chapter8-b - MEEG439 MACHINE DYNAMICS SPRING 2006 Example...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MEEG439 MACHINE DYNAMICS SPRING 2006 Example Design a double dwell cam to move a follower from 0 to 2 in in 60, dwell for 150, fall 2 in in 90 and dwell for the remainder. The cycle time is 4 sec. 1. Choose a suitable profile for the rise and fall to minimize acceleration. 2. Plot the s-v-a-j diagrams Tabulate the motion requirements Rise Dwell Fall Dwell 1 = 60 2=150 3 = 90 4 = 60 h1 = 2 h2 = 0 h3 = 2 h4 = 0 Curve which minimizes the acceleration is the modified trapezoid (fig. 8-18) See Fig. 8-18 Use the equations on pages 364 365 to define the curves See MathCAD file Ex1-cam.mcd Cam profile is Rise-Fall-Dwell (RFD) Can use same approach as for RDFD o May not be optimal o E.g. using a cycloidal profile See Fig. 8-23 Key Points: o Don't need acceleration to go to zero at transition from rise to fall o Just need to have the same value for acceleration on rise and fall curves o Change in acceleration rates (slopes) causes abrupt changes in jerk Can use a double harmonic curve o Two cosine curves of different frequency o Note that these curves have nonzero acceleration at one end, so they can't be used for a double dwell design h 1 - cos 2 1 2 1 - 4 - cos 8.4 Single Dwell Cam Design Rise equations s= v= h 1 sin - sin 2 2 2 2 h a = 2 cos cos - 2 2 j=- 3 h sin - 2 2sin 3 2 1 fa1c42148fbaa5c42fda8fe9d4973933700bea52.doc MEEG439 MACHINE DYNAMICS SPRING 2006 Fall equations s= v= a= 1 h 2 1 - cos - - cos 1 2 4 h sin 2 1 - 2 2 sin - 2 cos 2 h cos 2 2 j=- 3 h sin - 2 2sin 3 2 See Fig. 8-24 If rise and fall durations are the same, can use the double harmonic curve for the single dwell case 8.5 Polynomial Functions Can use polynomials for a wider variety of applications General form of the polynomial is s = C0 + C1 x + C2 x 2 + C3 x 3 + L + Cn x n 345 Polynomial C's are determined by the motion requirements We specify a number of conditions on s, v, a, j o These are the boundary conditions (BC) o Their number determines the degree of the polynomial For each BC we substitute into the s, v, a, or j equation to get some linear relations between the C's We then solve simultaneously for all the C's If there are k BC's the polynomial will have degree k-1 and there will be k C's from C0 to Ck-1 The motion requirement for the rise and fall are Rise s=0 =0 s=h = 1 Fall =0 = 2 s=h s=0 See Fig. 8-25 v=0 v=0 v=0 v=0 a=0 a=0 a=0 a=0 We want no discontinuities in s, v, a curves We have 6 BC's for the rise, so we need a polynomial of degree 5 s = C0 + C1 C2 + C3 + C4 + C5 + 2 3 4 5 2 fa1c42148fbaa5c42fda8fe9d4973933700bea52.doc MEEG439 MACHINE DYNAMICS SPRING 2006 We will also need the equations for velocity and acceleration 1 v = 1 + 2C2 C 2 3 4 + 3C3 + 4C4 + 5C5 2 3 + 12C4 + 20C5 1 a = 2 C2 + 6C3 2 Substituting in the BC's gives = 0, s = 0 = C0 C = 0, v = 0 = 1 2C = 0, a = 0 = 2 = , s = h = C3 + C 4 + C 5 1 = , v = 0 = [ 3C3 + 4C4 + 5C5 ] 1 = , a = 0 = 2 [ 6C3 + 12C4 + 20C5 ] C3 = 10h C4 = -15h C5 = 6h Solving for the remaining C's gives The resulting rise polynomial is 3 4 5 s = h - 15 + 6 10 Hence the name 3-4-5 polynomial, based on the powers of x in the equation See Fig. 8-26 4567 Polynomial Can generate a similar function for the fall part of the curve If we add a constraint that the jerk is also zero at the beginning and end of the rise o we get 8 BC's o a 7th degree polynomial o Use the same solution technique as for the 3-4-5 polynomial 4 5 6 7 s = h - 84 + 70 - 20 35 See Fig. 8-26 Get smoother jerk, but at the cost of higher accelerations 3 fa1c42148fbaa5c42fda8fe9d4973933700bea52.doc MEEG439 MACHINE DYNAMICS SPRING 2006 Single Dwell Application of Polynomials Can use only 2 segments instead of 3 by combining rise and fall into one polynomial Higher order polynomial may have desired behavior at specified points, but might oscillate from a desired path in between points o High order polynomials have large number of maxima and minima and inflection points which can cause these oscillations between specified points Motion requirements are s=0 v=0 a=0 =0 s=h = 1/2 s=0 v=0 a=0 = 1 We get a 6th degree polynomial The result is the 3-4-5-6 polynomial 3 4 5 6 s = h - 192 + 192 - 64 64 See Fig. 8-29 Polynomials can be used to specify many complex cam profiles Spline functions are another class of function that can be used to meet specified performance requirements For the single dwell profile, what if we also specify the velocity at the transition from rise to fall? The motion requirements are now s=0 v=0 a=0 =0 s=h v=0 = 1 s=0 = 1 + 2 We have a 7th degree polynomial s = C0 + C1 2 3 4 Example v=0 5 6 a=0 7 + C2 + C3 + C3 + C3 + C3 + C7 + 3C3 + 4C3 + 5C3 + 6C3 + 7C7 + 12C3 + 20C3 + 30C3 + 42C7 2 3 4 5 2 3 4 5 6 v = C1 + 2C2 a = 2C2 + 6C3 4 fa1c42148fbaa5c42fda8fe9d4973933700bea52.doc MEEG439 MACHINE DYNAMICS = 0, s = 0 = C0 C = 0, v = 0 = 1 2C = 0, a = 0 = 2 SPRING 2006 At = 1 s = h = C3 1 + C4 1 + C5 1 + C6 1 + C7 1 v = 0 = 3C3 1 + 4C4 1 + 5C5 1 + 6C6 1 + 7C7 1 2 3 4 5 6 3 4 5 6 7 At = = 1 + 2 s = 0 = C 3 + C 4 + C5 + C 6 + C 7 v = 0 = 3C3 + 4C4 + 5C5 + 6C6 + 7C7 a = 0 = 6C3 + 12C4 + 20C5 + 30C6 + 42C7 We put the 5 equations in matrix form and solve for the unknown C's At this point it is easier to solve numerically for given angles 1 and 2. A= 1 3 A 2 3 A 1 3 6 A4 A5 4 A3 5 A4 1 1 4 5 12 20 A6 6 A5 1 6 30 h A7 C3 6 0 7 A C4 = 0 1 C5 0 7 C6 0 42 C7 Invert matrix to get C's 5 fa1c42148fbaa5c42fda8fe9d4973933700bea52.doc ...
View Full Document

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.

Ask a homework question - tutors are online