Cory, Bertwin, et al.;The exact volume of an arbitrarily oriented, arbitrary 8-vertex hexahedron can be computed with a very elementary construct.
First, place an average-coordinate point at the center of the hexahedron; this point will be a common vertex point for a set of tetrahedrons whose construction follows.
Second, visiting the six faces of the hexahedron, place an average-coordinate point at the center of the quadrilateral polygonal face. For each of the four triangles, construct the four tetrahedrons using the center point above of the parent hexahedron.
The only limitation is that the arbitrary 8-vertex hexhedron must be star-convex with respect to the hexaheron's average-coordinate center point. This construct works for triangular prisms and pyramids with a quadrilateral base.
Once one has the coordinates of four vertex nodes defining a tetrahedron, the volume calculation of a tetrahedron is straight forward. Clearly, the total volume of the hexahedron is the sum of its 24 tetrahedral decomposition volumes.
If an n-vertex polyhedron finite element's mean quadrature gradient operator is constructed the same way as above, then the above construct works for the n-vertex polyhedron.
If an iso-surface transects the edges of an 8-node hexahedron and the resulting two sibling n-vertex polyhedrons are used as individual finite elements (which has been done), then the the above construct also provides exact volume fractions with respect to the two sibling n-vertex polyhedrons. The polygonal intersection surface need not be planar surface.
If multiple, non-intersecting iso-surfaces transect a hexahedron, then the respective volume fractions within the hexahedron can exactly be calculated inductively.
Cory, If you want to follow up off-line, I can provide a more lengthy write up with references plus example coding. Because of computational costs, I don't recommend this algorithm except in a user-invoked filter.
Samuel W Key FMA Development, LLC 1005 39th Ave NE Great Falls, Montana 59404 On 10/13/2014 9:36 AM, Cory Quammen wrote:
Bertwim, I'm not sure that the volume renderer can handle VTK_HEXAHEDRON elements. Try the "Tetrahedralize" filter on your source and see if the Volume representation works. Thanks, Cory On Mon, Oct 13, 2014 at 10:19 AM, B.W.H. van Beest <b...@xs4all.nl> wrote:Hi, I'm struggling to get a proper 3D view of my model system (yes, embarrassing!) but must admit my defeat. After stripping almost everything, keeping the minimum to exhibit the issue, I'm left with the following: I have created a simple box source. To get a 3D representation, I sub-classed the code for this Source from vtkUnstructuredGridAlgorithm. In the RequestData method, I define the 8 point of the unit cube. I added the points to the underlying unstructed grid, and I specified the cell topology. This all seems to work: when instantiating this box Source, I *do* get the expected cube in the representations (Surface, wireframe, Points). However, when I select the "Volume" representation, *the image disappears* What am I doing wrong? As the code is not too long and very simple, I take the freedom to paste it below. Kind regards. Bertwim ========================================= int sphBoxSourceC::RequestData( vtkInformation *vtkNotUsed(request), vtkInformationVector **vtkNotUsed(inputVector), vtkInformationVector *outputVector) { // Get the info object vtkInformation *outInfo = outputVector->GetInformationObject(0); vtkUnstructuredGrid *umesh = vtkUnstructuredGrid::SafeDownCast( outInfo->Get( vtkDataObject::DATA_OBJECT() ) ); // Pre-allocate some memory umesh->Allocate( 1024 ); // Specify points. double r0[] = { 0.0, 0.0, 0.0 }; double r1[] = { 1.0, 0.0, 0.0 }; double r2[] = { 0.0, 1.0, 0.0 }; double r3[] = { 1.0, 1.0, 0.0 }; double r4[] = { 0.0, 0.0, 1.0 }; double r5[] = { 1.0, 0.0, 1.0 }; double r6[] = { 0.0, 1.0, 1.0 }; double r7[] = { 1.0, 1.0, 1.0 }; // Collect the points in a vtk data structures. { vtkSmartPointer<vtkPoints> points = vtkSmartPointer<vtkPoints>::New(); points->SetDataType( VTK_DOUBLE ); points->InsertNextPoint( r0 ); points->InsertNextPoint( r1 ); points->InsertNextPoint( r2 ); points->InsertNextPoint( r3 ); points->InsertNextPoint( r4 ); points->InsertNextPoint( r5 ); points->InsertNextPoint( r6 ); points->InsertNextPoint( r7 ); // Transfer points to umesh. umesh->SetPoints( points ); } // Cell Topology vtkIdType vtx[8] = { 0, 1, 3, 2, 4, 5, 7, 6 }; umesh->InsertNextCell( VTK_HEXAHEDRON, 8, vtx ); return 1; } _______________________________________________ Powered by www.kitware.com Visit other Kitware open-source projects at http://www.kitware.com/opensource/opensource.html Please keep messages on-topic and check the ParaView Wiki at: http://paraview.org/Wiki/ParaView Follow this link to subscribe/unsubscribe: http://public.kitware.com/mailman/listinfo/paraview_______________________________________________ Powered by www.kitware.com Visit other Kitware open-source projects at http://www.kitware.com/opensource/opensource.html Please keep messages on-topic and check the ParaView Wiki at: http://paraview.org/Wiki/ParaView Follow this link to subscribe/unsubscribe: http://public.kitware.com/mailman/listinfo/paraview
_______________________________________________ Powered by www.kitware.com Visit other Kitware open-source projects at http://www.kitware.com/opensource/opensource.html Please keep messages on-topic and check the ParaView Wiki at: http://paraview.org/Wiki/ParaView Follow this link to subscribe/unsubscribe: http://public.kitware.com/mailman/listinfo/paraview
