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03-UnrealEngine/Math/四元数学习笔记.md
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@ -61,6 +61,20 @@ $$p \rightarrow q\cdot p \cdot q^-1$$
## 四元数规范化
$$x^2 + y^2 + z^2 =1$$
## 四元数乘法
$$q_1q_2=(a+bi+cj+dk)(e+fi+gi+hk)$$
<EFBFBD>1<EFBFBD>2=(<28>+<2B><>+<2B><>+<2B><>)(<28>+<2B><>+<2B><>+ℎ<>)
<EFBFBD>1<EFBFBD>2=<3D><>+<2B><><EFBFBD>+<2B><><EFBFBD>+<2B>ℎ<EFBFBD>+<2B><><EFBFBD>+<2B><><EFBFBD>2+<2B><><EFBFBD><EFBFBD>+<2B>ℎ<EFBFBD><E2848E>+<2B><><EFBFBD>+<2B><><EFBFBD><EFBFBD>+<2B><><EFBFBD>2+<2B>ℎ<EFBFBD><E2848E>+<2B><><EFBFBD>+<2B><><EFBFBD><EFBFBD>+<2B><><EFBFBD><EFBFBD>+<2B>ℎ<EFBFBD>2
使用 <EFBFBD>2=<3D>2=<3D>2=<3D><><EFBFBD>=−1 化简上述等式
<EFBFBD>1<EFBFBD>2=(<28><>−<EFBFBD><E28892>−<EFBFBD><E28892>−<EFBFBD>ℎ)+......(<28><>+<2B><>−<EFBFBD><E28892>+<2B>ℎ)<29>+.......(<28><>+<2B><>+<2B><>−<EFBFBD>ℎ)<29>+......(<28><>−<EFBFBD><E28892>+<2B><>+<2B>ℎ)<29>
可以看到前面的系数可以写成一个矩阵
![](https://pic3.zhimg.com/80/v2-93bec89f713eb33f8da672b44a96688e_720w.webp)
所以可以得到矩阵形式
## 四元数求反
```c++
inline QQuaternion QQuaternion::inverted() const
@ -77,6 +91,11 @@ inline QQuaternion QQuaternion::inverted() const
}
```
## 四元数复合旋转
$$q_{next}=q_2q_1$$
$$v'=q_2q_1vq_1^*q_2^*$$
`q*`为反函数。
# FBX
四元数旋转顺序 xyz
可以使用