Kinematic Model of Robot Manipulators

Kinematic Model of Robot Manipulators Claudio Melchiorri Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI) Universit` a di Bologna email: [email protected]

C. Melchiorri (DEIS)

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Summary 1

Kinematic Model

2

Direct Kinematic Model

3

Inverse Kinematic Model

4

Differential Kinematics

5

Statics - Singularities - Inverse differential kinematics

6

Inverse kinematics algorithms

7

Measures of performance


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Introduction

Kinematic Model In robotics, there are two main ‘kinematic’ problems: 1. Forward (direct) Kinematic Problem: once the joint position, velocity, acceleration are known, compute the corresponding variables of the end-effector in a given reference frame (e.g. a Cartesian frame). =⇒ Forward kinematic model: a function f defined between the joint space IRn and the work space IRm : x = f(q)


x ∈ IRm , q ∈ IRn

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Introduction

Kinematic Model 2. Inverse Kinematic Problem: computation of the relevant variables (positions, velocities, accelerations) from the work space to the joint space. =⇒ Inverse Kinematic Model: function

g = f −1

from IRm to IRn : q ∈ IRn , x ∈ IRm

q = g(x) = f −1 (x)

Some common (somehow arbitrary) definitions must be adopted ⇒ for the same manipulator, different (although equivalent) kinematic models can be defined.


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Introduction

Kinematic Model – Example: a 2 dof planar robot Forward kinematic model:

x

=

l1 cos θ1 + l2 cos(θ1 + θ2 )

y

=

l1 sin θ1 + l2 sin(θ1 + θ2 )

Inverse kinematic model:

cos θ2 =

x02 + y02 − l12 − l22 , 2l1 l2

sin θ2 = ±

q

(1 − cos2 θ2 )

θ2 = atan2(sin θ2 , cos θ2 ) φ

=

k1 = l1 + l2 cos θ2 ,

θ1 + θ2

sin θ1 =

An easy problem...

y0 k1 − x0 k2 , k2 + k2 1 2

k2 = l2 sin θ2 cos θ1 =

y0 − k1 sin θ1 k2

θ1 = atan2(sin θ1 , cos θ1 ) • The solution is not so simple. • Two possible solutions (sign of sin θ2 ).


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Kinematic Model

Introduction

Kinematic Model Homogeneous Transformations are used for the definition of the kinematic model. A robotic manipulator is a mechanism composed by a chain of rigid bodies, the links, connected by joints. A reference frame is associated to each link, and homogeneous transformations are used to describe their relative position/orientation.


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Introduction

Kinematic Model A convention for the description of robots. Each link is numbered from 0 to n, in order to be univocally identified in the kinematic chain: L0 , L1 , . . . , Ln . =⇒ Conventionally, L0 is the “base” link, and Ln is the final (distal) link. Each joint is numbered, from 1 to n, starting from the base joint: J1 , J2 , . . . , Jn . =⇒ According to this convention, joint Ji connects link Li −1 to link Li . A manipulator with n + 1 links has n joints.


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Kinematic Model

Introduction

Kinematic Model The motion of the joints changes the end-effector position/orientation in the work space. The position and the orientation of the end-effector result to be a (non linear) function of the n joint variables q1 , q2 , ..., qn , i.e. p = f(q1 , q2 , ..., qn ) = f(q) q = [q1 q2 . . . qn ]T is defined in the joint space IRn , p is defined in the work space IRm . Usually, p is composed by: some position components (e.g. x, y , z, wrt a Cartesian reference frame) some orientation components (e.g. Euler or RPY angles).


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Denavit-Hartenberg Parameters

Kinematic Model Need of defining a systematic and possibly unique method for the definition of the kinematic model of a robot manipulator: DENAVIT-HARTENBERG NOTATION A reference frame is assigned to each link, and homogeneous transformations matrices are used to describe the relative position/orientation of these frames. The reference frames are assigned according to a particular convention, and therefore the number of parameters needed to describe the pose of each link, and consequently of the robot, is minimized.


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Denavit-Hartenberg Parameters Problem: How to assign frames to the links in order to minimize the number of parameters? Generally speaking, 6 parameters are necessary to describe the position and the orientation of a rigid body in the 3D space (a rigid body has 6 dof), and therefore 6 parameters are required to describe Fb in Fa . Under some hypotheses, only 4 parameters are required: the Denavit-Hartenberg Parameters. Given two reference frames F0 and F6 in the 3D space, 4 cases are possible:


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Denavit-Hartenberg Parameters Most general case: Skew Axes. PROBLEM: Find a sequence of elementary homogeneous transformations relating two generic reference frames F0 e F6 , with skew axes z0 and z6 . SOLUTION: Infinite solutions are possible. It is desirable to define A sequence so that the kinematic model is defined univocally and using the minimum number of parameters.


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Denavit-Hartenberg Parameters - Procedure A common normal n exists among two skew z axes. Let us define: d the distance between the origin of F0 and the intersection point of z0 with n a the distance between z0 and z6 along n Apply the following sequence of translations/rotations: 1 Translate the origin of F0 along z0 for the quantity d: the frame F1 is obtained 2 Rotate (ccw) F1 about z1 by the angle θ until x1 is aligned with n:F2 is obtained 3 Translate F2 along x2 (= n) for a: F3 is obtained, with origin on the z6 axis 4 Rotate (ccw) F3 about x3 by α, so that z3 is aligned with z6 : F4 is obtained 5 Translate F4 along z4 for the quantity b until F6 : the frame F5 is obtained 6 Rotate F5 about z5 by the angle φ: F6 is reached


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Denavit-Hartenberg Parameters - Procedure


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Denavit-Hartenberg Parameters Six cyclic transformations have been employed to move from F0 to F6 : 3 translations and 3 rotations. There is a translation-rotation pattern: 0

T6 = Tras(z0 , d)Rot(z1 , θ)Tras(x2 , a)Rot(x3 , α)Tras(z4 , b)Rot(z5, φ)

(1)

The first 4 transformations are of particular interest: 2 couples of translations and rotations about two axes (note that z0 = z1 and x2 = x3 ): 0

H4

= =


Tras(z0 , d)Rot(z1 , θ)Tras(x2 , a)Rot(x3 , α)   Cθ −Sθ Cα Sθ Sα aCθ  Sθ Cθ Cα −Cθ Sα aSθ     0 Sα Cα d  0 0 0 1 Kinematic Model

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Denavit-Hartenberg Parameters Matrix 0 T6 can be expressed in terms of H matrices by adding to (1) a null translation along x6 , obtaining the frame F7 a null rotation about x7 , obtaining the frame F8 Therefore we have 0

T8

= Tras(z0 , d)Rot(z1 , θ)Tras(x2 , a)Rot(x3, α) Tras(z4 , b)Rot(z5 , φ)Tras(x6 , 0)Rot(x7 , 0)

that is expressed by cyclic transformations.


