Hi,
I am tracking down some issues in our code, and I am seeking some clarification on the way that Elmer handles the nonpenetrating condition at the basal boundary condition (i.e. the ice cannot penetrate through the bedrock) in the NavierStokes solver.
As I understand it, Elmer rotates the coordinate system of the nodes that are on the bottom boundary so that the xcoordinate is normal to the base. If you set the parameter "Back Rotate NT Solution = Logical True", it is supposed to rotate the coordinate system back to the original Cartesian coordinate system after DefaultSolve (and by extension SolveSystem) is finished executing. It is not clear, however, if the coordinate system of the input parameters prior calling DefaultSolve is supposed to be bednormal coordinate system or in Cartesian coordinates (regardless of the value of Back Rotate NT Solution). The reason I ask, is because I am setting the results of the Shallow Ice Approximation as an initial guess for the velocity of the ice, and this is stored in Cartesian coordinates in our program.
Another question is why it was set up like this. Wouldn't it be easier to rotate the nonpenetrating condition to Cartesian coordinates, rather than rotating the nodes to a local bednormal coordinate system? Then there would only have to be one rotation calculation at the start of the simulation and the coordinate system of all the nodes would remain consistent.
Cheers,
Evan
Rotated coordinate system for basal boundary condition
Re: Rotated coordinate system for basal boundary condition
Just to follow up, after investigating the problem I am having with our code, it seems that the BackRotate routine is rotating not only the bottom layer of nodes, but also the layer directly above it. However, after exit from DefaultSolve, the second layer is not rotated back. Below is a stack of nodes at a single location (the mesh has been extruded to six layers) before and after calling DefaultSolve. As you can see, the velocity values in the second layer are incorrect.
Code: Select all
vx (before) vy(before) vz(before) vx(after) vy(after) vz(after)
669.15 665.18 5.46 507.85 628.50 8.29
672.47 668.10 13.10 64.700 625.80 575.94
675.32 670.22 20.40 574.88 633.43 61.43
677.63 671.53 26.88 578.83 634.89 64.48
679.35 672.06 31.61 582.10 635.60 67.81
680.40 671.94 33.28 584.74 635.68 69.17

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Re: Rotated coordinate system for basal boundary condition
Hi Even
I try to clarify the 2nd question. The NT coordinate system does not rotate nodes. Instead it rotates the velocity vector related to the nodes on the boundary. The nodes on the NT coordinate system should be marked so that the operation is performed for only them. When you rotate the contribution to the matrix equation to the NT directions you can also set the Dirichlet conditions accordingly in the NT directions. This is convenient since we know that normal component should be zero and the tangential component can remain unknown. If you would not do the rotation you cannot set either component using Dirichlet conditions since both include part if the unknown tangential velocity. If the tangential velocity would be known it would be easy to prescribe the velocity for all components in the cartesian system.
Peter
I try to clarify the 2nd question. The NT coordinate system does not rotate nodes. Instead it rotates the velocity vector related to the nodes on the boundary. The nodes on the NT coordinate system should be marked so that the operation is performed for only them. When you rotate the contribution to the matrix equation to the NT directions you can also set the Dirichlet conditions accordingly in the NT directions. This is convenient since we know that normal component should be zero and the tangential component can remain unknown. If you would not do the rotation you cannot set either component using Dirichlet conditions since both include part if the unknown tangential velocity. If the tangential velocity would be known it would be easy to prescribe the velocity for all components in the cartesian system.
Peter
Re: Rotated coordinate system for basal boundary condition
Hi Peter,
Thank you for the explanation.
It turns out the problem I am having is not a problem with the rotation itself, but that we are using a reduced A matrix for DefaultSolve after the first time step. Somewhere in our code we have missed changing a pointer, so it is rotating too many nodes. Because the A matrix has dimensions that are about half what it was in the first time step, it looked like it was rotating two layers instead of one.
Thank you for the explanation.
It turns out the problem I am having is not a problem with the rotation itself, but that we are using a reduced A matrix for DefaultSolve after the first time step. Somewhere in our code we have missed changing a pointer, so it is rotating too many nodes. Because the A matrix has dimensions that are about half what it was in the first time step, it looked like it was rotating two layers instead of one.