# Stereographic Projection

(Difference between revisions)
 Revision as of 15:12, 24 June 2013 (edit)← Previous diff Revision as of 08:56, 25 June 2013 (edit) (undo)Next diff → Line 153: Line 153: ::$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$. ::$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$. + {{Anchor|Reference=Figure5|Link=[[Image:Plane-Sphere intersection.png|thumb|275px|right|Figure 5
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: :Any circle on the surface of a sphere is the intersection of a plane with the sphere:

## Revision as of 08:56, 25 June 2013

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

# Basic Description

Figure 1
Cross section of arbitrary points on a sphere being mapped to points on a plane.

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

Each point on the sphere is associated with a point on the plane. The process for determining the point on the plane is to draw a line from the north pole, letting it 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 a two-dimensional cross-section of the sphere. In a way, the figure is an example of the unit circle being mapped to the x axis. This page will examine the broader projection of the unit sphere onto the x-y plane.

The main image demonstrates stereographic projection. In this case, the plane is drawn under the sphere instead of cutting through its equator. The coloring demonstrates where regions of the sphere end up when they are mapped to the plane. The projection is still from the top of the sphere, but the bands of color are not centered around the vertical axis, so 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. You can rotate the sphere with the mouse, changing the orientation of the colors on the sphere which changes the projection on the plane. The sphere and projection point 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

## Coordinates

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

## Coordinates

A stereographic projection maps the points of a sphere onto a plane. Specifically, let's consider the unit sphere centered at the origin x2 + y2 + z2 = 1 and the x-y plane z = 0. We want to know how to map some point P (x, y, z) on the sphere to some point Q (X, Y, 0) on the 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.

Figure 2 to the right shows how these points are projected. A line is drawn from the pole T through some other point P on the sphere and some point Q on the plane; the point on the sphere is mapped to the point on the plane. All points on the sphere, besides T, can be mapped to the plane in this way.

Rectangular coordinates: The formula for the coordinates of Q, the point on the plane, is:

Eq. 1        $Q(X, Y, 0) = \left(\frac{x}{1 - z}, \frac{y}{1 - z}, 0 \right)$ or $Q(X, Y) = \left(\frac{x}{1 - z}, \frac{y}{1 - z} \right)$.

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.
$\blacksquare$

Inverse of rectangular coordinates: Since coordinates on the sphere are mapped uniquely, or one-to-one, to coordinates on the plane, this function is invertible. The explicit inverse is:

Eq. 2         $P(x, y, z) = \left(\frac{2 X}{X^2 + Y^2 + 1}, \frac{2 Y}{X^2 + Y^2 + 1}, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}\right)$.

Derivation of Eq. 2.

This inverse formula is derived by substituting the coordinates from Eq. 1 back into the equation for the unit sphere and solving for z.
$x^2 + y^2 + z^2 = 1$
$(X(1-z))^2 + (Y(1-z))^2 + z^2 = 1$
$(X^2 + Y^2)(z^2 - 2z + 1) + z^2 - 1 = 0$
$(X^2 + Y^2 + 1)z^2 - 2(X^2 + Y^2)z + (X^2 + Y^2 - 1) = 0$
$z = 1, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}$
The first solution may be discarded because T(0, 0, 1) is the pole of the circle; it is the one point on the sphere for which stereographic projection is not defined. (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 The plane tangent to the sphere at T is parallel to the x-y plane onto which we are projecting, so any line tangent to the sphere at T will never pass through the plane.) The second solution is what we expected.
In order to find inverse expressions for x and y, we return once again to the equation for our sphere.
$x^2 + y^2 + z^2 = 1$
$y^2 = 1 - x^2 - z^2$
$y = \sqrt{1 - X^2 \left(1- \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1} \right)^2 - \left(\frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1} \right)^2}$
This is certainly a mess to simplify. But in the end we obtain the expected result:
$y = \frac{2Y^2}{X^2 + Y^2 + 1}$
Since x and y are interchangeable for the purpose of these formulas, the same may be repeated for x to obtain:
$x = \frac{2X^2}{X^2 + Y^2 + 1}$
These are the coordinates in Eq. 2.
$\blacksquare$

Figure 3
The angles used in the spherical coordinate system. If radius ρ is drawn from the origin to P, then θ is the azimuthal angle between the positive x axis and ρ, and ϕ is the polar angle between the positive z axis and ρ.

