From chemistry-request@server.ccl.net Thu May 15 08:50:00 2003 Received: from dedalus.lcc.ufmg.br ([150.164.65.10]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FCnxJ02417 for ; Thu, 15 May 2003 08:49:59 -0400 Received: from dedalus.lcc.ufmg.br (loopback [127.0.0.1]) by dedalus.lcc.ufmg.br (8.12.8p1/8.12.8) with ESMTP id h4FChr3q047196; Thu, 15 May 2003 09:43:53 -0300 Received: from localhost (amolive@localhost) by dedalus.lcc.ufmg.br (8.12.8p1/8.12.8/Submit) with ESMTP id h4FChqlf041428; Thu, 15 May 2003 09:43:52 -0300 Date: Thu, 15 May 2003 09:43:51 -0300 (BSC) From: andre mauricio de oliveira X-X-Sender: amolive@dedalus To: Cesar Millan cc: CHEMISTRY@ccl.net Subject: Re: CCL:RNA 3D structure software In-Reply-To: <20030515012012.26973.qmail@web21010.mail.yahoo.com> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Dear Dr Millan, Program IsisDraw 2.2 or above (www.mdl.com) can perform it. Greets. On Wed, 14 May 2003, Cesar Millan wrote: > > Hi everyone! Does anyone knows a software that can > build a 3D structure of RNA from sequence or 2D > structure? All your help is really appreciate. > > Thanks. > > > __________________________________ > Do you Yahoo!? > The New Yahoo! Search - Faster. Easier. Bingo. > http://search.yahoo.com > > > -= This is automatically added to each message by mailing script =- > CHEMISTRY@ccl.net -- To Everybody | CHEMISTRY-REQUEST@ccl.net -- To Admins > Ftp: ftp.ccl.net | WWW: http://www.ccl.net/chemistry/ | Jan: jkl@ccl.net > > > > > > Andre Mauricio de Oliveira VOICE +55-031-374-1325 +55-031-499-5765 FAX +55-031-499-5700 Laboratorio de QSAR e Modelagem Molecular Nucleo de Estudos em Quimica Medicinal (NEQUIM) NEQUIM's Homepage: http://www.qui.ufmg.br/~nequim Departamento de Quimica ICEx Federal University of Minas Gerais Av Antonio Carlos 6627 Pampulha ZIP CODE 31270-901 Belo Horizonte MG Brazil From chemistry-request@server.ccl.net Thu May 15 10:11:39 2003 Received: from antivirus1.its.rochester.edu ([128.151.57.50]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FEBdJ07728 for ; Thu, 15 May 2003 10:11:39 -0400 Received: from antivirus1.its.rochester.edu (localhost [127.0.0.1]) by antivirus1.its.rochester.edu (8.12.9/8.12.4) with ESMTP id h4FEBcHs002894 for ; Thu, 15 May 2003 10:11:38 -0400 (EDT) Received: from mail.rochester.edu (mail1.ats.rochester.edu [128.151.224.31]) by antivirus1.its.rochester.edu (8.12.9/8.12.4) with SMTP id h4FEBW2b002862; Thu, 15 May 2003 10:11:32 -0400 (EDT) Received: from frenkel.chem.rochester.edu (frenkel.chem.rochester.edu [128.151.195.84]) by mail.rochester.edu (8.12.9/8.12.1) with ESMTP id h4FEBVDV011432; Thu, 15 May 2003 10:11:31 -0400 (EDT) Content-Type: text/plain; charset="us-ascii" From: wei Reply-To: weiz@mail.rochester.edu Organization: university of rochester To: Rick Venable , Rick Venable Subject: a question about design charge distribution in charmm Date: Thu, 15 May 2003 10:12:16 -0400 User-Agent: KMail/1.4.1 Cc: chemistry@ccl.net MIME-Version: 1.0 Message-Id: <200305151012.16286.weiz@mail.rochester.edu> Content-Transfer-Encoding: 8bit X-MIME-Autoconverted: from quoted-printable to 8bit by server.ccl.net id h4FEBdJ07729 Dear Rick: I need to do a simulation of a molecule which has an azobenzene group ( -phe-N=N-phe-),I need to set the partial charge for this group, I did a QM calculation with B3LYP/6-311G(d) for the azobenzene in vacuum and get the charges, now the question is can I use it directly in rtf file, as I remember, the charges in charmm are always smaller than reality, is it right? if I can not use it, what should I do? wei From chemistry-request@server.ccl.net Thu May 15 12:29:46 2003 Received: from mail.biosolveit.de ([194.8.222.62]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FGTjJ17221 for ; Thu, 15 May 2003 12:29:45 -0400 Received: from localhost (localhost [127.0.0.1]) by mail.biosolveit.de (Postfix) with ESMTP id 97F442C000 for ; Thu, 15 May 2003 18:29:33 +0200 (CEST) Received: by mail.biosolveit.de (Postfix, from userid 65534) id 350132C004; Thu, 15 May 2003 18:29:33 +0200 (CEST) Received: from nu.biosolveit.local (iota.biosolveit.local [192.168.9.10]) (using TLSv1 with cipher RC4-MD5 (128/128 bits)) (Client did not present a certificate) by mail.biosolveit.de (Postfix) with ESMTP id C46662813D for ; Thu, 15 May 2003 18:29:32 +0200 (CEST) From: BioSolve IT