Sphere Projection: Clarification Needed

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Carol_m
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Hello,

I was wondering if anyboday can clarify this for me. I am trying to project a sphere into a plane, I am using the stereogriphic projection which I believe in cartesian coordinates is:

x'=x/(R^2-z)
y'=y/(R^2-z)

where x' and y' are the coordinates in the plane, (x,y,z) the coordinates in the sphere and R is the radius of the sphere. Am I correct using this projection?
Also if I want to have x' and y' in terms of x and y ONLY can I rewrite z=sqrt(R^2-x^2-y^2)

Thank you in advance!
 
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Carol_m said:
Also if I want to have x' and y' in terms of x and y ONLY can I rewrite z=sqrt(R^2-x^2-y^2)

This only works if you wish to project only one half of the sphere, either the half with positive z or with negative z. But if you want to project the entire sphere, it won't work because there are two different points of the sphere that has the same x and y coordinates, the only difference being their z coordinate, that is +z for one point and -z for the another one.
 
Hi Coelho,

Thank you for pointing that out, I did not think about it!

On the other hand, do you know if the steregraphic projection formula that I wrote is correct?

Thanks for your help in advance :)
 
Yes, it's right. It's what you get if you sit the sphere on top of the plane, with the south pole of the sphere at the origin of the plane. Pick a point P on the sphere and draw a line through the north pole N and P. Where the line NP cuts the plane is the stereographic projection of P. This works for all points except the North pole z=R itself, which we might say projects to 'infinity' (though bear in mind this is a purely formal assignment). Draw a picture and make sure you can work out the equations you gave for yourself.

Beware that you might come across different conventions. For example, I prefer to use the projection with the plane cutting the equator of the sphere. Or if we want to include the north pole we can swap the role of the poles, so the projection is undefined (or infinity) for the south pole instead.
 
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