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07472 Comparing with the ``exact value'' 0746824 for the integral that we noted in Example 829, we find that we have achieved two decimal places of accuracy It is interesting to note that if we had chosen a partition with k = 6, as we did in Example 829, then Simpson's Rule would have guaranteed an accuracy of 12 15 M ( b a) 5 = 000005144 , 180 k4 180 64 or nearly four decimal places of accuracy

Thus the return spring rate, the linear frequency ratio, and the maximum re ected inertia ratio are kr = am 15(50) (0051)(3665) 2 = 0257 10 6 N m x maxq i2 = h 002 wl b w b 2pq i b ki ki 10, 000 = = = 931 2 2 I 2pq i ) I (2p 3665) 0005 (

p 2

using Simpson's Rule with a partition having four intervals What degree of accuracy does this represent

From Fig 1211 we obtain bm = 0006 The corresponding maximum driveshaft windup is p q i - q c = 0006 = 00094 rad ( 054 ) 2 and the associated dynamic windup torque is Tm = ks b m b = 10, 000 0006 p = 942 N-m 2

Switch to logarithms:

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Of course this example is parallel to Example 830, and you should compare the two examples Our function is f ( x) = 1/( 1 + x 2 ) and our partition is P = {0, 1/4, 2/4, 3/4, 1} The sum from Simpson's Rule is S= 1/4 { f ( 0) + 4 f ( 1/4) + 2 f ( 1/2) + 4 f ( 3/4) + f ( 1) } 3 1 1 1 +4 2 12 1+0 1 + ( 1/4) 2 +2 1 1 1 +4 + 1 + ( 1/2) 2 1 + ( 3/4) 2 1 + 12

Graphical plots of windup versus time and of radial contact force versus time are given as the dashed curves in Figs 1218 and 1219

4 Fv (k N)

1 {1 + 37647 + 16 + 256 + 05} 12

0 10 20 30 t (ms)

0785392 Comparing with Example 830, we see that this answer is accurate to four decimal places We invite the reader to do the necessary calculation with the Simpson's Rule error term to confirm that we could have predicted this degree of accuracy

If this design is unacceptable, we may replace the cycloidal cam curve with a simple harmonic curve, other modi ed harmonic, or modi ed trapezoidal curve, keeping the other condition intact By so doing we can calculate hp = 002 2b hp 2 xmax = = 004 2b 2 xmax = h will remain unchanged, but kr becomes equal to 02 106 N/m, and Qm becomes equal to approximately 40 From Fig 1211 we obtain bm = 00025 Thus qt - qc = 025 and its corresponding peak windup torque is Tm = 45 N-m Therefore, a change from a cycloidal rise to a harmonic rise reduces the forcing function by about 50 percent Charts of windup versus time and of radial contact force versus time for this harmonic cam system are given as the solid curves in Figs 1218 and 1219 1243 Two-Degree-of-Freedom System Let us consider a two-degree-of-freedom closed-track cam follower system with an elastic camshaft and an elastic follower In Fig 1220a we see a schematic of the mechanism In Fig 1220b it is observed that torque

Finally,

(1217)

using both the Trapezoid Rule and Simpson's Rule with a partition having six points Use the error term estimate to state what the accuracy prediction of each of your calculations is If the software Mathematica or Maple is available to you, check the answers you have obtained against those provided by these computer algebra systems

where T is the net torque accelerating the cam and the bottom of the follower Therefore, the torque T = Icq c + m1 yR tan f where m1 is the mass of the follower base, and rotational acceleration of the roller has been neglected Also the follower m1 has the following characteristics of displacement, velocity, and acceleration (see Chap 2 for symbols) y = y(q c ) y = y q

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