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6.3.3 Torsion and third order derivative vector

Since $ {\bf N}^A$ and $ {\bf N}^B$ lie on the normal plane, the terms $ \kappa'{\bf n} + \kappa\tau {\bf b}$ in (6.5) can be replaced by $ \gamma {\bf N}^A + \delta {\bf N}^B$ . Thus
$\displaystyle {\bf c}'''(s) =-\kappa^2 {\bf t} + \gamma {\bf N}^A + \delta {\bf N}^B\;.$     (6.35)

Now, if we project $ {\bf c}'''(s)$ onto the unit surface normal vector $ {\bf N}$ at $ P$ and denote by $ \lambda_n$ , we have
$\displaystyle \lambda_n^A = \gamma + \delta \cos\theta\;,$      
$\displaystyle \lambda_n^B = \gamma \cos\theta + \delta\;.$     (6.36)

Solving the linear system for $ \gamma$ and $ \delta$ , and substituting them into (6.35) yields
$\displaystyle {\bf c}''' = -\kappa^2 {\bf t} + \frac{\lambda_n^A - \lambda_n^B ... ...f N}^A + \frac{\lambda_n^B - \lambda_n^A \cos\theta}{\sin^2\theta} {\bf N}^B\;.$     (6.37)

Similar to the curvature vector case in Sect. 6.3.2, we need to provide $ \lambda_n^A$ and $ \lambda_n^B$ to evaluate $ {\bf c}'''$ . For a parametric surface, $ \lambda_n$ can be obtained by projecting $ {\bf c}'''$ , which is the third order derivative of the intersection curve as a curve on a parametric surface, i.e. (6.19), onto the unit surface normal vector $ {\bf N}$ , resulting in

$\displaystyle \lambda_n = {\bf c}'''\cdot{\bf N} = 3[Lu'u'' + M(u''v' + u'v'') + Nv'v''] + III\;,$     (6.38)

where
$\displaystyle III = {\bf r}_{uuu}\cdot{\bf N}(u')^3 + 3{\bf r}_{uuv}\cdot{\bf N... ...)^2v' + 3{\bf r}_{uvv}\cdot{\bf N}u'(v')^2 + {\bf r}_{vvv}\cdot{\bf N}(v')^3\;,$     (6.39)

and $ u''$ and $ v''$ in (6.38) are evaluated by taking the dot product on the both sides of (6.18) with $ {\bf r}_u$ and $ {\bf r}_v$ . Noting that $ {\bf c}'' = {\bf k}$ leads to a linear system
$\displaystyle Eu'' + Fv'' = {\bf k}\cdot{\bf r}_u - \frac{E_u}{2}(u')^2 - E_vu'v' - \left(F_v - \frac{G_u}{2}\right)(v')^2\;,$     (6.40)
$\displaystyle Fu'' + Gv'' = {\bf k}\cdot{\bf r}_v - \left({F_u-\frac{E_v}{2}}\right)(u')^2 - G_uu'v' - \frac{G_v}{2}(v')^2\;,$     (6.41)

which can be solved for $ u''$ and $ v''$ .

For an implicit surface, the projection of $ {\bf c}'''=(x''',y''',z''')$ onto the unit normal vector of the surface $ \frac{\nabla f}{\vert\nabla f\vert}$ can be obtained from (6.22) as

$\displaystyle \lambda_n = \frac{f_xx''' + f_yy''' + f_zz'''}{\sqrt{f_x^2 + f_y^2 + f_z^2}} = -\frac{F_1 + F_2 + F_3}{\sqrt{f_x^2 + f_y^2 + f_z^2}}\;,$     (6.42)

where
$\displaystyle F_1 = f_{xxx} (x')^3 + f_{yyy} (y')^3 + f_{zzz} (z')^3\;,$     (6.43)
$\displaystyle F_2 = 3(f_{xxy}(x')^2y' + f_{xxz}(x')^2z' + f_{xyy}x'(y')^2 + f_{yyz}(y')^2z' + f_{xzz}x'(z')^2 + f_{yzz}y'(z')^2 + 2f_{xyz}x'y'z')\;,$     (6.44)
$\displaystyle F_3 = 3(f_{xx}x'x'' + f_{yy}y'y'' + f_{zz}z'z'' +f_{xy}(x''y' + x'y'') + f_{yz}(y''z' + y'z'') + f_{xz}(x''z' + x'z''))\;,$     (6.45)

and $ (x'',y'',z'')$ are given by (6.28).

Finally, the torsion can be obtained from (6.6) and (6.37) as follows

$\displaystyle \tau = \frac{1}{\kappa\sin^2\theta}\{[\lambda_n^A - \lambda_n^B \... ...\bf N}^A) + [\lambda_n^B - \lambda_n^A \cos\theta] ({\bf b}\cdot{\bf N}^B)\}\;,$     (6.46)

where the binormal vector and curvature are evaluated by (2.40) and (6.34).
next up previous contents index
Next: 6.3.4 Higher order derivative Up: 6.3 Transversal intersection curve Previous: 6.3.2 Curvature and curvature   Contents   Index
December 2009