# Stereographic Projection

Stereographic Projection of a Sphere
Stereographic projection maps each point on a sphere onto a plane.

# Basic Description

Figure 1
Cross section of a sphere. Two arbitrary points on the sphere are mapped to the plane.

Stereographic projection is a method of mapping the surface of a sphere onto a plane.

A map corresponds each point on the sphere with a point on the plane. The process for mapping to the plane is to draw a line from the north pole, passing 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 a two-dimensional cross-section of the sphere. In a way, the figure is an example of the unit circle being stereographically projected onto the x axis. The rest of this page will examine the three-dimensional stereographic projection of the unit sphere onto the x-y plane.

The main image illustrates stereographic projection. Here, the plane is drawn under the sphere instead of cutting through its equator. The coloring demonstrates how regions of the sphere are mapped to corresponding the plane. The projection is still from the north pole of the sphere, and the bands of color are not centered around the vertical axis, the projection forms some interesting ellipses on the plane.

The following applet demonstrates how a sphere is projected onto a plane. A sphere with coaxial 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 changes the projection on the plane. The sphere and north pole remain fixed; only the colors are shifted.

If you can see this message, you do not have the Java software required to view the applet.

# 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
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.

Specifically, let S be the unit sphere centered at the origin; that is, the set of all points (x, y, z) that satisfy the equation x2 + y2 + z2 = 1. Let T be the north pole of the sphere, the point (0, 0, 1). We define H as the x-y plane, the horizontal plane that passes through the origin; 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 shows how these points are projected.

T could more generally 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). Throughout the page, we will derive any formulas using the conventions established here.

Our goal in the following two sections is to derive the formulas that map P to Q. 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 coordinates on the sphere are mapped uniquely, or one-to-one, to coordinates on the plane, Eq. 1 should be invertible. In other words, 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:
$y = \frac{2Y^2}{X^2 + Y^2 + 1}$
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 colatitude, 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.

We recall our definition of the unit sphere:
$S = \{ (x, y, z) \text{ } | \text{ } x^2 + y^2 + z^2 = 1 \}$.
A general 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. Some 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 it includes T and a circle on the plane when it 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 so can be written in terms of its corresponding point on the plane H, Q = (X, Y, 0). We do this by substituting Eq. 2 into Eq. 5:
$Ax + By + Cz + D = 0$
$A \frac{2X^2}{X^2 + Y^2 + 1} + B \frac{2Y^2}{X^2 + Y^2 + 1} + C \frac{X^2 + Y^2 -1}{X^2 + Y^2 + 1} + D = 0$
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 the X and Y (the coordinates of the point $Q \in H$) and A, B, C, and D (the coefficients that determine 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 $C = - D$, $T \in W$, and the coefficients of $X^2$ and $Y^2$ in Eq. 6 are 0, so Eq. 6 is the formula for a line in $H = \mathbb{R}^2$. Therefore, the stereographic projection of $W$ is a line.
• Case #2: When $C \neq - D$, $T \notin W$, 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 $H = \mathbb{R}^2$. Therefore, the stereographic projection of $W$ is a circle.
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 further.

In the former case where $C = -D$, we can simplify:
$(C + D)(X^2 + Y^2) + 2AX + 2BY -C + D = 0$
$2AX + 2BY = 2C$
$Y = - \frac{A}{B} X + \frac{C}{B}$
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 $\mathbb{R}^2$ 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 plane 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, as well as the page's main image, have not yet been rigorously justified. This property will justify such representations.

Here we will prove that stereographic projection scales with the distance of the plane H from T.

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 that is perpendicular to the axis that passes through T, the point from which we are projecting (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 relatively easy to demonstrate using basic trigonometry. Consider the two-dimensional cross section of a sphere in Figure 8 to the right. It is clear that triangles ATQ1, ATQ2, and ATQ3 are similar (since angle ATQ1 is shared by all three and the angles formed with the vertical axis, by construction, are right angles). Therefore,
$\frac{ \overline{AT} }{ \overline{AQ_1} } : \frac{ \overline{BT} }{ \overline{BQ_2} } : \frac{ \overline{CT} }{ \overline{CQ_3} } : \frac{k}{R}$.
In other words, k, the distance between T and the plane H is proportional to R, the distance of the point 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! For planes with $C \neq 0$, the radius of the resulting circle would be proportional to the distance of the plane from the equator. It would also be necessary to reformulate each of the equations for the coordinates of stereographic projection as maps from P = (x, y, z) to Q = (X, Y, C).

Since, under the convention established initially, it is still fully possible and quite easy to scale the resulting map, there is not much practical, mathematical utility to projecting onto any plane other than the x-y plane; however, Property 3 does justify visual representations of such.

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

Property 4: The stereographic projection is conformal, meaning that any angles formed on the surface of the sphere are the same as those projected onto the plane. This means that the projection preserves shape locally; the angles on the sphere and on the plane are the same at the point of intersection.

However, the stereographic projection does not preserve the area of regions on the sphere. This makes sense because the sphere has finite surface area, but is projected onto the whole of the x-y plane (the projection can be inverted and any point on the x-y plane can be mapped to a point on the sphere).

The image to the left shows this property. It is clear that the area of each enclosed region on the rectangular grid is not equal to that of its projection on the sphere. It also appears that the angles are preserved locally. These relationships can be proven.

An important mathematical application of stereographic projections in complex analysis is the Riemann Sphere. In the Riemann sphere, the north pole T of the sphere is the point at infinity (recall that T cannot be projected onto the plane, while the coordinates of the projection become much larger as z approaches 1). The Riemann sphere therefore constitutes the extended complex plane, the union of the complex numbers with infinity.

# 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 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.) Each type of map projection has its merit, but unfortunately, no map projection can be ideal; they cannot be both conformal and equiareal, but cartographers and geometers have left us many to choose from.

The stereographic projection is one such method of mapping. Stereographic projections are chosen because they are conformal and are not distorted much locally. By locally, we mean near the pole opposite of the projection.

The two stereographic maps in Figure X are polar stereographic projections. The center of the map is the north pole; the points on the sphere of the earth are projected from the south pole. (In other words, the point T from the More Mathematical Explanation is the south pole.)

Figure X
Polar stereographic projection. Source

Since these two stereographic maps are are both projected from the same point, they are technically the same projection. They appear different because the second is more "zoomed in." It is limited to the upper hemisphere. The red circle in the first map denotes the equator, outside of which (in the southern hemisphere) the map becomes rather distorted. The second, more local map is contained within the red circle on the first map.

This shows how the map can be distorted at points further from the pole. While the United States, Greenland, and Russia may appear fairly "normal" on these maps, one would probably not use it, for instance, around Chile or Antarctica. However, if one wanted to navigate the lower hemisphere, then one could create another polar stereographic map projection from the north pole. Such a map would be accurate around the south pole and inaccurate in the upper hemisphere.

Another type of stereographic projection is equatorial, as in the two maps of Figure Z.

Figure Z
Equatorial stereographic projection. Source

The process for creating a transverse stereographic projection is no different. One projects from some point on the equator rather than from one of the poles. The advantages of such a map are obvious: it is accurate in those places where most people live and travel. One might observe that the second map in Figure Z perhaps most closely resembles other common map projections.

This map is centered at 0°N 0°E. The projection is from 0°N 180°E, the antipode of 0°N 0°E; that is, the projection point T (refer to Definition) is at 0°N 180°E.

It may not be clear at first that the maps in Figure Z are stereographic! Consider the following, though:

Figure M
Equatorial stereographic projection.

One could obtain the sphere in this image by drawing a sphere with the longitudinal lines and then rotating it by 90° so its poles lie on the x-y plane. Figure M shows what the result would be if that sphere were projected onto the x-y plane; in particular, the lines of the longitude behave like those in Figure Z.

How do stereographic maps compare to other map projections? Shown below are two other map projections, the Mercator projection and the Gall-Peters projection, both of which have been popular in the past. Readers are perhaps familiar with the the Mercator projection, which, like a stereographic projection, is conformal. The Gall-Peters projection is not conformal. It is equiareal, so although the shapes of the landmasses appear distorted, their sizes are all properly relative.

 Figure AMercator projection. Figure BGall-Peters projection.

We can see that different 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. Compare the sizes of Greenland and Antarctica between these two maps.

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

Stereographic map projections are azimuthal. Azimuthal projections are so called because they preserve the azimuth of each point on the sphere. In other words, direction is preserved at the point around which the map is centered; great circles passing through the central point on the sphere are projected to lines passing through the central point on the plane.

Azimuthal maps may be thought of as projections from some "light source" on the 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.

Figure C compares stereographic projections to two other types of azimuthal projections. 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 light beams are parallel. As such, gnomonic projection and orthographic projection are both limited to mapping just one hemisphere to the plane, while stereographic projection can map the entire sphere. Furthermore, 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.

## 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.[1]

Such an image 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.

# About the Creator of this Image

Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967.

# References

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