Lecture-8.ppt

# Lecture-8 - Lecture-8 Structure from Motion Problem • Given optical flow or point correspondences compute 3-D motion(translation and rotation and

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Unformatted text preview: Lecture-8 Structure from Motion Problem • Given optical flow or point correspondences, compute 3-D motion (translation and rotation) and shape (depth). 1 3-D Rigid Motion (displacement) È X ¢˘ È X ˘ È 1 - a b ˘ È X ˘ ÈTX ˘ ÍY ¢ ˙ = R ÍY ˙ + T = Í a 1 - g ˙ ÍY ˙ + ÍT ˙ Í˙ Í˙ Í ˙Í ˙ Í Y ˙ ÍZ ¢ ˙ ÍZ ˙ Í- b g 1 ˙ ÍZ ˙ ÍTZ ˙ Î˚ Î˚ Î ˚Î ˚ Î ˚ X ¢ = X - aY + bZ + TX Y ¢ = aX + Y - gZ + TY Z ¢ = - bX + gY + Z + TZ Orthographic Projection (displacement model) X ¢ = X - aY + bZ + TX Y ¢ = aX + Y - gZ + TY Z ¢ = - bX + gY + Z + TZ x¢ = x - ay + bZ + TX y¢ = ax + y - gZ + TY 2 Perspective Projection (displacement) X ¢ = X - aY + bZ + TX Y ¢ = aX + Y - gZ + TY Z ¢ = - bX + gY + Z + TZ X - aY + bZ + TX -bX + gY + Z + TZ aX + Y - gZ + TY y¢ = -bX + gY + Z + TZ x¢ = TX Z x¢ = T - bx + gy + 1 + Z Z T ax + y - g + Y Z y¢ = T - bx + gy + 1 + Z Z x - ay + b + † Instantaneous Velocity Model Optical Flow 3 3-D Rigid Motion È X ¢ - X ˘ È 0 - a b ˘ È X ˘ ÈTX ˘ ÍY ¢ - Y ˙ = Í a 0 - g ˙ ÍY ˙ + ÍTY ˙ Í ˙Í ˙Í ˙ Í ˙ Í Z ¢ - Z ˙ Í- b g 0 ˙ Í Z ˙ ÍTZ ˙ Î ˚Î ˚Î ˚ Î ˚ È X ¢˘ È 1 - a b ˘ È X ˘ ÈTX ˘ ÍY ¢ ˙ = Í a 1 - g ˙ ÍY ˙ + ÍT ˙ Í˙Í ˙Í ˙ Í Y ˙ ÍZ ¢ ˙ Í- b g 1 ˙ ÍZ ˙ ÍTZ ˙ Î˚Î ˚Î ˚ Î ˚ È X ¢˘ Ê È 0 ÍY ¢ ˙ = Á Í a Í ˙ ÁÍ Í Z ¢ ˙ Á Í- b Î ˚ ËÎ -a 0 g & ÈX ˘ È 0 Í& ˙ Í ÍY ˙ = Í W 3 Í Z ˙ Í- W & 2 Î˚Î b ˘ È1 0 0˘ ˆ È X ˘ ÈTX ˘ ˜ - g ˙ + Í0 1 0˙ ˜ ÍY ˙ + ÍTY ˙ ˙Í ˙Í ˙ Í ˙ 0 ˙ Í0 0 1˙ ˜ Í Z ˙ ÍTZ ˙ ˚Î ˚ ¯Î ˚ Î ˚ - W3 0 W1 W 2 ˘ È X ˘ ÈV1 ˘ - W1 ˙ ÍY ˙ + ÍV2 ˙ ˙Í ˙ Í ˙ 0 ˙ Í Z ˙ ÍV3 ˙ ˚Î ˚ Î ˚ & X = W 2 Z - W 3Y + V1 & Y = W X - W Z +V 3 1 2 & Z = W1Y - W 2 X + V3 3-D Rigid Motion & X = W 2 Z - W 3Y + V1 & Y = W X - W Z +V 3 1 2 & Z = W1Y - W 2 X + V3 & X = W¥ X + V ÈX ˘ X = ÍY ˙, Í˙ ÍZ ˙ Î˚ ÈW1 ˘ W = ÍW 2 ˙ Í˙ ÍW 3 ˙ Î˚ Cross Product 4 Orthographic Projection & X = W 2 Z - W 3Y + V1 & Y = W X - W Z +V 3 1 y =Y x=X 2 & Z = W1Y - W 2 X + V3 & u = x = W 2 Z - W 3 y + V1 & v = y = W 3 x - W1Z + V2 (u,v) is optical flow Perspective Projection (arbitrary flow) fX Z fY y= Z & u=x= & X = W 2 Z - W 3Y + V1 & Y = W X - W Z +V 3 1 & & fZX - fXZ & & X Z -x Z2 Z Z & & & & fZY - fYZ Y Z & v= y= = f -y Z2 Z Z x= 2 & Z = W1Y - W 2 X + V3 =f & & W Z - W 3Y + V1 W Y - W 2 X + V3 X Z -x = f 2 -x 1 Z Z Z Z & & W X - W1Z + V2 W Y - W 2 X + V3 Y Z & v= y= f -y = f 3 -y 1 Z Z Z Z & u=x= f V V1 W W + W 2 ) - 3 x - W 3 y - 1 xy + 2 x 2 Z Z f f V V W