16 Conformal Mapping. Overview

Transcription

1 Algebraic Geodesy and Geoinformatics PART II APPLICATIONS 16 Conformal Mapping Overview First, the 3- point problem is discussed. A preliminary elimination of the translation vector reduces the size of system to 4 polynomial equations. This system can be solved in symbolic way, either with Dixon resultant or with reduced Groebner basis, both methods result in the same quartic univariate polynomial. Concerning numerical solution Extended NewtonRaphson method can be employed to solve all of the 9 equations in least square sense, utilizing the symbolic solution of any 7 equations as initial guess. For N- point problem Gauss- Jacobi solution improved by Extended Newton- Raphson method is very attractive. However, the General Procrustes algorithm is the fastest and precise as the direct global minimization technique requiring about 5 times longer computation time, than the General Procrustes. All numerical data of the examples are from the text book Problem definition Let us consider coordinates given in two systems A and B. The coordinates of the same physical point, Pi in system A are (Xi, Yi, Zi ), while its corresponding coordinates in system B are (xi, yi, zi ). We suppose that the relation between the two systems can be described by conformal mapping, namely Xi X0 xi yi = s R Yi + Y0 zi Zi Z0 This formula represents 3 elementary transformations, - scaling, with positive, real s, X0 - translation, with vector Y0, Z0 - rotation, with matrix R. The rotation matrix, R can be expressed by the skew matrix, S R = HI3 - SL-1 HI3 + SL where Clear@"Global *"D

2 2 ConformalMapping_16.nb S 0 c b c 0 a b a 0 ; and I 3 IdentityMatrix 3 ; Then the rotation matrix, R Inverse I 3 S. I 3 S Simplify; MatrixForm R 1 a 2 b 2 c 2 1 a 2 b 2 c 2 2 a b 2 c 1 a 2 b 2 c 2 2 a b c 1 a 2 b 2 c 2 1 a 2 b 2 c 2 1 a 2 b 2 c 2 2 b a c 1 a 2 b 2 c 2 2 b a c 1 a 2 b 2 c 2 2 a b c 1 a 2 b 2 c 2 2 a b c 1 a 2 b 2 c 2 1 a 2 b 2 c 2 1 a 2 b 2 c 2 which satisfies the following relation, I 3 R.Transpose R Simplify True The transformation has 7 parameters, (a, b, c, X 0, Y 0, Z 0, s) and to determine them, we need the coordinates of minimum 3 points in both systems (3 -Point Problem). The prototype equation for a point, P i, f 3 i 2 f 3 i 1 I 3 S. f 3 i x i y i z i s I 3 S. X i Y i Z i I 3 S. X 0 Y 0 Z 0 Expand; MatrixForm f 3 i 2 f 3 i 1 f 3 i Then for i = 1 x i X 0 s X i c y i c Y 0 c s Y i b z i b Z 0 b s Z i c x i c X 0 c s X i y i Y 0 s Y i a z i a Z 0 a s Z i b x i b X 0 b s X i a y i a Y 0 a s Y i z i Z 0 s Z i f 1 f 2 f 3 f 3 i 2 f 3 i 1 f 3 i. i 1 Expand; MatrixForm f 1 f 2 f 3 for i = 2 x 1 X 0 s X 1 c y 1 c Y 0 c s Y 1 b z 1 b Z 0 b s Z 1 c x 1 c X 0 c s X 1 y 1 Y 0 s Y 1 a z 1 a Z 0 a s Z 1 b x 1 b X 0 b s X 1 a y 1 a Y 0 a s Y 1 z 1 Z 0 s Z 1 f 4 f 5 f 6 f 3 i 2 f 3 i 1 f 3 i. i 2 Expand; MatrixForm f 4 f 5 f 6 for i = 3 x 2 X 0 s X 2 c y 2 c Y 0 c s Y 2 b z 2 b Z 0 b s Z 2 c x 2 c X 0 c s X 2 y 2 Y 0 s Y 2 a z 2 a Z 0 a s Z 2 b x 2 b X 0 b s X 2 a y 2 a Y 0 a s Y 2 z 2 Z 0 s Z 2