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Denavit-Hartenberg Parameters If another frame F12 is given, it is possible to move from F6 to F12 by means of a sequence similar to (1). Then, the transformation from F0 to F12 is 0

T12

=

0

H4 Tras(z4 , b)Rot(z5 , φ)Tras(z6 , d ′ )Rot(z7 , θ′ )Tras(x8 , a′ )Rot(x9 , α′ )

Tras(z10 , b ′ )Rot(z11 , φ′ )Tras(x12 , 0)Rot(x13 , 0)

Since a translation and a rotation about the same axis may commute, i.e. Rot(z5 , φ)Tras(z6 , d ′ ) = Tras(z6 , d ′ )Rot(z5 , φ) we have that Tras(z4 , b)Rot(z5 , φ)Tras(z6 , d ′ )Rot(z7 , θ′ ) = Tras(z4 , b)Tras(z6 , d ′ )Rot(z5 , φ)Rot(z7 , θ′ ) = Tras(z4 , b + d ′ )Rot(z5 , φ + θ′ )


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Denavit-Hartenberg Parameters In conclusion, the transformation between F0 and F12 is expressed by two DH transformations expressed by H matrices: the first one with parameters d, θ, a, α, the second one with parameters (b + d ′ ), (φ + θ′ ), a′ , α′ (and a third one with parameters b ′ , φ′ , 0, 0).


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Denavit-Hartenberg Parameters In general, only frames F0 and F4 are of interest, and not the intermediate ones (F1 -F3 ). Therefore, F4 will be indicated from now as F1 . The transformation 0 H4 is then indicated as 0 H1 . 0

H1

=

=

Tras(z0 , d)Rot(z1 , θ)Tras(x2 , a)Rot(x3 , α)   Cθ −Sθ Cα Sθ Sα aCθ  Sθ Cθ Cα −Cθ Sα aSθ     0 Sα Cα d  0 0 0 1

The frames associated to each link are used only for the definition of the kinematic model of the robot: usually their position/orientation may be freely assigned and do not depend by other constraints. Therefore, these frames are assigned in order to minimize the number of parameters required for the definition of the kinematic model. C. Melchiorri (DEIS)

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As a matter of fact, if F0 and F6 are two frames associated to two consecutive links, and the position and orientation of F6 are not constrained by other considerations, it is possible to choose F4 as the frame of the second link (NOT F6 ), reducing in this manner to 4 the number of parameters: b and φ are not necessary. Then, the transformation between two consecutive links is 0 H4 . C. Melchiorri (DEIS)

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Denavit-Hartenberg Parameters In conclusion: Although in general 6 parameters are necessary to specify the relative position and orientation of two frames F0 and F1 , only 4 parameters are sufficient (d, θ, a, α) by assuming that: 1 2

The axis x1 intersects z0 The axis x1 is perpendicular to z0

These parameters are known as the Denavit-Hartenberg Parameters.


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Denavit-Hartenberg Parameters Consider now a generic manipulator.


Li −1 , Li : consecutive links Ji ed Ji +1 i relative joints The motion axis of Ji defines the direction of zi −1 (frame Fi −1 ) associated to the proximal link zi (Fi ) is aligned with the motion axis of the following joint The origin of Fi is at the intersection of zi with the common normal ai between zi −1 and zi If a common normal does not exist (ai = 0), the origin of Fi is placed on zi −1 If the two axes intersect, the origin is placed at the intersection If the two axes coincide, also the origins of Fi −1 and Fi coincide xi (Fi ) is directed along the common normal yi is chosen in order to obtain a proper frame.

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Denavit-Hartenberg Parameters Conclusion: the position and the orientation of two consecutive frames, and therefore of the related links, may be defined by the four Denavit-Hartenberg parameters: ai = length of the common normal between the axes of two consecutive joints αi = ccw angle between zi −1 the axis of joint i, and zi , axis of joint i + 1 di = distance between the origin oi −1 of Fi −1 and the point pi , θi = ccw angle between the axis xi −1 and the common normal piˆoi about zi −1 .


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Denavit-Hartenberg Parameters The parameters ai and αi are constant and depend only on the link geometry: ai αi

is the link length is the link twist angle

between the joints’ axes. Considering the two other parameters, depending on the joint type one is constant and the other one may change in time: prismatic joint: rotational joint:


di is the joint variable and θi is the joint variable and

Kinematic Model

θi is constant; di is constant.

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Denavit-Hartenberg Parameters The homogeneous transformation matrix relating the frames Fi −1 and Fi is i −1

Hi = Trasl(zi −1 , di ) Rot(zi −1 , θi ) Trasl(xi , ai ) Rot(xi , αi ) 

1  0  = 0 0 

Cθi  Sθi =  0 0

0 1 0 0

0 0 1 0

  Cθi 0  Sθi 0     0 di  0 1

−Sθi Cαi Cθi Cαi Sαi 0

−Sθi Cθi 0 0

Sθi Sαi −Cθi Sαi Cαi 0

0 0 1 0

  1 0  0 0     0 0  0 1

0 1 0 0

0 0 1 0

  1 ai  0 0     0 0  0 1

0 Cαi Sαi 0

0 −Sαi Cαi 0

 0 0   0  1

 ai Cθi ai Sθi   di  1

known as Canonical Transformation. In literature, matrix C. Melchiorri (DEIS)

i −1

Hi is also indicated as Kinematic Model

i −1

Ai . 24 / 164

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Denavit-Hartenberg Parameters Each matrix i −1 Hi is a function of the i-th joint variable, di or θi depending on the joint type. For notational ease, the joint variable is generically indicated as qi , i.e.:

Therefore:

qi = di

for prismatic joints

q i = θi

for rotational joints

i −1

Hi =

i −1

Hi (qi ).

