Fitting of 2d data points with a function considering scaling, rotation and translation












3












$begingroup$


I have the following set of 2d data points:



data1=
{
{21.557, 801.607}, {5.84689, 800.425}, {50.9284, 770.49},
{46.4516, 750.192}, {32.9808, 671.931}, {48.8067, 673.198},
{3.59394, 671.167}, {18.1513, 671.949}, {64.1628, 670.801},
{13.1805, 652.588}, {55.6619, 651.298}, {26.9262, 650.35},
{41.4876, 650.752}, {5.45129, 635.602}, {20.3858, 633.391},
{64.1931, 632.506}, {33.9168, 631.006}, {58.7559, 613.401},
{36.0045, 612.007}, {23.5348, 608.289}, {54.6781, 598.251},
{26.4914, 548.723}, {65.0549, 531.442}, {82.9996, 514.631},
{74.4132, 479.425}, {58.3295, 458.015}, {27.1816, 413.334}
}


I want to apply ScalingTransform, TranslationTransform and RotationTransform to find the best fit to transform data1 into data2, whereby:



data2=
{
{1530.03, 790.2}, {1514.13, 789.}, {1559.17, 758.9},
{1554.5, 738.5}, {1540.5, 660.237}, {1556.15, 661.154},
{1511.34, 659.395}, {1525.63, 660.167}, {1572.13, 658.656},
{1520.66, 640.844}, {1562.55, 639.132}, {1533.79, 638.607},
{1548.37, 638.933}, {1512.62, 623.985}, {1526.88, 621.69},
{1571.44, 620.556}, {1540.44, 618.794}, {1565.69, 601.532},
{1543.06, 600.093}, {1530.22, 596.423}, {1560.9, 586.053},
{1532.93, 536.587}, {1571.9, 519.25}, {1590.15, 501.882},
{1580.39, 467.111}, {1564.73, 445.615}, {1532.8, 400.935}
}


The corresponding points of data1 that should be transformed into data2 are already sorted and at the same position of the lists.



Here are plots of the two data sets:



plot1 = ListPlot[data1, PlotRange -> {{1, 91}, {300, 900}}, 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

plot2 = ListPlot[data2, PlotRange -> {{1510, 1600}, {300, 900}},
PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

GraphicsColumn[{plot1, plot2}, ImageSize -> Large,
Spacings -> {{0, 0}, {0, 50}}]


enter image description here



I use the following naming:



s = ScalingTransform[{sx, sy}, {psx, psy}];
t = TranslationTransform[{vecx, vecy}];
r = RotationTransform[theta, {prx, pry}];


The combined transformation for each point {x, y} of data1 is:



combinedTransformation = s.t.r;


and finally :



combinedTransformation[{x, y}] =

{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


The fitting parameters are: sx, sy, vecx, vecy, theta.



The scaling is centered at the point {psx, psy} and the 2d rotation is around the point {prx, pry}.



I would set {psx, psy} = {1, 1} and {prx, pry} = {1, 1}.



How can I transform data1 best into data2 and how can I obtain the fitting paramaters and their errors?



ADDENDUM:



I already tried the same as what is proposed below by Ulrich Neumann and Carl Lange.




  • The problem with FindGeometricTransform is, it is not described how the error is obtained - I need this for a paper. See this question.


  • Second FindGeometricTransform does not give me the rotation angle and scaling factor in x and y separately, which are not exactly the same.
    FindGeometricTransform shows only the transformation function (or matrix) which is not enough for me.











share|improve this question











$endgroup$








  • 1




    $begingroup$
    Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 11:43










  • $begingroup$
    How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
    $endgroup$
    – JimB
    Jan 9 at 14:28










  • $begingroup$
    If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
    $endgroup$
    – JimB
    Jan 9 at 14:32






  • 1




    $begingroup$
    I answered the question about what the error is in this question of yours.
    $endgroup$
    – Carl Lange
    Jan 9 at 17:57










  • $begingroup$
    Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
    $endgroup$
    – mrz
    yesterday


















3












$begingroup$


I have the following set of 2d data points:



data1=
{
{21.557, 801.607}, {5.84689, 800.425}, {50.9284, 770.49},
{46.4516, 750.192}, {32.9808, 671.931}, {48.8067, 673.198},
{3.59394, 671.167}, {18.1513, 671.949}, {64.1628, 670.801},
{13.1805, 652.588}, {55.6619, 651.298}, {26.9262, 650.35},
{41.4876, 650.752}, {5.45129, 635.602}, {20.3858, 633.391},
{64.1931, 632.506}, {33.9168, 631.006}, {58.7559, 613.401},
{36.0045, 612.007}, {23.5348, 608.289}, {54.6781, 598.251},
{26.4914, 548.723}, {65.0549, 531.442}, {82.9996, 514.631},
{74.4132, 479.425}, {58.3295, 458.015}, {27.1816, 413.334}
}


I want to apply ScalingTransform, TranslationTransform and RotationTransform to find the best fit to transform data1 into data2, whereby:



data2=
{
{1530.03, 790.2}, {1514.13, 789.}, {1559.17, 758.9},
{1554.5, 738.5}, {1540.5, 660.237}, {1556.15, 661.154},
{1511.34, 659.395}, {1525.63, 660.167}, {1572.13, 658.656},
{1520.66, 640.844}, {1562.55, 639.132}, {1533.79, 638.607},
{1548.37, 638.933}, {1512.62, 623.985}, {1526.88, 621.69},
{1571.44, 620.556}, {1540.44, 618.794}, {1565.69, 601.532},
{1543.06, 600.093}, {1530.22, 596.423}, {1560.9, 586.053},
{1532.93, 536.587}, {1571.9, 519.25}, {1590.15, 501.882},
{1580.39, 467.111}, {1564.73, 445.615}, {1532.8, 400.935}
}


The corresponding points of data1 that should be transformed into data2 are already sorted and at the same position of the lists.



