[Nek5000-users] New Recycling BC method?

[nek5000-users at lists.mcs.anl.gov]

Hi Michael,

Access to the connectivity array wouldn't help in this case, given
that the recycling bc needs to establish a _new_ connectivity that
has nothing to do with the existing one.   The examples use this
approach of establishing a new connectivity.

I'm confident that we can make the existing approach for any mesh
that can legitimally employ the recycling bc - i.e., any mesh that
has the same cross-sectional distribution of elements at the two
planes of interest.

An alternative is to use the general purpose interpolation routines.
There may (or may not) be a performance issue -- we're looking into
this w/ the help of a summer student.

Please let me know if I can assist you w/ your specific case.

Paul


> I am still trying to figure this idea out, but I think I would need to
> access the connectivity/boundary condition information that is contained
> in the rea file. Does anyone know the array, or what to call in order to
> access this in the usr file?  Thanks again for any help!


>
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> - Michael M.
>
>
> ----- Original Message -----
> From: nek5000-users at lists.mcs.anl.gov
> To: "Nekton User List" <nek5000-users at lists.mcs.anl.gov>
> Sent: Tuesday, June 15, 2010 9:02:00 PM GMT -06:00 US/Canada Central
> Subject: [Nek5000-users] New Recycling BC method?
>
>
>
>
> Hello All,
>
>
>
> I am wondering if there is another technique to set up a recycling boundary condition that does not depend on the elements being in a certain order (say based on element coordinates for instance)?   I would like to pick the recycling plane based on a given offset distance rather than a number of elements.
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> Also, the outflow plane faces may not all necessarily be face 6 (and face 5 for the inflow plane).
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> I have tried going through the examples where a recycling BC is used however I am finding it difficult to understand as my knowledge of Fortan is still growing. Thanks for any help anyone can provide with this idea.
>
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> Regards,
>
> Michael M.
>
>
> Â 
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