Given width and height rectangle, how can I calculate diagonal when it's superimposed on a round surface (e.g. Earth)

Question

I have an width (1,250 km) and height (624 km) of a rectangle. Assuming that Earth is perfectly round, how can I calculate the length of the diagonal when the rectangle is superimposed onto the planet?

Answer 1

As mentioned in the comments you have to give some explanations when you draw a rectangle $R$ on the surface of the (spherical) earth. A simple concept is the following:

Let the horizontal midline of $R$ lie on the equator $\epsilon$. Measuring off $625$ km on both sides of the midpoint $M$ draw two meridians through the points $P_1$, $P_2\in\epsilon$, and measure off $312$ km north and south along both of them. Then connect the resulting two pairs of points with horizontal latitude arcs. The "rectangle" obtained in this way has the desired symmetries and right angles at the corners, but two of its edges are not geodesics on $S^2_{\rm earth}$. It is reasonable to draw the diagonals as arcs of great circles through two diagonally opposite vertices $V_i$. By symmetry these diagonals will intersect at $M$. Now $\triangle(MP_1 V_1)$ is a right spherical triangle. In order to obtain the length of the hypotenuse we have to reduce the given lengths of the legs to radius $1$ of the sphere. Then compute the length of the hypotenuse using the relevant formula of spherical trigonometry, and scale up the result by the earth radius, then multiply by $2$.