# 16 Conformal Mapping

```Algebraic Geodesy and Geoinformatics - 2009 - 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.
16- 1 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
[email protected]&quot;Global‘*&quot;D
2
ConformalMapping_16.nb
S=
0 -c b
c
0 -a ;
-b a
0
and
I3 = IdentityMatrix @3D;
Then the rotation matrix,
R = [email protected] - SLD .HI3 + SL  Simplify; [email protected]
1+a2 -b2 -c2
2 a b-2 c
2 Hb+a cL
1+a2 +b2 +c2
1+a2 +b2 +c2
2 Ha b+cL
1+a2 +b2 +c2
1+a2 +b2 +c2
2 H-b+a cL
1+a2 +b2 +c2
2 Ha+b cL
1-a2 -b2 +c2
1+a2 +b2 +c2
1+a2 +b2 +c2
1+a2 +b2 +c2
1-a2 +b2 -c2
-
2 Ha-b cL
1+a2 +b2 +c2
which satisfies the following relation,
I3 == [email protected]  Simplify
True
The transformation has 7 parameters, (a, b, c, X0 , Y0 , Z0 , 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, Pi ,
f3 i-2
f3 i-1
f3 i
xi
= HI3 - SL. yi
zi
Xi
- s HI3 + SL. Yi
Zi
X0
- HI3 - SL. Y0
Z0
xi - X0 - s Xi + c yi - c Y0 + c s Yi - b zi + b Z0 - b s Zi
- c xi + c X0 - c s Xi + yi - Y0 - s Yi + a zi - a Z0 + a s Zi
b xi - b X0 + b s Xi - a yi + a Y0 - a s Yi + zi - Z0 - s Zi
Then for i = 1
f1
f2
f3
=
f3 i-2
f3 i-1
f3 i
. i &reg; 1  Expand;
f1
MatrixFormB f2 F
f3
x1 - X0 - s X1 + c y1 - c Y0 + c s Y1 - b z1 + b Z0 - b s Z1
- c x1 + c X0 - c s X1 + y1 - Y0 - s Y1 + a z1 - a Z0 + a s Z1
b x1 - b X0 + b s X1 - a y1 + a Y0 - a s Y1 + z1 - Z0 - s Z1
for i = 2
f4
f5
f6
=
f3 i-2
f3 i-1
f3 i
. i &reg; 2  Expand;
f4
MatrixFormB f5 F
f6
x2 - X0 - s X2 + c y2 - c Y0 + c s Y2 - b z2 + b Z0 - b s Z2
- c x2 + c X0 - c s X2 + y2 - Y0 - s Y2 + a z2 - a Z0 + a s Z2
b x2 - b X0 + b s X2 - a y2 + a Y0 - a s Y2 + z2 - Z0 - s Z2
for i = 3
f3 i-2
 Expand; MatrixFormB f3 i-1 F
f3 i
ConformalMapping_16.nb
f7
f8
f9
=
f3 i-2
f3 i-1
f3 i
. i &reg; 3  Expand;
f7
MatrixFormB f8 F
f9
x3 - X0 - s X3 + c y3 - c Y0 + c s Y3 - b z3 + b Z0 - b s Z3
- c x3 + c X0 - c s X3 + y3 - Y0 - s Y3 + a z3 - a Z0 + a s Z3
b x3 - b X0 + b s X3 - a y3 + a Y0 - a s Y3 + z3 - Z0 - s Z3
Translation parameters, (X0 , Y0 , Z0 ) can be eliminated by differencing,
f14 = f1 - f4  Simplify
x1 - x2 - s X1 + s X2 + c y1 - c y2 + c s Y1 - c s Y2 - b z1 + b z2 - b s Z1 + b s Z2
f25 = f2 - f5  Simplify
- c x1 + c x2 - c s X1 + c s X2 + y1 - y2 - s Y1 + s Y2 + a z1 - a z2 + a s Z1 - a s Z2
f39 = f3 - f9  Simplify
b x1 - b x3 + b s X1 - b s X3 - a y1 + a y3 - a s Y1 + a s Y3 + z1 - z3 - s Z1 + s Z3
f69 = f6 - f9  Simplify
b x2 - b x3 + b s X2 - b s X3 - a y2 + a y3 - a s Y2 + a s Y3 + z2 - z3 - s Z2 + s Z3
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 = 8x12 &reg;
z12 &reg; z1 - z2 ,
X12 &reg; X1 - X2 ,
Y12 &reg; Y1 - Y2 ,
Z12 &reg; Z1 - Z2 ,
x1 - x2 , x13 &reg;
z13 &reg; z1 - z3 ,
X13 &reg; X1 - X3 ,
Y13 &reg; Y1 - Y3 ,
Z13 &reg; Z1 - Z3 ,
x1 - x3 , x23 &reg; x2 - x3 , y12 &reg; y1 - y2 , y13 &reg; y1 - y3 , y23 &reg; y2 - y3 ,
z23 &reg; z2 - z3 ,
X23 &reg; X2 - X3 ,
Y23 &reg; Y2 - Y3 ,
Z23 &reg; Z2 - Z3 &lt;;
Then our system becomes,
sys = 8x12 - 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&lt;;
Let us check it,
8f14 , f25 , f39 , f69 &lt; - sys . newvars  Simplify
80, 0, 0, 0&lt;
16- 2 Symbolic solution
16- 2- 1 Dixon Resultant
&lt;&lt; Resultant‘Dixon‘
We eliminate the linear parameters, a, b, and c, in order to get an univariate polynomial of s,
3
4
ConformalMapping_16.nb
AbsoluteTiming @solsdx = DixonResultant @sys, 8a, b, c&lt;, 8U, V, W&lt;D  Simplify;D
80.5781250, Null&lt;
solsdx
x122 I- x23 Hy13 + s Y13L + x13 y23 + s H- X23 y13 + X13 y23 + x13 Y23L + s2 H- X23 Y13 + X13 Y23LM +
y12 H- x23 Hy12 y13 + z12 z13L + x13 Hy12 y23 + z12 z23LL +
s IX13 y122 y23 + x13 y122 Y23 + X12 y23 z12 z13 - X23 y12 Hy12 y13 + z12 z13L -
x23 Iy122 Y13 + Y12 z12 z13 + y12 Z12 z13 - y12 z12 Z13M + X13 y12 z12 z23 +
x13 Y12 z12 z23 - X12 y13 z12 z23 + x13 y12 Z12 z23 - x13 y12 z12 Z23M + x12 Hz12 + s Z12L
