Eduard, You seem to be saying that you checked out branches/version-...@3546, which was before the changes that I made (r3549). Just checkout a clean version of branches/version-2_1 without specifying any revision number. You may be able to simply do "svn up" in your version-2_1 working copy and this will get you the required changes, but check the revision number using "svn info" before installing. Hopefully, this will do the trick. Cheers.
On Wed, May 19, 2010 at 9:16 PM, Eduard Manley <[email protected]> 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: [email protected] >> To: [email protected] >> 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 >> <[email protected]> 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 <[email protected]> >> > 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: [email protected] >> >>> To: [email protected] >> >>> 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 <[email protected]> >> >>> 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: [email protected] >> >>> >> To: [email protected] >> >>> >> 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 >> >>> >> <[email protected]> >> >>> >> 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: [email protected] >> >>> >> >> To: [email protected] >> >>> >> >> 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 >> >>> >> >> <[email protected]> >> >>> >> >> 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: [email protected] >> >>> >> >> >> To: [email protected] >> >>> >> >> >> 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 >> >>> >> >> >> <[email protected]> >> >>> >> >> >> 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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