Reply-To: workshop@biosolveit.de Organization: BioSolveIT GmbH To: chemistry@ccl.net Subject: flexx docking workshop 2003 Date: Thu, 15 May 2003 18:29:32 +0200 User-Agent: KMail/1.5.1 MIME-Version: 1.0 Content-Disposition: inline Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Message-Id: <200305151829.32070.workshop@biosolveit.de> X-Spam-Status: No, hits=-100.5 required=5.0 tests=AWL,USER_AGENT_KMAIL,USER_IN_WHITELIST version=2.54 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.54 (1.174.2.17-2003-05-11-exp) X-Virus-Scanned: by AMaViS snapshot-20020531 Docking Workshop 2003 FlexX: Professional Usage for Virtual Screening and Structure-based Drug Design October 9-10, 2003 Center for BioInformatics Hamburg (ZBH) University of Hamburg Hamburg, Germany Organized by BioSolveIT GmbH St. Augustin, Germany in cooperation with Matthias Rarey at ZBH http://www.biosolveit.de/workshops/2003 scope ---------------------------------------------------------- The workshop will give an in-depth introduction to the molecular docking program FlexX. We will cover both the efficiency and accuracy of docking calculations. You will learn how to set up and optimize your use of the stand-alone version of FlexX. Start working in parallel and automate your docking runs using scripts and optimized parameters. Benefit from our new Python module py-FlexX. Add-ons and new developments will also be included in the training: FlexE, FlexX-Pharm and more. During dinner the evening before the workshop you may also like to take the opportunity to learn more about other 'BioSolve IT' tools. target users ---------------------------------------------------------- The course is designed for novices and moderately experienced users. A basic knowledge of standard UNIX commands is assumed. The workshop will be held in English. venue ---------------------------------------------------------- The venue is equipped with PCs so the major part of the workshop will be devoted to hands-on training. tentative agenda ---------------------------------------------------------- The training sessions will be held on Thursday (9th) from 9.00 to 18.00, on Friday from 9.00 to 16.00. Sufficient time for breaks and lunches will be provided. fees ---------------------------------------------------------- The fees include all learning materials, lunches and refreshments during course times plus the welcome dinner on the evening of 8th (proposed arrival date). Early bird rate: (closes on June 30th) 1000 EUR Regular fee: (from July 1st) 1400 EUR academic discount --------------------------------------------- For a limited number of academic participants who register before June 30th, we offer a 40% discount on the early bird rate. All applications coming in later than June 30th have to be charged the regular fee. registration --------------------------------------------- Starting today (5/15) we will be happy to register you on a first come, first served basis. Please note that for this hands-on training, space is quite limited so please register early. If you want to attend the workshop please register through: http://www.biosolveit.de/workshops/2003/registration.html From chemistry-request@server.ccl.net Thu May 15 11:01:03 2003 Received: from Lilly.com ([40.33.1.1]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FF12J11424 for ; Thu, 15 May 2003 11:01:02 -0400 Received: from LILLY.COM ([40.25.2.46]) by INET.D48.LILLY.COM (PMDF V6.1-1 #30673) with ESMTP id <01KVX0C02M6Y002MUQ@INET.D48.LILLY.COM> for CHEMISTRY@ccl.net; Thu, 15 May 2003 10:01:00 -0500 (EST) Received: from mcntusmail25.d51.lilly.com (mcntusmail25.d51.lilly.com [40.1.134.25]) by CLAVIN.D51.LILLY.COM (PMDF V6.1-1 #30673) with ESMTP id <01KVX0BXAUM2001L82@CLAVIN.D51.LILLY.COM>; Thu, 15 May 2003 10:00:58 -0500 (EST) Content-return: prohibited Date: Thu, 15 May 2003 10:00:57 -0500 From: WU_GUOSHENG@Lilly.com Subject: Re: CCL:Error estimation of gibbs free energy To: Ioana Cozmuta Cc: CHEMISTRY@ccl.net Message-id: MIME-version: 1.0 X-Mailer: Lotus Notes Release 5.07a May 14, 2001 Content-type: multipart/alternative; boundary="=_alternative 00527CB105256D27_=" X-MIMETrack: Serialize by Router on USMAIL25/AM/LLY(Release 5.0.10 |March 22, 2002) at 05/15/2003 10:00:58 AM, Serialize complete at 05/15/2003 10:00:58 AM This is a multipart message in MIME format. --=_alternative 