[GetGlobalPlacement] Issue with the scale computation #159

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1. galou commented on Feb 27, 2023

   

   galou

   on Feb 27, 2023

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   Maybe a solution, given a Matrix m with rotation and scaling, Rotation(m).toMatrix().inverse() * m returns a matrix whose diagonal elements are the scaling factors. The method is not very accurate though (+/- 0.0003) and I don't know how generic it is.
2. Jolbas commented on Feb 27, 2023

   

   Jolbas

   on Feb 27, 2023

   Last edited by Jolbas

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   I don't know how to do the polar decomposition yet. But I know for sure there is no way to distinguish between a matrix representing no scaling at all and one that is a combination of scale(-1,-1,1) and a rotation of 180 degrees. Only option is to assume that all scale factors have the same sign. There is work going on to fix related issues in the creation of a rotation from a matrix. Maybe following code will work for now.

       return_type_link_matrix = 6  # Cf. DocumentObjectPyImp.cpp::getSubObject (l.417).
       matrix = object.getSubObject(subobject_fullpath, return_type_link_matrix,
                                    transform=True)
       if matrix is None:
           return
       scale_type = matrix.hasScale(1e-7)
       if scale_type == ScaleType.Other:
           app.Console.PrintWarning('Too complex transformation to find rotation\n')
           return
       if scale_type == app.ScaleType.NoScaling:
           return app.Placement(matrix), app.Vector(1.0, 1.0, 1.0)
       position = matrix.col(3)
       matrix.setCol(3, app.Vector())
       scale_type = matrix.hasScale(1e-5)
       if scale_type == ScaleType.NonUniformRight:
           matrix.transpose()
       scale_vec = App.Vector(*(m.row(i).Length for i in range(3)))
       matrix.scale(*(1 / s for s in scale_vec))
       if scale_type == ScaleType.NonUniformRight:
           matrix.transpose()
       if matrix.determinant() < 0:
           matrix.scale(-1)
           scale_vec *= -1
       matrix.setCol(3, position)
       return app.Placement(matrix), scale_vec