3 ms·
> They'll be scaled such that the point not on a plane will evaluate to 1 when plugged into the equation. Evaluating a plane equation with a point will give a
by thechao 2y ago
> They'll be scaled such that the point not on a plane will evaluate to 1 when plugged into the equation.
Evaluating a plane equation with a point will give a signed distance from the point to the plane. However, unless the equation is normalized, the distance could be scaled in unusual ways that are less than useful. I think a better way to say this is that a point who's shortest straight line distance to the line is 1 will evaluate to 1, after taking the absolute value.
The LRBni ISA had a whole set of instructions designed to take advantage of signed distance fields defined in both plane equation & barycentric forms. (That's what the bit mask for the lanes was for, in part.)
- a1369209993 2y ago> However, unless the equation is normalized, the distance could be scaled in unusual ways that are less than useful. > > []the[] point not on a plane will evaluate to 1 They mean the vertex (of the tetrahedron). The signed distance function is scaled such that the (single remaining) vertex that isn't at distance 0 (by construction) will instead be at distance 1. So the distance is always scaled in a unique well-defined way (assuming not all four points are coplanar). Whether that way of scaling is useful depends on your use case, of course.