Manipulator Dynamics (1) Read Chapter 6
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1 Manipulator Dynamics (1) Read Capter 6
2 Wat is dynamics? Study te force (torque) required to cause te motion of robots just like engine power required to drive a automobile Most familiar formula: f = ma f: force, m: mass, and a: acceleration Manipulator will be more complicated since eac link can move and rotate, not just a point mass. f Translational motion τ2 τ1 Rotational motion
3 Translational motion of a robotic link (two-dimensional case) m a F = ma Newton s equation F Center of Mass
4 Angular motion of a robotic link (two-dimensional case) F l I: inertia ωω I ωω = F ll = N Euler s equation N: torque NN = II In 3D: ωω + ωω IIωω
5 Acceleration of a rigid body (3D) is wat we need to calculate manipulator dynamics (1) linear acceleration Linear velocity of a vector P in frame {B} can be expressed in frame {A} as Because dd dddd AA VV PP = AA RR BB BB VV PP + AA ΩΩ BB AA RR BB BB PP AA RR BB BB PP = AA RR BB dd dddd ( BB PP) + dd dddd ( AA RR BB ) BB PP Acceleration can be expressed as AA ΩΩ BB {B} {A} {B} AA PP BBBBBBBB BB PP AA VV PP = AA RR BB BB VV PP +2 AA ΩΩ BB AA RR BB BB VV PP + AA ΩΩ BB AA RR BB BB PP + AA ΩΩ BB ( AA ΩΩ BB AA RR BB BB PP) Wen te origin of {B} is moving AA VV PP = AA VV BBOOOOOO + AA RR BB BB VV PP +2 AA ΩΩ BB AA RR BB BB VV PP + AA ΩΩ BB AA RR BB BB PP + AA ΩΩ BB ( AA ΩΩ BB AA RR BB BB PP) Complicated? due to: a. rigid body, and b. bot translational and rotational motions of {B} and P.
6 Acceleration of a rigid body (2) angular acceleration If tere are tree frames {A}, {B} and {C}, one may ave: Ten we obtain by taking te derivative of bot sides wit respect to time: AA wic turns out to be AA ΩΩ cc = AA ΩΩ BB + AA RR BB BB ΩΩ cc ΩΩ cc = AA ΩΩ BB + dd dddd ( AA RR BB BB ΩΩ cc ) AA ΩΩ cc = AA ΩΩ BB + AA BB RR BB ΩΩ cc + AA ΩΩ BB AA RR BB BB ΩΩ cc Angular acceleration in one frame can be expressed in anoter frame in an iterative way but could be complicated. Tat will be used in calculating manipulator dynamics
7 (3) Mass distribution For rigid body we ave to consider bot mass and moment of inertia since a rigid body is free to move in te space wit translational and rotational motions. Different configurations need different torques to acieve accelerations: consider τ 1 below τ 1 Z e Z 1 D 1 X e X 1 Z 2 X 2 D 2 D 0 Z 0 X 0 Te moment of inertia is related to te mass distribution of te links and its motions in particular coordinate frames
8 Inertia tensor Using inertia to describe mass distribution wit respect to a coordinate Inertia tensor in {A} can be expressed as AA II = II xxxx II xxyy II xxzz II xxyy II yyyy II yyyy II xxzz II yyyy II zzzz II xxxx, II yyyy, II zzzz are called mass moments of inertia II xxxx, II xxzz, II yyyy are called mass products of inertia distance mass II xxxx = VV (yy 2 +zz 2 )ρdv, II yyyy = VV (xx 2 +zz 2 )ρdv, II zzzz = VV (xx 2 +yy 2 )ρdv II jjjj = VV (jjjj)ρdv i, j = x, y, or z; i j ρ is te density of te material If we cose te frame in suc a way tat te products of inertias are all zero, te axes are called principal axes, and mass moments are called te principal moments of inertia
9 Inertia tensor examples (1) zz zz yy yy xx xx Wic one as greater II xxxx, II yyyy, or II zzzz? Calculate te inertia tensor for te following object and te attaced frame. Assuming density to be even zz ll xx ww yy
10 ll Inertia tensor examples (2) II xxxx = 0 2 ww ll 0 (yy 2 +zz 2 )ρdxdydz ll = 0 ll (yy 2 +zz 2 )ρwdydz = 0 ( ll zz2 ll)ρwdz =( ll3 w +3 lw)ρ 12 3 = mm 12 (l ) ll II xxyy = 0 2 ww ll 0 xxxxρdxdydz 2 = 0 xx Wat if we remove te frame to a new place as sown below? How can we make te mass products of inertia all zero? zz ll ww yy
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