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I have two corresponding image points (2D) visualized by the same camera with intrinsic matrix K each coming from different camera poses (R1, t1, R2, t2). If I triangulate the corresponding image points to a 3D point and then reproject it back to the original cameras it only closely matches the original image point in the first camera. Can someone help me understand why? Here is a minimal example showing the issue:

import cv2
import numpy as np

# Set up two cameras near each other

K = np.array([
    [718.856 ,   0.  ,   607.1928],
    [  0.  ,   718.856 , 185.2157],
    [  0.  ,     0.   ,    1.    ],
])

R1 = np.array([
    [1., 0., 0.],
    [0., 1., 0.],
    [0., 0., 1.]
])

R2 = np.array([
    [ 0.99999183 ,-0.00280829 ,-0.00290702],
    [ 0.0028008  , 0.99999276, -0.00257697],
    [ 0.00291424 , 0.00256881 , 0.99999245]
])

t1 = np.array([[0.], [0.], [0.]])

t2 = np.array([[-0.02182627], [ 0.00733316], [ 0.99973488]])

P1 = np.hstack([R1.T, -R1.T.dot(t1)])
P2 = np.hstack([R2.T, -R2.T.dot(t2)])

P1 = K.dot(P1)
P2 = K.dot(P2)

# Corresponding image points
imagePoint1 = np.array([371.91915894, 221.53485107])
imagePoint2 = np.array([368.26071167, 224.86262512])

# Triangulate
point3D = cv2.triangulatePoints(P1, P2, imagePoint1, imagePoint2).T
point3D = point3D[:, :3] / point3D[:, 3:4]
print(point3D)

# Reproject back into the two cameras
rvec1, _ = cv2.Rodrigues(R1)
rvec2, _ = cv2.Rodrigues(R2)

p1, _ = cv2.projectPoints(point3D, rvec1, t1, K, distCoeffs=None)
p2, _ = cv2.projectPoints(point3D, rvec2, t2, K, distCoeffs=None)

# measure difference between original image point and reporjected image point 

reprojection_error1 = np.linalg.norm(imagePoint1 - p1[0, :])
reprojection_error2 = np.linalg.norm(imagePoint2 - p2[0, :])

print(reprojection_error1, reprojection_error2)

The reprojection error in the first camera is always good (< 1px) but the second one is always large.

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1 回答 1

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记住你是如何用旋转矩阵的转置和平移向量的负值来构造投影矩阵的。当你把它放进去时，你必须做同样的事情cv2.projectPoints。

因此，对旋转矩阵进行转置并将其放入cv2.Rodrigues. 最后，将平移向量的负数提供为cv2.projectPoints：

# Reproject back into the two cameras
rvec1, _ = cv2.Rodrigues(R1.T) # Change
rvec2, _ = cv2.Rodrigues(R2.T) # Change

p1, _ = cv2.projectPoints(point3D, rvec1, -t1, K, distCoeffs=None) # Change
p2, _ = cv2.projectPoints(point3D, rvec2, -t2, K, distCoeffs=None) # Change

这样做我们现在得到：

[[-12.19064      1.8813655   37.24711708]]
0.009565768222768252 0.08597237597736622

可以肯定的是，以下是相关变量：

In [32]: p1
Out[32]: array([[[371.91782052, 221.5253794 ]]])

In [33]: p2
Out[33]: array([[[368.3204979 , 224.92440583]]])

In [34]: imagePoint1
Out[34]: array([371.91915894, 221.53485107])

In [35]: imagePoint2
Out[35]: array([368.26071167, 224.86262512])

我们可以看到前几个有效数字匹配，并且我们预计精度会略有下降，因为这是对点三角剖分的最小二乘法求解。

于 2019-06-07T21:53:05.597 回答