36
Ei = D2C2b + 2aCxC2cb 2CyCzSabCab 2DiCzCab +
Cy2S2,b + D12 + 2D1CySab
E2 = -4(aCyC2ab + CxCzSabCab D1CxSab CxCyS2,b)
E3 = 2(D2C2ab 2D1CzCab + 2Cx2S2ab Cy2S2cb + D12)
E4 = -4(aCyC2ab + CxCzSabCab D1CxSab + CxCyS2ab)
E5 = D2C2ab 2aCxC2ab + 2CyCzSabCab 2DlCzCab +
Cy2S2ab + D12 2DiCySab
Since Eq. (2.11) is a fourth-degree polynomial
equation, there are up to four possible solutions for
variable X (or 8a). Back substituting 8a into Eqs. (2.7)
and (2.9), we can obtain up to four possible sets of
solutions of 9b and da, respectively. Therefore, the
subchain (R-L)-R-S has a fourth-degree polynomial equation
in inverse kinematics.
Numerical example. The given parameters are as
follows:
a = 2", b = 12", sb = 8", ab = 720 and C = [-4.86,
-11.60, 3.97]T
The four possible solutions are computed as
Solutions 8a da 9b
(deg.) (in.) (deg.)
1 29.932 1.557 180.299
2 -50.609 12.297 -71.132
3 17.534 7.618 212.427
4 -87.785 -5.592 38.407