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6.3.2 Curvature and curvature vector

The curvature vector of the intersection curve at $ P$ , being perpendicular to $ {\bf t}$ , must lie in the normal plane spanned by $ {\bf N}^A$ and $ {\bf N}^B$ . Thus we can express it as
$\displaystyle {\bf k} = \alpha {\bf N}^A + \beta {\bf N}^B\;,$     (6.24)

where $ \alpha$ and $ \beta$ are the coefficients that we need to determine. The normal curvature at $ P$ in direction $ {\bf t}$ is the projection of the curvature vector $ {\bf k}$ onto the unit surface normal vector $ {\bf N}$ at $ P$ given by
$\displaystyle \kappa_n = {\bf k}\cdot{\bf N}\;.$     (6.25)

By projecting (6.24) onto the normals of both surfaces (see Fig. 6.1) we have
$\displaystyle \kappa_n^A = \alpha + \beta \cos\theta\;,$      
$\displaystyle \kappa_n^B = \alpha \cos\theta + \beta\;,$     (6.26)

where $ \theta$ is the angle between $ {\bf N}^A$ and $ {\bf N}^B$ and is evaluated by
$\displaystyle \cos\theta = {\bf N}^A\cdot{\bf N}^B\;.$     (6.27)

Solving the coefficients $ \alpha$ and $ \beta$ from linear system (6.26), and substituting into (6.24) yields

$\displaystyle {\bf k} = \frac{\kappa_n^A - \kappa_n^B cos\theta}{sin^2\theta} {\bf N}^A + \frac{\kappa_n^B - \kappa_n^A cos\theta}{sin^2\theta} {\bf N}^B\;.$     (6.28)

It follows that if we can evaluate the two normal curvatures $ \kappa_n^A$ and $ \kappa_n^B$ at $ P$ , we are able to obtain the curvature vector of the intersection curve at $ P$ from (6.28). Note that (6.28) does not depend on the type of surfaces. Let us first derive the normal curvature for a parametric surface. Recall that the curvature vector of the intersection curve is also given by (6.18) considered as a curve on the parametric surface. The normal curvature is obtained by projecting (6.18) onto the unit surface normal
$\displaystyle \kappa_n = L(u')^2 + 2 Mu'v' + N(v')^2\;,$     (6.29)

where $ L$ , $ M$ , $ N$ are the second fundamental form coefficients (3.28).

We still need to evaluate $ u'$ , $ v'$ to compute (6.29). Since we know the unit tangent vector of the intersection curve from (6.23), we can find $ u'$ and $ v'$ by taking the dot product on both hand sides of (6.17) with $ {\bf r}_u$ and $ {\bf r}_v$ , which leads to a linear system

$\displaystyle Eu' + Fv' = {\bf r}_u\cdot{\bf t}\;,$     (6.30)
$\displaystyle Fu' + Gv' = {\bf r}_v\cdot{\bf t}\;,$     (6.31)

where $ E$ , $ F$ , $ G$ are the first fundamental form coefficients given in (3.12). Thus,
$\displaystyle u' = \frac{({\bf r}_u\cdot{\bf t})G - ({\bf r}_v\cdot{\bf t})F}{E... ...uad v' = \frac{({\bf r}_v\cdot{\bf t})E - ({\bf r}_u\cdot{\bf t})F}{EG- F^2}\;,$     (6.32)

where $ EG-F^2\neq0$ , since we are assuming regular surfaces (see (6.16)). Similarly we can compute the normal curvature of the implicit surface by using (6.21). The projection of curvature vector $ {\bf c}'' = (x'', y'', z'')$ onto the unit normal vector $ \frac{\nabla f}{\vert\nabla f\vert}$ of the surface, from (6.21), is given by
$\displaystyle \kappa_n = \frac{f_xx'' + f_yy'' + f_zz''}{\sqrt{f_x^2 + f_y^2 + ... ...')^2 +2(f_{xy}x'y' + f_{yz}y'z' + f_{xz}x'z')}{\sqrt{f_x^2 + f_y^2 + f_z^2}}\;,$     (6.33)

where $ x'$ , $ y'$ , $ z'$ are the three components of $ {\bf t}$ given by (6.23).

Consequently, the curvature of the intersection curve $ {\bf c}$ at $ P$ can be calculated using (6.3), (6.27) and (6.28) as follows:

$\displaystyle \kappa = \sqrt{{\bf k}\cdot{\bf k}} = \frac{1}{\vert\sin\theta\ve... ...\sqrt{(\kappa_n^ A)^2 + (\kappa_n^B)^2 - 2 \kappa_n^A \kappa_n^B \cos\theta}\;.$     (6.34)

next up previous contents index
Next: 6.3.3 Torsion and third Up: 6.3 Transversal intersection curve Previous: 6.3.1 Tangential direction   Contents   Index
December 2009