如何解决评估由相机校准产生的内部/外部参数的正确性
我正在使用 OpenCV
进行相机校准并设法获取所有相机参数,但现在我不确定我是否正确完成了所有操作。
这是我使用的图像
我在图像上使用了 6 个点(4 个球场角和两个在球网接触球场线的中间)imgPts = [[577,303],[1333,[495,517],[1422,[366,857],[1562,857]]
假设左上角是原点,我以米为单位构建了相应的世界坐标(23.77m x 10.97m):objPts = [[0,0],[10.97,[0,11.8,23.77,0]]
以下是我获取camera_matrix、旋转平移向量的代码:
objPts = np.array(objPts)
objPts = objPts.astype('float32')
imgPts = np.array(imgPts)
imgPts = imgPts.astype('float32')
w = frame.shape[1]
h = frame.shape[0]
size = (w,h)
camera_matrix = cv2.initCameraMatrix2D([objPts],[imgPts],size)
rms,camera_matrix,dist_coefs,rvecs,tvecs = cv2.calibrateCamera([objPts],size,None,None)
print(rms)
2.2659039195846487
print(camera_matrix)
[[7.29904054e+04 0.00000000e+00 7.70590422e+02]
[0.00000000e+00 3.27820311e+03 1.05708724e+02]
[0.00000000e+00 0.00000000e+00 1.00000000e+00]]
print(dist_coefs)
[[-4.60113019e+00 1.52353355e+03 -1.11809613e+00 7.20674734e-02
-2.28959021e+04]]
print(rvecs[0])
[[ 0.48261931]
[-4.87671221]
[ 0.28501516]]
print(tvecs[0])
[[ -0.69935398]
[ 15.30349325]
[189.46509398]]
如何检查这些值/矩阵/向量是否正确?
解决方法
我用你的数字得到了奇怪的结果。坐标似乎与您分享的图片中可识别的任何内容都不匹配。
我根据您分享的 1366x768 图片进行了自己的测量。结果看起来很合理。
然而,由于稍微不同的imgPts
,我得到了截然不同的结果。这意味着您需要进行更多测量才能获得准确度。
图片来自在半径约 70 米的 Arthur Ashe 体育场举行的一场比赛。在距离中心约 30 米处,有一条环形路径,这台相机可能就在那里。
#!/usr/bin/env python3
import os
import sys
import numpy as np
import cv2 as cv
np.set_printoptions(suppress=True)
# https://en.wikipedia.org/wiki/Tennis_court#Dimensions
court_width = 10.97 # meters
court_length = 23.77 # meters
objPts = np.float32([
[-0.5,+0.5,0],# far left
[+0.5,# far right
# center of court is 0,0
[+0.5,-0.5,# near right
[-0.5,# near left
]) * np.float32([court_width,court_length,0])
# points centered on the outside lines
# imgPts = np.float32([
# [ 346,245],# far left
# [ 988,244],# far right
# [1188,607],# near right
# [ 142,611],# near left
# ])
# points on the outsides of the outside lines (one variant)
# imgPts = np.float32([
# [ 345,# far left
# [ 989,243],# far right
# [1192,609],# near right
# [ 139,612],# near left
# ])
# points on the outsides of the outside lines (other variant)
imgPts = np.float32([
[ 344,# far left
[ 989,# far right
[1192,# near right
[ 138,613],# near left
])
#im = cv.imread("vxUZD.jpg")
#height,width = im.shape[:2]
width,height = 1366,768
print(f"image size:\n\t{width} x {height}")
C = cv.initCameraMatrix2D([objPts],[imgPts],(width,height))
print("camera matrix:")
print(C)
fx = C[0,0]
# fx * tan(hfov/2) == width/2
hfov = np.arctan(width/2 / fx) * 2
print(f"horizontal FoV:\n\t{hfov / np.pi * 180:.2f} °")
# x? mm focal length -> 36 mm horizontal (24 vertical)?
fd = 36 / (np.tan(hfov/2) * 2)
print(f"focal length (35mm equivalent):\n\t{fd:.2f} mm")
(rv,rvec,tvec) = cv.solvePnP(objPts,imgPts,C,distCoeffs=None)
print("tvec [m]:")
print(tvec)
结果:
image size:
1366 x 768
camera matrix:
[[1850.17197043 0. 682.5 ]
[ 0. 1850.17197043 383.5 ]
[ 0. 0. 1. ]]
horizontal FoV:
40.52 °
focal length (35mm equivalent):
48.76 mm
tvec [m]:
[[-0.2618669 ]
[-0.45430541]
[30.2741125 ]]
这是一个更加充实的脚本,它使用 calibrateCamera
并确定了各种参数。这似乎会导致更稳定的结果。
https://gist.github.com/crackwitz/0d1e401b597b435bcc5e65349cbca870
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。