# chap11 - 107 Chapter 11 Synthesis of Linkages 11.1 A...

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Unformatted text preview: 107 Chapter 11 Synthesis of Linkages 11.1 A function varies from 0 to 10. Find the Chebychev spacing for two, three, four, five, and six precision positions. With 0.0 x = and 1 10.0 n x + = , Eq. (11.5) becomes: (2 1) 5.0 5.0cos 1,2,..., 2 j j x j n n- π =- = j \ n 2 3 4 5 6 1 1.46447 0.66987 0.38060 0.24472 0.17037 2 8.53553 5.00000 3.08658 2.06107 1.46447 3 9.33013 6.91342 5.00000 3.70591 4 9.61940 7.93893 6.29409 5 9.75528 8.53553 6 9.82963 11.2 Determine the link lengths of a slider-crank linkage to have a stroke of 600 mm and a time ratio of 1.20. After laying out the distance B 1 B 2 = 600 mm, we see that the time ratio is ( 29 ( 29 180 180 1.20 Q α α = ° + ° - = and, from this, our design must have 16.40 . α = ° Therefore we construct the point C such that the central angle 1 2 2 32.80 . B CB α = = ° R Using this point C as the center of a circle assures that any point O 2 on this circle will have the angle 1 2 2 16.40 B O B α ≡ = ° R and thus will be a possible solution point. One typical solution uses the point O 2 shown. Choosing the point O 2 shown we measure the two distances 108 2 1 3 2 1 324.956 mm O B r r = + = and 2 2 3 2 801.946 mm O B r r =- = and from these we find 2 261.50 mm r = Ans. 3 1 063.45 mm r = Ans. 11.3 Determine a set of link lengths for a slider-crank linkage such that the stoke is 16 in and the time ratio is 1.25. After laying out the distance B 1 B 2 = 16.00 in, we see that the time ratio is ( 29 ( 29 180 180 1.25 Q α α = ° + ° - = and, from this, our design must have 20.00 . α = ° Therefore we construct the point C such that the central angle 1 2 2 40.00 . B CB α = = ° R Using this point C as the center of a circle assures that any point O 2 on this circle will have the angle 1 2 2 20.00 B O B α ≡ = ° R and thus will be a possible solution point. One typical solution uses the point O 2 shown. Choosing the point O 2 shown we measure the two distances 2 1 3 2 29.82 in O B r r = + = and 2 2 3 2 15.69 in O B r r =- = and from these we find 2 7.06 in r = Ans. 3 22.75 in r = Ans. 11.4 The rocker of a crank-rocker linkage is to have a length of 500 mm and swing through a 109 total angle of 45 o with a time radio of 1.25. Determine a suitable set of dimensions for 1 2 3 , , and r r r . After laying out the angle 1 4 2 45 B O B φ ≡ = ° R with BO 4 = 500 mm, we see that the time ratio is ( 29 ( 29 180 180 1.25 Q α α = ° + ° - = and, from this, we find that 20.00 . α = ° Therefore we construct the point C such that the central angle 1 2 2 40.00 . B CB α = = ° R Then, using this point C as the center of a circle assures that any point O 2 on this circle will have the angle 1 2 2 20.00 B O B α ≡ = ° R and thus will be a possible solution point....
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chap11 - 107 Chapter 11 Synthesis of Linkages 11.1 A...

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