I don't have any pictures, but ponder the following:Mate Kovacs said:Could you show me some pictures of a (shaded) surface made of n patches of 4th order and it's approximation with 4n patches of 2nd order?Simon F said:Compute the normals, do some shading and then tell me if you can't see the difference between C0, C1 and perhaps C2....
I'm really curious by now.
Quadratic and even cubic triangular Bezier patches cannot guarantee C1 continuity, only C0, with their neighbouring (triangular) patches. This means that you will get some creases in the model. You have to go to quartic triangular patches (i.e. 15 control points) in order to guarantee C1.
With rectangular (aka tensor product) patches you might be able to get away with a bi-quadratic but you are certainly safe with a bi-cubic.
I suspect that the main reasons some like them areMfA said:What's so great about triangular patches anyway?
- Any line though the parameter space always maps to the same order curve on the patch, e.g. all cubics for the case of a cubic Bezier triangle. With rectangular patches, only lines parallel to the parameter axes have this property. For example, the diagonals across a bi-cubic patch I think are 6th order.
- They can model arbitrary topology more easily, although Catmull-Clark or Doo-Sabin subdivision surfaces might have reduced this advantage slightly.