问题
How can I detect the rotation angle from a perspective matrix?
I wrote this code ,but the resulted angle doesn't exceed 40 ...
Mat mmat;
mmat.create(3,3,CV_32FC1);
mmat=getPerspectiveTransform(templPoints,imgPoints);
cout<< mmat<<endl<<endl;
float angle=acos(mmat.at<double>(0,0));
angle=(angle*180)/3.14;
cout<<"angle is"<<angle<<endl;
回答1:
getPerspectiveTransform returns an homography matrix which can be decomposed like this:
[R11,R12,T1]
[R21,R22,T2]
[ P , P , 1]
R represents a rotation matrix, T represents a translation, and P represents a perspective warp.
More info on what rotation matrix represents:
http://en.wikipedia.org/wiki/Rotation_matrix
http://mathworld.wolfram.com/RotationMatrix.html
回答2:
The process is as follows. Consider a 3D point X, represented in homogeneous coordinates by a column vector [ x1, x2, x3, 1]' (where I use the prime symbol "'" to indicate transposition). Consider then a 2D point u = [u1, u2, 1]' that is the image of X in the (perspective) camera. Then the camera's projection is represented by a 3x4 projection matrix P such that
u = P * X .
where "*" means row-by-column product: u1 = P11*x1 + P12*x2 + P13*x3 + P14, etc.
Matrix P can be decomposed in the product of two matrices:
P = K * Q
where K is a 3x3 upper-diagonal matrix
K = [ fx s cx
0 fy cy
0 0 1 ]
that represents the camera, and Q is a 3x4 matrix
Q = [R | t]
where R is the 3x3 rotation matrix of the camera, and t is the 3x1 translation vector.
If you are given the project matrix P, the procedure for recovering the K, R and t is as follows:
- Compute the "RQ decomposition" of P
- Extract R from Q (just read its leftmost 3 columns)
来源：https://stackoverflow.com/questions/10969170/rotation-angle-of-the-perspective-matrix