstract image may help us understand the bizarre map in Figure 18. This figure has the meridians drawn in, but has been rotated 90° (so the projection is equatorial). The meridians no longer map to straight lines but to circles that all converge to the projections of the two poles, much like the meridians in Figure 18.

Oblique

The last type of stereographic projection, which we won't show here, is oblique. This would merely refer to a stereographic projection that is centered neither on a pole nor on a point along the equator. In principle, we can center a stereographic projection on any point.

A circumstance in which such a map is advantageous was mentioned after the proof for Property 4 in the More Mathematical Explanation: all straight lines through the point on which a stereographic map is centered correspond to great circles passing through that point on the sphere, so all lines on the map passing through the center represent the shortest distances to other points. This has obvious utility in navigation: if someone frequently travels from a certain geographical point (in the case of, say, an airport), then centering a stereographic map there would allow us to find the shortest route to any other point.

### Other Azimuthal Map Projections

Stereographic map projections are azimuthal. Azimuthal projections are so called because they preserve the of each point on the sphere. To confirm this fact, one might find it useful to refer to the introduction of the spherical coordinate system in the More Mathematical Explanation. It would also be useful to refer to Eq. 3 in that same section.

Figure 20
Stereographic projection (left), gnomonic projection (center), and orthographic projection (right). Note that the orthographic projection here is not centered at a pole. It is "oblique."

The three basic azimuthal maps, shown in Figure 20, may be thought of as projections from some "light source" on the vertical axis of the sphere onto a flat piece of paper tangent to the sphere. The light source, in stereographic projection, is at one of the poles of the sphere, the projection point. The main image of the page is a good example of this notion of a light source.

In gnomonic projection, the light source is at the center of the sphere. In orthographic projection, the light source is at an "infinite" distance, so the "rays of light" (or the lines that establish a correspondence between points on the sphere and points on the plane) are parallel. Figure 22 below shows this characterization of three basic azimuthal projections in terms of rays of light. The image cycles through examples of the projection to compare their respective "light sources."

Figure 21
The earth, as viewed from outer space, is actually an orthographic representation since the rays of light reaching our eyes are virtually parallel.

How do these azimuthal projections compare?

• Gnomonic projection and orthographic projection are both limited to mapping just one hemisphere to the plane. In the gnomonic projection, a line drawn from the sphere's center to a point in the upper hemisphere would not intersect the plane. In the orthographic projection, every line (excepting those passing through points on the equator) passes through two points, so we must choose to map just one hemisphere. The stereographic projection is useful in that it can map the entire sphere (although, as discussed above, it will inevitably begin to look distorted in the hemisphere opposite of its central point).
• Gnomonic projection, like stereographic projection, can map to the whole of the x-y plane, but orthographic projection only maps within the unit circle on the x-y plane. The gnomonic projection, for points near the equator, behaves similarly to the stereographic projection near the projection point. The orthographic projection, by constrast, appears "clustered" near the equator, which is the edge of what it maps.

Only the stereographic projection is conformal. Then what's the use of these other two projections?

• An orthographic map is actually how the earth appears from outer space or when we look at a globe, when we are sufficiently far away from it that the rays of light reaching our eyes are virtually parallel. The earth appears as a circle, and depth is more or less eliminated from our perspective. We can't see much around the edges, and we certainly can't see anything on the other side! While such maps may not be as useful as maps that show more of the earth, they certainly put things in perspective.
• The gnomonic projection is useful in that all shortest distances on the sphere are mapped to shortest distances on the plane. This is similar to Property 4 of the stereographic projection, but it is even more useful. The stereographic projection only preserves as lines on the plane the shortest distance of paths through its antipode (the point around which the map is centered). By contrast, any straight line on a gnomonic map (not just those passing through its center) corresponds to a great circle on the sphere, and so corresponds to the shortest distance on the sphere. This makes the gnomonic map even more useful for navigation, at least in some cases (it lacks some utility since it is not conformal). Note that this does not mean that the every path is equidistant, or that the length of every line segment is proportional to the arc length of its corresponding great circle on the sphere. It just means that every line on the plane is a segment of a great circle on the sphere, and lines and great circles represent the shortest distances between points in their respective spaces.

Figure 22
Comparison of the three azimuthal map projections: gnomonic, stereographic, orthographic.

Figure 23
Azimuthal equidistant projection (left) and Lambert equiareal projection (right).

There are two other azimuthal projections which cannot be thought of in terms of light sources.

There is a fourth azimuthal map projection, the azimuthal equidistant projection. Like all azimuthal projections (see Property 4 and the above discussion of oblique stereographic projections), the lines passing through the center of the equidistant projection represent the shortest distance to any other point on the sphere. However, the equidistant projection is called equidistant because, in addition to this fact, the distances to those other points are also preserved. In other words, the length on the sphere from the central point of the map to any other point is preserved. (Although the length between any two points are not preserved.) This makes the equidistant projection especially useful for navigation. It is, however, a mathematical construction and cannot be thought of in strictly geometrical terms.

A fifth azimuthal map projection is the Lambert equiareal projection. Necessarily, it is not conformal, but can be useful in cases when we would like to represent the sizes of landmasses in relation to each other. Like the equidistant projection, it is a mathematical construction, and we cannot represent it in geometrical terms like we can with the stereographic, gnomonic, or orthographic projections.

### Other Non-Azimuthal Map Projections

The other azimuthal maps are still pretty similar to the stereographic map, at least in appearance. How do stereographic maps compare to other map projections? The Mercator projection and the Gall-Peters projection, both of which have been popular in the past, are shown below. The Mercator projection, like a stereographic projection, is conformal. The Gall-Peters projection is equiareal (and so not conformal), so it should preserve the relative sizes of landmasses. They are both rectangular maps, unlike the azimuthal maps which are, in principle, circular (although the map can be "cut off" anywhere in practice).

 Figure AMercator projection. File:Gall–Peters projection.jpg Figure BGall-Peters projection.

We can see that these two map projections produce very different maps! Depending on what the map is being used for, these differences can be very relevant. Conformal maps tend to be useful for navigation but do not give the best perspective of the relative size of landmasses. The Mercator projection portrays Greenland and Africa as being about the same size; Greenland actually comes out slightly larger. But the Gall-Peters projection shows that Africa is, in fact, vastly larger than Greenland. To be more specific, Africa is about 14 times the size of Greenland, but the non-equiareal projection makes them look almost identical!

This conundrum highlights some of the political implications of map-making. Equiareal maps may not look too appealing because they distort the shape of continents like Greenland, but maps like the Mercator projection, which was the most popular map for a long time, can have a large impact on the way people perceive the world.

## Photography

Figure Z
Stereographic projection in photography. A comparison between a stereographic photograph and the panoramic photograph used to create it. Source.

Stereographic projections are used in photography to produce impressive, beautiful images.

An image such as the one to the left is created by taking a panoramic photo. This involves taking a photo, rotating by some fixed angle, and taking another photo. When the photos cover 360°, they may be strung together to form a panoramic photo. It is possible then to map the photo as a texture on a sphere, which can then be projected onto a plane to produce the image on the right of the figure. Various photographers, like Paul Bourke, have used this method to create "little planets," and they have shown more fully how to reproduce this process.[1]

# About the Creator of this Image

Paul Nylander produces a variety of math artwork to show the intersection between computer science and graphics. His website is bugman123.com.

# References

1. Bourke, Paul (2011). "Little Planet" Photographs. University of Western Australia.