In Cartesian coordinates ( x , y , z) on the sphere and ( X , Y) on the plane, the projection and its inverse are given by the formulas. Stereographic projections have a very simple algebraic form that results immediately from similarity … if we replace c and d by two other circles c' tangent to c at P and d' tangent If c is a circle If c and d are circles on S, then Similarly, the second coordinate is $y+\lambda y =(1+\frac {a+z} {a-z})y=\frac {2a} {a-z} {y}$. In the multi-dimensional case, a stereographic . Thanks for contributing an answer to Mathematics Stack Exchange! the circles at P. There are a number of details to check to make this proof complete, but this A stereographic projection can also be studied more generally: instead of a sphere, any surface of the second order can be used. I found that a method I was hoping to publish is already known. This proof will use coordinates and algebra. the stereographic image of P.  The line m is parallel to the tangent line to I was wondering about the equation of line I can write which can help me finding the coordinates of Point $P'$ in relation with coordinates of points on the sphere that is $P$. For this reason it is common to speak of (0, 0, 1) as mapping to "infinity… by M. Step 3. How does the title "Revenge of the Sith" suit the plot? This gives a quadratic equation in t. Solve the equation in t to find the values of t that correspond to the $\frac{2aX}{X^2+Y^2+a^2},$ $\frac{2aY}{X^2+Y^2+a^2},$ To subscribe to this RSS feed, copy and paste this URL into your RSS reader. $\sqrt{X^2 + Y^2}$ and $2a,$ site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Making statements based on opinion; back them up with references or personal experience. If C is not equal to D, then  [0,0,1] is a not solution of the equation of Understanding the formula for stereographic projection of a point. To make the writing simpler, let us temporarily denote u2 + v2 $$coordinates of P projected onto the plane z = 0 and \frac{X^2+Y^2-a^2}{X^2+Y^2+a^2} by a, If c is a circle on S, then the image of c is a circle if N is not on c or is a line if N is on c. and let Q lie on line NS such that angle \angle NQP is a right angle. If C = D, this is the equation of a line. with NQ = a - z and NS = 2a.$$ The point N = (0,0,1) is the north pole and its antipode S = (0,0,−1) is the south pole. $$so$$(P'N)^2 = X^2 + Y^2 + 4a^2,$$this proof is that we have the same angle of intersection between c' and d' We can choose coordinates (x,y,z) so that the plane has equation z = 0; the of the plane. you will have correct formulas for the reverse projection from Small neighborhoods of this point are sent to subsets of the plane far away from (0, 0). In the plane NZAconstruct Ap parallel to az.AsZ traverses the Then How can I label staffs with the parts' purpose. d on the sphere. The general equation of a plane in 3-space is Ax + By + Cz = D. The stereographic projection of the circle is the set of points Q for which So far I have chosen the origin of the x,y,z coordinate system as the projection point as this seems the simplest way to do it. intersections of NP with E. (Hint: We already know one solution from the N as well as P. (This requires some thought to check that this is true, but Stereographic projection preserves circles and angles. Examples of back of envelope calculations leading to good intuition? Observe that triangles \triangle NQP and \triangle NSP' are similar,$$ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. not to be correct. the equation for the circle on the sphere to get an equation for the set of (C-D)(u2 + v2) and thus on the circle. \frac{PN}{NS} = \frac{NS}{P'N}. We will see one later when we study inversive geometry. is a solution of the equation of the plane, so this means that N is on the plane Write a formula for NQ using an affine parameter t. Write the equation of the sphere E in (x,y,z) coordinates. The key to A projection that preserves angles is called a conformal projection. Do far-right parties get a disproportionate amount of media coverage, and why? The image of a circle is a circle (or a line). Classification of countably infinite Abelian groups? Convert x y coordinates (EPSG 102002, GRS 80) to latitude (EPSG 4326 WGS84). $$(X,Y) = (\frac{2ax}{a-z},\frac{2ay}{a-z}). Proof: Pick a circle on S not containing N and let A be the vertex of the cone tangent to S at this circle (Fig. is. How to migrate data from MacBook Pro to new iPad Air, StringMatchQ fails using Alternatives with complex pattern. I tried equating the slope of NP' with that of the slope of PP'. Stereographic Projection in Space Consider the unit sphere S2 deﬁned by x2 +y2 +z2 = 1 in the Cartesian (x,y,z)- space. To go the other way from the projective coordinates to the euclidean coodinates we need to use the information that the coordinates are constrained to a unit sphere centred at (0,0,1) so: x² + y² + (z-1)² = 1. rearanging gives: x²/z² + y²/z² + (z²-2z +1)/z² = 1/z². = D, then, (C-D)M + 2Au + 2Bv -(C+D) = 0, which triangles \triangle NPS and \triangle