chapter4-b_6 - MEEG439 MACHINE DYNAMICS SPRING2006 4.6...

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MEEG439 MACHINE DYNAMICS SPRING2006 4.6 Four-bar Slider-Crank Position Solution O 2 B θ 4 θ 3 θ 2 a b c d Offset slider crank – slider axis does not pass through the crank pivot Need only 3 vectors r 2 , r 3 , r s , but we will use r 4 , r 5 in place of r s to simplify expressions o Assume x axis is parallel to direction of slide o Coupler’s angle is measured at slider Loop equation 2 3 4 1 0 - - - = r r r r Vector equation 2 2 3 3 (cos sin ) (cos sin ) 0 a b c d θ + - + - - = i j i j j i o Assumes θ 4 = 90° and slide direction is along the x axis Component equations 2 3 2 3 cos cos 0 sin sin 0 a b d a b c - - = - - = This gives 1 2 3 2 3 sin sin cos cos a c b d a b - - = = - o Two possible solutions for θ 3 4.7 Inverted Slider-crank Position Solution 1 dc01c3ec1be4d2465e030cf7dfeebdba48a94995.doc
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MEEG439 MACHINE DYNAMICS SPRING2006 O 2 O 4 B 4 θ 4 θ 4 a d θ 3 γ Offset slider – slide direction does not pass through the pivot point Set up xy axes with x axis through ground pivots Angle γ is fixed o It defines the slide direction relative to the link O 4 B Loop equation 2 3 4 1 0 - - - = r r r r Vector equation 2 2 3 3 4 4 (cos sin ) (cos sin ) (cos sin ) 0 a b c d θ + - + - + - = i j i j i j i This gives 2 3 4 2 3 4 cos cos cos 0 sin sin sin 0 a b c d a b c - - - = - - = o We appear to have 3 variables ( b , θ 3 , θ 4 ) and only 2 equations o There is a relation between θ 3 and θ 4 3 4 γ = o + - for the open and crossed configurations We can then solve to get 2 2 2 2 2 sin 2 cos cos b a d c ad c = + - - - 2 dc01c3ec1be4d2465e030cf7dfeebdba48a94995.doc
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MEEG439 MACHINE DYNAMICS SPRING2006 2 2 2 2 2 1 2 2 2 2 4 2 2 2 2 2 1 2 2 2 2 2 sin 2 cos sin sin ( cos )cos 2tan ( cos )sin sin cos 2 sin 2 cos sin sin ( cos (
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This note was uploaded on 02/06/2012 for the course MEEG 439 taught by Professor Scf during the Spring '11 term at The Petroleum Institute.

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chapter4-b_6 - MEEG439 MACHINE DYNAMICS SPRING2006 4.6...

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