😊 thanks

Thanks

BA is a least squares approach and yes, you can exploit more points including uncertainties (I hope that answers the question).

DLT works without an initial guess, which is a big advantage. Least squares will need one. However, LS will allow you to consider the non-linear parameters and will allow to use uncertainties properly into account. So, starting with DLT for the initial guess and then go further with a LS approach could be your way to go (assuming you do not know your calibration params already).

Sorry, just found that the default qr decomposition of matlab cannot guarantee positive elements in R, so I manually fix those negative signs( for Q and R both) and everything looks good now. Still not sure whether this is a correct method because I just tested on several samples...

Yes

I guess Im quite off topic but do anyone know a good place to watch new series online?

@Chandler Roman flixportal :)

@Jon Jett Thanks, I signed up and it seems like a nice service :) Appreciate it!!

@Chandler Roman no problem :)

i think checkerboard uses different algorithm.

No, you need to fix 11 DoF, so 6 points are needed (as each point generates a 2D observation vector, the pixel coordinates)

Do 3X3 matrix infront? 3X3 * 3X4?

diag(f,f,1) is 3x3 diagonal matrix. i is 3x3 identity matrix (no rotation). 0 is 3x1 zero matrix (no translation). K is 3x3 intrinsic matrix. R is 3x3 rotation matrix, t is 3x1 translation matrix(or column)

See my lecture on camera parameters here: kzfaq.info/get/bejne/q65xo6eirZOchXk.html

@@CyrillStachniss Thank you very much.

You can completely skip your 3D imagination and resolve it in a different way: Decomposing H^-1 leads to K^-1, the inverse calibration matrix with positive elements on the main diagonal and thus also for K. We, however, want the camera constant to be negative (so that we are in the coordinate system we want to use). So what can we do in order to achieve that? We can introduce a 3x3 matrix, let's call it B, which has zero off-diagonal elements and -1,-1,1 on the main diagonal, i.e.m B = Diag(-1,-1,1). Note that B is a rotation matrix (called R(z,pi) in the video). Also, note that B = B^T. Now we can write H = K R = K B B^T R = K B B R. Our final calibration matrix, let's call it K', can now be written as K' = K B. The matrix R' = B R is rotation matrix as B and R are rotation matrices. Thus, through the introduction of B, we created a change in the orientation of the camera (R -> R') so that the calibration matrix now has the correct sign for the camera constant (K -> K'). That's it. By design, the "magic" matrix B, which does that job for us, can be seen as a rotation by pi around the z-axis.

@@CyrillStachniss Makes sense. Thanks alot for such an amazing course and providing the explanation.

See video @24:30

You will simply have a rank deficiency in your matrix defining the linear system.

DLT estimates the ex and intrinsics. In case you know them already, you do not need the DLT. Go for PnP solutions

So PnP gives more accurate results for extrinsic matrix compared to DLT? If not, there is no point in using intrinsics.