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6.2 More differential geometry of curves

Let $ x=x(s)$ , $ y=y(s)$ , $ z=z(s)$ or in vector form $ {\bf r}={\bf
c}(s)$ be the intersection curve with arc length parametrization. Then from (2.5) and (2.20), we have
$\displaystyle {\bf c}'(s) = {\bf t}\;,$     (6.1)
$\displaystyle {\bf c}''(s) = {\bf k}=\kappa{\bf n}\;,$     (6.2)

where $ {\bf t}$ is the unit tangent vector and $ {\bf k}$ is the curvature vector, which is the rate of change of the tangent vector. From (6.2) it follows that
$\displaystyle \kappa^2 = {\bf k}\cdot{\bf k}= {\bf c}''\cdot{\bf c}''\;.$     (6.3)

Now let us evaluate the third derivative $ {\bf c}'''(s)$ by differentiating (6.2)

$\displaystyle {\bf c}'''(s) =\kappa'{\bf n} + \kappa{\bf n}'\;,$     (6.4)

where we can replace $ {\bf n}'$ by the second equation of the Frenet-Serret formulae (2.56) yielding
$\displaystyle {\bf c}'''(s) =-\kappa^2 {\bf t} + \kappa'{\bf n} + \kappa\tau {\bf b}
\;.$     (6.5)

Since the vectors $ {\bf t}$ , $ {\bf n}$ , $ {\bf b}$ are a right-handed orthonormal triplet, the torsion can be obtained from (6.5) as
$\displaystyle \tau = \frac{{\bf b}\cdot{\bf c}'''}{\kappa}\;,$     (6.6)

provided that the curvature does not vanish.

Classical differential geometry textbooks [412,206,444,76] do not cover the case $ \kappa=0$ , which is addressed below following Ye and Maekawa [458]. When $ \kappa=0$ (6.2) does not define the unit principal normal vector. To obtain the principal normal vector at points where $ \kappa=0$ , higher order derivatives of the curve are involved. If $ \kappa \equiv 0$ , then the curve is a straight line, and the Frenet frame of the curve is not defined. We assume here that $ \kappa=0$ occurs only at isolated points. In such case, (2.56) is valid. If $ \kappa=0$ and $ \kappa'\neq 0$ , the third order derivative (6.4) reduces to

$\displaystyle {\bf c}'''(s) =\kappa'{\bf n}\;,$     (6.7)

which defines the unit principal normal vector where $ \kappa'$ is obtained from $ (\kappa')^2={\bf c}'''\cdot{\bf c}'''$ . If $ \kappa = \kappa' = 0$ and $ \kappa'' \neq 0$ , we need to evaluate the fourth order derivative by differentiating (6.4) yielding
$\displaystyle {\bf c}^{(4)}(s) =\kappa''{\bf n}\;,$     (6.8)

where $ (\kappa'')^2={\bf c}^{(4)}\cdot{\bf c}^{(4)}$ . In general, if $ \kappa = \kappa' = \cdots = \kappa^{(j-1)} = 0$ and $ \kappa^{(j)} \neq
0$ , then
$\displaystyle {\bf c}^{(j+2)}(s) = \kappa^{(j)}{\bf n}\;,$     (6.9)

where $ (\kappa^{(j)})^2= {\bf c}^{(j+2)}\cdot{\bf c}^{(j+2)}$ .

The evaluation of torsion when the curvature vanishes can be performed as follows. If $ \kappa=0$ and $ \kappa'\neq 0$ , we need to evaluate the fourth order derivative of $ {\bf c}(s)$ , i.e. $ {\bf
c}^{(4)}(s)$ . This can be obtained by differentiating (6.5) and replacing $ {\bf t}'$ , $ {\bf n}'$ , $ {\bf b}'$ using the Frenet-Serret formulae which results in:

$\displaystyle {\bf c}^{(4)}(s) = -3 {\kappa} {\kappa'} {\bf t} + (\kappa'' - \kappa
\tau^2 - \kappa^3) {\bf n} + (2 \kappa' \tau + \kappa \tau') {\bf b}\;.$     (6.10)

In this case (6.10) further reduces to
$\displaystyle {\bf c}^{(4)}(s) = \kappa'' {\bf n} + 2 \kappa' \tau {\bf b}\;,$     (6.11)

thus
$\displaystyle \tau = \frac{ {\bf b} \cdot {\bf c}^{(4)}} {2\kappa'}\;.$     (6.12)

Similarly we have
$\displaystyle {\bf c}^{(5)}(s) = (-4 \kappa \kappa''-3(\kappa')^2 +\kappa^4+
\k...
...appa'' \tau + 3\kappa'
\tau'-\kappa^3\tau-\kappa\tau^3+\kappa\tau'') {\bf b}\;.$     (6.13)

and hence, if $ \kappa = \kappa' = 0$ and $ \kappa'' \neq 0$ , then $ \tau$ becomes
$\displaystyle \tau = \frac{ {\bf b} \cdot {\bf c}^{(5)}} {3\kappa''}\;.$     (6.14)

