next up previous contents index
Next: 11. Offset Curves and Up: 10. Geodesics Previous: 10.7 Geodesic offsets   Contents   Index


10.8 Geodesics on developable surfaces

In this section it is shown that all two point BVPs for solving geodesics on developables can be reduced to IVPs using the differential geometry properties introduced in Sect. 9.7.1. Since a geodesic on a developable surface maps to a straight line on the developed plane, there is only one solution to the system (10.17)$ -$ (10.20) on a developable surface. Here we exclude periodic surfaces such as cylinders where there can be more than one solution. The basic procedure is to map the two desired points on the developable surface to a plane, draw the straight line between them and determine the angle between the generator and the geodesic line at one of the end points. The angle can be used to determine the initial direction $ (u',v')=(p,q)$ . Thus, all the information required for an IVP is available.
Figure 10.13: Geodesic on a degree (3,1) developable surface (adapted from [251])
\begin{figure}\centerline{ \psfig{figure=fig/geo_plane.ps,height=2.5in} \psfig{f... ...o_surf.ps,height=3in} \psfig{figure=fig/geo_surf1.ps,height=3in} }\end{figure}
Given two points $ A$ and $ B$ on the developable surface $ {\bf r}(u,v)$ as shown in Fig. 10.13(c), the corresponding points $ (X_A,Y_A)$ and $ (X_B,Y_B)$ in the developed planar surface are required. The Frenet-Serret formulae (2.56) state that $ {\bf t}' = -\kappa {\bf n}$ where $ {\bf t}$ is the unit tangent vector to a curve, $ {\bf n}$ is the unit normal vector to a curve and $ \kappa$ is the curvature. The minus sign ensures that $ \kappa$ is positive when $ \bf n$ points away from the center of curvature (see Table 3.2). For a planar curve in the $ (X,Y)$ plane, we can define the unit normal vector as $ {\bf n} = {\bf t} \times {\bf e_z}$ where $ {\bf e_z} = (0,0,1)$ . Substituting this equation into the Frenet-Serret formulae yields
$\displaystyle \frac{d^2X}{ds^2} + \kappa \frac{dY}{ds} = 0,\;\;\;\; \frac{d^2Y}{ds^2} - \kappa \frac{dX}{ds} = 0\;,$     (10.75)

where $ (X,Y)$ denote the 2D coordinates on the developed plane (X,Y). If we rewrite the first equation of (10.75) in terms of the parameter $ u$ , we obtain
$\displaystyle \frac{d^2X}{ds^2} + \kappa \frac{dY}{ds} = \frac{d^2X}{du^2}\left... ...X}{du}\left(\frac{d^2u}{ds^2}\right) + \kappa \frac{dY}{du}\frac{du}{ds} = 0\;.$     (10.76)

Similarly the second equation of (10.75) can be rewritten in terms of the parameter $ u$ . Since $ \frac{du}{ds}\neq 0$ , (10.75) reduce to
$\displaystyle \frac{d^2X}{du^2} + \frac{\left(\frac{d^2u}{ds^2}\right)} {\left(... ... \frac{dY}{du} - \frac{\kappa}{\left(\frac{du}{ds}\right)} \frac{dX}{du} = 0\;.$     (10.77)

The development is based on the fact that curves on isometric surfaces have the same geodesic curvatures. Therefore, $ \kappa$ in (10.77) can be replaced by $ \kappa_g$ , the geodesic curvature of the curve on the developable surface [116]. If we choose the curve on the developable surface to be $ {\bf r}(u,v_n)$ , an iso-parametric curve in terms of $ u$ , we can replace $ \frac{du}{ds}$ and $ \frac{d^2u}{ds^2}$ in (10.77) by

$\displaystyle \frac{du}{ds} = \frac{1}{\vert{\bf r}_u(u,v_n)\vert}, \;\;\;\;\;\... ...,v_n) \cdot {\bf r}_{uu}(u,v_n)} {({\bf r}_u(u,v_n)\cdot{\bf r}_u(u,v_n))^2}\;.$     (10.78)

Thus we have
$\displaystyle \frac{dX}{du} = p,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \frac{dY}{du} = q\;,$     (10.79)
$\displaystyle \frac{dp}{du} = p\frac{({\bf r}_u\cdot{\bf r}_{uu})}{({\bf r}_u \... ...cdot{\bf r}_{uu})}{({\bf r}_u\cdot{\bf r}_u)} + p\kappa_g\vert{\bf r}_u\vert\;.$     (10.80)