Iy23 Hz13 - s Z13L - y13 z23 + s HY23 z13 - Y13 z23 + y13 Z23L + s2 H- Y23 Z13 + Y13 Z23LM +
s2 I- X23 y122 Y13 - x13 Y122 y23 + X122 Hx23 y13 - x13 y23L +
X13 y122 Y23 - X23 Y12 z12 z13 - X23 y12 Z12 z13 + X23 y12 z12 Z13 +
x23 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M + 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 HY23 z12 z13 + y23 Z12 z13 - y23 z12 Z13 - Y13 z12 z23 - y13 Z12 z23 + y13 z12 Z23LM +
s3 Ix23 Y122 Y13 - X13 Y122 y23 - x13 Y122 Y23 + X122 HX23 y13 + x23 Y13 - X13 y23 - x13 Y23L +
x23 Y12 Z12 Z13 + X23 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M +
X13 Y12 Z12 z23 - X13 Y12 z12 Z23 - X13 y12 Z12 Z23 - x13 Y12 Z12 Z23 +
X12 HY23 Z12 z13 - Y23 z12 Z13 - y23 Z12 Z13 - Y13 Z12 z23 + Y13 z12 Z23 + y13 Z12 Z23LM +
s4 IX122 HX23 Y13 - X13 Y23L + X12 H- Y23 Z12 Z13 + Y13 Z12 Z23L +
Y12 HX23 HY12 Y13 + Z12 Z13L - X13 HY12 Y23 + Z12 Z23LLM
The result is a quartic polynomial of s,
Exponent[solsdx, {s,a,b,c}]
84, 0, 0, 0&lt;
The coeffients,
q0s = [email protected], s, 0D  Simplify
x122 H- x23 y13 + x13 y23L + x12 z12 Hy23 z13 - y13 z23L +
y12 H- x23 Hy12 y13 + z12 z13L + x13 Hy12 y23 + z12 z23LL
q1s = [email protected], sD  Simplify
- x23 y122 Y13 + X13 y122 y23 + x13 y122 Y23 + x122 H- X23 y13 - x23 Y13 + X13 y23 + x13 Y23L x23 Y12 z12 z13 + X12 y23 z12 z13 - x23 y12 Z12 z13 - X23 y12 Hy12 y13 + z12 z13L + 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 HY23 z12 z13 + y23 Z12 z13 - y23 z12 Z13 - Y13 z12 z23 - y13 Z12 z23 + y13 z12 Z23L
q2s = CoefficientAsolsdx, s2 E  Simplify
- x122 X23 Y13 - X23 y122 Y13 - x13 Y122 y23 + X122 Hx23 y13 - x13 y23L + x122 X13 Y23 +
X13 y122 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 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M +
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 HY23 z12 z13 + y23 Z12 z13 - y23 z12 Z13 - Y13 z12 z23 - y13 Z12 z23 + y13 z12 Z23L
q3s = CoefficientAsolsdx, s3 E  Simplify
x23 Y122 Y13 - X13 Y122 y23 - x13 Y122 Y23 + X122 HX23 y13 + x23 Y13 - X13 y23 - x13 Y23L +
x23 Y12 Z12 Z13 - x12 Y23 Z12 Z13 + X23 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M +
X13 Y12 Z12 z23 - X13 Y12 z12 Z23 - X13 y12 Z12 Z23 - x13 Y12 Z12 Z23 + x12 Y13 Z12 Z23 +
X12 HY23 Z12 z13 - Y23 z12 Z13 - y23 Z12 Z13 - Y13 Z12 z23 + Y13 z12 Z23 + y13 Z12 Z23L
ConformalMapping_16.nb
q4s = CoefficientAsolsdx, s4 E  Simplify
X122 HX23 Y13 - X13 Y23L + X12 H- Y23 Z12 Z13 + Y13 Z12 Z23L +
Y12 HX23 HY12 Y13 + Z12 Z13L - X13 HY12 Y23 + Z12 Z23LL
Let us check the result,
solsdx == q4s s4 + q3s s3 + q2s s2 + 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,
[email protected]@&eth;, 8a, b, c&lt;D &amp;, sysD
880, 1, 1&lt;, 81, 0, 1&lt;, 81, 1, 0&lt;, 81, 1, 0&lt;&lt;
Therefore, we do not take it into consideration when computing a,
soladx = DixonResultant @[email protected], 81, 1&lt;D, 8b, c&lt;, 8V, W&lt;D  Simplify
Hx12 + s X12L
Ia Is X23 y13 + s2 X23 Y13 + x23 Hy13 + s Y13L - x13 y23 - s X13 y23 - s x13 Y23 - s2 X13 Y23M x23 z13 - s X23 z13 + s x23 Z13 + s2 X23 Z13 + x13 z23 + s X13 z23 - s x13 Z23 - s2 X13 Z23M
Similarly, we leave out equations, which does not contain the corresponding parameter in case of b,
solbdx = DixonResultant @[email protected], 82, 2&lt;D, 8a, c&lt;, 8U, W&lt;D  Simplify
Hy12 + s Y12L
Ib Is X23 y13 + s2 X23 Y13 + x23 Hy13 + s Y13L - x13 y23 - s X13 y23 - s x13 Y23 - s2 X13 Y23M y23 z13 - s Y23 z13 + s y23 Z13 + s2 Y23 Z13 + y13 z23 + s Y13 z23 - s y13 Z23 - s2 Y13 Z23M
and c,
solcdx = DixonResultant @[email protected], 83, 3&lt;D, 8a, b&lt;, 8U, V&lt;D  Simplify
Hz12 + s Z12L Ic x23 y12 + y12 y23 + x12 Hx23 + s X23 - c y23 - c s Y23L + z12 z23 +
s Hc X23 y12 + c x23 Y12 - Y12 y23 - X12 Hx23 + c y23L + y12 Y23 + Z12 z23 - z12 Z23L s2 H- c X23 Y12 + Y12 Y23 + X12 HX23 + c Y23L + Z12 Z23LM
16- 2- 2 Reduced Groebner Basis
The similar result can be achieved by reduced Groebner basis.
