[vtkusers] vtkImageReslice : scalar value range altered? (by cubic interpolation)
dgobbi at atamai.com
Mon Jan 24 14:00:27 EST 2005
That's just how vtkImageReslice's cubic interpolation works. If you
have 4 nodes and you draw a cubic curve that passes through them, it
should be immediately apparent that sometimes the range of the curve
will exceed the range of the nodes.
Currently vtkImageReslice does not support B-spline interpolation,
which would solve your problem. This is something I've been thinking
of adding for a long time (it wouldn't be much work, only the testing
is a pain) so it will probably make it into VTK 5.0.
On Mon, 24 Jan 2005, Marius S Giurgi wrote:
> Does anyone know why the scalar value range (min/max) is altered by the
> vtkImageReslice when using the cubic interpolation?
> I have a volume (181x217x181 double) whose scalar value range is (0,
> 1). I perform a rigid transformation on it (rotation, translation etc).
> The linear interpolation preserves the scalar value range (0, 1) but
> the cubic interpolation does not. E.g. for a particular transformation
> (RotateX 20) the scalar range becomes (-0.2167, 1.1854).
> Remark: Even if the scalars are cropped by the given transformation
> (thus losing some values), the interpolated values should still remain
> within the original range (0, 1). Is this a bug or should certain
> parameters be set in vtkImageReslice to avoid such scalar value
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