The A…H vectors are constants based on the desired transformation. You are right that they are derived from the corners of the skewed box.

You can figure them out by solving a system of linear equations. There are 8 unknown vectors, with three components each, so 24 unknowns in total. If, for each corner of the box, you substitute the appropriate s, t, u on the right and and x, y, z values on the left in the above equations, you end up with a total of 24 linear equations in the components of A…H. This system can then be solved with Gaussian elimination, or your linear solver of choice.

Linear systems like this are a fairly common occurrence in graphics papers; I guess the author figured most readers would recognize it as linear and already know how to solve such things.

I am puzzled by something I just read in a paper and am trying to understand.

It’s called “parametric trilinear transformation” and is basically a way by the author to transform the coordinates of a unit box into any skewed box. Problem is, all the author tells me other than that are these three equations:

X(s, t, u) = Ax * s + Bx * t + Cx * u + Dx * st + Ex * tu + Fx * su + Gx * stu + Hx

Y(s, t, u) = Ay * s + By * t + Cy * u + Dy * st + Ey * tu + Fy * su + Gy * stu + Hy

Z(s, t, u) = Az * s + Bz * t + Cz * u + Dz * st + Ez * tu + Fz * su + Gz * stu + Hz

where 0 <= s, t, u <= 1 are the coordinates in the unit cube and x, y, z are the real world coordinates after transforming into the skewed box.

There isn’t even an explanation what A, B, …, H are, though I very much suppose they are the cube’s eight points in some way.

It’s rather obvious that the vector P = A * s + B * t + C * u will be the point for any input s, t and u if we had only an unskewed box, as it simply means moving along the sides. The rest of the equations have to be taking care of the skewing. But that I cannot figure out.

Perhaps someone knows his stuff around this topic? I’m rather confused by the lack of information.