两个 3d 矢量之间的欧拉角 [英] Euler angles between two 3d vectors
问题描述
如何找到 2 个 3D 向量之间的 3 个欧拉角?当我有一个 Vector 并且想要获取它的旋转时，通常可以使用此链接:计算旋转以查看 3D 点?
但是在相互计算它们时我该怎么做?
How do you find the 3 euler angles between 2 3D vectors?
When I have one Vector and I want to get its rotation, this link can be usually used: Calculate rotations to look at a 3D point?
But how do I do it when calculating them according to one another?
推荐答案
正如其他人已经指出的，您的问题应该修改.让我们称您的向量为 a 和 b.我假设 length(a)==length(b) >0 否则我无法回答问题.
As others have already pointed out, your question should be revised. Let's call your vectors a and b. I assume that length(a)==length(b) > 0 otherwise I cannot answer the question.
计算向量的叉积v = axb;v 给出旋转的轴.通过计算点积，您可以得到角度的余弦strong> 你应该用 cos(angle)=dot(a,b)/(length(a)length(b)) 旋转，并且用 acos 你可以唯一确定角度(@Archie 感谢指出我之前的错误).此时，您拥有旋转的轴角表示.
Calculate the cross product of your vectors v = a x b; v gives the axis of rotation. By computing the dot product, you can get the cosine of the angle you should rotate with cos(angle)=dot(a,b)/(length(a)length(b)), and with acos you can uniquely determine the angle (@Archie thanks for pointing out my earlier mistake). At this point you have the axis angle representation of your rotation.
剩下的工作就是把这个表示转换成你要找的表示:欧拉角.将轴角转换为欧拉是一种方法它，正如你发现的那样.当 v = [ 0, 0, 0] 时，即角度为 0 或 180 度时，您必须处理退化情况.
The remaining work is to convert this representation to the representation you are looking for: Euler angles. Conversion Axis-Angle to Euler is a way to do it, as you have found it. You have to handle the degenerate case when v = [ 0, 0, 0], that is, when the angle is either 0 or 180 degrees.
我个人不喜欢欧拉角，它们会破坏应用程序的稳定性，并且不适合插值，另见
I personally don't like Euler angles, they screw up the stability of your app and they are not appropriate for interpolation, see also
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