[vtkusers] unstructured grid + tetrahedrons
Jérôme
jerome.velut at gmail.com
Fri Dec 11 07:31:16 EST 2009
David,
You brought me into deep doubts (if plural exists for this noun...): I went
back to vtkPolyData documentation (
http://www.vtk.org/doc/nightly/html/classvtkPolyData.html#_details) and I
noted that 0D, 1D and 2D cells are represented by separate vtkCellArray's.
The flexibility you mention is better exposed in vtkUnstructuredGrid:
"vtkUnstructuredGrid<http://www.vtk.org/doc/nightly/html/classvtkUnstructuredGrid.html>is
a data object that is a concrete implementation of
vtkDataSet <http://www.vtk.org/doc/nightly/html/classvtkDataSet.html>.
vtkUnstructuredGrid<http://www.vtk.org/doc/nightly/html/classvtkUnstructuredGrid.html>represents
any combinations of any cell types. This includes 0D (e.g.,
points), 1D (e.g., lines, polylines), 2D (e.g., triangles, polygons), and 3D
(e.g., hexahedron, tetrahedron).".
Sorry for not confirming! Maybe a VTK developer could correct me if I
misunderstood something...
Jerome
2009/12/11 David Doria <daviddoria+vtk at gmail.com<daviddoria%2Bvtk at gmail.com>
>
> On Fri, Dec 11, 2009 at 4:45 AM, Jérôme <jerome.velut at gmail.com> wrote:
> > You are right, polydata is intended to describe 0D (vertices), 1D (lines)
> > and 2D (polygons) cell. Tetrahedrons are the "simplest" 3D cells, such as
> > triangles are the "simplest" 2D cells, but vtkPolyData cannot have 3D
> cells.
> > However, a tetrahedron is composed of 4 vertices, 6 edges (lines) and 4
> > triangles (polygons) that you can *represents* thanks to a vtkPolyData.
> when
> > you render a tetrahedral mesh through a polydata, you don't "see" the
> > tetrahedrons, but their bounds.
> >
> >
> > Jerome
>
> Jerome -
>
> I'm not sure that is accurate - I believe the whole point of a
> polydata is that it is the most flexible data type in that it is the
> only one that DOES allow 3D cells. I'd have to double check the VTK
> book on that, but maybe someone else can verify this? I've been
> planing to add a brief description of each of these types to the wiki
> for a while so you may have motivated me to get around to it :)
>
> Thanks,
>
> David
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