In case of a manipulator with n joints, the relationship between frame F0 and frame Fn is: 0

Tn =

0

H1 (q1 ) 1 H2 (q2 )...n−1 Hn (qn )

This equation expresses the position and orientation of the last link wrt the base frame, once the joint variables q1 , q2 , . . . , qn are known. This equation is the kinematic model of the manipulator. C. Melchiorri (DEIS)

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Reference Configuration of a Canonical Transformation A generic homogenous transformation 0 Tn may be expressed as a function of n canonical transformations 0

Tn =

n Y

i −1

Hi

i =1

If all the rotational joint variables are null, i.e. θi = 0, and all the prismatic joints variables are at the minimum value, i.e. dj = min(dj ) (with θj = 0), the so-called Reference Configuration for 0 Tn is obtained. Note that for prismatic joints the value θj may be imposed by the manipulator structure (and be not null). Also in these cases, it is arbitrarily considered null. A similar consideration holds also for rotational joints (θi = 0): The reference configuration may be non physically reachable by the manipulator. C. Melchiorri (DEIS)

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Kinematic Model In the reference configuration, the matrices i −1

i −1

Hi =

i −1

Hi =

i −1

Hi |θi =0;

Hi |θi =0

di =min(di )



=

=

i −1



1  0  0 0

1  0  0 0

0 Cαi Sαi 0

Hi are:

0 Cαi Sαi 0 0 −Sαi Cαi 0

0 −Sαi Cαi 0

 ai 0  di  1 

ai 0  min(di )  1

rotational joints

prismatic joints

The rotational part of these matrices indicates a rotation about the xi axis. Therefore, by composing all the orientation does not change).

i −1

Hi , the xi axes results only translated (their

In this configuration, all the xi axes have the same direction (they are aligned).


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Procedure for assigning frames

Kinematic Model of Robot Manipulators Direct Kinematic Model Claudio Melchiorri Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI) Universit` a di Bologna email: [email protected]


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Kinematic Model A procedure to assign frames to the links of a manipulator Need of common conventions, in order to define univocally the kinematic equations. First step: definition of the base frame F0 . In this case it is usually posible to consider not only the kinematic configuration of the manipulator but also other considerations, related e.g. to the work space. However, according to the DH convention, usually F0 is chosen so that z0 coincides with the motion axis of J1 .

F0 = ?

Fn = ?

Also Fn is assigned considering not only the robot’s kinematics, since a motion axis for the last link does not exist. =⇒ F0 and Fn are arbitrarily chosen! C. Melchiorri (DEIS)

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Kinematic Model The Denavit-Hartenberg convention does not define univocally the frames associated to the links. As a matter of fact, the frames may be assigned with some arbitrariness in the following cases: 1

F0 : only the direction of z0 may be univocally defined, while in general the origin o0 and the orientation of x0 and y0 are not assigned;

2

Fn : only xn is constrained to be perpendicular to zn−1 (i.e. to Jn ). Since the joint n + 1 does not exist, it is not possible to define the other elements;

3

parallel consecutive axes: it is not defined univocally the common normal line;

4

intersecting consecutive axes: the direction of xi is not defined;

5

prismatic joint: only zi is defined.

In these cases, it is possible to exploit the arbitrariness in order to simplify the kinematic model, for example by posing the origin of consecutive frames in the same point, or aligning axes of consecutive frames, and so on. C. Melchiorri (DEIS)

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Procedure for assigning the frames The frames are assigned to the links with the following procedure: 1. The joints and links are numbered (joints from 1 to n; links from 0 to n). Links Li −1 and Li are adjacent (proximal and distal, respectively), connected by the joint Ji (whose variable is qi ); 2. Definition of the axes zi , i = 0, . . . , n − 1, aligned with the joint motion directions (rotation/translation); (N.B. zi is the motion direction of joint Ji +1 : z0 → J1 ; z1 → J2 ; . . .) 3. Definition of F0 , with origin in any point of z0 , and axes x0 , y0 ‘properly’ chosen;


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Procedure for assigning the frames Steps 4 - 6 are repeated for i = 1, . . . , n − 1 4. Definition of Fi . Three cases are possible: a) the axes of joints Ji and Ji +1 have a common normal: the origin of Fi is placed at the intersection point of zi with the common normal between zi −1 and zi b) the axes of Ji e Ji +1 intersect: the origin of Fi is placed at the intersection point between zi −1 and zi c) the axes of joints Ji and Ji +1 are coincident or parallel: if Ji is rotational, the origin of Fi is chosen so that di = 0 if Ji is prismatic, the origin of Fi is placed at a joint limit

5. Definition of xi along the common normal between zi −1 and zi (if exists) with positive direction from Ji to Ji +1 ; if zi −1 intersects zi , then the following joints are considered; 6. Definition of yi in order to obtain a proper frame. C. Melchiorri (DEIS)

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Procedure for assigning the frames Finally: 7. Define on coincident with on−1 ; 8. Define xn perpendicular to zn−1 ; 9. If Jn is rotational, choose zn parallel to zn−1 ; If Jn is prismatic, it is possible to choose zn freely; 10. Define yn in order to obtain a proper frame. Note that: The position of on and its orientation zn are arbitrary In this manner, the frame Fn is different wrt the frame of the end-effector t T (with axes n, s, a). Therefore, it is in general necessary to define a constant homogeneous transform matrix to take into account this difference.


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Procedure for assigning the frames Once the n frames Fi (i = 1, . . . , n) are defined, the corresponding 4n DH parameters di , ai , αi , θi can be easily determined, and therefore also the matrices i −1 Hi can be computed. The kinematic model is then obtained. Then: a) Define the DH Parameters Table b) Compute the homogeneous transformation matrices

i −1

Hi ,

i = 1, . . . , n

c) Compute the direct kinematic function 0


Tn =

0

H1 1 H2 . . .

