Re: [fluka-discuss]: Area of a complicated surface

From: Alessandro Calamida <>
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
> 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 Wed Mar 18 2020 - 13:53:39 CET

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