Don't completely reinstall everything. It won't help. Did you do a clean svn checkout of branches/version-2_1 and point your PYTHONPATH at this? Did you also try putting a print statement in the new method created in the cyclindrial grid 1D class?
On Wed, May 19, 2010 at 9:16 PM, Eduard Manley <celez1...@hotmail.com> wrote: > Tried uninstall and did a new install using branches v. 2.1 @3546 but I get > the same result... > > Tomorrow I'll try a completely clean install of python, fipy and all > packages on another pc.... > > Thanks > > >> Date: Tue, 18 May 2010 18:46:05 -0400 >> From: daniel.wheel...@gmail.com >> To: fipy@nist.gov >> Subject: Re: Problem in solving Poisson equation with 1D Cylindrical mesh >> >> >> Make that branches/version-2_1 rather than trunk/. >> >> On Tue, May 18, 2010 at 5:28 PM, Daniel Wheeler >> <daniel.wheel...@gmail.com> wrote: >> > >> > It might be easier to do a fresh checkout of trunk rather than copying >> > files around. >> > >> > What happens if you put a print statement here >> > >> > <http://matforge.org/fipy/browser/trunk/fipy/meshes/numMesh/cylindricalGrid1D.py#L78>? >> > The method wasn't in the older version.. >> > >> > On Tue, May 18, 2010 at 3:37 PM, Eduard Manley <celez1...@hotmail.com> >> > wrote: >> >> I get again the first one.... >> >> As said before I copied the files in the dir fipy/meshes/numMesh and >> >> did a >> >> new install. It's wrong or there is something else.... >> >> >> >> Thanks >> >> >> >>> Date: Mon, 17 May 2010 10:10:45 -0400 >> >>> From: daniel.wheel...@gmail.com >> >>> To: fipy@nist.gov >> >>> Subject: Re: Problem in solving Poisson equation with 1D Cylindrical >> >>> mesh >> >>> >> >>> >> >>> An older version of fipy (tr...@3480, for example) gives, >> >>> >> >>> In [1]: from fipy import * >> >>> m = CylindricalGrid1D(dx=(1.,2.)) >> >>> >> >>> >> >>> /users/wd15/Documents/python/fipy/tr...@3480/fipy/viewers/gnuplotViewer/gnuplot1DViewer.py:76: >> >>> Warning: 'with' will become a reserved keyword in Python 2.6 >> >>> In [2]: m = CylindricalGrid1D(dx=(1.,2.)) >> >>> In [3]: >> >>> In [4]: >> >>> In [5]: m.getCellVolumes() >> >>> Out[5]: array([ 0.5, 16. ]) >> >>> >> >>> The latest version of trunk gives, >> >>> >> >>> In [1]: from fipy import * >> >>> In [2]: m = CylindricalGrid1D(dx=(1.,2.)) >> >>> In [3]: m.getCellVolumes() >> >>> Out[3]: array([ 0.5, 4. ]) >> >>> >> >>> What do you get? >> >>> >> >>> On Sun, May 16, 2010 at 2:52 PM, Eduard Manley <celez1...@hotmail.com> >> >>> wrote: >> >>> > Hi, >> >>> > thanks for your reply. >> >>> > >> >>> > I copied the files in the install dir (right subdir) and did a new >> >>> > install >> >>> > of fipy (completely new install on other pc), it should be ok but I >> >>> > tried >> >>> > running the code and the result is the same as before and so is >> >>> > wrong. >> >>> > >> >>> > I am doing some error? >> >>> > >> >>> > Thanks >> >>> > >> >>> >> Date: Thu, 13 May 2010 16:51:39 -0400 >> >>> >> From: daniel.wheel...@gmail.com >> >>> >> To: fipy@nist.gov >> >>> >> Subject: Re: Problem in solving Poisson equation with 1D >> >>> >> Cylindrical >> >>> >> mesh >> >>> >> >> >>> >> >> >>> >> With any luck this has now been dealt with. Latest versions of >> >>> >> branches/version-2_1 and trunk/ should now have the fixes for >> >>> >> <http://matforge.org/fipy/ticket/294> >> >>> >> >> >>> >> On Tue, May 11, 2010 at 5:44 PM, Eduard Manley >> >>> >> <celez1...