What is the meaning of J_aux?

[RichardChe]

Hello, When I was reading the source code, src\stereoFrameHandler.cpp, I found a variable J_aux. According to the comments and the corresponding paper, I guess that it is a variable for jacobian of the projection error. But after referring "A tutorial on SE(3) transformation parameterizations and on-manifold optimization", I am still confused about the result of J_aux:

        Vector6d J_aux;
        J_aux << + fgz2 * dx * gz,
                 + fgz2 * dy * gz,
                 - fgz2 * ( gx*dx + gy*dy ),
                 - fgz2 * ( gx*gy*dx + gy*gy*dy + gz*gz*dy ),
                 + fgz2 * ( gx*gx*dx + gz*gz*dx + gx*gy*dy ),
                 + fgz2 * ( gx*gz*dy - gy*gz*dx );
        J_aux = J_aux / std::max(Config::homogTh(),err_i_norm);

How does this result come out? Thanks!

[rubengooj]

It is the simplified expression of the composition of all the Jacobians involved in the projection error of a point observed in the first stereo frame to the next frame, as expressed in equations (1) and (2) from the paper (in the case of line segments you just have to further compose the Jacobians of both endpoints with the line equation).

[RichardChe]

According to equation (1), it seems that the error is a 2-dimension vector so the corresponding Jacobian is a 2x6 matrix. But J_aux is a 6-dimension vector, which means the error is a scalar (I guess it is the l2-norm of the equation (1)). In that case, I am confused about how to get this simplified form. Would it be possible for you to show me that? Thanks!

[rubengooj]

Yes, that one also stands for the norm of the error in eq. (1), which I programmed like this as I'm solving the problem in eq. (3) in a robust form, as expressed in Section III.C.

[RichardChe]

So it is the jacobian norm of the error with respect to the Lie algebra xi? But how to get this simplified form? I am still confused about this.

[rubengooj]

It is the Jacobian of the parameter 's', which is the squared norm of the error, with respect to 'xi'.

[RichardChe]

OK, I have a better understanding about this. Thx!