Robotics - Projective Geometry and Camera model. Matteo Pirotta

Transcription

1 Robotics - Projective Geometry and Camera model Matteo Pirotta pirotta@elet.polimi.it Dipartimento di Elettronica, Informazione e Bioingegneria Politecnico di Milano 14 March 2013 Inspired from Simone Ceriani s slides Como 2012)

2 2/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

3 3/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

4 4/84 Projective Transformations - Recall Definition A projectivity is an invertible mapping h( ) : P 2 P 2 such that x 1,x 2,x 3 lie on the same line h(x 1),h(x 2),h(x 3) do i.e., a projectivity maintains collinearity Fundamental theorem of projective geometry A mapping h( ) : P 2 P 2 is a projectivity a non-singular 3 3 matrix H such that p P 2, x h(p) = H p

5 5/84 Projective Transformations - Recall - Cont d Projective Transformation x = H x x y = w h 11 h 12 h 13 x h 21 h 22 h 23 y h 31 h 32 h 33 w Notes H p 3 Map plane to plane p 4 p 3 p 4 It s a linear transformation p 2 in homogeneous coordinates It s homogeneous too λh H p 1 p 2 H 1 p 1

6 6/84 Projective Transformations - Central projection x x / O y / / x π / y x π Central projection may be represented by linear mapping x = Hx Direct application of the fundamental theorem of projective geometry

7 7/84 Projective Transformations - Image Rectification - 1 Homography Estimation Take four point on first image x i = [ x i, y i, w i ] T Map on four known destination points x i = [ ] x i, y i T Rewrite: x i = h 11x i + h 12y i + h 13w i y i = h 21x i + h 22y i + h 23w i w i = h 31x i + h 32y i + h 33w i In cartesian: Fix h 33 = 1 and rewrite { x i = h 11x i +h 12 y i +h 13 w i h 31 x i +h 32 y i +h 33 w i y i = h 21x i +h 22 y i +h 23 w i h 31 x i +h 32 y i +h 33 w i { x i (h 31x i + h 32y i + w i ) = h 11x i + h 12y i + h 13w i y i (h 31x i + h 32y i + w i ) = h 21x i + h 22y i + h 23w i

8 8/84 Projective Transformations - Image Rectification - 2 { xi h 11 + y i h 12 + w i h 13 x i x i h 31 x i y i h 32 = x i w i Expand and separate x i h 21 + y i h 22 + w i h 23 y i x i h 31 y i y i h 32 = y i w i Matrix form (2-lines for each point) x 1 y 1 w x 1x 1 x 1y 1 h 11 x 1w x 1 y 1 w 1 x 1x 1 x 1y 1 h 12 y 1w 1 x 2 y 2 w x 2x 2 x 2y 2 h 13 x 2w x 2 y 2 w 2 x 2x 2 x 2y 2 h 21 x 3 y 3 w x 3x 3 x 3y 3 h 22 = y 2w 2 x 3w x 3 y 3 w 3 x 3x 3 x 3y 3 x 4 y 4 w x 4x 4 x 4y x 4 y 4 w 4 x 4x 4 x 4y 4 h 23 h 31 h 32 System Ax = b e.g. in Matlab solved with x = A\b y 3w 3 x 4w 4 y 4w 4

9 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Projective Transformations - Image Rectification - Example Original Image x: {179,525}, {187,73}, {690,307}, {698,467} x0 : {0,180}, {0,0}, {822,0}, {822,180} H = Image reference system x y 9/84

10 10/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

11 Hierarchy of transformations Projective transformations form a group, name projective linear group (PL(n)). PL(n): set of n n matrices with real elements related by a scalar multiplier. Consider the subgroup PL(3), it contains the: affine group last row is [0, 0, 1] Euclidean group subgroup of affine group upper left 2 2 matrix is orthogonal oriented Euclidean group subgroup of affine group upper left 2 2 has determinant 1 An alternative is to characterize the transformations in terms of elements that are preserved under transformation (invariants). e.g., Euclidean transformations (rotation and translation) leave distances unchanged. Similarity Affine Projective 11/84

12 12/84 Transformations - Recall Rototranslation Homography p 3 p 3 p 4 p 3 p 4 p 4 p 3 p 4 p 2 p 2 p 1 p 2 p = p 1 [ ] R t p 0 1 p 1 p 2 p 1 h 11 h 12 h 13 p = h 21 h 22 h 23 p h 31 h 32 h 33

13 13/84 Class I - Isometries - i.e., Rototranslations ξ cos(θ) sin(θ) t x p = ξ sin(θ) cos(θ) t y p H when ξ = 1 = [ ] R t p, 0 T 1 RR T = I iso: same, metric: measure ξ = +1 orientation preserving ξ = 1 orientation reversing 3 DoF (2 translation, 1 rotation) Special cases: Pure rotation Pure translation Invariants Length Area Angle H 1

14 14/84 Class II - Similarities s cos(θ) s sin(θ) t x p = s sin(θ) s cos(θ) t y p [ ] sr t = p 0 T 1 Isometry + scale factor 4 DoF (2 translation, 1 rotation, 1 scale) det(sr) = s H H 1 Invariants Shape Ratios of length Ratios of areas Angle Parallel lines

15 15/84 Class III - Affine transformations a 11 a 11 t x p = a 21 a 22 t y p = [ ] A t p 0 T 1 H Non-isotropic scaling 6 DoF (2 translation, 2 rotation, 2 scale) a11 a11 A = = UDV T a 21 a 22 UDV T = ( UV ) ( T VDV ) T U, V orthogonal, D diagonal H 1 Invariants Parallel lines Ratios of parallel segment lengths Ratios of areas R(θ) (R( φ)dr(φ)) rotation on scaled axis

16 16/84 Class IV - Projective transformations (Homographies) h 11 h 12 h 13 p = h 21 h 22 h 23 p h 31 h 32 h 33 [ ] A t = p, v = (v v T v 1, v 2) T H Mapping plane to plane linear in homogeneous coordinates 8 DoF 2 translation, 2 rotation, 2 scale, 2 for l H 1 Invariants Collinearities Cross-ratio of four points on a line

17 17/84 2D Transformations overview

18 18/84 Projective Transformations - Lines and Conics Points p = Hp Lines l = H T l Proof l T p = 0 l T p = 0 l T Hp = 0 (H T l) T Hp = 0 l T H 1 Hp = 0 Conics C = H T CH 1 Proof p T Cp = 0 p T C p = 0 (Hp) T H T CH 1 Hp = 0

19 19/84 Improper points and the l Projective (Homography) [ ] [ ] x A t x y = A y v T w 0 v 1x + v 2y Improper points mapped on finite 0 l = H T l 0 1 Affine [ ] [ ] x A t x y = A y 0 w 0 0 Improper points remain at infinity but they change! [ ] A l = H T 1 A 1 T t l = l 0 1 l = l = [ ] T Vanishing point: where world parallel lines converge in image

20 20/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

21 21/84 Cross Ratio Given 4 collinear points p i Distances d ij = (p i x p j x ) 2 +(p i y p j y ) 2 1D case CR(p 1, p 2, p 3, p 4) = p 1,p 3 p 2,p 4 p 2,p 3 p 1,p 4 p i, p j = det p i1 p i2 p j1 p j2 CR(p 1, p 2, p 3, p 4) = Property d 12 d 13 d 24 = d 12 d 34 d 13 d 24 d 34 Invariant under any projective transformation x = Hx CR(p 1, p 2, p 3, p 4) = CR(p 1, p 2, p 3, p 4) p 1 p 2 p 3 p 4 p 5 p 6 p 7 p 8