@hotmail.com> >> >>> >> wrote: >> >>> >> > Hi, >> >>> >> > thanks for your reply. >> >>> >> > >> >>> >> > You can find attached a sample problem. >> >>> >> > >> >>> >> > The equation is a heat equation, variable is temperature,other >> >>> >> > and >> >>> >> > transient >> >>> >> > and diffusion term, there is a spatially varying source term (q) >> >>> >> > computed >> >>> >> > by analytical expression which is and hyperbola. >> >>> >> > >> >>> >> > We should reach about 100°C in 30-35 sec and the shape of the >> >>> >> > solution >> >>> >> > should be similiar to and hyperbola. >> >>> >> > >> >>> >> > As it is now the file works fine (mesh declared with: mesh = >> >>> >> > CylindricalGrid1D(dr=dr, nr=(len(DR))) + (r_int,) ) but if you >> >>> >> > define >> >>> >> > the >> >>> >> > mesh with ( mesh = CylindricalGrid1D(dx=DR) + (r_int,) ) the >> >>> >> > result >> >>> >> > you >> >>> >> > obtain are wrong. >> >>> >> > >> >>> >> > The 2 declaration of CylindricalGrid1D should be the same, the >> >>> >> > face >> >>> >> > and >> >>> >> > cell >> >>> >> > centers are the same, but the result are different. >> >>> >> > >> >>> >> > You can observe the same problem also when solving the >> >>> >> > same equation >> >>> >> > but without transient term >> >>> >> > >> >>> >> > Thanks >> >>> >> > Eduard >> >>> >> > >> >>> >> > >> >>> >> > >> >>> >> >> Date: Tue, 11 May 2010 10:45:24 -0400 >> >>> >> >> From: daniel.wheel...@gmail.com >> >>> >> >> To: fipy@nist.gov >> >>> >> >> Subject: Re: Problem in solving Poisson equation with 1D >> >>> >> >> Cylindrical >> >>> >> >> mesh >> >>> >> >> >> >>> >> >> >> >>> >> >> Can you put together the simplest script possible that >> >>> >> >> demonstrates >> >>> >> >> the problem and I'll try and debug it? Thanks. >> >>> >> >> >> >>> >> >> On Thu, May 6, 2010 at 2:01 PM, Eduard Manley >> >>> >> >> <celez1...@hotmail.com> >> >>> >> >> wrote: >> >>> >> >> > Thanks for your reply. >> >>> >> >> > >> >>> >> >> > I probably found the reason of the problem. >> >>> >> >> > >> >>> >> >> > As said before I'm trying to solve an eq of this type: >> >>> >> >> > >> >>> >> >> > A(d phi/d t) = div (D grad phi) + q >> >>> >> >> > >> >>> >> >> > where A and D are costant coefficient and q is a spatially >> >>> >> >> > varying >> >>> >> >> > heat >> >>> >> >> > source. >> >>> >> >> > >> >>> >> >> > The problem is in how I create the cylindrical 1D mesh (origin >> >>> >> >> > of >> >>> >> >> > the >> >>> >> >> > mesh >> >>> >> >> > is not in 0.). >> >>> >> >> > >> >>> >> >> > It doesn't matter if the discretization is logarithmic or >> >>> >> >> > uniform >> >>> >> >> > but >> >>> >> >> > how I >> >>> >> >> > declare it: >> >>> >> >> > >> >>> >> >> > (using a uniform spacing:) >> >>> >> >> > >> >>> >> >> > ** mesh = CylindricalGrid1D(dx=dr, nx=(len(DR))) + (r_int,) >> >>> >> >> > ** >> >>> >> >> > SHOULD BE EQUAL TO: >> >>> >> >> > >> >>> >> >> > ** mesh = CylindricalGrid1D(dx=DR) + (r_int,) ** >> >>> >> >> > >> >>> >> >> > [dr = 5e-04, nr=58, r_int=0.00125, DR is a list which contains >> >>> >> >> > the >> >>> >> >> > various >> >>> >> >> > dx(58 elements of value dr for uniform grid)] >> >>> >> >> > >> >>> >> >> > BUT It is NOT. >> >>> >> >> > >> >>> >> >> > The mesh (cell centers, facecenters) is ok but the results are >> >>> >> >> > NOT. >> >>> >> >> > The results are right only if I create the mesh using >> >>> >> >> > dx=dr and >> >>> >> >> > nx=. >> >>> >> >> > . >> >>> >> >> > And this is why before I thought the problem was the >> >>> >> >> > logaritmic >> >>> >> >> > discretization (must use DR=[...]). >> >>> >> >> > >> >>> >> >> > As said before with Cylindrical 2D mesh results are >> >>> >> >> > instead correct. >> >>> >> >> > Is this a bug? >> >>> >> >> > >> >>> >> >> >> Date: Thu, 6 May 2010 11:11:10 -0400 >> >>> >> >> >> From: daniel.wheel...