3 ConformalMapping_16.nb f 7 f 8 f 9 f 3 i 2 f 3 i 1 f 3 i. i 3 Expand; MatrixForm f 7 f 8 f 9 x 3 X 0 s X 3 c y 3 c Y 0 c s Y 3 b z 3 b Z 0 b s Z 3 c x 3 c X 0 c s X 3 y 3 Y 0 s Y 3 a z 3 a Z 0 a s Z 3 b x 3 b X 0 b s X 3 a y 3 a Y 0 a s Y 3 z 3 Z 0 s Z 3 Translation parameters, (X 0, Y 0, Z 0 ) can be eliminated by differencing, f 14 f 1 f 4 Simplify x 1 x 2 s X 1 s X 2 c y 1 c y 2 c s Y 1 c s Y 2 b z 1 b z 2 b s Z 1 b s Z 2 f 25 f 2 f 5 Simplify c x 1 c x 2 c s X 1 c s X 2 y 1 y 2 s Y 1 s Y 2 a z 1 a z 2 a s Z 1 a s Z 2 f 39 f 3 f 9 Simplify b x 1 b x 3 b s X 1 b s X 3 a y 1 a y 3 a s Y 1 a s Y 3 z 1 z 3 s Z 1 s Z 3 f 69 f 6 f 9 Simplify b x 2 b x 3 b s X 2 b s X 3 a y 2 a y 3 a s Y 2 a s Y 3 z 2 z 3 s Z 2 s Z 3 Now, we have four equations and four unknown parameters (a, b, c, s). The nonlinearity is represented by the variable s only. Let us introduce new variables, newvars x12 x 1 x 2, x13 x 1 x 3, x23 x 2 x 3, y12 y 1 y 2, y13 y 1 y 3, y23 y 2 y 3, z12 z 1 z 2, z13 z 1 z 3, z23 z 2 z 3, X12 X 1 X 2, X13 X 1 X 3, X23 X 2 X 3, Y12 Y 1 Y 2, Y13 Y 1 Y 3, Y23 Y 2 Y 3, Z12 Z 1 Z 2, Z13 Z 1 Z 3, Z23 Z 2 Z 3 ; Then our system becomes, sys x12 s X12 c y12 c s Y12 b z12 b s Z12, c x12 c s X12 y12 s Y12 a z12 a s Z12, b x13 b s X13 a y13 a s Y13 z13 s Z13, b x23 b s X23 a y23 a s Y23 z23 s Z23 ; Let us check it, f 14, f 25, f 39, f 69 sys. newvars Simplify 0, 0, 0, Symbolic solution Dixon Resultant Resultant Dixon We eliminate the linear parameters, a, b, and c, in order to get an univariate polynomial of s,

4 4 ConformalMapping_16.nb AbsoluteTiming solsdx DixonResultant sys, a, b, c, U, V, W Simplify; , Null solsdx x12 2 x23 y13 s Y13 x13 y23 s X23 y13 X13 y23 x13 Y23 s 2 X23 Y13 X13 Y23 y12 x23 y12 y13 z12 z13 x13 y12 y23 z12 z23 s X13 y12 2 y23 x13 y12 2 Y23 X12 y23 z12 z13 X23 y12 y12 y13 z12 z13 x23 y12 2 Y13 Y12 z12 z13 y12 Z12 z13 y12 z12 Z13 X13 y12 z12 z23 x13 Y12 z12 z23 X12 y13 z12 z23 x13 y12 Z12 z23 x13 y12 z12 Z23 x12 z12 s Z12 y23 z13 s Z13 y13 z23 s Y23 z13 Y13 z23 y13 Z23 s 2 Y23 Z13 Y13 Z23 s 2 X23 y12 2 Y13 x13 Y12 2 y23 X12 2 x23 y13 x13 y23 X13 y12 2 Y23 X23 Y12 z12 z13 X23 y12 Z12 z13 X23 y12 z12 Z13 x23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 z12 z23 X13 y12 Z12 z23 x13 Y12 Z12 z23 X13 y12 z12 Z23 x13 Y12 z12 Z23 x13 y12 Z12 Z23 X12 Y23 z12 z13 y23 Z12 z13 y23 z12 Z13 Y13 z12 z23 y13 Z12 z23 y13 z12 Z23 s 3 x23 Y12 2 Y13 X13 Y12 2 y23 x13 Y12 2 Y23 X12 2 X23 y13 x23 Y13 X13 y23 x13 Y23 x23 Y12 Z12 Z13 X23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 Z12 z23 X13 Y12 z12 Z23 X13 y12 Z12 Z23 x13 Y12 Z12 Z23 X12 Y23 Z12 z13 Y23 z12 Z13 y23 Z12 Z13 Y13 Z12 z23 Y13 z12 Z23 y13 Z12 Z23 s 4 X12 2 X23 Y13 X13 Y23 X12 Y23 Z12 Z13 Y13 Z12 Z23 Y12 X23 Y12 Y13 Z12 Z13 X13 Y12 Y23 Z12 Z23 The result is a quartic polynomial of s, Exponent[solsdx, {s,a,b,c}] 4, 0, 0, 0 The coeffients, q0s Coefficient solsdx, s, 0 Simplify x12 2 x23 y13 x13 y23 x12 z12 y23 z13 y13 z23 y12 x23 y12 y13 z12 z13 x13 y12 y23 z12 z23 q1s Coefficient solsdx, s Simplify x23 y12 2 Y13 X13 y12 2 y23 x13 y12 2 Y23 x12 2 X23 y13 x23 Y13 X13 y23 x13 Y23 x23 Y12 z12 z13 X12 y23 z12 z13 x23 y12 Z12 z13 X23 y12 y12 y13 z12 z13 x23 y12 z12 Z13 X13 y12 z12 z23 x13 Y12 z12 z23 X12 y13 z12 z23 x13 y12 Z12 z23 x13 y12 z12 Z23 x12 Y23 z12 z13 y23 Z12 z13 y23 z12 Z13 Y13 z12 z23 y13 Z12 z23 y13 z12 Z23 q2s Coefficient solsdx, s 2 Simplify x12 2 X23 Y13 X23 y12 2 Y13 x13 Y12 2 y23 X12 2 x23 y13 x13 y23 x12 2 X13 Y23 X13 y12 2 Y23 X23 Y12 z12 z13 X23 y12 Z12 z13 x12 Y23 Z12 z13 X23 y12 z12 Z13 x12 Y23 z12 Z13 x12 y23 Z12 Z13 x23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 z12 z23 X13 y12 Z12 z23 x13 Y12 Z12 z23 x12 Y13 Z12 z23 X13 y12 z12 Z23 x13 Y12 z12 Z23 x12 Y13 z12 Z23 x13 y12 Z12 Z23 x12 y13 Z12 Z23 X12 Y23 z12 z13 y23 Z12 z13 y23 z12 Z13 Y13 z12 z23 y13 Z12 z23 y13 z12 Z23 q3s Coefficient solsdx, s 3 Simplify x23 Y12 2 Y13 X13 Y12 2 y23 x13 Y12 2 Y23 X12 2 X23 y13 x23 Y13 X13 y23 x13 Y23 x23 Y12 Z12 Z13 x12 Y23 Z12 Z13 X23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 Z12 z23 X13 Y12 z12 Z23 X13 y12 Z12 Z23 x13 Y12 Z12 Z23 x12 Y13 Z12 Z23 X12 Y23 Z12 z13 Y23 z12 Z13 y23 Z12 Z13 Y13 Z12 z23 Y13 z12 Z23 y13 Z12 Z23

