我曾经从事使用OpenCP和Python进行三维重建和距离测量的工作。我生成了左相机的视差图,然后使用以下公式来获取距离:
D=(f*b/disp)
其中f
是焦距,b
是两个相机之间的距离,disp
是视差图矩阵。
我的问题是:
我得到的数字,它们是否应该是图片中每个点的距离?
使用此方法,我能获得的最大距离是多少(例如,在我的项目中,我得到的最大数字是110)?
img_L = cv2.pyrDown( cv2.imread(Li) )
img_R = cv2.pyrDown( cv2.imread(Ri) )
'''h, w = img_L.shape[:2]
diff=(6,6,6)
mask=np.zeros((h+2,w+2),np.uint8)
window_size = 3
min_disp = 16
num_disp = 112-min_disp
stereo = cv2.StereoSGBM(minDisparity = min_disp,
numDisparities = num_disp,
SADWindowSize = window_size,
uniquenessRatio = 10,
speckleWindowSize = 100,
speckleRange = 32,
disp12MaxDiff = 1,
P1 = 8*3*window_size**2,
P2 = 32*3*window_size**2,
fullDP = False
)
print "computing disparity..."
disp = stereo.compute(img_L, img_R).astype(np.float32) / 16.0
print "generating 3d point cloud..."
h, w = img_L.shape[:2]
f = 0.8*w # guess for focal length
points = cv2.reprojectImageTo3D(disp, Mat)
colors = cv2.cvtColor(img_L, cv2.COLOR_BGR2RGB)
mask = disp > disp.min()
cv2.imshow('left', img_L)
disparity=(disp-min_disp)/num_disp
cv2.imshow('disparity',disparity )
b=6.50
D=b*f/disp
cv2.waitKey()
cv.DestroyAllWindows()
return D
reprojectImageTo3D
(链接)? - BConic