# OSU ECE 5463 - Manipulator Dynamics-2 (13 pages)

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## Manipulator Dynamics-2

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- Pages:
- 13
- School:
- Ohio State University
- Course:
- Ece 5463 - Introduction to Real Time Robotics Systems

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Manipulator Dynamics 2 Read Chapter 6 Dynamic equations for each link For each link there are two equations to describe the effects of force and torque to the motion Newton equation Euler s equation 2 Ci has its origin at the center of mass of the link and has the same orientation as the link frame i 1 xc1 x1 Dynamic equations for all the links Use the iterative Newton Euler algorithm iF zi zi 1 xi 1 xi in i 1f if i 1n iN Between links there are action and reaction forces torques For link n we have n 1 f 0 and n 1n 0 therefore the equations can start from the last link and go inwards Compute forces and torques Look the top figure in the previous slide one has From the above one can obtain the following iterative equations Note that n 1 f 0 and n 1n 0 so we can start the iteration from link n Since every joint has only one torque applied we have Question how the above equation should be changed for a prismatic joint The iterative Newton Euler dynamics algorithm 1 The iterative algorithm takes two steps First step perform outward iterations to compute velocities and accelerations Second step perform inward iterations to computer forces and torques First step Angular acceleration Linear acceleration 2 For the center of the link The iterative Newton Euler dynamics algorithm 2 Second step Draw the Newton Euler equations Inward iteration from link n to link 1 Iterative algorithm example Use the OSU hexapod as an example which is similar to the manipulator shown below to simplify the problem let m1 and m2 be point mass and located at the tip of the link 1 Xe D1 Z1 m1 X1 X2 Z2 D0 Z0 Ze m2 X0 D2 Use the two iterations Determine the values needed in the algorithm The position vectors of the center of mass The inertia tensor at the center of mass 0 0 Rotational matrices 0 0 0 0 1 0 0 0 1 0 Outward iteration For link 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 For link 2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2 2 0 0 0 Inward iteration For link 2 0 2 0 0 For link 1 0 0 0 1 0 2 Figure out the details Final dynamic equation Since every joint has only one degree of freedom one has Ultimately one can obtain M V G where Lagrangian formulation Another approach is using the Lagrangian formulation Step 1 obtain the Lagrangian of a manipulator where is the kinetic energy of the manipulator and is the potential energy of the manipulator The dynamic equation can be expressed as Becomes the same as the one shown earlier M V G

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