5 ConformalMapping_16.nb q4s Coefficient solsdx, s 4 Simplify X12 2 X23 Y13 X13 Y23 X12 Y23 Z12 Z13 Y13 Z12 Z23 Y12 X23 Y12 Y13 Z12 Z13 X13 Y12 Y23 Z12 Z23 Let us check the result, solsdx q4s s 4 q3s s 3 q2s s 2 q1s s q0s Simplify True The other parameters can be computed as function of the nonlinear parameter s. The first equation does not contain the parameter a, Map Exponent, a, b, c &, sys 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0 Therefore, we do not take it into consideration when computing a, soladx DixonResultant Drop sys, 1, 1, b, c, V, W Simplify x12 s X12 a s X23 y13 s 2 X23 Y13 x23 y13 s Y13 x13 y23 s X13 y23 s x13 Y23 s 2 X13 Y23 x23 z13 s X23 z13 s x23 Z13 s 2 X23 Z13 x13 z23 s X13 z23 s x13 Z23 s 2 X13 Z23 Similarly, we leave out equations, which does not contain the corresponding parameter in case of b, and c, solbdx DixonResultant Drop sys, 2, 2, a, c, U, W Simplify y12 s Y12 b s X23 y13 s 2 X23 Y13 x23 y13 s Y13 x13 y23 s X13 y23 s x13 Y23 s 2 X13 Y23 y23 z13 s Y23 z13 s y23 Z13 s 2 Y23 Z13 y13 z23 s Y13 z23 s y13 Z23 s 2 Y13 Z23 solcdx DixonResultant Drop sys, 3, 3, a, b, U, V Simplify z12 s Z12 c x23 y12 y12 y23 x12 x23 s X23 c y23 c s Y23 z12 z23 s c X23 y12 c x23 Y12 Y12 y23 X12 x23 c y23 y12 Y23 Z12 z23 z12 Z23 s 2 c X23 Y12 Y12 Y23 X12 X23 c Y23 Z12 Z Reduced Groebner Basis The similar result can be achieved by reduced Groebner basis. AbsoluteTiming solsgb GroebnerBasis sys, s, a, b, c, a, b, c Simplify; , Null

6 6 ConformalMapping_16.nb solsgb x12 2 s X23 y13 s 2 X23 Y13 x23 y13 s Y13 x13 y23 s X13 y23 s x13 Y23 s 2 X13 Y23 y12 x23 y12 y13 z12 z13 x13 y12 y23 z12 z23 s X13 y12 2 y23 x13 y12 2 Y23 X12 y23 z12 z13 X23 y12 y12 y13 z12 z13 x23 y12 2 Y13 Y12 z12 z13 y12 Z12 z13 y12 z12 Z13 X13 y12 z12 z23 x13 Y12 z12 z23 X12 y13 z12 z23 x13 y12 Z12 z23 x13 y12 z12 Z23 x12 z12 s Z12 y23 z13 s Z13 y13 z23 s Y23 z13 Y13 z23 y13 Z23 s 2 Y23 Z13 Y13 Z23 s 2 X23 y12 2 Y13 x13 Y12 2 y23 X12 2 x23 y13 x13 y23 X13 y12 2 Y23 X23 Y12 z12 z13 X23 y12 Z12 z13 X23 y12 z12 Z13 x23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 z12 z23 X13 y12 Z12 z23 x13 Y12 Z12 z23 X13 y12 z12 Z23 x13 Y12 z12 Z23 x13 y12 Z12 Z23 X12 Y23 z12 z13 y23 Z12 z13 y23 z12 Z13 Y13 z12 z23 y13 Z12 z23 y13 z12 Z23 s 3 x23 Y12 2 Y13 X13 Y12 2 y23 x13 Y12 2 Y23 X12 2 X23 y13 x23 Y13 X13 y23 x13 Y23 x23 Y12 Z12 Z13 X23 Y12 2 y13 Y12 Z12 z13 Y12 z12 Z13 y12 Z12 Z13 X13 Y12 Z12 z23 X13 Y12 z12 Z23 X13 y12 Z12 Z23 x13 Y12 Z12 Z23 X12 Y23 Z12 z13 Y23 z12 Z13 y23 Z12 Z13 Y13 Z12 z23 Y13 z12 Z23 y13 Z12 Z23 s 4 X12 2 X23 Y13 X13 Y23 X12 Z12 Y23 Z13 Y13 Z23 Y12 X23 Y12 Y13 Z12 Z13 X13 Y12 Y23 Z12 Z23 This is the same result as the result of the Dixon resultant, but the computation time is considerably longer. solsgb solsdx Simplify True The determination of the other parameters are similar. Again, we consider the parameter s as a constant parameter, solagb GroebnerBasis Drop sys, 1, 1, a, b, c, b, c a x23 y13 a s X23 y13 a s x23 Y13 a s 2 X23 Y13 a x13 y23 a s X13 y23 a s x13 Y23 a s 2 X13 Y23 x23 z13 s X23 z13 s x23 Z13 s 2 X23 Z13 x13 z23 s X13 z23 s x13 Z23 s 2 X13 Z23 The Dixon solution has two factors. The reduced Groebner basis gives the second one, solagb soladx 2 Simplify True Similarly solbgb GroebnerBasis Drop sys, 2, 2, a, b, c, a, c b x23 y13 b s X23 y13 b s x23 Y13 b s 2 X23 Y13 b x13 y23 b s X13 y23 b s x13 Y23 b s 2 X13 Y23 y23 z13 s Y23 z13 s y23 Z13 s 2 Y23 Z13 y13 z23 s Y13 z23 s y13 Z23 s 2 Y13 Z23 solbgb solbdx 2 Simplify True solcgb GroebnerBasis Drop sys, 3, 3, a, b, c, a, b x12 x23 s X12 x23 s x12 X23 s 2 X12 X23 c x23 y12 c s X23 y12 c s x23 Y12 c s 2 X23 Y12 c x12 y23 c s X12 y23 y12 y23 s Y12 y23 c s x12 Y23 c s 2 X12 Y23 s y12 Y23 s 2 Y12 Y23 z12 z23 s Z12 z23 s z12 Z23 s 2 Z12 Z23

