Here's my explanation of BSP trees, written as much for my own benefit as anybody else's (maybe it will help you): To start off with, we have a basic geometric

Here's my explanation of BSP trees, written as much for my own benefit as anybody else's (maybe it will help you): To start off with, we have a basic geometric fact: if you have a surface that divides space into separate regions (this can be a closed surface like a sphere, or an infinite surface like a plane), a line of sight can't cross between the two partitions of space without passing through the surface. If the surface is convex, a line of sight can only cross it once. (That's one definition of convexity.) Now, suppose that you have a lot of objects, and none of them are on both sides of the boundary. You can ensure that condition by splitting any object that crosses the boundary in two. No matter where the line of sight begins, and no matter what direction it moves in, the boundary then offers a definite, albeit partial ordering of events: objects on the same side of the boundary may intersect, then the line will intersect the boundary, then the objects on the other side may intersect. If you find an intersection with an opaque object on the same side of the boundary, you never need to check any objects on the other side, because the order of events above. That is a big benefit when about half of the objects in front of you are on your side of the boundary: if you are right in front of the boundary, or if there's nothing on the other side of it, it doesn't help a lot. Checking all of the objects on one side of the boundary is the same kind of problem as checking all the objects, and that means you can speed those up too by having sub-boundaries for the sub-problems. That gives you the recursion and makes it a tree. If the reasoning based on boundaries continues until there is only one object in a sub-problem, you don't need to sort lists based on distance - meaning that the whole ordering problem can be solved just with partition trees. Choosing a good tree is a difficult problem because, unlike the fact that the ordering problem can always be solved, the amount that an individual boundary division helps depends on where you and the objects are in relation to it. Planes are generally used instead of other things like spheres because splitting a polygon along the edge of a sphere doesn't result in another polygon, while splitting it along a plane does.