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KINEMATICS 31
Velocity:
˙
˙ x −l 1 sin θ 1 − l 2 sin(θ 1 + θ 2 ) − l 2 sin(θ 1 + θ 2 ) θ 1
˙
X = = (2.3)
˙ y l 1 cos θ 1 + l 2 cos(θ 1 + θ 2 ) l 2 cos(θ 1 + θ 2 ) θ 2
˙
˙
˙
or, in vector form, X = Jθ,where the 2 × 2matrix J is called the system’s
Jacobian (see, e.g., Refs. 6 and 7).
Acceleration:
¨
¨ x −l 1 sin θ 1 − l 2 sin(θ 1 + θ 2 ) θ 1
=
¨ y l 1 cos θ 1 l 2 cos(θ 1 + θ 2 ) θ 1 + θ 2
¨
¨
2
l 1 cos θ 1 l 2 cos(θ 1 + θ 2 ) θ 1
˙
− (2.4)
l 1 sin θ 1 l 2 sin(θ 1 + θ 2 ) (θ 1 + θ 2 )
˙ 2
˙
Inverse Transformation (Inverse Kinematics). From Figure 2.2, obtain the
position and velocity of the arm joints as a function of the arm endpoint Cartesian
coordinates:
Position:
2
2
2
x + y − l − l 2 2
1
cos θ 2 =
2l 1 l 2
y −1 l 2 sin θ 2
−1
θ 1 = tan − tan (2.5)
x l 1 + l 2 cos θ 2
Velocity:
˙ l 2 cos(θ 1 + θ 2 ) l 2 sin(θ 1 + θ 2 )
θ 1 1
=
˙ l 1 l 2 sin θ 2 −l 1 cos θ 1 − l 2 cos(θ 1 + θ 2 ) −l 1 sin θ 1 − l 2 sin(θ 1 + θ 2 )
θ 2
˙ x
× (2.6)
˙ y
Obtaining equations for acceleration takes a bit more effort; for these and
for other details on equations above, one is referred, for example, to Ref. 8.
In general, for each point (x, y) in the arm workspace there are two (θ 1 ,θ 2 )
solutions: One can be called “elbow up,” while the other can be called “elbow
down” (Figure 2.3a). This is not always so—one should remember special cases
and degeneracies:
• Any point on the workspace boundaries—that is, when θ 2 = 0or θ 2 =
π —has only one solution (Figure 2.3b).