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Next: 11.2.3 Computation of singularities Up: 11.2 Planar offset curves Previous: 11.2.1 Differential geometry   Contents   Index


11.2.2 Classification of singularities

There are two types of singularities on the offset curves of a regular progenitor curve, irregular points and self-intersections. Irregular points include isolated points and cusps. A point $ P$ on a curve $ C$ is called an isolated point of $ C$ if there is no other point of $ C$ in some neighborhood of $ P$ . This point occurs when the progenitor curve with radius $ R$ is a circle and the offset is $ d = -R$ . A cusp is an irregular point on the offset curve where the tangent vector vanishes. Cusps at $ t=t_c$ can be further subdivided into ordinary cusps when $ \dot{\kappa}(t_c)\neq 0$ and extraordinary points when $ \dot{\kappa}(t_c)= 0$ and $ \ddot{\kappa}(t_c)\neq 0$ [102]. An isolated point and a cusp occur when $ \vert\dot{\hat{\bf r}}(t)\vert=0$ , which using (11.6) reduces to
$\displaystyle \kappa (t) = - \frac{1}{d}\;.$     (11.9)

Note that $ (1 + \kappa d)/\vert 1 +\kappa d\vert$ in (11.6) and (11.7) changes abruptly from -1 to 1 when the parameter $ t$ passes through $ t=t_c$ at an ordinary cusp, while at extraordinary points $ (1 + \kappa d)/\vert 1 +\kappa d\vert$ does not change its value (see Fig. 11.9 (b)).

Figure 11.9: (a) Offsets to a parabola $ {\bf r}=(t, t^2)^T$ , $ -2\leq t \leq 2$ (thick solid line) with offsets $ d$ =-0.3, -0.5, -0.8 (adapted from [94]), (b) at $ d$ =-0.3, -0.5 the tangent and normal vectors of the offset have the same sense as the progenitor, while at $ d=-0.8$ they flip directions
\begin{figure}\centering {
\hspace*{-20mm}
\mbox{{\psfig{file=fig/off_par_farouk...
..._par_tn_flip.PS,height=3.0in}}}}
\mbox{(a) \hspace*{8cm}\quad (b)}\end{figure}

Offset curve/surface may self-intersect locally when the absolute value of the offset distance exceeds the minimum radius of curvature in the concave regions (see Fig. 11.10 (a)). Also the offset curve/surface may self-intersect globally when the distance between two distinct points on the curve/surface reaches a local minimum (i.e. the presence of a constriction of the curve/surface as illustrated in Fig. 11.11). These local and global self-intersections can be visualized as machining a part using a cylindrical/spherical cutter whose radius is too large for $ 2\frac {1}{2}$ -D/3-D milling. It is an essential task for many practical applications to detect all components of the self-intersection points/curves correctly and generate the trimmed offset curve/surface. If the cutter follows the trimmed offset, there will be no overcut or gouging, however we are left with undercut regions which must be milled with a smaller size cutter (see Fig. 11.10 (b)).

Self-intersections of offset curves include nodes and tacnodes. A node $ P$ is a point of curve $ C$ where two arcs of $ C$ pass through $ P$ and the arcs have different tangents. A tacnode is a special case of a node whose two tangents coincide, as illustrated in Fig. 11.11. Self-intersections of an offset curve can be obtained by seeking pairs of distinct parameter values $ \sigma\neq t$ such that

$\displaystyle {\bf r}(\sigma) + d{\bf n}(\sigma) = {\bf r}(t) + d{\bf n}(t)\;.$     (11.10)

Figure 11.10: Self-intersection of the offset curve of a parabola (adapted from [254]): (a) offsets to the parabola $ {\bf r}(t)=(t,t^2)^T$ with $ d=-0.8$ and cutter path with gouging, (b) trimmed offsets to the parabola $ {\bf r}(t)=(t,t^2)^T$ with $ d=-0.8$ and cutter path with undercut
\begin{figure}\vspace*{0.5in}
\centering {
\mbox{{\psfig{file=fig/off_08.eps,hei...
...eps,height=2.2in,width=2.2in}}}}
\mbox{(a) \hspace*{5cm}\quad (b)}\end{figure}

