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TABLE 4.1 Standard Trigonometric Follower Motions
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4.8 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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FIGURE 4.9 Example 1: (a) displacement diagram, in; (b) geometric velocity diagram, in/rad; (c) geometric acceleration diagram, in/rad2.
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lower and return spring for the following requirements: The speed of the cam is constant and equal to 150 r/min. Motion of the follower consists of six segments (Fig. 4.9): 1. 2. 3. 4. 5. 6. Accelerated motion to 1,end = 25 in/s (0.635 m/s) s Motion with constant velocity 25 in/s, lasting for 1.25 in (0.03175 m) of rise Decelerated motion (segments 1 to 3 describe rise of the follower) Return motion Return motion Dwell, lasting for t 0.085 s
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The total lift of the follower is 3 in (0.0762 m). Solution. Angular velocity = 150 /30 = 15.708 radians per second (rad/s). The cam rotation for 1.25 in of rise is equal to 2 = 1.25 in/s2 = 1.25 in/1.592 in/rad = 0.785 rad = 45 , where s2 = 25/15.708 = 1.592 in/rad. The following decisions are quite arbitrary and depend on the designer: 1. Use motion 1; then s1 = 0.5 in, smax = 0.05 / 2 = 0.5 /(0.628)2 = 4 in/rad2 (0.1016 1 m/rad2). s1,end = 2(0.5)/ 1; so 1 = 1/1.592 = 0.628 rad, or 36 . 2. For the motion with constant velocity, s2 = 1.592 in/rad (0.4044 m/rad); s2 = 1.25 in. 3. Motion type 2: s3 = s2 = 1.25 in, s3,init = s3 /(2 3) = 1.592 in/rad; therefore 3 = 1.25 /[2(1.592)] = 1.233 rad 71 , and s3,min = (1.25 2)/[4(1.233)2] = 2 in/rad2. (Points 1 through 3 describe the rise motion of the follower.) 4. Motion type 3: s 4,init = s4 2/(4 2 ) = 2 in/rad2 (the same value as that of s , ), s4,end = r 3,min s4/(2 4), and s4 + s5 = 3 in.
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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
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5. Motion type 4: s5,max = s5/ 2, and s5,init = s4,end = 2s5/ 5. We have here the four 5 unknowns 4, s4, 5, and s5. Assuming time t6 = 0.85 s for the sixth segment (a dwell), we can find 6 = t6 = 15.708(0.08) = 1.2566 rad, or 72 . Therefore 4 + 5 = 136 , or 2.374 rad (Fig. 4.9).Three other equations are s4 + s5 = 3, s4 2/(4 2 ) = 2, and 4 s4/(2 4) = 2s5/ 5. From these we can derive the quadratic equation in 4. 0.696 2 + 6.044 4 12 = 0 4 Solving it, we find 4 = 1.665 848 rad 95.5 and 5 = 40.5 . Since s4/s5 = 4 4/( 5) = 3.000 76, it is easy to find that s5 = 0.75 in (0.019 05 m) and s4 = 2.25 in (0.057 15 m). Maximum geometric acceleration for the fifth segment s5,max = 4.7 in/rad2 (0.0254 m/rad2), and the border (matching) geometric velocity s4,end = s5,init = 2.12 in/rad (0.253 m/rad). Example 2. Now let us consider a cam mechanism with spring loading of the type D-R-D-R (Fig. 4.7a). The rise part of the follower motion might be constructed of three segments (1, 2, and 3) described by standard follower motions 1, 2, and 3 (Fig. 4.8). The values of constants c and d in Table 4.1 are no longer zero and should be found from the boundary conditions. (They are zero only in the motion case R-R-D, shown in Fig. 4.7b, where there is no dwell between the rise and return motions.) For a given motion specification for the rise motion, the total follower stroke s0, and the total angular displacement of the cam 0, we have eight unknowns: 1, s1, 2, s2, 3, s3, and constants c and d. The requirements of matching the displacement derivatives will give us only six equations; thus two more must be added to get a unique solution. Two additional equations can be written on the basis of two arbitrary decisions: 1. The maximum value of the acceleration in segment 1, s 1,max, should be greater than that in segment 2 because of spring loading. So s 1,max = as2,min, where s2,min is the minimum value of the second-segment acceleration and a is any assumed number, usually greater than 2. 2. The end part of the rise (segment 3), the purpose of which is to avoid a sudden drop in a negative accelerative curve, should have a smaller duration than the basic negative part (segment 2). Therefore we can assume any number b (greater than 5) and write 2 = b 3.The following formulas were found after all eight equations for the eight unknowns were solved simultaneously: 1 = 0 1 + a + a/b b2 b ( + 4a) + 4a(2a + 1)
3 = 0
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