Similarity transformation in 3D between two matched points patterns.

Transcription

1 Similarity transformation in 3D between two matched points patterns. coordinate system 1 coordinate system 2 The first 3D coordinate system is transformed through a three-dimensional R followed by a translation t, which we will take into account as O (2) O (1), to obtain the second coordinate system. The coordinate systems and transformation are righthanded. The rotation have to satisfy { 1 i = j RR = R R = I 3 R = [ r 1 r 2 r 3 ] with r i r j = δ ij = i, j = 1, 2, 3 0 i j which gives six independent quadratic constaints on the three angles. However, in order to restrict the movements to a 3D space we must also have detr = 1 that is, R must be a proper orthonormal matrix. Taking the first coordinate system as A = I 3 and the rotation B = R, this change of coordinates in 3D is y [B] = Ry [A] where we project it in the second coordinate system. The translation and the three-dimensional point P, have the relation in the second coordinate system O (2) P = O (2) O (1) + O (1) P or in coordinates y (2) o = t + Ry (1) o

2 Each point the first coordinate system corresponds to the correct point in the second coordinate system, so only inliers are considered. The measurements are y (k) = y (k) o + δy (k) δy (k) GI(0, σ 2 I 3 ) k = 1, 2 E[δy (1) δy (2) ] = 0 and the noise in the two coordinate systems are independent. In the general case the estimation also have a scale change c subject to [ ĉ, ˆt, ˆR ] = arg min c,t,r y (2) io = cry(1) io + t y (2) i cry (1) i t 2 2 i = 1,..., n and the rotational constraints from above. This is a total least squares (TLS) estimation problem with the rotation depending nonlinearly on the six-dimensional {y (1) i, y (2) i } input. We start by eliminating the translation from the objective function. In each of the two frames the average is ȳ (k) = 1 n y (k) i k = 1, 2 and constraint on the averages between the two frames is ȳ (2) = ĉ ˆRȳ (1) + ˆt where we assumed separability of the three estimates. So, if we already know ĉ, ˆR, the translation can be found ˆt = ȳ (2) ĉ ˆRȳ (1) Next we estimate the scale between the two frames. The centered input variables are ỹ (k) i = y (k) i ȳ (k) k = 1, 2 2

3 and the objective function becomes after subtituting ˆt J = ỹ (2) i crỹ (1) i 2 = ỹ (2) i 2 2c ỹ (1 ) i R ỹ (2) i +c 2 Rỹ (1) i 2 because the middle term is a scalar and can be written in two ways. The zero mean sample variances are n ˆσ 2(k) = ỹ(k) i 2 k = 1, 2 The rotation matrix in an orthonormal matrix with norm one Rỹ (1) i = ỹ (1) i The sample covariance between the two frames is a symmetric positive semidefinite matrix K = 1 ỹ (2) i ỹ (1) i = 1 ỹ (1) i ỹ (2) i and the objective function becomes J = ()ˆσ 2 2 2c()trace[R K] + c 2 ()ˆσ 2 1 This is a quadratic equation in c and the solution is dj dc = 0 which is a minimum since d 2 J dc 2 = 2()ˆσ2 1 > 0 Thus, if we find ˆR, the scale will be also known ĉ = trace[ ˆR K] ˆσ 2 1 The objective function is further simplified arg min J = arg max R R (trace[r K]) A theorem... to help us. The Cauchy-Schwarz inequality can be written for two n n matrices trace[a B] A B 3

