LectureNotes for AMathematicalIntroductionto ...summerschool2009.robotik-bs.de/downloads/lecture_notes/summer... · Example:SCARA manipulator. g. st (ý) = ... kinematics Inverse

Chapter 3ManipulatorKinematics

Forwardkinematics

InverseKinematics

ManipulatorJacobian

Redundantand ParallelManipulators

Summer School-Math. Methods in [email protected] 13-31 July 2009

Chapter 3 Manipulator Kinematics

1

Lecture Notes

for

AMathematical Introduction to

Robotic Manipulation

By

Z.X. Li and Y.Q. Wu#

Dept. of ECE, Hong Kong University of Science & Technology#School of ME, Shanghai Jiaotong University

23 July 2009


Forwardkinematics

InverseKinematics

ManipulatorJacobian


Summer School-Math. Methods in [email protected] 13-31 July 2009

Chapter 3 Manipulator Kinematics

2

Main Concepts

1 Forward kinematics

2 Inverse Kinematics

3 Manipulator Jacobian

4 Redundant and Parallel Manipulators


Forwardkinematics

InverseKinematics

ManipulatorJacobian


3.1 Forward kinematicsChapter 3 Manipulator Kinematics

3

Adept Cobra i600x y

z

x y

z

Lower Pair Joints: Forward kinematics:

Revolute: S1 SO(2)Prismatic: R T(1) Link 0: Stationary(Base)

Link 1: First movable linkLink n: End-eector attached

adept.aviMedia File (video/avi)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



4

Joint space:

Revolute joint: S1, i S1 or i [ , ]Prismatic joint: R

Joint Space: Q S1 S1no. of R joint

R Rno. of P joint

Adept Q S1 S1 S1 RElbow Q = 6 S1 S1

6

Reference (nominal) joint cong: = (0, 0, . . . , 0) QReference (nominal) end-eector cong: gst(0) SE(3)Arbitrary cong.: gst(), or gst : Q SE(3)

gst()


Forwardkinematics

InverseKinematics

ManipulatorJacobian



5

Classical Approach:gst(1 , 2) = gst(1) gl1l2 gl2tProblem:

A coordinate frame for each link

e product of exponentials formula:Consider Fig 3.2 again.

TL2

L1

2

1

S

Figure 3.2: A two degree of freedom manipulator


Forwardkinematics

InverseKinematics

ManipulatorJacobian



6

Step 1: Rotating about 2 by 22 = [ 2 q22 ]gst(2) = e22 gst(0)

Step 2: Rotating about 1 by 11 = [ 1 q11 ]gst(1, 2) = e11 e22 gst(0)

oset (0, 0) (0, 2) (1 , 2)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



7

What if another route is taken?

(0, 0) (1 , 0) (1 , 2)Step 1: Rotating about 1 by 1

1 = [ 1 q11 ]gst(1) = e11 gst(0)

Step 2: Rotating about 2 by 2Let e11 = [ R1 p10 1 ]

2 = R1 2q2 = p1 + R1 q2


Forwardkinematics

InverseKinematics

ManipulatorJacobian



8

2=[ 2 q22 ] =[ R12RT1 (p1 + R1q2)R12

]= [ R1 p1R10 R1 ] [ 2 q22 ] = Ade 1 1 2 2 = e11 2 e11

gst(1, 2) = e22 e11 gst(0)= ee 1 1 22 e1 1 e11 gst(0)= e11 e22 e11 e11 gst(0)= e11 e22 gst(0)Independent of the route taken


Forwardkinematics

InverseKinematics

ManipulatorJacobian



9

Procedure for forward kinematic map:Identify a nominal cong.:

= (10 , . . . , n0) = 0, gst(0) gst(10 , . . . , n0)Simplication of forward kinematics mapping:

Revolute joint: i = [ qii ]Choose qi s.t. i is simple.