Here are plots of the two data sets:



plot1 = ListPlot[data1, PlotRange -> {{1, 91}, {300, 900}}, 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

plot2 = ListPlot[data2, PlotRange -> {{1510, 1600}, {300, 900}},
PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

GraphicsColumn[{plot1, plot2}, ImageSize -> Large,
Spacings -> {{0, 0}, {0, 50}}]


enter image description here



I use the following naming:



s = ScalingTransform[{sx, sy}, {psx, psy}];
t = TranslationTransform[{vecx, vecy}];
r = RotationTransform[theta, {prx, pry}];


The combined transformation for each point {x, y} of data1 is:



combinedTransformation = s.t.r;


and finally :



combinedTransformation[{x, y}] =

{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


The fitting parameters are: sx, sy, vecx, vecy, theta.



The scaling is centered at the point {psx, psy} and the 2d rotation is around the point {prx, pry}.



I would set {psx, psy} = {1, 1} and {prx, pry} = {1, 1}.



How can I transform data1 best into data2 and how can I obtain the fitting paramaters and their errors?



ADDENDUM:



I already tried the same as what is proposed below by Ulrich Neumann and Carl Lange.




  • The problem with FindGeometricTransform is, it is not described how the error is obtained - I need this for a paper. See this question.


  • Second FindGeometricTransform does not give me the rotation angle and scaling factor in x and y separately, which are not exactly the same.
    FindGeometricTransform shows only the transformation function (or matrix) which is not enough for me.











share|improve this question











$endgroup$








  • 1




    $begingroup$
    Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 11:43










  • $begingroup$
    How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
    $endgroup$
    – JimB
    Jan 9 at 14:28










  • $begingroup$
    If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
    $endgroup$
    – JimB
    Jan 9 at 14:32






  • 1




    $begingroup$
    I answered the question about what the error is in this question of yours.
    $endgroup$
    – Carl Lange
    Jan 9 at 17:57










  • $begingroup$
    Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
    $endgroup$
    – mrz
    yesterday
















3












3








3





$begingroup$


I have the following set of 2d data points:



data1=
{
{21.557, 801.607}, {5.84689, 800.425}, {50.9284, 770.49},
{46.4516, 750.192}, {32.9808, 671.931}, {48.8067, 673.198},
{3.59394, 671.167}, {18.1513, 671.949}, {64.1628, 670.801},
{13.1805, 652.588}, {55.6619, 651.298}, {26.9262, 650.35},
{41.4876, 650.752}, {5.45129, 635.602}, {20.3858, 633.391},
{64.1931, 632.506}, {33.9168, 631.006}, {58.7559, 613.401},
{36.0045, 612.007}, {23.5348, 608.289}, {54.6781, 598.251},
{26.4914, 548.723}, {65.0549, 531.442}, {82.9996, 514.631},
{74.4132, 479.425}, {58.3295, 458.015}, {27.1816, 413.334}
}


I want to apply ScalingTransform, TranslationTransform and RotationTransform to find the best fit to transform data1 into data2, whereby:



data2=
{
{1530.03, 790.2}, {1514.13, 789.}, {1559.17, 758.9},
{1554.5, 738.5}, {1540.5, 660.237}, {1556.15, 661.154},
{1511.34, 659.395}, {1525.63, 660.167}, {1572.13, 658.656},
{1520.66, 640.844}, {1562.55, 639.132}, {1533.79, 638.607},
{1548.37, 638.933}, {1512.62, 623.985}, {1526.88, 621.69},
{1571.44, 620.556}, {1540.44, 618.794}, {1565.69, 601.532},
{1543.06, 600.093}, {1530.22, 596.423}, {1560.9, 586.053},
{1532.93, 536.587}, {1571.9, 519.25}, {1590.15, 501.882},
{1580.39, 467.111}, {1564.73, 445.615}, {1532.8, 400.935}
}


The corresponding points of data1 that should be transformed into data2 are already sorted and at the same position of the lists.



Here are plots of the two data sets:



plot1 = ListPlot[data1, PlotRange -> {{1, 91}, {300, 900}}, 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

plot2 = ListPlot[data2, PlotRange -> {{1510, 1600}, {300, 900}},
PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

GraphicsColumn[{plot1, plot2}, ImageSize -> Large,
Spacings -> {{0, 0}, {0, 50}}]


enter image description here



I use the following naming:



s = ScalingTransform[{sx, sy}, {psx, psy}];
t = TranslationTransform[{vecx, vecy}];
r = RotationTransform[theta, {prx, pry}];


The combined transformation for each point {x, y} of data1 is:



combinedTransformation = s.t.r;


and finally :



combinedTransformation[{x, y}] =

{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


The fitting parameters are: sx, sy, vecx, vecy, theta.



The scaling is centered at the point {psx, psy} and the 2d rotation is around the point {prx, pry}.



I would set {psx, psy} = {1, 1} and {prx, pry} = {1, 1}.



How can I transform data1 best into data2 and how can I obtain the fitting paramaters and their errors?



ADDENDUM:



I already tried the same as what is proposed below by Ulrich Neumann and Carl Lange.




  • The problem with FindGeometricTransform is, it is not described how the error is obtained - I need this for a paper. See this question.


  • Second FindGeometricTransform does not give me the rotation angle and scaling factor in x and y separately, which are not exactly the same.
    FindGeometricTransform shows only the transformation function (or matrix) which is not enough for me.











share|improve this question











$endgroup$




I have the following set of 2d data points:



data1=
{
{21.557, 801.607}, {5.84689, 800.425}, {50.9284, 770.49},
{46.4516, 750.192}, {32.9808, 671.931}, {48.8067, 673.198},
{3.59394, 671.167}, {18.1513, 671.949}, {64.1628, 670.801},
{13.1805, 652.588}, {55.6619, 651.298}, {26.9262, 650.35},
{41.4876, 650.752}, {5.45129, 635.602}, {20.3858, 633.391},
{64.1931, 632.506}, {33.9168, 631.006}, {58.7559, 613.401},
{36.0045, 612.007}, {23.5348, 608.289}, {54.6781, 598.251},
{26.4914, 548.723}, {65.0549, 531.442}, {82.9996, 514.631},
{74.4132, 479.425}, {58.3295, 458.015}, {27.1816, 413.334}
}