AbsoluteTiming @8solsgb&lt; = GroebnerBasis @sys, 8s, a, b, c&lt;, 8a, b, c&lt;D  Simplify;D
815.2656250, Null&lt;
5
6
ConformalMapping_16.nb
solsgb
x122 Is X23 y13 + s2 X23 Y13 + x23 Hy13 + s Y13L - x13 y23 - s X13 y23 - s x13 Y23 - s2 X13 Y23M +
y12 Hx23 Hy12 y13 + z12 z13L - x13 Hy12 y23 + z12 z23LL +
s I- X13 y122 y23 - x13 y122 Y23 - X12 y23 z12 z13 + X23 y12 Hy12 y13 + z12 z13L +
x23 Iy122 Y13 + Y12 z12 z13 + y12 Z12 z13 - y12 z12 Z13M - X13 y12 z12 z23 -
x13 Y12 z12 z23 + X12 y13 z12 z23 - x13 y12 Z12 z23 + x13 y12 z12 Z23M + x12 Hz12 + s Z12L
Iy23 H- z13 + s Z13L + y13 z23 - s HY23 z13 - Y13 z23 + y13 Z23L + s2 HY23 Z13 - Y13 Z23LM +
s2 IX23 y122 Y13 + x13 Y122 y23 + X122 H- x23 y13 + x13 y23L -
X13 y122 Y23 + X23 Y12 z12 z13 + X23 y12 Z12 z13 - X23 y12 z12 Z13 x23 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M - 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 H- Y23 z12 z13 - y23 Z12 z13 + y23 z12 Z13 + Y13 z12 z23 + y13 Z12 z23 - y13 z12 Z23LM +
s3 I- x23 Y122 Y13 + X13 Y122 y23 + x13 Y122 Y23 + X122 H- X23 y13 - x23 Y13 + X13 y23 + x13 Y23L x23 Y12 Z12 Z13 - X23 IY122 y13 - Y12 Z12 z13 + Y12 z12 Z13 + y12 Z12 Z13M -
X13 Y12 Z12 z23 + X13 Y12 z12 Z23 + X13 y12 Z12 Z23 + x13 Y12 Z12 Z23 +
X12 H- Y23 Z12 z13 + Y23 z12 Z13 + y23 Z12 Z13 + Y13 Z12 z23 - Y13 z12 Z23 - y13 Z12 Z23LM +
s4 IX122 H- X23 Y13 + X13 Y23L + X12 Z12 HY23 Z13 - Y13 Z23L +
Y12 H- X23 HY12 Y13 + Z12 Z13L + X13 HY12 Y23 + Z12 Z23LLM
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,
8solagb&lt; = GroebnerBasis @[email protected], 81, 1&lt;D, 8a, b, c&lt;, 8b, c&lt;D
9- a x23 y13 - a s X23 y13 - a s x23 Y13 - a s2 X23 Y13 +
a x13 y23 + a s X13 y23 + a s x13 Y23 + a s2 X13 Y23 + x23 z13 + s X23 z13 s x23 Z13 - s2 X23 Z13 - x13 z23 - s X13 z23 + s x13 Z23 + s2 X13 Z23=
The Dixon solution has two factors. The reduced Groebner basis gives the second one,
solagb  - [email protected]@2DD  Simplify
True
Similarly
8solbgb&lt; = GroebnerBasis @[email protected], 82, 2&lt;D, 8a, b, c&lt;, 8a, c&lt;D
9- b x23 y13 - b s X23 y13 - b s x23 Y13 - b s2 X23 Y13 +
b x13 y23 + b s X13 y23 + b s x13 Y23 + b s2 X13 Y23 + y23 z13 + s Y23 z13 s y23 Z13 - s2 Y23 Z13 - y13 z23 - s Y13 z23 + s y13 Z23 + s2 Y13 Z23=
solbgb  - [email protected]@2DD  Simplify
True
8solcgb&lt; = GroebnerBasis @[email protected], 83, 3&lt;D, 8a, b, c&lt;, 8a, b&lt;D
9- x12 x23 + s X12 x23 - s x12 X23 + s2 X12 X23 - c x23 y12 - c s X23 y12 c s x23 Y12 - c s2 X23 Y12 + c x12 y23 + c s X12 y23 - y12 y23 + s Y12 y23 + c s x12 Y23 +
c s2 X12 Y23 - s y12 Y23 + s2 Y12 Y23 - z12 z23 - s Z12 z23 + s z12 Z23 + s2 Z12 Z23=
ConformalMapping_16.nb
solcgb  - [email protected]@2DD  Simplify
True
16- 2- 3 Computation of the translation vector
The translation vector can be computed from the original system consisting of 9 equations,
sysO = [email protected] , 8i, 1, 9&lt;D
8x1 - X0 - s X1 + c y1 - c Y0 + c s Y1 - b z1 + b Z0 - b s Z1 ,
- c x1 + c X0 - c s X1 + y1 - Y0 - s Y1 + a z1 - a Z0 + a s Z1 ,
b x1 - b X0 + b s X1 - a y1 + a Y0 - a s Y1 + z1 - Z0 - s Z1 ,
x2 - X0 - s X2 + c y2 - c Y0 + c s Y2 - b z2 + b Z0 - b s Z2 ,
- c x2 + c X0 - c s X2 + y2 - Y0 - s Y2 + a z2 - a Z0 + a s Z2 ,
b x2 - b X0 + b s X2 - a y2 + a Y0 - a s Y2 + z2 - Z0 - s Z2 ,
x3 - X0 - s X3 + c y3 - c Y0 + c s Y3 - b z3 + b Z0 - b s Z3 ,
- c x3 + c X0 - c s X3 + y3 - Y0 - s Y3 + a z3 - a Z0 + a s Z3 ,
b x3 - b X0 + b s X3 - a y3 + a Y0 - a s Y3 + z3 - Z0 - s Z3 &lt;
Concerning the translation parameters, we have only 3 unknown parameters, but 9 equations. The translation vector,
X0
Xi
Αi
xi
Y0 = yi - s R Yi = Βi
Γi
zi
Z0
Zi
for i = 1, 2, 3. Therefore the coefficient matrix has special structure,
1
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
1
X0
Y0
Z0
Α1
Β1
Γ1
Α2
= Β2
Γ2
Α3
Β3
Γ3
namely
A = [email protected]@IdentityMatrix @3D, 83&lt;D, 1D; [email protected]
1
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
1
the pseudoinverze of A,
7
8
ConformalMapping_16.nb
pIA = PseudoInverse @AD; pIA  MatrixForm
1
1
0 0
3
1
0
3
0 0
3
1
0 0
0 0
1
3
0 0
3
1
0 0
3
1
0 0
1
3
3
0 0
0
1
3
Therefore, the least square solution is a simple averaging, see Gauss-Jacobi combinatorial solution, see Chapter 7. The
solution,
Α1
Β1
Γ1
Α2
pIA. Β2
Γ2
Α3
Β3
Γ3
::
Α1
3
+
Α2
+
3
Α3
3
&gt;, :
Β1
3
+
Β2
3
+
Β3
3
&gt;, :
Γ1
3
+
Γ2
3
+
Γ3
3
&gt;&gt;
or in detailed form,
1
solXYZ0 =
3
::
3 I1 +
x1
y1
z1
X1
- s R. Y1
Z1
1
a2
+ b2 + c2 M
+
x2
y2
z2
X2
- s R. Y2
Z2
+
x3
y3
z3
X3
- s R. Y3
Z3
 Simplify
II1 + a2 + b2 + c2 M x1 + I1 + a2 + b2 + c2 M x2 + x3 +
a2 x3 + b2 x3 + c2 x3 - s X1 - a2 s X1 + b2 s X1 + c2 s X1 - s X2 - a2 s X2 + b2 s X2 +
c2 s X2 - s X3 - a2 s X3 + b2 s X3 + c2 s X3 - 2 a b s Y1 + 2 c s Y1 - 2 a b s Y2 + 2 c s Y2 -
2 a b s Y3 + 2 c s Y3 - 2 b s Z1 - 2 a c s Z1 - 2 b s Z2 - 2 a c s Z2 - 2 b s Z3 - 2 a c s Z3 M&gt;,
