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Geometria della telecamera. the image of a point P belongs to the line (P,O) p P O p = image of P = image plane ∩ line(O,P) interpretation line of p:

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Presentation on theme: "Geometria della telecamera. the image of a point P belongs to the line (P,O) p P O p = image of P = image plane ∩ line(O,P) interpretation line of p:"— Presentation transcript:

1 Geometria della telecamera

2 the image of a point P belongs to the line (P,O) p P O p = image of P = image plane ∩ line(O,P) interpretation line of p: line(O,p) = locus of the scene points projecting onto image point p image plane r  f Hp: Z >> a

3 p P O Z Y X c y x perspective projection f -nonlinear -not shape-preserving -not length-ratio preserving

4 Point [x,y] T expanded to [u,v,w] T Any two sets of points [u 1,v 1,w 1 ] T and [u 2,v 2,w 2 ] T represent the same point if one is multiple of the other [u,v,w] T  [x,y] with x=u/w, and y=v/w [u,v,0] T is the point at the infinite along direction (u,v) In 2D: add a third coordinate, w Homogeneous coordinates

5 Transformations translation by vector [ d x,d y ] T scaling (by different factors in x and y) rotation by angle 

6 Homogeneous coordinates In 3D: add a fourth coordinate, t Point [X,Y,Z] T expanded to [x,y,z,t] T Any two sets of points [x 1,y 1,z 1,t 1 ] T and [x 2,y 2,z 2,t 2 ] T represent the same point if one is multiple of the other [x,y,z,t] T  [X,Y,Z] with X=x/t, Y=y/t, and Z=z/t [x,y,z,0] T is the point at the infinite along direction (x,y,z)

7 Transformations scaling translation rotation Obs: rotation matrix is an orthogonal matrix i.e.: R -1 = RTRT

8 with Scene->Image mapping: perspective transformation With “ad hoc” reference frames, for both image and scene

9 Let us recall them O Z Y X c y x f scene reference - centered on lens center - Z-axis orthogonal to image plane - X- and Y-axes opposite to image x- and y-axes image reference - centered on principal point - x- and y-axes parallel to the sensor rows and columns - Euclidean reference

10 O Z Y X c y x f scene reference - not attached to the camera image reference - centered on upper left corner - nonsquare pixels (aspect ratio)  noneuclidean reference Actual references are generic

11 Scene-image relationship wrt actual reference frames image scene normally, s=0

12 K upper triangular : intrinsic camera parameters scene-camera tranformation extrinsic camera parameters orthogonal (3D rotation) matrix P: 10-11 degrees of freedom (10 if s=0)

13 Radial distortion

14 i.e., defining x = [x, y, z] T with and

15 The locus of the points x whose image is u is a straight line through o having direction is independent of u o is the camera viewpoint (perspective projection center) line(o, d) = Interpretation line of image point u Interpretation of o: u is image of x if i.e., if

16 Intrinsic and extrinsic parameters from P M  K and R QT-decomposition of a matrix: as the product between an orthogonal matrix and an upper triangular matrix M and m  t

17 Camera calibration

18 from scene-point to image point correspondence… …to projection matrix

19 Basic equations

20 Basic equations ctd. with(12x1) singular matrix

21 minimal solution over-determined solution  5½ correspondences needed (say 6) P has 11 dof, 2 independent eq./points n  6 points minimize subject to constraint p : eigenvector of A T A associated to its smallest eigenvalue

22 Degenerate configurations More complicate than 2D case (see Ch.21) (i)Camera and points on a twisted cubic (ii)Points lie on plane or single line passing through projection center

23 Data normalization (i)translate origin to gravity center (ii)(an)isotropic scaling

24 from line correspondences Extend DLT to lines (back-project image line) (2 independent eq.)

25 Geometric error

26 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)DLT: (ii)Minimization of geometric error: using the linear estimate as a starting point minimize the geometric error: (iii)Denormalization: ~~ ~

27 Calibration example (i)Canny edge detection (ii)Straight line fitting to the detected edges (iii)Intersecting the lines to obtain the images corners typically precision <1/10 (HZ rule of thumb: 5 n constraints for n unknowns

28 Exterior orientation Calibrated camera, position and orientation unkown  Pose estimation 6 dof  3 points minimal (4 solutions in general)

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30 short and long focal length Radial distortion

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33 Correction of distortion Choice of the distortion function and center Computing the parameters of the distortion function (i)Minimize with additional unknowns (ii)Straighten lines (iii)…

34 Properties of perspective transformations 1) vanishing points V image of the unproper point along direction d the interpretation line of V is parallel to d

35 O V d The images of parallel lines are concurrent lines

36 2) cross ratio invariance Given four colinear points letbe their abscissae Properties of perspective transformations ctd.

37 Cross ratio invariance under perspective transformation a point on the line y=0=z its image belongs to a line its coordinate u

38 Object localization 1: three colinear points geometric model of an object a perspective image of the object position and orientation of the object ? A B C C’ A’ B’ O  calibrated camera: A B C  known  known interpretation lines

39 A B C C’ A’ B’ O  ∞ V a) orientation Cross ratio invariance: solve for V (image of ∞) V: vanishing point of the direction of (A,B,C) interpretation line of V parallel to (A,B,C)direction

40 b) position (e.g., distance(O,A)) A B C C’ A’ B’ O  V interpretation lines angles  and   

41 Object localization 2: four coplanar points O (i)orientation of (A,E,C) (ii)orientation of (B,E,D) (iii)distance (O,A) E A B C D

42 a’ b’ b” a” Find vanishing point of the field-bottom direction Off-side images of symmetric segments

43 a and b : abscissae of the endpoints of a segment c=(a+b)/2 : abscissa of segment midpoint, d=∞ : point at the infinite along the segment direction ac bd ( a’,b’ ) and ( a”, b” ) are image of symmetric segments  same image of the midpoint c’, same vanishing point d’ Harmonic 4-tuple ( a,b,c,d )

44  solve { for c’, d’  system of two linear equations in ( c’d’ ) and ( c’+d’ )  two degree equation, whose solutions are c’ and d’ among the two solutions, the one for d’ is the value external to the range [a’,b’]

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