I want to apply ScalingTransform, TranslationTransform and RotationTransform to find the best fit to transform data1 into data2, whereby:



data2=
{
{1530.03, 790.2}, {1514.13, 789.}, {1559.17, 758.9},
{1554.5, 738.5}, {1540.5, 660.237}, {1556.15, 661.154},
{1511.34, 659.395}, {1525.63, 660.167}, {1572.13, 658.656},
{1520.66, 640.844}, {1562.55, 639.132}, {1533.79, 638.607},
{1548.37, 638.933}, {1512.62, 623.985}, {1526.88, 621.69},
{1571.44, 620.556}, {1540.44, 618.794}, {1565.69, 601.532},
{1543.06, 600.093}, {1530.22, 596.423}, {1560.9, 586.053},
{1532.93, 536.587}, {1571.9, 519.25}, {1590.15, 501.882},
{1580.39, 467.111}, {1564.73, 445.615}, {1532.8, 400.935}
}


The corresponding points of data1 that should be transformed into data2 are already sorted and at the same position of the lists.



Here are plots of the two data sets:



plot1 = ListPlot[data1, PlotRange -> {{1, 91}, {300, 900}}, 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

plot2 = ListPlot[data2, PlotRange -> {{1510, 1600}, {300, 900}},
PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large];

GraphicsColumn[{plot1, plot2}, ImageSize -> Large,
Spacings -> {{0, 0}, {0, 50}}]


enter image description here



I use the following naming:



s = ScalingTransform[{sx, sy}, {psx, psy}];
t = TranslationTransform[{vecx, vecy}];
r = RotationTransform[theta, {prx, pry}];


The combined transformation for each point {x, y} of data1 is:



combinedTransformation = s.t.r;


and finally :



combinedTransformation[{x, y}] =

{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


The fitting parameters are: sx, sy, vecx, vecy, theta.



The scaling is centered at the point {psx, psy} and the 2d rotation is around the point {prx, pry}.



I would set {psx, psy} = {1, 1} and {prx, pry} = {1, 1}.



How can I transform data1 best into data2 and how can I obtain the fitting paramaters and their errors?



ADDENDUM:



I already tried the same as what is proposed below by Ulrich Neumann and Carl Lange.




  • The problem with FindGeometricTransform is, it is not described how the error is obtained - I need this for a paper. See this question.


  • Second FindGeometricTransform does not give me the rotation angle and scaling factor in x and y separately, which are not exactly the same.
    FindGeometricTransform shows only the transformation function (or matrix) which is not enough for me.








fitting






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 9 at 12:38







mrz

















asked Jan 9 at 10:48









mrzmrz

5,65221243




5,65221243








  • 1




    $begingroup$
    Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 11:43










  • $begingroup$
    How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
    $endgroup$
    – JimB
    Jan 9 at 14:28










  • $begingroup$
    If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
    $endgroup$
    – JimB
    Jan 9 at 14:32






  • 1




    $begingroup$
    I answered the question about what the error is in this question of yours.
    $endgroup$
    – Carl Lange
    Jan 9 at 17:57










  • $begingroup$
    Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
    $endgroup$
    – mrz
    yesterday
















  • 1




    $begingroup$
    Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 11:43










  • $begingroup$
    How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
    $endgroup$
    – JimB
    Jan 9 at 14:28










  • $begingroup$
    If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
    $endgroup$
    – JimB
    Jan 9 at 14:32






  • 1




    $begingroup$
    I answered the question about what the error is in this question of yours.
    $endgroup$
    – Carl Lange
    Jan 9 at 17:57










  • $begingroup$
    Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
    $endgroup$
    – mrz
    yesterday










1




1




$begingroup$
Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
$endgroup$
– Ulrich Neumann
Jan 9 at 11:43




$begingroup$
Try FindGeometricTransform . It 's not necessary to require a scaling point and/or a rotationpoint , that is the task ogf the fitting procedure.
$endgroup$
– Ulrich Neumann
Jan 9 at 11:43












$begingroup$
How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
$endgroup$
– JimB
Jan 9 at 14:28




$begingroup$
How do you define "the error"? Is that the mean distance between each point in data2 to the nearest point in the transformed data1? Or the square root of the mean of the square of those distances? Or something else? If the former, then (using @CarlLange 's code) data1Transformed = transform@data1; data2Nearest = Flatten[Nearest[data2, #] & /@ data1Transformed, 1]; Mean[Norm[#] & /@ (data1Transformed - data2)]might do it.
$endgroup$
– JimB
Jan 9 at 14:28












$begingroup$
If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
$endgroup$
– JimB
Jan 9 at 14:32




$begingroup$
If by "their errors" you mean the errors in the individual parameters, you'd need to specify a probabilistic model that generates the transformation parameters. Much like in a linear regression you need not just $y=a+bx$ but $y=a+bx+error$.
$endgroup$
– JimB
Jan 9 at 14:32




1




1




$begingroup$
I answered the question about what the error is in this question of yours.
$endgroup$
– Carl Lange
Jan 9 at 17:57




$begingroup$
I answered the question about what the error is in this question of yours.
$endgroup$
– Carl Lange
Jan 9 at 17:57












$begingroup$
Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
$endgroup$
– mrz
yesterday






$begingroup$
Please see this follow up question: mathematica.stackexchange.com/questions/189592/…
$endgroup$
– mrz
yesterday












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votes


















4












$begingroup$

Try FindGeometricTransform



trafo = FindGeometricTransform[data2, data1 ];
F = TransformationMatrix[trafo[[2]]]


F[[{1, 2}, 3]] is the offset. Matrix



T= F[[{1, 2}, {1,2}]] 


describes rotation and scaling .



