细节改进
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@@ -197,7 +197,7 @@ z_1z_2&=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
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$$
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#### 3.4.2 四元数定义
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四元数([Quaternion](https://en.wikipedia.org/wiki/Quaternion))可以视为对复数的扩充。 四元数有1个实部,3个虚部,形式如下
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四元数([Quaternion](https://en.wikipedia.org/wiki/Quaternion))可以视为对复数的扩充,有1个实部,3个虚部,形式如下
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$$
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q=s+xi+yj+zk
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$$
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@@ -231,14 +231,14 @@ q_1-q_2&=[s_1-s_2,\quad \vec{v_1}-\vec{v_2}]\\
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q_1q_2&=[s_1s_2-\vec{v_1}\cdot\vec{v_2},\quad s_1\vec{v_2}+s_2\vec{v_1}+\vec{v_1}\times\vec{v_2}]
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\end{aligned}
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$$
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四元数的乘法满足满足结合律,但不满足交换律
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四元数的乘法满足结合律,但不满足交换律
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$$
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\begin{aligned}
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(q_1q_2)q_3&=q_1(q_2q_3)\\
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q_1q_2&\neq q_2q_1
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\end{aligned}
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$$
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但如果$q_1,q_2$中的两个向量部分平行,由于$\vec{v_1}\times\vec{v_2}=0$,易证
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但如果$q_1,q_2$中的向量部分平行,由于$\vec{v_1}\times\vec{v_2}=0$,易证
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$$
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q_1q_2=q_2q_1\quad(when\quad \vec{v_1}\parallel\vec{v_2})
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\tag{3.4.3.1}
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@@ -272,7 +272,7 @@ $$
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(q_1q_2)^{-1}=q_2^{-1}q_1^{-1}
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$$
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#### 3.4.5 四元数与向量
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如果一个四元数的实部为0,那么称之为纯四元数(Pure Quaternion),由于纯四元数仅有3个虚部,可以将一个3D向量转换为一个纯四元数。设$q_v=[0,\vec{v}]$,那么
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如果一个四元数的实部为0,那么称之为纯四元数(Pure Quaternion),由于纯四元数仅有3个虚部,可以将一个3D向量转换为一个纯四元数。记$q_v=[0,\vec{v}]$,那么
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$$\begin{aligned}
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\lambda q_v&=[0, \lambda\vec{v}]\\
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q_{u}\pm q_{v}&=[0, \vec{u}\pm\vec{v}]
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@@ -305,7 +305,7 @@ $$\begin{aligned}
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$$
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R_n(\vec{\boldsymbol{v}}_2)=\vec{\boldsymbol{v}}_2\cos(\theta)+(\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}})\sin(\theta)\tag{3.4.6.1}
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$$
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设两个纯四元数$q_n, q_v$
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设两个纯四元数$q_n, q_{v2}$
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$$\begin{aligned}
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q_n&=[0, \vec{n}]\\
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q_{v2}&=[0, \vec{v_2}]
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@@ -319,8 +319,7 @@ q_nq_{v2}&=[-\vec{n}\cdot\vec{v_2},\quad\vec{n}\times\vec{v_2}]\\
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所以$q_nq_{v2}$是一个纯四元数
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由于3.4.6.1中都是线性计算,将3.4.6.2和3.4.6.3代入其中,可以得到$R_n(\vec{\boldsymbol{v}}_2)$的四元数形式
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$$\begin{aligned}
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R_n(\vec{\boldsymbol{v}}_2)&=\vec{\boldsymbol{v}}_2\cos\theta+(\vec{\boldsymbol{n}}\times\vec{\boldsymbol{v_2}})\sin\theta\\
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&= q_v\cos\theta+q_nq_{v2}\sin\theta\\
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\quad[0, R_n(\vec{\boldsymbol{v}}_2)]&= q_{v2}\cos\theta+q_nq_{v2}\sin\theta\\
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&=(\cos\theta+q_n\sin\theta)q_{v2}\\
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&=[\cos\theta, \vec{n}\sin\theta]q_{v2}\\
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&=q(\theta, \vec{n})q_{v2}
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@@ -340,7 +339,7 @@ $$
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将3.4.6.4带入3.3.2.4中,可以得到罗德里格旋转公式的四元数形式为
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$$\begin{aligned}
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R_n(\vec{\boldsymbol{v}})&=\vec{\boldsymbol{v}}_1+R_n(\vec{\boldsymbol{v}}_2)\\
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&=q_{v1}+q(\theta, \vec{n})q_{v2}\\
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\quad[0,R_n(\vec{\boldsymbol{v}})]&=q_{v1}+q(\theta, \vec{n})q_{v2}\\
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&=p\overline{p}q_{v1}+ppq_{v2}
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\end{aligned}\tag{3.4.6.6}
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$$
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@@ -354,7 +353,7 @@ q_{v2}\overline{p}&=[0, \vec{v_2}][\cos\frac{\theta}{2},-\vec{n}\sin\frac{\theta
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\end{aligned}$$
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带入3.4.6.6,可以得到
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$$\begin{aligned}
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R_n(\vec{\boldsymbol{v}})&=pq_{v1}\overline{p}+pq_{v2}\overline{p}\\
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\quad[0,R_n(\vec{\boldsymbol{v}})]&=pq_{v1}\overline{p}+pq_{v2}\overline{p}\\
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&=p(q_{v1}+q_{v2})\overline{p}\\
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&=pq_v\overline{p}
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\end{aligned}\tag{3.4.6.7}
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