# Image Formation and Camera Calibration

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1 EECS 432-Advanced Computer Vision Notes Series 2 Image Formation and Camera Calibration Ying Wu Electrical Engineering & Computer Science Northwestern University Evanston, IL Contents inhole Camera Model 2 erspective rojection 2 2 Weak-erspective rojection 2 3 Orthographic rojection 2 2 Homogeneous Coordinates 3 3 Coordinate System Changes and Rigid Transformations 3 3 Translation 3 32 Rotation 4 33 Rigid Transformation 4 34 Summary 4 4 Image Formation (Geometrical) 5 5 Camera Calibration Inferring Camera arameters 6 5 The Setting of the roblem 6 52 Computing the rojection Matrix 7 53 Computing Intrinsic and Extrinsic arameters 7 54 Questions to Think Over 8

2 inhole Camera Model erspective rojection O Π f k O j pinhole =(x y ) p =(x y ) i Figure : Illustration of perspective projection of a pinhole camera From Figure, we can easily figure out: { x = f x/ y = f y/ () Note: Thin lenses cameras has the same geometry as pinhole cameras Thin lenses camera geometry: = f 2 Weak-erspective rojection { x = (f / 0 )x y = (f / 0 )y When the scene depth is small comparing to the average distance from the camera, ie, the depth of the scene is flat, weak-perspective projection is a good approximation of perspective projection It is also called scaled-orthographic projection 3 Orthographic rojection (2) { x = x y = y (3) 2

3 When camera always remains at a roughly constant distance from the scene, and scene centers the optic axis, orthographic projection could be used as an approximation Obviously, orthographic projection is not real 2 Homogeneous Coordinates A 3D point is = and a plane can be written by ax + by + c d = 0 We use homogeneous coordinates to unify the representation for points and lines by adding one dimensionality, ie, we use x = y for points and Π = x y a b c d for planes, where the plane Π is defined up to a scale We have Π = 0 3 Coordinate System Changes and Rigid Transformations For convenience, We use the Craig notation frame F F means the coordinates of point in 3 Translation B = A + B O A (4) where B O A is the coordinate of the origin O A of frame A in the new coordinate system B 3

4 32 Rotation B AR = ( B i A B j A B k A ) = A i T B A j T B A k T B (5) where B i A is the coordinate of the axis i A of frame A in the new coordinate system B Then we have, B = B A R A 33 Rigid Transformation B = B A R A + B O A (6) Let s see here an advantage of the homogeneous coordinates If we make two consecutive rigid transformation, ie, from A B C, the coordinate of point in frame C will be written by: C = C B R( B AR A + B O A ) + C O B = C B R B AR A + ( C BR B O A + C O B ) It looks very awkward If we present it by homogeneous coordinates, it looks very concise [ B [ BA [ R = B O A A 0 T and So, [ C = [ C = [ CB R C O B 0 T [ CB R C O B 0 T [ B [ BA R B O A 0 T [ A 34 Summary We could write a transformation by: m m 2 m 3 m 4 T = m 2 m 22 m 23 m 24 m 3 m 32 m 33 m 34 m 4 m 42 m 43 m 44 We call it a projective transformation If we can write: [ A t T = 0 T 4

5 It becomes an affine transformation If A = R, ie, a rotation matrix (R T R = I), [ R t T = 0 T it becomes a Euclidean transformation or a rigid transformation Obviously, Euclidean transformation preserves both parallel lines and angles, but affine preserves parallel lines but not angles 4 Image Formation (Geometrical) In this section, we discuss the process of image formation in terms of geometry For a 3D point p w = [x w, y w, w T in the world coordinate system, its will be mapped to a camera coordinate system (C) from the world coordinate system (F), then map to the physical retina, ie, the physical image plane, and get image coordinates [u, v T We shall ask how this 3D point is mapped to its image coordinate y world coordin v v p p camera coordin x u image coordin u normalied image plane physical retina Figure 2: Illustration of the geometry of the image formation process under perspective projection of a pinhole camera For convenience, we introduce a normalied image plane located at the focal length f = In such a normalied image plane, the pinhole (c) is mapped to the origin of the image plane 5

6 (ĉ), and p is mapped to ˆp = [û, ˆv T û ˆp = ˆv = [ I c [ p c 0 = [ I 0 c x c y c c And we also have u v = kf 0 u 0 x c 0 lf v c 0 y c = kf 0 u 0 0 lf v 0 0 c c 0 [ I Let α = kf and β = lf We call these parameters α,β,u 0 and v 0 intrinsic parameters, which present the inner camera imaging parameters We can write x u v = α 0 u 0 0 c 0 β v c 0 0 y c c = α 0 u 0 0 [ x w 0 β v c 0 0 R t y w T w We call R and t extrinsic parameters, which represent the coordinate transformation between the camera coordinate system and the world coordinate system So, we can write, u v = M M c 2 p w = c Mpw (7) We call M the projection matrix x c y c c 5 Camera Calibration Inferring Camera arameters 5 The Setting of the roblem We are given () a calibration rig, ie, a reference object, to provide the world coordinate system, and (2) an image of the reference object The problem is to solve (a) the projection matrix, and (b) the intrinsic and extrinsic parameters Mathematically, given [x w i, yi w, i w T, i =,, n, and [u i, v i t, i =,, n, we want to solve M and M 2, st, x u w i v i = i x w M i c M 2 yi w i w = i M y w i i c i w, i 6

