On 7/9/07, Anders Logg <logg@xxxxxxxxx> wrote:
>
>
> Matthew Knepley wrote:
> > On 7/9/07, Anders Logg <logg@xxxxxxxxx> wrote:
> >> It looks to me like the (quad) cell which first has vertices 1-2-5-4
> >> after the surgery has vertices 1-(0)-(2)-3. What are then the
> >
> > Yes.
> >
> >> coordinates of vertices (0) and (2)? Same as for 0 and 2, or same as
> >> original 2 and 5?
> >
> > Same as (0) and (2) since no other vertices actually exist in the mesh.
> > Those are vertices 0 and 2. I put them in parens to emphasize that I
> > was writing them twice, but they exist only once. Next time I will just
> > write the Sieves. Much clearer :)
> >
> > Matt
>
> It still doesn't make sense to me, but maybe I'm just getting tired...
>
> If the coordinates of (0) and (2) are the same as those of 0 and 2,
> won't we get the wrong geometry for the cells on the boundary? Imagine
> having a much finer grid. Then the cells on the right boundary will
> stretch all the way back to touch the left boundary.
Yes, but that is why I put in the explanation for calculation of J. With the
exception of coordinate functions (which we can talk about after this),
you only need J or functions of it. Now, the Jacobian only depends on
coordinate differences. So you just need something like
dx = (x_1 - x_0) mod L
where L is the periodic length.
Matt
