给定顶点坐标，在 3D 中找到三角形的旋转角度 [英] Find the rotation angles of a triangle in 3D, given the coordinates of its vertices

问题描述

我尝试在 3D 中旋转和平移一个等边三角形，直到他的顶点到达某个坐标.

I try to rotate and translate an equilateral triangle in 3D until his vertices reach some coordinates.

顶点坐标 F,G,H 和 F',G',H' 是已知的:

The vertices coordinates F,G,H and F',G',H' are known :

我能够像这样找到新的质心 c' 坐标:

I was able to find the new centroid c' coordinates like this :

c'.x = ( F'.x + G'.x + H'.x ) / 3
c'.y = ( F'.y + G'.y + H'.y ) / 3
c'.z = ( F'.z + G'.z + H'.z ) / 3

所以翻译三角形没问题.但我找不到一种方法来计算将 F'G'H' 三角形放在正确位置所需的旋转......

So no problem to translate the triangle. But I can't find a way to calculate the rotations needed to put F'G'H' triangle in the right position...

我必须知道三角形 F'G'H' 必须围绕每个轴 (x,y,z) 旋转多少度，知道初始三角形的旋转为 0°.

通过每个轴的旋转，我在谈论这个:

By rotation for each axis, I'm talking about this:

有什么想法吗?

推荐答案

技巧是使用叉积 b4 和旋转后找到三角形的法向量

trick is to find the normal vectors of the triangles using cross product b4 and after rotations

v1 = (F.x - G.x, F.y - G.y, F.z - G.z)
v2 = (F.x - H.x, F.y - H.y, F.z - H.z)
n  = cross_prod(v1, v2) # see http://en.wikipedia.org/wiki/Cross_product
n  = n / norm(n) # normalize to unit vector

v'1 = (F'.x - G'.x, F'.y - G'.y, F'.z - G'.z)
v'2 = (F'.x - H'.x, F'.y - H'.y, F'.z - H'.z)
n'  = cross_prod(v'1, v'2)
n'  = n' / norm(n')

rot = arc_cos(n.x * n'.x + n.y * n'.y + n.z * n'.z)

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