22 22/84 Parametric Lines Line l = p 1 p 2 Direction y p 2 + p 1 d + l d 12 = p 2 p 1 p i normalized d w = 0: improper point or direction d = d d Parametric Line p θ = p 1 + θd e.g., θ = d p 2 e.g., θ = 0 p Parametric Distance Consider p θ1, p θ2 d 12 = p θ2 p θ1 = p 2 + θ 2d p 1 θ 1d = (θ 2 θ 1) 2 d 2 x + (θ 2 θ 1) 2 d 2 y ( ) = (θ 2 θ 1) 2 d 2 x + d 2 y x = θ 2 θ 1

23 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Cross Ratio Example - 1 Image source Questions Identify the vanishing points Calculate the l0 Identify the vertical middle line Identify the field bottom line Calculate relative player position Identify vanishing point of the diagonal 23/84

24 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Cross Ratio Example - 2 Vanishing points - step 1 Identify 4 points on a rectangle in the world plane Calculate l1 = p1 p2 l2 = p3 p4 l3 = p1 p3 l4 = p2 p4 24/84

25 25/84 Cross Ratio Example - 3 Vanishing points - step 2 Given l 1, l 2, l 3, l 4 Calculate p h = l 1 l 2 p v = l 3 l 4 l = p h p v

26 26/84 Cross Ratio Example - 4 Vertical middle line - wrong way Middle point of lines p m1 = 1 (p1 + p2) 2 p m2 = 1 (p3 + p4) 2 l m = p m1 p m2 Wrong l m has to pass for p v is not the middle line Homography doesn t preserve ratios, length,...

27 27/84 Cross Ratio Example - 5 Vertical middle line - the right way! In the image CR(p 1, p m1, p 2, p h ) using parametric line = CR(0, θ m, θ 2, θ h ) = θm(θ h θ 2 ) θ 2 (θ h θ m) Equation CR(p 1, p m1, p 2, p h ) = CR(0, a, 2a, ) θ m(θ h θ 2) θ 2(θ h θ m) = 1/2 In the world CR(0, a, 2a, ) = a 2a = 1 2 a is the (unknow) half-length Solution θ m = θ 2θ h 2θ h θ 2 p m1 = p 1 + θ m d 12 do the same for p m2

28 Cross Ratio Example - 6 Identify field bottom line In the image Get the p 50 point (field middle) CR(p v, p 3, p 50, p end ) = CR(0, θ 3, θ m, θ end ) = θ 3(θ end θ m) θ m(θ end θ 3 ) Equation CR(p v, p 3, p 50, p end ) = CR(, 0, a, 2a) θ 3(θ end θ m) θ m(θ end θ 3) = 1/2 In the world CR(, 0, a, 2a) = a 2a = 1 2 a is the (unknow) half-length Solution θ end = θmθ 3 2θ 3 θ m p end = p v + θ end d v3 l end = p end p h 28/84

29 29/84 Cross Ratio Example - 7 Calculate relative player P position Origin in p 3 x towards p 4 y towards p 1 Calculate P x = (P p v ) (p 3 p 4) P y = (P p h ) (p 1 p 3) Cross Ratio CR(p 3, P x, p mid2, p 4) = CR(0, x, 1 2, 1) θ x (θ 4 θ mid2 ) θ mid2 (θ 4 θ x ) = x 1 x CR(p 3, P y, p 50, p end ) = CR(0, y, 1 2, 1) θ y (θ end θ 50 ) θ 50 (θ end θ y ) = y 1 y

30 30/84 Cross Ratio Example - 8 Calculate vanishing of the diagonal Calculate l d = p end p 4 p d = l d l

31 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Last step - Affine reconstruction Affine transformation T l = 0, 0, 1 invariant but not point-wise! T Consider l0 = lx0, ly0, lz0 image of l Consider H = lx0 ly0 lz0 Source image Could be verified that l = H T l0 i.e., paff = H pimg, H map points of the image to a affine transformation of the world Affine reconstruction 31/84

32 32/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

33 33/84 Projective Geometry - 3D Points X Points p e = Y R 3 Z in Cartesian coordinates x p h = y z R4 w in homogeneous coordinates X = x/w Y = y/w Z = z/w w 0 i.e., there is an arbitrary scale factor Planes a Planes π = b c R4 d [ ] T a, b, c n = ] T [ a, b, c unitary normal to the plane p h π p T h π = π T p h = 0 π = [ 0, 0, 0, 1 ] T : plane at infinity contains all improper points

34 34/84 Quadrics Definition Quadratic polynomial equation Quadric surface Matrix form equation x T Qx = 0 Q is 4 4 symmetric Q is homogeneous too, i.e., 10 parameters, 9 D.O.F.