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n−1

Hn

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Examples

Example Let us consider the following 3 dof manipulator:


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Example Step 1: Assign numbers to joints and links


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Example Step 2: Choice of the zi axes (joint rotation/translation motion axes)


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Example Step 3: Choice of F0


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Example Steps 4 - m: Definition of F1 ... Fn


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Example Finally (optional): choice of the tool frame


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Examples

Example Let’s consider a 2 dof planar manipulator:

Denavit-Hartenberg parameters L1 L2

The

i −1

d 0 0

θ θ1 θ2

a a1 a2

α 0o 0o

Hi matrices result: 

C1  S1 0  H1 =  0 0 C. Melchiorri (DEIS)

−S1 C1 0 0

0 0 1 0



 a1 C1 a1 S 1   0  1

C2  S2 1  H2 =  0 0 Kinematic Model

−S2 C2 0 0

0 0 1 0

 a2 C2 a2 S 2   0  1 41 / 164


Examples

Example

Then 0

T2 =

0

H1 1 H2 =

n s 0 0

a p 0 1



C12  S12 =  0 0

−S12 C12 0 0

0 0 1 0

 a1 C1 + a2 C12 a1 S1 + a2 S12    0 1

The vectors n, s, a express the orientation of the manipulator (rotation about z0 ), while p defines the end effector position (plane x0 − y0 ). C. Melchiorri (DEIS)

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Example zi −1 axes aligned with the motion direction of Ji Note that: di = 0: distances among common normal lines ai : distances among the joint axes Ji αi : angle between zi −1 and zi about xi

With the DH convention, the origin of obtains:  C12 −S12 0 a1 C1 C12 0 a1 S 1  S 0 T2 =  012 0 1 0 0 0 0 1 C. Melchiorri (DEIS)

F2 is coincident with F1 . In this case, one   

Then

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

1  0 2 Tt =  0 0

0 1 0 0

0 0 1 0

 a2 0  0  1

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Example: 3 dof anthropomorphic manipulator

Table of the Denavit-Hartenberg parameters L1 L2 L3

Matrices 

C1  S1 0  H1= 0 0

i −1

0 0 1 0

d 0 0 0

θ θ1 θ2 θ3

a 0 a2 a3

α 90o 0o 0o

Hi S1 −C1 0 0


  0  0  ,1 H2=  0 1

C2 S2 0 0

−S2 C2 0 0

0 0 1 0

Kinematic Model

  a2 C2  a2 S 2   ,2 H3=  0  1

C3 S3 0 0

−S3 C3 0 0

0 0 1 0

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Examples

Example: 3 dof anthropomorphic manipulator Kinematic model: 

C1 C23  S1 C23 0 T3 =   S23 0

−C1 S23 −S1 S23 C23 0

S1 −C1 0 0

 C1 (a2 C2 + a3 C23 ) S1 (a2 C2 + a3 C23 )    a2 S2 + a3 S23 1

The orientation of z3 depends only on the first joint J1 ; pz does not depend on θ1 .


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Examples

Example: 3 dof anthropomorphic manipulator Check if the model is correct z0

With θ1 = θ2 = θ3 = 0o   1 0 0 a2 + a3 0  0 0 −1  0 T3 =  0 1  0 0 0 0 0 1

y3 y0 F0 x0 a2

z3

a3

With θ1 = θ2 = θ3 = 90o   0 0 0 1 −1 0 0 −a  0 3  T3 =  0 −1 0 a2  0 0 0 1 C. Melchiorri (DEIS)

a3

x3

F3 y3

F3

a2 z3 z0 y0 F0

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Examples

Example: 3 dof anthropomorphic manipulator Another choice for the last frame could be

In this case, the last frame does not respect the DH convention: =⇒ x3 does not intersect z2 !

There are two possible manners to obtain the kinematic model.


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Example: 3 dof anthropomorphic manipulator There are two possible manners to obtain the kinematic model: 1) Use the previous model and add a constant  0 0 1  1 0 0 T=  0 1 0 0 0 0

and then

0

T3,new



−C1 S23  −S 1 C23 = 0 T3 T =   C23 0

S1 −C1 0 0

rotation, in this case  0 0   0  1 C1 C23 S1 C23 S23 0

 C1 (a2 C2 + a3 C23 ) S1 (a2 C2 + a3 C23 )    a2 S2 + a3 S23 1

The unit vector s depends only on the first joint. The position pz does not depend on θ1 . C. Melchiorri (DEIS)

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Examples

Example: 3 dof anthropomorphic manipulator 2) Use the DH convention by adding suitable (fictitious)

i −1

Hi matrices.

In this case, it is necessary to add a rotation of π/2 about z and a rotation of π/2 about x and therefore the ‘DH’ angles θ = α = π/2. y2

x′3

y3

x′′3

L3 x2 z2 J3


x3

y3′

z′′3 z′3

z3

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y3′′

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Examples

Example: 3 dof anthropomorphic manipulator The new DH parameters table (the joint angle θ3 and the new angle θ are defined about the same axis and then it is possible to simply add them together):

L1 L2 L3

The

i −1



d 0 0 0

θ θ1 θ2 θ3 + 90o

a 0 a2 a3

α 90o 0o 90o

Hi matrices are

C1  S1 0  H1= 0 0

0 0 1 0

S1 −C1 0 0


  0  0  ,1 H2=  0 1

C2 S2 0 0

−S2 C2 0 0

0 0 1 0

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  a2 C2 −S3  a2 S 2   ,2 H3= C3  0 0  1 0

0 0 1 0

C3 S3 0 0

 a3 C3 a3 S 3   0  1

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Examples

Example: 3 dof anthropomorphic manipulator The kinematic model results:

0

T3,new



−C1 S23  −S1 C23 =  C23 0

S1 −C1 0 0

C1 C23 S1 C23 S23 0

 C1 (a2 C2 + a3 C23 ) S1 (a2 C2 + a3 C23 )    a2 S2 + a3 S23 1

The unit vector s depends only on the first joint. The position pz does not depend on θ1 .