00527CB105256D27_= Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Dear Ioana, Could you please point out if there is any reference for your equation? I am sure there must be such kind of thing in some data analysis books,=20 but it's just inconvenient for me to find it at this moment. It seems to me the straightforwad way to get an error estimation is by the basis calculus, like following (with similar presentation to=20 yours): for f =3D f(x1, x2, ...), (del=5Ff)=3D (df/dx1)*(del=5Fx1) + (df/dx2)*(del=5Fx2) + ... Then, (del=5Ff)^2 can be the next form, if one has the interest: (del=5Ff)^2=3D (df/dx1)^2 * (del=5Fx1)^2 + (df/dx2)^2 * (del=5Fx2)^2 + 2*(df/dx1)* (del=5Fx1) * (df/dx2) * (del=5Fx2) + ... So for your example if f=3D x+y then (del=5Ff) =3D (del=5Fx)+(del=5Fy) if f =3D x*y then simply=20 del=5Ff =3D x * (del=5Fy) + y * (del=5Fx) If one like, (del=5Ff)/f =3D (del=5Fx)/x + (del=5Fy)/y=20 or [(del=5Ff)/f]^2 =3D (del=5Fx/x)^2 + (del=5Fy/y)^2=20 + 2 * (del=5Fx)*(del=5Fy)/(x*y) Of course, for practical usage, one may overlook one or more terms if they are much smaller than others. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -=20 Thanks for your attention. Sincerely, Guosheng Guosheng Wu, Ph.D. Eli Lilly & Company Ioana Cozmuta Sent by: Computational Chemistry List 05/14/2003 01:32 PM =20 To: "'CHEMISTRY@ccl.net'" cc:=20 Subject: CCL:Error estimation of gibbs free energy Hi, The most general formula from which you can always estimate the error del=5Ff of a function f(x1, x2, x3, ...xn) is (del=5Ff)^2=3D (df/dx1)^2*del=5Fx1^2+(df/dx2)^2*del=5Fx2^2+... where df/dxi are partial derivatives of the function f with respect to xi (i=3D1,n) and del=5Fxi are the individual errors. So for example if f=3D x+y then (del=5Ff)^2 =3D (del=5Fx1)^2+(del=5Fx2)^2 if f =3D x*y then (delf/f)^2=3D (del=5Fx/x)^2+(del=5Fy/y)^2 If x1, ...xn are functions of other variables I find the general formula above (with derivatives) much easier to use. Ioana On Wed, 14 May 2003, VITORGE Pierre 094605 wrote: > d(dG) =3D square root([d(dH)]^2 + [d(dS)]^2) > cannot be true, if dh=3D0 and T is constant the above Eq. gives > d(dG) =3D d(dS) > while it is of course > d(dG) =3D T d(dS) > > You probably mean: > d(DG) =3D square root([d(DH)]^2 + [(Td(DS)+(DS)dT)]^2)? > or > d(DG) =3D square root([d(DH)]^2 + [Td(DS)]^2 + [(DS)dT]^2)? > > Pierre Vitorge > CEA DEN Saclay DPC/SECR/LSRM & UMR 8587 (CEA-CNRS-Universite d'Evry) > Universite d'Evry UMR 8587 > pierre.vitorge@cea.fr > http://perso.club-internet.fr/vitorgen/pierre/pierre.html > > > -----Message d'origine----- > De: Dr. Richard L. Wood [mailto:rlw28@cornell.edu] > Date: lundi 12 mai 2003 15:58 > =C0: Wong Lai Ho > Cc: CHEMISTRY@ccl.net > Objet: CCL:Error estimation of gibbs free energy > > > Wouldn't the following be true > > d(dG) =3D square root([d(dH)]^2 + [d(dS)]^2)? > > dG is delta G, d(dG) is the error in delta G, dH is delta H, d(dH) is=20 the > error > in delta H, dS is delta S and d(dS) is the error in delta S. > > This is taken from a simple error propagation analysis as taught in some > physical/analytical chemistry laboratories. > > Richard > > Wong Lai Ho wrote: > > > Dear CCLers, > > > > I have a question regarding error estimation of the experimental=20 values of > > gibbs free energy. > > As we knows that: > > delta G(formation) > > =3D delta H(formation) > > - Temp * [entropy(molecule)-sum(entropy(atoms to form the=20 molecules))] > > > > Can I claim that the experimental error bar of the gibbs free=20 energy(delta > > G) is equal to the error bar of the enthalpy change of formation(delta = H)? > > (since the error introduced by the entropy is usually very small) > > Is there any papers or books talking about this error? > > > > I would be grateful if you can give some comments on this claim. > > > > Larry > > > > -- > Richard L. Wood, Ph. D. > Physical/Computational Chemist > Post-doctoral Associate > Department of Chemistry > Trinity University, San Antonio, TX 78212 > > > > > > > > > > > > > > > > > -=3D This is automatically added to each message by mailing script =3D- CHEMISTRY@ccl.net -- To Everybody | CHEMISTRY-REQUEST@ccl.net -- To=20 Admins Ftp: ftp.ccl.net | WWW: http://www.ccl.net/chemistry/ | Jan: jkl@ccl.net --=_alternative 00527CB105256D27_= Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Dear Ioana,

