将 2D 图像坐标转换为 z = 0 的 3D 世界坐标

Posted 2023-02-22

【中文标题】将 2D 图像坐标转换为 z = 0 的 3D 世界坐标

【英文标题】：Transforming 2D image coordinates to 3D world coordinates with z = 0

【发布时间】：2017-05-22 04:23:37

【问题描述】：

OpenCV => 3.2 操作系统/平台 => Windows 64 位 编译器 => Visual Studio 2015

我目前正在开展我的项目，该项目涉及车辆检测和跟踪以及估计和优化车辆周围的长方体。为此，我已经完成了车辆的检测和跟踪，我需要找到车辆边界框边缘的图像点的 3-D 世界坐标，然后估计长方体和项目边缘的世界坐标它回到图像来显示它。

所以，我是计算机视觉和 OpenCV 的新手，但据我所知，我只需要图像上的 4 个点并且需要知道这 4 个点的世界坐标，并在 OpenCV 中使用 solvePNP 来获取旋转和平移向量（我已经有了相机矩阵和失真系数）。然后，我需要使用 Rodrigues 将旋转向量转换为旋转矩阵，然后将其与平移向量连接以获得我的外在矩阵，然后将外在矩阵与相机矩阵相乘以得到我的投影矩阵。由于我的 z 坐标为零，所以我需要从投影矩阵中取出第三列，它给出了将 2D 图像点转换为 3D 世界点的单应矩阵。现在，我找到了单应矩阵的逆矩阵，它给出了 3D 世界点与 2D 图像点之间的单应矩阵。之后，我将图像点 [x, y, 1]t 与逆单应矩阵相乘得到 [wX, wY, w]t 并将整个向量除以标量 w 得到 [X, Y, 1]给我世界坐标的 X 和 Y 值。

我的代码如下所示：

#include "opencv2/opencv.hpp"
#include <stdio.h>
#include <iostream>
#include <sstream>
#include <math.h> 
#include <conio.h>

using namespace cv;
using namespace std;

Mat cameraMatrix, distCoeffs, rotationVector, rotationMatrix, 
translationVector,extrinsicMatrix, projectionMatrix, homographyMatrix, 
inverseHomographyMatrix;


Point point;
vector<Point2d> image_points;
vector<Point3d> world_points;

int main()

FileStorage fs1("intrinsics.yml", FileStorage::READ);

fs1["camera_matrix"] >> cameraMatrix;
cout << "Camera Matrix: " << cameraMatrix << endl << endl;

fs1["distortion_coefficients"] >> distCoeffs;
cout << "Distortion Coefficients: " << distCoeffs << endl << endl;



image_points.push_back(Point2d(275, 204));
image_points.push_back(Point2d(331, 204));
image_points.push_back(Point2d(331, 308));
image_points.push_back(Point2d(275, 308));

cout << "Image Points: " << image_points << endl << endl;

world_points.push_back(Point3d(0.0, 0.0, 0.0));
world_points.push_back(Point3d(1.775, 0.0, 0.0));
world_points.push_back(Point3d(1.775, 4.620, 0.0));
world_points.push_back(Point3d(0.0, 4.620, 0.0));

cout << "World Points: " << world_points << endl << endl;

solvePnP(world_points, image_points, cameraMatrix, distCoeffs, rotationVector, translationVector);
cout << "Rotation Vector: " << endl << rotationVector << endl << endl;
cout << "Translation Vector: " << endl << translationVector << endl << endl;

Rodrigues(rotationVector, rotationMatrix);
cout << "Rotation Matrix: " << endl << rotationMatrix << endl << endl;

hconcat(rotationMatrix, translationVector, extrinsicMatrix);
cout << "Extrinsic Matrix: " << endl << extrinsicMatrix << endl << endl;

projectionMatrix = cameraMatrix * extrinsicMatrix;
cout << "Projection Matrix: " << endl << projectionMatrix << endl << endl;

double p11 = projectionMatrix.at<double>(0, 0),
    p12 = projectionMatrix.at<double>(0, 1),
    p14 = projectionMatrix.at<double>(0, 3),
    p21 = projectionMatrix.at<double>(1, 0),
    p22 = projectionMatrix.at<double>(1, 1),
    p24 = projectionMatrix.at<double>(1, 3),
    p31 = projectionMatrix.at<double>(2, 0),
    p32 = projectionMatrix.at<double>(2, 1),
    p34 = projectionMatrix.at<double>(2, 3);


homographyMatrix = (Mat_<double>(3, 3) << p11, p12, p14, p21, p22, p24, p31, p32, p34);
cout << "Homography Matrix: " << endl << homographyMatrix << endl << endl;

inverseHomographyMatrix = homographyMatrix.inv();
cout << "Inverse Homography Matrix: " << endl << inverseHomographyMatrix << endl << endl;

