rotation - 四元数绕轴旋转的分量

Question

我无法找到有关此主题的任何有用信息。基本上我想找到四元数旋转的分量，即围绕给定轴(不一定是 X、Y 或 Z - 任何任意单位向量)。有点像将四元数投影到向量上。因此，如果我要求围绕与四元数轴平行的某个轴旋转，我会得到相同的四元数。如果我要求绕与四元数轴正交的轴旋转，我会得到一个恒等四元数。介于两者之间......好吧，这就是我想知道如何解决的:)

Answer 1

这个问题有一个优雅的解决方案，特别适合四元数。它被称为“摆动扭曲分解”:

伪代码

/**
   Decompose the rotation on to 2 parts.
   1. Twist - rotation around the "direction" vector
   2. Swing - rotation around axis that is perpendicular to "direction" vector
   The rotation can be composed back by 
   rotation = swing * twist

   has singularity in case of swing_rotation close to 180 degrees rotation.
   if the input quaternion is of non-unit length, the outputs are non-unit as well
   otherwise, outputs are both unit
*/
inline void swing_twist_decomposition( const xxquaternion& rotation,
                                       const vector3&      direction,
                                       xxquaternion&       swing,
                                       xxquaternion&       twist)
{
    vector3 ra( rotation.x, rotation.y, rotation.z ); // rotation axis
    vector3 p = projection( ra, direction ); // return projection v1 on to v2  (parallel component)
    twist.set( p.x, p.y, p.z, rotation.w );
    twist.normalize();
    swing = rotation * twist.conjugated();
}

这段代码的长答案和推导可以在这里找到 http://www.euclideanspace.com/maths/geometry/rotations/for/decomposition/