Rodrigues: vector representation of rotation matrices

目的：最近看论文,遇到Rodrigues公式,There has never been any derivation.So deduce it yourself.

在理解Rodrigues公式,Before you need to understand how the rotation vector is expressed.
The vector representation of the rotation matrix is ​​derived based on Euler's theorem,它有三个参数.And it can be converted to a rotation matrix.This way of converting to a rotation matrix is ​​calledRodrigues公式.

Rotation Vectors
in common cognition,旋转矩阵是$3\times 3$的矩阵.Such matrices have numerous constraints,The specific constraints can be found in the information（Fill up later）.It has only three degrees of freedom.Just know those three parameters,剩下的6parameters can be calculated,It's easy to express.in general optimization,使用 Rotation matrix不方便,Calculation of matrices is complex and time-consuming.

定义的来源：
in geometric rotation,Any rotation can be expressed as a rotation around an axis by a certain angle.So the expression can be expressed as :A rotated vector$\overrightarrow{u}$,and its rotation angle$\theta$.因此它可以用4个参数来决定.
假设向量$\overrightarrow{r}$is the principal axis vector,It's rotated from vector from vector$\overrightarrow{p}$,旋转角度为$\theta$.如下图：

因为$\overrightarrow{o}$为原点.简化表达$\overrightarrow{op}=\overrightarrow{p}$,Others use similar expressions.

从图上可以看到,向量$\overrightarrow{p}$绕着轴$\overrightarrow{u}$旋转的时候,Only and perpendicular to the vector$\overrightarrow{u}$有关.So follow the vector$\overrightarrow{u}$把它拆分为$\overrightarrow{a}$,$\overrightarrow{b}$,They are respectively plane to the vector$\overrightarrow{u}$and perpendicular to the vector$\overrightarrow{u}$.Two vectors are split,Two vectors need to be calculated.假设$\overrightarrow{u}$和$\overrightarrow{p}$的夹角为$\sigma$,因此可以得到
$$cos\sigma =\frac{\overrightarrow{u}\cdot \overrightarrow{p}}{|\overrightarrow{u}||\overrightarrow{p}|}(1)$$
因为$\overrightarrow{u}$为单位向量.因此可以得到
$$\overrightarrow{a}=\overrightarrow{u}|\overrightarrow{p}|cos\sigma =\frac{\overrightarrow{u}\cdot \overrightarrow{p}}{|\overrightarrow{u}||\overrightarrow{p}|}|\overrightarrow{p}|\overrightarrow{u}=(\overrightarrow{u}\cdot \overrightarrow{p})\overrightarrow{u}=\overrightarrow{u}(\overrightarrow{u}\cdot \overrightarrow{p})(2)$$
By unpacking and merging$(2)$Written in the form of a vector can be obtained：
$$\overrightarrow{a}=\overrightarrow{u}(\overrightarrow{u}\cdot \overrightarrow{p})=\overrightarrow{u}{\overrightarrow{u}}^T\overrightarrow{p}(3)$$

for perpendicular to the vector$\overrightarrow{u}$的分量$\overrightarrow{b}$
$$\overrightarrow{b}=\overrightarrow{p}-\overrightarrow{a}=(1-\overrightarrow{u}{\overrightarrow{u}}^T)\overrightarrow{p}(4)$$

Because from the vector$\overrightarrow{b}$旋转到向量$\overrightarrow{b^{\prime }}$,Another vector is required$\overrightarrow{c}$,It is perpendicular to the vector$\overrightarrow{u}$和向量$\overrightarrow{p}$所在的平面,and its length and vector$\overrightarrow{b}$一样.
满足：$|\overrightarrow{b}|=|\overrightarrow{c}|$
Obtained by the triangle line vertical formula：
$$|\overrightarrow{b}|=|\overrightarrow{u}||\overrightarrow{p}|sin\sigma (5)$$
通过叉乘,Vector can be designed$\overrightarrow{c}$:
$$\overrightarrow{c}=\overrightarrow{u}\times \overrightarrow{p}=(|\overrightarrow{u}||\overrightarrow{p}|sin\sigma )\frac{\overrightarrow{c}}{|\overrightarrow{c}|}=(|\overrightarrow{u}||\overrightarrow{p}|sin\sigma )\overrightarrow{n}(6)$$
其中$\overrightarrow{n}=\frac{\overrightarrow{c}}{|\overrightarrow{c}|}$单位向量.因此$|\overrightarrow{c}|=|\overrightarrow{u}||\overrightarrow{p}|sin\sigma$
by calculating their modulus,found that they were equal.
上面的公式$(6)$,将在下一篇blog介绍.Update their links later

因此可以得到$\overrightarrow{p}$The resulting vector after rotation$\overrightarrow{p^{\prime }}$为：
$$\overrightarrow{b^{\prime }}=\overrightarrow{b}cos\theta +\overrightarrow{c}sin\theta (7)$$
Then by vector addition
$$\begin{gathered} \overrightarrow{p^{\prime }}=\overrightarrow{a}+\overrightarrow{b^{\prime }}=\overrightarrow{a}+\overrightarrow{b}cos\theta +\overrightarrow{c}sin\theta \\ =\overrightarrow{u}{\overrightarrow{u}}^T\overrightarrow{p}+(1-\overrightarrow{u}{\overrightarrow{u}}^T)\overrightarrow{p}cos\theta +(\overrightarrow{u}\times \overrightarrow{p})sin\theta \\ =[Icos\theta +(1-cos\theta )\overrightarrow{u}{\overrightarrow{u}}^T+\overrightarrow{u}\times sin\theta ]\overrightarrow{p}(8) \end{gathered}$$

通过公式$(8)$可以得到一个向量$\overrightarrow{p}$around the vector$\overrightarrow{u}$得到旋转$\theta$新的向量$\overrightarrow{p^{\prime }}$的公式,其中$R=Icos\theta +(1-cos\theta )\overrightarrow{u}{\overrightarrow{u}}^T+\overrightarrow{u}\times sin\theta$.它也是Rodrigues公式.
还有一种特殊情况,如果$\overrightarrow{u}=[0,0,0{]}^T$,The resulting rotation matrix is ​​$I$

The above formula can also be reversed（因为它不太常用）,后面再补上.

参考资料为：
Vector Representation of rotations的论文.