Could you please point out if there = is any reference for your equation?

I am sure there must be such kind of= thing in some data analysis books,
but it's just inconvenient for me to= find it at this moment.

It seems to me the straightforwad wa= y to get an error estimation is
by the basis calculus, like followin= g (with similar presentation to yours):

for  f =3D f(x1, x2, ...),

(del=5Ff)=3D (df/dx1)*(del=5Fx1) + (df/dx2)*(del=5Fx2) + ...

Then, (del=5Ff)^2 can be the next f= orm, if one has the interest:

(del=5Ff)^2=3D (df/dx1)^2 * (del=5Fx1)^2 + (df/dx2)^2 * (del=5Fx2)^2
         += 2*(df/dx1)* (del=5Fx1) * (df/dx2) * (del=5Fx2) + ...

So for your example if f=3D x+y the= n (del=5Ff) =3D (del=5Fx)+(del=5Fy)
if f =3D x*y then simply
          =       del=5Ff =3D x * (del=5Fy) + y * (del=5Fx)

If one like, (del=5Ff)/f =3D (del= =5Fx)/x + (del=5Fy)/y

    or   [(del=5Ff)/= f]^2 =3D (del=5Fx/x)^2 + (del=5Fy/y)^2
          =              + 2 * (del=5Fx)*(del=5Fy)/(= x*y)

Of course, for practical usage, one= may overlook one or more terms if they
are much smaller than others.
- - - - - - - - - - - - - - - - - -= - - - - - - - - - - - - - - -
Thanks for your attention.

Sincerely,

Guosheng

Guosheng Wu, Ph.D.
Eli Lilly & Company



Ioana Cozmuta <ioana@nas.nasa.= gov>
Sent by: Computational Chemistry Lis= t <chemistry-request@ccl.net>

05/14/2003 01:32 PM

       
        To: &nbs= p;      "'CHEMISTRY@ccl.net'" <CHEMISTRY@ccl.ne= t>
        cc: &nbs= p;      
        Subject:=        CCL:Error estimation of gibbs free energy



Hi,

The most general formula from which you can always estimate the error
del=5Ff of a function f(x1, x2, x3, ...xn) is

(del=5Ff)^2=3D (df/dx1)^2*del=5Fx1^2+(df/dx2)^2*del=5Fx2^2+...

where df/dxi are partial derivatives of the function f with respect to xi (i=3D1,n) and del=5Fxi are the individual errors.

