next up previous contents index
Next: 2.3 Binormal vector and Up: 2. Differential Geometry of Previous: 2.1 Arc length and   Contents   Index


2.2 Principal normal and curvature

If $ {\bf r}(s)$ is an arc length parametrized curve, then $ {\bf r}'(s)$ is a unit vector (see (2.5)), and hence $ {\bf r}'\cdot{\bf r}'=1$ . Differentiating this relation, we obtain
$\displaystyle {\bf r}'\cdot{\bf r}''=0\;,$     (2.16)

which states that $ {\bf r}''$ is orthogonal to the tangent vector, provided it is not a null vector. This fact can be also interpreted from the definition of the second derivative $ {\bf r}''(s)$
Figure 2.4: Derivation of the normal vector of a curve (adapted from [455])
\begin{figure}\centerline{\psfig{figure=fig/normal.eps,height=2.5in}}\end{figure}

$\displaystyle {\bf r}''(s) = \lim_{\Delta s \rightarrow 0}\frac{{\bf r}'(s+\Delta s) - {\bf r}'(s)}{\Delta s}\;.$     (2.17)

As shown in Fig. 2.4, the direction of $ {\bf r}'(s+\Delta s) - {\bf r}'(s)$ becomes perpendicular to the tangent vector as $ \Delta s \rightarrow 0$ . The unit vector

$\displaystyle {\bf n} = \frac{{\bf r}''(s)}{\vert{\bf r}''(s)\vert} = \frac{{\bf t}'(s)}{\vert{\bf t}'(s)\vert}\;,$     (2.18)

which has the direction and sense of $ {\bf t}'(s)$ is called the unit principal normal vector at $ s$ . The plane determined by the unit tangent and normal vectors $ {\bf t}(s)$ and $ {\bf n}(s)$ is called the osculating plane at $ s$ . It is also well known that the plane through three consecutive points of the curve approaching a single point defines the osculating plane at that point [412].

When $ {\bf r}'(s+\Delta s)$ is moved from $ Q$ to $ P$ , then $ {\bf r}'(s)$ , $ {\bf r}'(s+\Delta s)$ and $ {\bf r}'(s+\Delta s) - {\bf r}'(s)$ form an isosceles triangle (see Fig. 2.4), since $ {\bf r}'(s+\Delta s)$ and $ {\bf r}'(s)$ are unit tangent vectors. Thus we have $ \vert{\bf r}'(s+\Delta s) - {\bf r}'(s)\vert=\Delta\theta\cdot1 = \Delta\theta =\vert{\bf r}''(s)\Delta s\vert$ as $ \Delta s \rightarrow 0$ and hence

$\displaystyle \vert{\bf r}''(s)\vert = \lim_{\Delta s \rightarrow 0}\frac{\Delt... ... 0}\frac{\Delta\theta}{\varrho\Delta\theta}= \frac{1}{\varrho} \equiv \kappa\;.$     (2.19)

$ \kappa$ is called the curvature , and its reciprocal $ \varrho$ is called the radius of curvature at $ s$ . It follows that
$\displaystyle {\bf r}'' = {\bf t}' = \kappa {\bf n}\;.$     (2.20)

The vector $ {\bf k} = {\bf r}'' = {\bf t}'$ is called the curvature vector, and measures the rate of change of the tangent along the curve. By definition $ \kappa$ is nonnegative, thus the sense of the normal vector is the same as that of $ {\bf r}''(s)$ .

The curvature for arbitrary speed (non-arc-length parametrized) curve can be obtained as follows. First we evaluate $ \dot{\bf r}$ and $ \ddot{\bf r}$ by the chain rule

$\displaystyle \dot{\bf r} = \frac{d{\bf r}}{ds}\frac{ds}{dt}= {\bf t}v \;,$     (2.21)
$\displaystyle \ddot{\bf r} = \frac{d}{dt}[{\bf t}v] = \frac{d{\bf t}}{ds}v^2 + {\bf t}\frac{dv}{dt} = \kappa{\bf n}v^2 + {\bf t}\frac{dv}{dt}\;,$     (2.22)

where $ v=\frac{ds}{dt}$ is the parametric speed. Taking the cross product of $ \dot{\bf r}$ and $ \ddot{\bf r}$ we obtain
$\displaystyle \dot{\bf r} \times \ddot{\bf r} = \kappa v^3 {\bf t}\times{\bf n} \;.$     (2.23)

For the planar curve, we can give the curvature $ \kappa$ a sign by defining the normal vector such that $ ({\bf t}, {\bf n}, {\bf e}_z)$ form a right-handed screw, where $ {\bf e}_z=(0,0,1)^T$ as shown in Fig. 2.5. The point where the curvature changes sign is called an inflection point (see also Fig. 8.3).

Figure 2.5: Normal and tangent vectors along a 2D curve
\begin{figure}\centerline{\psfig{figure=fig/normal2d.eps,height=2.3in}}\end{figure}

According to this definition the unit normal vector of the plane curve is given by

$\displaystyle {\bf n} = {\bf e}_z \times {\bf t} = \frac{(-\dot{y},\dot{x})^T}{\sqrt{\dot{x}^2 + \dot{y}^2}}\;,$     (2.24)

and hence from (2.23) we have


$\displaystyle \kappa = \frac{(\dot{\bf r} \times \ddot{\bf r})\cdot{\bf e}_z}{v... ...ac{\dot{x}\ddot{y} - \dot{y}\ddot{x}}{(\dot{x}^2 + \dot{y}^2)^{\frac{3}{2}}}\;.$     (2.25)

For a space curve, by taking the norm of (2.23) and using (2.4), we obtain
$\displaystyle \kappa = \frac{\vert\dot{\bf r} \times \ddot{\bf r}\vert}{\vert\dot{\bf r}\vert^3}\;.$     (2.26)

The normal vector for the arbitrary speed curve can be obtained from $ {\bf n} = {\bf b}\times{\bf t}$ , where $ {\bf b}$ is the unit binormal vector which will be introduced in Sect. 2.3 (see (2.41)).

