6.5.1 Transversal intersection of parametric-implicit surfaces

In this example, the parametric surface is a hyperbolic paraboloid given by

and the implicit surface is a cone given by

Figure 6.4 shows the two intersecting surfaces with two intersection curves, one of which coincides with the -axis.

Figure 6.4: Transversal intersection of parametric-implicit surfaces (adapted from [458])

From (6.78) we have , , , and the partial derivatives of order higher than two are all zero. The first and second fundamental form coefficients are readily given by , , , , .

Similarly we have , , , , , and the partial derivatives of order higher than two are all zero.

The unit normal vectors of surfaces and and their dot product are given by

Hence, the unit tangent vector of the intersection curve becomes

To evaluate the normal curvature of parametric surface in the direction of , we start by computing using (6.32) yielding

and hence

The normal curvature of the implicit surface in the direction can be obtained from (6.34)

By substituting , , , and into (6.28) we obtain the curvature vector, and hence the curvature .

The projection of the third order derivatives onto the unit surface normal vector can be computed using (6.38) and (6.42) for the parametric and implicit surfaces, respectively, yielding

where are obtained by solving the linear system (6.40) and (6.41)

Knowing , , , , , , and , the third order derivative is obtained from (6.37) and the torsion from (6.46).