Question

我使用双目相机从2d图片重建3d中的点，我通过双目相机拍摄了许多图片和重建点（已经找到了特征点），但我发现我重建的3d模型不在同一个坐标中

我不知道外在参数（顺便说一下，我想知道如何得到这个参数，因为我已经从校准得到了内在矩阵）
所以，我计算E矩阵（8点算法）并假设camera1的项目矩阵P1是P [I | 0]并且通过P1和E计算P2
最后一步是通过三角测量计算3d中的点数。

代码：

def compute_normalized_image_to_image_matrix(p1, p2, compute_essential=False):
    """ Computes the fundamental or essential matrix from corresponding points
        using the normalized 8 point algorithm.
    :input p1, p2: corresponding points with shape 3 x n
    :returns: fundamental or essential matrix with shape 3 x 3
    """
    n = p1.shape[1]
    if p2.shape[1] != n:
        raise ValueError('Number of points do not match.')

    # preprocess image coordinates
    p1n, T1 = scale_and_translate_points(p1)
    p2n, T2 = scale_and_translate_points(p2)

    # compute F or E with the coordinates
    F = compute_image_to_image_matrix(p1n, p2n, compute_essential)

    # reverse preprocessing of coordinates
    # We know that P1' E P2 = 0
    F = np.dot(T1.T, np.dot(F, T2))

    return F / F[2, 2]

def compute_fundamental_normalized(p1, p2):
    return compute_normalized_image_to_image_matrix(p1, p2)


def compute_essential_normalized(p1, p2):
    return compute_normalized_image_to_image_matrix(p1, p2, compute_essential=True)

def scale_and_translate_points(points):
    """ Scale and translate image points so that centroid of the points
        are at the origin and avg distance to the origin is equal to sqrt(2).
    :param points: array of homogenous point (3 x n)
    :returns: array of same input shape and its normalization matrix
    """
    x = points[0]
    y = points[1]
    center = points.mean(axis=1)  # mean of each row
    cx = x - center[0] # center the points
    cy = y - center[1]
    dist = np.sqrt(np.power(cx, 2) + np.power(cy, 2))
    scale = np.sqrt(2) / dist.mean()
    norm3d = np.array([
        [scale, 0, -scale * center[0]],
        [0, scale, -scale * center[1]],
        [0, 0, 1]
    ])

    return np.dot(norm3d, points), norm3d

def compute_P_from_fundamental(F):
    """ Compute the second camera matrix (assuming P1 = [I 0])
        from a fundamental matrix.
    """
    e = compute_epipole(F.T)  # left epipole
    Te = skew(e)
    return np.vstack((np.dot(Te, F.T).T, e)).T


def compute_P_from_essential(E):
    """ Compute the second camera matrix (assuming P1 = [I 0])
        from an essential matrix. E = [t]R
    :returns: list of 4 possible camera matrices.
    """
    U, S, V = np.linalg.svd(E)

    # Ensure rotation matrix are right-handed with positive determinant
    if np.linalg.det(np.dot(U, V)) < 0:
        V = -V

    # create 4 possible camera matrices (Hartley p 258)
    W = np.array([[0, -1, 0], [1, 0, 0], [0, 0, 1]])
    P2s = [np.vstack((np.dot(U, np.dot(W, V)).T, U[:, 2])).T,
          np.vstack((np.dot(U, np.dot(W, V)).T, -U[:, 2])).T,
          np.vstack((np.dot(U, np.dot(W.T, V)).T, U[:, 2])).T,
          np.vstack((np.dot(U, np.dot(W.T, V)).T, -U[:, 2])).T]

    return P2s

def linear_triangulation(p1, p2, m1, m2):
    """
    Linear triangulation (Hartley ch 12.2 pg 312) to find the 3D point X
    where p1 = m1 * X and p2 = m2 * X. Solve AX = 0.
    :param p1, p2: 2D points in homo. or catesian coordinates. Shape (2 x n)
    :param m1, m2: Camera matrices associated with p1 and p2. Shape (3 x 4)
    :returns: 4 x n homogenous 3d triangulated points
    """
    num_points = p1.shape[1]
    res = np.ones((4, num_points))

    for i in range(num_points):
        A = np.asarray([
            (p1[0, i] * m1[2, :] - m1[0, :]),
            (p1[1, i] * m1[2, :] - m1[1, :]),
            (p2[0, i] * m2[2, :] - m2[0, :]),
            (p2[1, i] * m2[2, :] - m2[1, :])
        ])

        _, _, V = np.linalg.svd(A)
        X = V[-1, :]
        res[:, i] = X / X[3]

    return res

那我怎么解决这个问题呢？我希望我所有的重建点都在同一个坐标系中，你能告诉我吗？非常感谢你！

为什么我重建的点不在同一个坐标中？

0 个答案: