Calculating a spherical angle
Given a sphere with a specified radius, and two perpendicular arcs produced by angles, phi and theta.
To be clear, phi and theta are the angles which give rise to the arcs which meet at a right angle. I placed the labels, phi and theta, on the actual arcs because it gets crowded near the center of the sphere. I am looking for the interior angle which gives rise to the oblique arc on the sphere's surface.
I read these links:
How do I measure distance on a globe?
relationship between a great circle arc and a latitude circle arc at a given latitude
but the derivation of an answer eludes me.
I realize that spending time to answer a question so fundamental and which has probably been answered clearly in some text is a waste of bandwidth. So maybe someone knows of a clear, simple exposition of this problem to which I can be referred?
$\endgroup$2 Answers
$\begingroup$The unit vectors in the directions of the sides of the unknown angle are (in a convenient basis)$$ v_1=(\cos\phi,0,\sin\phi),\qquad v_2=(\cos\theta,\sin\theta,0) $$Therefore, the cosine of the angle between them is their dot product$$ \langle v_1,v_2\rangle=\cos\phi\cos\theta $$That is, the angle is$$ \cos^{-1}(\cos\phi\cos\theta). $$
$\endgroup$ 2 $\begingroup$The spherical right triangle in non-euclidean geometry this relation can be written as$$ \cos\gamma= \cos\phi\cdot \cos\theta $$
In series expansion upto second degree$$ 1- \gamma^2/2!\approx (1- \phi^2/2!)\cdot(1- \theta^2/2!) $$
It can be readily derived using spherical trig Law of Cosines $ \cos \ \alpha \approx 1- \alpha^2$ etc., it should be memorized. It is a sort of deeper non-linear hypotenuse.
Reducible in Euclidean geometry to
$$ \gamma^2\approx \phi^2+\theta^2\;;\;c^2=a^2+b^2.$$
[ Although not relevant here, tempted to mention in hyperbolic geometry its mirroring relation.. $ \cosh\gamma= \cosh\phi\cosh\theta $].
$\endgroup$ 1