Example 11.2.1. (see Figs. 11.9 and 11.10)
Given a parabola $ {\bf r}=(t, t^2)^T$ , $ -2\leq t \leq 2$ , the unit tangent and normal vectors are given by

$\displaystyle {\bf t} = \frac{d{\bf r}}{ds}=\frac{d{\bf r}}{dt}\frac{dt}{ds}=\f...
...,\;\;\;
{\bf n} = {\bf t} \times {\bf e}_z = \frac{(2t,-1)^T}{\sqrt{1+4t^2}}\;.$      

The curvature and its first and second derivatives are given by
$\displaystyle \kappa(t) = \frac{(\dot{{\bf r}} \times \ddot{{\bf r}}) \cdot {\bf
e}_z}{\vert\dot{{\bf r}}\vert^3} = \frac{2}{(1+4t^2)^{\frac{3}{2}}}>0\;,$      
$\displaystyle \dot{\kappa}(t) = \frac{-24t(1+4t^2)^{\frac{1}{2}}}{(1+4t^2)^3},\;\;\;
\ddot{\kappa}(t) = \frac{24(16t^2-1)}{(1+4t^2)^{\frac{7}{2}}}\;.$      

Thus a stationary point of curvature occurs at $ t=0$ . Since $ \ddot{\kappa}(0)=-24<0$ , $ \kappa(0)$ is a maximum with a curvature value $ \kappa(0) = 2$ . It is evident that the offset distance $ d$ has to be negative to have a cusp, since $ \kappa (t)$ is always positive for any $ t$ . Now let us solve $ \kappa(t)=-1/d$ for $ t$ which yields
$\displaystyle t = \pm \frac{\sqrt{^{3}\sqrt{4d^2}-1}}{2}\;.$      

We can easily see that if $ d>-\frac{1}{2}$ , there is no real root. This means that there is no singularity as long as the magnitude of the offset distance is smaller than $ \frac{1}{2}$ . If $ d=-\frac{1}{2}$ , there exists a double root $ t=0$ , while if $ d <
-1/2$ there exist two symmetric values of $ t$ . When $ d=-\frac{1}{2}$ , at $ t=0$ , we have $ \dot{\kappa}(0)=0$ , $ \ddot{\kappa}(0)\neq0$ , therefore $ t=0$ is an extraordinary point, while when $ d<-\frac{1}{2}$ , $ \dot{\kappa}(\pm t_c)\neq0$ , so at points $ t=\pm t_c$ there are ordinary cusps on the offset curve.

The offset to the parabola $ {\bf r}=(t, t^2)^T$ is given by

$\displaystyle \hat{\bf r} = (t, t^2)^T + d\frac{(2t, -1)^T}{\sqrt{1+4t^2}}\;.$      

Therefore the equations for self-intersection of offset curve to the parabola in $ x$ and $ y$ components become
$\displaystyle \sigma + \frac{2d\sigma}{\sqrt{1+4\sigma^2}} = t + \frac{2dt}{\sqrt{1+
4t^2}}\;,$      
$\displaystyle \sigma^2 -\frac{d}{\sqrt{1+4\sigma^2}} = t^2 - \frac{d}{\sqrt{1+
4t^2}}\;.$      

It is readily observed that the offset is symmetric with respect to $ y$ -axis, which implies that the pair of distinct parameter values forming the self-intersection must satisfy $ \sigma=-t$ . The $ y$ component results in identity, while the $ x$ component yields
$\displaystyle t\left(1+\frac{2d}{\sqrt{1+4t^2}}\right)=0\;,$      

Finally, the non-trivial solutions are $ t=\pm\frac{\sqrt{4d^2-1}}{2}$ .


next up previous contents index
Next: 11.2.3 Computation of singularities Up: 11.2 Planar offset curves Previous: 11.2.1 Differential geometry   Contents   Index
December 2009