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Examples

Example: 3 dof spherical manipulator

Denavit-Hartenberg parameters: L1 L2 L3

d 0 d2 d3

θ θ1 θ2 0

a 0 0 0

α −90o 90o 0o

The Denavit-Hartenberg matrices are: 

C1  S1 0  H1 =  0 0

0 0 −1 0


−S1 C1 0 0

  0  0   , 1 H2 =   0  1

C2 S2 0 0

0 0 1 0

Kinematic Model

S2 −C2 0 0

  0  0   , 2 H3 =   d2  1

1 0 0 0

0 1 0 0

0 0 1 0

 0 0   d3  1 52 / 164


Examples

Example: 3 dof spherical manipulator The kinematic model results:

0

T3 =

n s 0 0

a p 0 1



C1 C2  C2 S1 =  −S2 0

−S1 C1 0 0

C1 S2 S1 S2 C2 0

 −d2 S1 + d3 C1 S2 d2 C1 + d3 S1 S2    d3 C2 1

The third joint J3 does not affect the orientation, s depends only on J1 .


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Examples

Example: 3 dof spherical manipulator z3 If θ1 = θ2 = 0o , d3  1 0  0 1 0 T3 =  0 0 0 0

F3 =d 0 0 1 0

If θ1 = θ2 = 90o , d3  0 −1 0  0 0 T3 =  −1 0 0 0


 0 d2  d  1

d y0 d2

F0 x0

y3

d

=d 0 1 0 0

y3

x3

z0

 −d2 d  0  1

z0

F3 z3

−d2 x3 y0 F0

x0 Kinematic Model

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Examples

Example: 3 dof spherical wrist Denavit-Hartenberg parameters L4 L5 L6

d 0 0 d6

θ θ4 θ5 θ6

a 0 0 0

α −90o 90o 0o

Note the numbers starting from 4...

Then 

C4  S4 3  H4 =  0 0

0 0 −1 0


−S4 C4 0 0

  0  0  ,4 H5 =   0 1

C5 S5 0 0

0 0 1 0

S5 −C5 0 0

Kinematic Model

  0  0  ,5 H6 =   0 1

C6 S6 0 0

−S6 C6 0 0

0 0 1 0

 0 0   d6  1

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Examples

Example: 3 dof spherical wrist The kinematic model is:  C4 C5 C6 − S4 S6  S4 C5 C6 + C4 S6 3 T6 =   −S5 C6 0

−S4 C6 − C4 C5 S6 C4 C6 − S4 C5 S6 S5 S6 0

C4 S5 S4 S5 C5 0

 d6 C4 S5 d6 S4 S5   d6 C5  1

In this case, the rotation matrix has the same expression as an Euler ZYZ rotation matrix. REuler (φ, θ, ψ)

= Rot(z0 , φ)Rot(y1 , θ)Rot(z2 , ψ)  Cφ Cθ Cψ − Sφ Sψ −Cφ Cθ Sψ − Sφ Cψ =  Sφ Cθ Cψ + Cφ Sψ −Sφ Cθ Sψ + Cφ Cψ −Sθ Cψ Sθ Sψ

 Cφ Sθ Sφ Sθ  Cθ

It means that the manipulator’s joints θ4 , θ5 and θ6 are equivalent to the Euler ZYZ angles. C. Melchiorri (DEIS)

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Examples

Example: 3 dof spherical wrist If θ1 = θ2 = θ3 = 0o  1  0 3 T6 =   0 0 If θ1 = θ2 = θ3 = 90o  −1  0 3 T6 =   0 0


0 1 0 0

0 0 1 0

0 0 1 0

0 1 0 0

 0 0   d6  1

 0 d6   0  1

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Examples

Example: Stanford manipulator

By composing the 3 dof spherical manipulator with the spherical wrist, the so-called “Stanford manipulator” is obtained, a 6 dof robot. Since the frames have been defined in a consistent manner, the kinematic model is simply obtained by multiplying the matrices 0 T3 of the arm and 3 T6 of the wrist.

Then 0


T6 =

0

T3 3 T6 =

n 0

Kinematic Model

s a 0 0

p 1

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Examples

Example: Stanford manipulator where  −S1 (S4 C5 C6 + C4 S6 ) + C1 (C2 (C4 C5 C6 − S4 S6 ) − S2 S5 C6 ) n =  C1 (S4 C5 C6 + C4 S6 ) + S1 (−S2 S5 C6 + C2 (C4 C5 C6 − S4 S6 ))  −C2 S5 C6 − S2 (C4 C5 C6 − S4 S6 )   −S1 (C4 C6 − S4 C5 S6 ) + C1 (−C2 (S4 C6 + C4 C5 S6 ) + S2 S5 S6 ) o =  C1 (C4 C6 − S4 C5 S6 ) + S1 (S2 S5 S6 − C2 (S4 C6 + C4 C5 S6 ))  C2 S5 S6 + S2 (S4 C6 + C4 C5 S6 )   −S1 S4 S5 + C1 (S2 C5 + C2 C4 S5 ) a =  C1 S4 S5 + S1 (S2 C5 + C2 C4 S5 )  C2 C5 − S2 C4 S5   −d2 S1 + d3 C1 S2 + d6 (C1 S2 C5 + C1 C2 C4 S5 − S1 S4 S5 ) p =  d2 C1 + d3 S1 S2 + d6 (S1 S2 C5 + S1 C2 C4 S5 + C1 S4 S5 )  d3 C2 + d6 (C2 C5 − S2 C4 S5 ) 


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Examples

Example: PUMA 260 Joint variables θi are defined about the zi −1 axes; a2 is the distance between z1 and z2 (in this case parallel), d3 is the offset between the origins of F2 and F3 , and d4 is the offset between the origins of F3 and F4 . Frames F4 and F5 coincides. The αi angles are either 0o or ±90o .

