3D 数学:根据 Look 和 Up 正交向量计算倾斜(滚动)角度 [英] 3D Math: Calculate Bank (Roll) angle from Look and Up orthogonal vectors

问题描述

我希望这是问这个问题的正确位置，即 与此相同，但表示为纯数学而不是图形(至少我希望我将问题正确地转换为数学).

I hope this is the proper location to ask this question which is the same as this one, but expressed as pure math instead of graphically (at least I hope I translated the problem to math correctly).

考虑:

• 两个正交的向量:Up (ux, uy, uz) 和 Look (lx, ly, lz)
• 垂直于Look的平面P(因此包括Up)
• Y1 是 Y(垂直轴)沿 Look 到 P 的投影

问题:Y1 和 Up 之间的夹角是多少?

Question: what is the value of the angle between Y1 and Up?

正如数学家会同意的那样，这是一个非常基本的问题，但我已经挠头至少两个星期了，但无法想象如何将 Y 投影到 P 上……也许现在太老了，无法找到解决方案学校练习.

As mathematicians will agree, this is a very basic question, but I've been scratching my head for at least two weeks without being able to visualize how to project Y onto P... maybe now too old for finding solutions to school exercises.

我正在寻找三角解法，而不是使用矩阵的解法.谢谢.

I'm looking for the trigonometric solution, not a solution using a matrix. Thanks.

编辑:我发现我需要确定角度的符号，相对于必须是 Look 的旋转轴.我在我的链接问题上发布了最终代码(见上面的链接).感谢那些提供帮助的人.感谢您的时间.

Edit: I found that I needed to determine the sign of the angle, relative to a rotation axis which had to be Look. I posted the final code on my linked question (see link above). Thanks to those who helped. I appreciate your time.

推荐答案

我只是在纸上做这件事.我希望它是对的.

I'm just doing this on paper. I hope it's right.

假设 Up 和 Look 被归一化，即长度为 1.假设平面 P 包含原点，L 是其法线.Y 是 (0, 1, 0)

Let's assume Up and Look are normalized, that is, length 1. Let's say that plane P contains the origin, and L is its normal. Y is (0, 1, 0)

要将 Y 投影到 P 上，求它到 P 的距离...

To project Y onto P, find its distance to P...

d = Y dot L = ly

...然后通过 -d 缩放法线以获得 Y1(即 Y 在 P 上的投影)

...and then scale the normal by -d to get the Y1 (that is, the projection of Y on P)

Y1 = (lx * ly, ly * ly, lz * ly)

现在对 Y1 进行归一化，即按 (1/长度) 对其进行缩放.如果它的长度为 0，那么你就不走运了.

Now normalize Y1, that is, scale it by (1 / length). If its length was 0 then you're out of luck.

Y1 和 Up 的点积 = 角度的余弦.所以

The dot product of Y1 and Up = the cosine of the angle. So

angle = acos(Y1 dot Up)

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