<17652047.3034261377919188455.JavaMail.coremail_at_mailweb>,<30507624.3107761378035584867.JavaMail.coremail_at_mailweb>,<30844077.3141761378087717403.JavaMail.coremail_at_mailweb>
In-Reply-To: <30844077.3141761378087717403.JavaMail.coremail_at_mailweb>
Accept-Language: en-GB, en-US
Content-Language: en-GB
X-MS-Has-Attach:
X-MS-TNEF-Correlator:
x-originating-ip: [192.91.242.129]
Content-Type: text/plain; charset="iso-2022-jp"
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
Sender: owner-fluka-discuss_at_smtp2.mi.infn.it
Dear Chenyuan,=0A=
=0A=
(1) Indeed you can have many USRBIN scorings in the same simulation, for di=
fferent quantities and/or different spatial binnings.=0A=
=0A=
(2) Good.=0A=
=0A=
(3,4) For the scoring, it think you should really ask yourself what you wan=
t to get out of the simulation. I know now your physical problem, but I don=
't know what you want to calculate (still, consider my comments below for t=
he y binning). Concerning the cuts I would say that you should maybe consid=
er adapting them e.g. according to measurement conditions (if this is the a=
im of your study). Detection of low-energy electrons can for example be lim=
ited by the entrance window of a detector (or by a voltage bias applied) an=
d hence, if you want to reproduce real conditions, I would maybe consider s=
uch detection limits.=0A=
=0A=
(5,6) Your input file crashes. I advice you to at least test it first befor=
e sending it to the mailing list. The problem is with the EMFCUT card: you =
expressed the electron cut in terms of the total electron energy (WHAT(1)>0=
), however I suppose you aimed for the kinetic energy, hence WHAT(1) should=
be <0. Secondly, you misplaced the lower and upper region boundary (the sh=
ould be on WHAT(4) and (5) and not on (5) and (6)). Finally, I would also l=
ower the cut for protons. In summary, if you want to apply a cut of 1 keV, =
your card should look as follows:=0A=
EMFCUT -1.0E-06 1.0E-06 target0 target10=0A=
=0A=
On the other hand, the definition of your MULSOPT seems fine (now you apply=
single scattering close to boundaries, for too short steps and for energie=
s where Moliere theory is not applicable). As you have a very thin layer on=
ly, you might even want to consider applying single scattering in the entir=
e layer (see Example 3 in the MULSOPT entry of the manual) if your CPU time=
s are still acceptable (you can test that).=0A=
=0A=
Cheers, Anton =0A=
=0A=
=0A=
________________________________________=0A=
From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] on be=
half of yyc2011_at_mail.ustc.edu.cn [yyc2011_at_mail.ustc.edu.cn]=0A=
Sent: 02 September 2013 04:08=0A=
To: fluka-discuss_at_fluka.org=0A=
Subject: Re: RE: RE: The maximum number of binnings in USRBIN=0A=
=0A=
Dear Anton,=0A=
=0A=
Thanks for your time.Thanks for your reply.The attachment is my new input f=
ile.=0A=
=0A=
According to your reply,the followings are my questions.=0A=
=0A=
=1B$B!J=1B(B1=1B$B!K=1B(BJudging from your reply=1B$B!$=1B(Bin one input fi=
le,We can have many USRBIN scorings at the same time.In other words,We can =
get different physical quantities just from one input file.Right?=0A=
=0A=
=1B$B!J=1B(B2=1B$B!K=1B(BI understand what you said.=0A=
=0A=
=1B$B!J=1B(B3=1B$B!K=1B(BI want to get the electrons distribution on the de=
tector plane.I want to study electron radiography,just like proton radiogra=
phy.For each electron energy=1B$B!J=1B(Bthe material is certain=1B$B!K=1B(=
B,there is a corresponding range in this material.If the range is too near =
the total target thickness,the electron beam will become sensitive to the =
target thickness.In this case, the electron beam can be used to diagnose mi=
nute variations about the target thickness.This is what I want to do.=0A=
=0A=
=1B$B!J=1B(B4=1B$B!K!J=1B(B5=1B$B!K!J=1B(B6=1B$B!K=1B(BCould you see my new=
input file?Could you give me any advice?Perhaps EMFCUT and MULSOPT card ar=
e the key to the problem.I'm very rusty to the two cards.So,I wish you cou=
ld give me more suggestions and some examples.=0A=
=0A=
Yours,=0A=
Chenyuan=0A=
=0A=
=0A=
=0A=
=0A=
=0A=
=0A=
=0A=
=0A=
=0A=
> -----Original E-mail-----=0A=
> From: "Anton Lechner" <Anton.Lechner_at_cern.ch>=0A=
> Sent Time: 2013-9-2 0:29:48=0A=
> To: "yyc2011_at_mail.ustc.edu.cn" <yyc2011_at_mail.ustc.edu.cn>, "fluka-discuss=
_at_fluka.org" <fluka-discuss_at_fluka.org>=0A=
> Cc:=0A=
