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\[d^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\]
change to spherical coordinates and simplify ?

So represent all the points like \((\rho, \theta, \phi\)) and see if I can get something?

Exactly! Also \(\rho \) = constant since we are "on" the sphere

No wait, that wont work. that formula still gives the direct distance between those two points

Oh. Lol, okay.

|dw:1438756659753:dw|

|dw:1438756776294:dw|

So we're looking at a triangle with a vertex at the origin, right?

Yep, isosceles triangle

|dw:1438757317770:dw|

\(d^{2} = R^{2} + R^{2} - 2R^{2}\cos(t)\)
\(\frac{2R^{2}-d^{2}}{2R^{2}} = \cos(t)\)

|dw:1438757456725:dw|

|dw:1438757852912:dw|
So if we need \(\phi\) we want \(\psi = \frac{\pi}{2} - \phi\) I guess.

that conversion from regular spherical to lat/long looks good to me

A few things simplify if you expand that out, but its still nasty as a direct distance.

Okay, Im with ya there. So that's definitely close to what we need.

we may simply equate that to another \(d^2\) that we get using cosine law in the triangle

Ah. With the angle t still or do we need \(\theta\) and/or \(\psi\)?

|dw:1438759481827:dw|

our goal is to get \(Rt\) in terms of angles

looks good!
\(s = \text{shortest distance between A,B over sphere}=|AB|\)

does this make sense ?
http://mathworld.wolfram.com/SphericalDistance.html

https://en.wikipedia.org/wiki/Spherical_law_of_cosines