[PDF] - Ch. 4: Velocity Kinematics Velocity Kinematics We want to relate PDF Document

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Ch. 4: Velocity Kinematics

Velocity Kinematics

We want to relate end-effector linear and angular velocities with

the joint velocities

First we will discuss angular velocities about a fixed axis
Second we discuss angular velocities about arbitrary (moving)

axes

We will then introduce the Jacobian

– Instantaneous transformation between a vector in Rn representing joint velocities to a vector in R6 representing the linear and angular velocities of the end-effector

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2 Introduction

Q1: What is the linear velocity of O1?
Q2: What is the angular velocity of O1?

y0 x1 y1 a x0 θ

Introduction (2)

Q1: What is the linear velocity of O2?
Q2: What is the angular velocity of O2?

y0 x1 y1 θ a x2 y2 b x0

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3 Angular velocity: arbitrary axis

Skew-symmetric matrices

– Definition: a matrix S is skew symmetric if: – i.e. L t th l t f S b d t d th b d fi iti

= + S ST

– Let the elements of S be denoted sij, then by definition: – Thus there are only three independent entries in a skew symmetric matrix

S ST − = j i s j i s s

ij ji ij

= = ≠ − = for for

Angular velocity: arbitrary axis

Properties of skew-symmetric matrices
1. The operator S is linear

( ) ( ) ( )

R R , ∈ ∈ ∀ + = + β α β α β α , , ,

3

b a b S a S b a S

2. The operator S is known as the cross product operator

– This can be seen by the definition of the cross product:

( )

3

R p a, , ∈ ∀ × = p a p a S

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4 Angular velocity: arbitrary axis

Properties of skew-symmetric matrices
3. For

– This can be shown for and R as follows:

( )

3

3 R , ∈ ∈ a SO R – This can be shown by direct calculation. Finally:

( ) ( )

3

SO R a,b Rb Ra b a R ∈ ∈ ∀ × = × , R ,

( ) ( )

Ra S R a RS

T =

Angular velocity: arbitrary axis

Derivative of a rotation matrix

– let R be an arbitrary rotation matrix which is a function of a single variable θ

R(θ)R(θ)T I

R(θ)R(θ)T=I
Differentiating both sides (w/ respect to θ) gives:

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5 Angular velocity: arbitrary axis

Example: let R=Rx,θ, then using the previous results we have:

Angular velocity: arbitrary axis

Now consider that we have an angular velocity about an

arbitrary axis

Further, let R = R(t)
Now the time derivative of R is:
Where ω(t) is the angular velocity of the rotating frame

( ) ( ) ( ) ( )

t R t S t R ω = &

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6 Addition of angular velocities

For most manipulators we will want to find the angular velocity
f one frame due to the rotations of multiple frames

– We assume that there are no translational components: all coordinate frames have coincident origins g – Consider three frames: o0, o1, o2: – To illustrate how the rotation of multiple frames is determined by the rotations of the individual frames, take the derivative of this rotation matrix:

( ) ( ) ( )

t R t R t R

1 2 1 2

=

Addition of angular velocities

Now the first term can be derived as follows:

( ) ( )

2 1 , 1 2 1 1 , 1 2 1

R S R R S R R ω ω = = &

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7 Addition of angular velocities

Further:
And since

( ) ( ) ( ) [ ]

2 1 2 , 1 1 1 , 2 2 ,

R R S S R S ω ω ω + =

Linear velocities

The linear velocity of any point on a rigid body is the sum of the

linear velocity of the rigid body and the velocity of the particle due to rotation of the rigid body

– First, the position of a point p attached to a rigid body is: First, the position of a point p attached to a rigid body is: – Where o is the origin of the o1 frame expressed in the inertial frame – To find the velocity, take the derivative as follows:

Rp

p + =

1

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8 The Jacobian

Now we are ready to describe the relationship between the joint

velocities and the end effector velocities.

