将四元数从右手坐标系转换为左手坐标系

Question

我的3d程序，对象的旋转由四元数表示，如[0.130526, 0.0, 0.0, 0.991445]。该程序使用右手坐标系，Z轴朝上（如3ds max）：

另一方面，我的应用程序使用左手坐标系，Y轴向上：

如何将四元数从一个坐标系转换为另一个坐标系，并考虑哪个轴已经到位？

Answer 1

围绕轴（u，v，w）的角度x的旋转可以用实部cos（x / 2）和虚部sin（x / 2）*（u，v，w）的四元数表示。 / p>

如果原始三面体中的轴坐标为（u，v，w），则它们将在您的三面体中为（u，w，v）。

因此，如果原始四元数是（a，b，c，d） - a + ib + jc + kd - 四元数必须转换为三面体中的（a，b，d，c）。

修改

但是因为你的三面体是左撇子，所以角度也必须颠倒，所以最终可以用你的三面体中的四元数（a，-b，-d，-c）表示相同的旋转。

Answer 2

This is a condensed version of an answer to a slightly different question.

The problem you ask about arises even if the two coordinate systems are same-handed; it turns out that handedness flips don't make the problem significantly harder. Here is how to do it in general. To change the basis of a quaternion, say from ROS (right-handed, Z up) to Unity (left-handed, Y up):

mat3x3 ros_to_unity = /* construct this by hand by mapping input axes to output axes */;
mat3x3 unity_to_ros = ros_to_unity.inverse();
quat q_ros = ...;
mat3x3 m_unity = ros_to_unity * mat3x3(q_ros) * unity_to_ros;
quat q_unity = mat_to_quat(m_unity);

Lines 1-4 are simply the method of https://stackoverflow.com/a/39519079/194921: "How do you perform a change-of-basis on a matrix?"

Line 5 is interesting; not all matrices convert to quats, but if ros_to_unity is correct, then this conversion will succeed.

Note that this will give you a correct result, but it goes through a lot of work -- conversion to and from a matrix, some multiplies, an inversion. But you can examine its results and then write a special-case version that rearranges or flips axes, like the one aka.nice derived.