# Edit Create an Image Page: Stereographic Projection

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 Image Title*: Upload a Math Image Stereographic projection maps each point on a sphere onto a plane. {{Anchor|Reference=Figure1|Link=[[Image:Stereographic diagram.png|thumb|360px|left|Figure 1
An example of a stereographic projection. Two points, ''P''1 in the upper hemisphere and ''P''2 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 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, thereby changing the projection on the plane. The sphere and north pole remain fixed; only the colors are shifted. {{HideThis|1=Interactive Applet|2=}} ==Definition== A '''stereographic projection''' maps the points of a sphere onto a plane. {{Anchor|Reference=Figure2|Link=[[Image:2dimensional.png|right|thumb|350px|Figure 2
Cross section of a sphere. Two arbitrary points on the sphere are mapped to the 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 $x^2 + y^2 + z^2 = 1$. :Let ''T'' be the north pole of the sphere, the point (0, 0, 1), and let ''T'' ' be the antipode 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. [[#Figure2|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|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, :{{EquationRef2|Eq. 1}} $(X, Y, 0) = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right)$. {{SwitchPreview|HideMessage=Click here to hide.|ShowMessage=Click here to show. |PreviewText=Derivation of Eq. 1.|FullText=: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 {{EquationNote|Eq. 1}}.
$\blacksquare$
}} Since points on the sphere are mapped ''uniquely'', or ''one-to-one'', to points on the plane, {{EquationNote|Eq. 1}} should be invertible. 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, :{{EquationRef2|Eq. 2}} $(x, y, z) = \left(\frac{2 X}{X^2 + Y^2 + 1}, \frac{2 Y}{X^2 + Y^2 + 1}, \frac{X^2 + Y^2 - 1}{X^2 + Y^2 + 1}\right)$. {{SwitchPreview|HideMessage=Click here to hide.|ShowMessage=Click here to show. |PreviewText=Derivation of Eq. 2.|FullText=:This inverse function is derived by substituting the coordinates from {{EquationNote|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 ''z''1 = 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 ''x''2 + ''bx'' + ''c'' = 0 must be equal to the product of the roots. As noted, ''z''1 = 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 {{EquationNote|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 {{EquationNote|Eq. 2}}.
$\blacksquare$
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 {{EquationNote|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, :{{EquationRef2|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. {{SwitchPreview|HideMessage=Click here to hide.|ShowMessage=Click here to show. |PreviewText=Derivation of Eq. 3.|FullText={{Anchor|Reference=Figure4|Link=[[Image:2dim for spherical proof.png|thumb|350px|right|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''. :[[#Figure4|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 {{EquationNote|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, :{{EquationRef2|Eq. 4}} $(1, \phi, \theta) = \left(1, 2 \arctan {1 \over R}, \Theta \right)$. {{SwitchPreview|HideMessage=Click here to hide.|ShowMessage=Click here to show. |PreviewText=Derivation of Eq. 4.|FullText=: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 {{EquationNote|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). {{SwitchPreview|HideMessage=Click here to hide proof.|ShowMessage=Click here to show proof. |PreviewText=|FullText=:The equation for a circle is ''R'' 2 = ''X'' 2 + ''Y'' 2. We will use the coordinates in {{EquationNote|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 [[#Prop1|Property 1]].
$\blacksquare$
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. {{SwitchPreview|HideMessage=Click here to hide proof.|ShowMessage=Click here to show proof. |PreviewText=|FullText=: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 ''x''2 + ''y''2 + ''z''2 = 1." :In general, a plane in $\mathbb{R}^3$ is defined as follows: ::$V = \{ (x, y, z) \text{ } | \text{ } Ax + By + Cz + D = 0, \text{ where } A, B, C, D \text{ are constants} \}$. {{Anchor|Reference=Figure6|Link=[[Image:Plane-Sphere intersection.png|thumb|300px|right|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, [[#Figure6|Figure 6]]. So the set of all points comprising a circle on the unit sphere is given by: ::{{EquationRef2| 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 [[#Prop2|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 {{EquationNote|Eq. 2}} into {{EquationNote|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: ::{{EquationRef2|Eq. 6}} $(C + D)(X^2 + Y^2) + 2AX + 2BY - C + D = 0$ {{Anchor|Reference=Figure7|Link=[[Image:Stereographic circles.png|thumb|350px|right|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]]}} :{{EquationNote|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 {{EquationNote|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 {{EquationNote|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 {{EquationNote|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 {{EquationNote|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 {{EquationNote|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 [[#Prop2|Property 2]].