S = MatrixPower[ Transpose[T].T , 1/2]  (* scaling matrix*)
(*{{0.970832, -0.00629071}, {-0.00629071, 1.00107}}*)
R = Inverse[Transpose[T]].S (* rotation matrix *)
(*{{0.999918, 0.0128058}, {-0.0128058, 0.999918}}*)
T - R.S // Chop (*T==R.S*)


The scaling factors are given by the eigenvalues of S.
The rotation angle can be obtained by



J = #.# &[Flatten[RotationMatrix[[CurlyPhi]] - R]];
NMinimize[{J, 0 <= [CurlyPhi] <= 2 Pi }, [CurlyPhi]]
(*{4.36514*10^-15, {[CurlyPhi] -> 6.27038}}*)
[CurlyPhi]/Degree /. %[[2]] (* angle in degree*)
(*359.266*)





share|improve this answer











$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
    $endgroup$
    – mrz
    Jan 9 at 13:11






  • 1




    $begingroup$
    Angle is around -.38 Degree.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 13:19








  • 1




    $begingroup$
    That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 15:19






  • 1




    $begingroup$
    @mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
    $endgroup$
    – Ulrich Neumann
    Jan 15 at 11:54



















3












$begingroup$

Per Ulrich Neumann's comment, FindGeometricTransform will do the job very nicely.



We get the transform by doing



transform = FindGeometricTransform[data2, data1][[2]]


This gives us a TransformationFunction, in this case:



$$
text{TransformationFunction}left[left(
begin{array}{ccc}
0.970671 & 0.00652924 & 1502.57 \
-0.0187224 & 1.00107 & -12.4938 \
-0.0000212516 & -text{2.8535460791719293$grave{ }$*${}^{wedge}$-7} & 1. \
end{array}
right)right]
$$



Now we can apply that TransformationFunction to our data and plot the result:



ListPlot[{transform@data1, data2}]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44



















2












$begingroup$

First change the function combinedTransformation to



combinedTransformation[{x_, y_}] = 
{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


and then try



v   = Map[combinedTransformation, data1] - data2;
err = Sum[v[[k]].v[[k]], {k, 1, Length[v]}];
sol = NMinimize[err, {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}]


and the result



enter image description here



The plot was produced as



plot1 = ListPlot[Map[combinedTransformation, data1] /. sol[[2]], 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

plot2 = ListPlot[data2, PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

Show[plot1, plot2]





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
    $endgroup$
    – mrz
    Jan 9 at 13:32












  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44






  • 1




    $begingroup$
    @mrz Included the plotting script
    $endgroup$
    – Cesareo
    Jan 9 at 14:43






  • 1




    $begingroup$
    Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
    $endgroup$
    – Carl Lange
    Jan 10 at 18:10






  • 1




    $begingroup$
    Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
    $endgroup$
    – Carl Lange
    Jan 10 at 23:01



















2












$begingroup$

An alternative approach is to use NonlinearModelFit after reorganizing your data:



ClearAll[trans, model]
trans[sx_, sy_, tx_, ty_, θ_] := Composition[ScalingTransform[{sx, sy}, {1, 1}],
TranslationTransform[{tx, ty}], RotationTransform[θ, {1, 1}]]
model[sx_, sy_, tx_, ty_, θ_][x_] := Module[{h, v},
Flatten@Transpose@CoefficientArrays[trans[sx, sy, tx, ty, θ][{h, v}], {h, v}]. Array[x, 6]]

designmat = ArrayFlatten[{{#, 0}, {0, #}}] &@(Prepend[#, 1] & /@ data1);
response = Join @@ Transpose[data2];
nlm = NonlinearModelFit[Join[designmat, List /@ response, 2],
{model[sx, sy, tx, ty, θ][x], 0 <= θ <= 2 Pi}, {sx, sy, tx, ty, {θ, Pi}},
Array[x, 6]];

Row[{ListPlot[{data2, Transpose[Partition[#, Length[#]/2] &@nlm["PredictedResponse"]]},
PlotStyle -> {Directive[PointSize[Medium], Blue],
Directive[PointSize[.03], Opacity[.4], Red]}, ImageSize -> 400],
MapAt[Style[#, 16] &, nlm["ParameterTable"], {1}]}, Spacer[10]]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    Great. How do you receive this value from theta?
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, theta / Degree transforms theta (in radians) to degrees.
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
    $endgroup$
    – mrz
    yesterday











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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Try FindGeometricTransform



trafo = FindGeometricTransform[data2, data1 ];
F = TransformationMatrix[trafo[[2]]]


F[[{1, 2}, 3]] is the offset. Matrix



T= F[[{1, 2}, {1,2}]] 


describes rotation and scaling .



S = MatrixPower[ Transpose[T].T , 1/2]  (* scaling matrix*)
(*{{0.970832, -0.00629071}, {-0.00629071, 1.00107}}*)
R = Inverse[Transpose[T]].S (* rotation matrix *)
(*{{0.999918, 0.0128058}, {-0.0128058, 0.999918}}*)
T - R.S // Chop (*T==R.S*)


The scaling factors are given by the eigenvalues of S.
The rotation angle can be obtained by



J = #.# &[Flatten[RotationMatrix[[CurlyPhi]] - R]];
NMinimize[{J, 0 <= [CurlyPhi] <= 2 Pi }, [CurlyPhi]]
(*{4.36514*10^-15, {[CurlyPhi] -> 6.27038}}*)
[CurlyPhi]/Degree /. %[[2]] (* angle in degree*)
(*359.266*)





share|improve this answer











$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
    $endgroup$
    – mrz
    Jan 9 at 13:11






  • 1




    $begingroup$
    Angle is around -.38 Degree.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 13:19








  • 1




    $begingroup$
    That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 15:19






  • 1




    $begingroup$
    @mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
    $endgroup$
    – Ulrich Neumann
    Jan 15 at 11:54
















4












$begingroup$

Try FindGeometricTransform



trafo = FindGeometricTransform[data2, data1 ];
F = TransformationMatrix[trafo[[2]]]


F[[{1, 2}, 3]] is the offset. Matrix



T= F[[{1, 2}, {1,2}]] 


describes rotation and scaling .



