Expressions for curvature of parametric surfaces

If we define a curve $ \alpha(t) = (u(t), v(t))$ in parameter space, (and thus $ (x(u(t)), x(v(t))$ on the surface) and restrict the normal field to the curve, then we get a function of normals along the curve $ N(t)$ . The derivative $ N'(t) = N_u u' + N_v v'$ is precisely $ dN(v)$ , where $ v = \alpha'(t)$ . Futhermore, given a parameterization of the surface $ x$ , we have a basis for vectors in the tangent plane $ \{x_u, x_v\}$ and $ N_u$ , $ N_v$ can be represented in this basis:

$\displaystyle N_u$ $\displaystyle = a_{11} x_u + a_{21} x_v$ (2.41)
$\displaystyle N_v$ $\displaystyle = a_{12} x_u + a_{22} x_v$ (2.42)

Labouring on, we have

$\displaystyle N'(t)$ $\displaystyle = (a_{11} x_u + a_{21} x_v) u' + (a_{12} x_u + a_{22} x_v) v'$ (2.43)
  $\displaystyle = (a_{11} u' + a_{12} v') x_u + (a_{21} u' + a_{22} x_v) v',$ (2.44)

or in vector form,

$\displaystyle dN \begin{pmatrix}u' \\ v'\end{pmatrix} = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix}u' \\ v' \end{pmatrix}.$ (2.45)

So in the basis $ \{x_u, x_v\}$ , $ dN$ is the linear map given by the $ a_{ij}$ . This is convienient, and we now want to find expressions for the $ a_{ij}$ in terms of the parameterization $ x$ and its derivatives.

The shape operator can be expressed in terms of the parameterization by invoking it on the tangent to a curve $ \alpha(t)$ :

$\displaystyle -\ensuremath{\langle dN_p(\alpha'(t)), \alpha'(t) \rangle}$ $\displaystyle = \ensuremath{\langle N'(t), x_u u' + x_v v' \rangle}$ (2.46)
  $\displaystyle = -\ensuremath{\langle N_u u' + N_v v', x_u u' + x_v v' \rangle}$ (2.47)
  $\displaystyle = -\ensuremath{\langle N_u, x_u \rangle} u'\,^2 - \ensuremath{\la...
...{\langle N_v, x_u \rangle} u' v' - \ensuremath{\langle N_v, x_v \rangle} v'\,^2$ (2.48)
  $\displaystyle = e u'\,^2 + 2 f u' v' + g v'\,^2,$ (2.49)

where we have defined

$\displaystyle e$ $\displaystyle = -\ensuremath{\langle N_u, x_u \rangle} = \ensuremath{\langle N, x_{uu} \rangle}$ (2.50)
$\displaystyle f$ $\displaystyle = -\ensuremath{\langle N_u, x_v \rangle} = \ensuremath{\langle N,...
...\ensuremath{\langle N, x_{vu} \rangle} = -\ensuremath{\langle N_v, x_u \rangle}$ (2.51)
$\displaystyle g$ $\displaystyle = -\ensuremath{\langle N_v, x_v \rangle} = \ensuremath{\langle N, x_{vv} \rangle}.$ (2.52)

The transformations producing the second derivatives of $ x$ are generated by noticing that since, say, $ x_u$ is in the tangent plane, it must be true that $ \ensuremath{\langle N, x_u \rangle} = 0$ , hence $ \ensuremath{\langle N_u, x_u \rangle} + \ensuremath{\langle N, x_{uu} \rangle} = 0$ .

Substituting (2.41)-(2.42) into (2.50)-(2.51) generates a set of relations between $ a_{ij}$ , $ \{e,f,g\}$ , and $ \{E,F,G\}$ , for example, $ -e = \ensuremath{\langle N_u, x_u \rangle} = a_{11} E + a_{21} F$ . When solved, these give

$\displaystyle \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} =...
...e & f \\ f & g \end{pmatrix} \begin{pmatrix}E & F \\ F & G \end{pmatrix} ^{-1}.$ (2.53)

Hence, given a parameterization $ x$ of a surface, we can write down the differential or shape operator in the basis $ \{x_u, x_v\}$ .

Copyright © 2005 Adrian Secord. All rights reserved.