Some time ago I was curious about a question related to my bath towel, which I hang on a rope to have fun (you can use your own towel to do this experiment in bath-o if you want): 'There is this rectangle with sides $a<b$. The rectangle is bent along a line that passes through the center of the rectangle. At which angle $\alpha$ (should we bend the rectangle in order to get the minimum area of crossing intersection?' - the solution is posted in OEIS
By now I'd like to find a solution for a similar problem but related to a spheric rectangle. I know I can't bend the rectangle on sphere. However, I guess it's possible to use reflection of a relatively large circle to find coordinates of vertices in Cartesian coordinates. My problem is that I have not realized how to do it and how to calculate the area of the self-intersection on a sphere with radius $R$ centered at the origin?
Firstly, I considered the area of the spheric rectangle to see how it works for a small area$S=R^{2}(\theta+\theta+\theta+\theta - 2\pi)=R^{2}(4\theta - 2\pi)$. Let denote the vertices of the spheric rectangle by $$ v_{1} = (A, B, C),\qquad v_{2} = (A, -B, C),\qquad v_{3} = (A, -B, -C),\qquad v_{4} = (A, B, -C). $$ Of course, $A^{2} + B^{2} + C^{2} = R^{2}$. (Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.) This resulted in the area $S= R^{2}(4\theta - 2\pi) = R^{2}(2\pi - 4\arccos(\tan\frac{a}{2R} \tan\frac{b}{2R}))$.
So, Tylor series (if $R=1$): $S=2\pi - 4(\pi/2-\frac{a}{2} \frac{b}{2})=ab$ like for area of rectangle on plane.
Any ideas on finding area of the spherical polygon while 'folding' spherical rectangle are highly welcomed.