This question is inspired by this answer I gave, where I promised I would do a computation. The computation in question is

Parametrize the unit sphere by spherical coordinates $(\theta,\phi)$, where $\theta$ is the latitude (with the north pole given by $\theta=0$ and the south pole by $\theta = \pi$) and $\phi$ is the longitude. Given a point $(\theta,\phi)$, take the circle of fixed angle $\epsilon\ll\pi$ centered in $(\theta,\phi)$ and find the ratio of circumference that is below a given meridian $\theta' =\text{const}$.

Notice that without loss of generality, $\phi=0$ by symmetry. I have two possible ways I could use to proceed:

  1. Apply a rotation $R\in SO(3)$ that sends $(\theta,\phi)$ to the south pole, so that the circle of radius $\epsilon$ is sent to a meridian. Now project to $\mathbb{R}^2$ via stereographic projection, find the equation for the image of the rotated meridian, and then find the intersection points of the two curves.
  2. Directly apply stereographic projection, find the curves corresponding to the great circles passing from the center of the circle and the intersection of the circle with the meridian, then compute the angle of intersection of these curves. As the stereographic projection is conformal, the angles we find give us the ratios we needed.

However, both these ways would involve a pretty big amount of calculations (also, there are some special cases to consider when $\theta$ is close enough to $0,\pi$ that the circle "contains" a pole). Does anyone know of a faster or smarter way do do this?


Let $O$ be the sphere center, $P$ the center of the circle of radius $\epsilon$. Perform a rotation carrying $P$ to the north pole ($P=(0,0,1)$) and the north pole to $N=(\sin\theta, 0, \cos\theta)$. A point $A$ on the circle will then have coordinates $A=(\sin\epsilon\cos\phi,\sin\epsilon\sin\phi,\cos\epsilon)$.

The parallel $\theta'=\ $const. after the rotation is transformed into a circle of center $N$ and radius $\theta'$. To find the intersections between such circle and the one centered at $P$, observe that if $A$ belongs to the intersection then $AN=2\sin(\theta'/2)$. This leads to the equation

$$ (\sin\epsilon\cos\phi-\sin\theta)^2+ (\sin\epsilon\sin\phi)^2+(\cos\epsilon-\cos\theta)^2 =4\sin^2(\theta'/2) $$

By expanding and solving for $\cos\phi$ one then finds

$$ \cos\phi= {\cos\theta'-\cos\epsilon\cos\theta\over\sin\epsilon\sin\theta}, $$ where I also used the identity $2\sin^2(\theta'/2)=1-\cos\theta'$.

If $\phi$ is the solution of the above equation in the range $[0,\pi]$, then the angular part of circle "below" the given parallel is simply $2\pi-2\phi$.


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