九乡新闻网非常61砸金蛋音效:欧拉角转成四元数的公式
来源：百度文库 编辑：九乡新闻网 时间：2024/05/05 23:36:28
书上的公式是这样的：
|cos(h/2) |
| 0 |
h=|sin(-h/2) |
| 0 |
|cos(p/2) |
|-sin(p/2) |
p=| 0 |
| 0 |
|cos(b/2) |
| 0 |
b=| 0 |
|-sin(b/2) |
| cos(h/2)cos(p/2)cos(b/2)+sin(h/2)sin(p/2)sin(b/2)| (w)
|cos(h/2)sin(p/2)cos(b/2)+sin(h/2)cos(p/2)sin(b/2) | (x)
q惯性-物体(h,p,b) = |sin(h/2)cos(p/2)cos(b/2)-cos(h/2)sin(p/2)sin(b/2) | (y)
|cos(h/2)cos(p/2)sin(b/2)-sin(h/2)sin(p/2)cos(b/2) | (z)
上面的代码与公式比较以后，发现有点像这样：
Quaternion[0] = y;
Quaternion[1] = x;
Quaternion[2] = z;
Quaternion[3] = w;
这样明显是不对的，而四元数转矩阵的时候，用的四元数应该是
Quaternion[0] = x;
Quaternion[1] = y;
Quaternion[2] = z;
Quaternion[3] = w;
颠倒顺序乘也都不对
今天想起了前些天摘录的文章《Quaternion Powers》
里面提到的欧拉角转换成四元数的公式：
Euler to Quaternion
Converting from Euler angles to a quaternion is slightly more tricky, as the order of operations must be correct. Since you can convert the Euler angles to three independent quaternions by setting the arbitrary axis to the coordinate axes, you can then multiply the three quaternions together to obtain the final quaternion.
So if you have three Euler angles (a, b, c), then you can form three independent quaternions
Qx = [ cos(a/2), (sin(a/2), 0, 0)]
Qy = [ cos(b/2), (0, sin(b/2), 0)]
Qz = [ cos(c/2), (0, 0, sin(c/2))]
And the final quaternion is obtained by Qx * Qy * Qz.
算出来的结果跟代码中的是一样的。竟然是两种截然不同的转换公式，狂晕一个先。具体的原因还不知道 ：（，搞清楚了再贴上来。
公式如下：
| cos(a/2)*cos(b/2)*cos(c/2)+sin(a/2)*sin(b/2)*sin(c/2) | (w)
| sin(a/2)*cos(b/2)*cos(c/2)-cos(a/2)*sin(b/2)*sin(c/2) | (x)
q=| cos(a/2)*sin(b/2)*cos(c/2)+sin(a/2)*cos(b/2)*sin(c/2) | (y)
| cos(a/2)*cos(b/2)*sin(c/2)-sin(a/2)*sin(b/2)*cos(c/2) | (z)