@gmail.com >> >>> >> >> >> To: fipy@nist.gov >> >>> >> >> >> Subject: Re: Problem in solving Poisson equation with 1D >> >>> >> >> >> Cylindrical >> >>> >> >> >> mesh >> >>> >> >> >> >> >>> >> >> >> >> >>> >> >> >> Edward, This may be to do with having a very small volume (or >> >>> >> >> >> area >> >>> >> >> >> or >> >>> >> >> >> line or point) for the inner most element of the domain. It >> >>> >> >> >> should >> >>> >> >> >> be >> >>> >> >> >> the same whether one is using a 1D or 2D mesh. Since you are >> >>> >> >> >> getting >> >>> >> >> >> differences in the 1D and 2D case, it should be relatively >> >>> >> >> >> easy >> >>> >> >> >> to >> >>> >> >> >> debug and figure out what's going on It could also be that >> >>> >> >> >> the >> >>> >> >> >> boundary condition on the inner boundary as zero area and >> >>> >> >> >> this is >> >>> >> >> >> causing issues. Try shifting the grid by a small value away >> >>> >> >> >> from >> >>> >> >> >> the >> >>> >> >> >> zero point and see if things are improved I have always had >> >>> >> >> >> this >> >>> >> >> >> issue with cylindrical grids and have never really had a >> >>> >> >> >> satisfactory >> >>> >> >> >> solution (other than shifting away from the zero point). If >> >>> >> >> >> you >> >>> >> >> >> discover a better way to handle this, let me know. Cheers. >> >>> >> >> >> >> >>> >> >> >> >> >>> >> >> >> If you can't debug it, then send me the most minimalist >> >>> >> >> >> scripts >> >>> >> >> >> that >> >>> >> >> >> show the issue and I'll give it a shot. >> >>> >> >> >> >> >>> >> >> >> On Wed, May 5, 2010 at 9:53 PM, Eduard Manley >> >>> >> >> >> <celez1...@hotmail.com> >> >>> >> >> >> wrote: >> >>> >> >> >> > Problem partially solved: >> >>> >> >> >> > >> >>> >> >> >> > I'm using a logarithmic discretization (first dr= 5e-04 and >> >>> >> >> >> > next >> >>> >> >> >> > dr increasing as 1.05)(with internal radius= 0.00125, >> >>> >> >> >> > external >> >>> >> >> >> > radius=0.03) >> >>> >> >> >> > and, >> >>> >> >> >> > for some unknown reason this create problems and wrong >> >>> >> >> >> > result with >> >>> >> >> >> > cylindrical 1D mesh. I tried using a uniform discretization >> >>> >> >> >> > (dr=5e-04 >> >>> >> >> >> > nx=58) >> >>> >> >> >> > and now the result is correct. >> >>> >> >> >> > >> >>> >> >> >> > However I need to use the logarithmic discr. so after some >> >>> >> >> >> > hours >> >>> >> >> >> > of >> >>> >> >> >> > sleep >> >>> >> >> >> > I'll think about the reason..... >> >>> >> >> >> > >> >>> >> >> >> > ________________________________ >> >>> >> >> >> > Hotmail: Trusted email with powerful SPAM protection. Sign >> >>> >> >> >> > up >> >>> >> >> >> > now. >> >>> >> >> >> >> >>> >> >> >> >> >>> >> >> >> >> >>> >> >> >> -- >> >>> >> >> >> Daniel Wheeler >> >>> >> >> >> >> >>> >> >> >> >> >>> >> >> > >> >>> >> >> > ________________________________ >> >>> >> >> > Your E-mail and More On-the-Go. Get Windows Live Hotmail Free >> >>> >> >> > Sign >> >>> >> >> > up >> >>> >> >> > now. >> >>> >> >> >> >>> >> >> >> >>> >> >> >> >>> >> >> -- >> >>> >> >> Daniel Wheeler >> >>> >> >> >> >>> >> >> >> >>> >> > >> >>> >> > ________________________________ >> >>> >> > Hotmail: Trusted email with Microsoft’s powerful SPAM protection. >> >>> >> > Sign >> >>> >> > up >> >>> >> > now. >> >>> >> >> >>> >> >> >>> >> >> >>> >> -- >> >>> >> Daniel Wheeler >> >>> >> >> >>> >> >> >>> > >> >>> > >> >>> > ________________________________ >> >>> > Hotmail: Free, trusted and rich email service. Get it now >> >>> >> >>> >> >>> >> >>> -- >> >>> Daniel Wheeler >> >>> >> >>> >> >> >> >> ________________________________ >> >> Hotmail: Powerful Free email with security by Microsoft. Get it now. >> > >> > >> > >> > -- >> > Daniel Wheeler >> > >> > >> > >> >> >> >> -- >> Daniel Wheeler >> >> > > ________________________________ > Hotmail: Trusted email with powerful SPAM protection. 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