The last two Eqs. form a linear system in S1 and C1 and provide a unique value for el.
Eqs. (4.13) and (4.14) along with (9.1) provide a way to solve for 84 and 85. By solving for C5 in (9.1) and substituting in
tx = C1 C23 C4S5 + S1 S4S5 C1 S23 C5 (9.7)
and
ty = S1 C23 C4S5 C1 S4S5 S1 S23 C5, (9.8)
we obtain
C1 C4S5 + S1 C23 S4S5 = tx C23 + tz C1 S23 (9.9)
S1 C4S5 C1 C23 S4S5 = ty C23 + tz S1 S23. (9.10)
This linear system can be solved for the products C4S5 and S4S5 uniquely. When S5 is not 0, two solutions for e4 are then obtained by
94 = Atan2(S4S5,C4S5) or 94 = Atan2(-S4S5,-C4S5).
When S5=0, joint axes z3 and z5 are aligned and the manipulator loses one DOF. Only the sum 84+86 can be found by use of Eqs. (4.3) and (4.4).
With e4 known, the tx and ty equations above constitute a linear system of equations which yields a unique solution for E5. The last joint variable 86 can then be obtained from 2 more equations from (4.3) such as the nz and bz equations.
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