Imagine that you have a sphere in $\mathbb{R}^3$ and a plane (that is parallel to the x,y plane) through the sphere.
Now you want to have a clockwise rotation in the x/y plane that does the following:
The point that ends up after the rotation at the top of the sphere should have been rotated by an angle $\alpha$. The one that ends up at the bottom of the sphere should have been rotated by an angle $\beta$. The ones that end up at the intersection between plane and sphere should have been rotated by $\frac{\alpha + \beta}{2}$. My question is: Can anybody here construct an interpolation of this rotation?(Or even give me a linear transformation that does that?)
Note that although my painting could suggest that the plane goes through the center of the sphere, this is not necessarily the case!
A few details aren't clear (to me) from context: (i) Are $\alpha$ and $\beta$ specified in advance? (ii) Is the "height" of the cutting plane specified in advance, or can it be chosen once $\alpha$ and $\beta$ are known? (iii) Do all four points in question lie on a great circle perpendicular to the cutting plane (in which case the word "sphere" may as well be replaced by "circle" in the question)?
In any case, assuming I understand the spirit of the question: If you let $P$ be the plane containing your four points and you introduce an angle coordinate $\theta \in [-\pi, \pi]$ in $P$, measured from the center of the sphere and with $\theta = 0$ at the top of the sphere, and if $\pm\theta_0$ denotes the angle where $P$ intersects the cutting plane, you're looking for a monotone function $f:[-\pi, \pi] \to \mathbf{R}$ whose values at four points satisfy $$f\bigl(-\theta_0 - (\alpha+\beta)/2\bigr) = -\theta_0,\quad f(-\alpha) = 0,\quad f\bigl(\theta_0 - (\alpha+\beta)/2\bigr) = \theta_0,\quad f(\pi - \beta) = \pi.$$ This should be a straightforward interpolation (subject to any other criteria you wish).