Projection in terms of spherical coordinates: A point on the sphere has rectangular coordinates $P(x, y, z) = P(\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$, where $\phi$ is the polar angle formed from the positive z axis and $\theta$ is the azimuthal angle formed from the positive x axis. In spherical coordinates, we say equivalently $P(1, \theta, \phi)$ (radius 1 for unit sphere). We can write the coordinates of the projection in terms of these spherical coordinates:

Eq. 3        $Q(R, \Theta) = \left( {\frac{\sin \phi}{1 - \cos \phi}}, \theta \right)$ where $(R, \Theta)$ are polar coordinates of the stereographic projection.

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

Derivation of Eq. 3.

By substituting spherical coordinates into the Cartesian coordinates of Eq. 1, we can use the Pythagorean Theorem to find R:
$R = \sqrt{X^2 + Y^2} = \sqrt{\frac{\cos^2 \theta \sin^2 \phi}{(1 - \cos \phi)^2} + \frac{\sin^2 \theta \sin^2 \phi}{(1 - \cos \phi)^2}} = \sqrt{\frac{\sin^2 \phi (\sin^2 \theta + \cos^2 \theta)}{(1 - \cos \phi)^2}} = \frac{\sin \phi}{1 - \cos \phi}$
That the spherical coordinate $\theta$ is the same as the polar coordinate $\Theta$ should be unsurprising if we visualize the projection. Both are the azimuthal angle.
We have obtained the coordinates in Eq. 3.
$\blacksquare$

Inverse of polar coordinates: The inverse can be formulated in terms of spherical coordinates as well.

Eq. 4         $P(1, \phi, \theta) = \left(1, 2 \arctan {1 \over R}, \Theta \right)$

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
$R = \frac{\sin \phi}{1 - \cos \phi} = \frac{1}{\tan {\phi \over 2}}$.
This can easily be rewritten:
$\phi = 2 \arctan {1 \over R}$.
Again, we complete our derivation by noting that the radius of the spherical point must be 1, and that $\theta = \Theta$.
$\blacksquare$

## Properties

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

We prove this by rewriting the right-hand side of the equation R 2 = X 2 + Y 2 in terms of z using the coordinates in Eq. 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 the projection falls outside of the unit circle. For negative z, the denominator is greater than the numerator, so the projection falls inside of the unit circle.
This completes the proof.
$\blacksquare$

Figure 4
A circle on the surface of the sphere is projected to a circle on the plane. Note that the lines drawn from the pole T to the projected points Q form an oblique circular cone.

The stereographic projection preserves circles. We distinguish between two possible cases:

• The circle on the sphere does not contain T. Then the projection of the circle onto the x-y plane is a circle.
• This may be understood intuitively, if one visualizes a cone with its apex at T. The cone's intersection with the sphere would form a circle, as would the cone's intersection with the plane (although the two circles are generally not of the same radius).
• 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.
• This case is probably less obvious. Recall that we've said that the stereographic projection of T is undefined. This fact can be seen in a variety of ways; for instance, the coordinates for X and Y in Eq. 1 would have denominators of 0 with z = 1. Geometrically, it is apparent that points very close to T will be projected very far from the origin on the x-y plane. So we would expect projections of circles on the sphere containing T to extend very far away from the origin. Rather than circles, they are lines which extend infinitely.

Here we will also prove this theorem analytically.

We recall that our sphere is defined:
$S_2 = \{ (x, y, z) \text{ } | \text{ } x^2 + y^2 + z^2 = 1 \}$.
Planes in $\mathbb{R}^3$ are defined:
$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$.

