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3.6 Euler's theorem and Dupin's indicatrix

The normal curvatures of a surface in an arbitrary direction (in the tangent plane) at point $ P$ can be expressed in terms of principal curvatures $ \kappa_{1}$ and $ \kappa_{2}$ at point $ P$ and the angle $ \Phi$ between the arbitrary direction and the principal direction corresponding to $ \kappa_1$ , namely,
$\displaystyle \kappa_n = \kappa_1\cos^2\Phi + \kappa_2\sin^2\Phi\;.$     (3.87)

This is known as Euler's theorem. For simplicity, we assume that the iso-parametric curves of a surface are lines of curvature, which leads to $ F=M=0$ (see (3.60)). Now (3.26) takes the form

$\displaystyle \kappa_n = \frac{Ldu^2 + N dv^2}{Edu^2 + G dv^2}\;.$     (3.88)

For $ v=const$ iso-parametric lines $ dv=0$ and for $ u=const$ iso-parametric lines $ du=0$ , thus the principal curvatures $ \kappa_1$ and $ \kappa_2$ are given by:
$\displaystyle \kappa_1 = \frac{L}{E},\;\;\;\;\;\; \kappa_2 = \frac{N}{G}\;.$     (3.89)

The angle $ \Phi$ between the direction $ \frac{dv}{du}$ and the principal direction corresponding to $ \kappa_1$ ($ dv_1=0$ , $ u_1$ arbitrary) is evaluated by (3.17) as

$\displaystyle \cos\Phi=E\frac{du}{ds}\frac{du_1}{ds_1}\;.$     (3.90)

Since $ ds_1=\sqrt{Edu_1^2}$ and $ ds=\sqrt{Edu^2 + Gdv^2}$ we deduce
$\displaystyle \cos\Phi = \sqrt{E}\frac{du}{ds},\;\;\;\;\;\;\sin\Phi = \sqrt{G}\frac{dv}{ds}\;.$     (3.91)

As a consequence, we have the Euler's theorem (3.88).

Next we explain Euler's theorem in a more simple way. Let us consider a section of the surface cut by a plane parallel to the tangent plane at the point $ P$ , and at an infinitesimal distance $ h>0$ from it [441]. We also consider a plane through $ P$ containing the normal vector. If we denote the intersection points of the surface and the two planes by $ Q$ and $ Q'$ , the signed radius of curvature of this normal section by $ \varrho$ , and the length of $ QQ'$ by $ 2R$ as shown in Fig. 3.11, we have the relation

$\displaystyle (\vert\varrho\vert - h)^2 + R^2 = \vert\varrho\vert^2\;,$     (3.92)

thus
$\displaystyle R^2 = 2h\vert\varrho\vert\;,$     (3.93)

to the first order. If $ \Phi$ is the inclination of this normal section to the principal direction corresponding to $ \kappa_1$ , Euler's theorem provides
$\displaystyle \kappa_1\cos^2\Phi + \kappa_2\sin^2\Phi= \frac{1}{\varrho}= \pm \frac{2h}{R^2}\;.$     (3.94)

If we set
$\displaystyle \xi=R\cos\Phi,\;\;\;\;\;\;\eta=R\sin\Phi\;,$     (3.95)

we obtain
$\displaystyle \frac{\xi^2}{2h\varrho_1} + \frac{\eta^2}{2h\varrho_2} = \pm 1\;,$     (3.96)

where $ \varrho_1$ and $ \varrho_2$ are principal radius of curvatures. Consequently a section of the surface cut by a plane parallel to the tangent plane at the point $ P$ , and at an infinitesimal distance is a conic section. If we scale the $ \xi$ -$ \eta$ coordinates as follows
$\displaystyle X= \frac{\xi}{\sqrt{2h}}=\frac{R}{\sqrt{2h}}\cos\Phi=\sqrt{\vert\varrho\vert}\cos\Phi,$     (3.97)
$\displaystyle Y= \frac{\eta}{\sqrt{2h}}=\frac{R}{\sqrt{2h}}\sin\Phi=\sqrt{\vert\varrho\vert}\sin\Phi\;,$     (3.98)

we obtain
$\displaystyle \frac{X^2}{\varrho_1} + \frac{Y^2}{\varrho_2} = \pm 1\;.$     (3.99)

This equation determines a conic section called Dupin's indicatrix as shown in Fig. 3.12. If $ P$ is an elliptic point, both principal curvatures have the same sign, and the indicatrix is an ellipse, while if it is a hyperbolic point, the principal curvatures have different sign and the indicatrix consists of a pair of hyperbolas with asymptotic lines $ Y=\pm\sqrt{\frac{\vert\varrho_2\vert}{\vert\varrho_1\vert}}X$ . If one of the principal curvatures vanishes, it is a parabolic point and the indicatrix yields a pair of parallel lines.

