四元数乘法

首先来看一下四元数的乘法：
对两个四元数q和p：
$\begin{array}{l} {q=w_{1}+x_{1} i+y_{1} j+z_{1} k} \\ {p=w_{2}+x_{2} i+y_{2} j+z_{2} k} \end{array}$
他们的乘法可以写成：
$\begin{aligned} q p= +\left(x_{1} * w_{2}+w_{1} * x_{2}+z_{1} * y_{2}-z_{1} * y_{2}\right) i\\ +\left(y_{1} * w_{2}+w_{1} * y_{2}+z_{1} * x_{2}-x_{1} * z_{2}\right) j\\ +\left(z_{1} * w_{2}+w_{1} * z_{2}+x_{1} * y_{2}-x_{2} * y_{1}\right) k\\ +\left(w_{1} * w_{2}-x_{1} * x_{2}-y_{1} * y_{2}-z_{1} * z_{2}\right)\\ \end{aligned}$

当四元数实部为0，即：
$\begin{array}{l} {q=0+x_{1} i+y_{1} j+z_{1} k} \\ {p=0+x_{2} i+y_{2} j+z_{2} k} \end{array}$
他们的乘法可以写成：
$\begin{aligned} q ×p= +\left(z_{1} * y_{2}-z_{1} * y_{2}\right) i\\ +\left(z_{1} * x_{2}-x_{1} * z_{2}\right) j\\ +\left(x_{1} * y_{2}-x_{2} * y_{1}\right) k\\ +\left(-x_{1} * x_{2}-y_{1} * y_{2}-z_{1} * z_{2}\right)\\ \end{aligned}$

四元数的时间导数

再看一下四元数的时间导数：
$\begin{aligned} \frac{d \mathbf{q}(t)}{d t} &=\lim _{t \rightarrow 0} \frac{\mathbf{q}(t+\delta t)-\mathbf{q}(t)}{\delta t} \\ &=\lim _{t \rightarrow 0} \frac{\delta \mathbf{q} \otimes \mathbf{q}(t)-\mathbf{q}(t)}{\delta t} \\ &=\lim _{t \rightarrow 0} \frac{(\delta \mathbf{q}-1) \otimes \mathbf{q}(t)}{\delta t}\\ &=\lim _{t \rightarrow 0} \frac{\left[\begin{array}{l} {0} \\ {\frac{j \theta}{2}} \end{array}\right] \otimes \mathbf{q}(t)}{\delta t}=\frac{1}{2}\left[\begin{array}{l} {0} \\ {\omega} \end{array}\right] \otimes \mathbf{q} \end{aligned}$

实际上就是把三维角速度用四元数的方式表示，Ω与四元数相乘，实际上就是四元数之间的乘法。

VINS-MONO里的相关公式

看下VINS-MONO里面和四元数相关的函数：
$\mathbf{q}_{b_{k+1}}^{w}=\mathbf{q}_{b_{k}}^{w} \otimes \int_{t \in\left[t_{k}, t_{k+1}\right]} \frac{1}{2} \mathbf{\Omega}\left(\hat{\omega}_{t}-\mathbf{b}_{w_{t}}-\mathbf{n}_{w}\right) \mathbf{q}_{t}^{b_{k}} d t$

$\boldsymbol{\Omega}(\boldsymbol{\omega})=\left[\begin{array}{cc} {-\lfloor\boldsymbol{\omega}\rfloor_{\times}} & {\boldsymbol{\omega}} \\ {-\boldsymbol{\omega}^{T}} & {0} \end{array}\right],\lfloor\boldsymbol{\omega}\rfloor_{\times}=\left[\begin{array}{ccc} {0} & {-\omega_{z}} & {\omega_{y}} \\ {\omega_{z}} & {0} & {-\omega_{x}} \\ {-\omega_{y}} & {\omega_{x}} & {0} \end{array}\right]$

把Ω完整地写出来之后：
$\Omega(\omega)=\left[\begin{array}{cccc} {0} & {\omega_{z}} & {-\omega_{y}} &{\omega_x}\\ {-\omega_{z}} & {0} & {\omega_{x}} &{\omega_y}\\ {\omega_{y}} & {-\omega_{x}} & {0} &{\omega_z} \\ {{-\omega_x}} & {-\omega_{y}} &{-\omega_{z}} &0 \end{array}\right]$

我们发现Ω和形式为 $[ k_x,k_y,k_z,0]$的四元数相乘：
$\Omega(\omega)=\left[\begin{array}{cccc} {\omega_{z}}*k_y - {\omega_{y}}*k_z \\ {-\omega_{z}*k_x} + {\omega_{x}}*k_z \\ {\omega_{y}*k_x - \omega_{x}*k_y} \\ {-\omega_{x}*k_x} -{\omega_{y}*k_y}-{\omega_{z}*k_z} \end{array}\right]$

这是不是和四元数乘法很像？于是我们就知道Ω和四元数乘法其实就是一样的东西。

$\frac{1}{2}\left[\begin{array}{l} {0} \\ {\omega} \end{array}\right] \otimes \mathbf{q} = \frac{1}{2}\Omega(\omega)q$
用四元数表示的角速度与 $b_k$时刻的四元数相乘，表示当前时刻各个方向的角速度，然后在 $k$到 $k+1$间隔求积分，得到 $k+1$时刻相对于 $k$时刻的旋转。

参考：
https://www.cnblogs.com/youzx/p/6387740.html
https://mp.weixin.qq.com/s/FDMDleH8wszsD9HkBXwAnQ