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t Answers for method 2 problem: A=110 B=110+150+20=280 C=280+50+20=350 D=350+250+20=620 E=620+475+20=1115 Detailing and Dimensioning Training ? Dimensioning Methods C A M C o v e r S t u dR e t a ine rC A M C o v e r S t u dR e t a ine rC A M C o v e r S t u dR e t a ine rC A M C o v e r S t u dR e t a ine rR e a r E n g ineR e t a ine r A t t a c h m e n t P o intP C M 5 0 W a y310335 275520760P o int C E ngine Harne s s E x a mpleP C M 7 0 W a yExample of Method 1 and 2 being used in conjunction to achieve a desired result. Detailing and Dimensioning Training ? Tolerance StackUp ? Review ESF65B14A121AA ? Single wire tolerance ? Multiple wire tolerance ? Retainers ? Tape Ties ? Reviewing Tolerance StackUp on Critical Lengths ? Determine Critical area to be reviewed ? Calculate length using worse case ? Adjust dimensions to acmodate tolerance Detailing and Dimensioning Training ? Conduct Grey Beard Review ? Review all principle wire assemblies to ensure all fit requirements have been addressed ? Include principle fit characteristics in initial print reviews with AFL Manufacturing ? Verify principle fit characteristics during master sample sign off Detailing and Dimensioning Training ? Questions and Answers