35 35/84 Quadrics - Summmary

36 36/84 Quadrics & conics Intersection Q π conic Conics are planar sections of quadrics

37 37/84 Hierarchy of transformations Projective 15 dof [ A ] t V T v Intersection and tangency Affine 12 dof [ A ] t 0 T 1 Parallellism of planes, Volume ratios, centroids, The plane at infinity π Similarity 7 dof [ sr ] t 0 T 1 The absolute conic Ω Euclidean 6 dof [ R ] t 0 T 1 Volume

38 38/84 Vanishing points Vanishing points π contains all the directions All the lines with the same direction intersect on π at the same point The vanishing point is the image of this intersection Vanishing lines Parallel planes intersect π in a common line The vanishing line is the image of this intersection e.g., the horizon line is the image of the intersection of the set of horizontal planes {π H } with π

39 Art & Perspective - 1 Fresco in Pompeii - I B.C. Partially correct perspective The skill was lost during the middle ages, it did not reappear in paintings until the Renaissance 39/84

40 40/84 Art & Perspective - 2 The school of Athens - Raffaello Sanzio Correct perspective

41 41/84 Vanishing points example - 1 Question Find the three vanishing point in the image Compute the horizon line in the image Compute others vanishing lines

42 42/84 Vanishing points example - 2

43 43/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

44 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Reconstruction example - 1 Flagellazione di Cristo - Piero della Francesca /84

45 47/84 Reconstruction example - 2 Trinity - Masaccio

46 49/84 Reconstruction example - 3 A simple photo

47 Nice stuff with Projective geometry - 1 Felice Varini - Details can be found here /84

48 Nice stuff with Projective geometry - 2 Felice Varini /84

49 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Nice stuff with Projective geometry - 3 Felice Varini /84

50 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Nice stuff with Projective geometry - 4 Felice Varini /84

51 Nice stuff with Projective geometry - 5 Felice Varini /84

52 55/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

53 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Exercise 1 - Tiles Image source Questions Identify the vanishing points using cross ratio i.e., without use parallel lines 56/84

54 57/84 Exercise 1 - Tiles - Solution Horizontal CR(p o1, p o23, p o2, p o ) = CR(0, a2/3, a, ) CR(0, θ 23, θ 3, θ o) = 2/3 θ o = θ 23θ 3 2θ 3 3θ 2 3 p o = p o1 + θ od o Vertical CR(p v1, p v12, p v2, p v ) = CR(0, a, 2a, ) CR(0, θ 12, θ 12, θ v ) = 1/2 θ v = θ 12(θ v θ 2 ) θ 2 (θ v θ 12 ) p v = p v1 + θ v d v

55 58/84 Exercise 1 - Tiles - Check Magenta lines only for check correctness

56 59/84 Exercise 2 - Soccer field Image source Find Center of the goal-line Vanishing point of the goal-line Solution 4 symmetric points a, a + and b, b + { CR(0, a, a, ) = CR(θc, θ a, θ a+, θ v ) CR(0, b, b, ) = CR(θ c, θ b, θ b+, θ v ) 2 equations, 2 unknown 4 solutions, only 2 are are valid

57 60/84 Exercise 2 - Soccer field Magenta lines only for check correctness

58 61/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

59 62/84 What is an image Image Two-dimensional brightness array: I 3 two-dimensional array: I R, I G, I B RGB: Red, Green, Blue others: YUV, HSV, HSL, Ideal: I : Ω R 2 R + Discrete: I : Ω N 2 R + e.g., Ω = [0, 639] [0, 479] N 2 e.g., Ω = [1, 1024] [1, 768] N 2 e.g., R + = [0, 255] N e.g., R + = [0, 1] R I(x, y) is the intensity I result of 3D 2D projection: flat we lose: angles, distances (lengths)

