Is that twice differentiable for real or everywhere except irregular vertices?

So we use the left figure's weight to decide the position of the vertex that we newly inserted in, and use the right figure's weight to decide the position of the vertex that originally existed in the polygon, is that right?

@ceviri The limit surface is twice differentiable away from irregular vertices.

@Osoii Yes, to get the new vertex positions you're only taking weighted combinations of the original vertex positions.