> Maybe you're right, and for some cases you can find an ellipsoid
> which could accomodate the three normals (even if I have no idea of
> how to find the ellipsoid parameters; using a unit sphere is easier
> for the math).
> But a linear transform cannot transform a positive curvature into
> a negative one. So the ellipsoid won't solve all the cases.
>
> Given also the troubling case where two normals are parallel (which
> cannot be solve on someting like a sphere or even an ellipsoid),
> I wonder if the curved triangle shouldn't be on a torus instead (*).
>
> At least for the two parallel normals case, it seems that two vertices
> would be on the very top circle of the torus with the third vertex
somewhere
> so that its normals is ok. Currently, I can only figure the math to
> position the first vertex, the circle where the second vertex is and
> where the third vertex might be.
> Maybe using also the true normal of the initial triangle might help to
> limit the position of the second and third vertices.
>
> Once the triangle on the torus is defined, we still go back to the
classical
> transformation of three points from one space to another, nothing really
difficult!
> But maybe the torus is not even the right solution.
>
>
I would assume that for a curved triangle that had two parallel normals,
that the entire edge between those two normals would have the same normal.
for a torus, it is true that we can make a triangle that has two parallel
normals (ie, on the top of the torus) but it is quite easy to see that
between these two points the edge will curve (or even disappear in some
cases).
I don't think that simple primitives have the solution.  There must be some
way of creating quadratic type surfaces without having a discrete mesh.
Perhaps isosurfaces could be accomodated, but I would think that would come
at great CPU expense.

-tgq

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