The unit principal normal vector and curvature for implicit curves can be obtained as follows. For the planar curve the normal vector can be deduced by combining (2.14) and (2.24) yielding

$\displaystyle {\bf n} = {\bf e}_z \times {\bf t} = \frac{(f_x, f_y)^T}{\sqrt{f_x^2 + f_y^2}} = \frac{\nabla f}{\vert\nabla f\vert}\;,$     (2.27)

where only the $ +$ sign of $ {\bf t}$ was used (although it is not necessary).

We will introduce a derivative operator with respect to arc length so that the derivation becomes simple. If we rewrite the plane implicit curve as $ f(x(s), y(s))=0$ where $ s$ is arc length along the implicit curve, the total derivative with respect to the arc length becomes

$\displaystyle \frac{df}{ds} = \frac{\partial f}{\partial x}\frac{dx}{ds} + \frac{\partial f}{\partial y}\frac{dy}{ds}\;.$     (2.28)

Now if we replace $ \frac{dx}{ds}$ and $ \frac{dy}{ds}$ by using (2.5) and (2.14) (+ sign), we obtain the derivative operator with respect to arc length


$\displaystyle \frac{d}{ds} = \frac{1}{\vert\nabla f\vert}\left(f_y\frac{\partial }{\partial x}- f_x\frac{\partial }{\partial y}\right)\;.$     (2.29)

By applying the operator (2.29) to (2.14) (+ sign) and equating with $ \kappa{\bf n}$ (using (2.20) and (2.27)), we obtain
$\displaystyle \kappa = -\frac{f_{xx}f_y^2 - 2f_{xy}f_xf_y + f_x^2f_{yy}}{(f_x^2 + f_y^2)^{\frac{3}{2}}}\;.$     (2.30)

For a 3-D implicit curve, we can deduce a derivative operator [444] similar to (2.29),


$\displaystyle \frac{d}{ds} = \frac{1}{\vert\mbox{\boldmath $\alpha$}\vert}\left... ..._2\frac{\partial }{\partial y} + \alpha_3\frac{\partial }{\partial z}\right)\;,$     (2.31)

where $ \alpha$ is the tangent vector of the 3-D implicit curve (see (2.15)) given by
$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle = (\alpha_1, \alpha_2, \alpha_3) = \nabla f\times\nabla g\;,$     (2.32)

and
$\displaystyle \alpha_1 = \frac{\partial f}{\partial y}\frac{\partial g}{\partial z} - \frac{\partial g}{\partial y}\frac{\partial f}{\partial z}\;,$     (2.33)
$\displaystyle \alpha_2 = \frac{\partial g}{\partial x}\frac{\partial f}{\partial z} - \frac{\partial f}{\partial x}\frac{\partial g}{\partial z}\;,$     (2.34)
$\displaystyle \alpha_3 = \frac{\partial f}{\partial x}\frac{\partial g}{\partial y} - \frac{\partial g}{\partial x}\frac{\partial f}{\partial y}\;.$     (2.35)

By applying the derivative operator (2.31) to $ \vert$$ \alpha$ $ \vert{\bf t}=$   $ \alpha$ we obtain

$\displaystyle \frac{d\vert\mbox{\boldmath$\alpha$}\vert{\bf t}}{ds} = \frac{1}{... ...al y} + \alpha_3\frac{\partial \mbox{\boldmath$\alpha$} }{\partial z}\right)\;,$     (2.36)

which gives
$\displaystyle \vert$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \vert^2\kappa{\bf n} + \vert$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \vert\vert$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \vert'{\bf t} = \left(\alpha_1\frac{\partial \mbox{\boldmath$\alp... ...al y} + \alpha_3\frac{\partial \mbox{\boldmath$\alpha$} }{\partial z}\right)\;.$     (2.37)

Taking the cross product of $ \vert$$ \alpha$ $ \vert{\bf t}=$   $ \alpha$ and (2.37) yields
$\displaystyle \vert$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \vert^3\kappa{\bf b} =$   $\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \times \left(\alpha_1\frac{\partial \mbox{\boldmath$\alpha$}}{\pa... ...al y} + \alpha_3\frac{\partial \mbox{\boldmath$\alpha$} }{\partial z}\right)\;.$     (2.38)

Thus,
$\displaystyle \kappa = \frac{\left\vert\mbox{\boldmath$\alpha$} \times \left(\a... ...ha$} }{\partial z}\right)\right\vert} {\vert\mbox{\boldmath$\alpha$}\vert^3}\;.$     (2.39)

A different derivation of the curvature of a 3-D implicit curve is given in Sect. 6.3.2.

next up previous contents index
Next: 2.3 Binormal vector and Up: 2. Differential Geometry of Previous: 2.1 Arc length and   Contents   Index
December 2009