4 where the norm can be 2-norms or Frobenius norms. We will use the Frobenius norm. Equality appears when A = βb and without loss of generality (w.l.g.) we can take β = 1. Assume A = BQ where Q is an orthonormal matrix QQ = Q Q = I n. Then, due to the orthonormality of Q, the inequality is trace[qb B] B 2 F = trace[b B] If BQ = B = A, that is Q = I n, the inequality reaches the upper bound and becomes an equality. The D = B B is always a symmetric positive semidefinite matrix. We did prove traced trace[qd] Coming back to the objective function, the svd of the sample covariance is K = UΣV and we can write (after substituting the rotation with the estimated rotation) trace[ ˆR UΣV ] = trace[v ˆR UΣ] = trace[qd] where Q = V ˆR U is an orthonormal matrix and D = Σ is a symmetric positive semidefinite (diagonal) matrix. A upper bound (the maximum) is reached when Q = V ˆR U = I3 or ˆR = UV But still we did not take into account that the rotation has to be a proper orthonormal matrix. The proper matrix have to be 1 ˆR = U... 1 V = UZV det(uv ) where the determinant det(uv ), which has the value 1 or 1, is taken accordingly so det ˆR = 1. The scale estimate, ĉ can be rewritten ĉ = trace[(uzv ) UΣV ] ˆσ 2 1 = trace[σz] ˆσ 2 1 4

5 in the second coordinated system. The physical meaning of the (inlier) similarity transformation can be seen if we rearrange the changing part of the objective function c trace[r K] = c trace[kr] = c trace[rk] where K is symmetric and positive semidefinite. Than 1 c Rỹ (1) i ỹ (2 ) i = (c Rỹ (1) i ) (ỹ (2 ) i ) Points in the first coordinate system are transformed into the second coordinate system. The scale is taken into account and maximum alignment is sought. In p > 3 dimensions, given two n p matrices A, B we want to find an orthonormal p p matrix Q such that min Q A BQ F subject QQ = Q Q = I p detq = 1 The proper rotation that transforms B to A is known in the literature as orthogonal Procrutes problem. The solution (like above) is max Q (trace[q B A]) = max Q (trace[q (UΣV )]) = max Q (trace[(v Q U)Σ]) from where the p p matrix in ˆQ = UZV with Z having ones on the diagonal and det(uv ) on the last row. Conditions for ˆR to be unique. Dimension of the space equal three. If the covariance matrix K is full rank three, the rotation also will be full rank. In general this is the case. If the rank of the matrix K is two, the third coordinate of the rotation, e.g., r 3, still can be recovered due to the orthonormality. Example. The rotation is around i 3 and the three points, P 1, P 2, P 3, move from the plane (i 1, i 3 ) to the plane (i 2, i 3 ). In this case, no line beside the rotation axis i 3 is satisfactory. If the rank of the matrix K is one, only a coordinate of the rotation, e.g. r 2, is defined. In this case, there are multiple solutions for the rotation. Example. The three points are now on the i 1 axis and after the rotation move to the i 2 axis. Any line through the origin and the bisector plane is satisfactory. 5

6 In general, only rank K = p, p 1 give unique solutions to orthogonal Procrutes problem. Pose of an object in the coordinate system of the camera. Three 3D vectors, m 1, m 2, m 3, are given, which are independent but not orthonormal. We have to find the rotation relative to I 3 such that Developing the sum min R 3 m i r i = min R M R 2 F M R 2 F = M 2 F 2 trace[r M] + R 2 F but M 2 F is a constant of the problem and R 2 F is equal to three. The rotation matrix is found max R trace[r M] as before... Note we have a unique solution even when one of the m vectors is unknown. Orthogonal frame reconstruction. Appears in 3D motion problems and object recognition. Given three independent vectors in a plane T = [ t 1 t 2 t 3 ], which represent the 2D projection of a 3D 6

7 rotation matrix R RR = R R = I 3 detr = ±1 find the rotation matrix and the unit normal to the plane n. The unit normal n is perpendicular to the plane which is the range of T. The svd of T is rank two. The projection in the range of T is P = I 3 nn = u 1 u 1 + u 2 u 2 from where ˆn is easy to recover, ˆn = u 3. The 3D rotation matrix is the minimum in 2D of arg min R T PR 2 F = arg max R trace[r P T] = arg max R trace[r T] = because P T = T and the solution is like before. The matrix Q = I 3 2nn is the reflection of the projection PQ = I 3 3nn + 2nn nn = P If for one 3D vector we know a priori the orientation (away/toward the image plane), we have a unique solution. 7