《图像处理与计算机视觉 Image Processing and Computer Vision》课程教学资源（PPT课件讲稿）Chapter 11 Bundle adjustment Structure reconstruction SFM from N-frames

SFM1: find pose phase Pose estimation discussed in the last chapter Pose estimation vo.a 6

SFM1 : find pose phase Pose estimation discussed in the last chapter Pose estimation V0.a 6

SFM2: Model finding by the iterative method Similar to pose estimation In pose estimation model is known, pose is unknown Here model finding by the iterative method) Assume pose is known model is unknown The algorithms are similar Pose estimation vo.a

SFM2: Model finding by the iterative method • Similar to pose estimation. – In pose estimation: model is known, pose is unknown. – Here (Model finding by the iterative method) Assume pose is known, model is unknown. – The algorithms are similar. Pose estimation V0.a 7

From the pose estimation slides (chapter iv 10-pose estimation) Problem f= focal length. Assume for pose{RT}θ=[1a2@TT写 Formulation T, T,, T, are translations along,Y, Z axes repectively (@i, a )are rotation angles around X, Y, Z axes respectively, M,=x, y zI is the i model point(i =1, 2, N), *=u, v] the image point of P, rewrite(la),1(b) F1X1+n121+1321+T1 与X,+12+h32+F,(,M) -(2a) F21X1+2+r2321+72 v;h+r2y+n32+r8,(O,M) -(2b In this chapter Problem definition: There are t=1, 2, Iimage frames, and the object i=1, 2,.N features Asume the pose 0,=1.2.r=/91,92,B3, T,T2,T]i and image measurements xi. v,I are known SFM2 task: From the given image frames and pose 8, 12r Find Ml2N=x, Y ZP Pose estimation vo.a

Problem Formulation • Pose estimation V0.a 8     ( )         i N T i N i i i t , ,..,Γ T i,t i i t T t , ,..,Γ t v i i i i i i i i i u i i i i i i i i i i T i i i T th i i i i T M X Y Z θ x u v θ [ , , ,T ,T ,T ] g M b r X r Y r Z T r X r Y r Z T v f g M a r X r Y r Z T r X r Y r Z T u f P M X Y Z i x u v T T T f R,T θ T T T 1,2,.. 1,2,.. 1 2 1 2 1 2 3 1 2 3 , 31 32 33 3 21 22 23 2 , 31 32 33 3 11 12 13 1 1 2 3 1 2 3 1 2 3 1 2 3 Find SFM2 task :From the given image frames and pose and image measurements are known. Asume the pose , the object i 1,2,..., N features Problem definition :There are t 1,2,, image frames, and ---------------In this chapter -------------------------- ( , ) (2 ) ( , ) (2 ) the image point of , rewrite (1a),1(b) is the model point (i 1,2,, N), is , , are rotation angles around X,Y,Z axesrespectively, , , are translations along X,Y,Z axesrepectively, focal length.Assume for pose From the pose estimation slides(chapter iv10 - pose estimation): = = = = = = = = =  = − − − − − − − − + + + + + + = = − − − − − − − − + + + + + + = = = = = =           

Derivation for the model partial derivatives(note f= focal length en=g,(9,,,,T,T,x,,2)= Given u=81(,,,7,7,T,Xx,F,2)= D2z +T x+1+2+=2/-x++乙++n=若=厂 X1-3Y1+中21+7 r31 r321+73 n2X1+1+21+73 112 1 1≈1,2≈-g3,1≈2 R=r 1 31≈-n2,32≈内, Find I Find aX F1X1+F21+132+7 F1X1+F2x1+13z1+T r31X1+ry21+32z1+73 r,X+PY+、Z+T Y aX aX aX (xA)1 X ar aX aX o(x2) a(rr )f az ay ar Z aZ. aX OX Z X+r (z X (z) ZR ation vo.a

