旋转矩阵不精确以将向量与轴对齐

Question

几个小时以来，我一直在用头撞墙，但我似乎无法弄清楚自己做错了什么。

我正在尝试生成一个旋转矩阵，它将向量与特定轴对齐[=29=]（我最终会转换更多数据，因此拥有旋转矩阵很重要).

我觉得我的方法是正确的，如果我在各种向量上测试它，它工作 相当好，但是转换后的向量总是 有点偏.

这是我用来测试该方法的完整代码示例：

import numpy as np import matplotlib.pyplot as plt import mpl_toolkits.mplot3d import matplotlib as mpl def get_rotation_matrix(i_v, unit=None): # From http://www.j3d.org/matrix_faq/matrfaq_latest.html#Q38 if unit is None: unit = [1.0, 0.0, 0.0] # Normalize vector length i_v = np.divide(i_v, np.sqrt(np.dot(i_v, i_v))) # Get axis u, v, w = np.cross(i_v, unit) # Get angle phi = np.arccos(np.dot(i_v, unit)) # Precompute trig values rcos = np.cos(phi) rsin = np.sin(phi) # Compute rotation matrix matrix = np.zeros((3, 3)) matrix[0][0] = rcos + u * u * (1.0 - rcos) matrix[1][0] = w * rsin + v * u * (1.0 - rcos) matrix[2][0] = -v * rsin + w * u * (1.0 - rcos) matrix[0][1] = -w * rsin + u * v * (1.0 - rcos) matrix[1][1] = rcos + v * v * (1.0 - rcos) matrix[2][1] = u * rsin + w * v * (1.0 - rcos) matrix[0][2] = v * rsin + u * w * (1.0 - rcos) matrix[1][2] = -u * rsin + v * w * (1.0 - rcos) matrix[2][2] = rcos + w * w * (1.0 - rcos) return matrix # Example Vector origv = np.array([0.47404573, 0.78347482, 0.40180573]) # Compute the rotation matrix R = get_rotation_matrix(origv) # Apply the rotation matrix to the vector newv = np.dot(origv.T, R.T) # Get the 3D figure fig = plt.figure() ax = fig.gca(projection='3d') # Plot the original and rotated vector ax.plot(*np.transpose([[0, 0, 0], origv]), label="original vector", color="r") ax.plot(*np.transpose([[0, 0, 0], newv]), label="rotated vector", color="b") # Plot some axes for reference ax.plot([0, 1], [0, 0], [0, 0], color='k') ax.plot([0, 0], [0, 1], [0, 0], color='k') ax.plot([0, 0], [0, 0], [0, 1], color='k') # Show the plot and legend ax.legend() plt.show()

我已链接找到方法 here。为什么这产生的转换总是 稍微偏离 ???

Answer 1

您需要规范 uvw 才能正常工作。所以替换

u, v, w = np.cross(i_v, 单位)

有

uvw = np.cross(i_v, unit)
uvw /= np.linalg.norm(uvw)

这与您已有的 i_v = np.divide(i_v, np.sqrt(np.dot(i_v, i_v))) 行基本相同。

不过你可以做得更好，完全避免触发：

def get_rotation_matrix(i_v, unit=None):
    # From http://www.j3d.org/matrix_faq/matrfaq_latest.html#Q38
    if unit is None:
        unit = [1.0, 0.0, 0.0]
    # Normalize vector length
    i_v /= np.linalg.norm(i_v)

    # Get axis
    uvw = np.cross(i_v, unit)

    # compute trig values - no need to go through arccos and back
    rcos = np.dot(i_v, unit)
    rsin = np.linalg.norm(uvw)

    #normalize and unpack axis
    if not np.isclose(rsin, 0):
        uvw /= rsin
    u, v, w = uvw

    # Compute rotation matrix - re-expressed to show structure
    return (
        rcos * np.eye(3) +
        rsin * np.array([
            [ 0, -w,  v],
            [ w,  0, -u],
            [-v,  u,  0]
        ]) +
        (1.0 - rcos) * uvw[:,None] * uvw[None,:]
    )

最后一个表达式是来自维基百科页面的方程式：

旋转矩阵不精确以将向量与轴对齐

Imprecision with rotation matrix to align a vector to an axis

python

numpy

rotation

linear-algebra

rotational-matrices