Robotics & Automation
Lecture 23
Closed Kinematic Chain and Parallel Mechanisms
John T. Wen
November 10, 2008 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I
Examples of closed kinematic chains
• Constrained robot, e.g., robot in contact with environment (polishing, assembly, etc.). • Multiple interacting robots, e.g., multi-finger grasp, multiple cooperative robots. • Parallel mechanism, e.g., 4-bar linkage, slider-crank, Stewart-Gough Platform, Delta robot. qq 2
qq 1
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Kinematics
Forward kinematics and inverse kinematics are the same as before: given joint positions, find the task frame, and vice versa – subject to the closed chain constraints. Usually: Serial mechanism has complicated geometry, inverse kinematics is more dif- ficult than the forward kinematics. Parallel mechanisms usually are based on imposing kinematic constraints to simple serial mechanisms. Therefore, inverse kinematics is usu- ally easy. The forward kinematics needs to take into account of constraints, and is more difficult to solve.
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Constrained mechanisms
Consider a serial kinematic chain with n joints (θ ∈ Rn). Suppose the chain is constrained:
φ(θ) = 0, φ : Rn → Rk.
Then the mechanism has n − k (unconstrained) DOF. Example: • 3-DOF Planar arm with tip constrained to move along a line (2 dof) • 4-bar linkage (1 dof)
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Greubler’s Formula
Consider a planar mechanism. Let N = # of links, g = # of joints, fi = # of DOF of ith joint. DOF: 3N g Constraints: 3g − ∑i=1 fi. g Net # of DOF: 3(N − g) + ∑i=1 fi for spatial mechanisms (bodies are free to translate and rotate in Cartesian space), we have g Net # of DOF: 6(N − g) + ∑ fi. i=1 Example: planar platform with three legs.
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Kinematics: 4-bar linkage Example
rotation: R = exp(zˆθ1)exp(zˆθ3) = exp(zˆθ2)exp(zˆθ4) (1 constraint)
position: p14 = exp(zˆθ1)(p13 + exp(zˆθ3)p34) = p12 + exp(zˆθ2)p24 (2 constraints)
Inverse kinematics: given R (equiv. θE = θ1 + θ3 = θ2 + θ4), find (θ1,θ2,θ3,θ4).
Forward kinematics: Suppose θ1 is the active joint. Given θ1, find θE . Typically, inverse kinematics is easy (since each leg is simple), but forward kinematics hard (solving the constraint equations).
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Slider Crank
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qq1
qq3
Structure equation (kinematics + constraints):
(θ1+θ2+θ3)zˆ R03 = R01R12R23 = e
p0T = R01 p12 + R01R12 p23 + R01R12R23 p3T
R03 = I T y p0T = 0.
November 10, 2008Copyrighted by John T. Wen Page 6