:
3 I1 +
1
a2
+ b2 + c2 M
I- 2 Ha b + cL s X1 - 2 Ha b + cL s X2 - 2 a b s X3 - 2 c s X3 + y1 +
a2 y1 + b2 y1 + c2 y1 + y2 + a2 y2 + b2 y2 + c2 y2 + y3 + a2 y3 + b2 y3 + c2 y3 - s Y1 +
a2 s Y1 - b2 s Y1 + c2 s Y1 - s Y2 + a2 s Y2 - b2 s Y2 + c2 s Y2 - s Y3 + a2 s Y3 -
b2 s Y3 + c2 s Y3 + 2 a s Z1 - 2 b c s Z1 + 2 a s Z2 - 2 b c s Z2 + 2 a s Z3 - 2 b c s Z3 M&gt;,
:
1
3 I1 + a2 + b2 + c2 M
I2 Hb - a cL s X1 + 2 Hb - a cL s X2 + 2 b s X3 - 2 a c s X3 - 2 a s Y1 -
2 b c s Y1 - 2 a s Y2 - 2 b c s Y2 - 2 a s Y3 - 2 b c s Y3 + z1 + a2 z1 + b2 z1 + c2 z1 +
z2 + a2 z2 + b2 z2 + c2 z2 + z3 + a2 z3 + b2 z3 + c2 z3 - s Z1 + a2 s Z1 + b2 s Z1 c2 s Z1 - s Z2 + a2 s Z2 + b2 s Z2 - c2 s Z2 - s Z3 + a2 s Z3 + b2 s Z3 - c2 s Z3 M&gt;&gt;
16- 3 Numerical Example for the 3-Point Problem
Let us consider the following data of 3 physical points,
dataC3 =
X2 -&gt;
X3 -&gt;
x1 -&gt;
x2 -&gt;
x3 -&gt;
8X1 -&gt; 423.09, Y1 -&gt; 500.42, Z1 -&gt; 800.02,
423.78, Y2 -&gt; 766.61, Z2 -&gt; 379.80,
465.59, Y3 -&gt; 267.41, Z3 -&gt; 367.23,
183.00, y1 -&gt; 500.40, z1 -&gt; 800.02,
181.53, y2 -&gt; 749.31, z2 -&gt; 402.84,
223.76, y3 -&gt; 279.16, z3 -&gt; 393.28&lt;;
Let us compute the parameter s. The quartic polynomial for s,
ConformalMapping_16.nb
Let us compute the parameter s. The quartic polynomial for s,
eqs = s4 q4s + s3 q3s + s2 q2s + s q1s + q0s . newvars . dataC3
- 2.15763 &acute; 109 - 4.66636 &acute; 109 s - 9.09524 &acute; 107 s2 + 5.25536 &acute; 109 s3 + 2.83912 &acute; 109 s4
Let us normalize it,
eqs = eqs  [email protected], s, 4D  Expand
- 0.759966 - 1.6436 s - 0.0320354 s2 + 1.85105 s3 + 1. s4
The roots,
sols = [email protected], sD  Flatten
8s &reg; - 0.952173, s &reg; - 0.942298, s &reg; - 0.89888, s &reg; 0.942298&lt;
The admissible solution should be positive, real,
s0 = [email protected], [email protected]&eth;@@2DDD  0 &igrave; &eth;@@2DD &gt; 0 &amp;D
8s &reg; 0.942298&lt;
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,
a0 = [email protected] . newvarsL  0, aD  Simplify
99a &reg; I- s X2 z1 + s X3 z1 + x1 z2 + s X1 z2 - s X3 z2 x1 z3 - s X1 z3 + s X2 z3 + s2 X2 Z1 - s2 X3 Z1 - s x1 Z2 - s2 X1 Z2 + s2 X3 Z2 +
x3 Hz1 - z2 - s Z1 + s Z2 L + s x1 Z3 + s2 X1 Z3 - s2 X2 Z3 + x2 H- z1 + z3 + s Z1 - s Z3 LM 
I- s X2 y1 + s X3 y1 + x1 y2 + s X1 y2 - s X3 y2 - x1 y3 - s X1 y3 + s X2 y3 - s2 X2 Y1 +
s2 X3 Y1 + s x1 Y2 + s2 X1 Y2 - s2 X3 Y2 + x3 Hy1 - y2 + s Y1 - s Y2 L s x1 Y3 - s2 X1 Y3 + s2 X2 Y3 + x2 H- y1 + y3 - s Y1 + s Y3 LM==
then
a0 = a0 . s0 . dataC3  Flatten
8a &reg; - 0.00241637&lt;
similarly
b0 = [email protected] . newvarsL  0, bD  Simplify
99b &reg; I- s Y2 z1 + s Y3 z1 + y1 z2 + s Y1 z2 - s Y3 z2 y1 z3 - s Y1 z3 + s Y2 z3 + s2 Y2 Z1 - s2 Y3 Z1 - s y1 Z2 - s2 Y1 Z2 + s2 Y3 Z2 +
y3 Hz1 - z2 - s Z1 + s Z2 L + s y1 Z3 + s2 Y1 Z3 - s2 Y2 Z3 + y2 H- z1 + z3 + s Z1 - s Z3 LM 
I- s X2 y1 + s X3 y1 + x1 y2 + s X1 y2 - s X3 y2 - x1 y3 - s X1 y3 + s X2 y3 - s2 X2 Y1 +
s2 X3 Y1 + s x1 Y2 + s2 X1 Y2 - s2 X3 Y2 + x3 Hy1 - y2 + s Y1 - s Y2 L s x1 Y3 - s2 X1 Y3 + s2 X2 Y3 + x2 H- y1 + y3 - s Y1 + s Y3 LM==
b0 = b0 . s0 . dataC3  Flatten
8b &reg; - 0.000146775&lt;
9
10
ConformalMapping_16.nb
c0 = [email protected] . newvarsL  0, cD  Simplify
99c &reg; Ix22 - s x3 X1 + s x3 X2 + s2 X1 X2 - s2 X22 - s2 X1 X3 + s2 X2 X3 - x2 Hx3 - s X1 + s X3 L +
x1 H- x2 + x3 - s X2 + s X3 L - y1 y2 + y22 + y1 y3 - y2 y3 + s y2 Y1 - s y3 Y1 - s y1 Y2 + s y3 Y2 +
s2 Y1 Y2 - s2 Y22 + s y1 Y3 - s y2 Y3 - s2 Y1 Y3 + s2 Y2 Y3 - z1 z2 + z22 + z1 z3 - z2 z3 - s z2 Z1 +
s z3 Z1 + s z1 Z2 - s z3 Z2 + s2 Z1 Z2 - s2 Z22 - s z1 Z3 + s z2 Z3 - s2 Z1 Z3 + s2 Z2 Z3 M 
Is X2 y1 - s X3 y1 - x1 y2 - s X1 y2 + s X3 y2 + x1 y3 + s X1 y3 - s X2 y3 + s2 X2 Y1 s2 X3 Y1 - s x1 Y2 - s2 X1 Y2 + s2 X3 Y2 + x3 H- y1 + y2 - s Y1 + s Y2 L +
s x1 Y3 + s2 X1 Y3 - s2 X2 Y3 + x2 Hy1 - y3 + s Y1 - s Y3 LM==
c0 = c0 . s0 . dataC3  Flatten
8c &reg; 0.00447553&lt;
The translation vector,
XYZ0 =
[email protected]&eth;1 &reg; &eth;2 &amp;, 88X0 , Y0 , Z0 &lt;, solXYZ0 . dataC3 . s0 . a0 . b0 . c0  Flatten&lt;D
8X0 &reg; - 211.663, Y0 &reg; 21.644, Z0 &reg; 48.3429&lt;
The residium of the equations of the original system,
rs = sysO . dataC3 . s0 . a0 . b0 . c0 . XYZ0
90.460014, 0.0228701, - 1.42109 &acute; 10-14 , 0.460014,
0.0228701, 4.54747 &acute; 10-13 , - 0.920027, - 0.0457402, - 3.2685 &acute; 10-13 =
The residium of the solution,
[email protected]%D
1.12819
This method is implemented as a Mathematica function, Conform3DV7, in the GeoAlgebra package,
&lt;&lt; GeoAlgebra‘Conform3DV7‘
? Conform3DV7
Solves the 7-Parameter Datum Transformation Problem computing
the 'best fitting' parameters of the linear transform, between
systems 8X,Y,Z&lt; -&gt; 8x,y,z&lt; in form
8x,y,z&lt; = s R Ha, b, cL 8X,Y,Z&lt; + 8X0,Y0,Z0&lt;.
The inputs:
- xyz =88x1,y1,z1&lt;, 8x2,y2,z2&lt;, 8x3,y3,z3&lt;&lt;
- XYZ =88X1,Y1,Z1&lt;, 8X1,Y1,Z1&lt;,8X1,Y1,Z1&lt;&lt;.
The result:
8s, a, b, c, X0, Y0, Z0&lt;
- scale,s,
- elements of the skew matrix a, b, c,
- translation vector, X0, Y0, Z0
Let us employ the function,
ConformalMapping_16.nb
11
AbsoluteTiming @sol = [email protected], 500.40, 800.02&lt;,
8181.53, 749.31, 402.84&lt;,
8223.76, 279.16, 393.28&lt;&lt;,
88423.09, 500.42, 800.02&lt;,
8423.78, 766.61, 379.80&lt;,
8465.59, 267.41, 367.23&lt;&lt;D  [email protected]&eth;, 12D &amp;;D
80., Null&lt;
sol
80.942297664317 , -0.0024163658149 , -0.000146774979062 ,
0.00447552595506 , -211.662662588 , 21.6440005507 , 48.3428540977 &lt;
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
8x1 - X0 - s X1 + c y1 - c Y0 + c s Y1 - b z1 + b Z0 - b s Z1 ,
- c x1 + c X0 - c s X1 + y1 - Y0 - s Y1 + a z1 - a Z0 + a s Z1 ,
b x1 - b X0 + b s X1 - a y1 + a Y0 - a s Y1 + z1 - Z0 - s Z1 ,
x2 - X0 - s X2 + c y2 - c Y0 + c s Y2 - b z2 + b Z0 - b s Z2 ,
- c x2 + c X0 - c s X2 + y2 - Y0 - s Y2 + a z2 - a Z0 + a s Z2 ,
b x2 - b X0 + b s X2 - a y2 + a Y0 - a s Y2 + z2 - Z0 - s Z2 ,
x3 - X0 - s X3 + c y3 - c Y0 + c s Y3 - b z3 + b Z0 - b s Z3 ,
- c x3 + c X0 - c s X3 + y3 - Y0 - s Y3 + a z3 - a Z0 + a s Z3 ,
b x3 - b X0 + b s X3 - a y3 + a Y0 - a s Y3 + z3 - Z0 - s Z3 &lt;
Take the first 7 equations,
sysOr = [email protected], 81, 7&lt;D
8x1 - X0 - s X1 + c y1 - c Y0 + c s Y1 - b z1 + b Z0 - b s Z1 ,
- c x1 + c X0 - c s X1 + y1 - Y0 - s Y1 + a z1 - a Z0 + a s Z1 ,
b x1 - b X0 + b s X1 - a y1 + a Y0 - a s Y1 + z1 - Z0 - s Z1 ,
x2 - X0 - s X2 + c y2 - c Y0 + c s Y2 - b z2 + b Z0 - b s Z2 ,
- c x2 + c X0 - c s X2 + y2 - Y0 - s Y2 + a z2 - a Z0 + a s Z2 ,
b x2 - b X0 + b s X2 - a y2 + a Y0 - a s Y2 + z2 - Z0 - s Z2 ,
x3 - X0 - s X3 + c y3 - c Y0 + c s Y3 - b z3 + b Z0 - b s Z3 &lt;
The variable list
X = 8a, b, c, s, X0 , Y0 , Z0 &lt;;
We have two solutions and the first one has considerable &quot;error&quot;,
solabcs = [email protected] . newvars . dataC3, XD
88a &reg; 455.301, b &reg; 38.2629, c &reg; 23.6376, s &reg; - 0.942298,
X0 &reg; 730.149, Y0 &reg; 105.142, Z0 &reg; 97.1707&lt;, 8a &reg; - 0.00241786, b &reg; 0.000764641,
c &reg; 0.00302899, s &reg; 0.942298, X0 &reg; - 213.952, Y0 &reg; 22.807, Z0 &reg; 49.0663&lt;&lt;
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
&quot;error&quot; 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!
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
&quot;error&quot; 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,
&lt;&lt; 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,
X0 = [email protected]&eth;@@2DD &amp;, [email protected]@2DDD
8- 0.00241786, 0.000764641, 0.00302899, 0.942298, - 213.952, 22.807, 49.0663&lt;
The result is somewhat different from the symbolic solution,
AbsoluteTiming @
soln = [email protected]&eth;1 &reg; &eth;2 &amp;, 8X, NewtonExtended @sysO . dataC3, X, X0 D  Last&lt;D;D
80.3281250, Null&lt;
soln
8a &reg; - 0.00244041, b &reg; 0.000757418, c &reg; 0.0030362,
s &reg; 0.942205, X0 &reg; - 213.898, Y0 &reg; 22.8437, Z0 &reg; 49.154&lt;
However, the residium,
rn = sysO . dataC3 . soln
80.00242036, - 0.0293541, 0.00173727, 0.000252836,
0.0130756, - 0.0259387, - 0.00267319, 0.0162785, 0.0242014&lt;
is considerably smaller, see Fig.16.1,
[email protected]%D
0.0507172
and it is distributed nearly uniformly,
ConformalMapping_16.nb
13
[email protected], rn&lt;, Joined &reg; True, PlotRange &reg; All, Frame &reg; TrueD
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
0
2
4
6
8
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,
X0 = [email protected]&eth;@@2DD &amp;, [email protected], a0, b0, c0, XYZ0DD
80.942298, - 0.00241637, - 0.000146775, 0.00447553, - 211.663, 21.644, 48.3429&lt;
then applying Extended Newton- Raphson method,
soln = [email protected]&eth;1 &reg; &eth;2 &amp;, 8X, NewtonExtended @sysO . dataC3, X, X0 D  Last&lt;D
8a &reg; - 0.00244041, b &reg; 0.000757418, c &reg; 0.0030362,
s &reg; 0.942205, X0 &reg; - 213.898, Y0 &reg; 22.8437, Z0 &reg; 49.154&lt;
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
16- 5- 1 The system of equations and data structures
In this case, there are data for more than 3 points at our disposal.
The prototype equation,
xi
e = HI3 - SL. yi
zi
Xi
- s HI3 + SL. Yi
Zi
X0
- HI3 - SL. Y0
Z0
 Expand  Flatten
8xi - X0 - s Xi + c yi - c Y0 + c s Yi - b zi + b Z0 - b s Zi ,
- c xi + c X0 - c s Xi + yi - Y0 - s Yi + a zi - a Z0 + a s Zi ,
b xi - b X0 + b s Xi - a yi + a Y0 - a s Yi + zi - Z0 - s Zi &lt;
In our illlustrative example there are seven points. The system in case of these 7 points,
14
ConformalMapping_16.nb
sys = [email protected], 8i, 1, 7&lt;D  Flatten
8x1 - X0 - s X1 + c y1 - c Y0 + c s Y1 - b z1 + b Z0 - b s Z1 ,
- c x1 + c X0 - c s X1 + y1 - Y0 - s Y1 + a z1 - a Z0 + a s Z1 ,
b x1 - b X0 + b s X1 - a y1 + a Y0 - a s Y1 + z1 - Z0 - s Z1 ,
x2 - X0 - s X2 + c y2 - c Y0 + c s Y2 - b z2 + b Z0 - b s Z2 ,
- c x2 + c X0 - c s X2 + y2 - Y0 - s Y2 + a z2 - a Z0 + a s Z2 ,
b x2 - b X0 + b s X2 - a y2 + a Y0 - a s Y2 + z2 - Z0 - s Z2 ,
x3 - X0 - s X3 + c y3 - c Y0 + c s Y3 - b z3 + b Z0 - b s Z3 ,
- c x3 + c X0 - c s X3 + y3 - Y0 - s Y3 + a z3 - a Z0 + a s Z3 ,
b x3 - b X0 + b s X3 - a y3 + a Y0 - a s Y3 + z3 - Z0 - s Z3 ,
x4 - X0 - s X4 + c y4 - c Y0 + c s Y4 - b z4 + b Z0 - b s Z4 ,
- c x4 + c X0 - c s X4 + y4 - Y0 - s Y4 + a z4 - a Z0 + a s Z4 ,
b x4 - b X0 + b s X4 - a y4 + a Y0 - a s Y4 + z4 - Z0 - s Z4 ,
x5 - X0 - s X5 + c y5 - c Y0 + c s Y5 - b z5 + b Z0 - b s Z5 ,
- c x5 + c X0 - c s X5 + y5 - Y0 - s Y5 + a z5 - a Z0 + a s Z5 ,
b x5 - b X0 + b s X5 - a y5 + a Y0 - a s Y5 + z5 - Z0 - s Z5 ,
x6 - X0 - s X6 + c y6 - c Y0 + c s Y6 - b z6 + b Z0 - b s Z6 ,
- c x6 + c X0 - c s X6 + y6 - Y0 - s Y6 + a z6 - a Z0 + a s Z6 ,
b x6 - b X0 + b s X6 - a y6 + a Y0 - a s Y6 + z6 - Z0 - s Z6 ,
x7 - X0 - s X7 + c y7 - c Y0 + c s Y7 - b z7 + b Z0 - b s Z7 ,
- c x7 + c X0 - c s X7 + y7 - Y0 - s Y7 + a z7 - a Z0 + a s Z7 ,
b x7 - b X0 + b s X7 - a y7 + a Y0 - a s Y7 + z7 - Z0 - s Z7 &lt;
The numerical data are,
xyz =
4 157 870.237
4 149 691.049
4 173 451.354
4 177 796.064
4 137 659.549
4 146 940.228
4 139 407.506
664 818.678
688 865.785
690 369.375
643 026.700
671 837.337
666 982.151
702 700.227
4 775 416.524
4 779 096.588
4 758 594.075
4 761 228.899 ;
4 791 592.531
4 784 324.099
4 786 016.645
XYZ =
4 157 222.543
4 149 043.336
4 172 803.511
4 177 148.376
4 137 012.190
4 146 292.729
4 138 759.902
664 789.307
688 836.443
690 340.078
642 997.635
671 808.029
666 952.887
702 670.738
4 774 952.099
4 778 632.188
4 758 129.701
4 760 764.800 ;
4 791 128.215
4 783 859.856
4 785 552.196
and
where xyz are the coordinates for the WGS-84 system, while XYZ are the coordinates for the local system. In rule form,
xyzR = [email protected]&eth;1 &reg; &eth;[email protected]@1DD, y&eth;1 &reg; &eth;[email protected]@2DD, z&eth;1 &reg; &eth;[email protected]@3DD&lt; &amp;, [email protected], xyz&lt;D  Flatten
9x1 &reg; 4.15787 &acute; 106 , y1 &reg; 664 819., z1 &reg; 4.77542 &acute; 106 , x2 &reg; 4.14969 &acute; 106 ,
y2
x4
y5
z6
&reg;
&reg;
&reg;
&reg;
688 866., z2 &reg; 4.7791 &acute; 106 , x3 &reg; 4.17345 &acute; 106 , y3 &reg; 690 369., z3 &reg; 4.75859 &acute; 106 ,
4.1778 &acute; 106 , y4 &reg; 643 027., z4 &reg; 4.76123 &acute; 106 , x5 &reg; 4.13766 &acute; 106 ,
671 837., z5 &reg; 4.79159 &acute; 106 , x6 &reg; 4.14694 &acute; 106 , y6 &reg; 666 982.,
4.78432 &acute; 106 , x7 &reg; 4.13941 &acute; 106 , y7 &reg; 702 700., z7 &reg; 4.78602 &acute; 106 =
XYZR = [email protected]&eth;1 &reg; &eth;[email protected]@1DD, Y&eth;1 &reg; &eth;[email protected]@2DD, Z&eth;1 &reg; &eth;[email protected]@3DD&lt; &amp;, [email protected], XYZ&lt;D  Flatten
9X1 &reg; 4.15722 &acute; 106 , Y1 &reg; 664 789., Z1 &reg; 4.77495 &acute; 106 , X2 &reg; 4.14904 &acute; 106 ,
Y2
X4
Y5
Z6
&reg;
&reg;
&reg;
&reg;
688 836., Z2 &reg; 4.77863 &acute; 106 , X3 &reg; 4.1728 &acute; 106 , Y3 &reg; 690 340., Z3 &reg; 4.75813 &acute; 106 ,
4.17715 &acute; 106 , Y4 &reg; 642 998., Z4 &reg; 4.76076 &acute; 106 , X5 &reg; 4.13701 &acute; 106 ,
671 808., Z5 &reg; 4.79113 &acute; 106 , X6 &reg; 4.14629 &acute; 106 , Y6 &reg; 666 953.,
4.78386 &acute; 106 , X7 &reg; 4.13876 &acute; 106 , Y7 &reg; 702 671., Z7 &reg; 4.78555 &acute; 106 =
16- 5- 2 Global Minimization
ConformalMapping_16.nb
15
16- 5- 2 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; [email protected], 10D
94.15787 &acute; 106 - 4.77542 &acute; 106 b + 664 819. c - 4.15722 &acute; 106 s - 4.77495 &acute; 106 b s +
664 789. c s - X0 - c Y0 + b Z0 , 664 819. + 4.77542 &acute; 106 a - 4.15787 &acute; 106 c - 664 789. s +
4.77495 &acute; 106 a s - 4.15722 &acute; 106 c s + c X0 - Y0 - a Z0 , 4.77542 &acute; 106 - 664 819. a +
4.15787 &acute; 106 b - 4.77495 &acute; 106 s - 664 789. a s + 4.15722 &acute; 106 b s - b X0 + a Y0 - Z0 ,
16, 702 700. + 4.78602 &acute; 106 a - 4.13941 &acute; 106 c - 702 671. s + 4.78555 &acute; 106 a s 4.13876 &acute; 106 c s + c X0 - Y0 - a Z0 , 4.78602 &acute; 106 - 702 700. a + 4.13941 &acute; 106 b 4.78555 &acute; 106 s - 702 671. a s + 4.13876 &acute; 106 b s - b X0 + a Y0 - Z0 =
The objective function,
obj = [email protected], [email protected]&eth; ^ 2 &amp;, sysnDD; [email protected], 10D
I4.79159 &acute; 106 - 671 837. a + 4.13766 &acute; 106 b -
4.79113 &acute; 106 s - 671 808. a s + 4.13701 &acute; 106 b s - b X0 + a Y0 - Z0 M +
2
I4.78602 &acute; 106 - 702 700. a + 4.13941 &acute; 106 b - 4.78555 &acute; 106 s 702 671. a s + 4.13876 &acute; 106 b s - b X0 + a Y0 - Z0 M +
2
17 + I4.17345 &acute; 106 - 4.75859 &acute; 106 b + 690 369. c - 4.1728 &acute; 106 s 4.75813 &acute; 106 b s + 690 340. c s - X0 - c Y0 + b Z0 M +
2
I4.13941 &acute; 106 - 4.78602 &acute; 106 b + 702 700. c - 4.13876 &acute; 106 s 4.78555 &acute; 106 b s + 702 671. c s - X0 - c Y0 + b Z0 M
2
The result,
AbsoluteTiming @solGM = [email protected], XD;D
80.5312500, Null&lt;
solGM
90.0835105, 9a &reg; 2.42043 &acute; 10-6 , b &reg; - 2.16637 &acute; 10-6 ,
c &reg; - 2.40732 &acute; 10-6 , s &reg; 1.00001, X0 &reg; 641.88, Y0 &reg; 68.6553, Z0 &reg; 416.398==
The rotation matrix,
Rn = R . [email protected]@2DD; [email protected]
4.81463 &acute; 10-6 - 4.33276 &acute; 10-6
1.
