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I'm trying to implement the traversal algorithm for binary rope trees suggested here, however, I'm stuck at finding a fast and precise way to determine the exit face of a ray-AABB intersection, with the purpose of finding the next rope to follow. The paper does not indicate a method for this, and neither do any articles I've read from the Internet.

Is there a better way to get the exit face of a ray-AABB intersection than checking the intersection of each plane with the ray? (even with dot product elimination, I still see this method as not very precise)

Thank you.

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I think I have discovered a satisfactory solution! Knowing the exit point, the exit plane is the one with the smallest distance to the exit point. It's as precise as the ray-AABB intersection algorithm and (being an AABB) it only requires a few subtractions, which makes it pretty fast.

I will mark the question as answered when I can actually test this method.