[Users] GRHydro::Con2Prim question
g.crotti1 at campus.unimib.it
Fri Nov 15 04:29:48 CST 2019
thank you for the explanation.
Actually, I did mean the fifth reflevel (e.g. level "4", counting from 0).
My thoughts were the following: if the finest reflevel is the eighth (that
is reflevel "7"), at this reflevel the code is executed at each iteration;
on reflevel "6", every two iterations; on reflevel "5", every four
iterations and on reflevel "4" (on which I get the crash) every eight
iterations. So everything should be correct, if I have understood you
Il giorno gio 14 nov 2019 alle ore 18:22 Haas, Roland <rhaas at illinois.edu>
> Hello Giulia,
> this has to do with the order of timesteps when doing subcycling in
> time (which is what Carpet does).
> cctk_iteration counts RHS evaluations, and is not directly related to
> cctk_time ie cctk_time != cctk_iteration * DELTA_TIME.
> Basically the coarsest reflevel has to step first, so that in iteration
> 1 (evaluating the RHS for the first time) all reflevels 0..7 step
> forward and evaluate the RHS in the EVOL bin.
> In the ANALYSIS bin code only runs once data has been properly
> restricted from the finer ones etc. Which happens at slightly different
> times than EVOL.
> The net result is that in *EVOL* a function runs in refinement levels
> whenever (cctk_iteration - 1) % reflevel_every == 0 while in *ANALYSIS*
> it is cctk_iteration % reflevel_every == 0. The strange rule in EVOL
> comes from asking "was the last iteration one where the timelevel was
> aligned so that I now need to compute a new RHS"?
> Thus 473 - 1 = 472 which is evenly divided by 8. NB you say reflevel 4,
> but do you mean the fourth reflevel (which is level "3" since they
> count from "0")?
> > Hi all,
> > I am using GRHydro in running a simulation, and I get a crash in
> > Con2Prim at iteration 473 on reflevel 4. Given that I have set 8
> > refinement levels, I'm wondering why Con2Prim is being executed at
> > iteration 473 on this reflevel: shouldn't it be executed only at an
> > iteration which is a multiple of 8? Maybe I'm missing something
> > fundamental. Thank you,
> > Cheers,
> > Giulia Crotti
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