- 4.81465 &acute; 10
-6
4.33274 &acute; 10-6
1.
- 4.84085 &acute; 10-6
4.84087 &acute; 10-6
1.
16- 5- 3 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
ConformalMapping_16.nb
mn = [email protected], nD
35
This is not a big number, therefore it is reasonable to use combinatorial solution, especially because the solution of the 3Point Problem is very fast.
These subsets are,
qs = [email protected]@&eth; &amp;, [email protected]@[email protected], 8n&lt;DDD, nD
881,
81,
81,
82,
83,
2,
3,
6,
4,
5,
3&lt;,
6&lt;,
7&lt;,
7&lt;,
6&lt;,
81,
81,
82,
82,
83,
2,
3,
3,
5,
5,
4&lt;,
7&lt;,
4&lt;,
6&lt;,
7&lt;,
81,
81,
82,
82,
83,
2,
4,
3,
5,
6,
5&lt;,
5&lt;,
5&lt;,
7&lt;,
7&lt;,
81,
81,
82,
82,
84,
2,
4,
3,
6,
5,
6&lt;,
6&lt;,
6&lt;,
7&lt;,
6&lt;,
81,
81,
82,
83,
84,
2,
4,
3,
4,
5,
7&lt;,
7&lt;,
7&lt;,
5&lt;,
7&lt;,
81,
81,
82,
83,
84,
3,
5,
4,
4,
6,
4&lt;,
6&lt;,
5&lt;,
6&lt;,
7&lt;,
81,
81,
82,
83,
85,
3,
5,
4,
4,
6,
5&lt;,
7&lt;,
6&lt;,
7&lt;,
7&lt;&lt;
The data values for the subsets can be generated as it follows,
dataGJ = [email protected]@&eth;D &amp;,
[email protected]@@[email protected]@i, jDDDD, [email protected]@[email protected]@i, jDDDD&lt;, 8i, 1, mn&lt;, 8j, 1, n&lt;DD;
[email protected], 10D
99994.15787 &acute; 106 , 664 819., 4.77542 &acute; 106 =, 94.14969 &acute; 106 , 688 866., 4.7791 &acute; 106 =,
94.17345 &acute; 106 , 690 369., 4.75859 &acute; 106 ==, 994.15722 &acute; 106 , 664 789., 4.77495 &acute; 106 =,
94.14904 &acute; 106 , 688 836., 4.77863 &acute; 106 =, 94.1728 &acute; 106 , 690 340., 4.75813 &acute; 106 ===, 33,
9994.13766 &acute; 106 , 671 837., 4.79159 &acute; 106 =, 94.14694 &acute; 106 , 666 982., 4.78432 &acute; 106 =,
94.13941 &acute; 106 , 702 700., 4.78602 &acute; 106 ==, 994.13701 &acute; 106 , 671 808., 4.79113 &acute; 106 =,
94.14629 &acute; 106 , 666 953., 4.78386 &acute; 106 =, 94.13876 &acute; 106 , 702 671., 4.78555 &acute; 106 ====
&lt;&lt; GeoAlgebra‘Conform3DV7‘ ;
The solution of the 35 subsets,
AbsoluteTiming @solGJ = [email protected]@&eth;@@1DD, &eth;@@2DDD &amp;, dataGJD;D
80.2031250, Null&lt;
ConformalMapping_16.nb
17
[email protected]  TableForm, 6D
0.999999 -7.89644 &acute; 10-7 1.11598 &acute; 10-6
0.999999 -5.35996 &acute; 10
-6
0.999999 3.53688 &acute; 10-7
0.999999 2.38348 &acute; 10
-6
1.96505 &acute; 10-8
643.095 22.6163
481.602
-6
517.556 -38.212
599.401
-2.24546 &acute; 10-6 -4.9477 &acute; 10-7
674.343 37.7954
452.134
730.196 64.7628
399.818
0.0000145529
-8.21312 &acute; 10
-6
0.999999 -2.55255 &acute; 10-6 6.29899 &acute; 10-6
1.00000
2.46743 &acute; 10-6
-6
5.09255 &acute; 10-6
-1.40804 &acute; 10
-6
8.12839 &acute; 10-7
594.52
-0.760895 527.04
-5.93527 &acute; 10-6 586.749 101.518
-9.99846 &acute; 10
-7
498.265
1.00000
-2.10379 &acute; 10
1.00000
-5.34206 &acute; 10-6 -7.71392 &acute; 10-6 2.49642 &acute; 10-6
1.00000
1.00001
1.00001
-1.92394 &acute; 10-7 7.30811 &acute; 10-7
0.0000151895
-0.0000121382
0.0000190266
-0.0000163347
-3.06353 &acute; 10-6 631.99 52.2648
-0.0000128003 703.346 273.884
-0.0000155323 739.81 333.255
465.536
293.595
253.605
1.00001
1.00001
-7.11025 &acute; 10-6 0.0000122498
0.0000380055
-0.0000184347
3.07746 &acute; 10-6
-0.0000277583
491.543 -71.1452
760.504 618.886
526.017
230.342
1.00001
2.38451 &acute; 10-6
680.297 41.6576
380.533
1.00000
4.93064 &acute; 10
-6
737.581 67.2038
369.354
1.00000
3.07903 &acute; 10-6
1.00000
3.22055 &acute; 10
-6
635.101 89.5105
436.099
1.00000
3.92191 &acute; 10-6
-7.58078 &acute; 10-7 -4.44507 &acute; 10-6 633.896 101.236
435.501
1.00000
1.54574 &acute; 10-6
-9.08449 &acute; 10-7 -2.39469 &acute; 10-6 638.116 61.5296
437.527
-6
-2.40361 &acute; 10
-6
2.07598 &acute; 10
-6.06425 &acute; 10-6 7.51972 &acute; 10-7
-9.85975 &acute; 10
-6
1.01197 &acute; 10
-6
664.69
-8.11419 &acute; 10-7 -3.71775 &acute; 10-6 635.47
-8.02463 &acute; 10
-1.68488 &acute; 10
-7
-6
-3.83987 &acute; 10
-6
87.135
442.016
402.17
436.22
1.00001
3.2616 &acute; 10
637.744 78.2913
420.066
1.00001
3.89241 &acute; 10-6
-2.71372 &acute; 10-6 -2.98945 &acute; 10-6 647.065 87.6438
410.659
1.00001
0.0000107567
-0.0000139093
-7.35337 &acute; 10-6 748.047 189.463
308.303
1.00001
4.25212 &acute; 10
-6
-7
618.438 128.159
421.246
1.00001
2.55219 &acute; 10-6
-3.20825 &acute; 10-6 -5.77505 &acute; 10-6 643.765 97.2313
403.623
1.00000
3.51389 &acute; 10-6
-3.3106 &acute; 10-6
418.37
1.00000
3.12668 &acute; 10
-6
638.548 70.9403
430.423
1.00000
3.17964 &acute; 10-6
-1.71742 &acute; 10-6 -1.73543 &acute; 10-6 644.127 71.1657
425.534
1.00000
3.18301 &acute; 10-6
-1.75421 &acute; 10-6 -1.73338 &acute; 10-6 644.487 71.2004
425.222
1.00001
4.93025 &acute; 10
-6
1.00001
2.54331 &acute; 10-6
-2.07665 &acute; 10-6 -3.94928 &acute; 10-6 638.758 82.5886
416.868
1.00000
3.09686 &acute; 10-6
