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From: Alessandro Calamida <alex.calamida2_at_tiscalinet.it>

Date: Wed, 18 Mar 2020 11:54:22 +0100

Dear FLUKA experts,

I tried with differents bodys and test surface. Unfortunately the method

that you suggested seems to not work for me.

The last attempt was with a sphere and it leads the results of At=624.86

cm^2 instead of an Ar=314.16 cm^2 as it should be. The sphere has a

radius of 5 cm.

Maybe I made some mistakes. What is the correct value given by the

usrbdx scoring that I have to use?

I attach on the email the input and the scoring file.

Best regards and thank you for your time, Alessandro Calamida.

Il 16/03/2020 10:27, Answers ha scritto:

*> 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.
*

*>>
*

*>>
*

*>>
*

*>> --
*

*>> Questa e-mail è stata controllata per individuare virus con Avast
*

*>> antivirus.
*

*>> https://www.avast.com/antivirus
*

*>>
*

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Received on Wed Mar 18 2020 - 13:53:39 CET

Date: Wed, 18 Mar 2020 11:54:22 +0100

Dear FLUKA experts,

I tried with differents bodys and test surface. Unfortunately the method

that you suggested seems to not work for me.

The last attempt was with a sphere and it leads the results of At=624.86

cm^2 instead of an Ar=314.16 cm^2 as it should be. The sphere has a

radius of 5 cm.

Maybe I made some mistakes. What is the correct value given by the

usrbdx scoring that I have to use?

I attach on the email the input and the scoring file.

Best regards and thank you for your time, Alessandro Calamida.

Il 16/03/2020 10:27, Answers ha scritto:

-- Questa e-mail è stata controllata per individuare virus con Avast antivirus. https://www.avast.com/antivirus

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- text/plain attachment: Test_Area.inp

- text/plain attachment: Test_Area_21_tab.lis

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