Question

我正在将3D现实点（X，Y，Z）米转换为图像坐标（u，v）像素。我可以访问相机矩阵K，相机方向（以度为单位）和平移矢量。

我使用下面的代码来计算像素坐标。对于(X,Y,Z) = (0.04379281, 0.15902013, 0.73328906)，我在尺寸为(u, v) = (-184.52735432, -249.19158505)的图像上获得了352x287 px。图像中像素坐标的原点是否还在左上角？如果是，则获得的像素位置在图像外部。是这样吗？

# Calculates Rotation Matrix given euler angles.
def eulerAnglesToRotationMatrix(theta):
R_x = np.array([[1, 0, 0],
            [0, math.cos(theta[0]), -math.sin(theta[0])],
            [0, math.sin(theta[0]), math.cos(theta[0])]
            ])

R_y = np.array([[math.cos(theta[1]), 0, math.sin(theta[1])],
            [0, 1, 0],
            [-math.sin(theta[1]), 0, math.cos(theta[1])]
            ])

R_z = np.array([[math.cos(theta[2]), -math.sin(theta[2]), 0],
            [math.sin(theta[2]), math.cos(theta[2]), 0],
            [0, 0, 1]
            ])

R = np.dot(R_z, np.dot(R_y, R_x))
return R

# x-roll, y-pitch, z-yaw (in degrees)
theta = [-128.41639709472657, -11.528900146484375, 53.37379837036133]

# rotation vector
rvec = eulerAnglesToRotationMatrix(theta)
# translation vector
tvec = np.array([[0.26409998536109927], [1.294700026512146], 
[0.017799999564886094]], np.float)
# camera matrix
cameraMatrix = np.array([[148.130859375, 0, 142.5562286376953],
                [0, 148.130859375, 179.04293823242188],
                [0, 0, 1]], np.float)


# Point in (X,Y,Z) meters
objPts = np.array([[0.04379281, 0.15902013, 0.73328906]], np.float)

distCoeffs = np.array([0, 0, 0, 0, 0], np.float)

imgPts, _ = cv2.projectPoints(objPts, rvec, tvec, cameraMatrix, distCoeffs)

对projectPoints（）函数的解释

0 个答案: