Re: Engineering requests
- To: mathgroup at smc.vnet.net
- Subject: [mg125483] Re: Engineering requests
- From: "McHale, Paul" <Paul.McHale at excelitas.com>
- Date: Thu, 15 Mar 2012 00:31:20 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201203130804.DAA12797@smc.vnet.net>
Paul McHale | Electrical Engineer, Energetics Systems | Excelitas Technologies Corp. Phone: +1 937.865.3004 | Fax: +1 937.865.5170 | Mobile: +1 937.371.2828 1100 Vanguard Blvd, Miamisburg, Ohio 45342-0312 USA Paul.McHale at Excelitas.com www.excelitas.com It's not the re-formulation I mind, it's there error I introduce when doing it :). We make an assumption that the max/min answer will exist when using some permutation of max/min variables. For the cases we deal most with, this is true. It would be easiest enough to add a nominal value to ensure max/min answer is some permutation of the max/min of the input variables. Not fool proof in higher order issues. If the test fails, then other techniques (monte carlo) would be used. The simplest answer would be: Eq=5*R2/(R1+R2) TolerancePermutations[Eq,R1->{min,nom,max}, R2->{min,nom,max}] Out: {Min -> Num1, Max -> Num2} Or >> peak value found at nominal tolerance, recommend other methods Out: {Min -> Num1, Max -> indeterminite} Paul Please consider the environment before printing this e-mail. This email message and any attachments are confidential and proprietary to Excelitas Technologies Corp. If you are not the intended recipient of this message, please inform the sender by replying to this email or sending a message to the sender and destroy the message and any attachments. Thank you -----Original Message----- From: Dana DeLouis [mailto:dana01 at me.com] Sent: Tuesday, March 13, 2012 4:04 AM To: mathgroup at smc.vnet.net Subject: [mg125483] Re: Engineering requests On Mar 12, 5:07 am, "McHale, Paul" <Paul.McH... at excelitas.com> wrote: > The interval function looks a little to foreign to this intended purpose. Slight changes in the representation you chose seems to break it quickly. <snip> > > (* using table (correct answers) *) > > R1=R1int; > > R2=R2int; > > Table[5/(Ra+Rb) Rb,{Ra,R1},{Rb,R2}]// Max > > Table[5/(Ra+Rb) Rb,{Ra,R1},{Rb,R2}]// Min > > > Out[3]= 1.34146 > > Out[4]= 0.985401 vs: 5./(1+r1/r2) Interval[{0.985401,1.34146}] Hi. Not sure what was changed, but I believe one needs to eliminate duplicate variables. I think you need to reduce the equation so that each interval occurs once. Another way might be... The numbers used are the min and max of the intervals involved. 5. * 27000 / (27000+110000) 0.985401 5.* 33000 / (90000+33000) 1.34146 = = = = = = = = = = HTH :>) Dana DeLouis = = = = = = = = = =
- References:
- Re: Engineering requests
- From: Dana DeLouis <dana01@me.com>
- Re: Engineering requests