Derivation for the model partial derivatives (Note: f= focal length) • Pose estimation V0.a 9 ( ) ( ) ( ) ( ) ( )         = − +         = − +         = −   +           =   +         =         =        + + + + + +  =               − − −            = − + + + − + + =  + + + + + + = = = 2 1 1 3 1 1 1 2 3 1 2 3 1 1 1 3 1 3 2 3 3 3 1 1 1 2 1 3 1 2 1 3 1 3 2 3 1 3 2 3 3 2 1 2 2 2 3 1 1 1 2 1 3 2 1 3 3 2 1 3 1 3 2 3 3 3 1 1 1 2 1 3 1 1 2 3 1 2 3 1 1 Find . 1 1 1 R Given ( , , , , , , , , ) R i i R i R i R i R i R i R i R i R i R R i i R i R i R i R i R R i i R i R i R i i i i i i i i i i i i i i i R i R i i i i i i i i u i i i Z r X Z r f Z r f Z r X f Z f r X r Z f Z f X X X X Z Z Z f X Z f X X X Z f X Z X f X r X r Y r Z T r X r Y r Z T f X u X u r r r r r r r r r X Y Z T X Y Z T f Z X f r X r Y r Z T r X r Y r Z T f u g T T T X Y Z              ( ) ( ) ( ) ( ) ( )         = − +         = − +         = −   +           =   +         =         =        + + + + + +  =      −     −  − + + + − + + =  + + + + + + = = = 2 1 2 3 2 1 2 2 3 2 2 3 2 1 2 3 1 3 2 3 3 3 1 1 1 2 1 3 1 3 1 2 3 2 1 3 3 1 1 1 2 3 1 3 2 2 1 3 3 2 1 3 1 3 2 3 3 3 1 1 1 2 1 3 1 1 2 3 1 2 3 1 1 Find . , , 1 so 1, , Given ( , , , , , , , , ) R i i R i R i R i R i R i R i R i R i R R i i R i R i R i R i R R i i R i R i R i i i i i i i i i i i i i i i R i R i i i i i i i i u i i i Z r X Z r f Z r f Z r X f Z f r X r Z f Z f Y X X Y Z Z Z f Y Z f X X Y Z f Y Z X f Y r X r Y r Z T r X r Y r Z T f Y u Y u r r r r r r X Y Z T X Y Z T f Z X f r X r Y r Z T r X r Y r Z T f u g T T T X Y Z           

CMSC5711: Exercise 11. 1: (Note: f= focal length) Given u;=gu(@1, P2,23,Ti, T2, T3,Xi,Y, Zi) F1X1+121+F1321+7 R f≈f B3Y+o2zi+ rX.tr.t +t 33 2X1+ny1+Z1+ Son1≈12≈-的,73 r31≈-n22≈,3 ≈1.Or 2 m的 R 33 21 Find Xtr.trz+t X R Xtr.trz R 33 +T, aZ aZ aZ Pose estimation vo.a

CMSC5711: Exercise11.1: (Note: f= focal length) • Pose estimation V0.a 10 ? Find 1 1 1 R , , 1, so 1, , Given ( , , , , , , , , ) 3 1 3 2 3 3 3 1 1 1 2 1 3 1 2 1 3 1 3 2 3 1 3 2 3 3 2 1 2 2 2 3 1 1 1 2 1 3 3 1 2 3 2 1 3 3 1 1 1 2 3 1 3 2 2 1 3 3 2 1 3 1 3 2 3 3 3 1 1 1 2 1 3 1 1 2 3 1 2 3 =           =          + + + + + +  =             − − −            =  −     −  − + + + − + + =  + + + + + + = = = Z Z X f Z r X r Y r Z T r X r Y r Z T f Z u r r r r r r r r r r r r or r r r X Y Z T X Y Z T f Z X f r X r Y r Z T r X r Y r Z T f u g T T T X Y Z R i R i i i i i i i i i i i i i i R i R i i i i i i i i u i i i                 