static force analysis: another role for the...
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Static Force Analysis:Another Role for the Jacobian
Portions abstracted from H. Asada and J.-J. E. Slotine, “Robot Analysis and Control,” Wiley, 1986
©2017 Max Donath
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Static Force Analysis: Balance of forces and moments
Consider forces and moments acting on link i.
Arm links are stationary.
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Exerted on the end effector by an external object
Special Cases
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Equivalent Joint Torques:
Actuator I exerts scalar torque or force tI between link I-1 and link I in the direction of joint axis I-1. Find the relationship between tI and F˚I-1 or N ˚I-1, I
Moment N ˚I-1, I and other components of F ˚I-1, I are supported by the joint mechanism internal force and moment
All the other components are supported by the joint structure.
Joint torque vector, N x 1
Endpoint force and moment, 6 x1
Definition:
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Consider the virtual work done by the N actuators and the environment on the arm linkage
Theorem:
The joint torques t required for a manipulator to exert a given force and moment F at its end point are given by
where friction in the joint mechanisms are ignored and gravity forces are not included. The matrix J is the (6 x N) Jacobian matrix associated with the differential relationship between joint displacements DQ and end point displacements DP:
Proof
Principle of Virtual Work
Let dQ and dP be virtual displacements of the joints and endpoint, respectively.
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According to the principle of virtual work, d Work must vanish for arbitrary geometrically admissible displacements, dQ and dP , when the arm linkage is in equilibrium
The joint torques t that balance with external force -F are called the equivalent joint torques.
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To resolve the endpoint force to joint torques, one does not need to compute the inverse of the Jacobian matrix.
Example 1: Find the equivalent joint torques to the endpoint force F
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The wrist force/moment sensormeasures forces and momentsFU, FV, Fw and NU, Nv, NW.
Find the force and moment actingon the tool at point T.
Note: U-V-W is aligned with X-Y-Z in this example
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DφU =DφX , DφV =DφY , DφW =DφZDU =DX + RZDφY −RY DφZDV =DY −RZDφX + RX DφZDW =DZ + RYDφX −RX DφY
€
DP~
=
DUDVDWDφUDφVDφW
#
$
% % % % % % %
&
'
( ( ( ( ( ( (
=
1 0 0 RZ −RY1 −RZ 0 RX
0 1 RY −RX 00 1 0
0 10 0 1
#
$
% % % % % % %
&
'
( ( ( ( ( ( (
DXDYDZDφXDφYDφZ
#
$
% % % % % % %
&
'
( ( ( ( ( ( (
= J~DQ~
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http://www.interfaceforce.com/
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Servo stiffness and endpoint compliance:
Consider a joint drive system with position feedback control.
Find the endpoint compliance assuming that arm links are rigid
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1) Equivalent joint torques to the external endpoint force/moment:
2) Joint deflections
3) Endpoint deflections:
Note A) C varies depending on the arm configuration Q
B) The magnitude of ∆P varies depending on the direction of the applied force F.
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Manipulator Dynamics
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So far only kinematics considered, what about
the manipulator dynamics and its control?
Problems:
A. Multimode structural resonance at each joint. In a serial manipulator each joint is supported by the preceding link.
Resonant frequency for many manipulators is approximately 10 HZ
Must sample and control at above 60 HZ
Result:
Computational time limit for on-line computer control … 16.7 msec
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B. Variation of link moments of inertia with manipulator motion and with payload
C. Large gravitational effects and its variation with manipulator orientation
D. Coordination of joint motion
E. Coriolis and centrifugal effects during reasonably fast motion
Possible Result:
Poor path control for unknown or varying loads