相机标定，焦距值似乎太大

Question

我尝试使用 python 和 opencv 进行相机校准来找到相机矩阵。我使用了此链接中的以下代码

https://automaticaddison.com/how-to-perform-camera-calibration-using-opencv/

import cv2 # Import the OpenCV library to enable computer vision
import numpy as np # Import the NumPy scientific computing library
import glob # Used to get retrieve files that have a specified pattern
 
# Path to the image that you want to undistort
distorted_img_filename = r'C:\Users\uid20832\3.jpg'
 
# Chessboard dimensions
number_of_squares_X = 10 # Number of chessboard squares along the x-axis
number_of_squares_Y = 7  # Number of chessboard squares along the y-axis
nX = number_of_squares_X - 1 # Number of interior corners along x-axis
nY = number_of_squares_Y - 1 # Number of interior corners along y-axis
 
# Store vectors of 3D points for all chessboard images (world coordinate frame)
object_points = []
 
# Store vectors of 2D points for all chessboard images (camera coordinate frame)
image_points = []
 
# Set termination criteria. We stop either when an accuracy is reached or when
# we have finished a certain number of iterations.
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 30, 0.001)
 
# Define real world coordinates for points in the 3D coordinate frame
# Object points are (0,0,0), (1,0,0), (2,0,0) ...., (5,8,0)
object_points_3D = np.zeros((nX * nY, 3), np.float32)       
 
# These are the x and y coordinates                                              
object_points_3D[:,:2] = np.mgrid[0:nY, 0:nX].T.reshape(-1, 2) 
 
def main():
     
  # Get the file path for images in the current directory
  images = glob.glob(r'C:\Users\Kalibrierung\*.jpg')
     
  # Go through each chessboard image, one by one
  for image_file in images:
  
    # Load the image
    image = cv2.imread(image_file)  
 
    # Convert the image to grayscale
    gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)  
 
    # Find the corners on the chessboard
    success, corners = cv2.findChessboardCorners(gray, (nY, nX), None)
     
    # If the corners are found by the algorithm, draw them
    if success == True:
 
      # Append object points
      object_points.append(object_points_3D)
 
      # Find more exact corner pixels       
      corners_2 = cv2.cornerSubPix(gray, corners, (11,11), (-1,-1), criteria)       
       
            # Append image points
      image_points.append(corners)
 
      # Draw the corners
      cv2.drawChessboardCorners(image, (nY, nX), corners_2, success)
 
      # Display the image. Used for testing.
      #cv2.imshow("Image", image) 
     
      # Display the window for a short period. Used for testing.
      #cv2.waitKey(200) 
                                                                                                                     
  # Now take a distorted image and undistort it 
  distorted_image = cv2.imread(distorted_img_filename)
 
  # Perform camera calibration to return the camera matrix, distortion coefficients, rotation and translation vectors etc 
  ret, mtx, dist, rvecs, tvecs = cv2.calibrateCamera(object_points, 
                                                    image_points, 
                                                    gray.shape[::-1], 
                                                    None, 
                                                    None)

但我认为我总是得到错误的参数。我的校准焦距在 x 和 y 方向上约为 1750。我认为这不可能是正确的，这几乎是正确的。相机文档称焦距在 4-7 毫米之间。但我不确定为什么校准后的值如此之高。这是我用于校准的一些照片。也许他们出了什么问题。我在镜头下以不同的方向、角度和高度移动棋盘。

我还想知道，为什么我不需要代码中正方形的大小。有人可以向我解释一下还是我在某处忘记了这个输入？

Answer 1

您的误解是关于“焦距”。这是一个超载的术语。

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• “焦距”（单位mm）：它描述了镜头平面和图像/传感器平面之间的距离，假设焦点为无限远
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• “焦距”（单位像素）：它描述了将现实世界映射到一定分辨率的图片的比例因子
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1750如果您有高分辨率图片（全高清或其他），则很可能是正确的。

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计算如下：

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f [像素] = (焦距 [mm]) / (像素间距 [\xc2\xb5m / 像素])

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（注意单位和前缀，1 mm = 1000 \xc2\xb5m）

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例子：Pixel 4a 手机的像素间距为 1.40 \xc2\xb5m，焦距为 4.38 mm，f = ~3128.57 (= fx = fy)。

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另一个例子：Pixel 4a 的对角线视野约为 77.7 度，分辨率为 4032 x 3024 像素，因此对角线有 5040 像素。您可以计算：

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f = (5040 / 2) / tan(~77.7\xc2\xb0 / 2)

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f = ~3128.6 [像素]

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您可以将该计算应用于已知视野和图片尺寸的任意相机。如果对角线分辨率不明确，请使用水平FoV 和水平分辨率。如果传感器不是 16:9 但您从中拍摄的视频被裁剪，则可能会发生这种情况为 16:9，则可能会发生这种情况……假设裁剪仅垂直裁剪，而只保留水平方向。

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为什么这段代码中不需要棋盘格子的大小？因为它只标定内在参数（相机矩阵和畸变系数）。这些不取决于到棋盘或场景中任何其他物体的距离。

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如果您要校准外部参数，即立体设置中摄像机的距离，那么您需要给出正方形的大小。

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