-1.39389 &acute; 10-6 -3.88772 &acute; 10-6 641.46
432.236
-6
-8.02273 &acute; 10
-1.14035 &acute; 10
-8.40752 &acute; 10
-7
-6
-7.54063 &acute; 10
-6
-3.91826 &acute; 10-6 661.998 93.4872
-1.76754 &acute; 10
-6.14995 &acute; 10
-6
-6
624.015 123.68
88.8769
-5.77091 &acute; 10
2.74527 &acute; 10-6
-2.28113 &acute; 10-6 -2.24433 &acute; 10-6 633.262 68.8019
403.579
1.00001
3.30644 &acute; 10-6
-2.7641 &acute; 10-6
404.564
1.00001
-6
-2.63389 &acute; 10
583.346 -66.0798
423.887
1.00001
-6
4.19823 &acute; 10
-6
1.00001
3.14131 &acute; 10
3.83191 &acute; 10
-6
-2.58841 &acute; 10
-6
16.8065
720.124 -43.1786
-3.07473 &acute; 10-6 641.79
-1.54197 &acute; 10
-6
81.87
618.544 63.6421
465.602
378.063
The average
solGJavg = [email protected]@&eth;D &amp;, [email protected]  [email protected]&eth;, 12D &amp;
91.00000432576 , 3.5648198311 &acute; 10-6 , -2.42990247217 &acute; 10-6 ,
-3.54841881878 &acute; 10-6 , 648.123456076 , 89.9350637311 , 418.714884865 =
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 = [email protected] . 8s &reg; &eth;@@1DD, a &reg; &eth;@@2DD,
b &reg; &eth;@@3DD, c &reg; &eth;@@4DD, X0 &reg; &eth;@@5DD, Y0 &reg; &eth;@@6DD, Z0 &reg; &eth;@@7DD&lt; &amp;, solGJD
80.692454, 2.97175, 0.465146, 0.80273, 1.10084, 1.98812, 0.450086, 1.72955, 0.450835,
4.31309, 6.65965, 2.26175, 23.2864, 0.65846, 1.3637, 0.201256, 0.189137, 0.228662,
0.126695, 0.101355, 0.12014, 1.56615, 0.585342, 0.3723, 0.163248, 0.153814, 0.164332,
0.147514, 0.392229, 0.190792, 0.188549, 2.16535, 0.127353, 0.111554, 0.434054&lt;
Select the solution, which has the minimal residual,
18
ConformalMapping_16.nb
Select the solution, which has the minimal residual,
solGJs = [email protected]@[email protected]@[email protected], [email protected];
solGJs  [email protected]&eth;, 12D &amp;
91.00000541154 , 3.26159701109 &acute; 10-6 , -1.68487595715 &acute; 10-6 ,
-2.58841274695 &acute; 10-6 , 637.744465387 , 78.2912587174 , 420.065675625 =
and do improve it with the Extended Newton- Raphson method, using it as initial guess,
16- 5- 4 Extended Newton- Raphson method
&lt;&lt; GeoAlgebra‘NewtonExtended‘
AbsoluteTiming @solNE = NewtonExtended @sysn, X, solGJsD  Last;D
80.1875000, Null&lt;
solNE
92.42043 &acute; 10-6 , - 2.16637 &acute; 10-6 , - 2.40732 &acute; 10-6 , 1.00001, 641.88, 68.6553, 416.398=
In order to demonstrate the robustness of the method, let us select the worst result,
solGJs = [email protected]@[email protected]@[email protected], [email protected];
solGJs  [email protected]&eth;, 12D &amp;
81.00000634587 , 0.0000380054712855 , -0.0000184346607387 ,
-0.0000277582684802 , 760.503916369 , 618.886376051 , 230.342364432 &lt;
Again, the solution fast and precise,
AbsoluteTiming @solNE = NewtonExtended @sysn, X, solGJsD  Last;D
80.1875000, Null&lt;
solNE
92.42043 &acute; 10-6 , - 2.16637 &acute; 10-6 , - 2.40732 &acute; 10-6 , 1.00001, 641.88, 68.6553, 416.398=
16- 5- 5 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,
&lt;&lt; GeoAlgebra‘GeneralProcrustes‘
? GeneralProcrustes
Solves the 7-Parameter Datum Transformation Problem computing
the scale parameter, s, the translation vector, 8X0, Y0, Z0&lt;
and the rotation matrix, R, as well as the norm of the error matrix, nEL.
The input data are:
Y1 - matrixHn &acute; 3L, the coordinates of the image points, Hxi, yi, ziL,
Y2 - matrix Hn &acute; 3L, the coordinates of the object points, HXi, Yi, ZiL,
W - weight matrix Hn &acute; nL.
Here n is the number of the pairs of points.
The output is a list, 8R, s, 8X0, Y0, Z0&lt;, nEL&lt;.
Now we employ identity matrix for the weigth matrix in order to compare the result with that of the other methods.
ConformalMapping_16.nb
19
Now we employ identity matrix for the weigth matrix in order to compare the result with that of the other methods.
W = IdentityMatrix @mD
881, 0, 0, 0, 0, 0, 0&lt;, 80, 1, 0, 0, 0, 0, 0&lt;, 80, 0, 1, 0, 0, 0, 0&lt;,
80, 0, 0, 1, 0, 0, 0&lt;, 80, 0, 0, 0, 1, 0, 0&lt;, 80, 0, 0, 0, 0, 1, 0&lt;, 80, 0, 0, 0, 0, 0, 1&lt;&lt;
AbsoluteTiming @solGP = GeneralProcrustes @xyz, XYZ, WD;D
80.0312500, Null&lt;
The rotation matrix,
[email protected]@1DD  MatrixForm
0.9999999999790232905
- 4.8146461561368 &acute; 10
4.3327360257663 &acute; 10
-6
-6
4.8146251818956 &acute; 10-6
- 4.3327593327616 &acute; 10-6
0.9999999999766926608 - 4.8408533143026 &acute; 10-6
4.8408741749041 &acute; 10-6
0.9999999999788966679
The scaling parameter,
[email protected]@2DD  [email protected]&eth;, 12D &amp;
1.00000558252
The translation vector,
[email protected]@3DD  MatrixForm
641.88
68.6553
416.398
The residual,
[email protected]@4DD
0.0835105
This is the same error norm as was computed with direct global minimization, (Section 16- 5- 2), 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.
```