[vtkusers] Intersection of two iso surfaces

Bill Lorensen bill.lorensen at gmail.com
Sat Dec 3 10:34:16 EST 2011


5b) Store the valuve of the second function in the pointdata scalars
of the isosurface. Then run vtkContourFilter

On Fri, Dec 2, 2011 at 3:29 PM, Luke <hazelnusse at gmail.com> wrote:
>> Something like:
>>
>> for each point in the extracted isosurface, evaluate the second
>> function and store the result in the pointdata of the extracted
>> isosurface.
>
> David Doria helped me figure out how to get the Points from the
> contour filter in a separate thread, so now I'm clear on how I would
> evaluate the second function at each point on the isosurface of the
> first function.
>
> Can you elaborate on what you mean about storing the result in the
> pointdata of the extracted isosurface?  I'm not sure where this extra
> scalar associated with each point of the isosurface would go.
> Ideally, it would be nice to generate contours of the second function
> on the isosurface of the first function, so storing it in some fashion
> that would make it work with the vtkContourFilter would be ideal.
>
> To reiterate my understanding of the process:
>
> 1) subclass vtkImplicitFunction to create an implicit function, call
> it f1.  This is the numerically cheap function to evaluate.
> 2) use vtkSampleFunction to sample f1 over the desired volume of
> interest, with the desired discretization.
> 3) use vtkContourFilter to generate the isosurface of interest (f1(x,y,z) = c).
> 4) use vtkContourFilter->GetOutput()->GetPoints() to obtain points
> (approximately) on the isosurface.
> 5) for each point in vtkPoints
>  a) Possibly enhance the accuracy by some method (perhaps
> Newton-Raphson, or ray casting?)
>  b) evaluate the second function, f2, at each point.  Where should I
> store this?
> 6) generate contours of the second function.  Each contour curve would
> represent points of constant f2 (and f1, since it is on the isosurface
> of f1).
> 9) Profit.
>
> It is becoming more clear what I need to do, just a few things are fuzzy still.
>
> Thank you for your time.
>
> ~Luke



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