From: Alberto Fasso' (fasso@SLAC.Stanford.EDU)
Date: Sun Jan 13 2008 - 03:56:38 CET
Sorry, I forgot to answer your last 2 questions.
> and what if i am scoring energy in log binning, in case of neutron i am
> using log binning.
> the same eq i have to use ?
First, whether you score in log or in linear binning, for what concerns
the neutrons with energy < 20 MeV it is the same thing. In that case you
have no choice: the neutrons of energy larger than 20 MeV will be binned
log or linear as you requested, but those below 20 MeV will IN BOTH CASES be
binned according to the Fluka group structure (see Chap. 10 of the manual),
and you cannot change it.
Again, if you use the results on the ..._tab.lis, you must integrate by
multiplying by the energy width of the interval:
IIIrd column X (IInd column - Ist column)
but keeping in mind that this time the width is different for each
interval. In practice, it is often useful to write a simple program
or to use a spreadsheet to do the above operation.
If, instead, you are for instance scoring in an ASCII file, you need to do
both the above operation AND to multiply by 2pi (or 4pi, if it is a 2-way
scoring).
> Also I want to know why to multiply by 2 pi, as i am using only single
> binning of angular distribution.
Because the Fluka results are differential (dPhi/(dE dOmega), independent
of how many bins you use for E or for Omega. It would be the same thing
if you had many angular bins and one single energy bin: each bin
must be multiplied by dE and by dOmega. For many applications (especially
plotting), the differential fluence is more convenient: if you want integral
fluence in each bin, you must calculate it yourself.
Kind regards,
Alberto
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