3
$\begingroup$

It appears to me that after repeated applications of Girard's theorem on the area of spherical triangles that we can obtain the surface area of a spherical polygon with interior angles $\theta_{1},...,\theta_{n}$ on a sphere of radius $R$ is $$\text{area}(\text{spolygon}(\theta_{1},...,\theta_{n}))=R^2 \left(\sum_{i=1}^{n} \theta_{i} - (n-2)\pi\right).$$ Does this make sense to anyone else? I wanted to check before I used the result in a paper I am writing. Even if this just holds for spherical squares, it is all I need.

$\endgroup$
4
$\begingroup$

Yeah, you can always break a spherical n-gon in n−2 spherical triangles... You know, an interesting fact about higher dimensional spherical and euclidean simplices is that not all things generalize. For instance there is no relation between angle sums and volume in spherical simplices in the n-dimensional sphere for odd n (see here), and the fact that the sum of the angles is π for euclidean triangles does not generalize (see here) even in 3 dimensions...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.