W v = f ( 2 - W1 ) + W 3 x - 3 y + 2 xy - 1 y 2 Z Z f f u= f( 5 Perspective Projection (optical flow) V V1 W W + W 2 ) - 3 x - W 3 y - 1 xy + 2 x 2 Z Z f f V V W W v = f ( 2 - W1 ) + W 3 x - 3 y + 2 xy - 1 y 2 Z Z f f u= f( fV1 - V3 x W W + fW 2 - W 3 y - 1 xy + 2 x 2 Z f f fV - V y W W v = 2 3 - fW1 + W 3 x + 2 xy - 1 y 2 Z f f u= Pure Translation (FOE) fV1 - V3 x Z fV - V y =23 Z u (T ) = v (T ) V3 Z V = ( y0 - y ) 3 Z u (T ) = ( x0 - x) v (T ) x0 = f V1 V , y0 = f 2 V3 V3 p0 = ( x 0 , y 0 ) † 6 Pure Translation (FOE) • p0 is the vanishing point of the direction of translation. • p0 is the intersection of the ray parallel to the translation vector with the image plane. Pure Translation (FOE) • If V3 is not zero, the flow field is radial, and all vectors point towards (or away from) a single point. • The length of flow vectors is inversely proportional to the depth, if V3 is not zero, then it is also proportional to the distance between p and p0. 7 Pure Translation (FOE) fV1 - V3 x Z fV - V y v (T ) = 2 3 Z fV u (T ) = 1 Z •If V =0, the flow field is parallel. fV2 (T ) v= Z u (T ) = 3 8 Structure From Motion ORTHOGRAPHIC PROJECTION Orthographic Projection (displacement) È X ¢˘ È X ˘ È 1 - a b ˘ È X ˘ ÈTX ˘ ÍY ¢ ˙ = R ÍY ˙ + T = Í a 1 g ˙ ÍY ˙ + ÍTY ˙ Í˙ Í˙ Í ˙Í ˙ Í ˙ Í˙ Í˚ Í- b g 1 ˙ Í Z ˙ ÍTZ ˙ ÎZ ¢ ˚ ÎZ ˙ Î ˚Î ˚ Î ˚ x ¢ = x - ay + bZ + TX y ¢ = ax + y - gZ + TY 9 Simple Method • Two Steps Method -Assume depth is known, compute motion x ¢ = x - ay + bZ + T X y ¢ = ax + y - gZ + TY Èa Í Íb È x¢ - x ˘ È - y Z 0 1 0 ˘ Í Í y¢ - y ˙ = Í x 0 - Z 0 1˙ Íg Î ˚Î ˚T Íx Í ÍTy Î ˘ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˚ Simple Method -Assume motion is known, refine depth x ¢ = x - ay + bZ + TX y ¢ = ax + y - gZ + TY È x¢ - x - ay - Tx ˘ Èb ˘ Í- g ˙[Z ]= Í y¢ - y - ax - T ˙ y˚ Î˚ Î 10 Tomasi and Kanade Orthographic Projection Assumptions • The camera model is orthographic. • The positions of “p” points in “f” frames (f>=3), which are not all coplanar, have been tracked. • The entire sequence has been acquired before starting (batch mode). • Camera calibration not needed, if we accept 3D points up to a scale factor. 11 Tomasi & Kanade Image point {u ( fp , v fp ) | f = 1, K , F , p = 1, K , P} È u11 K u1P ˘ Í ˙ M Í ˙ ÍuF1 K uFP ˙ W =Í ˙ v11 K v1P ˙ Í Í ˙ M Í ˙ Îv F1 K v FP ˚ ÈU ˘ W = Í- ˙ Í˙ ÍV ˙ Î˚ † Tomasi & Kanade 1P af = Âup P p =1 1P bf = Â vp P p =1 ~ u fP = u fP - a fP ~ v fP = v fP - b fP 12 s p = ( X p , YP , Z P ) u fP = i T ( sP - t f ) f v fP = j T ( sP - t f ) f 3D world point Orthographic projection kf = if ¥ jf † 13 ~ u fp = u fP - a f 1PT = i ( s p - t f ) - Â i f ( sq - t f ) P q =1 T f È 1P ˘ = i Í s P - Â sq ˙ P q =1 ˚ Î T f = iT sP f ~ = iT s u fP f P ~ v = jT s fP f P Origin of world is at the centroid of object points ~ ÈU ˘ ~ Í˙ W = Í- ˙ ~ ÍV ˙ Î˚ 14 ~ u fP = i T sP f ~ v = jT s fP f P ~ ÈU ˘ ~ Í˙ W = Í- ˙ ~ ÍV ˙ Î˚ È i1T ˘ Í˙ ÍM˙ T ~ Íif ˙ W = Í T ˙[s1 K sP ]= RS Í j1 ˙ 3XP ÍM˙ Í T˙ Rank of S is 3, because points in 3D space are not Í jf ˙ Î˚ 2FX3 Co-planar Rank Theorem Without noise, the registered ~ measurement matrix W is at most of rank three. È i1T ˘ Í˙ ÍM˙ T ~ Íi ˙ W = Í fT ˙[s1 K sP ]= RS Í j1 ˙ ÍM˙ 3XP Í T˙ Í jf ˙ Î˚ 2FX3 15 Translation ~ u fp = u fP - a f ~ u fp = u fP + a f ~ u fp = i T sP f u fp = i f s p + a f u fp = i T ( s p - t f ) f a f is projection of camera translation along x-axis Translation u fp = i f s p + a f v fp = j f s p + b f T W = RS + tep 2FX3 3XP 2FX1 1XP t = (a1 , K , a f , b1 , K , b f )T T e p = (1, K1) 16 Translation Projected camera translation can be computed: 1P T - if t f = af = Âup P p =1 1P - j t = bf = Â vp P p =1 T ff Noisy Measurements ~ • Without noise, the matrix W must be at most of rank 3. When noise corrupts the ~ images, however, W will not be rank 3. Rank theorem can be extended to the case of noisy measurements. 17 Approximate Rank SVD ~ W = O1SO2 2FXP PXP PXP Singular Value Decomposition (SVD) • For some linear systems Ax=b, Gaussian Elimination or LU decomposition does not work, because matrix A is singular, or very close to singular. SVD will not only diagnose for you, but it will solve it. 18 Singular Value Decomposition (SVD) Theorem: Any m by n matrix A, for which m ≥ n ,can be written as A = O1SO2 mxn mxn nxn is diagonal S O1 , O2 nxn are orthogonal T O1T O1 = O2 O2 = I Singular Value Decomposition (SVD) If A is square, then O1 , S, O2 are all square. O1-1 = O1T T O2 1 = O2 S-1 = diag( 1 ) wj A = O1SO2 † † A-1 = O2 diag( 1 )O1 wj † 19 Singular Value Decomposition (SVD) The condition number of a matrix is the ratio of the largest of the w j to the smallest of w j . A matrix is singular if the condition number is infinite, it is ill-conditioned if the condition number is too large. Singular Value Decomposition (SVD) Ax = b • If A is singular, some subspace of “x” maps to zero; the dimension of the null space is called “nullity”. • Subspace of “b” which can be reached by “A” is called range of “A”, the dimension of range is called “rank” of A. 20 Range and Null Space Ax=b x b b=0 Null space of A Dimension of Null space is Nullity Range of A Dimension of range is rank of