7 52 Computing the rojection Matrix We can write c i u i v i x m m 2 m 3 m w i 4 = m 2 m 22 m 23 m 24 y w i i m 3 m 32 m 33 m w 34 i c u i = m x w i + m 2 yi w + m 3 i w + m 4 i c v i = m 2 x w i + m 22 yi w + m 23 i w + m 24 i c = m 3 x w i + m 32 yi w + m 33 i w + m 34 Then x w i m + y w i m 2 + w i m 3 + m 4 u i x w i m 3 u i y w i m 32 u i w i m 33 = u i m 34 x w i m 2 + y w i m 22 + w i m 23 + m 24 v i x w i m 3 v i y w i m 32 v i w i m 33 = v i m 34 Then, x w y w w u x w u y w u w x w y w w v x w v y w v w x w n yn w n w u n x w n u n yn w u n n w x w n yn w n w v n x w n v n yn w v n n w m m 2 m 3 m 4 m 32 m 33 u m 34 v m 34 u 2 m 34 = v 2 m 34 u n m 34 v n m 34 (8) Obviously, we can let m 34 =, ie, the projection matrix is scaled by m 34 We have: Km = U (9) Where, K is a 2n matrix, m is a -D vector, and U is a 2n-D vector The least squares solution of Equation 9 is obtainted by: m = K U = (K T K) K T U (0) where K is the pseudoinverse of K Here, m and m 34 = constitute the projection matrix M This is a linear solution to the project matrix 53 Computing Intrinsic and Extrinsic arameters After computing the projection matrix M, we can compute the intrinsic and extrinsic parameters from M lease notice that the projection matrix M obtained from Equation 9 is a scaled version of the true M, since we have let m 34 = To decompose the projection 7

8 matrix in order to solve M and M 2, we need to take into account of m 34, ie, the true m 34 needs to be figured out Apparently, we have: m T r m 4 α 0 u 0 0 T t x m 34 M = m 34 m T 2 m 24 = 0 β v 0 0 r T 2 t y αr T m T r T 3 t = + u 0 r T 3 αt x + u 0 t x βr T 2 + v 0 r T 3 βt y + v 0 t x () 0 T r T 3 t x where m T i = [m i, m i2, m i3, M = {m ij } is computed from Equation 9, and r T i = [r i, r i2, r i3, R = {r ij } is the rotation matrix To see it clearly, we have: αr T + u 0 r T 3 βr T 2 + v 0 r T 3 r T 3 αt x + u 0 t x m 34 m T m 34 m 4 βt y + v 0 t x = m 34 m T 2 m 34 m 24 (2) m 34 m T 3 m 34 t x Obviously, by comparing these two matrices, it is easy to see m 34 m 3 = r 3 In addition, since R is a rotation matrix, we have r i = Then, we have m 34 = m 3 Then, it is easy to figure out all the other parameters: After that, it is also easy to get: r 3 = m 34 m 3 (3) u 0 = (αr T + u 0 r T 3 )r 3 = m 2 34m T m 3 (4) v 0 = (βr T 2 + v 0 r T 3 )r 3 = m 2 34m T 2 m 3 (5) α = m 2 34 m m 3 (6) β = m 2 34 m 2 m 3 (7) r = m 34 α (m u 0 m 3 ) (8) r 2 = m 34 β (m 2 v 0 m 3 ) (9) t = m 34 (20) t x = m 34 α (m 4 u 0 ) (2) t y = m 34 β (m 24 v 0 ) (22) Now, we have obtained all the intrinsic and extrinsic parameters For the analysis of skewed camera models (ie, θ 90 o ), please read chapter 6 of Forsyth&once 54 Questions to Think Over We v introduced a linear approach for camera calibration Let s think some questions over: 8

9 The matrix M, representing the intrinsic characteristics of cameras, has 4 independent variables; and the matrix M 2, representing the extrinsic characteristics, has 6 independent variables So the projection matrix M has 0 independent variables When letting m 34 =, we still have parameters to determine However, please note, these parameters are not independent! But our solution by Equation 9 did not address such dependency or constraints among these parameters Consequently, computation errors are not avoidable The question is that, if we can find an approach to take into account of that to enhance the accuracy? The method described above assumes that we know the 2D image coordinates How can we get these 2D image points? If the detection of these 2D image points contains noise, how does the noise affect the accuracy of the result? If we are not using point-corespondences, can we use lines? 9

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