Prismatic joint: i = [ vi0 ]Write gst() = e11 . . . enn gst(0) (product of exponential map-

ping)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



10

Example: SCARA manipulator

gst(0) =

I [ 0l1 + l2l0]

0 1

1 = 2 = 3 = [ 00

1]

x

z

y

1

l1

2

l0

l2

3

4

q3q2

T

Sq1

q1 = [ 001] , q2 = [ 0l1

1] , q3 = [ 0l1 + l2

0]

1 =

000001

, 2 =

l100001

, 3 =

l1 + l200001

, 4 = [ v40 ] =

001000

,


Forwardkinematics

InverseKinematics

ManipulatorJacobian



11

gst() = e11 e22 e33 e44 gst(0) = [ R() p()0 1 ]e11 =

c1 s1 0 0s1 c1 0 00 0 1 00 0 0 1

, e22 =

c2 s2 0 l1s2s2 c2 0 l1s20 0 1 00 0 0 1

,

e33 =c3 s3 0 (l1s2)s3s3 c3 0 (l1s2)v30 0 1 00 0 0 1

, e44 =

I [ 00

4]

0 1

gst() =

c123 s123 0 l1s1 l2s12s123 c123 0 l1c1 + l2c120 0 1 l0 + 40 0 0 1

in which, c123 = cos(1 + 2 + 3)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



12

Example: Elbow manipulator

gst(0) =

I [ 0l1 + l2l0]

0 1

q1

5

q2

2 3

1 4

l1 l26

l0

T

S


Forwardkinematics

InverseKinematics

ManipulatorJacobian



13

1 =

[ 001] [ 00

l0]

[ 001]

=

000001

,

2 =

[ 100] [ 00

l0]

[ 100]

=

0l00100

, 3 =

0l0l1100

,

4 =

l1 + l200001

, 5 =

0l0l1 + l2100

, 6 =

l000010

gst(1 , . . . 6) = e11 . . . e66 gst(0) = [ R() p()0 1 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



14

p() = s1(l2c2 + l2c23)c1(l1c2 + l2c23)l0 l1s2 l2s23

,R() = [rij]in which, r11 = c6(c1c4 s1c23s4) + s6(s1s23c5 + s1c23c4s5 + c1s4s5)

r12 = c5(s1c23c4 + c1s4) + s1s23s5r13 = c6(c5s1s23 (c23c4s1 + c1s4)s5) + (c1c4 c23s1s4)s6r21 = c6(c4s1 + c1c23s4) (c1c5s23 + (c1c23c4 s1s4)s5)s6r22 = c5(c1c23c4 s1s4) c1s23s5r23 = c6(c1c5s23 + (c1c23c4 s1s4)s5) + (c4s1 + c1c23s4)s6r31 = (c6s23s4) (c23c5 c4s23s5)s6r32 = (c4c5s23) c23s5r33 = c6(c23c5 c4s23s5) s23s4s6

Simplify forward Kinematics Map:

Choose base frame or ref. Cong. s.t. gst(0) = I


Forwardkinematics

InverseKinematics

ManipulatorJacobian



15

Manipulator Workspace:W = {gst() Q} SE(3)

Reachable Workspace:

WR = {p() Q} R3Dextrous Workspace:

WD = {p R2R SO(3), , gst() = (p,R)}


Forwardkinematics

InverseKinematics

ManipulatorJacobian



16

Example: A planar serial 3-bar linkage

(a) Workspace calculation:g = (x, y, )x = l1c1 + l2c12 + l3c123y = l1s1 + l2s12 + l3s123 = 1 + 2 + 3

(b) Construction of Workspace:

(c) Reachable Workspace:

(d) Dextrous Workspace:

(a) (b)

(c) (d)

l3

1

l2

3

l1

l1 l2 l3

2l3 2l3

2

Manipulators maximum workspace (Paden):Elbow manipulator and its kinematics inverse.


Forwardkinematics

InverseKinematics

ManipulatorJacobian


3.2 Inverse KinematicsChapter 3 Manipulator Kinematics

17

Given g SE(3), nd Q s.t.gst() = g , where gst Q SE(3)

Example: A planar example

l2

l1

2

1

(x, y)

B

r2

1

(x, y)

x

y

(a) (b)

x = l1 cos 1 + l2 cos (1 + 2)y = l1 sin 1 + l2 sin (1 + 2)Given (x, y), solve for (1 , 2)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



18

Review:Polar Coordinates: (r, ), r =x2 + y2Law of cosines:

2 = , = cos1 l21+l22r22l1 l2Flip solution: +

1 = atan2(y, x) , = cos1 r2 + l21 l222l1r

Hight Lights:

Subproblems

Each has zero, one or two solutions!


