[vtkusers] vtkImageReslice : scalar value range altered? (by cubic interpolation)

[David Gobbi]

Hi Marius,

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.

 - David



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
> alteration?
>
> thanks
> marius
>
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