From Eq. (2.31), we can find certain relationships by
equating corresponding elements of the two matrices on
either side of the equation. Thus we obtain the following
equations:
-axS8a + ayCOa
9bl = tan-1( -axSa + ayCS) (2.35)
axCea + aySqB
axC8a + aySea
Eb2 = tan-1( t C + ) (2.36)
-azCbl
sz
8b3 = tan-(--- ) (2.37)
-n"
There are two possible solutions for dc in Eq. (2.32),
since it is a quadratic polynomial equation. Back
substituting these solutions of dc into Eqs. (2.33) -
(2.37), respectively, we can obtain two possible sets of
solutions for 6a, da, 9bl, Bb2 and 8b3 from each equation.
Therefore, the subchain (R-L)-S-P has a second-degree
polynomial equation in inverse kinematics.
Numerical example. The given parameters are as
follows:
a = 3", b = 2", ac = 0, c = 0.25" and
0.7259 0.5900 0.3536 5.1151
0.4803 -0.8027 0.3536 3.4645
HO =
0.4924 -0.0868 -0.8660 0.4428
0 0 0 1