Hi,
Followed up the previous question, now I calculated the curvature everywhere
for the mesh and found the accuracy was not that within second order. Is my
mesh not stable?
Here is my mesh:
mesh = Grid2D(dx=1.0, dy=1.0, nx=100, ny=100)
var = DistanceVariable(name='Test',
mesh=mesh,
value=-1.0,
hasOld=1)
For initial value,
var.setValue(1.0, where = ((x-50.0)**2+(y-50.0)**2>625.0))
var.calcDistanceFunction()
For this case, what is the best way to calculate the curvature?
Thanks,
Regards,
Yunbo
On Jan 20, 2012, at 10:47 AM, Daniel Wheeler wrote:
> On Thu, Jan 19, 2012 at 6:34 PM, wang yunbo <[email protected]> wrote:
> I'm trying to calculate the curvature of a circle in 2D.
>
> Sounds reasonable.
>
> Firstly I set up a distance variable "var" to represent the circular shape.
> To calculate the curvature, I took the gradient of "var" and then the
> divergence as shown below:
> (var.getFaceGrad()).getDivergence()
>
> Should work with second order accuracy on a well structured mesh.
>
> My trouble is :
>
> My expectation is to obtain a matrix of curvature and
> at the boundary of circle (where var=0). However, the obtained result from
> my program showed
> at the boundary.
>
> The boundary of the domain assumes a zero gradient. It doesn't use any other
> information about the field. In version 2.1 there is nothing to be done other
> than ignore the cells close to the boundary. However, In FiPy 3.0 (not yet
> released) this will work:
>
> from fipy import *
> m = Grid2D(nx=4, ny=4)
> v = CellVariable(mesh=m, value=m.x**2 + m.y**2)
> X, Y = m.faceCenters
> v.faceGrad.constrain((2 * X, 2 * Y), where=m.exteriorFaces)
> print v.faceGrad.divergence
> [ 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4. 4.]
>
> Cheers.
>
> --
> Daniel Wheeler
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