Hello
the USRBDX scoring should be FLUENCE, not current (what(1)=111 , not 11)
( in an isotropic field, there is a factor two between fluence and
current, that is exactly what you find)
Regards
On Wed, 18 Mar 2020, Alessandro Calamida wrote:
> 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.
>>>
>>>
>>>
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>
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Received on Wed Mar 18 2020 - 17:03:32 CET