OpenCV Homography, Transform a point, what is this code doing?

谁说我不能喝 提交于 2019-12-29 14:31:31

问题


I'm working with a homography calculated by OpenCV. I currently use this homography to transform points using the function below. This function performs the task I require however I have no clue how it actually works.

Can anyone explain, line by line exactly, the logic/theory behind the last 3 lines of code, I understand that this transforms the point x,y but I'm unclear as to why this works:

Why are Z, px and py calculated in this way, what do the elements in h correspond to?

Your comments are greatly appreciated :)

double h[9];
homography = cvMat(3, 3, CV_64F, h);
CvMat ps1 = cvMat(MAX_CALIB_POINTS/2,2,CV_32FC1, points1);
CvMat ps2 = cvMat(MAX_CALIB_POINTS/2,2,CV_32FC1, points2);

cvFindHomography(&ps1, &ps2, &homography, 0);

...

// This is the part I don't fully understand
double x = 10.0;
double y = 10.0;
double Z = 1./(h[6]*x + h[7]*y + h[8]);
px = (int)((h[0]*x + h[1]*y + h[2])*Z);
py = (int)((h[3]*x + h[4]*y + h[5])*Z);

回答1:


cvFindHomography() returns a matrix using homogenous coordinates:

Homogeneous coordinates are ubiquitous in computer graphics because they allow common operations such as translation, rotation, scaling and perspective projection to be implemented as matrix operations

What's happening in the code: The cartesian point p_origin_cartesian(x,y) is transformed to homogenous coordinates, then the 3x3 perspective transformation matrix h is applied and the result is converted back to cartesian coordinates p_transformed_cartesian(px,py).

UPDATE

In detail:

Convert p_origin_cartesian to p_origin_homogenous:

(x,y)  =>  (x,y,1)

Do perspective transformation:

p_transformed_homogenous = h * p_origin_homogenous =

(h0,h1,h2)    (x)   (h0*x + h1*y + h2)   (tx)   
(h3,h4,h5)  * (y) = (h3*x + h4*y + h5) = (ty) 
(h6,h7,h8)    (1)   (h6*x + h7*y + h8)   (tz)

Convert p_transformed_homogenous to p_transformed_cartesian:

(tx,ty,tz)  =>  (tx/tz, ty/tz) 

Your code translated:

px = tx/tz;
py = ty/tz;
Z  = 1/tz;



回答2:


OpenCV Python implementation following @Ben answer

p = np.array((x,y,1)).reshape((3,1))
temp_p = M.dot(p)
sum = np.sum(temp_p ,1)
px = int(round(sum[0]/sum[2]))
py = int(round(sum[1]/sum[2]))


来源:https://stackoverflow.com/questions/9275567/opencv-homography-transform-a-point-what-is-this-code-doing

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!