Mat point2D = (Mat_<double>(3, 1) << image_points[0].x, image_points[0].y, 1);
cout << "First Image Point" << point2D << endl << endl;

Mat point3Dw = inverseHomographyMatrix*point2D;
cout << "Point 3D-W : " << point3Dw << endl << endl;

double w = point3Dw.at<double>(2, 0);
cout << "W: " << w << endl << endl;

Mat matPoint3D;
divide(w, point3Dw, matPoint3D);

cout << "Point 3D: " << matPoint3D << endl << endl;

_getch();
return 0;

我已经获得了四个已知世界点的图像坐标，并为简化起见对其进行了硬编码。 image_points 包含四个点的图像坐标，world_points 包含四个点的世界坐标。我正在考虑将第一个世界点作为世界轴上的原点 (0, 0, 0)，并使用已知距离计算其他四个点的坐标。现在在计算逆单应矩阵之后，我将它与与世界坐标 (0, 0, 0) 相关的 [image_points[0].x, image_points[0].y, 1]t 相乘。然后我将结果除以第三个分量 w 得到 [X, Y, 1]。但是打印出 X 和 Y 的值后，发现它们分别不是 0 和 0。做错了什么？

我的代码输出是这样的：

Camera Matrix: [517.0036881709533, 0, 320;
0, 517.0036881709533, 212;
0, 0, 1]

Distortion Coefficients: [0.1128663679798094;
-1.487790079922432;
0;
0;
2.300571896761067]

Image Points: [275, 204;
331, 204;
331, 308;
275, 308]

World Points: [0, 0, 0;
1.775, 0, 0;
1.775, 4.62, 0;
0, 4.62, 0]

Rotation Vector:
[0.661476468596541;
-0.02794460022559267;
0.01206996342819649]

Translation Vector:
[-1.394495345140898;
-0.2454153722672731;
15.47126945512652]

Rotation Matrix:
[0.9995533907649279, -0.02011656447351923, -0.02209848058392758;
 0.002297501163799448, 0.7890323093017149, -0.6143474069013439;
 0.02979497438726573, 0.6140222623910194, 0.7887261380159]

Extrinsic Matrix:
[0.9995533907649279, -0.02011656447351923, -0.02209848058392758, 
-1.394495345140898;
 0.002297501163799448, 0.7890323093017149, -0.6143474069013439, 
-0.2454153722672731;
 0.02979497438726573, 0.6140222623910194, 0.7887261380159, 
15.47126945512652]

Projection Matrix:
[526.3071813531748, 186.086785938988, 240.9673682002232, 4229.846989065414;
7.504351145361707, 538.1053336219271, -150.4099339268854, 3153.028471890794;
0.02979497438726573, 0.6140222623910194, 0.7887261380159, 15.47126945512652]

Homography Matrix:
[526.3071813531748, 186.086785938988, 4229.846989065414;
7.504351145361707, 538.1053336219271, 3153.028471890794;
0.02979497438726573, 0.6140222623910194, 15.47126945512652]

Inverse Homography Matrix:
[0.001930136511648154, -8.512427241879318e-05, -0.5103513244724983;
-6.693679705844383e-06, 0.00242178892313387, -0.4917279870709287
-3.451449134581896e-06, -9.595179260534558e-05, 0.08513443835773901]

First Image Point[275;
204;
1]

Point 3D-W : [0.003070864657310213;
0.0004761913292736786;
0.06461112415423849]

W: 0.0646111
Point 3D: [21.04004290792539;
135.683117651025;
1]

【问题讨论】：

如果您知道 3D 点位于平面上，则可以使用光线和平面之间的交点计算相机帧中的 3D 点。然后，使用相机位姿将相机帧中的 3D 点转换为物体帧。

嗨，只是检查你的代码，看起来不错，但我无法确定问题（这里是新手）。你发现问题了吗？您的 github 存储库 (github.com/indrajeetdatta/extrinsics) 上的代码似乎没有更多更新...

没关系。只是根据***.com/questions/12299870/… 以不同的方式制定它，请参阅我在github.com/rodolfoap/OpenCV-2Dto3D 上的fork。还是谢谢！

【参考方案1】：

你的推理是合理的，但是你在最后一个分区中犯了一些错误..还是我错过了什么？

你在W除之前的结果是：

Point 3D-W : 
[0.003070864657310213;
0.0004761913292736786;
0.06461112415423849]

现在我们需要通过将所有坐标除以 W（数组的第三个元素）来对其进行标准化，正如您在问题中所描述的那样。所以：

Point 3D-W Normalized = 
[0.003070864657310213 / 0.06461112415423849;
0.0004761913292736786 / 0.06461112415423849;
0.06461112415423849 / 0.06461112415423849]

结果：

Point 3D-W Normalized = 
[0.047528420183179314;
 0.007370113668614144;
 1.0]

这该死的接近 [0,0]。