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Distance of a 'North' and 'South' surface of a cell for 2D mesh |
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October 6, 2020, 11:22 |
Distance of a 'North' and 'South' surface of a cell for 2D mesh
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#1 |
Senior Member
René Thibault
Join Date: Dec 2019
Location: Canada
Posts: 114
Rep Power: 6 |
Hi everyone!
My problem could be easy for some people, but I personally don't have any idea on how to code this thing here. I would like to calculate, for a 2D mesh, the distance 'z' from the wall to the 'north' (Zn) and 'South' (Zs) faces of a cell (see image in attachment). Im able to obtain the distance for a center of a cell (ZN, ZP, ZS) with 'mesh_.C' like so: Code:
const vectorField& CellC = mesh_.C(); But I don't have any idea whatsoever how to code this thing. Can I have some example and some help? Thank you everyone! Last edited by Tibo99; October 6, 2020 at 15:52. |
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November 27, 2020, 06:58 |
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#2 |
New Member
Join Date: Feb 2019
Posts: 23
Rep Power: 7 |
Hi,
You can loop over all cell faces and take into account only the faces which have the z component of its normal vector different from 0. Then, see which face is the N and which is S. Finally, just take the distance from the wall (assuming z=0) to the face centers. The code could be something like (for cell ID = 10, for example): Code:
const cellList& cellFaces = mesh.cells(); const surfaceVectorField& Sf = mesh.Sf(); const surfaceVectorField& Cf = mesh.Cf(); const surfaceVectorField& C = mesh.C(); scalar zWall = 0.0; label cellI = 10; scalar zn = 0.0; scalar zs = 0.0; forAll(cellFaces[cellI],faceI) { label faceIG = cellFaces[cellI][faceI]; scalar zSf = Sf[faceIG][2]; if (zSf != 0.0) { if (Cf[faceIG][2] > C[cellI][2]) { zn = Cf[faceIG][2] - zWall; } if (Cf[faceIG][2] < C[cellI][2]) { zs = Cf[faceIG][2] - zWall; } } } |
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November 27, 2020, 10:06 |
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#3 |
Senior Member
René Thibault
Join Date: Dec 2019
Location: Canada
Posts: 114
Rep Power: 6 |
Thank you very much for replying!
The fact that the code don't take into account any boundary faces in the domain is not that a big deal for me since I use it, for the moment, for a 2D channel. My 1st question is, because I need to apply this from the 2nd cell to the 4th cells from the wall in the entire domain, how I could modify the loop to take this into account? The 2nd one is, is this option work with parallel computing? I was thinking using 'topoSet' to get the ID for these cells and then, I would use your loop inside a new loop in order to apply it to all of the cells that 'topoSet' has selected. If this approach is right, I don't know at this point how to get all of these cells ID, stored them in a constant ''IDList'' (for instance) and using it after in the new loop. I keep working on it. Even though, my main issue is pretty much solved with the loop you suggested. Thank you very much again and be safe. Regards, Last edited by Tibo99; November 29, 2020 at 20:32. |
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November 28, 2020, 07:04 |
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#4 |
Senior Member
Mark Olesen
Join Date: Mar 2009
Location: https://olesenm.github.io/
Posts: 1,714
Rep Power: 40 |
Your use case is fairly specific, but could still be worth generalizing. Starting with a simple struct with four label members (north, south, east, west) that will hold the cellId of the respective neighbours. You could also just use a FixedList of 4 labels for this purpose.
Create a List of these structs that is nCells long, with each element initialized to -1. Code:
enum compass { NORTH = 0, SOUTH, EAST, WEST }; typedef FixedList<label, 4> compassType; List<compassType> compassAddr(mesh.nCells(), compassType(-1)); Now that the addressing is complete, you can now handle everything else that you need. A simple example (untested) Code:
// cellCentres() for polyMesh const pointField& cc = mesh.C(); scalar dist = 0; label celli = someCellID; // walk four to the east for (label i=0; i < 4; ++i) { const label next = compassAddr[celli][compass::EAST]: if (next == -1) break; dist += mag(cc[next] - cc[celli]); celli = next; } |
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November 28, 2020, 07:16 |
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#5 |
Senior Member
Mark Olesen
Join Date: Mar 2009
Location: https://olesenm.github.io/
Posts: 1,714
Rep Power: 40 |
Just re-read your first post and noticed that you actually wanted the centre to face distances. In this case you need to store the face id instead of the cell id in the north/south/east/west slots. This will give you an immediate calculation of centre to face, but when walking to the next cell you will need to use the mesh face owner/neighbour information for the connectivity and to walk in the correct direction.
This is probably more flexible than what I originally posted. |
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November 29, 2020, 14:29 |
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#6 |
Senior Member
René Thibault
Join Date: Dec 2019
Location: Canada
Posts: 114
Rep Power: 6 |
Hi Olesen,
I maybe should have mentioned in the first post that I still considered myself at a beginner level with OF. Regarding the fact that the solution you suggested is generalized and probably the best one for my application, I have to admit, I don't exactly understand the code. I feel that this is an another level for me, which I'm not there yet. Sorry about that. But, here is a thread that I posted and it's related to. That should help everyone to understand why I asked how to code this. Applying correction to the k-e transport equation Thank you very much for your time and be safe! Last edited by Tibo99; November 30, 2020 at 11:57. |
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