> Subject: RE: RE: The maximum number of binnings in USRBIN=0A=
>=0A=
> Dear Chenyuan,=0A=
>=0A=
> For clarity, I inserted different numbers in your Email below to refer to=
your specific questions. Here are my answers:=0A=
>=0A=
> (1) When saying "number of USRBINs", I mean the number of USRBIN scoring=
s you have in your input file (always counting the USRBIN card and its corr=
esponding continuation card as *one* USRBIN) and *independently* which quan=
tity you are scoring (energy deposition, fluence, etc.). For example, if yo=
u have following cards in your input file, it means you have two USRBIN sco=
rings:=0A=
> USRBIN 10. ELECTRON -30. 0.4 0.4 0.0221flu=
Elec=0A=
> USRBIN -0.4 -0.4 0.022 40.0 40.0 1.&=
=0A=
> USRBIN 10. POSITRON -30. 0.4 0.4 0.0221flu=
Pos=0A=
> USRBIN -0.4 -0.4 0.022 40.0 40.0 1.&=
=0A=
>=0A=
> (2) Let me illustrate this by means of the example given above: each USRB=
IN has 40*40*1=3D1600 bins, but since you have two USRBINs your total numbe=
r of bins is 2*1600=3D3200. This means you need a memory allocation for 320=
0 bins. Of course, if you have further USRBINs, you need to determine the n=
umber of bins for each and add them up in the same way.=0A=
>=0A=
> (3) This really depends what you actually want to calculate, i.e. what yo=
u intend to learn from the simulation. Just a hint: consider that if your g=
eometry has a granularity of 6 mm in y and at the same time you have a rect=
angular homogeneous electron field of the same size (covering all 6 mm), th=
en a 200 um bin size in y will not give you new information for most of the=
(central) bins, but only around the region boundaries. You may want to co=
nsider to use a more fine-grained binning only around boundaries. But I rep=
eat myself, I really don't know what you actually want to get out of the si=
mulation.=0A=
>=0A=
> (4) I admit that I might have expressed myself not clear enough. It is no=
t necessarily the difference between incident electron energy and transport=
cut-off/production cut which matters, but it is the artificial cut on ele=
ctrons (in your case mostly on the secondary electrons) which you introduce=
by having these cuts. As you have a fluence scoring, it will of course cha=
nge your result if you have lower cuts (since each electron is counted equa=
lly independent of its energy). Which cut you should choose once more depen=
ds on your physical problem.=0A=
>=0A=
> (5-6) You did not attach any input file (in this context, please note tha=
t the manual is always a very valuable source of information)=0A=
>=0A=
> Cheers, Anton=0A=
>=0A=
>=0A=
> ________________________________________=0A=
> From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] on =
behalf of yyc2011_at_mail.ustc.edu.cn [yyc2011_at_mail.ustc.edu.cn]=0A=
> Sent: 01 September 2013 13:39=0A=
> To: fluka-discuss_at_fluka.org=0A=
> Subject: Re: RE: The maximum number of binnings in USRBIN=0A=
>=0A=
> Dear Anton,=0A=
>=0A=
> Thanks for your reply in detail.I'm very grateful to you.=0A=
>=0A=
> I read your mail carefully and I think about your suggestions all the day=
.But I still have some questions.=0A=
>=0A=
> First, you say "maximum number of binnings" refers to the number of USRBI=
Ns I can have.What is the meaning?It means we can have many=0A=
> usrbins,one of usrbin about energy deposition,one of usrbin about fluence=
,one of usrbin about neutron balance,and so on.I understand=0A=
> it,Right?=0A=
> (1)=0A=
>=0A=
> Second,you say "It is not important how many bins you have per USRBIN, bu=
t how many you have in total for all USRBINs".I don't=0A=
> understand what you say.Could you give me a example or a detailed explain=
?=0A=
> (2)=0A=
>=0A=
> Third,you say "your smallest geometric structure is 20 um in x and 6 mm i=
n y, while you aim for a bin size of less than 1 um in both=0A=
> x and y".At this point,I make some changes in my input file.In the new in=
put file,the smallest geometric structure is still 20 um in=0A=
> x and 6 mm in y,but my aim for a bin size of 4 um in x and 200 um in y.Do=
you think I was right in doing this?=0A=
> (3)=0A=
>=0A=
> Fourth, my primary electrons energy is 160 KeV.The transport cut-off ener=
gy is 100KeV for electrons.You ask me to consider if this=0A=
> difference between primary energy and cut is sufficient for my problem.At=
this point,I want to know how can I estimate if this=0A=
> difference between primary energy and cut is sufficient for my problem".=
