Projecting a band of a sphere onto a 2D surface For a craft project, I want to take a "band" of a sphere (i.e. the area between two latitudes) and project it onto a plane, so that I can fold the 2d shape onto the sphere and recreate the band in 3 dimensions.
The sphere I am working with has a radius of 13cm. The band I wish to project is the area between a great circle and a small circle with radius 11.2cm.
If my trig is correct, the distance between these two circles, following the surface of the sphere, is about 6.76cm.

I do not know how to perform a projection given this setup. From similar crafts I can see that one strategy is to create a sort of arced rectangle, which looks as though it could be constructed by aligning a circle and an offset of the same circle.

However I am unsure how the arcs could be calculated.
 A: As noted in the comments, there is no projection that makes it possible to exactly fold the 2d shape onto the sphere, without creasing, tearing or stretching. Depending on the materials used, it may or may not look nice.
An arced rectangle like you suggest can be constructed quite easily; it is a region bounded by two circle segments on opposing sides, and two straight line segments at the other two sides, with the sides meeting at right angles.
The lengths of these two straight line segments are both precisely the distance between the two circles along the sphere. The radius of the sphere is $13$, and so the circle of radius $11.2$ is at a latitude of
$$\arccos\left(\frac{11.2}{13}\right)\approx0.5325,$$
measured in radians. This means the distance between the circles along the sphere is
$$13\arccos\left(\frac{11.2}{13}\right)\approx6.92.$$
The lengths of the two circular arcs are the circumferences of the two circles, which are
$$11.2\cdot2\pi\approx70.37\qquad\text{and}\qquad 13\cdot2\pi\approx81.86.$$
These two circular arcs cover the same arc length, say $\theta$, but their radii differ by $6.92$. If we measure $\theta$ in radians and label the radii by $r_1$ and $r_2$, this means
$$r_1\theta=70.37,\qquad r_2\theta=81.86,\qquad r_2-r_1=6.92.$$
It follows that
$$81.86=r_2\theta=(r_1+6.92)\theta=r_1\theta+6.92\theta=70.37+6.92\theta,$$
from which it follows that $\theta\approx1.634$, and hence that
$$r_1\approx43.07\qquad\text{ and }\qquad r_2\approx50.00.$$
Note that now the difference between the radii is $6.93$, so depending on the desired accuracy, you may want to take care when rounding at intermediate steps.