Figure 3.11: Cross section of the surface cut by a normal plane: (a) normal curvature is positive, (b) normal curvature is negative (Here we followed the curvature convention (a); see Fig. 3.7)
\begin{figure}\centerline{ \psfig{file=fig/dupin_cut.eps,height=3.0in}}\end{figure}

Figure 3.12: Dupin's indicatrix for (a) elliptic point, (b) parabolic point, (c) hyperbolic point
\begin{figure}\centerline{ \psfig{file=fig/dupin.eps,height=2.0in}}\end{figure}


Table 3.2: A list of equations which involves a sign change due to the sign convention of curvature of the planar curve or the normal curvature of the surface (see Fig. 3.7). In sign convention (a) the center of curvature is on the same side of the normal vector, while in sign convention (b) it is on the opposite direction

Equation		 Convention (a) Convention (b)

(2.20)		 $ {\bf r}'' = {\bf t}' = \kappa {\bf n}$ $ {\bf r}'' = {\bf t}' = -\kappa {\bf n}$

(2.22)		 $ \ddot{\bf r} = \kappa{\bf n}v^2 + {\bf t}\frac{dv}{dt}$ $ \ddot{\bf r} = -\kappa{\bf n}v^2 + {\bf t}\frac{dv}{dt}$

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

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

(2.55)		 $ {\bf n}'= -\kappa {\bf t}\; (\tau=0)$ $ {\bf n}'= \kappa {\bf t}\; (\tau=0)$

(3.23)		 $ {\bf k}_n = \kappa_n {\bf N}$ $ {\bf k}_n = -\kappa_n {\bf N}$

(3.25)		 $ \kappa_n=\frac{d{\bf t}}{ds} \cdot {\bf N} = -{\bf t} \cdot \frac{d{\bf N}}{d s} $ $ \kappa_n=-\frac{d{\bf t}}{ds} \cdot {\bf N} = {\bf t} \cdot \frac{d{\bf N}}{d s} $
  $ = -\frac{d{\bf r}}{ds} \cdot \frac{d {\bf N}}{ds} =-\frac{d{\bf r}\cdot d{\bf N}}{d{\bf r}\cdot d{\bf r}} $ $ = \frac{d{\bf r}}{ds} \cdot \frac{d {\bf N}}{ds}=\frac{d{\bf r}\cdot d{\bf N}}{d{\bf r}\cdot d{\bf r}}$

(3.26)		 $ \kappa_n = \frac{Ldu^2 + 2 Mdudv + N dv^2}{Edu^2 + 2 F dudv + G dv^2} $ $ \kappa_n = -\frac{Ldu^2 + 2 Mdudv + N dv^2}{Edu^2 + 2 F dudv + G dv^2} $


(3.30)		 $ \kappa_n = \frac{II}{I} = \frac{L + 2 M \lambda + N \lambda^2}{E + 2 F \lambda + G \lambda^2}$ $ \kappa_n = -\frac{II}{I} = -\frac{L + 2 M \lambda + N \lambda^2}{E + 2 F \lambda + G \lambda^2}$


(3.38)		 $ \kappa_n = \frac{L + 2M\lambda + N\lambda^2}{E + 2F\lambda + G\lambda^2}$ $ \kappa_n = -\frac{L + 2M\lambda + N\lambda^2}{E + 2F\lambda + G\lambda^2}$
  $ = \frac{M + N\lambda }{F + G\lambda }$ $ = -\frac{M + N\lambda }{F + G\lambda }$

(3.41)		 $ (L - \kappa_n E)du + (M - \kappa_n F)dv $ $ (L + \kappa_n E)du + (M + \kappa_n F)dv $
  $ =0$ $ =0$
  $ (M - \kappa_n F)du + (N - \kappa_n G)dv $ $ (M + \kappa_n F)du + (N + \kappa_n G)dv$
  $ =0$ $ =0$

(3.42)		 $ \left\vert \begin{matrix} L - \kappa_n E & M - \kappa_n F \cr M - \kappa_n F & N - \kappa_n G \end{matrix} \right\vert = 0 $ $ \left\vert \begin{matrix} L + \kappa_n E & M + \kappa_n F \cr M + \kappa_n F & N + \kappa_n G \end{matrix} \right\vert = 0 $

(3.43)		 $ (EG - F^2)\kappa_n^2$ $ (EG - F^2)\kappa_n^2$
  $ - (EN + GL -2FM)\kappa_n $ $ + (EN + GL -2FM)\kappa_n $
  $ + (LN - M^2) =0$ $ + (LN - M^2) =0$

(3.47)		 $ H = \frac{EN + GL -2FM}{2(EG - F^2)}$ $ H = \frac{2FM -EN - GL}{2(EG - F^2)}$

(3.51)		 $ \lambda = - \frac{M-\kappa_nF}{N- \kappa_nG}$ $ \lambda = - \frac{M+\kappa_nF}{N+ \kappa_nG}$

(3.52)		 $ \lambda = - \frac{L-\kappa_nE}{M- \kappa_nF} $ $ \lambda = - \frac{L+\kappa_nE}{M+ \kappa_nF} $

(3.68)		 $ H=$ $ H=$
  $ \frac{ (1+h^2_x) h_{yy} - 2h_xh_yh_{xy} + (1+h^2_y) h_{xx}}{ 2(1 + h^2_x + h^2_y)^{3/2}}$ $ \frac{ 2h_xh_yh_{xy} -(1+h^2_x) h_{yy} - (1+h^2_y) h_{xx}}{2(1 + h^2_x + h^2_y)^{3/2}}$

(3.89)		 $ \kappa_n = \frac{Ldu^2 + N dv^2}{Edu^2 + G dv^2}$ $ \kappa_n = -\frac{Ldu^2 + N dv^2}{Edu^2 + G dv^2}$

(3.90)		 $ \kappa_1 = \frac{L}{E},\;\;\;\;\;\; \kappa_2 = \frac{N}{G}$ $ \kappa_1 = -\frac{L}{E},\;\;\;\;\;\; \kappa_2 = -\frac{N}{G}$

   

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Next: 4. Nonlinear Polynomial Solvers Up: 3. Differential Geometry of Previous: 3.5.2 Implicit surfaces   Contents   Index
December 2009