Forward Kinematics · Forward Kinematics 4 because the base frame and frame 1 have the same...

Forward Kinematics

1

Links and Joints

2

n joints, n + 1 links

link 0 is fixed (the base)

joint i connects link i – 1 to link i

link i moves when joint i is actuated

joint 1

joint 2

joint 3 joint 4 joint n-1joint n

link 0

link 1

link 2

link 3

link n-1

link n

.................

prismatic

revolute

i

i

id

q

Forward Kinematics

3

given the joint variables and dimensions of the links what is

the position and orientation of the end effector?

2

1

a1

a2

x0

y0

p0 ?

Forward Kinematics

4

because the base frame and frame 1 have the same

orientation, we can sum the coordinates to find the position

of the end effector in the base frame

2

1

a1

a2

x0

y0

( a1 cos 1 , a1 sin 1 )

1

x1

y1( a2 cos (1 + 2),

a2 sin (1 + 2) )

(a1 cos 1 + a2 cos (1 + 2),

a1 sin 1 + a2 sin (1 + 2) )

Forward Kinematics

5

from earlier in the course

2

1

a1

a2

x0

y01

x2 = (cos (1 + 2),

sin (1 + 2) )

y2 = (-sin (1 + 2),

cos (1 + 2) )

x2y2

p0 = (a1 cos 1 + a2 cos (1 + 2),

a1 sin 1 + a2 sin (1 + 2) )

Frames

6

2

1

a1

a2

x0

y0 x1

y1

x2y2

Forward Kinematics

7

using transformation matrices

11 ,,

0

1 axz DRT

22 ,,

1

2 axz DRT

1

2

0

1

0

2 TTT

Links and Joints

8

n joints, n + 1 links

link 0 is fixed (the base)

joint i connects link i – 1 to link i

link i moves when joint i is actuated

joint 1

joint 2

joint 3 joint 4 joint n-1joint n

link 0

link 1

link 2

link 3

link n-1

link n

.................

prismatic

revolute

i

i

id

q

Forward Kinematics

9

attach a frame {i} to link i

all points on link i are constant when expressed in {i}

if joint i is actuated then frame {i} moves relative to frame {i - 1}

motion is described by the rigid transformation

the state of joint i is a function of its joint variable qi (i.e., is a function of qi)

this makes it easy to find the last frame with respect to the

base frame

1i

iT

)(11 ii

i

i

i qTT

12

3

1

2

0

1

0 nnn TTTTT

Forward Kinematics

10

more generally

the forward kinematics problem has been reduced to matrix

multiplication

jiji

ji

T

I

TTT

Ti

j

j

j

i

j

i

i

i

j

if

if

if

1

11

21

Forward Kinematics

11

Denavit J and Hartenberg RS, “A kinematic notation for lower-

pair mechanisms based on matrices.” Trans ASME J. Appl. Mech,

23:215–221, 1955

described a convention for standardizing the attachment of frames

on links of a serial linkage

common convention for attaching reference frames on links of

a serial manipulator and computing the transformations

between frames

Denavit-Hartenberg

12

iiii xaxdzz

i

i RTTRT ,,,,1

1000

0 i

i

i

dcs

sascccs

casscsc

ii

iiiiii

iiiiii

anglejoint

offsetlink

link twist

lengthlink

i

i

i

i

d

a

Denavit-Hartenberg

13

Denavit-Hartenberg

14

notice the form of the rotation component

this does not look like it can represent arbitrary rotations

can the DH convention actually describe every physically

possible link configuration?

ii

iiiii

iiiii

cs

scccs

sscsc

0

Denavit-Hartenberg

15

yes, but we must choose the orientation and position of the

frames in a certain way

(DH1)

(DH2)

claim: if DH1 and DH2 are true then there exists unique

numbers

1ˆˆ ii zx

1ˆ intersects ˆ ii zx

,,,,0

1 such that ,,, xaxdzz RDDRTda

Denavit-Hartenberg

16

proof: on blackboard in class

Denavit-Hartenberg

17

DH Parameters

18

ai : link length

distance between zi-1 and zi measured along xi

i : link twist

angle between zi-1 and zi measured about xi

di : link offset

distance between oi-1 to the intersection of xi and zi-1 measured

along zi-1

i : joint angle

angle between xi-1 and xi measured about zi-1

Example with Frames Already Placed

19

Step 5: Find the DH parameters

20

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

Denavit-Hartenberg Forward Kinematics

21

RPP cylindrical manipulator

http://strobotics.com/cylindrical-format-robot.htm

http://strobotics.com/cylindrical-format-robot.htm

Denavit-Hartenberg Forward Kinematics

22

How do we place the frames?

