Page 44 - Cam Design Handbook
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THB2  8/15/03  12:48 PM  Page 32

          32                       CAM DESIGN HANDBOOK

          SIMPLE POLYNOMIAL CURVES The displacement equations of simple polynomial curves
          are of the form

                                         y =  Cq  2                     (2.12)
          where n = any number
                 C = a constant
          In  this  polynomial  family,  we  have  the  following  popular  curves  with  integer  powers:
          straight line, n = 1; parabolic or constant acceleration, n = 2; cubic or constant jerk,
          n = 3. High degree polynomial curves are shown in Chapter 3.
          TRIGONOMETRIC  CURVES The  curves  of  trigonometric  form  are:  simple  harmonic
          motion (SHM) or crank curve, which has a cosine acceleration curve; cycloidal, which
          has a sine acceleration curve; double harmonic; and elliptical. (Appendix B presents tab-
          ulated values of the simple harmonic motion curve and the cycloidal curve as a direct
          approach to follower characteristics.)

          OTHER  CURVES In  addition  to  these  two  families  are  the  miscellaneous,  little-used
          curves: modified straight-line circular arc and the circular arc curves (see Chapter 14).
          These are employed primarily as an improvement over the characteristics of the straight-
          line curve and for special design requirements.
          EXAMPLE A cam rotating at 120rpm has the positive acceleration part of its rise of 3/8
          in. in 40 degrees of cam rotation. A simple polynomial curve having n = 2.4 is used. Find
          the velocity and acceleration values at the end of 30 degrees of cam rotation.
                                                  2 p
                                                          7
                                                        .
          Solution The  angular  velocity  of  cam  = w  = 120  ¥  = 12 5 rad sec.  The  total  cam
                                                  60
          angle = 40p/180 = 0.698 radian. Substituting into Eq. (2.12) gives the displacement
                                          3        24 .
                                                .
                                  y =  Cq  n  =  =  C(0698 )
                                          8
          Solving yields C = 0.888. Therefore, the basic equation is
                                          .
                                       y = 0888q  24 .
          The velocity by differentiating Eq. (2.12) is
                                      ˙ y = wq  n-1  ips
                                        C n
          Thus, the velocity after 30 degrees of rotation is
                                          Ê     p  ˆ  . 14
                                                        . ips
                                    . )(2 4 30
                               .
                           ˙ y = 0888 (12 57  . )  ¥  =108
                                          Ë    180 ¯
          The acceleration, differentiating again is
                                           (
                                    ˙˙ y =  Cw  2 n n - )q1  n-2
          The acceleration after 30 degrees of rotation is
                                           Ê
                                         . )
                                 . ) (2 4
                       ˙˙ y = 0888 (12 57  2  . )(1 4 30  ¥  p  ˆ  . 04  = 364 in sec 2
                           .
                                           Ë    180 ¯
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