$\blacksquare$
:[[#Figure7|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 {{EquationNote|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. {{SwitchPreview|HideMessage=Click here to hide manipulation of Case #1.|ShowMessage=Click to show manipulation of Case #1. |PreviewText=|FullText=: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 ::$\frac{C}{B} = -\frac{D}{B}$. }} {{SwitchPreview|HideMessage=Click here to hide manipulation of Case #2.|ShowMessage=Click to show manipulation of Case #2. |PreviewText=|FullText=:In the latter case where $C \neq - D$, we can divide {{EquationNote|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 ::$\left( - \frac{A}{C + D}, -\frac{B}{C + D}\right)$ :with radius ::$\frac{\sqrt{A^2 + B^2 + C^2 - D^2}}{C + D}$. }} }}
'''Property 3'''
Maps onto planes other than the ''x''-''y'' are scaled in proportion to their distance from ''T''. In [[#Definition|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'': :[itex]H = \{ (x, y, z) \text{ } Paul Nylander produces a variety of math artwork to show the intersection between computer science and graphics. His website is [http://bugman123.com/index.html bugman123.com]. Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other None Algebra Analysis Calculus Dynamic Systems Fractals Geometry Graph Theory Number Theory Polyhedra Probability Topology Other 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 conformal and equiareal. 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 [[#Prop5|Property 5]] in the [[#MME|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''' {{Anchor|Reference=Figure17|Link=[[Image:PolarCompare.png|right|thumb|600px|Figure 17
Polar stereographic projection. [http://commons.wikimedia.org/wiki/User:RokerHRO Source]]]}} The two stereographic maps in [[#Figure17|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|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 [[#Figure17|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''' {{Anchor|Reference=Figure18|Link=[[Image:TransverseCompare.png|left|thumb|600px|Figure 18
Equatorial stereographic projection. [http://commons.wikimedia.org/wiki/User:RokerHRO 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 [[#Figure17|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 [[#Figure18|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. {{Anchor|Reference=Figure19|Link=[[Image:Equatorial stereographic.gif|right|thumb|300px|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 [[#Figure18|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 [[#Figure17|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 [[#Figure19|Figure 19]] to the right, however. This more abstract image may help us understand the bizarre map in [[#Figure18|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 [[#Figure18|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 [[#Prop4|Property 4]] in the [[#MME|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 azimuth of each point on the sphere. To confirm this fact, one might find it useful to refer to the introduction of the [[#Spherical Coordinates|spherical coordinate system]] in the [[#MME|More Mathematical Explanation]]. It would also be useful to refer to {{EquationNote|Eq. 3}} in that same section. {{Anchor|Reference=Figure20|Link=[[Image:Stereographic gnomonic orthographic.png|center|thumb|684px|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 [[#Figure20|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. [[#Figure22|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." {{Anchor|Reference=Figure21|Link=[[Image:Earth from space orthographic.jpg|right|thumb|300px|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, [[#Types of Stereographic Projections|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 [[#Prop4|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. {{Anchor|Reference=Figure22|Link=[[Image:Azimuthal comparison.gif|center|thumb|500px|Figure 22
Comparison of the three azimuthal map projections: gnomonic, stereographic, orthographic.]]}} {{Anchor|Reference=Figure23|Link=[[Image:Equidistant equiareal.png|right|thumb|456px|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 [[#Prop4|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). {{{!}}border="0" cellpadding=5 cellspacing=5 {{!}}{{Anchor|Reference=FigureA|Link=[[Image:Mercator projection.jpg|center|thumb|400px|Figure A