1. More on single-view geometry class 10
Multiple View Geometry
Comp
Marc Pollefeys

2. Multiple View Geometry course schedule (subject to change)
Jan. 7, 9
Intro & motivation
Projective 2D Geometry
Jan. 14, 16
(no class)
Jan. 21, 23
Projective 3D Geometry
Jan. 28, 30
Parameter Estimation
Feb. 4, 6
Algorithm Evaluation
Camera Models
Feb. 11, 13
Camera Calibration
Single View Geometry
Feb. 18, 20
Epipolar Geometry
3D reconstruction
Feb. 25, 27
Fund. Matrix Comp.
Structure Comp.
Mar. 4, 6
Planes & Homographies
Trifocal Tensor
Mar. 18, 20
Three View Reconstruction
Multiple View Geometry
Mar. 25, 27
MultipleView Reconstruction
Bundle adjustment
Apr. 1, 3
Auto-Calibration
Papers
Apr. 8, 10
Dynamic SfM
Apr. 15, 17
Cheirality
Apr. 22, 24
Duality
Project Demos

3. Single view geometry
Camera model
Camera calibration
Single view geom.

4. Gold Standard algorithm
Objective
Given n≥6 2D to 2D point correspondences {Xi↔xi’}, determine the Maximum Likelyhood Estimation of P
Algorithm
Linear solution:
Normalization:
DLT
Minimization of geometric error: using the linear estimate as a starting point minimize the geometric error:
Denormalization: ~ ~ ~

8. More Single-View Geometry
Projective cameras and
planes, lines, conics and quadrics.
Camera calibration and vanishing points, calibrating conic and the IAC

9. Action of projective camera on planes
The most general transformation that can occur between
a scene plane and an image plane under perspective
imaging is a plane projective transformation
(affine camera-affine transformation)

10. Action of projective camera on lines
forward projection
back-projection

11. Action of projective camera on conics
back-projection to cone
example:

12. Images of smooth surfaces
The contour generator G is the set of points X on S at which rays are tangent to the surface. The corresponding apparent contour g is the set of points x which are the image of X, i.e. g is the image of G
The contour generator G depends only on position of projection center, g depends also on rest of P

13. Action of projective camera on quadrics
back-projection to cone
The plane of G for a quadric Q is camera center C is given by P=QC (follows from pole-polar relation)
The cone with vertex V and tangent to the quadric Q is the degenerate Quadric:

14. The importance of the camera center

15. Moving the image plane (zooming)

16. Camera rotation
conjugate rotation

17. Synthetic view
Compute the homography that warps some a rectangle to the correct aspect ratio
warp the image

18. Planar homography mosaicing

19. close-up: interlacing
can be important problem!

20. Planar homography mosaicing
more examples

21. Projective (reduced) notation

22. Moving the camera center
motion parallax
epipolar line

23. What does calibration give?
An image l defines a plane through the camera center with normal n=KTl measured in the camera’s Euclidean frame

24. The image of the absolute conic
mapping between p∞ to an image is given by the planar homogaphy x=Hd, with H=KR
image of the absolute conic (IAC)
IAC depends only on intrinsics
angle between two rays
DIAC=w*=KKT
w  K (cholesky factorisation)
image of circular points

25. A simple calibration device
compute H for each square
(corners  (0,0),(1,0),(0,1),(1,1))
compute the imaged circular points H(1,±i,0)T
fit a conic to 6 circular points
compute K from w through cholesky factorization
(= Zhang’s calibration method)

26. Orthogonality = pole-polar w.r.t. IAC

27. The calibrating conic

28. Vanishing points

29. ML estimate of a vanishing point from imaged parallel scene lines

30. Vanishing lines

33. Orthogonality relation

38. Five constraints gives us five equations and can determine w

39. Assumes zero skew, square pixels and 3 orthogonal vanishing points
Calibration from vanishing points and lines
Principal point is the orthocenter of the trinagle made of 3 orthogonol vanishing lines
Assumes zero skew, square pixels and 3 orthogonal vanishing points

40. Assume zero skew, square pixels,  calibrating conic is a circle;
How to find it, so that we can get K?

41. Assume zero skew , square pixels, and principal point is at the image center
Then IAC is diagonal{1/f^2, 1/f^2,1) i.e. one degree of freedom need one more
Constraint to determine f, the focal length  two vanishing points corresponding
To orthogonal directions.

44. Next class: Two-view geometry Epipolar geometry