[Paraview] Disappearing volume rendering?

Mon Oct 13 14:29:34 EDT 2014

Samuel,

Thanks for taking the efforts to dive into this.
In fact, it is a good reminder to me to try a single tetraeder first. My
primary problem seems to be that
I cannot get volume rendering to work.

Kind Regards,
Bertwim

On 10/13/2014 07:17 PM, Samuel Key wrote:
> 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 <bwvb at 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;
>>> }
>>>
>>>
>>>
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