我正在尝试使用OpenCV来估计相机相对于另一个姿势的一个姿势,使用SIFT特征跟踪,FLANN匹配以及基本和基本矩阵的后续计算。在分解基本矩阵之后,我检查退化配置并获得"权利" R和t。
问题是,它们似乎永远不对。我包括几个图像对:
结果
结果
第二种情况下的平移向量似乎高估了Y中的运动并低估了X中的运动。转换为欧拉角时的旋转矩阵在两种情况下都给出了错误的结果。许多其他数据集也会发生这种情况。我已经尝试在RANSAC,LMEDS等之间切换基本矩阵计算技术,现在我正在使用RANSAC和仅使用8点法的内点进行第二次计算。更改特征检测方法也无济于事。极线看起来是正确的,基本矩阵满足x< .F.x = 0
我在这里错过了一些根本错误的东西吗?鉴于程序正确理解极线几何,可能会发生什么导致完全错误的姿势?我正在检查以确保点位于两个摄像头前。任何想法/建议都会非常有帮助。谢谢!
编辑:尝试使用两个不同的校准相机分开相同的技术;并将基本矩阵计算为K2' .F.K1,但翻译和旋转仍然有效。Code以供参考
import cv2
import numpy as np
from matplotlib import pyplot as plt
# K2 = np.float32([[1357.3, 0, 441.413], [0, 1355.9, 259.393], [0, 0, 1]]).reshape(3,3)
# K1 = np.float32([[1345.8, 0, 394.9141], [0, 1342.9, 291.6181], [0, 0, 1]]).reshape(3,3)
# K1_inv = np.linalg.inv(K1)
# K2_inv = np.linalg.inv(K2)
K = np.float32([3541.5, 0, 2088.8, 0, 3546.9, 1161.4, 0, 0, 1]).reshape(3,3)
K_inv = np.linalg.inv(K)
def in_front_of_both_cameras(first_points, second_points, rot, trans):
# check if the point correspondences are in front of both images
rot_inv = rot
for first, second in zip(first_points, second_points):
first_z = np.dot(rot[0, :] - second[0]*rot[2, :], trans) / np.dot(rot[0, :] - second[0]*rot[2, :], second)
first_3d_point = np.array([first[0] * first_z, second[0] * first_z, first_z])
second_3d_point = np.dot(rot.T, first_3d_point) - np.dot(rot.T, trans)
if first_3d_point[2] < 0 or second_3d_point[2] < 0:
return False
return True
def drawlines(img1,img2,lines,pts1,pts2):
''' img1 - image on which we draw the epilines for the points in img1
lines - corresponding epilines '''
pts1 = np.int32(pts1)
pts2 = np.int32(pts2)
r,c = img1.shape
img1 = cv2.cvtColor(img1,cv2.COLOR_GRAY2BGR)
img2 = cv2.cvtColor(img2,cv2.COLOR_GRAY2BGR)
for r,pt1,pt2 in zip(lines,pts1,pts2):
color = tuple(np.random.randint(0,255,3).tolist())
x0,y0 = map(int, [0, -r[2]/r[1] ])
x1,y1 = map(int, [c, -(r[2]+r[0]*c)/r[1] ])
cv2.line(img1, (x0,y0), (x1,y1), color,1)
cv2.circle(img1,tuple(pt1), 10, color, -1)
cv2.circle(img2,tuple(pt2), 10,color,-1)
return img1,img2
img1 = cv2.imread('C:\\Users\\Sai\\Desktop\\room1.jpg', 0)
img2 = cv2.imread('C:\\Users\\Sai\\Desktop\\room0.jpg', 0)
img1 = cv2.resize(img1, (0,0), fx=0.5, fy=0.5)
img2 = cv2.resize(img2, (0,0), fx=0.5, fy=0.5)
sift = cv2.SIFT()
# find the keypoints and descriptors with SIFT
kp1, des1 = sift.detectAndCompute(img1,None)
kp2, des2 = sift.detectAndCompute(img2,None)
# FLANN parameters
FLANN_INDEX_KDTREE = 0
index_params = dict(algorithm = FLANN_INDEX_KDTREE, trees = 5)
search_params = dict(checks=50) # or pass empty dictionary
flann = cv2.FlannBasedMatcher(index_params,search_params)
matches = flann.knnMatch(des1,des2,k=2)
good = []
pts1 = []
pts2 = []
# ratio test as per Lowe's paper
for i,(m,n) in enumerate(matches):
if m.distance < 0.7*n.distance:
good.append(m)
pts2.append(kp2[m.trainIdx].pt)
pts1.append(kp1[m.queryIdx].pt)
pts2 = np.float32(pts2)
pts1 = np.float32(pts1)
F, mask = cv2.findFundamentalMat(pts1,pts2,cv2.FM_RANSAC)
