clamp Spherical coordinates to latitude/longitude

Question

For usage in a shader I am looking for an efficient way to convert spherical coordinates to latitude/longitude values in $[0,\pi]$, $[0,2\pi]$ respectively. (Conversion to convential long/lat with these values would be trivial)

For the spherical coordinates I'm using the mathematical convention with $\phi$ corresponding to the latitude and $\phi=0$ at the north pole and $\theta$ corresponding to the longitude.

As a starting point one could write

$$lat = \phi$$ $$long = mod(\theta, 2\pi) $$

$r$ is left out as it is constant.

Now the difficulty I'm having is extending this to convert spherical coordinates with $\phi$ not in $[0,\pi]$. In this case spherical coordinates can be written as (again ignoring $r$)

$$\vec{r} = \phi\hat{\phi} + \theta\hat{\theta}$$

I am not sure how I should express the way in which the latitude $\phi$ influences the longitude if

$$\pi<mod(\phi, 2\pi)<2\pi$$

If I'm not mistaken, one way would be to write it as a piecewise function like this:

$$ \vec{r}_{longlat}=\begin{cases} mod(\phi, \pi)\hat{\phi} + mod(\theta, 2\pi)\hat{\theta} & mod(\phi, 2\pi)\in[0,\pi]\\ (\pi-mod(\phi, \pi))\hat{\phi} + mod(\theta+\pi, 2\pi)\hat{\theta} & mod(\phi, 2\pi)\notin[0,\pi] \\ \end{cases} $$

However, I'm wondering if there's a more elegant way to go about it.

Another way to do it would be to simplify the result of converting the spherical coordinates to carthesian coordinates and then getting the spherical coordinates back from that again. In that case one finds the carthesian coordinates,

$$ \vec{p} = \begin{bmatrix} cos(\theta)sin(\phi) \\ sin(\theta)sin(\phi) \\ cos(\phi) \end{bmatrix} $$

which can be converted back to get

$$ \vec{r}_{longlat} = \begin{bmatrix} arccos(cos(\phi)) \\ atan2(cos(\theta)sin(\phi), sin(\theta)sin(\phi)) \\ \end{bmatrix} $$

Then the $arccos(cos(\phi))$ gives the ping-pong behaviour and atan2 would handle the offset on $\theta$ based on $\phi$.

Which of these two makes more sense to use? Which would you expect to be more efficient? Are there any alternatives/well known formulae I'm not considering?

Answer 1

If you want efficiency, using $\text{arccos}(\cos \phi)$ is like driving 5 miles south to the ring-freeway, following it around to the north side of town, and driving another 5 miles south to visit the second house up the block from where you started. Similarly for $\text{atan2}$. You would be hard-pressed to come up with a more in-efficient solution.

Seriously - computers directly add, subtract, multiply, and divide. Anything else takes multiple steps to calculate - usually many such steps. It used to be that even multiplying and dividing required multiple steps, but then chip technology improved until the whole logic could be performed in one step. And yes, on the most advanced chips, some functions - square rooting in particular - are now incorporated into a single operation. But even if the trig functions were all incorporated, why would you bother when the problem has a much more simple computation?

A simple algorithm takes care of it:

ToLatitudeLongitude(phi, theta)    
   phi = mod(phi, 2*pi)
   theta = mod(theta, 2*pi)
   if (phi > pi) 
      phi = 2*pi - phi
      if (theta < pi) 
         theta = theta + pi
      else 
         theta = theta - pi

   return (phi, theta)