Dear Alessandro,
Sorry for interference, but I found this interesting and checked the
suggested method. I placed cube with 50 cm length in the center of
sphere with R = 300 cm. The result is 12 779 cm^3 (error 0.279% with 23
157 436 prim. in total, 10 threads), but actual surface area of cube is:
15 000 cm^3. Then I did another try with R = 100 cm and the result is:
11985 cm^3 (error 8.8e-2% with 25 320 511 prim. in total, 10 threads).
Maybe more statistics needed, but, it works indeed! For everyone
interested see attached .inp file.
Regards,
Ivan Gordeev
16.03.2020 12:27, Answers пишет:
> Dear Alessandro
>
> there is no obvious way for computing the surface of a complicated
> region.
>
> However there could be a possibility (never tested... ) using the FLOOD
> option in the BEAMPOS card.
>
> This possibility works perfcetly when computing volumes for
> complicated regions, on paper it could work for surfaces as well (no
> guarantee).
>
> Let me explain the logic (see also the manual for the BEAMPOS card):
>
> a) your problem should have for this purpose all regions (apart
> blackhole)
> filled with vacuum;
> b) you define a beam (whichever particle, say PHOTON's) and put a
> BEAMPOS card with SDUM=FLOOD;
> c) this will generate an isotropic and uniform fluence inside the sphere
> with radius = WHAT(1) of the BEAMPOS card, be sure such radius
> contains wholly your region;
> d) the fluence so generated will be equal to 1/(pi R^2), this is an
> exact analytical result;
> e) you define a USRBDX, fluence-like, two-ways, estimator between the
> region you want to know the surface of and the surrounding, with
> normalization surface=1. It would be highly preferable/simple if the
> surrounding is made of a single region (you could make an ad hoc
> run/geometry for this purpose);
> f) you run Fluka for sufficient primaries/cycles in order to get
> a negligible statistical error on the USRBDX result (let's call
> its result F);
> g) since you know that F/Area should be equal to 1/(pi R^2) you
> can easily derive Area
>
> Normally this method is used in order to compute volumes using
> a tracklength estimator instead of a USRBDX, and using the
> equation F_track_length/Volume = 1/(pi R^2). This works perfectly
> and it is sure to give the exact answer within the statistical
> errors. On paper I do not see an obvious reason why it should not
> work for computing an area as well, there will be numerical precision
> issues since for grazing incidence a fluence boundary crossing
> estimator would result into an infinite (or better a division by
> zero). This
> is protected in the code and the influence of the necessarily
> approximate protection should be small, however as I said before we
> never tried this method.
>
> Let us know if it works!
>
> On Fri, 13 Mar 2020, Alessandro Calamida wrote:
>
>> Dear FLUKA experts,
>>
>> In my geometry I have a regione that is the union of different bodies.
>> Calculating the surface of it is quite complicated and so I cannot put
>> the normalization factor in the USRBDX scoring.
>>
>> There is a FLUKA or Flair tool that allows to evaluate surfaces.
>>
>> I attach in the email the input of the file. The region from wich I need
>> the surface is the SUPP_TRG.
>>
>> Best regards and thank you for your time, Alessandro Calamida.
>>
>>
>>
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Received on Mon Mar 16 2020 - 16:23:59 CET