A Singular Value Decomposition (SVD) • If A is non-singular its rank is “n”. • If A is singular its rank <n. • Rank+nullity=n 21 Singular Value Decomposition (SVD) A = O1SO2 • SVD constructs orthonormal basses of null space and range. of O1 spans range. • Columns • Columns of O2 null space. with non-zero with zero wj w j spans Solution of Linear System • How to solve Ax=b, when A is singular? • If “b” is in the range of “A” then system has many solutions. • Replace 1 wj by zero if x = O2 [diag ( wj = 0 1 )]O1T b wj 22 Solution of Linear System If b is not in the range of A, above eq still gives the solution, which is the best possible solution, it minimizes: r ≡| Ax - b | Approximate Rank ~ W = O1SO2 ¢ ¢ O1SO2 = O1¢S¢O2 + O1¢¢S¢¢O2¢ P-3 3 O1 = [ 1¢ O1¢¢] O 3 2F P-3 È S¢ 0 ˘ S=Í ˙ Î 0 S¢¢˚ 3 P-3 ¢ ÈO2 ˘ 3 O2 = Í ˙ ¢ P-3 ÎO2¢˚ P 23 Approximate Rank ˆ ¢ W = O1¢S¢O2 The best rank 3 approximation to the ideal registered measurement matrix. Rank Theorem for noisy measurement The best possible shape and rotation estimate is obtained by considering only ~ 3 greatest singular values of W together with the corresponding left, right eigenvectors. 24 Approximate Rank ˆ = O¢[S¢]12 R 1 Approximate Rotation matrix ˆ = [S¢]12 O¢ Approximate Shape matrix S 2 ˆ ˆˆ W = RS This decomposition is not unique ˆ ˆ ˆ W = RQ Q-1S ( )( ) Q is any 3X3 invertable matrix † Approximate Rank ˆ R = RQ How to determine Q? ˆ ˆ i fT QQT i fT = 1 ˆ S =Q S -1 R and S are linear transformation of approximate Rotation and shape matrices ˆT QQT ˆT = 1 jf jf ˆ i fT QQT ˆT = 0 jf Rows of rotation matrix are unit vectors and orthogonal 25 How to determine Q: Newton’s Method ˆ ˆ f1 (q) = iiT QQT iiT - 1 = 0 f (q) = ˆ T QQT ˆ T - 1 = 0 j j MDq = e ˆ f 3 (q) = i QQ ˆ1T = 0 j Dq = [Dq1,K, Dq9 ] 2 1 1 T 1 T M M ij = ˆ ˆ f 3 f - 2 (q) = i fT QQT i fT - 1 = 0 f 3 f -1 (q) = ˆ T QQT ˆ T - 1 = 0 jf jf ∂f i ∂q j e is error ˆ f 3 f (q) = i fT QQT ˆ T = 0 jf † † Algorithm • Compute SVD of • define • Compute 1 ˆ R = O1¢[S¢] 2 ~ W = O1SO2 1 ˆ ¢ S = [S¢] 2 O2 Q • Compute R ˆ = RQ ˆ S = Q -1S 26 Hotel Sequence Results (rotations) 27 Selected Features Reconstructed Shape 28 Comparison House Sequence 29 Reconstructed Walls ../tomasiTr92Figures.pdf 30 Web Page • http://vision.stanford.edu/cgibin/svl/publication/publication1992.cgi 31 ...
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## This note was uploaded on 06/12/2011 for the course CAP 6411 taught by Professor Shah during the Spring '09 term at University of Central Florida.

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