20221005 Two rotations represented by quaternions

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Quaternion definition

The definition of quaternion comes from Euler's theorem, $$\mathbf{q}=[\cos (\frac{}{2}),\mathbf{n}\sin (\frac{}{2})]$$ where$n$ is the (vector) axis,$\varphi$ is aroundn \boldsymbol{n}Euler angles of$n\; .$

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Quaternion multiplication rule

The specific derivation of the quaternion multiplication rule will not be given here. The understanding is to perform two rotations. The specific rule is $${\mathbf{q}}^{\prime }\otimes q=\left[\begin{array}{l}{q}_0^{\prime }\ast q_0-q_v^{\prime }\cdot q_v\\ q_0^{\prime }\ast q_v+q_0\ast q_v^{\prime }+q_v^{\prime }\times q_v\end{array}\right]$$

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rotation transform

One of the reasons why readers are confused about the understanding of rotation is that different books use different rotation transformations.

There are generally two types of rotation transformations: (the second type is more commonly used in attitude description)
1. The coordinate system does not move, and the vector rotates;
2. The vector does not move, and the coordinate system rotates.

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Detailed explanation:

1. The coordinate system does not move, and the vector rotates.
For the first type, the initial vector is ${\mathbf{r}}_0$, becomes r 1 \boldsymbol{r}_1 after spatial rotation$r_1$, based on $r_0$Rotate to $r_1$The process is described as a quaternion $\mathbf{q}$。则有$$\mathrm{thing}(r_1)={\mathbf{q}}^{\ast }\otimes \mathrm{thing}\left(r_0\right)\otimes \mathbf{q}$$展开可得
$${\mathbf{r}}_1=\left((q_0^2-q_v^T{q}_v)\mathbf{I}+2{\mathbf{q}}_v{q}_v^T+2q_0{q}_v^{\times }\right)r_0$$

2. The vector does not move, and the coordinate system rotates.
For the second type, the fixed vector $The\; projection\; of\; r$ in the initial coordinate system is${\mathbf{r}}^r$ , after the initial coordinate system is rotated in space, the fixed vector$r$ is projected in the new coordinate system as$r^b$ , the process of rotating to a new coordinate system based on the initial coordinate system is expressed as$q$ . Pay attention to two points:1. The vector has not moved in space, but the coordinate system used for reference has changed; 2. The projection we use to express is just a mathematical representation in different coordinate systems, and does not mean that the vector has moved.
$$\mathrm{thing}(r^b)=\mathbf{q}\otimes \mathrm{thing}(r^r)\otimes q^{\ast }$$展开可得
$${\mathbf{r}}^b=\left((q_0^2-{\mathbf{q}}_v^T{q}_v)\mathbf{I}+2{\mathbf{q}}_v{q}_v^T-2q_0{q}_v^{\times }\right)r^r$$

注：$\mathrm{thing}\left(r\right)=[0,\mathbf{r}{]}^T$