I am a programmer, don't know much about advanced math. I would need the exact formula(s) that could achieve this, so I can translate it to my programming language. I am having a headache trying to figure out a way to do this.
Lets say we have a sphere, whose center is in P(0,0,0). It's radius is 1 unit. As far as I know, an infinite number of points form that sphere. A couple of those could be (0,1,0), (1,0,0), (-1,0,0),(0,-1,0), and so on.
I need to build a function that, given two integers U and V, each from 0 to 360 not included, will return a point of that sphere P(X,Y,Z) calculated the next form:
1) First we use a flat circle where all its points are z=0 (its flat on the ground). We start from a position on that circle, (maybe 0,1,0), and move U degrees along the circle. For example, if U = 90º, it would move to (1,0,0). If U = 45º, then I don't quite know. 2) Then from there, it will move V degrees upwards, over an imaginary circle that has been created from the previous point, and that has its center also in (0,0).
Finally it should return a point P(X,Y,Z), which, whatever the value of U and V, will always be part of the sphere of radius 1 and center (0,0,0).
It will be used on a 3D software, when the camera has to move to any position on the screen, with always the same length from where it is pointing. I know It's hard, but I really need this to be done. I whish I could use my programing-stackoverflow reputation to set a bounty to this question.
Where $\phi$ is latitude and $\theta$ is longitude, and your chosen coordinate setup:
$$(\cos\phi\times\sin\theta, \cos\phi\times\cos\theta, \sin\phi)$$
To see how this works, consider the circle that is the parallel of latitude. What $z$ coordinate does it have? What is its radius?
Major Warning: math libraries tend to work in radians, not degrees, for a variety of reasons that mostly involve simplicity of various calculations. A whole circle is $2\pi$ radians.
Minor Warning: When doing this, be aware that there is gimbal lock: if $\phi$ is $90°$ or $-90°$, $\theta$ will have no effect. This means in particular that in games like Half-Life, aiming at the sky and spinning around gets you still aiming at the sky. The alternative is arguably worse though: a more robust camera system can roll as well, and that can screw you up really spectacularly.