ABSTRACT
Let rl lie along the x axis. Equation (3.1) can be written as two scalar equations, corresponding to the x and y components of the rvectors. Thus,
(3.2a)
(3.2b)
Combining Eqs. (3.2a) and (3.2b), letting {}2 = ¢ and (}4 = rx + n, and simplifying yields Freudenstein's (1955) equation:
I R, cos(rx) - R2 cos(¢) + R3 - cos(rx - ¢) = 0
where
(3.3)
(3.4)
Consider the particular four-bar linkage specified by rj = 10, r2 = 6, r3 = 8, and r4 = 4, which is illustrated in Figure 3.1. Thus, R, = i, R2 = ~, R3 = ¥, and Eq. (3.3) becomes
I icos(rx) - ~cos(¢) +¥- cos(rx - ¢) = 0 (3.5) The exact solution of Eq. (3.5) is tabulated in Table 3.1 and illustrated in Figure 3.2. Table 3.1 and Figure 3.2 correspond to the case where links 2, 3, and 4 are in the upper halfplane. This problem will be used throughout Chapter 3 to illustrate methods of solving for the roots of nonlinear equations. A mirror image solution is obtained for the case where links 2, 3, and 4 are in the lower half-plane. Another solution and its mirror image about the x axis are obtained if link 4 is in the upper half plane, link 2 is in the lower half-plane, and link 3 crosses the x axis, as illustrated by the small insert in Figure 3.1.
