More on single-view geometry class 10

Transcription

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) Projective 2D Geometry Jan. 21, 23 Projective 3D Geometry (no class) Jan. 28, 30 Parameter Estimation 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 rifocal ensor Mar. 18, 20 hree View Reconstruction Multiple View Geometry Mar. 25, 27 MultipleView Reconstruction Bundle adjustment Apr. 1, 3 Auto-Calibration Papers Apr. 8, 10 Dynamic SfM Papers Apr. 15, 17 Cheirality Papers 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 {X i x i }, determine the Maximum Likelyhood Estimation of P Algorithm (i) Linear solution: (a) Normalization: (b) DL X ~ i = UX i x~ = x i i (ii) Minimization of geometric error: using the linear estimate as a starting point minimize the geometric error: ~ ~ ~ (iii) Denormalization: P = -1 ~ PU

5

6

7

8 More Single-View Geometry P CP = Q cone Projective cameras and * PQ P = planes, lines, conics and quadrics. C * Camera calibration and vanishing points, calibrating conic and the IAC

9 Action of projective camera on planes [ ] [ ] = = = 1 p p p 1 0 p p p p PX x Y X Y X he 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 ( μ) = P(A + μb) = PA + μpb = a μb X + back-projection Π = P l Π X = l PX

11 Action of projective camera on conics back-projection to cone Q co = P CP example: Q co = K 0 C [ K 0] = K CK 0 0 0

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

13 Action of projective camera on quadrics back-projection to cone C * = * * * Q l PQ P l 0 PQ P Π Π = = he plane of Γ for a quadric Q is camera center C is given by Π=QC (follows from pole-polar relation) he cone with vertex V and tangent to the quadric Q is the degenerate Quadric: Q CO = (V QV)Q - (QV)(QV) Q CO V = 0

14 he importance of the camera center P = P' = x' = KR[I C ~ ],P' = K' R' P' X = ( )-1 KR P K' R' x' = Hx with H = K' R'[I C ~ ] ( )-1 KR PX K' R' ( KR) -1 = x K' R' ( KR) -1

15 Moving the image plane (zooming) x = x' = K[I 0]X K'[I 0]X = K' ( )-1 K x k)x ( ) ~ -1 ki (1 0 H = K' K = 0 1 k = f / f ki (1 k)x ~ k)x ~ 0 ki (1 0 A K' = K = ka x~ 0 = = ki 0 K ' x~ 1 0

16 Camera rotation x = K[I 0]X x' = K[R 0]X = KRK -1 x -1 H = KRK conjugate rotation { e iθ μ,μ,μe iθ }

17 Synthetic view (i) Compute the homography that warps some a rectangle to the correct aspect ratio (ii) 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 ) (0,0,0,1,X (0,0,1,0),X (0,1,0,0),X (1,0,0,0) X = = = = ) (1,1,1 x, (0,0,1) x, (0,1,0) x, (1,0,0) x = = = = = d c d b d a P d c b a ),,, ( C =

22 Moving the camera center motion parallax epipolar line

23 What does calibration give? x = K[I 0] d 0 d = K 1 x d1 d 2 cos = = θ ( )( ) ( )( ) d d d d x (K K )x x (K K )x An image l defines a plane through the camera center with normal n=k l measured in the camera s Euclidean frame x 1 (K - 1 K -1 2 )x 2 2

24 he image of the absolute conic x = PX KR[I C ~ ] d = = 0 KRd mapping between π to an image is given by the planar homogaphy x=hd, with H=KR image of the absolute conic (IAC) ( ) KK K K ω = = (i) IAC depends only on intrinsics (ii) angle between two rays cos θ = (iii) DIAC=ω * =KK (iv) ω K (cholesky factorisation) (v) image of circular points ( 1 C a H CH ) ( x ωx )( x ωx ) 1 x 1 1 ωx 2 2 2

25 A simple calibration device (i) compute H for each square (corners (0,0),(1,0),(0,1),(1,1)) (ii) compute the imaged circular points H(1,±i,0) (iii) fit a conic to 6 circular points (iv) compute K from ω through cholesky factorization (= Zhang s calibration method)

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

27 he calibrating conic C = K 1 1 K 1 1

28 Vanishing points ( λ) = PX( λ) = PA + λpd = a λkd x + v = lim x λ v = PX = ( λ) = lim ( a + λkd) = Kd Kd λ

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

30 Vanishing lines

31

32

33 Orthogonality relation cos θ = ( )( ) v ωv v ωv 1 v 1 1 ωv 2 2 2

34

35

36

37

38 Five constraints gives us five equations and can determine w

39 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 hen 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 o orthogonal directions.

42

43

44 Next class: wo-view geometry Epipolar geometry