From chemistry-request -x- at -x- ccl.net Wed Apr 28 11:03:32 2004 Received: from mtiwmhc13.worldnet.att.net (mtiwmhc13.worldnet.att.net [204.127.131.117]) by server.ccl.net (8.12.8/8.12.8) with ESMTP id i3SG3VPG002335 for ; Wed, 28 Apr 2004 11:03:31 -0500 Received: from laptop.us.fujitsu.com (206.denver-06rh15rt.co.dial-access.att.net[12.73.181.206]) by worldnet.att.net (mtiwmhc13) with SMTP id <2004042816044511300bv88ce>; Wed, 28 Apr 2004 16:04:46 +0000 Message-Id: <6.0.1.1.2.20040428095243.0355f3e8*at*postoffice.worldnet.att.net> X-Sender: mrmopac*at*postoffice.worldnet.att.net X-Mailer: QUALCOMM Windows Eudora Version 6.0.1.1 Date: Wed, 28 Apr 2004 09:55:48 -0600 To: CHEMISTRY*at*ccl.net From: "James J. P. Stewart" Subject: Re: CCL:Calculation of an angle In-Reply-To: <1082573532.4086c2dc45b7a*at*webmail.u-picardie.fr> References: <1082573532.4086c2dc45b7a*at*webmail.u-picardie.fr> Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii"; format=flowed Given the three distances between each pair of atoms, use the familiar equation: c^2 = a^2 + b^2 - 2abCos(c) or Cos(c) = (a^2 + b^2 - c^2)/(2ab) Given Cos(c) take the inverse to get the angle c. Jimmy At 02:03 PM 4/21/2004, you wrote: >Dear All, > >I am looking for an equation to calculate an angle from the cartesian >coordinates of the 3 atoms making this angle. > ( .,at,. .,at,. ) .-------------oOOo----(_)----oOOo-------------------------------------. | James J. P. Stewart | | | Stewart Computational Chemistry | E-mail: jstewart*at*us.fujitsu.com | | 15210 Paddington Circle | 39/03/15 N, 104/49/29 W | | Colorado Springs CO 80921-2512 | | | USA .ooo0 | Phone: USA +(719) 488-9416 | | ( ) Oooo. | | .--------------------\ (----( )-------------------------------------. \_) ) / (_/