7 ConformalMapping_16.nb solcgb solcdx 2 Simplify True Computation of the translation vector The translation vector can be computed from the original system consisting of 9 equations, syso Table f i, i, 1, 9 x 1 X 0 s X 1 c y 1 c Y 0 c s Y 1 b z 1 b Z 0 b s Z 1, c x 1 c X 0 c s X 1 y 1 Y 0 s Y 1 a z 1 a Z 0 a s Z 1, b x 1 b X 0 b s X 1 a y 1 a Y 0 a s Y 1 z 1 Z 0 s Z 1, x 2 X 0 s X 2 c y 2 c Y 0 c s Y 2 b z 2 b Z 0 b s Z 2, c x 2 c X 0 c s X 2 y 2 Y 0 s Y 2 a z 2 a Z 0 a s Z 2, b x 2 b X 0 b s X 2 a y 2 a Y 0 a s Y 2 z 2 Z 0 s Z 2, x 3 X 0 s X 3 c y 3 c Y 0 c s Y 3 b z 3 b Z 0 b s Z 3, c x 3 c X 0 c s X 3 y 3 Y 0 s Y 3 a z 3 a Z 0 a s Z 3, b x 3 b X 0 b s X 3 a y 3 a Y 0 a s Y 3 z 3 Z 0 s Z 3 Concerning the translation parameters, we have only 3 unknown parameters, but 9 equations. The translation vector, for i = 1, 2, 3. Therefore the coefficient matrix has special structure, namely Flatten Table IdentityMatrix 3, 3, 1 ; MatrixForm the pseudoinverze of,

8 8 ConformalMapping_16.nb pi PseudoInverse ; pi MatrixForm Therefore, the least square solution is a simple averaging, see Gauss-Jacobi combinatorial solution, see Chapter 7. The solution, pi. Α 1 Β 1 Γ 1 Α 2 Β 2 Γ 2 Α 3 Β 3 Γ 3 Α 1 3 Α 2 3 Α 3 3, Β 1 3 Β 2 3 Β 3 3, Γ 1 3 Γ 2 3 Γ 3 3 or in detailed form, solxyz0 1 3 x 1 y 1 z 1 s R. X 1 Y 1 Z 1 x 2 y 2 z 2 s R. X 2 Y 2 Z 2 x 3 y 3 z 3 s R. X 3 Y 3 Z 3 Simplify 1 1 a 2 b 2 c 2 x 1 1 a 2 b 2 c 2 x 2 x a 2 b 2 c 2 a 2 x 3 b 2 x 3 c 2 x 3 s X 1 a 2 s X 1 b 2 s X 1 c 2 s X 1 s X 2 a 2 s X 2 b 2 s X 2 c 2 s X 2 s X 3 a 2 s X 3 b 2 s X 3 c 2 s X 3 2 a b s Y 1 2 c s Y 1 2 a b s Y 2 2 c s Y 2 2 a b s Y 3 2 c s Y 3 2 b s Z 1 2 a c s Z 1 2 b s Z 2 2 a c s Z 2 2 b s Z 3 2 a c s Z 3, 1 2 a b c s X 1 2 a b c s X 2 2 a b s X 3 2 c s X 3 y a 2 b 2 c 2 a 2 y 1 b 2 y 1 c 2 y 1 y 2 a 2 y 2 b 2 y 2 c 2 y 2 y 3 a 2 y 3 b 2 y 3 c 2 y 3 s Y 1 a 2 s Y 1 b 2 s Y 1 c 2 s Y 1 s Y 2 a 2 s Y 2 b 2 s Y 2 c 2 s Y 2 s Y 3 a 2 s Y 3 b 2 s Y 3 c 2 s Y 3 2 a s Z 1 2 b c s Z 1 2 a s Z 2 2 b c s Z 2 2 a s Z 3 2 b c s Z 3, 1 2 b a c s X 1 2 b a c s X 2 2 b s X 3 2 a c s X 3 2 a s Y a 2 b 2 c 2 2 b c s Y 1 2 a s Y 2 2 b c s Y 2 2 a s Y 3 2 b c s Y 3 z 1 a 2 z 1 b 2 z 1 c 2 z 1 z 2 a 2 z 2 b 2 z 2 c 2 z 2 z 3 a 2 z 3 b 2 z 3 c 2 z 3 s Z 1 a 2 s Z 1 b 2 s Z 1 c 2 s Z 1 s Z 2 a 2 s Z 2 b 2 s Z 2 c 2 s Z 2 s Z 3 a 2 s Z 3 b 2 s Z 3 c 2 s Z Numerical Example for the 3-Point Problem Let us consider the following data of 3 physical points, datac3 X , Y , Z , X , Y , Z , X , Y , Z , x , y , z , x , y , z , x , y , z ;