L1 L2 L3 L4 L5 L6


Kinematic Model

d 0 0 −d3 d4 0 d6

θ θ1 θ2 θ3 θ4 θ5 θ6

a 0 a2 0 0 0 0

α 90o 0o 90o −90o 90o 0o

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Examples

Example: PUMA 260 Canonical transformation matrices:



0 0 1 0



0 0 −1 0

C1  S1 0  H1= 0 0 C4  S4 3  H4= 0 0

S1 −C1 0 0 −S4 C4 0 0


  0  0  ,1 H2=  0 1

C2 S2 0 0

  0  0   ,4 H5=  d4  1

−S2 C2 0 0 C5 S5 0 0

0 0 1 0

0 0 1 0

  a2 C2  a2 S 2   ,2 H3=  0  1

C3 S3 0 0

0 0 1 0

  0  0  ,5 H6=  0 1

C6 S6 0 0

−S6 C6 0 0

S5 −C5 0 0

Kinematic Model

S3 −C3 0 0 0 0 1 0

 0 0   −d3  1  0 0   d6  1

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Examples

Example: PUMA 260 

C1 C2 C3 − C1 S2 S3  C2 C3 S1 − S1 S2 S3 0  T3 =  C3 S2 + C2 S3 0 

C4 C5 C6 − S4 S6  C5 C6 S4 + C4 S6 3 T6 =   −(C6 S5 ) 0 0


T6 = 0 T6 =

S1 −C1 0 0

C1 C3 S2 + C1 C2 S3 C3 S1 S2 + C2 S1 S3 −(C2 C3 ) + S2 S3 0

−(C6 S4 ) − C4 C5 S6 C4 C6 − C5 S4 S6 S5 S6 0

0

T3 3 T6 =

Kinematic Model

n s 0 0

C4 S5 S4 S5 C5 0 a p 0 1

 a2 C1 C2 − d3 S1 C1 d3 + a2 C2 S1    a2 S2 1  C4 d6 S5 d6 S4 S5   d4 + C5 d6  1 62 / 164


Examples

Example: PUMA 260 n=

"

S1 (C5 C6 S4 + C4 S6 ) + C1 (C2 (−(C6 S3 S5 ) + C3 (C4 C5 C6 − S4 S6 )) − S2 (C3 C6 S5 + S3 (C4 C5 C6 − S4 S6 ))) −(C1 (C5 C6 S4 + C4 S6 )) + S1 (C2 (−(C6 S3 S5 ) + C3 (C4 C5 C6 − S4 S6 )) − S2 (C3 C6 S5 + S3 (C4 C5 C6 − S4 S6 ))) S2 (−(C6 S3 S5 ) + C3 (C4 C5 C6 − S4 S6 )) + C2 (C3 C6 S5 + S3 (C4 C5 C6 − S4 S6 ))

#

" # S1 (C4 C6 − C5 S4 S6 )+C1 (C2 (S3 S5 S6 + C3 (−(C6 S4 ) − C4 C5 S6 )) − S2 (−(C3 S5 S6 ) + S3 (−(C6 S4 ) − C4 C5 S6 ))) s = −(C1 (C4 C6 − C5 S4 S6 ))+S1 (C2 (S3 S5 S6 + C3 (−(C6 S4 ) − C4 C5 S6 ))−S2 (−(C3 S5 S6 )+S3 (−(C6 S4 ) − C4 C5 S6 ))) S2 (S3 S5 S6 + C3 (−(C6 S4 ) − C4 C5 S6 )) + C2 (−(C3 S5 S6 )+S3 (−(C6 S4 ) − C4 C5 S6 ))

a=

p=

"

"

S1 S4 S5 + C1 (C2 (C5 S3 + C3 C4 S5 ) − S2 (−(C3 C5 ) + C4 S3 S5 )) −(C1 S4 S5 ) + S1 (C2 (C5 S3 + C3 C4 S5 ) − S2 (−(C3 C5 ) + C4 S3 S5 )) S2 (C5 S3 + C3 C4 S5 ) + C2 (−(C3 C5 ) + C4 S3 S5 )

#

S1 (−d3 + d6 S4 S5 ) + C1 (a2 C2 + C2 ((d4 + C5 d6 )S3 + C3 C4 d6 S5 ) − S2 (−(C3 (d4 + C5 d6 )) + C4 d6 S3 S5 )) −(C1 (−d3 + d6 S4 S5 )) + S1 (a2 C2 + C2 ((d4 + C5 d6 )S3 + C3 C4 d6 S5 ) − S2 (−(C3 (d4 + C5 d6 )) + C4 d6 S3 S5 )) a2 S2 + S2 ((d4 + C5 d6 )S3 + C3 C4 d6 S5 ) + C2 (−(C3 (d4 + C5 d6 )) + C4 d6 S3 S5 )


Kinematic Model

#

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Examples

Example: planar 4 dof manipulator (redundant) DH parameters

L1 L2 L3 L4

d 0 0 0 0

θ θ1 θ2 θ3 θ4

a a1 a2 a3 a4

α 0o 0o 0o 0o

All the i −1 Hi matrices have the same structure   Ci −Si 0 ai Ci  Si Ci 0 ai Si  i −1  Hi =   0 0 1 0  0 0 0 1


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Examples

Example: planar 4 dof manipulator (redundant) Then

0

T4

=

0



H1 1 H2 2 H3 3 H4 =

C1234  S1234 =   0 0

−S1234 C1234 0 0

0 0 1 0

n s 0 0

a 0

p 1

 a1 C1 + a2 C12 + a3 C123 + a4 C1234 a1 S1 + a2 S12 + a3 S123 + a4 S1234    0 1

The vectors n, s, a define the end-effector orientation (rotation about z), while p defines its position (on the x − y plane, pz = 0). =⇒ The procedure can be applied also to redundant manipulators.


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Kinematic Model of Robot Manipulators Inverse Kinematic Model Claudio Melchiorri Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI) Universit` a di Bologna email: [email protected]


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Introduction

Inverse Kinematic Model Direct Kinematic Model: The direct kinematic model consists in a function f(q) mapping the joint position variables q ∈ IRn to the position/orientation of the end effector. The definition of f(q) is conceptually simple, and a general approach for its computation has been defined.


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Introduction

Inverse Kinematic Model Inverse Kinematic Model: The inverse kinematics consists in finding a function g(x) mapping the position/orientation of the end-effector to the corresponding joint variables q: the problem is not simple! A general approach for the solution of this problem does not exist On the other hand, for the most common kinematic structures, a scheme for obtaining the solution has been found. Unfortunately The solution is not unique. In general we have: No solution (e.g. starting with a position x not in the workspace); A finite set of solutions (one or more); Infinite solutions.

We seek for closed form solutions, and not based on numerical techniques: The analytic solution is more efficient from the computational point of view; If the solutions are known analytically, it is possible to select one of them on the basis of proper criteria. C. Melchiorri (DEIS)

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Introduction

Inverse Kinematic Model In order to obtain a closed form solution to the inverse kinematic problem, two approaches are possible: An algebraic approach, i.e. elaborations of the kinematic equations until a suitable set of (simple) equations is obtained for the solution A geometric approach based, when possible, on geometrical considerations, dependent on the kinematic structure of the manipulator and that may help in the solution.


