6.5.2 Tangential intersection of implicit-implicit surfaces

In this example, the two implicit surfaces are ellipsoids given by

The partial derivatives of the two implicit functions are readily computed as , , , , , , , , , , , , , . The partial derivatives higher than order two are all zero. It is clear that at the intersection point (0, 0.8, 0), and become parallel, thus is a tangential intersection point (see Fig. 11.17). Now let us compute the tangential direction and curvature vector of the intersection curves at the tangential intersection point . Since at , we can express in terms of and

Furthermore reduces to zero, because at . At the normal curvatures of both implicit surfaces in direction are the same, thus from (6.33) we have

This is a quadratic equation in and that can be solved for in terms of . By substituting , , we have and hence is a branch point. Normalization gives the unit tangent vector .

Next, we evaluate the curvature vector as described in Sect. 6.4.2. The first linear equation in is obtained from (6.21)

The second equation is given by equating the projections of the third derivatives of the two implicit surfaces onto the normal vector of the surface

The third equation is due to the fact that the curvature vector is perpendicular to the tangent vector, i.e.

Solving the linear system (6.79), (6.80) and (6.81) at point , we have , thus the curvature of the intersection curve at is .