Notes
(0000596) Mathieu Malaterre (developer) 2004-02-02 10:08	John, I am not sure I understand your remark. The interpolation are evaluated in the 'VTK' parametric space. The derivatives are also -as I deduce them from the interpolation functions-. I use a c++ code like this to check they are ok: //------------------------------------------------------ #include "vtkQuadraticHexahedron.h" #include <assert.h> int main() { vtkQuadraticHexahedron aHex = vtkQuadraticHexahedron::New (); double sf[20]; double foo = aHex->GetParametricCoords(); for(int i=0;i<60;) { double point[3]; point[0] = foo[i++]; point[1] = foo[i++]; point[2] = foo[i]; // std::cout << "point(" << i/3 << "):" << point[0] << "," << point[1] << "," << point[2] << std::endl; aHex->InterpolationFunctions(point, sf); std::cout << "sf(" << i/3 << "):" << std::endl; for(int j=0;j<20;j++) { std::cout << sf[j] << "," ; if(j == (i/3)) assert( sf[j] == 1); else assert( sf[j] == 0); } std::cout << std::endl; i++; } aHex->Delete(); return 0; } //------------------------------------------------------ You can check that coordinate are from 0 to 1. Thanks for your time, Mathieu

(0000600) Mathieu Malaterre (developer) 2004-02-02 16:25	John answer me this: The first thing vtkQuadraticHexahedron::InterpolationDerivs() does is to transform the VTK parametric coordinates to the iso-P coordinates - double r = 2.0(pcoords[0]-0.5); double s = 2.0(pcoords[1]-0.5); double t = 2.0(pcoords[2]-0.5); The computed gradients therefore represent the change in interpolation function due to a change in iso-P coordinate, not VTK parametric coordinate. If you want to update VTK parametric coordinates using these gradients, you need to apply the inverse transformation. If you like, the interpolation gradients wrt to VTK parametric coordinates, p, are dN/dp = ( dN/dr )( dr/dp ) = 2 dN/dr where, r, are the iso-P coordinates and dN/dr are the gradients given by InterpolationDerivs(). This is the required form of the Jacobian matrix for the Newton correction in VTK parametric coordinates. Rather than scale all the derivatives by 2, the determinant, d, will increase by a factor of 8 but the determinant in the numerator only by a factor of 4 since it contains only 2 columns of derivatives - hence the 0.5. Similar transformations may be required elsewhere so it is probably better to change InterpolationDerivs(). Back to your numerical trial... 1. Take an arbitrary hex20 and evaluate the world coordinates, X1, at some internal point given by VTK parametric coordinates (p,q,r). 2. Chose a close point (p+dp,q,r) and evaluate the world coordinates, X2. 3. Create a finite difference approximation to the gradients - dx/dp = (x2-x1)/dp, dy/dp = (y2-y1)/dp, dz/dp = (z2-z1)/dp 4. Compute the shape function derivatives at (p,q,r) and transform to VTK parametric space by multiplying by 2. In the limit as dp -> 0, the gradients should become identical. I've a lot of explaining to do if this test doesn't work!!

(0000601) Mathieu Malaterre (developer) 2004-02-02 16:26	I'll commit the changes tomorrow.

(0000682) Mathieu Malaterre (developer) 2004-02-20 15:14	Ok, I have just commited the changes.

Issue History
Date Modified	Username	Field	Change
2010-11-29 17:59	Mathieu Malaterre	Source_changeset_attached	=> VTK master 1bd74903
2011-01-13 17:00		Source_changeset_attached	=> VTK master a2bd8391
2011-01-13 17:00		Source_changeset_attached	=> VTK master 020ef709
2011-06-16 13:11	Zack Galbreath	Category	=> (No Category)

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