[vtk-developers] vtkPolyData::ComputeBounds() problem

Sebastien BARRE sebastien at barre.nom.fr
Mon Jul 31 19:27:06 EDT 2000

At 15:40 31/07/00 -0400, Lorensen, William E (CRD) a écrit:
>I hope you ran all of the regression tests before checking in the changes...

It appears that my implementation of vtkPolyData::ComputeBounds() crashes 
the following regression tests :




A)  DelMesh.tcl, Delaunay2D.tcl, Delaunay2DAlpha.tcl, Delaunay3D.tcl

=> they all use either vtkDelaunay2D, or vtkDelaunay3D
=> I do not know why it crashes. May the implementors of theses classes 
contact me ?

B) financialField.tcl, financialField2.tcl, splatFace.tcl

=> they all use either vtkGaussianSplatter
=> I do not know why it crashes. Same stuff, may the implementors of this 
class contact me ?

B) capSphere.tcl

=> no problem. As I reported when asking for the CoumputeBounds() change in 
my first email :

>Rational : suppose that given a polygonal object A, I use a 
>vtkClipPolyData filter to extract two other polygonal objects B, and C, 
>which will have different topologies (and most probably different extents).
>If I query the bounds of B and C using GetBounds(), or use vtkOulineFilter 
>to display very simple bounding boxes, I'll get ... the bounds of A. This 
>is no surprise because both A, B and C share the *same* points 
>(vtkPoints), and as A, B, C are vtkPolyData derived from vtkPointSet, 
>GetBounds() use vtkPointSet::ComputeBounds(), which computes the bounds by 
>iterating over the *points*.

capSphere.tcl use vtkClipPolyData. It works correctly with the new 
vtkPolyData::ComputeBounds(), *but* as only one half of a sphere is 
displayed, the bounds of these resulting objects is really half of a 
sphere, not a whole sphere (the bounds of the original object that was 
cut). Therefore, as the camera position/FOV depends on the bounds of the 
object, the resulting picture is mathematically different, it appears 
larger on the screen, but the results is the same.

=> I guess we must create a new valid image, am I right ? (but solve A and 
B first).

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