Josep Bosch - 3 months ago 38

Python Question

I want to compute the epipolar lines of a stereo camera.

I know both camera intrinsics matrix as well as R and T.

I tried to compute the essential matrix as told in Learning Opencv book and wikipedia.

where [t]x is the matrix representation of the cross product with t.

so

I tried to implement this with python and then use the opencv function cv2.computeCorrespondEpilines to compute the epilines.

The problem is that the lines I get don't converge in a point as they should...

I guess I must have a problem computing F.

This is the relevant pice of code:

`T #Contains translation vector`

R #Rotation matrix

S=np.mat([[0,-T[2],T[1]],[T[2],0,-T[1]],[-T[1],T[0],0]])

E=np.mat(R)*S

M1=np.mat(self.getCameraMatrix(cam1))

M1_inv=np.linalg.inv(M1)

M2=np.mat(self.getCameraMatrix(cam2))

M2_inv=np.linalg.inv(M2)

F=(M2_inv.T)*E*M1_inv

The matrices are:

`M1=[[ 776.21275864 0. 773.70733324]`

[ 0. 776.21275864 627.82872456]

[ 0. 0. 1. ]]

M2=[[ 764.35675708 0. 831.26052677]

[ 0. 764.35675708 611.85363745]

[ 0. 0. 1. ]]

R=[[ 0.9999902 0.00322032 0.00303674]

[-0.00387935 0.30727176 0.9516139 ]

[ 0.0021314 -0.95161636 0.30728124]]

T=[ 0.0001648 0.04149158 -0.02854541]

The ouput F I get it's something like:

`F=[[ 4.75910592e-07 6.28777619e-08 -2.78886982e-04]`

[ -4.66942275e-08 -7.62837993e-08 -7.34825205e-04]

[ -8.86965149e-04 -6.86717269e-04 1.40633035e+00]]

EDITED:

The cross multiplication matrix was wrong, it has to be:

S=np.mat([[0,-T2,T1],[T2,0,-T[0]],[-T1,T[0],0]])

The epilines converge now at the epipole.

Answer

Hum, your F matrix seems wrong - to begin with, the rank is closer to 3 than 2. From your data I get:

```
octave:9> tx = [ 0 -T(3) T(2)
> T(3) 0 -T(1)
> -T(2) T(1) 0]
tx =
0.000000 0.028545 0.041492
-0.028545 0.000000 -0.000165
-0.041492 0.000165 0.000000
octave:11> E= R* tx
E =
-2.1792e-04 2.8546e-02 4.1491e-02
-4.8255e-02 4.6088e-05 -2.1160e-04
1.4415e-02 1.1148e-04 2.4526e-04
octave:12> F=inv(M1')*E*inv(M2)
F =
-3.6731e-10 4.8113e-08 2.4320e-05
-8.1333e-08 7.7681e-11 6.7289e-05
7.0206e-05 -3.7128e-05 -7.6583e-02
octave:14> rank(F)
ans = 2
```

Which seems to make more sense. Can you try that F matrix in your plotting code?