Thank you, Prof Stachniss. The entire course is clarity and the equation deduction is easy to understand. We like these series of online courses.

Thank you, Prof. Stachniss. Your courses are enjoyable.

Amazing series! This is extremely helpful for all students across the globe. If you are taking suggestions for future topics to make videos on, could you consider making a series on Visual SLAM? That would be very helpful. Thanks.

Thank you for this video.

Your series are the best i ever watched.

@CyrillStachniss

4 жыл бұрын

Thanks :-)

@sags

4 жыл бұрын

couldn't agree more. Kudos 🤟

@hssensana1678

3 жыл бұрын

absolutly !!!!!!

Thank you so much . You are are really good teacher

thank you so much for this video and the one dealing with the DLT , it helped me a lot to understand the Zhang paper.

@mahmodsahal3414

3 жыл бұрын

Hi, hssen, please could we talk?

@lordstinson8095

3 жыл бұрын

@@mahmodsahal3414 WTF MAHMOD!?

Thanks so much for that, it helped a lot!

I went through Zhangs paper....without this explanation, it wud have taken months for me to understand...thanks a lot

@CyrillStachniss

Жыл бұрын

My pleasure - and I may say that it also took me a while when I first tried to dive into it…

Question, you go over how to obtain the intrinsic matrix (K), but is it also possible to get the extrinsic matrix using this approach?

In an attempt to implement this, I used a single checkerboard image with known xy-coordinates of the crossings as well as the corresponding pixel coordinates. A gradient descent method based on least squares errors was computationally faster and gave me both the intrinsic as well as extrinsinc parameters for that picture. Is this coincidence? When taking computational effort and robustness into account how does zhangs compares to such a gradient descent m eethod?

I want to implement Zhang's Method and using your lecture to help with it. I noticed that on slide 40 the vij vector is not defined the same way as in Zhang's paper. The elements 3 and 4 are exchanged. So I'm asking myself: does the order of the elements matter? Is the exchange on purpose or just a mistake?

Hi! Thank you for the video! I have a doubt: what if we want to set zero-skew?

Hello Professor, The original paper on Zhang Calibration mentions that this calibration requires at least two images but you mentioned we need at least two. Is there something that I am missing ?

what kinda workouts you do?

thanks for your video! question, what if I also want to find out the rotation matrix and transition vector? what should I do to calculate them after the K matrix is found? would DLT work for finding out R and t? thanks again!

@TheTacticalDood

2 жыл бұрын

Yes, the direct linear transform method can be used to find the camera pose.

Hello, thank you very much for your explanation, please i want to know if we can use camera calibration to measuer a size of an object without knowing the distance lens/object ?

@CyrillStachniss

3 жыл бұрын

Not from a single image. Even if you have multiple images, you can only measure it if you know the length of the baseline vector.

I’m having trouble heat mapping to calibrate my checkerboard I don’t have enough sunlight. Any ideas how I solve this ?

thanks for your video! question, are you sure that K is invertible matrix? because si a projection matrix (from 3D to 2D)

@CyrillStachniss

2 жыл бұрын

K itself covers only the (linear) intrinsic calibration parameters is a 3x3 matrix and yes, it is invertible.

@poojakabra1479

Жыл бұрын

@@CyrillStachniss so if we multiplied pixel coordinates by inv(K) will we get 3D coordinates in camera coordinate system in the true scale?

How do we ensure the matrix B we obtain from the least squares solution is positive definite? The Cholesky decomposition will only work if it is positive (semi) definite

@pu239

Жыл бұрын

I got this exact problem :/ Did you figure anything out?

Short question for the matrix B: Why is B a symmetric Matrix?

@rohanjoshi523

3 жыл бұрын

Write down the expanded form of( K^-T * K^-1). Since K is upper triangular, the product will be symmetric.

@javocremona

3 жыл бұрын

@@rohanjoshi523 and why B is positive definite?

@rohanjoshi523

3 жыл бұрын

@@javocremona Try performing elimination and see the signs of the pivots. All the pivots should turn out to be positive. The sign of the pivots are the sign of the eigen values.

@javocremona

3 жыл бұрын

@@rohanjoshi523 Thanks!

I never realized a homography matrix was just a projection matrix with the Z column deleted.

@TyNguyen-hd2wm

3 жыл бұрын

It's not 100% correct though. Homography matrix can be formed by removing either r1, r2, r3 or t in the homogeneous transformation matrix. [R|t].

@rohanjoshi523

3 жыл бұрын

Any NxN non-singular matrix is a homography. Since we deal with a 2 dimensional projective space, we represent homographies as a 3x3 matrix that can transform points on one plane to another. Hence any non-singular 3x3 matrix is a homography.

Shouldn't the matrix H have 9 degrees of freedom? 5 intrinsics and 4 extrinsics

@TyNguyen-hd2wm

3 жыл бұрын

it's up to the scale so we can normalize all by element H_{33}, making it 8 Dof.

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