9 ConformalMapping_16.nb Let us compute the parameter s. The quartic polynomial for s, eqs s 4 q4s s 3 q3s s 2 q2s s q1s q0s. newvars. datac s s s s 4 Let us normalize it, eqs eqs Coefficient eqs, s, 4 Expand s s s 3 1. s 4 The roots, sols NSolve eqs, s Flatten s , s , s , s The admissible solution should be positive, real, s0 Select sols, Im & s The elements of the skew matrix can be directly computed from the symbolic result, for example let us take the result of the Dixon resultant, then a0 Solve soladx. newvars 0, a Simplify a s X 2 z 1 s X 3 z 1 x 1 z 2 s X 1 z 2 s X 3 z 2 x 1 z 3 s X 1 z 3 s X 2 z 3 s 2 X 2 Z 1 s 2 X 3 Z 1 s x 1 Z 2 s 2 X 1 Z 2 s 2 X 3 Z 2 x 3 z 1 z 2 s Z 1 s Z 2 s x 1 Z 3 s 2 X 1 Z 3 s 2 X 2 Z 3 x 2 z 1 z 3 s Z 1 s Z 3 s X 2 y 1 s X 3 y 1 x 1 y 2 s X 1 y 2 s X 3 y 2 x 1 y 3 s X 1 y 3 s X 2 y 3 s 2 X 2 Y 1 s 2 X 3 Y 1 s x 1 Y 2 s 2 X 1 Y 2 s 2 X 3 Y 2 x 3 y 1 y 2 s Y 1 s Y 2 s x 1 Y 3 s 2 X 1 Y 3 s 2 X 2 Y 3 x 2 y 1 y 3 s Y 1 s Y 3 a0 a0. s0. datac3 Flatten a similarly b0 Solve solbdx. newvars 0, b Simplify b s Y 2 z 1 s Y 3 z 1 y 1 z 2 s Y 1 z 2 s Y 3 z 2 y 1 z 3 s Y 1 z 3 s Y 2 z 3 s 2 Y 2 Z 1 s 2 Y 3 Z 1 s y 1 Z 2 s 2 Y 1 Z 2 s 2 Y 3 Z 2 y 3 z 1 z 2 s Z 1 s Z 2 s y 1 Z 3 s 2 Y 1 Z 3 s 2 Y 2 Z 3 y 2 z 1 z 3 s Z 1 s Z 3 s X 2 y 1 s X 3 y 1 x 1 y 2 s X 1 y 2 s X 3 y 2 x 1 y 3 s X 1 y 3 s X 2 y 3 s 2 X 2 Y 1 s 2 X 3 Y 1 s x 1 Y 2 s 2 X 1 Y 2 s 2 X 3 Y 2 x 3 y 1 y 2 s Y 1 s Y 2 s x 1 Y 3 s 2 X 1 Y 3 s 2 X 2 Y 3 x 2 y 1 y 3 s Y 1 s Y 3 b0 b0. s0. datac3 Flatten b

10 10 ConformalMapping_16.nb c0 Solve solcdx. newvars 0, c Simplify c x 2 2 s x 3 X 1 s x 3 X 2 s 2 X 1 X 2 s 2 X 2 2 s 2 X 1 X 3 s 2 X 2 X 3 x 2 x 3 s X 1 s X 3 x 1 x 2 x 3 s X 2 s X 3 y 1 y 2 y 2 2 y 1 y 3 y 2 y 3 s y 2 Y 1 s y 3 Y 1 s y 1 Y 2 s y 3 Y 2 s 2 Y 1 Y 2 s 2 Y 2 2 s y 1 Y 3 s y 2 Y 3 s 2 Y 1 Y 3 s 2 Y 2 Y 3 z 1 z 2 z 2 2 z 1 z 3 z 2 z 3 s z 2 Z 1 s z 3 Z 1 s z 1 Z 2 s z 3 Z 2 s 2 Z 1 Z 2 s 2 Z 2 2 s z 1 Z 3 s z 2 Z 3 s 2 Z 1 Z 3 s 2 Z 2 Z 3 s X 2 y 1 s X 3 y 1 x 1 y 2 s X 1 y 2 s X 3 y 2 x 1 y 3 s X 1 y 3 s X 2 y 3 s 2 X 2 Y 1 s 2 X 3 Y 1 s x 1 Y 2 s 2 X 1 Y 2 s 2 X 3 Y 2 x 3 y 1 y 2 s Y 1 s Y 2 s x 1 Y 3 s 2 X 1 Y 3 s 2 X 2 Y 3 x 2 y 1 y 3 s Y 1 s Y 3 c0 c0. s0. datac3 Flatten c The translation vector, XYZ0 MapThread 1 2 &, X 0, Y 0, Z 0, solxyz0. datac3. s0. a0. b0. c0 Flatten X , Y , Z The residium of the equations of the original system, rs syso. datac3. s0. a0. b0. c0. XYZ , , , , , , , , The residium of the solution, Norm This method is implemented as a Mathematica function, Conform3DV7, in the GeoAlgebra package, GeoAlgebra Conform3DV7? Conform3DV7 Solves the 7 Parameter Datum Transformation Problem computing the best fitting parameters of the linear transform, between systems X,Y,Z x,y,z in form x,y,z s R a, b, c X,Y,Z X0,Y0,Z0. The inputs: xyz x1,y1,z1, x2,y2,z2, x3,y3,z3 XYZ X1,Y1,Z1, X1,Y1,Z1, X1,Y1,Z1. The result: s, a, b, c, X0, Y0, Z0 scale,s, elements of the skew matrix a, b, c, translation vector, X0, Y0, Z0 Let us employ the function,

