= = = =

For the problem presented in Section 3.1, r, = 10, r2 = 6, r( = 8, and r4 = 4. Consider the case where 84 = 220.0 deg. Let 8~1) = 30.0 deg and 83') = 0.0 deg. From Eq. (3.173):

1(30.0,0.0) = 6.0 cos(30.0) + 8.0 cos(O.O) + 4.0 cos(220.0) - 10.0 = 0.131975 (3.1na)

g(30.0, 0.0) = 6.0 sin(30.0) + 8.0 sin(O.O) + 4.0 sin(220.0) = 0.428850 (3.1nb) Equations (3.175a) and (3.175b) give

= -6.0 sin(30.0) = -3.000000

gll2 = 6.0 cos(30.0) = 5.196152 and gllJ = 8.0 cos(O.O) = 8.0 Substituting these results into Eq. (3.176) gives

-3.00000011.82 + 0.011.83 = -0.131975

5.196152 11.82 + 8.0 11.83 = -0.428850 Solving Eq. (3.179) gives

11.82 = 0.043992(180/n) = 2.520530 deg

11.83 = -0.082180(180/n) = -4.708541 deg where the factor (180/n) is needed to convert radians to degrees. Thus,

82 = 32.520530 deg and 83 = -4.708541 deg

(3.178b)

(3.179a)

(3.179b)

(3.180a)

(3.180b)

(3.181) These results and the results of subsequent iterations are presented in Table 3.14. Four iterations are required to satisfy the convergence criteria 111.82 1 ::::: 0.000001 and 183 1 ::::: 0.000001. These results were obtained on a 13-digit precision computer. As illustrated in Figure 3.1, 82 = ¢. From Table 3.1, ¢ = 32.015180deg, which is the same as 82,

In the general case,

I f(x) = 0 I where f(X)T = [Ji (x) hex) . .. j;,(x)] and xT = [x, X2 (3.170) and (3.171) become

(3.182)

X,.]. In this case, Eqs.