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by: Paige Cruickshank


Paige Cruickshank
GPA 3.95

Scott Banks

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About this Document

Scott Banks
Class Notes
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This 5 page Class Notes was uploaded by Paige Cruickshank on Friday September 18, 2015. The Class Notes belongs to EML 6281 at University of Florida taught by Scott Banks in Fall. Since its upload, it has received 55 views. For similar materials see /class/206705/eml-6281-university-of-florida in Engineering Mechanical at University of Florida.

Popular in Engineering Mechanical


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Date Created: 09/18/15
JD Yamokoski PA10 Reverse Poisition Analysis Program Instructions To test the reverse analysis function written in MATLAB reverseipaIO simply run the MATLAB script file named PA107PosAnalysis S ample output Reverse position analysis for the PA1076C robot Phi71 Thetae2 Thetae3 Thetae4 Thetae5 Thetae6 ChecksOut Pitooiif Given 350 7450 500 1230 30 1 350 7450 500 123 0 130 7210 YES 2 350 7450 500 757 0 7130 1590 YES 3 350 1051 1300 12 9 1220 109 6 YES 4 350 1051 1300 7167 1 71220 7704 YES 5 2150 749 500 71671 1220 1096 YES 6 2150 749 500 12 9 71220 770 4 YES 7 2150 7135 0 1300 757 0 130 721 0 YES 8 2150 7135 0 1300 123 0 7130 159 0 YES lof7 JD Yamokoski Appendicies A Coordinate system definitions B Reverse Analysis 20f7 LDV Yamukuskx m a m m m m m a a JD Yamokoski Appendix B Reverse Analysis Solution for 01 and p1 The vector loop for the closed loop PA 10 is S1 1 023a23 S4 4 S6 6 S7 57 a7la71 6 Expressing the vectors of 1 in terms of set 14 of the table of direction cosines for the spatial heptagon and substituting known mechanism parameters yields 0 02 X5671 X71 X1 01 S1 512 023 52 S4 Y5671 S6 Y71 S7 Y1 a71 51012 0 012 0 Z5671 Z71 Z1 U12 0 02 X32 X71 X1 01 S1 512 023 52 S4 X32 S6 Y71 S7 Y1 6171 51012 0 2 012 0 3 Z71 Z1 51 The z component of 2 is S1012 S4Z3 S6Z71S7Z1a71U12 0 3 By definition Z3 023034 32333403 but 0123 0 and 0134 270 therefore Z 0 Further all2 90 so cm 2 0 and S12 2 1 So 3 can reduced to S6271 S7Z1 a71U12 0 4 Expanding Z71 Z1 and U12 yields S6s12 X751 Y7cl 012Z7 S7 13121371 51257101 a71slsu 0 5 Making substitutions again for C12 2 0 and S1 1 simplifies 5 to S6X7s1 Y7clS7 s7lcla71sl 0 6 Finally grouping the c1 and s1 terms yields Aclle D0 7 where A S6Y7 S73 B S6X7 a D 0 we can further manipulate 7 into the form 71 71 and D 0 Because this is a special case when 4of7


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