On 02/10/2016 05:31 PM, Berk Geveci wrote:
There is a better way. Build a multi-block structure that creates enough
copies with the appropriate transforms to advance the streamlines as many
periods as you need. Make sure to shallow copy all heavy data structures
such as data arrays. If you are using vtkImageData or vtkRectilinearGrid,
this is easy. If you are using vtkStructuredGrids or vtkUnstructuredGrids,
take a look at vtkPeriodicDataArray and its subclass
vtkAngularPeriodicDataArray. If you need other type of periodicity, you may
need to create a new subclass of vktPeriodicDataArray. Also take a look at
vtkPeriodicFilter that can create such a multi-block dataset.
AngularPeriodicFilter (since 4.4) exist that do the job, isn't it?


Best,
-berk

On Wed, Feb 10, 2016 at 9:16 AM, Andy Bauer <andy.ba...@kitware.com> wrote:

Hi Tim,

There isn't any such functionality that exists directly in PV like this. A
long way of doing this may be:

    - StreamTracer for first set of streamlines
    - Use threshold or cut filter to get last points before they exit the
    domain
    - Use the Transform filter to move these points to the entrance of the
    domain on the other periodic side
    - StreamTracer with Custom source using Transform filter as source to
    continue the streamlines
    - Calculator filter to adjust IntegrationTime and other field values
    (if this is important you'll probably need to do this one streamline at a
    time).

This may get a bit sloppy and unwieldy but that's the best way I can think
about doing what you want.

If you do this and find a better way, please let us know!

Cheers,

Andy




On Tue, Feb 9, 2016 at 5:34 PM, Tim Gallagher <tim.gallag...@gatech.edu>
wrote:

Hi,

I have a simulation that is periodic in 2 of the 3 directions and I would
like to plot some streamlines coming from specific points in the domain.
Some of these points are close to one of the periodic edges.

Is there a way to tell Paraview that the domain is periodic and that
streamline integration can continue from the opposite face?

Thanks,

Tim
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