With e123 known, Eqs. (5.36) and (5.37) become (elbow
equations)
Px a3 C123 = a2 C12 + al C1 (5.36')
py a3 S123 = a2 S12 + a1 S1 (5.37')
and can be solved for C2 by
C2 = [(Px a3 C123)2 + (py a3 S123) 2
a2 a12] / (2 a1 a2)
which is obtained by applying the cosine law to the triangle having links 1 and 2 as its sides. Two values of 92 follow from e2=atan2( /(1-C22), C2)
and a unique value of 81 can then be computed from Eqs. (5.36') and (5.37') which yield a linear system in S1 and C1 when S12 and C12 are expanded using sum of angles trigonometric identities. Joint variable 3, 83 is given by
63 = 6123 81 82
and the solution set is completed when the last angle 04 is. computed as shown earlier. This development proves that there can be at most 2 solution sets for a 4-DOF arm with this particular geometry.
Case 2: ai=a3=0. The first two joint axes are parallel and the last two joint axes are parallel. The reduced
system is
..