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Algebraic Approach

Algebraic Approach For a 6 dof manipulator, the kinematic model is described by the equation 0

T6 =

0

H1 (q1 ) 1 H2 (q2 ) . . . 5 H6 (q6 )

equivalent to 12 equations in the 6 unknowns qi , i = 1, . . . , 6. Example: spherical manipulator (only 3 dof) 

0.5868  0.5265 T= −0.5736 0

−0.6428 0.7660 0.0000 0

0.4394 0.3687 0.8192 0

  C1 C2 −0.4231 0.9504   C2 S1 =   −S2 0.4096 0 1

−S1 C1 0 0

C1 S2 S1 S2 C2 0

 −d2 S1 + d3 C1 S2 d2 C1 + d3 S1 S2   d3 C2 1

Since both the numerical values of 0 T6 and the structure of the i −1 Hi matrices are known, by suitable pre- / post-multiplications it is possible to obtain [ 0 H1 (q1 ) . . .i −1 Hi (qi )]−1 0 T6 [ j Hj+1 (qj+1 ) . . .5 H6 (q6 )]−1 = i Hi +1 (qi +1 ) . . .j−1 Hj (qj ), i < j

obtaining 12 new equations for each couple (i, j), i < j. By selecting the most simple equations among all those obtained, it might be possible to obtain a solution to the problem. C. Melchiorri (DEIS)

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Geometric Approach

Geometric Approach General considerations that may help in finding solutions with algebraic techniques do not exist, being these strictly related to the mathematical expression of the direct kinematic model. On the other hand, it is often possible to exploit considerations related to the geometrical structure of the manipulator.

PIEPER APPROACH Many industrial manipulators have a kinematically decoupled structure, for which it is possible to decompose the problem in two (simpler) sub-problems: 1

Inverse kinematics for the position

(x, y , z) → q1 , q2 , q3

2

Inverse kinematics for the orientation

R → q4 , q5 , q6 .


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Geometric Approach

Geometric Approach PIEPER APPROACH: Given a 6 dof manipulator, sufficient condition to find a closed form solution for the IK problem is that the kinematic structure presents: three consecutive rotational joints with axes intersecting in a single point or three consecutive rotational joints with parallel axes.


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Geometric Approach

Geometric Approach In many 6 dof industrial manipulators, the first 3 dof are usually devoted to position the wrist, that has 3 additional dof give the correct orientation to the end-effector. In these cases, it is quite simple to decompose the IK problem in the two sub-problems (position and orientation).


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Geometric Approach

Geometric Approach In case of a manipulator with a spherical wrist, a natural choice is to decompose the problem in 1

IK for the point pp (center of the spherical wrist)

2

solution of the orientation IK problem

Therefore: 1 The point pp is computed since 0 T6 is known (submatrices R and p): p p = p − d6 a pp depends only on the joint variables q1 , q2 , q3 ; 2

The IK problem is solved for q1 , q2 , q3 ;

3

The rotation matrix 0 R3 related to the first three joints is computed;

4

The matrix

5

The IK problem for the rotational part is solved (Euler). C. Melchiorri (DEIS)

3

R6 =

0

RT 3 R

is computed;

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Examples

Solution of the spherical manipulator Direct kinematic model: n s a p 0 T3 = 0 0 0 1  C1 C2 −S1 C1 S2  C2 S1 C1 S1 S2 =   −S2 0 C2 0 0 0

 −d2 S1 + d3 C1 S2 d2 C1 + d3 S1 S2    d3 C2 1

If the whole matrix 0 T3 is known, it is possible to compute:   θ1 = atan2 (−sx , sy ) Unfortunately, according to =⇒ the Pieper approach only p θ2 = atan2 (−nz , az )  d3 = pz / cos θ2 is known! C. Melchiorri (DEIS)

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Examples

Solution of the spherical manipulator

We known only the position vector p.

From

(0 H1 )−1 0 T3



C1  0 =   −S1 0

=


1

S1 0 C1 0

H2 2 H3

0 −1 0 0

0

T 3 = 0 H1 1 H2 2 H3

 nx 0  ny 0  0  nz 1 0

sx sy sz 0

Kinematic Model

ax ay az 0

we have

  px C2  S2 py  = pz   0 1 0

0 0 1 0

S2 −C2 0 0

 d3 S2 −d3 C2   d2  1

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Examples

Solution of the spherical manipulator By equating the position vectors,     px C1 + py S1 d3 S2 1  =  −d3 C2  −pz pp =  −px S1 + py C1 d2

The vector 1 pp depends only on θ2 and d3 ! Let’s define a = tan θ1 /2, then C1 =

1 − a2 1 + a2

S1 =

2a 1 + a2

By substitution in the last element of 1 pp 2

(d2 + py )a + 2px a + d2 − py = 0

=⇒

a=

−px ±

p

px2 + py2 − d22 d2 + py

Two possible solutions! ((px2 + py2 − d22 ) > 0??) Then θ1 = 2 atan2 (−px ± C. Melchiorri (DEIS)

q px2 + py2 − d22 , d2 + py )

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Examples

Solution of the spherical manipulator Vector 1 pp is defined as    d3 S2 px C1 + py S1 1  =  −d3 C2  −pz pp =  d2 −px S1 + py C1 

From the first two elements

d3 S2 px C1 + py S1 = −pz −d3 C2 from which θ2 = atan2 (px C1 + py S1 , pz ) Finally, if the first two elements are squared and added together q d3 = (px C1 + py S1 )2 + pz2 C. Melchiorri (DEIS)

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Examples

Solution of the spherical manipulator Note that two possible solutions exist considering the position of the end-effector (wrist) only. If also the orientation is considered, the solution (if exists) is unique. We have seen that the relation (px2 + py2 − d22 ) > 0 must hold:


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Examples

Solution of the spherical manipulator Numerical example: Given a spherical manipulator with d2 = 0.8 m in the pose θ1 = 20o , we have



0.8138  0.2962 0  T3 =  −0.5 0

θ2 = 30o , −0.342 0.9397 0 0

d3 = 0.5 m

0.4698 0.171 0.866 0

 −0.0387 0.8373   0.433  1

• The solution based on the whole matrix 0 T3 is: θ1 = 20o , θ2 = 30o , d3 = 0.5. • The solution based on the vector p gives: a) θ1 = 20o , θ2 = 30o , d3 = 0.5 b) θ1 = −14.7o , θ2 = −30o , d3 = 0.5


(with the “+” sign). (with the “-” sign).