S = MatrixPower[ Transpose[T].T , 1/2]  (* scaling matrix*)
(*{{0.970832, -0.00629071}, {-0.00629071, 1.00107}}*)
R = Inverse[Transpose[T]].S (* rotation matrix *)
(*{{0.999918, 0.0128058}, {-0.0128058, 0.999918}}*)
T - R.S // Chop (*T==R.S*)


The scaling factors are given by the eigenvalues of S.
The rotation angle can be obtained by



J = #.# &[Flatten[RotationMatrix[[CurlyPhi]] - R]];
NMinimize[{J, 0 <= [CurlyPhi] <= 2 Pi }, [CurlyPhi]]
(*{4.36514*10^-15, {[CurlyPhi] -> 6.27038}}*)
[CurlyPhi]/Degree /. %[[2]] (* angle in degree*)
(*359.266*)





share|improve this answer











$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
    $endgroup$
    – mrz
    Jan 9 at 13:11






  • 1




    $begingroup$
    Angle is around -.38 Degree.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 13:19








  • 1




    $begingroup$
    That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 15:19






  • 1




    $begingroup$
    @mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
    $endgroup$
    – Ulrich Neumann
    Jan 15 at 11:54














4












4








4





$begingroup$

Try FindGeometricTransform



trafo = FindGeometricTransform[data2, data1 ];
F = TransformationMatrix[trafo[[2]]]


F[[{1, 2}, 3]] is the offset. Matrix



T= F[[{1, 2}, {1,2}]] 


describes rotation and scaling .



S = MatrixPower[ Transpose[T].T , 1/2]  (* scaling matrix*)
(*{{0.970832, -0.00629071}, {-0.00629071, 1.00107}}*)
R = Inverse[Transpose[T]].S (* rotation matrix *)
(*{{0.999918, 0.0128058}, {-0.0128058, 0.999918}}*)
T - R.S // Chop (*T==R.S*)


The scaling factors are given by the eigenvalues of S.
The rotation angle can be obtained by



J = #.# &[Flatten[RotationMatrix[[CurlyPhi]] - R]];
NMinimize[{J, 0 <= [CurlyPhi] <= 2 Pi }, [CurlyPhi]]
(*{4.36514*10^-15, {[CurlyPhi] -> 6.27038}}*)
[CurlyPhi]/Degree /. %[[2]] (* angle in degree*)
(*359.266*)





share|improve this answer











$endgroup$



Try FindGeometricTransform



trafo = FindGeometricTransform[data2, data1 ];
F = TransformationMatrix[trafo[[2]]]


F[[{1, 2}, 3]] is the offset. Matrix



T= F[[{1, 2}, {1,2}]] 


describes rotation and scaling .



S = MatrixPower[ Transpose[T].T , 1/2]  (* scaling matrix*)
(*{{0.970832, -0.00629071}, {-0.00629071, 1.00107}}*)
R = Inverse[Transpose[T]].S (* rotation matrix *)
(*{{0.999918, 0.0128058}, {-0.0128058, 0.999918}}*)
T - R.S // Chop (*T==R.S*)


The scaling factors are given by the eigenvalues of S.
The rotation angle can be obtained by



J = #.# &[Flatten[RotationMatrix[[CurlyPhi]] - R]];
NMinimize[{J, 0 <= [CurlyPhi] <= 2 Pi }, [CurlyPhi]]
(*{4.36514*10^-15, {[CurlyPhi] -> 6.27038}}*)
[CurlyPhi]/Degree /. %[[2]] (* angle in degree*)
(*359.266*)






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 14 at 12:12

























answered Jan 9 at 12:30









Ulrich NeumannUlrich Neumann

7,960516




7,960516












  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
    $endgroup$
    – mrz
    Jan 9 at 13:11






  • 1




    $begingroup$
    Angle is around -.38 Degree.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 13:19








  • 1




    $begingroup$
    That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 15:19






  • 1




    $begingroup$
    @mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
    $endgroup$
    – Ulrich Neumann
    Jan 15 at 11:54


















  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
    $endgroup$
    – mrz
    Jan 9 at 13:11






  • 1




    $begingroup$
    Angle is around -.38 Degree.
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 13:19








  • 1




    $begingroup$
    That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
    $endgroup$
    – Ulrich Neumann
    Jan 9 at 15:19






  • 1




    $begingroup$
    @mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
    $endgroup$
    – Ulrich Neumann
    Jan 15 at 11:54
















$begingroup$
Please see the addendum.
$endgroup$
– mrz
Jan 9 at 12:33




$begingroup$
Please see the addendum.
$endgroup$
– mrz
Jan 9 at 12:33












$begingroup$
Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
$endgroup$
– mrz
Jan 9 at 13:11




$begingroup$
Thank you very much. What is the angle in degree (should be in the order of 0.5 degree)? The scaling factors should be about 1,004 in y and about 1.001 in x (data2 are larger by this factors). Do you understand what the error of FindGeometricTransform exactly means?
$endgroup$
– mrz
Jan 9 at 13:11




1




1




$begingroup$
Angle is around -.38 Degree.
$endgroup$
– Ulrich Neumann
Jan 9 at 13:19






$begingroup$
Angle is around -.38 Degree.
$endgroup$
– Ulrich Neumann
Jan 9 at 13:19






1




1




$begingroup$
That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
$endgroup$
– Ulrich Neumann
Jan 9 at 15:19




$begingroup$
That's because the general mapping used by GeometericTransform maps vectors x->y in the form y=(A.x+b)/(c.x+d) and describes the central projection completly. The affine case c=0,d=1 is only a rough approximation!
$endgroup$
– Ulrich Neumann
Jan 9 at 15:19




1




1




$begingroup$
@mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
$endgroup$
– Ulrich Neumann
Jan 15 at 11:54




$begingroup$
@mrz Usually the mean of nondiagonal elements of S describe shear: (S[[1,2]]+S[2,1])/2, (S[[1,3]]+S[3,1])/2,(S[[2,3]]+S[3,2])/2...
$endgroup$
– Ulrich Neumann
Jan 15 at 11:54











3












$begingroup$

Per Ulrich Neumann's comment, FindGeometricTransform will do the job very nicely.