In general, if $ \kappa = \kappa' = \cdots = \kappa^{(j-1)} = 0$ and $ \kappa^{(j)} \ne 0$ , then [458]

$\displaystyle \tau = \frac{ {\bf b} \cdot {\bf c}^{(j+3)}} {(j+1)\kappa^{(j)}}\;.$     (6.15)

Let $ x=x^A(u_{A}, v_{A})$ , $ y=y^A(u_{A}, v_{A})$ , $ z=z^A(u_{A},
v_{A})$ and $ x=x^B(u_{B}, v_{B})$ , $ y=y^B(u_{B}, v_{B})$ , $ z=z^B(u_{B}, v_{B})$ or in vector form, $ {\bf r}={\bf r}^A(u_{A},
v_{A})$ and $ {\bf r}={\bf r}^B(u_{B}, v_{B})$ , be the two parametric surfaces. Also, let us denote the two implicit surfaces as $ f^A(x,y,z)=0$ and $ f^B(x,y,z)=0$ . We assume that these surfaces are all regular. In other words

$\displaystyle {\bf r}^A_{u_{A}}\times{\bf r}^A_{v_{A}}\neq{\bf0},\;\;\;\;\;\;
{...
...neq{\bf0},\;\;\;\;\;\;
\nabla f^A\neq{\bf0},\;\;\;\;\;\;\nabla f^B\neq{\bf0}\;.$     (6.16)

The unit normal vector of a parametric surface and an implicit surface are given by (3.3) and (3.9).

So far, we have studied the intersection curve independent of the two intersecting surfaces. However, the intersecting curve can also be viewed as a curve on the two intersecting surfaces. A curve $ u=u(s)$ , $ v=v(s)$ in the $ uv$ -plane defines a curve $ {\bf r}= {\bf c}(s) ={\bf r}(u(s), v(s))$ on a parametric surface $ {\bf r}(u,v)$ , while a curve $ x=x(s), y=y(s), z=z(s)$ with constraint $ f(x(s),
y(s), z(s))=0$ defines a curve on an implicit surface $ f(x,y,z)=0$ .

We can easily derive the first three derivatives of the intersection curve $ {\bf c}'(s)$ , $ {\bf c}''(s)$ , $ {\bf c}'''(s)$ as a curve on a parametric surface using the chain rule:

$\displaystyle {\bf c}'(s) = {\bf r}_uu'+ {\bf r}_vv'\;,$     (6.17)
$\displaystyle {\bf
c}''(s) = {\bf r}_{uu}(u')^2 + 2{\bf r}_{uv}u'v' + {\bf
r}_{vv}(v')^2 + {\bf r}_uu''+ {\bf r}_vv''\;,$     (6.18)
$\displaystyle {\bf
c}'''(s) = {\bf r}_{uuu}(u')^3 + 3{\bf r}_{uuv}(u')^2v' + 3{...
...f
r}_{uv}(u''v' + u'v'') + {\bf r}_{vv}v'v'')
+ {\bf r}_uu'''+ {\bf r}_vv'''\;.$     (6.19)

Similarly we can evaluate $ \frac{df}{ds}$ , $ \frac{d^2f}{ds^2}$ and $ \frac{d^3f}{ds^3}$ as follows:

$\displaystyle \frac{df}{ds} = f_xx' + f_yy' + f_zz'=0\;,$     (6.20)
$\displaystyle \frac{d^2f}{ds^2} = f_{xx} (x')^2 + f_{yy} (y')^2 + f_{zz} (z')^2
+2(f_{xy}x'y' + f_{yz}y'z' + f_{xz}x'z')$     (6.21)
$\displaystyle +f_xx'' + f_yy'' + f_zz''=0\;,$      
$\displaystyle \frac{d^3f}{ds^3} = f_{xxx} (x')^3 + f_{yyy} (y')^3 + f_{zzz} (z')^3
+ 3(f_{xxy}(x')^2y' + f_{xxz}(x')^2z'$     (6.22)
$\displaystyle + f_{xyy}x'(y')^2 + f_{yyz}(y')^2z'
+ f_{xzz}x'(z')^2 + f_{yzz}y'(z')^2 + 2f_{xyz}x'y'z')$      
$\displaystyle +3(f_{xx}x'x'' + f_{yy}y'y'' + f_{zz}z'z'' +f_{xy}(x''y' + x'y'')$      
$\displaystyle + f_{yz}(y''z' + y'z'') + f_{xz}(x''z' + x'z'')) +
f_xx''' + f_yy''' + f_zz''' = 0\;.$      


next up previous contents index
Next: 6.3 Transversal intersection curve Up: 6. Differential Geometry of Previous: 6.1 Introduction   Contents   Index
December 2009