Epipolar Geometry Basics (Cyrill Stachniss)

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Cyrill Stachniss

4 жыл бұрын

Epipolar Geometry Basics
Cyrill Stachniss, Spring 202

Пікірлер: 15

@specimon 4 жыл бұрын

Thank you so much for this! Unbelievably well explained! If I could thumb up the video 10 times I would!

@squirrelbrains2197 4 жыл бұрын

Glad I'm not the only one who gets mixed up about mathematical notations (like the description about epipoles). You do a good job explaining the notation and the concepts, thank you for making and posting those videos.

@adisingh4422 3 жыл бұрын

Cyrill is the Computer Vision GOAT

@AstroMagi 4 жыл бұрын

Thanks for doing these videos! You are a great teacher and these are very helpful!

@michalrymland 3 жыл бұрын

Your videos are so interesting and helpful. Thanks a lot !

@mfatihaydogdu7 Жыл бұрын

Well presented

@blakeedwards3582 2 жыл бұрын

Thank you

@aswinp9129 3 жыл бұрын

Sir, How epipolar lines can be used in monocular SLAM. I mean to get the F matrix, we need data associations, right? if we already have the data associations, then why would we need epipolar lines. Is it just for scoring RANSAC iterations?

@saahilnayyer6865 3 жыл бұрын

Thank you so much for the video! However I don't understand how we determine that the line equation Fx'' is certainly the epipolar line. I mean, the point x' could lie on any line and the epipolar line is just one of the many lines it can lie on. We need at least two points to define a line, don't we?

@yurigansmith 3 жыл бұрын

But e'^T*F*x'' is always zero because e'^T*F = (F^T*e')^T, and F^T*e' = 0, since e' spans the nullspace of F^T. Hence, for arbitrary x'' in image 2, F*x'' always defines a line that goes through e'. Geometrically, for every x'' there's a ray from the origin O'' of camera system 2 through the image point x'', and this ray gets projected into system 1, resulting in a line in image plane 1. All these rays are different for different x'', but have one common element, namely the origin of camera system 2, which is O''. The projection of O'', which is equal for all rays through x'', is the epipole e' in image plane 1.

@yurigansmith 3 жыл бұрын

In order to complete your argumentation with the 2 points defining a line: Assume you have determined one x' fulfilling x'*F*x'' = 0 for a given x''. Then you automatically know, that this x' lies on the epipolar line, because you already know that e' also fulfills e'*F*x'' = 0. Now you have 2 points: The determined x', and e'.

@amortalbeing 2 жыл бұрын

If the epipolar line is just made of a single feature ray being projected in the other image plane, why do we have to search that line in first place? We can simply randomly choose any where in that line and find our feature point! as it seems to me that epipolar line is compose of just that featurepoint! I dont understand the notion of search here! Also if we assume that the epipolar line contains not only one single feature but several ones, and also know that we wouldnt have single features, but feature segments (a ray on the left image, can create a line segment on the right image plane if I'm not mistaken, as explained @6:45, then how are we going to choose between several same featurepint on that line? clearly it has direct effect on the estimated distance) can anyone kindly shed some light on this?

@kemalalperencetiner6181 10 ай бұрын

Hi, this is why simple template matching may fail while looking for the correspondence in 1D space. That is simply a shortcoming of this method, if such ill conditioned cases arise, one needs to apply other techniques like sift, orb etc.