# Stereographical projection?

1. Sep 21, 2007

### pivoxa15

I noticed that it dosen't project to the origin of the plane from the north pole. However the projection describes it as mapping to the whole equitorial plane which is wrong!

i.e take S^1 projecting to R.

From the north pole the projection formula is y=-(x-a)/a however a can't be 0. So the origin is never mapped.

Last edited: Sep 21, 2007
2. Sep 22, 2007

### benorin

Stereographic projection maps the plane to the unit sphere thus: Let the equator of the sphere be the unit circle in the plane. Call the north pole of the sphere N. The map any point P in the plane to the point Q on the sphere by constructing the ray $$\vec{NP}$$ and assigning the point Q to be the point of intersection of the sphere and the ray.
Clearly the origin of the plane is mapped to the south pole of the sphere, points inside the unit circle in the plane are mapped to the southern hemisphere of the sphere, and points outside the unit circle in plane are mapped to the northern hemisphere of the sphere. The so-called "point at infinity" is mapped to the north pole.

3. Sep 22, 2007

### futurebird

Whew! This was a *clear* response. I thought this was how it worked, but after reading the question I became confused.

The line through the north pole that will "go to infinity" will be tangent to the top of the sphere and parallel to the plane, as soon as you incline it, even slightly, it will go through the north pole and through one other point on the sphere.

But if the equator the sphere is the unit circle on the plane isn't the radius of the sphere going to be less than one?

If it's a unit sphere the unit circle would be mapped like this:

http://en.wikipedia.org/wiki/Image:Stereoprojnegone.svg

Right?

Last edited: Sep 22, 2007
4. Sep 22, 2007

### pivoxa15

I see what is going on now. It all works.