Forwardkinematics

InverseKinematics

ManipulatorJacobian



19

Paden-Kahan Subproblems:

Subproblem 1: Rotation about a single axis

Let be a zero-pitch twist, with unitmagnitude and twopoints p, q R3 . Find s.t. ep = q

u

v

(b)

r

q

p

v

u

(a)

Solution: Let r l, dene u = p r, v = q r, er = r


Forwardkinematics

InverseKinematics

ManipulatorJacobian



20

Also,

ep = q e (p r)u

= q rv

[ e 0 1

] [ u0 ] = [ v0 ]

eu = v { wTu = wTvu2 = v2

u

v

(b)

r

q

p

v

u

(a)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



21

u = (I T)u, v = (I T)ve solution exists only if { u2 = v2

Tu = TvIf u 0, then

u v = sin uvu v = cos uv

= atan2(T(u v), uTv)If u = 0, Innite number of solutions!


Forwardkinematics

InverseKinematics

ManipulatorJacobian



22

Subproblem 2: Rotation about two subsequent axes

let 1 and 2 be two zero-pitch, unit magnitude twists, with

intersecting axes, and p, q R3. nd 1 and 2 s.t. e11e22p = q.

r

p

c

q

2

1

2

1


Forwardkinematics

InverseKinematics

ManipulatorJacobian



23

Solution: If two axes of 1 and 2 coincide, then we get:Subproblem 1: 1 + 2 = If the two axes are not parallel, 1 2 0, then, let c satisfy:

e22p = c = e11qSet r l1 l2

e22 p ru

= c rz

= e11 (q r)v

, e22u = z = e11v

{ T2 u = T2 zT1 v = T1 z , u

2 = z2 = v2As 1,2 and 1 2 are linearly independent,

z = 1 + 2 + (1 2) z2 = 2 + 2 + 2T1 2 + 21 22


Forwardkinematics

InverseKinematics

ManipulatorJacobian



24

T2 u = T2 1 + T1 v = + T1 2

= (T1 2)T2 uT1 v(T1 2)21 = (T1 2)T1 vT2 u(T1 2)21

z2 = u2 2 = u2 2 2 2T1 21 22 ()() has zero, one or two solution(s):

Given z c { e22p = ce11p = c

for 1 and 2

1 Two solutions when the two circles intersect.

2 One solution when they are tangent

3 Zero solution when they do not intersect


Forwardkinematics

InverseKinematics

ManipulatorJacobian



25

Subproblem 3: Rotation to a given point

Given a zero-pitch twist , with unit magnitude and p, q R3, nd s.t. q ep = Dene: u = p r, v = q r, v eu2 = 2

u = u Tuv = v Tv

r

u

v

q

p

T (pq)

v

u

eu

0

u = u + Tu, v = v + Tv, 2 = 2 T(p q)2


Forwardkinematics

InverseKinematics

ManipulatorJacobian



26

(v + Tv) e(u + Tu)2 = 2v eu + T(v u)

T(qp)2 = 2

v eu2 = 2 T(p q)2 = 2,0 = atan2(T(u v), uTv), = 0 u2 + v2 2u v cos = 2, = 0 cos1 u + v 2

2u v ()Zero, one or two solutions!


Forwardkinematics

InverseKinematics

ManipulatorJacobian



27

Solving inverse kinematics using subproblems:

Technique 1: Eliminate the dependence on a joint

ep = p, if p l. Given e11e22e33 = g,select p l3 , p l1 or l2 , then

gp = e11e22pTechnique 2: subtract a common point

e11e22e33 = g , q l1 l

2

e11e22e33p q = gp qe33p q = gp q


Forwardkinematics

InverseKinematics

ManipulatorJacobian



28

Example: Elbow manipulator

52 3

1 4

l0

l1 l26

pbqw

S

gst() = e11e22e33e44e55e66gst(0) = gd e11e22e33e44e55e66 = gd g1st (0) = g1


Forwardkinematics

InverseKinematics

ManipulatorJacobian



29

Step 1: Solve for 3

Let e11 . . . e66q = g1 q e11e22e33q = g1 q

Subtract pb from g1q:

e11e22(e33q pb) = g1q pb e33q pb Subproblem 3

Step 2: Given 3, solve for 1 , 2

e11e22(e33q) = g1q, Subproblem 2 1 , 2


Forwardkinematics

InverseKinematics

ManipulatorJacobian



30

Step 3: Given 1, 2, 3, solve 4, 5

e44e55e66 = e33e22e11g1g2

let p l6 , p l4 or l5 , e44e55p = g2p,Subproblem 2 4 and 5.