So for example if f=3D x+y then (del=5Ff)^2 =3D (del=5Fx1)^2+(del=5Fx2)^2 if f =3D x*y then (delf/f)^2=3D (del=5Fx/x)^2+(del=5Fy/y)^2

If x1, ...xn are functions of other variables I find the general formula
above (with derivatives) much easier to use.

Ioana

On Wed, 14 May 2003, VITORGE Pierre 094605 wrote:

> d(dG) =3D square root([d(dH)]^2 + [d(dS)]^2)
> cannot be true, if dh=3D0 and T is constant the above Eq. gives
> d(dG) =3D d(dS)
> while it is of course
> d(dG) =3D T d(dS)
>
> You probably mean:
> d(DG) =3D square root([d(DH)]^2 + [(Td(DS)+(DS)dT)]^2)?
> or
> d(DG) =3D square root([d(DH)]^2 + [Td(DS)]^2 + [(DS)dT]^2)?
>
> Pierre Vitorge
> CEA DEN Saclay DPC/SECR/LSRM & UMR 8587 (CEA-CNRS-Universite d'Evr= y)
> Universite d'Evry UMR 8587
> pierre.vitorge@cea.fr
> http://perso.club-internet.fr/vitorgen/pierre/pierre.html
>
>
> -----Message d'origine-----
> De: Dr. Richard L. Wood [mailto:rlw28@cornell.edu]
> Date: lundi 12 mai 2003 15:58
> =C0: Wong Lai Ho
> Cc: CHEMISTRY@ccl.net
> Objet: CCL:Error estimation of gibbs free energy
>
>
> Wouldn't the following be true
>
> d(dG) =3D square root([d(dH)]^2 + [d(dS)]^2)?
>
> dG is delta G, d(dG) is the error in delta G, dH is delta H, d(dH) is = the
> error
> in delta H, dS is delta S and d(dS) is the error in delta S.
>
> This is taken from a simple error propagation analysis as taught in so= me
> physical/analytical chemistry laboratories.
>
> Richard
>
> Wong Lai Ho wrote:
>
> > Dear CCLers,
> >
> > I have a question regarding error estimation of the experimental = values of
> > gibbs free energy.
> > As we knows that:
> > delta G(formation)
> >    =3D delta H(formation)
> >    - Temp * [entropy(molecule)-sum(entropy(atoms to for= m the molecules))]
> >
> > Can I claim that the experimental error bar of the gibbs free ene= rgy(delta
> > G) is equal to the error bar of the enthalpy change of formation(= delta H)?
> > (since the error introduced by the entropy is usually very small)=
> > Is there any papers or books talking about this error?
> >
> > I would be grateful if you can give some comments on this claim.<= br> > >
> > Larry
> >
>
> --
> Richard L. Wood, Ph. D.
> Physical/Computational Chemist
> Post-doctoral Associate
> Department of Chemistry
> Trinity University, San Antonio, TX 78212
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>


-=3D This is automatically added to each message by mailing script =3D-
CHEMISTRY@ccl.net -- To Everybody  | CHEMISTRY-REQUEST@ccl.net -- To A= dmins
Ftp: ftp.ccl.net  |  WWW: http://www.ccl.net/chemistry/   | = Jan: jkl@ccl.net

