# What is the area of a spherical non-geodetic triangle?

The question relates to a general polygon, but I guess a single triangle is a reasonable reduction.

I can formulate the problem as follows: Given three points $v_1,v_2,v_3$ on the unit sphere, and three quaternions $q_{12}, q_{23}, q_{31}$ that transform the respective points to each other, what is the signed area enclosed by the interpolated curve? Assume it is supposed to be right-handed positive for an outward normal to the sphere.

The transformations are not necessarily geodetic (great circles), which mean we don't have that $q_{ij}=v_i^{-1}v_j$ (assuming $v$ as imaginary quaternions). This rules out the simple Girard formula for spherical excess (or does it?).

• Do you connect the vertices by interpolating the quaternions? – Justin Solomon Nov 13 '16 at 19:27
• A quaternion $q_{ij}$ defines a curve $v_{ij}(t), t \in [0,1]$ by an interpolation function: $v_{ij}(t)=q_{ij}^{-1}(t)v_iq_{ij}$, where $q_{ij}(t)=(q_{ij})^t$. – Amir Vaxman Nov 13 '16 at 19:28
• Why do you need quaternions if you have the three points $v_1$, etc.? – Tpofofn Nov 14 '16 at 3:11
• Because there is an extra degree of freedom, as these are not shortest rotations/great circles. See last paragraph. – Amir Vaxman Nov 14 '16 at 9:51
• As area on the sphere is proportional to angle excess, I'd guess one should be able to integrate the angle excess along each edge, then add the angle excess at each of the corners. – MvG Nov 14 '16 at 14:12