Re: Plotting USRBDX with one angular bin

From: Sebastien WURTH (wurth@ipno.in2p3.fr)
Date: Tue Sep 19 2006 - 17:00:10 CEST

  • Next message: Alberto Fasso': "Re: Plotting USRBDX with one angular bin"

    All right, apparently, it was not very clear for me, although I saw it
    (the fact that you must not multiply again by 2 pi or 4 pi) by
    experience on comparing some USRTRACK an "similar" USRBDX results...
    But then with the same example (one angular bin), in the *_tab.lis file
    when it says "integrated over solid angle" at the first lign, it refers
    to the differential Flux as a function of energy in (Part/GeV/cmq/pr) in
    *_sum.lis file, I do have to multiply by each energy bin value to obtain
    integrated flux in part/cmq/pr. Am I right this time ?

    Thank you.
    Regards.
    Sebastien.

    Alberto Fasso' a écrit :

    >Be careful: the manual says that you must multiply by 2 pi or 4 pi,
    >but assuming that you do the integration yourself. The usxsuw program
    >takes already care of that, so you must not do it again
    >
    >Alberto
    >
    >On Tue, 19 Sep 2006, Sebastien WURTH wrote:
    >
    >
    >
    >>Hello,
    >>
    >>In your example, you cannot have a neutron fluence, you put WHAT(2) =
    >>7.0 in your first USRBDX card, this would give you photon fluence (maybe
    >>you choose only the wrong example to illustrate).
    >>To answer your question (I was asking me the same one once), I quote the
    >>manual from USRBDX part :
    >>*********
    >>Notes
    >>1. IMPORTANT! The results of a USRBDX boundary crossing estimator are
    >>always given as double differential
    >>distributions of fluence (or current) in energy and solid angle, in
    >>units of cm−2 GeV−1 sr−1 per incident primary,
    >>even when only 1 interval (bin) has been requested, which is often the
    >>case for angular distributions.
    >>Thus, for example, when requesting a fluence or current energy spectrum,
    >>with no angular distribution, to
    >>obtain integral binned results (fluence or current in cm−2 per energy
    >>bin per primary) one must multiply the
    >>value of each energy bin by the width of the bin (even for logarithmic
    >>binning), and by 2 pi or 4 pi (depending
    >>on whether one-way or two-way scoring has been requested).[...]
    >>*********
    >>
    >>Best Regards.
    >>Sebastien.
    >>
    >>
    >>
    >>
    >>Nicole Patricia Lee Pratt-Boyden a écrit :
    >>
    >>
    >>
    >>>Hello,
    >>>
    >>>I'm using a USRBDX card to score neutron fluence (linear angular and
    >>>energy bins). The programme seems to be running fine, and I'm compiling
    >>>the data with the usxsuw.f programme.
    >>>
    >>>I asked for eight energy bins, and one angular bin.
    >>>
    >>>USRBDX 101.0 7.0 -48.0 3.0 4.0 400.0
    >>>PH-199m
    >>>USRBDX 6.5E-4 2.5E-4 8.0 0.0 1. &
    >>>
    >>>The *_tab file only returns the differential fluence integrated over
    >>>solid angle; there are no double differential values returned at all,
    >>>no mention of solid angle boundaries.
    >>>
    >>>Assuming this is fine, do I still need to multiply the differential
    >>>values by 2PI (one-way scoring) in order to get diff. fluence in
    >>>particles/GeV/cm^2/primary, or do I only do this if I want fluence
    >>>(/cm^2/pr)?
    >>>
    >>>Apologies if this is extremely simple, I was just thrown by the lack of
    >>>any information for the double differential! Best regards,
    >>>
    >>>Nicole Pratt-Boyden.
    >>>
    >>>
    >>>
    >>>
    >>
    >>
    >>
    >
    >
    >


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