《机器人导论》(英文版) MECH572-lecture4

+4 McGill Dept Of Mechanical Engineering MECH572 Introduction To robotics Lecture 4

MECH572 Introduction To Robotics Lecture 4 Dept. Of Mechanical Engineering

Review Concept of Screw Motion of a rigid body do=e d d A d w·p=w·a= constant (Qa-a Po 2(1-co6o) ford≠0

Review • Concept of Screw Motion of a rigid body O  Q P P' A A'  P0 dp e dA L

Review Use Plucker array to represent a line(screw axis) p1 xe Instantaneous motion Velocity Analysis: p=a+wx(p-a =a+92p-a) Angular velocity tensor S=QQ( Skew-symmetric W=vect()

Review • Use Plücker array to represent a line (screw axis) • Instantaneous Motion Velocity Analysis: Angular velocity tensor (Skew-symmetric)

Rigid-Body mechanics Instant Screw of rigid-Body Motion(cont'd) Define isa Consider 2 points a and p al+al P a=a“应=电 Ref Eq (26a&b) )4=2 Recall theorem 2.3. 4 If A 02=wwT-w|F1 Then A=-la 11+aa P 92a+(p-a)

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion (cont'd) Define ISA Consider 2 points A and P Recall Theorem 2.3.4 If Then Ref. Eq. (2.6a&b)

Rigid-Body mechanics Instant Screw of rigid-Body Motion(cont'd) Points on Isa p⊥=0 Pl‖ p ‖P2 P 92p=(a+ [> Ap=b p=0 A b=「a(1/|422a A Apo=A b ATA=S3+ww=-5+wT

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion (cont'd) Points on ISA O P'0

Rigid-Body mechanics Instant Screw of rigid-Body Motion(cont'd) ATA=w1, Ab=n(a-S2a a-na) wxa-wx Jw2 wl2

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion (cont'd)

Rigid-Body mechanics Instant Screw of rigid-body motion - example H Unit Cube H e B C E A Seek: L Q=00 100

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion - Example e'1 = e3 e'2 = -e1 Q = e'3 = -e2 z x A y B C D E F G H A' B' C' D' E' F' G' H' e1 e2 e3 e' 1 e' 3 e' 2 1 Unit Cube           − − 1 0 0 0 0 1 0 1 0 Seek: L

Rigid-Body mechanics Instant Screw of rigid-Body motion- example tr(Q)=1+2 cosd=0 φ=120 vect(Q)=sinde sino e To get equation for L, we still need a (Q-1)2(Qa-a) p0= 2(1-co6o) ford≠0

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion – Example tr(Q) = 1 + 2 cos = 0 cos = - vect(Q) = sine = sin = e = To get equation for L, we still need a: 2 1                 = −           − 3 1 3 1 3 1 2 3 1 1 1 2 1 2 3  = 120º                 − 3 1 3 1 3 1

Rigid-Body mechanics Instant Screw of Rigid-Body motion- Example(cont'd Use point a as reference point 000 Q-I 1010 (Q-1)(-a) 1-10|3=-3cp= 31+12 >x+1=y-3=z+

Rigid-Body Mechanics • Instant Screw of Rigid-Body Motion – Example (cont'd) Use point A as reference point a = a' = Q – I = (Q – I) (- a') = p0 = L: x + 1 = y – 3 = z + 2           0 0 0           −1 3 0           − − − − − 1 0 1 0 1 1 1 1 0           − − − =            −          − − − − − 2 3 1 1 3 0 0 1 1 1 1 0 1 0 1               − − − 3 2 1 3 1 T 3 1 3 2 3 1 1 3 1 3 1 + = − + = + z y x

Rigid-Body mechanics Twist of rigid-body Concept -decompose velocity into two parts V=-4×p+w Velocity perpendicular Velocity along l x p Define Angular velocity of the rigid P body Linear velocity of a point on the rigid body describes the velocity field of a rigid body (3-D figure)

Rigid-Body Mechanics • Twist of Rigid-Body Concept - decompose velocity into two parts Define describes the velocity field of a rigid body Velocity perpendicular to L' Velocity along L' L'' O P p Angular velocity of the rigid body Linear velocity of a point on the rigid body (3-D figure)