Step 4: Given (1, . . . , 5), solve for 6e66 = (e11 . . . e55)1 g1 g3Let p l6 e66p = g3 p = q Subproblem 1Maximum of solutions: 8


Forwardkinematics

InverseKinematics

ManipulatorJacobian



31

Example: Inverse Kinematics of SCARA

x

z

y

1

l1

2

l0

l2

3

4

q3q2

T

Sq1


Forwardkinematics

InverseKinematics

ManipulatorJacobian



32

gst() = e11 . . . e44gst(0) =c s 0 xs c 0 y0 0 1 z0 0 0 1

gd ,

p =0001

p() = [ l1s1 l2s12l1c1 + l2c12

l0 + 4 ] = [xyz] 4 = z l0

e11e22e33 = gdg1st (0)e44 g1


Forwardkinematics

InverseKinematics

ManipulatorJacobian



33

Let p l3 , q l1 e11e22p = g1p,e11(e22p q) = g1p q,

e22p q = Subproblem 3 to get 2 e11(e22p) = g1p 1 Subproblem 1 to get 1 e33 = e22e11gdg1st (0)e44 g2e33p = g2p, p l3

ere are a maximum of two solutions!


Forwardkinematics

InverseKinematics

ManipulatorJacobian


3.3 Manipulator JacobianChapter 3 Manipulator Kinematics

34

Given gst Q SE(3), (t) = (1(t) . . . n(t))T e11 . . . enngst(0)and (t) = (1(t) . . . n(t))T ,

What is the velocity of the tool frame?

Vsst = gst()g1st () =ni=1

(gsti

i)g1st ()= n

i=1

(gsti

g1st ())i Vsst =ni=1

(gsti

g1st ()) i= [(gst

1g1st ()) , . . . , (gstn g1st ())]

Jsst()R6n

1n


Forwardkinematics

InverseKinematics

ManipulatorJacobian



35

gst() = e11 . . . enngst(0)gst1

g1st () = (1e11 . . . enngst(0))(gst())1 = 1(gst1

g1st ()) = 1gst2

g1st () = (e11 2e22 . . . enngst(0))(gst())1= e11 2e22 . . . enngst(0)g1st () = e11 2e11 2


Forwardkinematics

InverseKinematics

ManipulatorJacobian



36

(gst2

g1st ()) = Ade 1 1 2 = 2

(gsti

g1st ()) = Ade 1 1 ...e i1 i1 i i Jsst() = [1, 2, . . . , n]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



37

Interpretation of i:i is only aected by 1 . . . i1e twist associated with joint i, at the present conguration.

Body jacobian:

Vbst = Jbst() Jbst() = [1 . . . n1, n]

i = Ad1e i+1 i+1 ...e n n gst(0)iJoint twist written with respect to the body frame at the current con-

guration!

Jsst() = Adgst() Jbst()If Jsst is invertible, (t) = (Jsst())1 Vsst(t)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



38

Given g(t), how to nd (t)?1) Vsst = g(t)g1(t)2) { (t) = (Jsst())1Vsst(t)

(0) = 0 (t)

Example: Jacobian for a SCARA manipulator

q1 = [ 000] , q2 = [ l1s1l1c1

0] , q3 = [ l1s1 l2s12l1c1 + l2c12

0] ,

1 = 2 = 3 = [0 0 1]T1 = [ 1 q11 ] = [ 0 0 0 0 0 0 ]T2 = [ 2 q22 ] = [ l1c1 l1s1 0 0 0 0 ]T3 = [ 3 q33 ] = [ l1c1 + l2c12 l1s1 + l2s12 0 0 0 1 ]T


Forwardkinematics

InverseKinematics

ManipulatorJacobian



39

S

T

l1

l2

1

23

4

l0

4 = [ v40 ] = [ 0 0 1 0 0 0 ]T

Jsst() =

0 l1c1 l1c1 + l1c12 00 l1s1 l1s1 + l1s12 00 0 0 10 0 0 00 0 0 01 1 1 0


Forwardkinematics

InverseKinematics

ManipulatorJacobian



40

Example: Jacobian of Stanford arm

1

2

l0

q1

45

6

3

l1

S

qw

q1 = q2 = [ 00l0], 1 = [ 00

1] 2 = [ c1s10 ]

1 = [ 1 q11 ] = [ 0 0 0 0 0 1 ]2 = [ 2 q22 ] = [ l0s1 = l0c1 0 c1 s1 0 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



41

3 =ez1 ex2 [ 00

1]

0

= [s1c2 c1c2 s2 0 0 0]T = [ v30 ]

q = [ 00l0] + ez1 ex2 [ 0l1 + 3

0] =(l1 + 3)s1c2(l1 + 3)c1c2l0 (l1 + 3)s2

4 = ez1 ex2 [ 00

1] = [ s1s2c1s2c2 ]

5 = ez1 ex2 ez4 [ 100] = [ c1c4 + s1c2s4s1c4 c1c2s4s2s4 ]

6 = ez1 ex2 ez4 ex5 [ 010] = [ c5(s1c2c4) + s1s2s5c5(c1c2c4 s1s4) c1s2s5s2c4c5 c2s5 ]

Jsst = [ 0 2 q1 v3 5 q 5 q 6 q1 2 0 4 5 6 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



42

End-eector force:

Ft = [ forcetorque ]W =

t2

t1Vbst Ftdt = t2

t1 dt = t2

t1T(Jbst())T Ftdt

= (Jbst)TFt = (Jsst)TFsGiven Ft , what is required to balance that force?

If we apply a set of joint torques, what is the resulting

end-eector wrench?


Forwardkinematics

InverseKinematics

ManipulatorJacobian



43

Structural force:Structural force: that produces no work on admissible velocity

space Vb

Fb Vb = 0,Vb ImJbst() Fb (ImJbst) Review:

A Rmn ,(ImA) = kerAT(kerA) = ImAT

(ImJbst) = ker(Jbst)T , = (Jbst)TFb 0,Fb ker(Jbst)T


Forwardkinematics

InverseKinematics

ManipulatorJacobian



44

Example: SCARA manipulator

TFN1

FN2S

4

3

2

1

(Jsst())T =

0 0 0 0 0 1l1c1 l1s1 0 0 0 1

l1c1 + l1c12 l1s1 + l1s12 0 0 0 10 0 1 0 0 0

ker((Jsst())T): spanned by

FN1 = [ 0 0 0 1 0 0 ]FN2 = [ 0 0 0 0 1 0 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



45

Singularities:

is called a singular conguration if there 0 s.t. Vsst = Jsst() =0

or...A singularity cong. is a point at which Jsst drops rank.Consequence: (n = 6)

1 Cant move in certain directions.

2 Large joint motion is required.

3 Large structural force.

4 Cant apply end-eector force in certain direction force!

Singularities for 6R-manipulators:

Jsst = [ 1 q1 2 q2 . . . 6 q61 2 . . . 6 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



46

Case 1: Two collinear revolute joints

J() is singular if there exists two joints1 = [ 1 q11 ] , 2 = [ 2 q22 ]

s.t.

1 e axes are parallel,1 = 22 e axes are collinear,i (q1 q2) = 0, i = 1, 2

Proof :

Elementary row or column operation do not change rank of J()J() = [ 1 q1 2 q2 1 2 ] R6n 1=2

J() [ 1 q1 2 (q2 q1) 1 0 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



47Case 2: Parallel coplanar revolute joint axes

J()is singular if there exists two joints s.t.1 e axes are parallel,i = j, i, j = 1, 2, 32 e axes are coplanar, i.e. there exists a plane with normal n

s.t. nTi = 0, nT(qi qj) = 0, i, j = 1, 2, 3

3

q2

2

1

x

y

q3

z

y2

y3

q1


Forwardkinematics

InverseKinematics

ManipulatorJacobian



48

Proof :

Using change of frame J AdgJ and assume

J() = [ 1 q1 2 (q2 q1) 1 2 ] ,

AdgJ() =

0 y2 y3 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1

Linearly independent

Examples are such as the Elbowmanipulator in its reference cong.


Forwardkinematics

InverseKinematics

ManipulatorJacobian



49

Case 3: Four intersecting revolute joints axes

J() is singular if there exists four concurrent revolute joints withintersection point q s.t.:

i (qi q) = 0, i = 1, . . . , 4Proof :

Choose the frame origin at q,

p = qi, i = 1, . . . , 4

J() = [ 0 0 0 0 1 2 3 4 ]


Forwardkinematics

InverseKinematics

ManipulatorJacobian



50

Manipulability:

xy

= J()l1

l2

l31

2

3

Singular value decomposition:Given A Rn Rm, assume: n m and rank(A) = r n, then

A = UVT = [u1 un] [ 1 2 n]

vT1

vTn

where U Rmn, V Rnn, Rnn, 1 r > r+1 = =n = 0 and U

TU = VTV = VVT = Inn.Moreover, Im(A) = {u1 , . . . , ur} and ker(A) = {vr+1 , . . . , vn}.


Forwardkinematics

InverseKinematics

ManipulatorJacobian



51

Review: Basic facts of a linear transformationRn= Im(AT) ker(A)

Rm= Im(A) ker(AT)

rank(A) = dim(Im(A))Im(A) = {u1 , . . . , ur}ker(A) = {vr+1, . . . , vn}Im(AT) = (kerA) = {v1 , . . . , vr}ker(AT) = (Im(A)) = {ur+1 , . . . , um}

(need to span {u1 , . . . , un} to a full orhogonal set {u1 , . . . , um})


Forwardkinematics

InverseKinematics

ManipulatorJacobian


3.4 Redundant and Parallel ManipulatorsChapter 3 Manipulator Kinematics

52

Redundant manipulator:No. of joints > Dimension of Task Space.

E.g.No. of joints = 3, Task space R2(dim = 2)No. of joints = 7, Task space SE(3)(dim = 6).

Purpose of Redundancy:

1. Avoid obstacles

2. Optimize some cost criteria.

gst() = e11 . . . e66gst(0),Jst() = [ 1 2 . . . n ] , n > p, p = 3, 6

Inverse Kinematis:

gst((t)) = gd Qs = { Qgst() = gd} self motion manifoldTQs { TQJsst() = 0} Internal motion


Forwardkinematics

InverseKinematics

ManipulatorJacobian



53

Given Vst = Jst() = Jst()JT(JJT)1 Moore-Penrose generalized inverse

Vst

{ l1c1 + l2c12 + l3c123 = xl1s1 + l2s12 + l3s123 = yJ =

p

= [ l1s1 l2s12 l3s123 l2s12 l3s123 l3s123l1c1 + l2c12 + l3c123 l2c12 + l3c123 l3c123 ] ,

vN = [ l2l3s3l2l3s3 l1l3s23l1l2s3 + l1l3s23 ]

(x, y)

3 2

1

1

2

3

VN

(a) (b)


Forwardkinematics

InverseKinematics

ManipulatorJacobian



54

Robot d.o.f mass load Repeatability loadmass

Adept 3 4 205 25 0.025 0.122Epson H803N-MZ 4 96 8 0.02 0.0833

GMF A-600 4 120 6 0.02 0.05Pana Robo HR50 4 52 5 0.02 0.096

Puma RS84 4 130 5 0.02 0.0384Sankyo Skilam SR-3C 4 120 6 0.02 0.05

Sony SRX-3CH 4 64 2 0.02 0.0625TABLE 1.1.Characteristics of industrial manipulators (SCARA type,

mass of the robot and load capacity in kg, repeatability in mm, ac-

cording to the manufacturers notice).


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55

Robot mass load Repeatability loadmass

Acma SR 400 430 10 0.1 0.0232Acma YR500 590 30 0.2 0.0508ABB IRB L6 145 6 0.2 0.0414ABB IRB 2000 370 10 0.1 0.027CM T3646 1500 22 0.25 0.0147CM T3786 2885 90 0.25 0.0312

GMF Arc Mate 120 5 0.2 0.0417GMF S 10 200 10 0.1 0.05

Hitachi M6100 405 6 0.2 0.0148Hitachi M6100 410 10 0.1 0.0243Kuka IR 163/65 1700 60 0.5 0.0353Kuka IR 363/15 290 8 0.1 0.0276

Puma 550 63 4 0.1 0.0634Puma 762 590 20 0.2 0.0338

TABLE 1.2.Characteristics of industrial manipulators (spherical

type, mass of the robot and load capacity in kg, repeatability in

mm, according to the manufacturers notice).


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56

Parallel Manipulator:

Delta manipulator Stewart manipulator

delta.aviMedia File (video/avi)

stewart.aviMedia File (video/avi)


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57

Structure equation:

gst() = e1111 . . . e1n1 1n1 gst(0)= e2121 . . . e2n2 2n2 gst(0)= ek1k1 . . . eknk knk gst(0)E = {(11 , . . . , 1n1 , . . . , k1 , . . . , knk)} (free cfg. space)Q = { Egst() = g1st(1) = gkst(k)} (actual cfg. space)

Computation of dimension of Q: (DOF)

Greublers formula:F = 6N g

i=1

(6 fi) = 6(N g) +gi=1

fi

g =No. of joints, fi =DOF of the ith joint, N = No. of links


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58

Using the above equation to the manipulator shown in the gure:

g = 9ft = 1N = 7 F = 3(7 9) + 9i=1

1 = 3(2) + 9 = 3 Parametrization of Q (the process ofdesignating active joints):

Velocity Relation: Vsst = J1s 1 = J2s 2 = = Jks k

[ J1 Jk]

1

k

= [ I

I]Vsst

J = I VsstJa a + Jp p = I Vsst


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59

Im(Js1): Allowed velocity by chain 1Im(Js1) Im(Jsk): Allowed velocity

g: Rank of structure equationg = dimki=1Im(Ji())Denition: Conguration Space Singularity

A point Q s.t.the structure equation drops rank.

J1(1)1 J2(2)2 = 0J1(1)1 J3(3)3 = 0J1(1)k Jk(k)k = 0

J1(1) J2(2)J1(1) 0 J3(3) J1(1) . . . Jk(k)

12k

= 0

J() = 0


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60

Howmany constraint equations?: p(k 1)E.g.

k = 1, p = 3 3k = 3, p = 3 6k = 6, p = 6 6 5 = 30

n = DOF, dimQ = dimQn

No. of independent constrainsDenition: Conguration Space Singularity

A point Q s.t. rank(J()) drops below its full rank. Example: 4-bar mechanism

r l1 sin 11 + r cos (11 + 12) = x = r l2 sin 21 r cos 21 + 22l1 cos 11 + r sin (11 + 12) = y = h + l2 cos 21 r sin (21 + 22)


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61

11 + 12 = = 21 + 22[ l1 cos 11 r sin 11 r sin 12 l2 cos 21 + r sin 21 r sin 22l1 sin 11 + r cos 12 r cos 12 l2 sin 21 + r cos 22 r cos 22

1 1 1 1 ] = 0

(a) Singular configuration

active joint

(b) Uncertainty configuration


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62

An alternative method:

l1 cos 1 + l2 cos 2 l3 cos 3 l4 = 0l1 sin 1 + l2 sin 2 l3 sin 3 = 0

Q = (1 , 2 ,3)l1

l2

l3

l4

1

2

3

[ l1 sin 1 l2 sin 2 l3 sin 3l1 cos 1 l2 cos 2 l3 cos 3 ]J()

123

= 0

J() drop rank!{ 1 2 = , or 02 3 = , or 01 3 = , or 0


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63

If J() is of normal rank, the conguration space of the system is:Q = f 1(0), where f = 0 is the close loop constraint

How to parameterizeQ by active joints: a,p

J1()a + J2()p = 0Actuator singularity if rank J2() < normal rank of J2().

Chapter 3 Manipulator KinematicsForward kinematicsInverse KinematicsManipulator JacobianRedundant and Parallel Manipulators

Documents

LectureNotes for AMathematicalIntroductionto ...summerschool2009.robotik-bs.de/downloads/lecture_notes/summer... · Example:SCARA manipulator. g. st (ý) = ... kinematics Inverse