= +'

Equation (3.80) is applied repetitively until either one or both of the following two convergence criteria are satisfied:

and/or (3.81)

Example 3.5. The secant method.

Let's solve the four-bar linkage problem presented in Section 3.1 by the secant method. Recall Eq. (3.3):

f(¢) = R, cos(a) - Rz cos(¢) + R3 - cos(a - ¢) = 0 (3.82) Thus, Eq. (3.80) becomes

where g'(¢;) is given by

For R, = ~, Rz = ~,R3 = ¥, and a = 40.0 deg, Eq. (3.82) yields f(¢) = ~cos(40.0) - ~cos(¢) + ¥- cos(40.0 - ¢)

(3.83)

(3.84)

(3.85)

For the first iteration, let ¢o = 30.0 deg and ¢I = 40.0 deg. Equation (3.85) gives f(¢o) = ~cos(40.0) - ~cos(30.0) +¥- cos(40.0 - 30.0) = -0.03979719 (3.86a) f(¢l) = ~cos(40.0) - ~ cos(40.0) + ¥- cos(40.0 - 40.0) = 0.19496296 (3.86b)

Substituting these results into Eq. (3.84) gives '(A.. ) = (0.19496296) - (-0.03979719) = 234 602

g 'f" 40.0 _ 30.0 0.0 7

Substituting g'(¢,) into Eq. (3.83) yields 0.19496296

¢z = 40.0 - 0.02347602 = 31.695228 deg

(3.87)

(3.88)

Substituting ¢z = 31.695228 deg into Eq. (3.85) gives f(¢z) = -0.00657688. These results and the results of subsequent iterations are presented in Table 3.8. The

convergence criterion, I¢i+' - ¢il :::: 0.000001 deg, is satisfied on the fifth iteration, which is one iteration more than Newton's method requires.