Question

我尝试使用OpenCV在Python中实现基本的运动结构管道，以便在给定两个图像中的对应点的情况下从两个图像生成点云。我设法从中获取3D点数，但这些地点确实没有意义。

import sys, cv2, numpy as np
from numpy import *

def valid_cameras(inliers1, inliers2, rot, trans):
    # check if the point correspondences are in front of both images
    rot_inv = rot
    for first, second in zip(inliers1, inliers2):
        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

# points1 point2 are corresponding pixel in the two images that match computer with SIFT and RANSAC
def point_cloud(points1, points2):
    F, inliers = cv2.findFundamentalMat(points2, points1, cv2.RANSAC)
    mask = np.where( inliers.flatten() )


    for x1, x2 in zip(np.int32(points1[mask]), np.int32(points2[mask])):
        x1 = x1.tolist()
        x2 = x2.tolist()

        x1.append(1)
        x2.append(1)

        x1 = np.array(x1)
        x2 = np.array(x2)

        #print x1.T.dot(F.dot(x2)) should be approxiatemly 0.0

    # FROM HERE
    # Is this correct for iPhone 6?
    focal  = 4.89 # mm EFL
    x, y   = 2448, 3264
    sx, sy = 24, 36
    fx, fy = focal * x / sx, focal * y / sy

    K  = np.array([
      [fx, 0.0, 0.0],
      [0.0, fy, 0.0],
      [0.0, 0.0, 1.0,],
    ])

    K_inv = np.linalg.inv(K)

    # Decompose into the essential matrix
    E = K.T.dot(F).dot(K)

    # Decompose essential matrix into R, t (See Hartley and Zisserman 9.13)
    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)
    # print .flatten()
    mask = mask[0].tolist()

    inliers1, inliers2 = [], []
    for i in range( len(mask) ):
        if mask[i]:
            # normalize and homogenize the image coordinates
            inliers1.append(K_inv.dot([points1[i][0], points1[i][1], 1.0]))
            inliers2.append(K_inv.dot([points2[i][0], points2[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 valid_cameras(inliers1, inliers2, R, T):
        # Second choice: R = U * W * Vt, T = -u_3
        T = - U[:, 2]
        if not valid_cameras(inliers1, inliers2, R, T):
            # Third choice: R = U * Wt * Vt, T = u_3
            R = U.dot(W.T).dot(Vt)
            T = U[:, 2]
            if not valid_cameras(inliers1, inliers2, R, T):
                # Fourth choice: R = U * Wt * Vt, T = -u_3
                T = - U[:, 2]

    T = T.reshape(1, 3)

    P1 = np.mat('1 0 0 0 ; 0 1 0 0 ; 0 0 1 0')
    P2 = np.bmat([[R, T.T]])

    points1 = np.hstack((points1, np.ones((points1.shape[0], 1))))
    points2 = np.hstack((points2, np.ones((points2.shape[0], 1))))

    X = cv2.triangulatePoints(P1[:3], P2[:3], points1.T[:2], points2.T[:2])

    # Remember to divide out the 4th row. Make it homogeneous
    X /= X[3]

    # Recover the origin arrays from PX
    x1 = dot(P1[:3],X)
    x2 = dot(P2[:3],X)

    # Again, put in homogeneous form before using them
    x1 /= x1[2]
    x2 /= x2[2]

    print X # 3d points

我担心＃FROM HERE评论中的逻辑是否正确？特别是iPhone 6的相机矩阵是正确的，并且是从2d对应计算3d点的逻辑吗？

你如何从两个图像点对应估计3D点？

0 个答案: