[Paraview] problems with volume rendering

[David E DeMarle]

David E DeMarle
Kitware, Inc.
R&D Engineer
28 Corporate Drive
Clifton Park, NY 12065-8662
Phone: 518-371-3971 x109



On Wed, Apr 22, 2009 at 5:28 AM, Natalie Happenhofer
<nataliehapp at hotmail.com> wrote:
> Hi!
> I´m trying to render the volume of a vector valued dataset. That´s what I
> tried (and what happens):
>
> 1) I read in a scalar valued dataset and applied the gradient filter.
> Viewing the output as a "surface" works well and the data look ok, but
> clicking on "Volume", Paraview get a segmentation fault.

Sounds like a bug that needs fixing. Please post a bug on the bug
tracker. I can reproduce it with the wavelet source.

>
> 2) I read in a scalar valued dataset and applied the gradient filter. Then I
> saved the data and opened it again with paraview. It seems, that the output
> of the gradient filter is a dataset with two associated arrays to it, the
> first scalar-valued and being the values of which the gradient was computed,
> and the second data-array consists of vectors, which are the computed
> gradients.
> Clicking now on "volume", I get a rather strange picture, it seems that in
> the first third of the dataset the scalars are displayed and then there are
> many little points on the rest of the cube.. if anyone wants I can mail a
> screenshot.

A screenshot would be helpful.

>
> 3) I opened a dataset with just a vector-valued array associated. Again,
> "Surface" is ok, but clicking on "volume", I get the error: "Cannot volume
> render since no point (or cell) data is available". (Well, the vector-array
> should be a point data array).
>
> Maybe it is not possible to render the volume of a vector -valued dataset?

I'm not sure if there are any volume rendering algorithms in VTK that
operate directly on vectors. What do you want the transfer function to
do with the vector exactly? The obvious choice is to compute the
vector magnitude and then volume render that, which you can by
filtering and then volume rendering the output of the magnitude
filter.

>
> thanks for any piece of advice,
> Natalie
>
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