60 63/84 Camera Optical system Set of lenses to direct light change in the direction of propagation CCD sensor integrate energy both in time (exposure time) in space (pixel size)

61 64/84 Thin lenses model Thin lenses Mathematical model Optical axis (z) Focal plane π f ( z) Optical center o π f f y Parameters o z Property f distance o, π f Parallel rays converge π f Rays through o undeflected

62 65/84 Rays from scene Image from a scene point P P = (Z, Y ) Ray through o undeflected y Ray parallel to z cross in ( f, 0) f Z Similarities P Blue triangles: h Y = r f f Green triangles: h Y = r Z Fresnel s Law 1 Z + 1 r = 1 f Note: Z r f h p r o Y z

63 66/84 The image plane Image plane π I Plane z at distance d Blur Circle If d r π I π f f y a Z P image of P is a circle C o z Diameter of C: a(d r) φ(c) = r a is the aperture C p d r Focused image φ(c) < pixel size Depth of field : range [Z 1, Z 2] : φ(c) < pixel size set of values that satisfies Fresnel s law

64 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Depth of field - Example 1 67/84

65 Projective Hierarchy Cross Ratio Geometry 3D Nice stuff Extras Camera Geometry Pin Hole Model Depth of field - Example 2 The same scene - different aperture 68/84

66 69/84 Aperture controls depth of field Why not make the aperture as small as possible? Less light gets through; Diffraction effects...

67 70/84 Aperture controls depth of field Changing the aperture size affects depth of field A smaller aperture increases the range in which the object is approximately in focus; But small aperture reduces amount of light need to increase exposure.

68 71/84 Outline 1 Projective 2 Hierarchy 3 Cross Ratio 4 Geometry 3D 5 Nice stuff 6 Extras 7 Camera Geometry 8 Pin Hole Model

69 72/84 Pinhole model - Definition Hypothesis Z a Z f r f Image of P y O (I) p x z Pin-Hole Model -f image plane O X camera centre Y P Z principal axis l Po : line that join P and o Frontal Pin-Hole Model X p = π I l Po Notes p is the image of P i l Po l Po : interpretation line of p O f camera centre Y x p O (I) y image plane P Z principal axis

70 73/84 Pinhole model - Geometry camera centre: centre of projection or optical centre (O); principal axis: line from the camera centre perpendicular to the image plane (z); principal point: intersection between image plane an principal axis (O (I ) ); principal plane: plane through the camera centre, parallel to the image plane. X O f camera centre Y p x O (I) y image plane P Z principal axis

71 74/84 Pinhole model - Geometry (Cont d) Given [ ] T P (O) = X, Y, Z, 1 [ P (O) = X, Y, Z, 1 [ ] T p (I) = x, y, 1 ] T Z = f, X = f X Z, Y = f Y Z x = X, y = Y Projection y = f Y Z y (O) Z x = f X Z look at the triangles P (O) Y Note y I O z (O) λp (O) projects on p (I) [ sx, sy, sz, 1] T projects on p (I), p (I) y (I) f s 0

72 75/84 Pinhole model - Frontal Camera Model The pinhole camera model is inconvenience that the focal length f is negative. The coordinate system: We will use the pinhole model as an approximation; Put the camera centre (O) at the origin; Put the image plane in front of the camera centre; The camera looks up the positive z axis. Projection equations Compute intersection with IP of ray from P to O; Derived using similar triangles x = f X Z, y = f Y Z Ignoring the final image coordinate: (X, Y, Z) T (f X Z, f Y Z )T

73 76/84 Pinhole model - Matrix Projection equations y = f Y Z x = f X Z In matrix form x X f y = 0 f Y w Z W = X f f 0 0 Y Z W Define f 0 0 K = 0 f 0 : intrinsic parameters p (I) = π P (O) π = [ K 0 ] : projection matrix

