Sphere parameterization with 6 patches I am looking for a parameterization of the sphere with 6 patches, like in http://www.image.ucar.edu/staff/rnair/research09.html 
and the inverse of this parameterization.
As well, I would need a method to determine, for a point $x$ (which coordinates are known of course) on the sphere, which patch (patches for boundaries) $x$ lies in (with patches defined by the above parametrization).
Thanks.
 A: Hint: you get the parametrization by composing "parametrisation" of the surface of the cube and normalizing points of that surface viewed as vectors in a euclidean space.
How to determine which patch point belongs to? We know that points on the sphere are normalized points from the surface of the cube. So it is enough to find such a way of determining a face of the cube that behaves well with respect to change of length (normalization). Can you find such a method?
EDIT. I'll show two methods of determining the patch. Let's assume we are talking about the unit sphere in $\mathbb R^3$ ($x^2+y^2+z^2=1$). The inscribed cube is $\left[-\frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right]^3$.
Method 1. The top patch is that patch where $z>0$ and $|z|=\max(|x|,|y|,|z|)$. (It's obviously a correct method for the cube surface and it relies on inequalities between coordinates, so behaves well with respect to normalizing.) The same with the other patches.
Method 2. Note that the cube is a ball in the maximum ($\ell_\infty$) metric. So our surface may be described by the formula $\max(|x|, |y|, |z|)=\frac{1}{\sqrt3}$.
Now: we have "parametrization" $g:\partial C\to S^2$ (from the cube surface to the sphere) given by normalization with respect to the euclidean metric. By the previous observation we can easily find its inverse - it's given by the formula (it's just normalizing with respect to the maximum metric + homothety):
$$g^{-1}(x,y,z) = \frac{(x,y,z)}{\sqrt3 \max(|x|,|y|,|z|)}.$$
So the second method is first use $g^{-1}$ and then just test which coordinate(s) are equal to $\pm\frac{1}{\sqrt3}$.
EDIT2 I realized that $\pi/4$ from your link may not be a random mistake. Maybe they want a different parametrization, not coming from the natural parametrization of the cube surface (but maybe more natural for the sphere). Consider the following (for the top patch):
$$\varphi'(\alpha, \beta) = (
\sin\alpha \cdot \cos\beta,
\cos\alpha \cdot \sin\beta,
\cos\alpha \cdot \cos\beta)$$
$$\varphi(\alpha, \beta)=\varphi'(\alpha, \beta)\cdot S\left(\varphi'(\alpha, \beta)\right)$$
Why this formula? $\pi/4$ is an angle - so I wanted the parametrization to be ingular in some sense. But I can't just take one vector and rotate it around one axis by $\alpha$ and then around another by $\beta$ - rotations are not be commutative and we want our solution to be symmetric.
It also translates nicely to the cube - if you normalize it with respect to the maximum metric (you can divide it by $\sqrt3$ if you prefer) you end up with a nonstandard parametrization of the top face of the cube:
$$(\tan\alpha, \tan\beta, 1).$$
EDIT3 (the OP asked in comments for a version for an ellipsoid)
Let's consider an ellipsoid given by the formula $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$. It is the image of the unit sphere under linear automorphism given in the standard basis by a diagonal matrix $D=\mathrm{diag}(a,b,c)$.
It seems that the requested projection is a projection from $[-a,a] \times [-b,b] \times [-c,c]$ (the case of projecting a cube was already discussed in the comments). But this case is actually trivial - if we apply $D^{-1}$ to our picture, we will get the known case of projecting the unit cube $[-1,1]^3$ onto the unit sphere.
All together
Actually we can handle all the cases (cube or cuboid, sphere or ellipsoid, the same parameters for cuboid and ellipsoid or different ones) in one go. We need to firstly: parametrize, secondly: determine the patch. The parametrization of the surface of a cube/cuboid is easy, the tricky thing is how to project it onto a sphere/ellipsoid. One of the ways of determining a patch is by projecting back onto the cube/cuboid. So all we need is projections.
Projections there and back are done by normalizing with respect to appropriate norm. So, projection to the surface of the cube $[-a,a] \times [-b,b] \times [-c,c]$ may be done with the formula $(x,y,z)\mapsto\frac{(x,y,z)}{\|(x,y,z)\|_\infty^{a,b,c}}$, where $\|(x,y,z)\|_\infty^{a,b,c}$ is given by the formula:
$$\|(x,y,z)\|_\infty^{a,b,c} = \left\|\left(\frac{x}{a},\frac{y}{b},\frac{z}{c}\right)\right\|_\infty.$$
Similarly, projection on the ellipse given by $\frac{x^2}{d^2} + \frac{y^2}{e^2} + \frac{z^2}{f^2}=1$ may be done with the formula $(x,y,z)\mapsto\frac{(x,y,z)}{\|(x,y,z)\|_2^{d,e,f}}$, where $\|(x,y,z)\|_\infty^{d,e,f}$ is given by:
$\|(x,y,z)\|_\infty^{d,e,f} = \left\|\left(\frac{x}{d},\frac{y}{e},\frac{z}{f}\right)\right\|_2.$
