0 = -CxSeaCab + CyCGaCab + CzSab daSab (2.15)
db = CxSeSaSb CyCeaSab + CCab daCab (2.16)
Let tan(8g/2) = X, and then substituting CGa = (1 X2)/(1 +
X2) and Se, = 2X/(1 + X2) into Eq. (2.14), we obtain
(a + b + Cx)X2 2CyX + a + b Cx = 0 (2.17)
There are up to two possible solutions of X in Eq. (2.17),
or up to two possible solutions for 8a. Back substituting
these two possible solutions of 8a into Eqs. (2.15) and
(2.16), we will have up to two possible solutions of da and
db from each equation. Thus it is seen that this subchain
has a second-degree polynomial equation in inverse
kinematics.
Numerical example. The given parameters are as
follows:
a = 3", b = 2", ab = 600 and CO = [5.85, -0.13, 4.25]T
The two possible solutions are computed as
Solutions Sa da db
(deg.) (in.) (in.)
1 30.024 2.495 3.510
2 -32.570 6.005 -3.510
2.5.3 Subchain (R-L)-S-R
In Fig. 2.13, a schematic diagram of the subchain
(R-L)-S-R is shown. The spherical pair is kinematically