Stereographic Projection

(Difference between revisions)
 Revision as of 16:40, 20 June 2013 (edit)← Previous diff Revision as of 10:11, 21 June 2013 (edit) (undo)Next diff → Line 135: Line 135: }} }} - Consider the intersection of a plane with the unit sphere to form a circle. We distinguish between two possible cases: + The stereographic projection preserves circles. We distinguish between two possible cases: - *The plane passes through the north pole ''T''. Then the stereographic projection of the intersected points onto the ''x''-''y'' plane is a line. + *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. - *The plane does not pass through ''T''. Then the projection of the intersected points onto the ''x''-''y'' plane is a circle. + **This may be easier to see with respect to great circles that pass through ''T''. From {{EquationNote|Eq. 3}}, we know that the azimuth of any point on the sphere is the projected point's polar angle. Great circles that pass through ''T'' would have the same azimuth for each point, and so the same polar angle, or bearing, for each projection onto the ''x''-''y'' plane. The points would have to form a line. + **The case is perhaps less obvious with respect to circles on the sphere that contain ''T'' but are not great circles. In this case, we note again that ''T'' itself cannot be projected onto the plane. The projections of points very close to ''T'' become very far from the origin. So any circle that contains ''T'', when projected, would have to extend without bound in both directions; it would have to be a line. + *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 tip 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). - Part of this result can be seen intuitively if one considers great circles that pass through the north pole. It also can be understood in light of {{EquationNote|Eq. 3}}, which establishes that the azimuth on the sphere is equal to the polar angle measurement of the stereographic projection. All points of a great circle that passes through the north pole will have the same azimuth; such a great circle, when projected onto the plane, will have the same polar angle for each point, so the projected points would form a line. + Here we will also prove this theorem analytically. - + - This theorem is less obvious, however, when one considers circles that pass through ''T'' but are not great circles. Here we will also prove this theorem analytically. + {{SwitchPreview|HideMessage=Click here to hide proof.|ShowMessage=Click here to show proof. {{SwitchPreview|HideMessage=Click here to hide proof.|ShowMessage=Click here to show proof. Line 149: Line 150: ::$S_2 = \{ (x, y, z) \text{ } | \text{ } x^2 + y^2 + z^2 = 1 \}$. ::$S_2 = \{ (x, y, z) \text{ } | \text{ } x^2 + y^2 + z^2 = 1 \}$. - :Planes are defined: + :Planes in $\mathbb{R}^3$ are defined: ::$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$. ::$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$. - :A circle on the sphere is the intersection of these sets: + :Any circle on the surface of a sphere is the intersection of a plane with the sphere: ::$W = S_2 \bigcap V$. ::$W = S_2 \bigcap V$.

Revision as of 10:11, 21 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
Stereographic projection of two points on a sphere. A line is drawn from the pole through some other point on the sphere and some point 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.

When P is in the upper hemisphere (as in the case of P1), Q, the point on the plane, is external to the sphere. When P is in the lower hemisphere (as in the case of P2), Q falls in the unit circle on the x-y plane.

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.
$\blacksquare$

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

• 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 may be easier to see with respect to great circles that pass through T. From Eq. 3, we know that the azimuth of any point on the sphere is the projected point's polar angle. Great circles that pass through T would have the same azimuth for each point, and so the same polar angle, or bearing, for each projection onto the x-y plane. The points would have to form a line.
• The case is perhaps less obvious with respect to circles on the sphere that contain T but are not great circles. In this case, we note again that T itself cannot be projected onto the plane. The projections of points very close to T become very far from the origin. So any circle that contains T, when projected, would have to extend without bound in both directions; it would have to be a line.
• 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 tip 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).

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:
$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$.
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         $(C + D)(X^2 + Y^2) + 2AX + 2BY - C + D = 0$

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.
$\blacksquare$

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

A difficult task for cartographers has always been 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, and each has different 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 close to the pole.

Figure X
Four examples of stereographic projections. The first two are polar stereographic projections; the second two are transverse stereographic projections.

Two polar stereographic projections. To the left, an "extreme" polar map projection maps much of the southern hemisphere (the red line denotes the equation). To the right, a more local polar map projection maps just the northern hemisphere. These are projections from the same point T; the right projection just projects fewer points and so is "zoomed in" and only includes the most accurate portions of the map. Source

Since the lengths of segments on the sphere are not preserved by the stereographic projection, stereographic map projections can become rather distorted at some points. Particularly, stereographic projections are most accurate around the pole opposite of T (see More Mathematical Explanation). Many stereographic maps are called polar for this reason; they are used to map one hemisphere and are often centered around one of the poles.

For this reason, multiple polar stereographic maps might be necessary to have an accurate picture of the whole world. The two stereographic projections to the left are both polar stereographic map projections, but the first does not seem to represent the whole world very well. The red circle denotes the equator, and the map become very distorted outside of the equator in the southern hemisphere.

Another way to map more of the world using a stereographic projection is with a transverse stereographic map.

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.