# Selecting only the inliers
pts1 = pts1[mask.ravel()==1]
pts2 = pts2[mask.ravel()==1]
F, mask = cv2.findFundamentalMat(pts1,pts2,cv2.FM_8POINT)
print "Fundamental matrix is"
print
print F
pt1 = np.array([[pts1[0][0]], [pts1[0][1]], [1]])
pt2 = np.array([[pts2[0][0], pts2[0][1], 1]])
print "Fundamental matrix error check: %f"%np.dot(np.dot(pt2,F),pt1)
print " "
# drawing lines on left image
lines1 = cv2.computeCorrespondEpilines(pts2.reshape(-1,1,2), 2,F)
lines1 = lines1.reshape(-1,3)
img5,img6 = drawlines(img1,img2,lines1,pts1,pts2)
# drawing lines on right image
lines2 = cv2.computeCorrespondEpilines(pts1.reshape(-1,1,2), 1,F)
lines2 = lines2.reshape(-1,3)
img3,img4 = drawlines(img2,img1,lines2,pts2,pts1)
E = K.T.dot(F).dot(K)
print "The essential matrix is"
print E
print
U, S, Vt = np.linalg.svd(E)
W = np.array([0.0, -1.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 1.0]).reshape(3, 3)
first_inliers = []
second_inliers = []
for i in range(len(pts1)):
# normalize and homogenize the image coordinates
first_inliers.append(K_inv.dot([pts1[i][0], pts1[i][1], 1.0]))
second_inliers.append(K_inv.dot([pts2[i][0], pts2[i][1], 1.0]))
# Determine the correct choice of second camera matrix
# only in one of the four configurations will all the points be in front of both cameras
# First choice: R = U * Wt * Vt, T = +u_3 (See Hartley Zisserman 9.19)
R = U.dot(W).dot(Vt)
T = U[:, 2]
if not in_front_of_both_cameras(first_inliers, second_inliers, R, T):
# Second choice: R = U * W * Vt, T = -u_3
T = - U[:, 2]
if not in_front_of_both_cameras(first_inliers, second_inliers, R, T):
# Third choice: R = U * Wt * Vt, T = u_3
R = U.dot(W.T).dot(Vt)
T = U[:, 2]
if not in_front_of_both_cameras(first_inliers, second_inliers, R, T):
# Fourth choice: R = U * Wt * Vt, T = -u_3
T = - U[:, 2]
# Computing Euler angles
thetaX = np.arctan2(R[1][2], R[2][2])
c2 = np.sqrt((R[0][0]*R[0][0] + R[0][1]*R[0][1]))
thetaY = np.arctan2(-R[0][2], c2)
s1 = np.sin(thetaX)
c1 = np.cos(thetaX)
thetaZ = np.arctan2((s1*R[2][0] - c1*R[1][0]), (c1*R[1][1] - s1*R[2][1]))
print "Pitch: %f, Yaw: %f, Roll: %f"%(thetaX*180/3.1415, thetaY*180/3.1415, thetaZ*180/3.1415)
print "Rotation matrix:"
print R
print
print "Translation vector:"
print T
plt.subplot(121),plt.imshow(img5)
plt.subplot(122),plt.imshow(img3)
plt.show()
答案 0 :(得分:3)
有很多因素可能导致从点对应中估计相机姿势的不准确。您需要考虑的一些因素: -
(*)8点法最小化代数误差(x&#39; .F.x = 0)。通常最好找到最小化有意义的几何误差的解决方案。例如,您可以在RANSAC实现中使用重新投影错误。
(*)从8点求解基本矩阵的线性算法对噪声敏感。亚像素精确点匹配,正确的数据归一化和精确的摄像机校准对于获得更好的结果都很重要。
(*)特征点定位和匹配导致噪声点匹配,因此通过求解代数方程x&#39; Fx得到的解决方案应该真正用作初始估计,并且需要进一步的参数优化等步骤用于改进解决方案。
(*)有两种视图相机配置可能导致模糊解决方案,因此需要进一步的方法(例如第三视图消除歧义)才能获得可靠的结果。
答案 1 :(得分:0)
如何获得相机内部参数K?在我看来,基本矩阵的计算是正确的,因为匹配点位于极线上。但是如果矩阵K不准确,你可能得到一个错误的基本矩阵,从而错误的R和t。
答案 2 :(得分:0)
在“使用Python编程计算机视觉”一书中,我们需要使用这样的代码来检查E的排名:
计算第二个相机矩阵(假设P1 = [I 0]) 从基本矩阵。输出是四个列表 可能的相机矩阵。
确保E排名2
U,S,V = np.linalg.svd(E)
if np.linalg.det(np.dot(U,V))<0:
V = -V
E = np.dot(U,np.dot(np.diag([1.0,1.0,0.0]),V))
我不确定这是否也能提高性能。 请告诉我。