Surface Tangent Vectors:

Illustrate that if the boundary curves of a bilinear Coons patch are coplanar, the resulting patch is also planar.

Solution

We have to show that the surface normal at any particular point on the surface is normal to the plane of the boundary curves. The surface tangent vectors are following :

P_u (u, v) = [P_u (u, 0) + P (1, v) - P (0, v) + P (0, 0) - P (1, 0)] + v [(P_u (u, 1)

- P_u (u, 0) + P (1, 0) - P (0, 0) - P (1, 1) + P (0, 1)) = A + v B and

P_v (u, v) = [P_v (0, v) + P (u, 1) - P (u, 0) + P (0, 0) - P (0, 1)] + u [P_v (1, v)

- P_v (0, v) - P (0, 0) + P (0, 0) + P (0, 1) + P (1, 0) - P (1, 1)] = C + u D

It may easily be illustrated that P_u (u, v) and P_v (u, v) lie in the plane of the boundary curves. Considering P_u (u, 0), P (1, 0), P (1, 0) - P (0, 0), and P (0, 1) - P (1, 1) also lie in the given plane. Thus, vectors A and B lie in the plane. As a result, the tangent vector P_u (u, v) to the surface at any point (u, v) lies in the plane of the boundary curves. The similar argument may be extended to P_v (u, v).

The surface normal is specified by :

N (u, v) = P_u × P_v = (A + v B) × (C + u D)

That is perpendicular to the plane of P_u and P_v and, thus, the plane of the boundary curves. Therefore, for any point on the surface, the direction of the surface normal is constant (the magnitude based on the point) or the unit normal is fixed in space. By Knowing that the plane surface is the merely surface that has a fixed unit normal, we conclude that a bilinear Coons patch degenerates to a plane if its boundary curves are coplanar. Therefore, this patch may be used to create planes with curved boundaries same to the bicubic surface covered. Note, though, that a bicubic Coons patch or any other patch that has nonlinear blending functions does not decrease to a plane while all its boundaries are coplanar.