As I was able to compute all but the four inner bezier controlpoints (or the 
hermite twist vectors) I searched the internet again and found the following 
equation to convert from bezier to hermite (or vice versa):

d^2P/dudv(0,0) = 9(P00-P011+P10-P11)

I'll have to try it, but it looks promising.

Anyway, thanks a lot for your help!
-Sascha


andrel wrote:
> Have you solved your ploblem yet.
> 
> Today in the train I realized that my suggestion
> is correct, but that it is too much influenced by
> the fact that I do have a program to solve the inverse
> of a 16 by 16.
> 
> Rethinking it from the perspective of: 'how would
> I solve this by hand?' often gives better results.
> - there are only three types of points in the (realized)
>  patch: corner points (4), side points (8) and center
>  poinys (4)
> - cornerpoints are in all representations the same, so
>  we do not have to apply any interpolation.
> - sidepoints are only influenced by the two cornerpoints
>  on that side and in case ot the bezier patch the control
>  points on that side, and in the hermite case by the
>  2 derivatives along the side. All in all we need 4
>  numbers only to do the conversion from one representation
>  to the other. We use symmetry to apply the same 4 to
>  every sidepoint of course.
> - For the center points we need only 16 numbers to
>  describe the conversions.
> All in al we need compute only 4+16 numbers to do
> all conversions. so the 16 by 16 matrix is a bit overkill
> (I will keep using it because it greatly somplifies
>  my code but that is another issue).
> 
>   Andrel
> Sascha Ledinsky wrote:
> 
>> I've got to delve into that.
>> Thanks a lot for your help!
>> -Sascha
> 
>

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