如何繪制3D坐標軸與OpenCV進行人臉姿勢估計？

Question

假設我有面部歐拉角（俯仰、偏航、滾動）。 如何繪制顯示面部姿勢的 3D 坐標軸？

這是來自此處的示例：

Answer 1

只要您具有相機參數，就可以在純OpenCV中完成此操作。 您應該能夠創建與x，y，z軸相對應的三個向量（基本上是[0,0,0] [1、0、0]，[0、1、0]，[0、0、1]點），您將隨后投影到圖像平面中，首先應根據您的偏航/俯仰/橫滾旋轉這些點（例如，將它們乘以旋轉矩陣）。

為了將3D點投影到圖像平面，請使用projectPoints函數。 它需要3D點，相機參數並生成2D圖像點。 有了圖像點后，您可以簡單地使用line函數在投影的中心點（3D中為[0,0,0]）與每個軸點投影之間繪制線。

Answer 2

一個簡單的例子：

def draw_axis(img, R, t, K):
    # unit is mm
    rotV, _ = cv2.Rodrigues(R)
    points = np.float32([[100, 0, 0], [0, 100, 0], [0, 0, 100], [0, 0, 0]]).reshape(-1, 3)
    axisPoints, _ = cv2.projectPoints(points, rotV, t, K, (0, 0, 0, 0))
    img = cv2.line(img, tuple(axisPoints[3].ravel()), tuple(axisPoints[0].ravel()), (255,0,0), 3)
    img = cv2.line(img, tuple(axisPoints[3].ravel()), tuple(axisPoints[1].ravel()), (0,255,0), 3)
    img = cv2.line(img, tuple(axisPoints[3].ravel()), tuple(axisPoints[2].ravel()), (0,0,255), 3)
    return img

Answer 3

我用了這段代碼。 它來自 Basel Face model(BFM)，您可以從他們的 web 站點找到 matlab 代碼

def draw_axis(img, yaw, pitch, roll, tdx=None, tdy=None, size = 100):

    pitch = pitch * np.pi / 180
    yaw = -(yaw * np.pi / 180)
    roll = roll * np.pi / 180

    if tdx != None and tdy != None:
        tdx = tdx
        tdy = tdy
    else:
        height, width = img.shape[:2]
        tdx = width / 2
        tdy = height / 2

    # X-Axis pointing to right. drawn in red
    x1 = size * (math.cos(yaw) * math.cos(roll)) + tdx
    y1 = size * (math.cos(pitch) * math.sin(roll) + math.cos(roll) * math.sin(pitch) * math.sin(yaw)) + tdy

    # Y-Axis | drawn in green
    #        v
    x2 = size * (-math.cos(yaw) * math.sin(roll)) + tdx
    y2 = size * (math.cos(pitch) * math.cos(roll) - math.sin(pitch) * math.sin(yaw) * math.sin(roll)) + tdy

    # Z-Axis (out of the screen) drawn in blue
    x3 = size * (math.sin(yaw)) + tdx
    y3 = size * (-math.cos(yaw) * math.sin(pitch)) + tdy

    cv2.line(img, (int(tdx), int(tdy)), (int(x1),int(y1)),(0,0,255),3)
    cv2.line(img, (int(tdx), int(tdy)), (int(x2),int(y2)),(0,255,0),3)
    cv2.line(img, (int(tdx), int(tdy)), (int(x3),int(y3)),(255,0,0),3)

    return img

Answer 4

對上面給出的代碼的一些說明

def draw_axis(img, rotation_vec, t, K, scale=0.1, dist=None):
    """
    Draw a 6dof axis (XYZ -> RGB) in the given rotation and translation
    :param img - rgb numpy array
    :rotation_vec - euler rotations, numpy array of length 3,
                    use cv2.Rodrigues(R)[0] to convert from rotation matrix
    :t - 3d translation vector, in meters (dtype must be float)
    :K - intrinsic calibration matrix , 3x3
    :scale - factor to control the axis lengths
    :dist - optional distortion coefficients, numpy array of length 4. If None distortion is ignored.
    """
    img = img.astype(np.float32)
    dist = np.zeros(4, dtype=float) if dist is None else dist
    points = scale * np.float32([[1, 0, 0], [0, 1, 0], [0, 0, 1], [0, 0, 0]]).reshape(-1, 3)
    axis_points, _ = cv2.projectPoints(points, rotation_vec, t, K, dist)
    img = cv2.line(img, tuple(axis_points[3].ravel()), tuple(axis_points[0].ravel()), (255, 0, 0), 3)
    img = cv2.line(img, tuple(axis_points[3].ravel()), tuple(axis_points[1].ravel()), (0, 255, 0), 3)
    img = cv2.line(img, tuple(axis_points[3].ravel()), tuple(axis_points[2].ravel()), (0, 0, 255), 3)
    return img