Figure 5
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:
$W = S_2 \bigcap V$.
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 a circle.
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 excludes T.
Recall that T (0, 0, 1), P (x, y, z), Q (X, Y, 0) lie on a line, so
$\overline{TP} = t ( \overline{TQ} )$ for some real, nonzero t.
In vector notation,
$\langle x, y, z - 1 \rangle = t \langle X, Y, -1 \rangle = \langle tX, tY, -t \rangle$
so
$x = tX$,    $y = tY$,    $1 - z = t$.
In the previous proof, we showed that
$R^2 = X^2 + Y^2 = \frac{1 + z}{1 - z}$ where R is the distance of the projected point Q from the origin, and Q (X, Y) are rectangular coordinates.
By manipulating this relationship, we get
\begin{align} R^2 = \frac{1 + z}{1 - z} & \Rightarrow R^2 - zR^2 = 1 + z \\ & \Rightarrow R^2 - 1 = z(R^2 + 1) \\ & \Rightarrow z = \frac{R^2 - 1}{R^2 + 1} \end{align}
and
\begin{align} R^2 = \frac{1 + z}{1 - z} & \Rightarrow R^2 + 1 = 1 + \frac{1 + z}{1 - z} = \frac{1 - z + 1 + z}{1 - z} = \frac{2}{1 - z} \\ & \Rightarrow R^2 + 1 = \frac{2}{t} \\ & \Rightarrow t = \frac{2}{R^2 + 1} \end{align}
Now we consider the equation of the plane which intersects the sphere:
$Ax + By + Cz + D = 0$
By substituting, this becomes:
$AtX + BtY + C \frac{R^2 - 1}{R^2 + 1} + D = 0$
$\frac{2AX}{R^2 + 1} + \frac{2BY}{R^2 + 1} + C \frac{R^2 - 1}{R^2 + 1} + D = 0$
Multiply by $R^2 + 1$:
$2AX + 2BY + C (R^2 - 1) + D (R^2 + 1) = 0$
$2AX + 2BY + (C + D)R^2 - C + D = 0$
Since $R^2 = X^2 + Y^2$,
Eq. 5         $(C + D)(X^2 + Y^2) + 2AX + 2BY - C + D = 0$

Figure 6
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

To conclude, note that when $C = - D$,   $T(0, 0, 1) \in V$ because $0A + 0B + 1C + D = 0$. Furthermore, $T \in V \Rightarrow T \in W$.
If $C = - D$, then coefficients of $X^2$ and $Y^2$ in Eq. 5 are 0, so Eq. 5 is the formula for a line in $\mathbb{R}^2$. Therefore, when $T \in W$, the stereographic projection of $W$ is a line.
If $C \neq - D$, then the coefficients of $X^2$ and $Y^2$ in Eq. 5 are the same, so Eq. 5 is the formula for a circle in $\mathbb{R}^2$. Therefore, when $T \notin W$, the stereographic projection of $W$ is a circle.
This completes the proof.
$\blacksquare$
Figure 6 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.

However, it is possible to rearrange Eq. 5 further. In the latter case where $C \neq - D$, we have
$X^2 + Y^2 + \frac{2A}{C + D}X + \frac{2B}{C + D}Y + \frac{-C + D}{C + D} = 0$
$X^2 + \frac{2A}{C + D}X + Y^2 + \frac{2B}{C + D}Y = \frac{C - D}{C + D}$
By completing the square, this becomes
$X^2 + \frac{2A}{C + D}X + \left( \frac{A}{C + D} \right)^2 + Y^2 + \frac{2B}{C + D}Y + \left( \frac{B}{C + D} \right)^2 = \frac{C - D}{C + D} + \left( \frac{A}{C + D} \right)^2 + \left( \frac{A}{C + D} \right)^2$
$\left( X + \frac{A}{C + D} \right)^2 + \left( Y + \frac{B}{C + D} \right)^2 = \frac{A^2 + B^2 + C^2 - D^2}{(C + D)^2}$
Therefore, the stereographic projection of $W = S_2 \bigcap V$ onto $\mathbb{R}^2$ is a circle centered at
$\left( - \frac{A}{C + D}, -\frac{B}{C + D}\right)$
$\frac{\sqrt{A^2 + B^2 + C^2 - D^2}}{C + D}$.

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

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. Various map projections, or projections of earth as an idealized three-dimensional sphere onto a two-dimensional plot, have been used for this purpose, and each has its merits. Unfortunately, map projections cannot 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 transverse, as in the two maps of Figure Z.

Figure Z
Transverse 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.

Mathematically, the general case discussed in the More Mathematical Explanation is a projection onto the plane z = 0. Often, for the purpose of map projections, it makes sense to project the sphere representing the earth onto a plane tangent to a pole of the sphere (for example, z = -1 or z = 1, assuming the unit sphere). An example of how this affects the projection is shown to the right. The formulas for the projection would have to be derived again.

## 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.