To find the points $ A$ and $ B$ in the plane, we first set $ (X_0, Y_0)$ as the $ (0,0)$ point in the plane corresponding to $ {\bf r}(0,0)$ on the surface. We integrate the system (10.79) and (10.80) along the directrix that corresponds to $ v=0$ to determine the point $ C=(u_A,0)$ , shown in Fig. 10.13(a). Since isometric maps are conformal, the angle between the directrix and the generator at $ (u_A,0)$ is the same in both representations and can be found by $ \cos\theta = \frac{{\bf r}_u(u_A,0)}{\vert{\bf r}_u(u_A,0)\vert} \cdot \frac{{\bf c}}{\vert{\bf c}\vert}$ , where $ {\bf c}=\overrightarrow{CA}$ is a vector whose direction corresponds to the iso-parametric line $ {\bf r}(u_A,v)$ which is a straight line on the surface. Therefore it is a geodesic and will be developed into the plane as a straight line. The distance is given by $ \vert{\bf c}\vert = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2 + (z_A - z_C)^2}$ . The point $ A$ on the plane is found using $ C$ , $ \vert{\bf c}\vert$ and $ \theta$ . The point $ B$ on the plane is found by following the same procedure, and the points are connected as shown in Fig. 10.13(a). The angle $ \omega$ between $ {\bf c}$ and $ {\bf b}=\overrightarrow{AB}$ is given by $ \cos\omega = \frac{{\bf b} \cdot{\bf c} }{\vert{\bf b}\vert\vert{\bf c}\vert}$ . The angles $ \omega$ and $ \theta$ are shown in Figs. 10.13(a) and (c). This angle $ \omega$ is preserved between the iso-parametric line $ {\bf r}(u_A,v)$ and the geodesic curve $ {\bf g}(s)$ on the developable surface at point $ A$ . Thus we have $ \cos\omega = \frac{{\bf r}_v \cdot {\bf g}'(s)}{\vert{\bf r}_v\vert\vert{\bf g}'(s)\vert}$ , where the tangent vector to the geodesic is given by

$\displaystyle {\bf g}'(s) = {\bf r}_u\frac{du}{ds} + {\bf r}_v\frac{dv}{ds}\;.$     (10.81)

Multiplying (10.81) by $ {\bf r}_v$ yields
$\displaystyle {\bf r}_v \cdot {\bf g}'(s) = {\bf r}_u \cdot {\bf r}_v\frac{du}{... ...f r}_v\frac{dv}{ds} = \cos (\omega) \vert{\bf r}_v\vert\vert{\bf g}'(s)\vert\;,$     (10.82)

which (since $ \vert{\bf g}'(s)\vert = 1$ ) can be reduced to
$\displaystyle \frac{dv}{ds} = \frac{\cos (\omega)}{\sqrt{G}} - \frac{F}{G}\left(\frac{du}{ds}\right)\;,$     (10.83)

where $ F = {\bf r}_u \cdot {\bf r}_v$ and $ G = {\bf r}_v \cdot {\bf r}_v$ (coefficients of the first fundamental form). From the first fundamental form,
$\displaystyle {\bf g}'(s) \cdot {\bf g}'(s) = E \left(\frac{du}{ds}\right)^2 + 2F\frac{du}{ds}\frac{dv}{ds} + G \left(\frac{dv}{ds}\right)^2 = 1\;.$     (10.84)

Plugging (10.83) into (10.84) and solving for $ \frac{du}{ds}$ yields
$\displaystyle \frac{du}{ds} = \pm \sqrt{\frac{\sin^2(\omega) G}{EG-F^2}}\;,$     (10.85)

and thus (10.83) reduces to
$\displaystyle \frac{dv}{ds} = \frac{\cos (\omega)}{\sqrt{G}} \pm \frac{1}{\sqrt{G}}\frac{F}{\sqrt{EG-F^2}}\sin(\omega)\;.$     (10.86)

Evaluating (10.85) and (10.86) at the initial point, we have all the initial conditions required to solve the IVP ((10.17) to (10.20)) for a geodesic. The solution to the IVP yields the $ uv$ parametric values for the geodesic that are graphed in Fig. 10.13(b). The corresponding three-dimensional coordinate values are shown in Figs. 10.13(c) and (d). The geodesic runs from $ (u_A, v_A)$ = (0.1, 0.3) to $ (u_B, v_B)$ = (0.9, 0.8).

next up previous contents index
Next: 11. Offset Curves and Up: 10. Geodesics Previous: 10.7 Geodesic offsets   Contents   Index
December 2009