[vtkusers] Cut a triangulated surface with a line

[sebastian ordas]

Have a look at vtkvmtkPolyDataScissors in www.vmtk.org

good luck!
sebastian


On Thu, Jul 18, 2013 at 10:42 AM, Martin Vymazal
<martin.vymazal at vki.ac.be>wrote:

> **
>
> Hello Alex,
>
>  thank you for your reply. My surface is a 3D surface, but I want to
> split it into several patches given by this line(s). That's why I'm
> speaking about cutting by line and not plane/box/sphere etc. The (poly)line
> is already given as a sequence of starting and ending points of several
> straight segments. Basically all I'm asking is whether vtk could do proper
> modification to the connectivity tables because the cut is going to modify
> some existing and introduce new triangles. From your answer I have the
> feeling that this is unfortunately not possible.
>
>
>
> Best regards,
>
>
>
> Martin Vymazal
>
>
>
> On Wednesday 17 of July 2013 14:36:19 Alex Malyushytskyy wrote:
>
> Cutter will change the dimensions of the cells when Clipping does not..
> Assuming you work with polydata you should check vtkClipPolyData<http://www.vtk.org/doc/nightly/html/classvtkClipPolyData.html>
>
> http://www.vtk.org/doc/nightly/html/classvtkClipPolyData.html
> It is important to remember that in VTK you are always dealing with 3d data
>
> so you define plans not line as a clipping function.
>
> Regards,
>
>    Alex
>
>
>
> On Wed, Jul 17, 2013 at 3:46 AM, Martin Vymazal <martin.vymazal at vki.ac.be>
> wrote:
>
> Hello,
>
>  I would like to ask if  I can slice elements in triangulation. Suppose my
> input is a line on a triangulated surface (given as a sequence of vertex
> pairs
> indicating which mesh edges the line should cut, for example), I would
> like to
> split the triangles that the line passes through, eventually introducing
> new
> triangles in the mesh. Apparently there is the class vtkCutter, but I'm not
> sure that that's what should be used. Could you please point me to some
> examples if there are any? How should I feed my input line to vtk? Create a
> polyline perhaps?
>
> Thank you very much,
>
>  Martin Vymazal
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