Python cv2.triangulatePoints不是很精确吗？

摘要

我正在尝试从两张图片中对点进行三角剖分，但根本得不到准确的结果

详细信息

以下是我正在做的：

在真实世界坐标中测量我的16个对象点

确定每个图像16个对象点的像素坐标

使用cv2.solvePnP（）获取每个摄像头的TVEC和rvecs

使用cv2.projectPoints验证tvecs和rvecs是否将给定的3D点重新投影到正确的图像坐标（它确实有效）。例如：

img_point_right = cv2.projectPoints(np.array([[0,0,39]], np.float), 
                                    right_rvecs, 
                                    right_tvecs,
                                    right_intrinsics,
                                    right_distortion)

left_undist = cv2.undistortPoints(left_points, 
                                   cameraMatrix=left_intrinsics,
                                   distCoeffs=left_distortion)

# Transpose to get into OpenCV's 2xN format.
left_points_t = np.array(left_undist[0]).transpose()
right_points_t = np.array(right_undist[0]).transpose()
# Note, I take the 0th index of each points matrix to get rid of the extra dimension, 
# although it doesn't affect the output.

triangulation = cv2.triangulatePoints(left_projection, right_projection, left_points_t, right_points_t)
homog_points = triangulation.transpose()

euclid_points = cv2.convertPointsFromHomogeneous(tri_homog)

验证后，使用以下公式获得旋转矩阵：

left_rotation, jacobian = cv2.Rodrigues(left_rvecs)
right_rotation, jacobian = cv2.Rodrigues(right_rvecs)

RT = np.zeros((3,4))
RT[:3, :3] = left_rotation
RT[:3, 3] = left_translation.transpose()
left_projection = np.dot(left_intrinsics, RT)

RT = np.zeros((3,4))
RT[:3, :3] = right_rotation
RT[:3, 3] = right_translation.transpose()
right_projection = np.dot(right_intrinsics, RT)

然后是投影矩阵：

left_rotation, jacobian = cv2.Rodrigues(left_rvecs)
right_rotation, jacobian = cv2.Rodrigues(right_rvecs)

RT = np.zeros((3,4))
RT[:3, :3] = left_rotation
RT[:3, 3] = left_translation.transpose()
left_projection = np.dot(left_intrinsics, RT)

RT = np.zeros((3,4))
RT[:3, :3] = right_rotation
RT[:3, 3] = right_translation.transpose()
right_projection = np.dot(right_intrinsics, RT)

在进行三角剖分之前，请使用cv2.UndortPoints取消对点的扭曲。例如：

img_point_right = cv2.projectPoints(np.array([[0,0,39]], np.float), 
                                    right_rvecs, 
                                    right_tvecs,
                                    right_intrinsics,
                                    right_distortion)

left_undist = cv2.undistortPoints(left_points, 
                                   cameraMatrix=left_intrinsics,
                                   distCoeffs=left_distortion)

# Transpose to get into OpenCV's 2xN format.
left_points_t = np.array(left_undist[0]).transpose()
right_points_t = np.array(right_undist[0]).transpose()
# Note, I take the 0th index of each points matrix to get rid of the extra dimension, 
# although it doesn't affect the output.

triangulation = cv2.triangulatePoints(left_projection, right_projection, left_points_t, right_points_t)
homog_points = triangulation.transpose()

euclid_points = cv2.convertPointsFromHomogeneous(tri_homog)

对这些点进行三角测量。例如：

img_point_right = cv2.projectPoints(np.array([[0,0,39]], np.float), 
                                    right_rvecs, 
                                    right_tvecs,
                                    right_intrinsics,
                                    right_distortion)

left_undist = cv2.undistortPoints(left_points, 
                                   cameraMatrix=left_intrinsics,
                                   distCoeffs=left_distortion)

# Transpose to get into OpenCV's 2xN format.
left_points_t = np.array(left_undist[0]).transpose()
right_points_t = np.array(right_undist[0]).transpose()
# Note, I take the 0th index of each points matrix to get rid of the extra dimension, 
# although it doesn't affect the output.

triangulation = cv2.triangulatePoints(left_projection, right_projection, left_points_t, right_points_t)
homog_points = triangulation.transpose()

euclid_points = cv2.convertPointsFromHomogeneous(tri_homog)

不幸的是，当我得到最后一步的输出时，我的点甚至没有一个正的Z方向，尽管我试图重现的3D点有一个正的Z位置

作为参考，正Z是向前的，正Y是向下的，正X是向右的

例如，3D点

（0,0,39）

-想象你前面39英尺的一个点-给出

（4.47，-8.77，-44.81）的三角测量输出

问题

这是对点进行三角测量的有效方法吗

如果是这样的话，cv2.triangulatePoints是否不是一种很好的方法，通过它可以对点进行三角剖分，以及对备选方案有何建议

谢谢您的帮助。
事实证明，如果在调用
三角点
函数之前我没有调用
无畸变点
函数，那么我会得到合理的结果。这是因为
不失真点
在执行不失真时使用内在参数对点进行规格化，但是我仍然使用解释内在参数的投影矩阵调用
三角形点

但是，我可以通过不失真点，然后使用使用单位矩阵作为内在矩阵构建的投影矩阵调用
三角形点
来获得更好的结果

问题解决了
 前一天我和你有同样的问题。结果表明，如果传递
P
矩阵，则
undistortPoints
将按预期工作，因此它将以像素为单位返回结果（否则它将假定
P
为标识并返回标准化）：
left\u undist=cv2.undistortPoints（左切点，cameraMatrix=左切点，DistCoefs=左切点，R=左切点）

这样，你就不需要弄乱本质，结果也是一样的

另外，确保在传递给三角点的参数中使用float

projMat1 = mtx1 @ cv2.hconcat([np.eye(3), np.zeros((3,1))]) # Cam1 is the origin
projMat2 = mtx2 @ cv2.hconcat([R, T]) # R, T from stereoCalibrate

# points1 is a (N, 1, 2) float32 from cornerSubPix
points1u = cv2.undistortPoints(points1, mtx1, dist1, None, mtx1)
points2u = cv2.undistortPoints(points2, mtx2, dist2, None, mtx2)

points4d = cv2.triangulatePoints(projMat1, projMat2, points1u, points2u)
points3d = (points4d[:3, :]/points4d[3, :]).T