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6.5.2 Tangential intersection of implicit-implicit surfaces

In this example, the two implicit surfaces are ellipsoids given by
$\displaystyle f^A(x,y,z)= \frac{x^2}{0.6^2}+ \frac{y^2}{0.8^2} +\frac{z^2}{1^2}-1=0\;,$      
$\displaystyle f^B(x,y,z)= \frac{x^2}{0.45^2}+ \frac{y^2}{0.8^2} +\frac{z^2}{1.25^2}-1=0\;.$      

The partial derivatives of the two implicit functions are readily computed as $ f_x^A(x,y,z)=\frac{x}{0.18}$ , $ f_y^A=3.125y$ , $ f_z^A=2z$ , $ f_{xx}^A= \frac{1}{0.18}$ , $ f_{yy}^A=3.125$ , $ f_{zz}^A=2$ , $ f_{xy}^A=f_{yz}^A=f_{xz}^A=0$ , $ f_x^B(x,y,z)=\frac{x}{0.10125}$ , $ f_y^B=3.125y$ , $ f_z^B=1.28z$ , $ f_{xx}^B=\frac{1}{0.10125}$ , $ f_{yy}^B=3.125$ , $ f_{zz}^B=1.28$ , $ f_{xy}^B=f_{yz}^B=f_{xz}^B=0$ . The partial derivatives higher than order two are all zero. It is clear that at the intersection point $ P$ (0, 0.8, 0), $ \nabla f^A$ and $ \nabla f^B$ become parallel, thus $ P$ 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 $ P$ . Since $ f_y^A\neq 0$ at $ P$ , we can express $ y'$ in terms of $ x'$ and $ z'$
$\displaystyle y'=-\frac{f_x^Ax' + f_z^Az'}{f_y^A}\;.$      

Furthermore $ y'$ reduces to zero, because $ f_x^A=f_y^A=0$ at $ P$ . At $ P$ the normal curvatures of both implicit surfaces in direction $ \bf t$ are the same, thus from (6.33) we have
$\displaystyle \frac{f_{xx}^A(x')^2 + f_{zz}^A(z')^2}{f_y^A}= \frac{f_{xx}^B(x')^2 + f_{zz}^B(z')^2}{f_y^B}\;.$      

This is a quadratic equation in $ x'$ and $ z'$ that can be solved for $ z'$ in terms of $ x'$ . By substituting $ x=z=0$ , $ y=0.8$ , we have $ z'=\pm\frac{25}{27}\sqrt{7}x'$ and hence $ P$ is a branch point. Normalization gives the unit tangent vector $ (x', y', z') = \left(\frac{27}{4\sqrt{319}}, 0, \pm\frac{25\sqrt{7}}{4\sqrt{319}}\right)$ .

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

$\displaystyle f_x^Ax'' + f_y^Ay''+ f_z^Az''=-f_{xx}^A(x')^2 - f_{yy}^A(y')^2 - f_{zz}^A(z')^2\;.$     (6.79)

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
$\displaystyle \frac{f_{xx}^Ax'x'' + f_{yy}^Ay'y'' + f_{zz}^Az'z''}{\sqrt{(f_x^A... ...' + f_{yy}^By'y'' + f_{zz}^Bz'z''}{\sqrt{(f_x^B)^2 + (f_y^B)^2 + (f_z^B)^2}}\;.$     (6.80)

The third equation is due to the fact that the curvature vector is perpendicular to the tangent vector, i.e.
$\displaystyle x'x''+y'y''+z'z''=0\;.$     (6.81)

Solving the linear system (6.79), (6.80) and (6.81) at point $ P$ , we have $ (x'', y'', z'')=(0,-\frac{320}{319}, 0)$ , thus the curvature of the intersection curve at $ P$ is $ \kappa=\frac{320}{319}$ .

next up previous contents index
Next: 7. Distance Functions Up: 6.5 Examples Previous: 6.5.1 Transversal intersection of   Contents   Index
December 2009