Assume that we have an n-link manipulator with joint variables

q1, q2, …, qn q1, q2, , qn

– Our homogeneous transformation matrix that defines the position and orientation of the end-effector in the inertial frame is: – We can call the angular velocity of the tool frame ω0,n

0 and:

( ) ( )

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 1 q

q

R T

n n

n

– Call the linear velocity of the end-effector:

The Jacobian

Therefore, we want to come up with the following mappings:

q J q J v

n v n

& &

ω

ω = = – Thus Jv and Jω are 3xn matrices q

n ω

ω

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9 Deriving Jω

Remember that each joint i rotates around the axis zi-1
Thus we can represent the angular velocity of each frame with

respect to the previous frame

– If the ith joint is revolute, this is:

Deriving Jω

Now Jω can simply be written as follows:

– There are n columns, each is 3x1, thus Jω is 3xn

[ ]

1 1 2 1 −

⋅ ⋅ ⋅ =

n nz

z z J ρ ρ ρ

ω

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10 Deriving Jv

Linear velocity of the end effector:

∑

= ∂

∂ =

n i i i n n

q q

1

& &

i

Deriving Jv

End-effector velocity due to prismatic joints

– Assume all joints are fixed other than the prismatic joint di – The motion of the end-effector is pure translation along zi-1

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11 Deriving Jv

End-effector velocity due to revolute joints

– Assume all joints are fixed other than the revolute joint θi – The motion of the end-effector is given by:

The complete Jacobian

The ith column of Jv is given by:

( )

⎩ ⎨ ⎧ − × =

− − −

prismatic for revolute for

1 1 1

i z i

z

J

i i n i vi

The ith column of Jω is given by:

⎩ ⎨ ⎧ =

−

prismatic for revolute for

1

i i z J

i

ω

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12 Example: two-link planar manipulator

Calculate J for the following manipulator:

– Two joint angles, thus J is 6x2 ( )

⎩ ⎨ ⎧ − × =

− − −

prismatic for revolute for

1 1 1

i z i

z

J

i i n i vi

⎩ ⎨ ⎧ =

−

prismatic for revolute for

1

i i z J

i

ω

⎩ p

Example: velocity of an arbitrary point

We can also use the Jacobian to calculate the velocity of any

arbitrary point on the manipulator

( )

⎩ ⎨ ⎧ − × =

− − −

prismatic for revolute for

1 1 1

i z i

z

J

i i n i vi

⎩ ⎨ ⎧ =

−

prismatic for revolute for

1

i i z J

i

ω

⎩

−

p

1 i

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13 Example: Stanford manipulator

The configuration of the Stanford manipulator allows us to make

the following simplifications:

( )

⎩ ⎨ ⎧ − × =

− − −

prismatic for revolute for

1 1 1

i z i

z

J

i i n i vi

⎩ ⎨ ⎧ =

−

prismatic for revolute for

1

i i z J

i

ω

Example: Stanford manipulator

From the forward kinematics of the Stanford manipulator, we

calculated the homogeneous transformations for each joint: calculated the homogeneous transformations for each joint:

⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ − = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ − = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − = 1 1 1 1 1 1 1 1 1 1 1

6 6 6 6 6 5 5 5 5 5 4 4 4 4 4 3 3 2 2 2 2 2 2 1 1 1 1 1

d c s s c A c s s c A c s s c A d A d c s s c A c s s c A , , , , ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − 1 1 1 1 1 1

6

d

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14 Example: Stanford manipulator

To complete the derivation of the Jacobian, we need the

following quantities: z0, z1, … , z5, o0, o1, o3, o6

– o3 is o and o0 = [0 0 0]T

Example: Stanford manipulator

And the oi terms are given as:

( ) ( ) ( )

⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ + + + − − + + − = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ + − = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ =

2 1 5 5 1 4 2 5 4 1 6 2 1 3 2 1 5 4 1 2 5 1 5 4 2 1 6 2 1 3 2 1 6 2 1 3 2 1 2 1 3 2 1 3 1

, , , s s c s s c c s s c d d c d s s s s s s c c s c c c d d s d s c

d

c d s s d s d s c

Finally, the Jacobian can be assembled as follows:

( )

⎥ ⎦ ⎢ ⎣ − + ⎥ ⎦ ⎢ ⎣ ⎥ ⎦ ⎢ ⎣ ⎥ ⎦ ⎢ ⎣

5 2 4 5 2 6 3 2 3 2 2

s s c c c d d c d c d

( ) ( ) ( ) ( )

⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ − − − + − − − + − = ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ − − = ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡− =

4 1 4 2 1 , 3 2 , 3 1 1 , 3 2 , 3 2 1 2 2 1 2 1 1 1 1 1

c s s c c

d

c

d

s c

d

c

d

s s J c s s s c J d c d s d s d c J d d J

x x z z y y z z x y z z x y

⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ − = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ = , , , 1

4 2 4 3 1 1 2 1

s s J J c s J J

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15 Example: Stanford manipulator

Finally, the Jacobian can be assembled as follows:

( )( )

( )

⎥ ⎤ ⎢ ⎡ − − − + −

, 3 4 2 , 3 4 1 4 2 1

d

s s

d

c c s c s

y y z z

( )( ) ( ) ( )(

) (

)( ) ( )(

) (

)(

)

( )( ) ( )( )⎥

⎥ ⎤ ⎢ ⎢ ⎡ − − + − + − − − − + − + + ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ − − − − + − − − − + − + − =

, 3 5 2 5 4 2 , 3 5 2 1 5 4 1 5 4 2 1 , 3 5 2 5 4 2 , 3 5 2 1 5 4 1 5 4 2 1 4 2 4 1 4 2 1 , 3 4 1 4 2 1 , 3 4 1 4 2 1 , 3 4 2 , 3 4 1 4 2 1 5

d

c c s c s

d

c s c s s s s c c c

d

c c s c s

d

c s s s s c s c c s s s c s c c c

d

c c s c s

d

c s s c c

d

s s

d

c s s c c J

x x z z y y y y x x y y x x z z

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ + − + + + − =

5 2 5 4 2 2 1 5 4 1 5 4 2 1 5 2 1 5 4 1 5 4 2 1 6

c c s c s c s s s s c s c c s c s c s s s s c c c J

Example: SCARA manipulator

Jacobian will be a 6x4 matrix

( ) ( ) ( )

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − × − × − × =

3 1 3 4 3 2 1 4 1 4

z z z

z

z

z
z

J

Thus we will need to determine the following quantities: z0, z1,

… , z3, o0, o1, o2, o4

– Since all the joint axes are parallel, we can see the following:

( ) ( )

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − × − × =

3 1 2 1 4 1 4

z z z z

z
z

– From the homogeneous transformation matrices we can determine the origins of the coordinate frames

k z z k z z ˆ , ˆ

3 2 1

− = = = =

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16 Example: SCARA manipulator

Thus o0, o1, o2, o4 are given by:

⎥ ⎥ ⎤ ⎢ ⎢ ⎡ + + = ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ = ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ =

12 2 1 1 12 2 1 1 4 1 1 1 1 1

, , s a s a c a c a

s

a c a

Finally, we can assemble the Jacobian:

⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − + ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ ⎥ ⎥ ⎦ ⎢ ⎢ ⎣

4 3 12 2 1 1 4 1 1 1

, , d d s a s a

s

a

⎥

⎥ ⎤ ⎢ ⎢ ⎡ + − − −

12 2 12 2 1 1 12 2 12 2 1 1

c a c a c a s a s a s a ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ − − = 1 1 1 1

12 2 12 2 1 1

J

Next class…

Formal definition of singularities
Tool velocity
manipulability