欧拉角转成四元数的公式

最新推荐文章于 2024-02-27 15:04:07 发布

pizi0475

最新推荐文章于 2024-02-27 15:04:07 发布

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分类专栏： 3D数学文章标签： float 图形游戏 c

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这两天看MDL模型的显示代码，参考《3D数学基础：图形与游戏开发》中的公式，发现欧拉角转成四元数的公式和书中提到的公式不一致，狂晕！~~~

void AngleQuaternion( const vec3_t Angles, vec4_t Quaternion)

{

float Angle = Angles[0] * 0.5f;

float Sin0 = (float)sin(Angle);

float Cos0 = (float)cos(Angle);

Angle = Angles[1] * 0.5f;

float Sin1 = (float)sin(Angle);

float Cos1 = (float)cos(Angle);

Angle = Angles[2] * 0.5f;

float Sin2 = (float)sin(Angle);

float Cos2 = (float)cos(Angle);

Quaternion[0] = Sin0 * Cos1 * Cos2 - Cos0 * Sin1 * Sin2;

Quaternion[1] = Cos0 * Sin1 * Cos2 + Sin0 * Cos1 * Sin2;

Quaternion[2] = Cos0 * Cos1 * Sin2 - Sin0 * Sin1 * Cos2;

Quaternion[3] = Cos0 * Cos1 * Cos2 + Sin0 * Sin1 * Sin2;

}

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)