Step 1: Choose the z-axis for each frame

23

recall the DH transformation matrix

iiii xaxdzz

i

i RTTRT ,,,,1

1000

0 i

i

i

dcs

sascccs

casscsc

ii

iiiiii

iiiiii

1ˆ iix

1ˆ iiy

1ˆ iiz


24

axis of actuation for joint i+1

iẑ

link i link ilink i+1 link i+1

joint i+1 joint i+1

iẑ

iẑ


25

Warning: the picture is deceiving. We do not yet know the origin of the

frames; all we know at this point is that each zi points along a joint axis

Step 2: Establish frame {0}

26

place the origin o0 anywhere on z0 often the choice of location is obvious

choose x0 and y0 so that {0} is right-handed

often the choice of directions is obvious

Step 2: Establish frame {0}

27

Step 3: Iteratively construct {1}, {2}, ... {n-1}

28

using frame {i-1} construct frame {i}

DH1: xi is perpendicular to zi-1

DH2: xi intersects zi-1

3 cases to consider depending on the relationship between zi-1and zi


29

Case 1

zi-1 and zi are not coplanar (skew)

i angle from zi-1 to zi measured about xi

iẑ

1ˆiz

(out of page)shortest line between

and1

ˆiz iẑ

ix̂

io point of intersection

ia


30

Case 2

zi-1 and zi are parallel ( i = 0 )

notice that this choice results in di = 0

iẑ1ˆ iz

ix̂

io point of intersection

1ioia


31

Case 3

zi-1 and zi intersect ( ai = 0 )

iẑ

1ˆiz

ix̂io point of intersection

(out of page)


32


33

Step 4: Place the end effector frame

34

“approach”

“sliding”

“normal”

Step 4: Place the end effector frame

35


36

ai : distance between zi-1 and zi measured along xi

i : angle between zi-1 and zi measured about xi di : distance between oi-1 to the intersection of xi and zi-1

measured along zi-1

i : angle between xi-1 and xi measured about zi-1


37

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

More Denavit-Hartenberg Examples

38


39

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

Step 6: Compute the transformation

40

once the DH parameters are known, it is easy to construct

the overall transformation

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

1000

100

00

00

1

11

11

,,,,

0

1 1111 d

cs

sc

RTTRT xaxdzz


41

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

1000

010

0100

0001

2

,,,,

1

2 2222 dRTTRT xaxdzz


42

Link ai i di i

1 0 0 d1 1*

2 0 -90 d2* 0

3 0 0 d3* 0

* joint variable

1000

100

0010

0001

3

,,,,

2

3 3333 dRTTRT xaxdzz


43

1000

010

0

0

21

3111

3111

2

3

1

2

0

1

0

3dd

dccs

dssc

TTTT

Spherical Wrist

44

Spherical Wrist

45

Spherical Wrist: Step 1

46


47


48


49

Step 5: DH Parameters

50

Link ai i di i

4 0 -90 0 4*

5 0 90 0 5*

6 0 0 d6 6*

* joint variable


51

1000

6556565

654546465464654

654546465464654

5

6

4

5

3

4

3

6dccsscs

dssssccscsscccs

dscsccssccssccc

TTTT

RPP + Spherical Wrist

52

RPP + Spherical Wrist

53

1000

333231

232221

131211

3

6

0

3

0

6

z

y

x

drrr

drrr

drrr

TTT

21654

651641654111

dddssd

csssscccccr

z

Stanford Manipulator + Spherical Wrist

54

Link ai i di i

1 0 -90 0 1*

2 0 90 d2 2*

3 0 0 d3* 0

4 0 -90 0 4*

5 0 90 0 5*

6 0 0 d6 6*

* joint variable

SCARA + 1DOF Wrist

55

Link ai i di i

1 a1 0 d1 1*

2 a2 180 0 2*

3 0 0 d3* 0

4 0 0 d4 4* * joint variable

Forward Kinematics · Forward Kinematics 4 because the base frame and frame 1 have the same...

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Transcript of Forward Kinematics · Forward Kinematics 4 because the base frame and frame 1 have the same...