# Lecture-19 - X0 =(X07 Y07 Z0)T 3—D coordinates of a point...

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Unformatted text preview: X0 = (X07 Y07 Z0)T: 3—D coordinates of a point at time to T : (T1, T2, T3)T: Translation vector spherical coordinates: slant, QT, and tilt7 ngT: T = (sinBTcosgzﬁT7 sinBTsingzﬁT7 cos6T)T. q = ((10, C117 C127 013V: Quaternion q : (sin %n7 cos %)T Rotation Matrix: qg — q? — q; + 1% 2(Qoqi + q2t13) 2(qoq2 — Inga) R 2 2(Q0611 — 61293) —q3 + (Ii — (13+ (132, 2(Q1CI2 + (Jags) 2((10612 + qu3) 2(Q1CJ2 — (10613) —q§ — (Ii + (1% + ‘19? 2’ 27 2 w=ﬂw§+w§+w§andw7é0 T 01(t) : ( ““0 sin w my sin 7” “’2 sin 7”, cos w—t ) 7 w w w 2 3—D motion: Xt R(q(t))X0 + tT. Perspective projection: X : (X17X27X3)7 Unknowns: a : (6T, ¢T,Wx,wy,wz, Z0)T. Problem: determine a7 given {Xs7 XS+1, - ~ ~ , XSJFN}7 Where Xt = (Xt,yt)T,t =s,-~-7s+N7 8 Z 0 Let ht(a) be the 3—D coordinates of a point at time t, given parameter a: ht(a) = R(q(t))X0 + tT. Given image coordinates xt, and depth Zt, its corresponding 3—D coordinates are: Xt = (ZtXt/f7 Zth/ﬂ Zt)T~ Let p = (0, 07 1)7 then Zt = p ht(a). Novv7 Xt becomes \$t/f Xt : p ht(a) yt/f The objective function to be minimized is given by: s+N E2(a) = Z H Xi — Ma) Ila \$t/f Xt : p ht(a) yt/f 1 xt/f 8 8T 8T . a—Eli(Xt—ht(a)):tp8—ai yt/f —t8—ai’ 1:1727 1 \$t/f 8 (9R (9R 8—ai(Xt_ht(a)):pa—aiX0 yt/f —6—aiX0 12345 1 3150/10 JCt/f ﬁve/f 8 a—aJXt — W” = PR yo/f yt/f — R yo/f 1 1 1 Where 8T _ aT _ 8&1 — 891“ — T = (sin6TcosqﬁT, sin6TsinqﬁT, COSQT)T. cos 6T cos qST 8T _ 8T _ COSHT sinng 7 Fag—@— — Sin 6T — sin 6T sin qu sin 6T cos ng q = ((107 (117 (12; Q3)T: Quaternion T q(t) : ““0 sin w my sin 7” “’2 sin 7” cos w—t 7 w w 2 7 w 2 7 2 2 7 Rotation Matrix: qg — q? — q; + 1% 2(Qoqi + q2t13) 2(qoq2 — (Ma) R 2 2(Q0611 — 61293) —q3 + 61% — (13+ (132, 2(Q1CI2 + (Jags) 2(Q092 + (11613) 2(Q1Q2 — 61093) —(1(i— (1% + (13+ €19? w=,/w%+w§+w§,andw7é0. 3 3 Where Dk : aR/aqk, k : 0,1,2,3, are given by QO Q1 (12 —611 QO —Q3 D0 2 2 (11 —q0 q3 ’ D1 : 2 go (11 q2 7 Q2 —Q3 —610 (13 Q2 —91 —Q2 Q3 (10 (J3 q2 —(11 D2 : 2 —C]3 —C]2 Q1 ’ D3 2 2 —Q2 (13 QO C10 (11 Q2 (11 —(10 —Q3 7 q = ((107 (117 (12; Q3)T: Quaternion “’2 sin 7” cos w—t 7 w 2 Rotation Matrix: 2 go — q? — q; + 1% 2((10Q1 + q2t13) 2(qoq2 — (Ma) R 2 2(Q0611 — 61293) —q3 + (I? — (13+ (132, 2(Q1CI2 + (Jags) 2(Q092 + (11613) 2(Q1Q2 — 61093) —qg — (1%4’ (13+ £19? (15) w=,/w%+w§+w§,andw7é0. Now, aqk/aai, k: = 0, 1, 2,3, 2‘ = 3,4, 5, are given by wt 8(11 8mm:(i—::—§)Sin—+%Z—§CO 08—7 8me—w—w3ySiD—tw g%—— ysinw7t——%w\$wyc0s2,g% (i—Z—f’ﬁln 35; = sin 353;, 5 ysm 331: : —w—3WZ sin 1% — 511130111; cos—“1,35: 93” sin 1% 35;: = (i Mm : 5: 55: sin + t wmwy 2 COSw — wt COS 7, s+N =2 || Xt— an? (17) Then the ﬁrst partial derivatives of the function E2 with respect to ai are given by: 8E2 s+N a . 8a =2Z(X h(t ))Taai(Xt—ht(a))7 1:1,27~~-76. (18) The second partial derivatives are obtained by ignoring the second derivatives of the model function: 82E2 s+N a a ~2 X—h T—X—h 7 ’7'21,2,-~~,6. 19 8511an E 8a1( t t(a)) 8aj( t t(a)) 1 J ( ) Let ﬁl dif —%%"§:, and or” dif %ai%2j the elements in matrix [oz ] . The minimization problem is reduced to iteratively solving the following linear equation: 2 akléal = 5k, (20) 1:1 where m is the number of unknown parameters. Here m : 6. Algorithm: TrajFit 1. Compute E2(a) in Equation (17). Set A : 0.001. 2. Compute 52‘ and 05277 Where z',j : 17 2, - - - 7 6, using equations (18) and (19)7 respectively. 3. Compute matrix [04] by augmenting its diagonal elements: Gag-j = arm-(1 + /\)7 and I — i 4. Solve Equation (20) for 6(a) and evaluate E2(a + 6a). 5. If E2(a + 6a) 2 E2(a), increase A by a factor and go back to 2. 6. If E2(a + 6a) < E2(a), decrease /\ by a factor, update the trial solution a <— a + 6a, and go back to 2. ...
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