Assume I have a camera defined by its position and direction, and a box defined by its center and extents (three orthogonal vectors from the box center to face centers). Face is visible when its outer surface is facing the camera and invisible when its inner surface is facing it.
It seems obvious that depending on box position and orientation there may be 1-3 faces of the box visible. Is there some clever way how to determine which faces are visible? An obvious solution would be to compute 6 dot-products of the face normal against the face-camera vector for each face. Is there a better way?
Note: perspective projection will be used, but I do not think it matters, the property of “facing camera” seems independent to a projecting.
I believe the method you described is the normal way to do this. It’s a very fast calculation so you shouldn’t be worried too much about speed. This is the same method they use to reduce the number of calculations for ray-triangle intersection algorithms. If the front of the face isn’t visible, the method doesn’t continue calculations for that face. See this paper for a c++ implementation of this algorithm. It’s in the first half of the calculations. http://jgt.akpeters.com/papers/MollerTrumbore97/code.html