UE5中的四元数

我们知道，四元数是除了欧拉角，旋转矩阵之外，主要用来描述旋转的量。四元数直观的定义就是
$$q=[cos(\frac{\theta }{2}),sin(\frac{\theta }{2})N]$$
其中，N表示旋转轴单位向量，$\theta$表示旋转的角度。那么，给定一个任意向量V，绕N旋转$\theta$角度之后的V’可以用四元数乘法表示。令四元数v = [0, V]，v’ = [0, V’]，那么
$$v^{\prime }=qv{q}^{\ast }$$
我们先来推导一下这个公式。

绕任意轴旋转

首先，四元数的乘法公式为
$$q_1=[s,\text{v}]$$

$$q_2=[t,\text{u}]$$

$$q_1{q}_2=[st-\text{v}\cdot \text{u},s\text{u}+t\text{v}+\text{v}\times \text{u}]$$

那么
$$v{q}^{\ast }=[sin(\frac{\theta }{2})N\cdot V,cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V]$$

$$\begin{gathered} qv{q}^{\ast }=[sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\cdot V-sin(\frac{\theta }{2})N\cdot (cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V), \\ cos(\frac{\theta }{2})(cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V)+si{n}^2(\frac{\theta }{2})(N\cdot V)N+sin(\frac{\theta }{2})N\times (cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V)] \end{gathered}$$

我们先对结果的实部进行化简，因为点乘满足分配律，所以有
$$\begin{gathered} sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\cdot V-sin(\frac{\theta }{2})N\cdot (cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V) \\ =sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\cdot V-sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\cdot V-si{n}^2(\frac{\theta }{2})N\cdot (N\times V) \\ =-si{n}^2(\frac{\theta }{2})N\cdot (N\times V) \end{gathered}$$

N和V的叉乘显然与N垂直，所以点积为0，因此

$$-si{n}^2(\frac{\theta }{2})N\cdot (N\times V)=0$$

再看虚部，同样叉乘也满足分配律，所以有

$$\begin{gathered} cos(\frac{\theta }{2})(cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V)+si{n}^2(\frac{\theta }{2})(N\cdot V)N+sin(\frac{\theta }{2})N\times (cos(\frac{\theta }{2})V+sin(\frac{\theta }{2})N\times V) \\ =co{s}^2(\frac{\theta }{2})V+sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\times V+si{n}^2(\frac{\theta }{2})(N\cdot V)N+sin(\frac{\theta }{2})cos(\frac{\theta }{2})N\times V+si{n}^2(\frac{\theta }{2})N\times (N\times V) \end{gathered}$$

注意，叉乘并不满足结合律，且有
$$\text{a}\times (\text{b}\times \text{c})=(\text{a}\cdot \text{c})\text{b}-(\text{a}\cdot \text{b})\cdot \text{c}$$
所以上式进一步化简得到
c o s 2 ( θ 2 ) V + 2 s i n ( θ 2 ) c o s ( θ 2 ) N × V + s i n 2 ( θ 2 ) ( N ⋅ V ) N + s i n 2 ( θ 2 ) ( ( N ⋅ V ) N − V ) = ( c o s 2 ( θ 2 ) − s i n 2 ( θ 2 ) ) V + 2 s i n 2 ( θ 2 ) ( ( N ⋅ V ) N + 2 s i n (