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Examples

Solution of the spherical manipulator • The solution based on the vector p gives: a) θ1 = 20o , θ2 = 30o , d3 = 0.5 (with the “+” sign). b) θ1 = −14.7o , θ2 = −30o , d3 = 0.5


(with the “-” sign).

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Examples

Solution of the 3 dof anthropomorphic manipulator From the kinematic structure, one obtains θ1 = atan2 (py , px ) Concerning θ2 and θ3 , note that the arm moves in a plane defined by θ1 only. One obtains C3 =

px2 + py2 + pz2 − a22 − a32 2a2 a3

S3 = ±

q

1 − C32

θ3 = atan2 (S3 , C3 )

Moreover, by geometrical arguments, it is possible to state that: θ2 = atan2 (pz , C. Melchiorri (DEIS)

q

px2 + py2 ) − atan2 (a3 S3 , a2 + a3 C3 ) Kinematic Model

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Examples

Solution of the 3 dof anthropomorphic manipulator Note that also the solution is valid θ2′ = π − θ2

θ1 = π + atan2 (py , px ), Then, FOUR possible solutions exist: shoulder right - elbow up; shoulder left - elbow up;

shoulder right - elbow down; shoulder left - elbow down;

Same position, but different orientation!

Note that the conditions px 6= 0, py 6= 0 C. Melchiorri (DEIS)

must hold (singular configuration).

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Examples

Solution of the spherical wrist Let us consider the matrix   nx s x ax 3 R6 =  ny sy ay  nz s z az

From the direct kinematic equations one obtains 3



C4 C5 C6 − S4 S6 R6 =  S4 C5 C6 + C4 S6 −S5 C6


−S4 C6 − C4 C5 S6 C4 C6 − S4 C5 S6 S5 S6

Kinematic Model

 C4 S5 S4 S5  C5

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Examples

Solution of the spherical wrist 3

R6 =

"

C4 C5 C6 − S4 S6 S4 C5 C6 + C4 S6 −S5 C6

−S4 C6 − C4 C5 S6 C4 C6 − S4 C5 S6 S5 S6

C4 S5 S4 S5 C5

#

The solution is then computed as (ZYZ Euler angles): θ5 ∈ [0, π]: θ4

=

θ5

=

θ6

=

atan2 (ay , ax ) q atan2 ( ax2 + ay2 , az ) atan2 (sz , −nz )

θ5 ∈ [−π, 0]:


θ4

=

θ5

=

θ6

=

atan2 (−ay , −ax ) q atan2 (− ax2 + ay2 , az ) atan2 (−sz , nz ) Kinematic Model

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Examples

Solution of the spherical wrist Numerical example: Let us consider a spherical wrist in the pose θ4 = 10o Then 3

R6 =

"

θ5 = 20o ,

0.7146 0.6337 −0.2962

θ6 = 30o

−0.6131 0.7713 0.1710

0.3368 0.0594 0.9397

#

Therefore, if • θ5 ∈ [0, π] • θ5 ∈ [−π, 0]

θ4 = 10o θ4 = −170o

θ5 = 20o ,

θ6 = 30o

θ5 = −20o ,

θ6 = −30o

Note that the PUMA has an anthropomorphic structure (4 solutions) and a spherical wrist (2 solutions): =⇒ 8 different configurations are possible! C. Melchiorri (DEIS)

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Kinematic Model of Robot Manipulators Differential Kinematics Claudio Melchiorri Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI) Universit` a di Bologna email: [email protected]


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Introduction

Differential Kinematics: the Jacobian matrix In robotics it is of interest to define, besides the mapping between the joint and workspace position/orientation (i.e. the kinematic equations), also The relationship between the joints and end-effector velocities: v ⇐⇒ q˙ ω The relationship between the force applied on the environment by the manipulator and the corresponding joint torques f ⇐⇒ τ n These two relationships are based on a linear operator, a matrix J, called the Jacobian of the manipulator. C. Melchiorri (DEIS)

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Introduction

The Jacobian matrix In robotics, the Jacobian is used for several purposes: To define the relationship between joint and workspace velocities To define the relationship between forces/torques between the spaces To study the singular configurations To define numerical procedures for the solution of the IK problem To study the manipulability properties.


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The Jacobian Matrix

Velocity domain The translational and rotational velocities are considered separately. Let us consider two frames F0 (base frame) and F1 (integral with the rigid body).

The translational velocity of point p of the rigid body, with respect to F0 , is ˙ defined as the derivative (with respect to time) of p, denoted as p: p˙ = C. Melchiorri (DEIS)

d p dt

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The Jacobian Matrix

Velocity domain With respect to the rotational velocity, two different definitions are possible: 1

2

A triple γ ∈ IR3 giving the orientation of F1 with respect to F0 (Euler, RPY,... angles) is adopted, and its derivative is used to define the rotational velocity γ˙ dγ γ˙ = dt An angular velocity vector ω is defined, giving the rotational velocity of a third frame F2 with origin coincident with F0 and axes parallel to F1 .

The velocity vector ω is placed in the origin, and its direction coincides with the instantaneous rotation axis of the rigid body.


Kinematic Model

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The Jacobian Matrix

Analytical and Geometrical expressions of the Jacobian The two descriptions lead to different results concerning the expression of the Jacobian matrix, in particular in the part relative to the rotational velocity. One obtains respectively the: Analytic Jacobian JA Geometric Jacobian JG Two problems: 1) Integration of the rotational velocity Z γdt ˙ =⇒ γ: orientation of the rigid body Z ωdt =⇒ ??


Kinematic Model

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The Jacobian Matrix

Analytical and Geometrical expressions of the Jacobian Example: Let’s consider a rigid body and the following rotational velocities Case a) ω

= [π/2, 0, 0]T

0≤t≤1

ω

= [0, π/2, 0]T

1

Kinematic Model of Robot Manipulators

Kinematic Model of Robot Manipulators

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