11 ConformalMapping_16.nb AbsoluteTiming sol Conform3DV , , , , , , , , , , , , , , , , , NumberForm, 12 &; 0., Null sol , , , , , , The function is very fast, thanks for the symbolic solution! 16-4 Numerical Solutions First, we start with the Global Numerical Solver. Using it as a global method, any 7 equations can be solved from the 9 ones. The system of the 9 equations, syso x 1 X 0 s X 1 c y 1 c Y 0 c s Y 1 b z 1 b Z 0 b s Z 1, c x 1 c X 0 c s X 1 y 1 Y 0 s Y 1 a z 1 a Z 0 a s Z 1, b x 1 b X 0 b s X 1 a y 1 a Y 0 a s Y 1 z 1 Z 0 s Z 1, x 2 X 0 s X 2 c y 2 c Y 0 c s Y 2 b z 2 b Z 0 b s Z 2, c x 2 c X 0 c s X 2 y 2 Y 0 s Y 2 a z 2 a Z 0 a s Z 2, b x 2 b X 0 b s X 2 a y 2 a Y 0 a s Y 2 z 2 Z 0 s Z 2, x 3 X 0 s X 3 c y 3 c Y 0 c s Y 3 b z 3 b Z 0 b s Z 3, c x 3 c X 0 c s X 3 y 3 Y 0 s Y 3 a z 3 a Z 0 a s Z 3, b x 3 b X 0 b s X 3 a y 3 a Y 0 a s Y 3 z 3 Z 0 s Z 3 Take the first 7 equations, sysor Take syso, 1, 7 x 1 X 0 s X 1 c y 1 c Y 0 c s Y 1 b z 1 b Z 0 b s Z 1, c x 1 c X 0 c s X 1 y 1 Y 0 s Y 1 a z 1 a Z 0 a s Z 1, b x 1 b X 0 b s X 1 a y 1 a Y 0 a s Y 1 z 1 Z 0 s Z 1, x 2 X 0 s X 2 c y 2 c Y 0 c s Y 2 b z 2 b Z 0 b s Z 2, c x 2 c X 0 c s X 2 y 2 Y 0 s Y 2 a z 2 a Z 0 a s Z 2, b x 2 b X 0 b s X 2 a y 2 a Y 0 a s Y 2 z 2 Z 0 s Z 2, x 3 X 0 s X 3 c y 3 c Y 0 c s Y 3 b z 3 b Z 0 b s Z 3 The variable list a, b, c, s, X 0, Y 0, Z 0 ; We have two solutions and the first one has considerable "error", solabcs NSolve sysor. newvars. datac3, a , b , c , s , X , Y , Z , a , b , c , s , X , Y , Z

12 12 ConformalMapping_16.nb Remark : Do not mix the error of the method and the error of the technique, which means how to use it! The reason of the "error" here, is the fact that the system of the 7 equations mathematically has two solutions! Would be the model perfect and the data without error, then the additional two equations were redundant! However, it is generally not true, therefore these solutions do not represent the least square solution of the overdetermined system of 9 equations. In addition again, we can consider the first solution as a parasitic solution, but not as an error of the computation or the method. This phenomenon can be considered as a side effect of the algebraic solution! Now, let us use these results as initial values for the Extended Newton- Raphson method employed for solving the overdetermined system, GeoAlgebra NewtonExtended? NewtonExtended Computes the solution of an overdetermined nonlinear system. Input parameters: f list of functions of the system, x list of variables, x0 list of the initial values, eps error limit for the iteration, default value: 10^ 12 n maximum number of the iterations, default value: 100. Output: list of the iterative solutions We select the second solution as initial values, 0 Map 2 &, solabcs , , , , , , The result is somewhat different from the symbolic solution, AbsoluteTiming soln MapThread 1 2 &,, NewtonExtended syso. datac3,, 0 Last ; , Null soln a , b , c , s , X , Y , Z However, the residium, rn syso. datac3. soln , , , , , , , , is considerably smaller, see Fig.16.1, Norm and it is distributed nearly uniformly,

