Position Analysis: Review (Chapter 2) Objective: Given the geometry of a mechanism and the input motion, find the output motion
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1 Position Anlysis: Review (Chpter Ojetive: Given the geometry of mehnism n the input motion, fin the output motion Grphil pproh Algeri position nlysis Exmple of grphil nlysis of linges, four r linge. Given - n fin n Open Crosse
2 Moifie Prolem: Given: - n θ Fin: position of -r linge A θ O
3 Oservtion: Joint is on irle otine y shifting the pth of the tip of the rn y vetor O O O A θ O
4 Solution: Drw pth of tip of rn, A. This is irle with rius n enter the pivot of the rn. Swift the irle esriing the pth of the rn y O O. Drw irle with enter O n rius the length of the roer C. Fin the lotion of joint y fining the intersetion of the two irles on whih is lote. O A θ O O
5 Algeri position nlysis Use trigonometry to fin positions of lins n joints. Exmple: four r linge: O A O Proeure: 1.Fin tringle O AO from,,..fin tringle AO from,, AO. 5
6 Complex numer metho for position nlysis Ie: represent lins s position vetors n represent these vetors s omplex numers. Why omplex numers? In mny prolems, it is more strightforwr to erive the equtions for position nlysis using omplex numers Complex numer: Retngulr form Im R y R R x jr y R x Re 6
7 Complex numer: Polr form Im R y R (mgnitue (ngle R R e j R x Re Complex onjugte z x jy Euler s theorem: e j os j sin z x jy z os j z sin retngulr form z = z z e j polr form os e j e j Re( e j sin e j e j j Im( e j 7
8 Complex numer lger: z x jy w u jv Equlity: z = w x = u n y = v or z = w n ngle of w = ngle of v Aition, sutrtion: z w x u j( y v Multiplition, Division, Powers Multiplition of two numers: multiply mgnitues, phse ngles. Division of two numers: ivie mgnitues, sutrt phse ngles Rising omplex numer into power: rise mgnitue into the power, multiply phse ngle y the power Use retngulr form when ing or sutrting Use polr from when multiplying, iviing or rising into power Complex eqution solving: f(z = 0, where z n f re omplex quntities. Solve for z. 8
9 Solution Rel(f(z = 0 or Rel[f(x+jy] = 0 Im(f(z = 0 or Im[f(x+jy] = 0 Solving the ove two inepenent equtions for x n y we fin z. Exmple: (+jz+5-j = 0, where z = x + jy Solution Sustituting z = x + jy into the eqution we otin: (+j(x+jy+5-j = 0 or x+y j +xj-y+5-j=0 or x-y+5+j(y+x-=0 oth rel n imginry prts shoul e zero: Rel prt = 0 x - y+5=0 Imginry prt = 0 y+x-=0 Solving the two equtions for x n y we otin the rel n imginry prts of omplex numer z: x = n y = Therefore: z = j
10 Exmple: rn-roer four r linge R A R R 1 R O O Prolem: Given,,, n, fin n Solution: Vetor loop eqution: R R R R1 0 j j j e e e j e 0 Rel prt = 0 Imginry prt = 0 Two equtions with two unnowns, n Importnt efinitions: Open Grshof mehnism: If 0 / then the two lins jent to the shortest lin (rn o not ross eh other. Crosse Grshof mehnism: If 0 / then the two lins jent to the shortest lin (rn ross eh other. 10
11 11 C A A AC / os( 1 ( sin os( os( ( tn F E D D DF E E / 1 os ( sin os( os( ( tn Open solution for negtive squre root, rosse solution for positive squre root Open solution for negtive squre root, rosse solution for positive squre root
12 Exmple: Slier-rn mehnism Prolem: Given -, fin,, Open solution first R, R, R 1, R, Steps: R R R R1 0 Finl result: Fin ngle y solving numerilly or lgerilly the following eqution: sin sin os os( 0 tn( Algeri solution: 1
13 Open solution tn 1 T T SU S where S R-Q, T P, U Q R P sin sin ( os os Q sin os ( os sin R sin Inlintion ngle of oupler: En of open solution Crosse mehnism: sin sin os os( 0 tn( Algeri solution: T tn 1 T S SU Fin oeffiients T, S n U from the equtions for the open solution. 1
14 Inlintion ngle of oupler: 1
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