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MultipleImages - More than one camera(or image Image...

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1 Image Processing : 3. 3. Stereo & Stereo & Structure from Motion Structure from Motion Aleix M. Martinez [email protected] More than one camera (or image) In many applications, we can make use of more than one camera or of a sequence of images. These two problems are very similar (although not identical). In this section, we will develop the fundamental tools needed to recover the 3D position of a set of points (or objects). These algorithms can also be used to calibrate cameras from data. Epipolar geometry One of the most important concepts we will discuss is that of epipolar geometry. Given a 3D point P and its image on a first camera A , p , it is know that the P can only project onto a line in the image plane of a second camera B. I.e., p B is on l’ . Epipolar Geometry Epipolar Epipolar Plane Plane Epipoles Epipoles Epipolar Lines Baseline These three vectors define a plane Image A Image B Epipolar Epipolar Constraint Potential matches for p have to lie on the corresponding epipolar line l’ . Potential matches for p’ have to lie on the corresponding epipolar line l . Epipolar Constraint: Calibrated Case Essential Matrix (Longuet-Higgins, 1981)
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2 Properties of the Essential Matrix E p’ is the epipolar line associated with p’. E p is the epipolar line associated with p. E e’=0 and E e=0. E is singular. E has two equal non-zero singular values (Huang and Faugeras, 1989). T T Epipolar Epipolar Constraint: Uncalibrated Case Constraint: Uncalibrated Case Fundamental Matrix (Faugeras and Luong, 1992) Properties of the Fundamental Matrix F p’ is the epipolar line associated with p’. F p is the epipolar line associated with p. F e’=0 and F e=0. F is singular. T T | F | =1. Minimize: under the constraint 2 The 8-point Algorithm Linear LS The other classical method is to solve the overconstrained problem using linear LS; when n>8. Here, we assume 2 1 ' n i i T F Error i p p . 1 F F Epipolar line: ) , ( ) , ( b a b a n direction normal Unit T c b a l ) , , ( ) , ( b a c : line the to origin of Distance T v u p ) 1 , , ( l = F p is the epipolar line of p ’ in the first image. l =( a , b , c ) T . Recall that the equation of a line is au + bv + c =0. Here, p =( u,v ,1) T . Hence, p T l =0 We want to know how far p is from l . For this we need to draw a l in the first image.
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3 % Assume p, p_prime are two corresponding image points. % p=(u,v,1); p_prime=(u_prime,v_prime,1); epipolar_line = F * p_prime; a = epipolar_line(1); b = epipolar_line(2); c = epipolar_line(3); % Consider a region around p in two cases due to different slope % ‘length’ is half the line segment length you want to draw if(abs(a)<abs(b)) d = length/sqrt((a/b)^2+1); drawpoint = [u-d u+d ; ((-c - a*(u-d))/b (-c - a*(u+d))/b ]; else d = length/sqrt((b/a)^2+1); drawpoint = [(-c - b*(v-d))/a (-c - b*(v+d))/a ; v-d v +d ]; end plot(drawpoint(1,:), drawpoint(2,:)); Epipolar line: p u+d u-d (-c - a*(u-d) ) /b (-c - a*(u+d) ) /b To draw the line, it is convenient to choose an interval.
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