13 ConformalMapping_16.nb ListPlot rs, rn, Joined True, PlotRange All, Frame True Fig.16.1 Distribution of the residiums in case of symbolic (blue) and numeric solution of the 9 equations in least square sense (maroon) Remark: Probably, the best strategy is to employ symbolic solution first then to improve it with Extended Newton- Raphson method applied to the overdetermined system, to 9 equations. Let us do it! Employing the solution of the symbolic method as initial values, 0 Map 2 &, Join s0, a0, b0, c0, XYZ , , , , , , then applying Extended Newton- Raphson method, soln MapThread 1 2 &,, NewtonExtended syso. datac3,, 0 Last a , b , c , s , X , Y , Z Remark: Generally the best strategy is to combine symbolic and robust local numeric methods to get unique, precise solution without quessing initial values and with just a few iterations! 16-5 N-Point Problem The system of equations and data structures In this case, there are data for more than 3 points at our disposal. The prototype equation, e I 3 S. x i y i z i s I 3 S. X i Y i Z i I 3 S. X 0 Y 0 Z 0 x i X 0 s X i c y i c Y 0 c s Y i b z i b Z 0 b s Z i, c x i c X 0 c s X i y i Y 0 s Y i a z i a Z 0 a s Z i, b x i b X 0 b s X i a y i a Y 0 a s Y i z i Z 0 s Z i Expand Flatten In our illlustrative example there are seven points. The system in case of these 7 points,

14 14 ConformalMapping_16.nb sys Table e, i, 1, 7 Flatten x 1 X 0 s X 1 c y 1 c Y 0 c s Y 1 b z 1 b Z 0 b s Z 1, c x 1 c X 0 c s X 1 y 1 Y 0 s Y 1 a z 1 a Z 0 a s Z 1, b x 1 b X 0 b s X 1 a y 1 a Y 0 a s Y 1 z 1 Z 0 s Z 1, x 2 X 0 s X 2 c y 2 c Y 0 c s Y 2 b z 2 b Z 0 b s Z 2, c x 2 c X 0 c s X 2 y 2 Y 0 s Y 2 a z 2 a Z 0 a s Z 2, b x 2 b X 0 b s X 2 a y 2 a Y 0 a s Y 2 z 2 Z 0 s Z 2, x 3 X 0 s X 3 c y 3 c Y 0 c s Y 3 b z 3 b Z 0 b s Z 3, c x 3 c X 0 c s X 3 y 3 Y 0 s Y 3 a z 3 a Z 0 a s Z 3, b x 3 b X 0 b s X 3 a y 3 a Y 0 a s Y 3 z 3 Z 0 s Z 3, x 4 X 0 s X 4 c y 4 c Y 0 c s Y 4 b z 4 b Z 0 b s Z 4, c x 4 c X 0 c s X 4 y 4 Y 0 s Y 4 a z 4 a Z 0 a s Z 4, b x 4 b X 0 b s X 4 a y 4 a Y 0 a s Y 4 z 4 Z 0 s Z 4, x 5 X 0 s X 5 c y 5 c Y 0 c s Y 5 b z 5 b Z 0 b s Z 5, c x 5 c X 0 c s X 5 y 5 Y 0 s Y 5 a z 5 a Z 0 a s Z 5, b x 5 b X 0 b s X 5 a y 5 a Y 0 a s Y 5 z 5 Z 0 s Z 5, x 6 X 0 s X 6 c y 6 c Y 0 c s Y 6 b z 6 b Z 0 b s Z 6, c x 6 c X 0 c s X 6 y 6 Y 0 s Y 6 a z 6 a Z 0 a s Z 6, b x 6 b X 0 b s X 6 a y 6 a Y 0 a s Y 6 z 6 Z 0 s Z 6, x 7 X 0 s X 7 c y 7 c Y 0 c s Y 7 b z 7 b Z 0 b s Z 7, c x 7 c X 0 c s X 7 y 7 Y 0 s Y 7 a z 7 a Z 0 a s Z 7, b x 7 b X 0 b s X 7 a y 7 a Y 0 a s Y 7 z 7 Z 0 s Z 7 The numerical data are, xyz ; and XYZ ; where xyz are the coordinates for the WGS-84 system, while XYZ are the coordinates for the local system. In rule form, xyzr MapThread x 1 2 1, y 1 2 2, z &, Range 7, xyz Flatten x , y , z , x , y , z , x , y , z , x , y , z , x , y , z , x , y , z , x , y , z XYZR MapThread X 1 2 1, Y 1 2 2, Z &, Range 7, XYZ Flatten X , Y , Z , X , Y , Z , X , Y , Z , X , Y , Z , X , Y , Z , X , Y , Z , X , Y , Z

15 ConformalMapping_16.nb Global Minimization The global minimization is the most simple and robust method, but quite time consuming. The system equations in numerical form, sysn sys. xyzr. XYZR; Short sysn, b c s b s c s X 0 c Y 0 b Z 0, a c s a s c s c X 0 Y 0 a Z 0, a b s a s b s b X 0 a Y 0 Z 0, 16, a c s a s c s c X 0 Y 0 a Z 0, a b s a s b s b X 0 a Y 0 Z 0 The objective function, The result, obj Apply Plus, Map ^2 &, sysn ; Short obj, a b s a s b s b X 0 a Y 0 Z a b s a s b s b X 0 a Y 0 Z b c s b s c s X 0 c Y 0 b Z b c s b s c s X 0 c Y 0 b Z 0 2 AbsoluteTiming solgm NMinimize obj, ; , Null solgm , a , b , c , s , X , Y , Z The rotation matrix, Rn R. solgm 2 ; MatrixForm Rn Gauss-Jacobi solution Now, we have 7 points and any 3 of them form a subset, n 3; m 7; The number of the subsets