74 77/84 Distortions from Physical Lenses Pinhole camera model assumes: Perfect pinhole camera lenses; Image x = I (p) R 2 be measured in infinite accuracy; Principal point is at the center of the image. Physical camera lenses give us: Distorted imaging projections; Finite resolutions dened by the sensing devices in digital cameras; Offset between image center and optical center. p I y x I p x O y z x y

75 78/84 Pinhole model - Image coordinates - 1 Reference system on image Metric I: origin centered on z (O) π I I : origin centered top-left image [ ] T c (I ) = c x, c y : position of I in I I metric I in pixel c (I ) in pixel Definition [ ] T (I) [ ] T (I ) 0, 0 c x, c y : principal point Image of the optical center (o) or z (O) π I I y x y I x

76 79/84 Pinhole model - Image coordinates - 2 The origin of coordinates in image planes is not [ ] T (I ) at the principal point c x, c y : I x (X, Y, Z) T (f X Z + cx, f Y Z + cy )T y c (I ) x X f 0 c x 0 y Y = 0 f c y 0 w Z W p (I ) = π P (O) I y x

77 80/84 Pinhole model - Image coordinates - 3 (CCD Camera) The pinhole camera model assumes that image coordinate Euclidean (equal scales in both axial direction); In CCD there is the possibility of non-square pixels; Image coordinates measured in pixel introduces unequal scale factors in each direction. If the number of pixels per unit distance coordinates are m x and m y in the x and y direction: m x 0 0 f 0 c x 0 π = 0 m y 0 0 f c y where (c x, c y ) is the principal point in terms of pixel dimensions.

78 81/84 Pinhole model - Image coordinates - 4 Meters to pixels Consider I : origin on I, in pixel Scale meters to pixels p (I ) x p (I ) y = s x p (I) x = s y p (I) y s x = 1 d x, d x: width of a pixel [m] s y = 1 d y, d y : height of a pixel [m] s x = s y : square pixel s x 0 0 p (I ) = 0 s y 0 p (I) Translation 1 0 c x p (I ) = 0 1 c y p (I ) I y x c (I ) I I y y x x

79 82/84 Pinhole model - Intrinsic camera matrix Consider f p (I) = 0 f 0 0 P (O) s x 0 0 p (I ) = 0 s y 0 p (I) c x p (I ) = 0 1 c y p (I ) In one step s xf 0 c x 0 p (I ) = 0 s y f c y 0 P (O) p (I ) = πp (O) = K [ I 0 ] P (O) The intrinsic camera matrix or calibration matrix f x s c x K = 0 f y c y f x, f y : focal lenght (in pixels) f y /f x = s y /s x = a: aspect ratio s: skew factor pixel not orthogonal usually 0 in modern cameras c x, c y : principal point (in pixel) usually half image size due to misalignment of CCD

80 83/84 Camera rotation and translation Consider p (I ) = πp (O) point is expressed in terms of world coordinate frame: P (W ). W and O are related via a rotation and a translation. ) P (O) = T (O) OW P(W ) = R (O) OW (P (W ) O (W ) extrinsic camera matrix One step Note p (I ) = π = [ ] [ ] R RC K 0 [ KR 0 1 ] KRC complete projection matrix R is R (O) OW, C is O(O) OW P (W ) (camera centre) i.e., the position and orientation of W in O X O Y W Z p (W ) R, t T (O) OW X W Z p (O) O Y

81 84/84 Camera rotation and translation - 2 General pinhole camera [ ] p (I ) = KR KRC [ ] π = KR I C P (W ) C is the camera centre in world reference 9 (11) dof: 3 (5) for K, 3 for R and 3 for C parameters in K are called internal parameters in R and C are called external Other formulation do not make the camera center explicit [ ] [ ] p (I ) R t = K 0 P (W ) 0 1 [ ] π = KR Kt t = RC The intrinsic parameters f x s c x K = 0 f y c y The extrinsic parameters R = R (φ, θ, ρ) C = [c 1, c 2, c 3] T