We get the transform by doing



transform = FindGeometricTransform[data2, data1][[2]]


This gives us a TransformationFunction, in this case:



$$
text{TransformationFunction}left[left(
begin{array}{ccc}
0.970671 & 0.00652924 & 1502.57 \
-0.0187224 & 1.00107 & -12.4938 \
-0.0000212516 & -text{2.8535460791719293$grave{ }$*${}^{wedge}$-7} & 1. \
end{array}
right)right]
$$



Now we can apply that TransformationFunction to our data and plot the result:



ListPlot[{transform@data1, data2}]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44
















3












$begingroup$

Per Ulrich Neumann's comment, FindGeometricTransform will do the job very nicely.



We get the transform by doing



transform = FindGeometricTransform[data2, data1][[2]]


This gives us a TransformationFunction, in this case:



$$
text{TransformationFunction}left[left(
begin{array}{ccc}
0.970671 & 0.00652924 & 1502.57 \
-0.0187224 & 1.00107 & -12.4938 \
-0.0000212516 & -text{2.8535460791719293$grave{ }$*${}^{wedge}$-7} & 1. \
end{array}
right)right]
$$



Now we can apply that TransformationFunction to our data and plot the result:



ListPlot[{transform@data1, data2}]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44














3












3








3





$begingroup$

Per Ulrich Neumann's comment, FindGeometricTransform will do the job very nicely.



We get the transform by doing



transform = FindGeometricTransform[data2, data1][[2]]


This gives us a TransformationFunction, in this case:



$$
text{TransformationFunction}left[left(
begin{array}{ccc}
0.970671 & 0.00652924 & 1502.57 \
-0.0187224 & 1.00107 & -12.4938 \
-0.0000212516 & -text{2.8535460791719293$grave{ }$*${}^{wedge}$-7} & 1. \
end{array}
right)right]
$$



Now we can apply that TransformationFunction to our data and plot the result:



ListPlot[{transform@data1, data2}]


enter image description here






share|improve this answer









$endgroup$



Per Ulrich Neumann's comment, FindGeometricTransform will do the job very nicely.



We get the transform by doing



transform = FindGeometricTransform[data2, data1][[2]]


This gives us a TransformationFunction, in this case:



$$
text{TransformationFunction}left[left(
begin{array}{ccc}
0.970671 & 0.00652924 & 1502.57 \
-0.0187224 & 1.00107 & -12.4938 \
-0.0000212516 & -text{2.8535460791719293$grave{ }$*${}^{wedge}$-7} & 1. \
end{array}
right)right]
$$



Now we can apply that TransformationFunction to our data and plot the result:



ListPlot[{transform@data1, data2}]


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 9 at 12:21









Carl LangeCarl Lange

2,7061727




2,7061727












  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44


















  • $begingroup$
    Please see the addendum.
    $endgroup$
    – mrz
    Jan 9 at 12:33










  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44
















$begingroup$
Please see the addendum.
$endgroup$
– mrz
Jan 9 at 12:33




$begingroup$
Please see the addendum.
$endgroup$
– mrz
Jan 9 at 12:33












$begingroup$
Please read also my last comment to Ulrich Neumann.
$endgroup$
– mrz
Jan 9 at 13:44




$begingroup$
Please read also my last comment to Ulrich Neumann.
$endgroup$
– mrz
Jan 9 at 13:44











2












$begingroup$

First change the function combinedTransformation to



combinedTransformation[{x_, y_}] = 
{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


and then try



v   = Map[combinedTransformation, data1] - data2;
err = Sum[v[[k]].v[[k]], {k, 1, Length[v]}];
sol = NMinimize[err, {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}]


and the result



enter image description here



The plot was produced as



plot1 = ListPlot[Map[combinedTransformation, data1] /. sol[[2]], 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

plot2 = ListPlot[data2, PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

Show[plot1, plot2]





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
    $endgroup$
    – mrz
    Jan 9 at 13:32












  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44






  • 1




    $begingroup$
    @mrz Included the plotting script
    $endgroup$
    – Cesareo
    Jan 9 at 14:43






  • 1




    $begingroup$
    Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
    $endgroup$
    – Carl Lange
    Jan 10 at 18:10






  • 1




    $begingroup$
    Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
    $endgroup$
    – Carl Lange
    Jan 10 at 23:01
















2












$begingroup$

First change the function combinedTransformation to



combinedTransformation[{x_, y_}] = 
{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


and then try



v   = Map[combinedTransformation, data1] - data2;
err = Sum[v[[k]].v[[k]], {k, 1, Length[v]}];
sol = NMinimize[err, {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}]


and the result



enter image description here



The plot was produced as



plot1 = ListPlot[Map[combinedTransformation, data1] /. sol[[2]], 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

plot2 = ListPlot[data2, PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

Show[plot1, plot2]





share|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
    $endgroup$
    – mrz
    Jan 9 at 13:32












  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44






  • 1




    $begingroup$
    @mrz Included the plotting script
    $endgroup$
    – Cesareo
    Jan 9 at 14:43






  • 1




    $begingroup$
    Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
    $endgroup$
    – Carl Lange
    Jan 10 at 18:10






  • 1




    $begingroup$
    Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
    $endgroup$
    – Carl Lange
    Jan 10 at 23:01














2












2








2





$begingroup$

First change the function combinedTransformation to



combinedTransformation[{x_, y_}] = 
{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


and then try



v   = Map[combinedTransformation, data1] - data2;
err = Sum[v[[k]].v[[k]], {k, 1, Length[v]}];
sol = NMinimize[err, {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}]


and the result



enter image description here



The plot was produced as



plot1 = ListPlot[Map[combinedTransformation, data1] /. sol[[2]], 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

plot2 = ListPlot[data2, PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

Show[plot1, plot2]





share|improve this answer











$endgroup$



First change the function combinedTransformation to



combinedTransformation[{x_, y_}] = 
{sx (prx (-Cos[theta]) + prx + pry Sin[theta]) + psx (-sx) + psx +
sx x Cos[theta] - sx y Sin[theta] + sx vecx,
sy (-(prx Sin[theta]) + pry (-Cos[theta]) + pry) + psy (-sy) + psy +
sy x Sin[theta] + sy y Cos[theta] + sy vecy}


and then try



v   = Map[combinedTransformation, data1] - data2;
err = Sum[v[[k]].v[[k]], {k, 1, Length[v]}];
sol = NMinimize[err, {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}]


and the result



enter image description here



The plot was produced as



plot1 = ListPlot[Map[combinedTransformation, data1] /. sol[[2]], 
PlotStyle -> Red, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

plot2 = ListPlot[data2, PlotStyle -> Blue, Frame -> True,
FrameLabel -> {{"y", ""}, {"x", "data1 and data2"}},
BaseStyle -> {FontWeight -> "Bold", FontSize -> 15,
FontFamily -> "Calibri"}, ImageSize -> Large]