16 16 ConformalMapping_16.nb mn Binomial m, n 35 This is not a big number, therefore it is reasonable to use combinatorial solution, especially because the solution of the 3- Point Problem is very fast. These subsets are, qs Partition Map &, Flatten Subsets Range m, n, n 1, 2, 3, 1, 2, 4, 1, 2, 5, 1, 2, 6, 1, 2, 7, 1, 3, 4, 1, 3, 5, 1, 3, 6, 1, 3, 7, 1, 4, 5, 1, 4, 6, 1, 4, 7, 1, 5, 6, 1, 5, 7, 1, 6, 7, 2, 3, 4, 2, 3, 5, 2, 3, 6, 2, 3, 7, 2, 4, 5, 2, 4, 6, 2, 4, 7, 2, 5, 6, 2, 5, 7, 2, 6, 7, 3, 4, 5, 3, 4, 6, 3, 4, 7, 3, 5, 6, 3, 5, 7, 3, 6, 7, 4, 5, 6, 4, 5, 7, 4, 6, 7, 5, 6, 7 The data values for the subsets can be generated as it follows, datagj Map Transpose &, Table xyz qs i, j, XYZ qs i, j, i, 1, mn, j, 1, n ; Short datagj, , , , , , , , , , , , , , , , , , , 33, , , , , , , , , , , , , , , , , , Loading the symbolic solution of the 3 - point problem, GeoAlgebra Conform3DV7 ; The solution of the 35 subsets, AbsoluteTiming solgj Map Conform3DV7 1, 2 &, datagj ; , Null

17 ConformalMapping_16.nb The average NumberForm solgj TableForm, solgjavg Map Mean &, Transpose solgj NumberForm, 12 & , , , , , , which is a quite bad result. Now, instead of using weighting technique, we use a different method to improve this result. Let us compute the value of objective function for every subset solution, objgj Map obj. s 1, a 2, b 3, c 4, X 0 5, Y 0 6, Z 0 7 &, solgj , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

18 18 ConformalMapping_16.nb Select the solution, which has the minimal residual, solgjs solgj First Flatten Position objgj, Min objgj ; solgjs NumberForm, 12 & , , , , , , and do improve it with the Extended Newton- Raphson method, using it as initial guess, Extended Newton- Raphson method GeoAlgebra NewtonExtended AbsoluteTiming solne NewtonExtended sysn,, solgjs Last; , Null solne , , , , , , In order to demonstrate the robustness of the method, let us select the worst result, solgjs solgj First Flatten Position objgj, Max objgj ; solgjs NumberForm, 12 & , , , , , , Again, the solution fast and precise, AbsoluteTiming solne NewtonExtended sysn,, solgjs Last; , Null solne , , , , , , General Procrustes method Another numerical solution technique, the General Procrustes method is also a good candidate for solving the problem, see Chapter 9. We also implemented it as a function in the GeoAlgebra package, GeoAlgebra GeneralProcrustes? GeneralProcrustes Solves the 7 Parameter Datum Transformation Problem computing the scale parameter, s, the translation vector, X0, Y0, Z0 and the rotation matrix, R, as well as the norm of the error matrix, nel. The input data are: Y1 matrix n 3, the coordinates of the image points, xi, yi, zi, Y2 matrix n 3, the coordinates of the object points, Xi, Yi, Zi, W weight matrix n n. Here n is the number of the pairs of points. The output is a list, R, s, X0, Y0, Z0, nel.

19 ConformalMapping_16.nb Now we employ identity matrix for the weigth matrix in order to compare the result with that of the other methods. W IdentityMatrix m 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1 AbsoluteTiming solgp GeneralProcrustes xyz, XYZ, W ; , Null The rotation matrix, solgp 1 MatrixForm The scaling parameter, solgp 2 NumberForm, 12 & The translation vector, The residual, solgp 3 MatrixForm solgp This is the same error norm as was computed with direct global minimization, (Section ), but the computation time is considerably less in that case. Conclusions In case of 3- point problem, Dixon resultant and reduced Groebner basis give the same symbolic result, although the Dixon method is faster than the Groebner method. The computation of the translation vector in least square sense - 3 unknowns and 9 equations - can be achieved by simple averaging because of the linearity of the problem. Employing symbolic solution, a very fast Mathematica function, Conform3DV7 was implemented in the GeoAlgebra package to solve conformal mapping in case of 3-points. Direct numerical solution with Extended Newton- Raphson method is also fast and precise, but it needs initial guess, which can be provided by the Global Numerical Solver, solving any 7 equations from the 9 ones. In all, the best strategy to solve 3-point problem, is the symbolic solution, Conform3DV7 and considering its result as a guess value for the function NewtonExtended. In case of N-point problem, direct global minimization in least square sense is alwasy a good choice, but it can be time consuming. Gauss-Jacobi combinatorial solution is reasonable, if the number of the points - therefore the number of the triplets- is not too high and the technique to solve a triplet is fast. However, the results of the different triplets can differ from each other very considerably. It seems to be the a good strategy to solve a triplet selected randomly, then the result can provide an initial guess for the Extended Newton- Raphson method. Employing General Procrustes algorithm is probably the best choice. It is precise and at least faster than any other methods.