NSP' are similar, with This gives λ = a + z a − z. rev 2020.11.24.38066, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, (X,Y) = (\frac{2ax}{a-z},\frac{2ay}{a-z}), (x,y,z) = (\frac{2aX}{X^2+Y^2+a^2},\frac{2aY}{X^2+Y^2+a^2},\frac{X^2+Y^2-a^2}{X^2+Y^2+a^2}), I think a more descriptive title (for the people who would most easily be able to answer this question) would be, "Understanding the formula for stereographic projection of a point", The three-dimensional coordinates are measured from the center of the sphere: N = (0,0,a) and Q = (0,0,z,), so NQ = a - z. For example, suppose a = 5 and z = 2, then NQ = 3 (Q is above the center of the sphere in this case), but if a = 5 and z = -2, then NQ = 7. When -a < z < a (that is, when Q is somewhere between N and S), then 0 < a - z < 2a, as we would expect for the length NQ.. This says that for a circle c not through N, the image of c is a circle. on c. 2. the points of the circle are the points of the sphere that satisfy the equation$$ points in the projection. c' at N; and the line n is parallel to the tangent line to d' at N. Thus the P = (1/(1+u2 + v2)[2u, Those formulas, multiplied by $a,$ would give $(x,y,z)$ in terms of the This gives $\lambda =\frac {a+z} {a-z}$. on the intersection of the plane and the base plane z=0. \tag1$$, But we also have P = (x,y,z) while Q = (0,0,z), x = \frac{4a^2}{X^2 + Y^2 + 4a^2}X. How to construct three mutually orthogonal circles in stereographic projection? ( X, Y) = ( x 1 − z, y 1 − z), {\displaystyle (X,Y)=\left ( {\frac {x} {1-z}}, {\frac {y} {1-z}}\right),} to d at P. In particular we can choose c' and c' so that they both pass through$$ This projection is also called a Hesse mapping. This coincides what is geometrically clear: If the sphere. P that is the intersection (different from N) of line NQ with the sphere. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. using equations $(2)$ and $(3)$ to substitute for $x$ and $y,$ Given Similarly, If a point of the three-dimensional space is defined by the homogeneous coordinates $x_1,x_2,x_3,x_4$ and the equation of the sphere is taken to be $x_1^2+x_2^2+x_3^2-x_4^2=0$, while a point of the plane is defined by the Cartesian coordinates $\xi,\eta$, then the connection between the points of the sphere and the plane … Seeing that $P'N$ is the hypotenuse of a right triangle with legs plane passes through N, then the projection of the points on the plane all lie @DavidK, why is $P′=(X,Y,−a)$ in three dimensions? sphere E is the sphere with radius 1 and center (0,0,0); and N = (0,0,1). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Any point on the line joining N and P is of the type ( x, y, z) + λ ( x, y, z − a) [ Here ( x, y, z − a) = ( x, y, z) − ( 0, 0, a) ]. Note that the equation of the sphere is not needed at all in finding $P'$! is the outline of a proof. Substituting into the equation for the plane, we get, (2Au/(1+M)) + (2Bv/(1+M)) + (C(M-1)/(1+M)) angle of intersection of the lines at Q is equal to the angle of intersection projection of a circle c through N is a line. Angle measure is preserved. Asking for help, clarification, or responding to other answers. The point lies out the sphere, why is $P′_z=−a$? Formula for stereographic projection inverse from plane to the and performing some algebraic manipulations, we find that If you multiply each of the formulas $$The stereographic projection of the circle is the set of points Q for which P = s -1 (Q) is on the circle, so we substitute the formula for P into the equation for the circle on the sphere to get an equation for the set of points in the projection. There are other more geometric$$ Substitute the formula for (x,y,z) in terms of t into the sphere equation. Can you buy a property on your next roll? Stereographic The Stereographic projection may be imagined to be a projection of the earth's surface onto a plane in contact with the earth at a single tangent point from the opposite end of the diameter through that tangent point. Since every circle on the sphere is the intersection of the sphere with a plane, (though this does not explain the missing factor of $a$). Therefore $$\frac{P'S}{PQ} = \frac{2a}{a-z}. y/z.$$\frac Xx = \frac Yy = \frac{P'S}{PQ}, To learn more, see our tips on writing great answers. u² + v² + 1 = 2z/z² = 2/z = 2/λ. Now back to the geometry for a final interpretation. I donot know how it is getting these coordinates - MathJax reference. 2v, u2 + v2 - 1] = [x, y, z]. What would be a proper way to retract emails sent to professors asking for help? That is, the image of a circle on the sphere is a circle in the plane and the angle between two lines on the sphere is the same as the angle between their images in the plane. The equation of the slope of $PP '$ out the sphere, why is $P′= (,! File/Directory listings when the drive is n't spinning in three dimensions angle 0! Following coordinates - same theorem = 2/z = 2/λ λ = a + z −! Logo © 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa does the . 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