--=_alternative 00527CB105256D27_=-- From chemistry-request@server.ccl.net Thu May 15 14:35:06 2003 Received: from marcy.nas.nasa.gov ([129.99.113.17]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FIZ5J20576 for ; Thu, 15 May 2003 14:35:06 -0400 Received: from localhost (ioana@localhost) by marcy.nas.nasa.gov (8.11.7+Sun/8.11.6/NAS 8.11.6-5n) with ESMTP id h4FIZ5q20355; Thu, 15 May 2003 11:35:05 -0700 (PDT) Date: Thu, 15 May 2003 11:35:05 -0700 (PDT) From: Ioana Cozmuta To: WU_GUOSHENG@Lilly.com cc: CHEMISTRY@ccl.net Subject: Re: CCL:Error estimation of gibbs free energy In-Reply-To: Message-ID: References: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=X-UNKNOWN Content-Transfer-Encoding: 8bit X-MIME-Autoconverted: from QUOTED-PRINTABLE to 8bit by server.ccl.net id h4FIZ6J20577 Hi Guosheng, > From the top of my head, that formula comes from a Taylor expansion in which you neglect the higher-order terms. Honestly, the best info I have on error propagation are my hand written notes from the courses I followed during undergrad/grad studies. However, if you do a search on google with the following terms: general formula error propagation it gives lots of references. The most frequently mentioned reference I see is Taylor, J. R., "An introduction to Error Analysis", University Science Books, 1982. I also think that any book on numerical analysis should have at least a chapter that covers error propagation. See also: http://mulliken.chem.hope.edu/~polik/Chem345-2000/errorpropagation.htm Regards, Ioana **************************************************************************** * Ioana Cozmuta, PhD * * * NASA-AMES Research Center * "Gravitation can not be held responsible* * Mail Stop 230-3 * for people falling in love" * * Moffet Field,CA 94035 * * * phone: (650) 604-0993 * Albert Einstein* * fax: (650) 604-0350 * (1879-1955) * **************************************************************************** On Thu, 15 May 2003 WU_GUOSHENG@Lilly.com wrote: > Dear Ioana, > > Could you please point out if there is any reference for your equation? > > I am sure there must be such kind of thing in some data analysis books, > but it's just inconvenient for me to find it at this moment. > > It seems to me the straightforwad way to get an error estimation is > by the basis calculus, like following (with similar presentation to > yours): > > for f = f(x1, x2, ...), > > (del_f)= (df/dx1)*(del_x1) + (df/dx2)*(del_x2) + ... > > Then, (del_f)^2 can be the next form, if one has the interest: > > (del_f)^2= (df/dx1)^2 * (del_x1)^2 + (df/dx2)^2 * (del_x2)^2 > + 2*(df/dx1)* (del_x1) * (df/dx2) * (del_x2) + ... > > So for your example if f= x+y then (del_f) = (del_x)+(del_y) > > if f = x*y then simply > del_f = x * (del_y) + y * (del_x) > > If one like, (del_f)/f = (del_x)/x + (del_y)/y > > or [(del_f)/f]^2 = (del_x/x)^2 + (del_y/y)^2 > + 2 * (del_x)*(del_y)/(x*y) > > Of course, for practical usage, one may overlook one or more terms if they > are much smaller than others. > - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - > Thanks for your attention. > > Sincerely, > > Guosheng > > Guosheng Wu, Ph.D. > Eli Lilly & Company > > > > > > Ioana Cozmuta > Sent by: Computational Chemistry List > 05/14/2003 01:32 PM > > > To: "'CHEMISTRY@ccl.net'" > cc: > Subject: CCL:Error estimation of gibbs free energy > > > Hi, > > The most general formula from which you can always estimate the error > del_f of a function f(x1, x2, x3, ...xn) is > > (del_f)^2= (df/dx1)^2*del_x1^2+(df/dx2)^2*del_x2^2+... > > where df/dxi are partial derivatives of the function f with respect to xi > (i=1,n) and del_xi are the individual errors. > > So for example if f= x+y then (del_f)^2 = (del_x1)^2+(del_x2)^2 > if f = x*y then (delf/f)^2= (del_x/x)^2+(del_y/y)^2 > > If x1, ...xn are functions of other variables I find the general formula > above (with derivatives) much easier to use. > > Ioana > > On Wed, 14 May 2003, VITORGE Pierre 094605 wrote: > > > d(dG) = square root([d(dH)]^2 + [d(dS)]^2) > > cannot be true, if dh=0 and T is constant the above Eq. gives > > d(dG) = d(dS) > > while it is of course > > d(dG) = T d(dS) > > > > You probably mean: > > d(DG) = square root([d(DH)]^2 + [(Td(DS)+(DS)dT)]^2)? > > or > > d(DG) = square root([d(DH)]^2 + [Td(DS)]^2 + [(DS)dT]^2)? > > > > Pierre Vitorge > > CEA DEN Saclay DPC/SECR/LSRM & UMR 8587 (CEA-CNRS-Universite d'Evry) > > Universite d'Evry UMR 8587 > > pierre.vitorge@cea.fr > > http://perso.club-internet.fr/vitorgen/pierre/pierre.html > > > > > > -----Message d'origine----- > > De: Dr. Richard L. Wood [mailto:rlw28@cornell.edu] > > Date: lundi 12 mai 2003 15:58 > > À: Wong Lai Ho > > Cc: CHEMISTRY@ccl.net > > Objet: CCL:Error estimation of gibbs free energy > > > > > > Wouldn't the following be true > > > > d(dG) = square root([d(dH)]^2 + [d(dS)]^2)? > > > > dG is delta G, d(dG) is the error in delta G, dH is delta H, d(dH) is > the > > error > > in delta H, dS is delta S and d(dS) is the error in delta S. > > > > This is taken from a simple error propagation analysis as taught in some > > physical/analytical chemistry laboratories. > > > > Richard > > > > Wong Lai Ho wrote: > > > > > Dear CCLers, > > > > > > I have a question regarding error estimation of the experimental > values of > > > gibbs free energy. > > > As we knows that: > > > delta G(formation) > > > = delta H(formation) > > > - Temp * [entropy(molecule)-sum(entropy(atoms to form the > molecules))] > > > > > > Can I claim that the experimental error bar of the gibbs free > energy(delta > > > G) is equal to the error bar of the enthalpy change of formation(delta > H)? > > > (since the error introduced by the entropy is usually very small) > > > Is there any papers or books talking about this error? > > > > > > I would be grateful if you can give some comments on this claim. > > > > > > Larry > > > > > > > -- > > Richard L. Wood, Ph. D. > > Physical/Computational Chemist > > Post-doctoral Associate > > Department of Chemistry > > Trinity University, San Antonio, TX 78212 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -= This is automatically added to each message by mailing script =- > CHEMISTRY@ccl.net -- To Everybody | CHEMISTRY-REQUEST@ccl.net -- To > Admins > Ftp: ftp.ccl.net | WWW: http://www.ccl.net/chemistry/ | Jan: jkl@ccl.net > > > > > > > > From chemistry-request@server.ccl.net Thu May 15 18:13:47 2003 Received: from kafka.net.nih.gov ([165.112.130.10]) by server.ccl.net (8.11.6/8.11.0) with ESMTP id h4FMDkJ26850 for ; Thu, 15 May 2003 18:13:46 -0400 Received: from pollux.cber.nih.gov (pollux.cber.nih.gov [128.231.52.5]) by kafka.net.nih.gov (8.12.9/8.12.9) with ESMTP id h4FMDix9002440 for ; Thu, 15 May 2003 18:13:44 -0400 (EDT) Date: Thu, 15 May 2003 17:47:09 -0400 From: Rick Venable To: chemistry@ccl.net Subject: molecular animation tools In-Reply-To: <200305151012.16286.weiz@mail.rochester.edu> Message-ID: References: <200305151012.16286.weiz@mail.rochester.edu> ReplyTo: Rick_Venable@nih.gov MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII I've been making molecular animations from CHARMM trajectories in MPEG1 format using mpeg_encode* for a while, but I'm having trouble finding good MPEG1 players for M$ Windows and MacOS these days. The only 2 that were decent were VMPEG for M$, and Sparkle on the Mac; in particular, they allow the user to set the frame rate for playback. However, these haven't been updated in some time; I don't think there's a native OSX version of Sparkle, for instance. * from a series of still frames rendered via POV-Ray In contrast, newer apps either don't play MPEG1 at all (M$ Media Player, for instance) and others will play the MPEG1 files, but offer no means of controlling the frame rate (QuickTime, RealPlayer). My questions are: Are there other inexpensive or free media players for M$ and Mac that can handle MPEG1 and can provide a way to set the playback speed? Can anyone recommend other inexpensive animation software for making "movies" in a common format from a series of still frames? Is anyone aware of conversion tools, that might convert MPEG1 to QuickTime or other common format? I prefer Linux tools, but I'll use whatever works. Thanks. =+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+= Rick Venable 29/500 FDA/CBER/OVRR Biophysics Lab 1401 Rockville Pike HFM-419 Rockville, MD 20852-1448 U.S.A. (301) 496-1905 Rick_Venable@nih.gov ALT email: rvenable@speakeasy.org ------------------------------------- "Don't blame me, I voted for Kang." Homer =+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=