Rookie‘s view about the derivative of Rotation Matrix(简明旋转矩阵求导推导)

Preface:This article is written for a nifty girl who I cherish.

(0)Pre-acknowledge

(0.0)Skew-symmetric matrix

$a\times b=[a{]}^{\times }b$

Among which $[a{]}^{\times }$:

$\left[\begin{array}{ccc}0& -a_z& a_y\\ a_z& 0& -a_x\\ -a_y& a_x& 0\end{array}\right]$

• Meaning: 用反对称矩阵乘代替了外积，使运算方便
• Proving:
  We know that:
  $(x_1,y_1,z_1)\times (x_2,y_2,z_2)$=$(\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ i& j& k\end{array}\right|)$
  =
  $\left[\begin{array}{c}{y}_1{z}_2-y_2{z}_1\\ z_1{x}_2-z_2{x}_1\\ x_1{y}_2-x_2{y}_1\end{array}\right]$=$\left[\begin{array}{c}0\ast x_2-z_1{y}_2+y_1{z}_2\\ z_1{x}_2+0\ast y_2-x_1{z}_2\\ -y_1{x}_2+x_1{y}_2+0\ast z_2\end{array}\right]$
  =
  $\left[\begin{array}{ccc}0& -z_1& y_1\\ z_1& 0& -x_1\\ -y_1& x_1& 0\end{array}\right]\ast \left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right]$

(0.1)The formula of line velocity (Basic Principle)

$\frac{d\overrightarrow{r_e}}{dt}=\overrightarrow{w_e}\times \overrightarrow{r_e}$

• Comment: 在$e$坐标系下对向量$\overrightarrow{r_e}$求导，得到的是向量$\overrightarrow{r_e}$端点的线速度，该公式既是某一向量的线速度等于角速度叉乘该向量，可见角速度方向是右手系中旋转轴的方向，角速度叉乘向量得到的方向又恰好是向量端点的运动方向
• Example: $\overrightarrow{w_e}=(-sin(t),cos(t),0{)}^T$

(0.2)The perspective point Multiply

• jargon: 左乘行变换右乘列变换

• new perspective:

  $\left[\begin{array}{ccc}{x}_{11}& x_{12}& x_{13}\\ x_{21}& x_{22}& x_{23}\\ x_{31}& x_{32}& x_{33}\end{array}\right]\ast \left[\begin{array}{c}a\\ b\\ c\end{array}\right]=a\ast \left[\begin{array}{c}{x}_{11}\\ x_{21}\\ x_{31}\end{array}\right]+b\ast \left[\begin{array}{c}{x}_{12}\\ x_{22}\\ x_{32}\end{array}\right]+c\ast \left[\begin{array}{c}{x}_{13}\\ x_{23}\\ x_{33}\end{array}\right]$

• comment: 相当于向量$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$右乘矩阵$\left[\begin{array}{ccc}{x}_{11}& x_{12}& x_{13}\\ x_{21}& x_{22}& x_{23}\\ x_{31}& x_{32}& x_{33}\end{array}\right]$，所以看起来是列变换，我们可以把矩阵的三列看作三个基，对于旋转矩阵，由于其正交性质，且叉乘保留原始方向，矩阵的三列可以看作三个正交基，所以上面的矩阵乘以一个向量叫做正交基变换；由于三个基是相互垂直的，在对坐标进行基变换时相互不干扰，所以我们可以把三个基分别拿出来当作三组独立的向量，对接下来的求导可以更方便理解

(0.3)One principle of cross mutiply

$(R\vec{a})\times (R\vec{b})=R(\vec{a}\times \vec{b})$

• comment:叉乘的性质有叉乘得出向量的大小$||a\times b||=||a||\ast ||b||\ast sin\theta$与方向符合右手坐标系可以证明此性质

(1)Topic

• notation define:

symbol	describe
$e$	大地坐标系
$\overrightarrow{e_n}$	大地坐标系的坐标轴（$n$可为$x,y,z$）
$R_b^e$	从坐标系$b$到坐标系$e$的旋转变换
$b_n^e$	在$e$坐标系下观察到$b$坐标系的$n$轴方向的单位向量（$n$可为$x,y,z$）
$\overrightarrow{w_e}$	在$e$坐标系下观察到绕旋转轴旋转的角速度
$\overrightarrow{w_b}$	在$b$坐标系下观察到绕旋转轴旋转的角速度

• start:
  $\frac{d{R}_b^e}{dt}=\frac{d[b_x^e{b}_y^e{b}_z^e]}{dt}=[\overrightarrow{w_e}\times b_x^e\overrightarrow{w_e}\times b_y^e\overrightarrow{w_e}\times b_z^e]$
  =
  $[R_b^e\overrightarrow{w_b}\times b_x^e{R}_b^e\overrightarrow{w_b}\times b_y^e{R}_b^e\overrightarrow{w_b}\times b_z^e]$
  =
  $[R_b^e\overrightarrow{w_b}\times (R_b^e\overrightarrow{e_x})R_b^e\overrightarrow{w_b}\times (R_b^e\overrightarrow{e_y})R_b^e\overrightarrow{w_b}\times (R_b^e\overrightarrow{e_z})]$
  =
  $R_b^e[\overrightarrow{w_b}\times \overrightarrow{e_x}\overrightarrow{w_b}\times \overrightarrow{e_y}\overrightarrow{w_b}\times \overrightarrow{e_z}]$
  =
  $R_b^e[\overrightarrow{w_b}{]}^{\times }e$
  =
  $R_b^e[\overrightarrow{w_b}{]}^{\times }$
  get the answer:
  $R_b^e[\overrightarrow{w_b}{]}^{\times }$
• perspective: 本问题求解了在大地坐标系中观察某个坐标系旋转的角速度，而在某个坐标系中观察大地坐标系的结果是$R_b^{e-1}[\overrightarrow{w_e}{]}^{\times }$

(2)Reference

https://blog.youkuaiyun.com/weixin_44382195/article/details/110677858

https://blog.youkuaiyun.com/weixin_38632538/article/details/106085426