=0A=
> (4)=0A=
>=0A=
> Meanwhile,I add the EMFCUT card in my new input file.Please=0A=
> see my new input file.And,I will very happy if you give me some suggestio=
ns.=0A=
> (5)=0A=
>=0A=
> Fifth,you say "if I have such low primary electron energies and um geomet=
ries, I should consider applying single scattering."In my=0A=
> new input file,I add the MULSOPT card.This is the first time I use this c=
ard.Could you see my input file?Is it wrong?Maybe You can=0A=
> give me some examples.=0A=
> (6)=0A=
>=0A=
> Thanks for your time.Thanks for any suggestions.=0A=
>=0A=
> Yours,=0A=
> Chenyuan=0A=
>=0A=
>=0A=
>=0A=
>=0A=
>=0A=
>=0A=
>=0A=
>=0A=
> > -----Original E-mail-----=0A=
> > From: "Anton Lechner" <Anton.Lechner_at_cern.ch>=0A=
> > Sent Time: 2013-8-31 19:30:33=0A=
> > To: "yyc2011_at_mail.ustc.edu.cn" <yyc2011_at_mail.ustc.edu.cn>, "fluka-discu=
ss_at_fluka.org" <fluka-discuss_at_fluka.org>=0A=
> > Cc:=0A=
> > Subject: RE: The maximum number of binnings in USRBIN=0A=
> >=0A=
> > Dear Chenyuan,=0A=
> >=0A=
> > You misunderstood the comment in the manual: "maximum number of binning=
s" refers to the number of USRBINs you can have and not to the number of bi=
ns (see Manual: "A 'binning' is a regular spatial mesh"). The number of bin=
s you can have is limited by the memory available for scorings:=0A=
> > *) See following answer on the FLUKA discuss list: http://www.fluka.org=
/web_archive/earchive/new-fluka-discuss/5260.html: the memory allocation is=
hard-wired in the library and is 360 MB=0A=
> > *) It is not important how many bins you have per USRBIN, but how many =
you have in total for all USRBINs=0A=
> > *) Your mesh of 10000*10000*1 occupies 800 MB and is hence far beyond t=
he limit of allocated memory=0A=
> >=0A=
> > Apart from this, I have some further comments:=0A=
> > *) Even if such a large number of bins would be possible, you should as=
k yourself if it would make sense. First, your smallest geometric structure=
is 20 um in x and 6 mm in y, while you aim for a bin size of less than 1 =
um in both x and y. Secondly, such a large number of bins would probably be=
challenging in terms of statistical error.=0A=
> > *) Your primary electrons have an energy of 160 keV, but the transport =
cut-off and production threshold set by PRECISIO is 100 keV. You should ask=
yourself if this difference between primary energy and cut is sufficient =
for your problem. You can set lower cut values with the EMFCUT card.=0A=
> > *) Note that if you have such low primary electron energies and um geom=
etries, you should consider applying single scattering. I also quote the FL=
UKA manual: "The minimum recommended energy for PRIMARY electrons is about =
50 to 100 keV for low-Z materials and 100-200 keV for heavy materials, unle=
ss the single scattering algorithm is used. Single scattering transport all=
ows to overcome most of the limitations at low energy for the heaviest mate=
rials at the price of some increase in CPU time."=0A=
> >=0A=
> > Cheers, Anton=0A=
> >=0A=
> >=0A=
> >=0A=
> > ________________________________________=0A=
> > From: owner-fluka-discuss_at_mi.infn.it [owner-fluka-discuss_at_mi.infn.it] o=
n behalf of yyc2011_at_mail.ustc.edu.cn [yyc2011_at_mail.ustc.edu.cn]=0A=
> > Sent: 31 August 2013 05:19=0A=
> > To: fluka-discuss_at_fluka.org=0A=
> > Subject: The maximum number of binnings in USRBIN=0A=
> >=0A=
> > Dear Fluka experts,=0A=
> >=0A=
> > I have a question about the maximum number of binnings that I can defin=
e in USRBIN.It seems to go wrong when I set the number of=0A=
> > binnings=3D10000 or 8000 in my input file.But when I set the number of =
binnings=3D1000,it runs normal. Meanwhile,I note in the Fluka=0A=
> > manual"The maximum number of binnings that the user can define is 400. =
This value can be changed by modifying the parameter MXUSBN=0A=
> > in member USRBIN of the flukapro library or directory and then re-compi=
ling and linking Fluka." In the manual,It says the maximum=0A=
> > number of binnings that the user can define is 400.But when I set the n=
umber of binnings=3D1000,I do not modify the parameter=0A=
> > MXUSBN.But the fluka runs normal.The attachments are my input file and =
.out file.Could you help me?=0A=
> >=0A=
> > Thanks for any suggestions.=0A=
> >=0A=
> > Yours=0A=
> > Chenyuan=0A=
> >=0A=
>=0A=
>=0A=
Received on Mon Sep 02 2013 - 16:48:07 CEST