像素坐标到3D线（opencv） [英] Pixel coordinates to 3D line (opencv)

问题描述

我有一个图像显示在屏幕上，通过cvInitUndistortMap& cvRemap（已经完成相机校准），并且用户点击图像中的特征。所以我有（u，v）像素坐标的特征，我也有内在矩阵和失真矩阵。

I have an image displayed on screen which is undistorted via cvInitUndistortMap & cvRemap (having done camera calibration), and the user clicks on a feature in the image. So I have the (u,v) pixel coordinates of the feature, and I also have the intrinsic matrix and the distortion matrix.

我要找的是用户点击的特征必须位于的相机/真实世界坐标中的3D线的方程。我已经有相机的图像平面和特征之间的垂直距离，因此我可以将它与上述方程组合，给出空间中特征的（X，Y，Z）坐标。

What I'm looking for is the equation of the 3D line in camera/real-world coordinates on which the feature the user clicked must lie. I already have the perpendicular distance between the camera's image plane and the feature, so I can combine that with the aforementioned equation to give me the (X,Y,Z) coordinate of the feature in space.

听起来很容易（反向内在矩阵或某事？），但我不能在任何地方找到一步一步的说明。 C ++或C＃代码。

Sounds easy (inverse intrinsic matrix or something?) but I can't find step-by-step instructions anywhere. C++ or C# code preferred.

推荐答案

这是一个有点旧的问题，但仍然可能对某人有用。
所有行都通过点（0,0,0），因此：

This is a bit old question but still might be usefull for someone. All lines go through the point (0,0,0), so:

line.x0 = 0;
line.y0 = 0;
line.z0 = 0;

line.x0 = 0; line.y0 = 0; line.z0 = 0;

方向向量如下：
line.A =（u / fx） - （cx / fx） ;
line.B =（v / fy） - （cy / fy）;
line.C = 1;

direction vector is as follows: line.A = (u/fx) - (cx/fx); line.B = (v/fy) - (cy/fy); line.C = 1;

cx，cy，fx，fy是来自相机矩阵的参数。
方程式在Learning OpenCv书中解释。

cx,cy,fx,fy are parameters from camera matrix. Equations are explained in "Learning OpenCv" book.

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