Show[plot1, plot2]






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 9 at 14:42

























answered Jan 9 at 12:04









CesareoCesareo

3014




3014












  • $begingroup$
    Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
    $endgroup$
    – mrz
    Jan 9 at 13:32












  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44






  • 1




    $begingroup$
    @mrz Included the plotting script
    $endgroup$
    – Cesareo
    Jan 9 at 14:43






  • 1




    $begingroup$
    Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
    $endgroup$
    – Carl Lange
    Jan 10 at 18:10






  • 1




    $begingroup$
    Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
    $endgroup$
    – Carl Lange
    Jan 10 at 23:01


















  • $begingroup$
    Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
    $endgroup$
    – mrz
    Jan 9 at 13:32












  • $begingroup$
    Please read also my last comment to Ulrich Neumann.
    $endgroup$
    – mrz
    Jan 9 at 13:44






  • 1




    $begingroup$
    @mrz Included the plotting script
    $endgroup$
    – Cesareo
    Jan 9 at 14:43






  • 1




    $begingroup$
    Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
    $endgroup$
    – Carl Lange
    Jan 10 at 18:10






  • 1




    $begingroup$
    Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
    $endgroup$
    – Carl Lange
    Jan 10 at 23:01
















$begingroup$
Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
$endgroup$
– mrz
Jan 9 at 13:32






$begingroup$
Thanks a lot for this solution. You get sx -> 1.00395, sy -> 1.00219, which is what I expect. Theta (theta -> -0.00683116) is probably in radian which is 0.39 degree, also what I expect. Only the translation surprises me a little bit: vecx -> 1513.99, vecy -> -7.23374, compared to the results of Ulrich Neumann and Carl Lange (vecx ca. 1503 and vecy ca. -13). I trust their vecy value because of this estimation vecy = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95. On the other hand: vecx=Mean[data2[[All, 1]] - data1[[All,1]]]=1507.11 is between your result and their.
$endgroup$
– mrz
Jan 9 at 13:32














$begingroup$
Please read also my last comment to Ulrich Neumann.
$endgroup$
– mrz
Jan 9 at 13:44




$begingroup$
Please read also my last comment to Ulrich Neumann.
$endgroup$
– mrz
Jan 9 at 13:44




1




1




$begingroup$
@mrz Included the plotting script
$endgroup$
– Cesareo
Jan 9 at 14:43




$begingroup$
@mrz Included the plotting script
$endgroup$
– Cesareo
Jan 9 at 14:43




1




1




$begingroup$
Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
$endgroup$
– Carl Lange
Jan 10 at 18:10




$begingroup$
Consider using FindMinimum instead of NMinimize: {er, ru} = FindMinimum[ Mean[MapThread[ EuclideanDistance, {combinedTransformation /@ data1, data2}]], {prx, pry, psx, psy, sx, sy, vecx, vecy, x, y, theta}, MaxIterations -> 200] gives roughly the same error as FindGeometricTransform.
$endgroup$
– Carl Lange
Jan 10 at 18:10




1




1




$begingroup$
Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
$endgroup$
– Carl Lange
Jan 10 at 23:01




$begingroup$
Well, the mean euclidean distance is minimized to the same amount as the FindGeometricTransform one. I don't know that the values are strange, I suppose you have better knowledge. You could consider giving the various variables default conditions if you have a decent idea of what the values might be. It's unclear to me that the message is specifically an error message - it looks like it's informational.
$endgroup$
– Carl Lange
Jan 10 at 23:01











2












$begingroup$

An alternative approach is to use NonlinearModelFit after reorganizing your data:



ClearAll[trans, model]
trans[sx_, sy_, tx_, ty_, θ_] := Composition[ScalingTransform[{sx, sy}, {1, 1}],
TranslationTransform[{tx, ty}], RotationTransform[θ, {1, 1}]]
model[sx_, sy_, tx_, ty_, θ_][x_] := Module[{h, v},
Flatten@Transpose@CoefficientArrays[trans[sx, sy, tx, ty, θ][{h, v}], {h, v}]. Array[x, 6]]

designmat = ArrayFlatten[{{#, 0}, {0, #}}] &@(Prepend[#, 1] & /@ data1);
response = Join @@ Transpose[data2];
nlm = NonlinearModelFit[Join[designmat, List /@ response, 2],
{model[sx, sy, tx, ty, θ][x], 0 <= θ <= 2 Pi}, {sx, sy, tx, ty, {θ, Pi}},
Array[x, 6]];

Row[{ListPlot[{data2, Transpose[Partition[#, Length[#]/2] &@nlm["PredictedResponse"]]},
PlotStyle -> {Directive[PointSize[Medium], Blue],
Directive[PointSize[.03], Opacity[.4], Red]}, ImageSize -> 400],
MapAt[Style[#, 16] &, nlm["ParameterTable"], {1}]}, Spacer[10]]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    Great. How do you receive this value from theta?
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, theta / Degree transforms theta (in radians) to degrees.
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
    $endgroup$
    – mrz
    yesterday
















2












$begingroup$

An alternative approach is to use NonlinearModelFit after reorganizing your data:



ClearAll[trans, model]
trans[sx_, sy_, tx_, ty_, θ_] := Composition[ScalingTransform[{sx, sy}, {1, 1}],
TranslationTransform[{tx, ty}], RotationTransform[θ, {1, 1}]]
model[sx_, sy_, tx_, ty_, θ_][x_] := Module[{h, v},
Flatten@Transpose@CoefficientArrays[trans[sx, sy, tx, ty, θ][{h, v}], {h, v}]. Array[x, 6]]

designmat = ArrayFlatten[{{#, 0}, {0, #}}] &@(Prepend[#, 1] & /@ data1);
response = Join @@ Transpose[data2];
nlm = NonlinearModelFit[Join[designmat, List /@ response, 2],
{model[sx, sy, tx, ty, θ][x], 0 <= θ <= 2 Pi}, {sx, sy, tx, ty, {θ, Pi}},
Array[x, 6]];

Row[{ListPlot[{data2, Transpose[Partition[#, Length[#]/2] &@nlm["PredictedResponse"]]},
PlotStyle -> {Directive[PointSize[Medium], Blue],
Directive[PointSize[.03], Opacity[.4], Red]}, ImageSize -> 400],
MapAt[Style[#, 16] &, nlm["ParameterTable"], {1}]}, Spacer[10]]


enter image description here






share|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    Great. How do you receive this value from theta?
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, theta / Degree transforms theta (in radians) to degrees.
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
    $endgroup$
    – mrz
    yesterday














2












2








2





$begingroup$

An alternative approach is to use NonlinearModelFit after reorganizing your data:



ClearAll[trans, model]
trans[sx_, sy_, tx_, ty_, θ_] := Composition[ScalingTransform[{sx, sy}, {1, 1}],
TranslationTransform[{tx, ty}], RotationTransform[θ, {1, 1}]]
model[sx_, sy_, tx_, ty_, θ_][x_] := Module[{h, v},
Flatten@Transpose@CoefficientArrays[trans[sx, sy, tx, ty, θ][{h, v}], {h, v}]. Array[x, 6]]

designmat = ArrayFlatten[{{#, 0}, {0, #}}] &@(Prepend[#, 1] & /@ data1);
response = Join @@ Transpose[data2];
nlm = NonlinearModelFit[Join[designmat, List /@ response, 2],
{model[sx, sy, tx, ty, θ][x], 0 <= θ <= 2 Pi}, {sx, sy, tx, ty, {θ, Pi}},
Array[x, 6]];

Row[{ListPlot[{data2, Transpose[Partition[#, Length[#]/2] &@nlm["PredictedResponse"]]},
PlotStyle -> {Directive[PointSize[Medium], Blue],
Directive[PointSize[.03], Opacity[.4], Red]}, ImageSize -> 400],
MapAt[Style[#, 16] &, nlm["ParameterTable"], {1}]}, Spacer[10]]


enter image description here






share|improve this answer









$endgroup$



An alternative approach is to use NonlinearModelFit after reorganizing your data:



ClearAll[trans, model]
trans[sx_, sy_, tx_, ty_, θ_] := Composition[ScalingTransform[{sx, sy}, {1, 1}],
TranslationTransform[{tx, ty}], RotationTransform[θ, {1, 1}]]
model[sx_, sy_, tx_, ty_, θ_][x_] := Module[{h, v},
Flatten@Transpose@CoefficientArrays[trans[sx, sy, tx, ty, θ][{h, v}], {h, v}]. Array[x, 6]]

designmat = ArrayFlatten[{{#, 0}, {0, #}}] &@(Prepend[#, 1] & /@ data1);
response = Join @@ Transpose[data2];
nlm = NonlinearModelFit[Join[designmat, List /@ response, 2],
{model[sx, sy, tx, ty, θ][x], 0 <= θ <= 2 Pi}, {sx, sy, tx, ty, {θ, Pi}},
Array[x, 6]];

Row[{ListPlot[{data2, Transpose[Partition[#, Length[#]/2] &@nlm["PredictedResponse"]]},
PlotStyle -> {Directive[PointSize[Medium], Blue],
Directive[PointSize[.03], Opacity[.4], Red]}, ImageSize -> 400],
MapAt[Style[#, 16] &, nlm["ParameterTable"], {1}]}, Spacer[10]]


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









kglrkglr

179k9199410




179k9199410












  • $begingroup$
    This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    Great. How do you receive this value from theta?
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, theta / Degree transforms theta (in radians) to degrees.
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
    $endgroup$
    – mrz
    yesterday


















  • $begingroup$
    This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    Great. How do you receive this value from theta?
    $endgroup$
    – mrz
    yesterday








  • 1




    $begingroup$
    @mrz, theta / Degree transforms theta (in radians) to degrees.
    $endgroup$
    – kglr
    yesterday










  • $begingroup$
    All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
    $endgroup$
    – mrz
    yesterday
















$begingroup$
This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
$endgroup$
– mrz
yesterday






$begingroup$
This is a very interesting solution. Thanky you. Only th angle I do not understand: In the two solutions below both get Ulrich Neumann and Cesareo get an angle of about 0.38 to 0.39 degree.
$endgroup$
– mrz
yesterday






1




1




$begingroup$
@mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
$endgroup$
– kglr
yesterday




$begingroup$
@mrz, the estimated angle is 359.607 Degrees (6.27632/Degree).
$endgroup$
– kglr
yesterday












$begingroup$
Great. How do you receive this value from theta?
$endgroup$
– mrz
yesterday






$begingroup$
Great. How do you receive this value from theta?
$endgroup$
– mrz
yesterday






1




1




$begingroup$
@mrz, theta / Degree transforms theta (in radians) to degrees.
$endgroup$
– kglr
yesterday




$begingroup$
@mrz, theta / Degree transforms theta (in radians) to degrees.
$endgroup$
– kglr
yesterday












$begingroup$
All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
$endgroup$
– mrz
yesterday




$begingroup$
All values which you receive are similar as of Ulrich Neumann and Cesareo except for the translation. Ulrich Neumann got {1502.57, -13.1302} and Cesareo got {1513.99, -7.23374}. You got {1496.06, -13.01}. My approximate estimation for the vertical translation is = Mean[data2[[All, 2]] - data1[[All, 2]]]=-11.95 and for the horizontal translation = Mean[data2[[All, 1]] - data1[[All,1]]]=1507.1.Which values of the there solutions are most reliable?
$endgroup$
– mrz
yesterday


















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