On Tue, Jun 19, 2012 at 1:04 PM, Jonathan Guyer <[email protected]> wrote:
> > On Jun 19, 2012, at 12:20 PM, Kendall Boniface wrote: > > > I am having a bit of trouble manipulating a 2D cylindrical mesh and was > wondering if anyone has any helpful advice? > > > > mesh = CylindricalGrid2D(dx=dx, dy=dy, nx=nx, ny=ny) + ((0.0046,),) > > print mesh.getCellCenters() > > > > I want the z axis to go from 0 to 1.5 > > The second term in the mesh line seems to translate the r AND z > coordinates outward when all I want it to do is translate the r > coordinates. Is there a way to do this the way I would like? > > The displacement should be a vector. You're trying to translate a 2D mesh > by a 1D vector. As it happens, NumPy interprets that to mean displacing the > r and z coordinates by the same amount, but what you want is > > mesh = CylindricalGrid2D(dx=dx, dy=dy, nx=nx, ny=ny) + ((0.0046,),(0.,)) > > Thank you Jonathan, that works perfectly. > > > Is there a way to access the radial distances from the origin for every > cell? That isn't the best way to word the question, but to make it more > clear, I would like to be able to do something like: pi * (r_outer**2 - > r_inner**2) for each cell. I have tried as many of the mesh.get functions > as I could get my hands on, but haven't been successful yet. Can anyone > suggest something for me to use? > > A mesh.getFaceCenters()[..., mesh._getCellFaceIDs()] would get you an > array of the coordinates of the faces for each cell. You'd probably have to > do something with mesh._getFaceNormals() to figure out which are the > "inner" and "outer" faces. > > For the actual calculation you present, though, it seems like > mesh.getCellVolumes() is what you want. > I'm a bit confused about the getCellVolumes() function. Since I am using the 2D cylindrical grid, I assume it isn't actually giving me "volumes" so to speak. I've manually played around with some of my numbers and can't seem to figure out what the numbers I get from getCellVolumes() actually represent. Can you clarify that for me? Thank you very much for for all your help! Kendall
_______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
