假设我们有几个点位于三维空间中。现在，我们指定两个点，命名为point1和point2。为了简化问题，我们将这两点视为一条线。现在我们将 point1 移动到原点。这种平移使得所有点都相应地移位。现在，所有点都必须旋转，以便 point2 恰好位于 Y 轴 上，这意味着它的 x 和 z 坐标应该为零。但是需要执行哪些具体计算来确定所有这些点应旋转的方向，以便 point2 精确定位在Y 轴上？
我非常感谢您的帮助。
如果我想更精确地问我的问题，Ax、Ay和Az的值应该如何计算？
import numpy as np
points = np.array([[0.36266313,0.70320135,0.88975275],
[0.26227164,0.32188661,0.39514979],
[0.26100571,0.63643259,0.20245194],
[0.25701545,0.59125069,0.80146842],
[0.23185588,0.19422526,0.68689653]])
point1, point2 = points[0], points[2]
points = points - point1
point2 = point2 - point1
point1 = point1 - point1
def rotate_points_around_point(points, angle_degrees, axis, center_point):
translated_points = points - center_point
angle_radians = np.radians(angle_degrees)
rotation_matrix = {
'x': np.array([[1, 0, 0],
[0, np.cos(angle_radians), -np.sin(angle_radians)],
[0, np.sin(angle_radians), np.cos(angle_radians)]]),
'y': np.array([[np.cos(angle_radians), 0, np.sin(angle_radians)],
[0, 1, 0],
[-np.sin(angle_radians), 0, np.cos(angle_radians)]]),
'z': np.array([[np.cos(angle_radians), -np.sin(angle_radians), 0],
[np.sin(angle_radians), np.cos(angle_radians), 0],
[0, 0, 1]])
}
rotated_points = np.dot(translated_points, rotation_matrix[axis].T)
rotated_points = rotated_points + center_point
return rotated_points
# ======== Question ========
Ax = 96
Ay = -3.0626
Az = -8.38675
# ======== Question ========
points = rotate_points_around_point(points, Ax, "x", point1)
points = rotate_points_around_point(points, Ay, "y", point1)
points = rotate_points_around_point(points, Az, "z", point1)
print(points)
结果：
[[ 0. 0. 0. ]
[-0.004306 0.538135 -0.332422]
[ 0. 0.697979 0. ]
[-0.084459 0.11303 -0.107608]
[-0.066404 0.267493 -0.49128 ]]
我很重视您提供的见解和指导。
平移这些点，使您的第一个参考点位于原点。
设B为平移后第二参考点的位置向量。
关键步骤：取 B 的向量叉积和 y 方向上的单位向量（按顺序）。得到的向量将具有最终旋转轴的方向和幅值norm(B)x1xsin(theta)，其中theta是旋转角度。
您可以根据该轴和旋转角度创建旋转矩阵。这是您需要的single旋转矩阵。
以下代码执行此操作。请注意，浮点运算会在点 [2] 中留下 1e-16 阶的 x 和 z 坐标：实际上是 0。
import math
import numpy as np
import matplotlib.pyplot as plt
def rotationMatrix( a, theta ): # rotation matrix for a rotation of angle theta about axis a
R = np.zeros( ( 3, 3 ) )
n = a / np.linalg.norm( a )
C = math.cos( theta )
S = math.sin( theta )
R[0,0] = C + n[0] * n[0] * ( 1.0 - C )
R[0,1] = n[0] * n[1] * ( 1.0 - C ) - n[2] * S
R[0,2] = n[0] * n[2] * ( 1.0 - C ) + n[1] * S
R[1,1] = C + n[1] * n[1] * ( 1.0 - C )
R[1,2] = n[1] * n[2] * ( 1.0 - C ) - n[0] * S
R[1,0] = n[1] * n[0] * ( 1.0 - C ) + n[2] * S
R[2,2] = C + n[2] * n[2] * ( 1.0 - C )
R[2,0] = n[2] * n[0] * ( 1.0 - C ) - n[1] * S
R[2,1] = n[2] * n[1] * ( 1.0 - C ) + n[0] * S
return R
points = np.array([[0.36266313,0.70320135,0.88975275],
[0.26227164,0.32188661,0.39514979],
[0.26100571,0.63643259,0.20245194],
[0.25701545,0.59125069,0.80146842],
[0.23185588,0.19422526,0.68689653]])
A = points[0] # First reference point
points -= A # Translate such that A goes to origin
B = points[2] # B is the point you want to rotate onto the y axis (did you mean [2]?)
Y = np.array( [0.0, 1.0, 0.0 ] ) # Unit vector along y axis
axis = np.cross( B, Y ) # Axis of rotation (not normalised)
mag = np.linalg.norm( axis )
if ( abs(mag) < 1.0e-20 ): axis = np.array([1.0,0,0]) # Already on y axis, so rotation axis is irrelevant
sintheta = mag / np.linalg.norm( B ) # Sine of angle of rotation
theta = math.asin( sintheta )
if np.dot( B, Y ) < 0: theta = math.pi - theta # Appropriate angle to rotate to the POSITIVE y axis
R = rotationMatrix( axis, theta ) # Find rotation matrix
points = ( R @ (points.T) ).T # Rotate points
print( points )
输出：
[[ 0.00000000e+00 0.00000000e+00 0.00000000e+00]
[-7.51354446e-02 5.38134329e-01 -3.23848081e-01]
[-3.85256819e-17 6.97979012e-01 -2.79186184e-16]
[-1.05473850e-01 1.13030002e-01 -8.71090761e-02]
[-1.69698494e-01 2.67492963e-01 -4.65798002e-01]]