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Questions in category: 曲线曲面论 (Curve and surface theory).

## 测地曲率、法曲率、测地挠率(geodesic curvature, normal curvature, geodesic curvature)

Posted by haifeng on 2017-08-12 23:05:41 last update 2017-08-12 23:41:43 | Answers (0) | 收藏

$\begin{pmatrix} T\\ t\\ u\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & \sin\alpha\\ 0 & -\sin\alpha & \cos\alpha \end{pmatrix} \begin{pmatrix} T\\ N\\ B\end{pmatrix}$

$d\begin{pmatrix} T\\ t\\ u\end{pmatrix}= \begin{pmatrix} 0 & 0 & 0\\ 0 & -\sin\alpha & \cos\alpha\\ 0 & -\cos\alpha & -\sin\alpha \end{pmatrix}d\alpha\cdot \begin{pmatrix} T\\ N\\ B\end{pmatrix}+ \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & \sin\alpha\\ 0 & -\sin\alpha & \cos\alpha \end{pmatrix} \begin{pmatrix}dT\\ dN\\ dB\end{pmatrix}$

$\begin{pmatrix}\dot{T}(s)\\ \dot{N}(s)\\ \dot{B}(s)\end{pmatrix}= \begin{pmatrix} 0 & \kappa & 0\\ -\kappa & 0 & \kappa\\ 0 & -\tau & 0 \end{pmatrix} \begin{pmatrix}T\\ N\\ B\end{pmatrix}$

$\begin{pmatrix} T\\ N\\ B\end{pmatrix}= \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\alpha & -\sin\alpha\\ 0 & \sin\alpha & \cos\alpha \end{pmatrix} \begin{pmatrix} T\\ t\\ u\end{pmatrix}$

$d\begin{pmatrix} T\\ t\\ u\end{pmatrix}= \begin{pmatrix} 0 & \kappa\cos\alpha ds & -\kappa\sin\alpha ds\\ -\kappa\cos\alpha ds & 0 & \tau ds+d\alpha\\ \kappa\sin\alpha ds & -\tau ds-d\alpha & 0 \end{pmatrix} \begin{pmatrix} T\\ t\\ u\end{pmatrix}$

$d\begin{pmatrix} T\\ t\\ u\end{pmatrix}= \begin{pmatrix} 0 & \kappa_g ds & \kappa_n ds\\ -\kappa_g ds & 0 & \tau_{\gamma}ds\\ -\kappa_n ds & -\tau_{\gamma}ds & 0 \end{pmatrix} \begin{pmatrix} T\\ t\\ u\end{pmatrix}$

References:

https://en.wikipedia.org/wiki/Darboux_frame#Geodesic_curvature.2C_normal_curvature.2C_and_relative_torsion