Robot Modeling and the Forward Kinematic Solution

ME 3230

R. R. Lindeke

Looking Closely at the T0n Matrix

At the end of our discussion of “Robot Mapping” we found that the T0

n matrix related the end of the arm frame (n) to its base (0) – Thus it must contain information related to

the several joints of the robot It is a 4x4 matrix populated by complex

equations for both position and orientation (POSE)

Looking Closely at the T0n Matrix

To get at the equation set, we will choose to add a series of coordinate frames to each of the joint positions

The frame attached to the 1st joint is labeled 0 – the base frame! – while joint two has Frame 1, etc.

The last Frame is the end or n Frame

Looking Closely at the T0n Matrix

As we have seen earlier, we can define a HTM (T(i-1)

i) to the transformation between any two SO3 based frames

Thus we will find that the T0n is

given by a product formed by:1 2

0 0 1 1n n

nT T T T g gL g

Looking Closely at the T0n Matrix

For simplicity, we will redefine each of the of these transforms (T(i-1)

i) as Ai

Then, for a typical 3 DOF robot Arm, T0

n = A1*A2*A3

While for a full functioned 6 DOF robot (arm and wrist) would be: T0

n = A1*A2*A3*A4*A5*A6

A1 to A3 ‘explain’ the arm joint effect while A4 to A6 explain the wrist joint effects

Looking At The Frame To Frame Arrangements – Building A Modeling Basis

When we move from one frame to another, we will encounter: Two translations (in a controlled sense) Two rotations (also in a controlled sense)

A rotation and translation WRT the Zi-1

These are called the Joint Parameters

A rotation and translation WRT the Xi

These are called the Link Parameters

A model of the Joint Parameters

NOTE!!!

A model of the Link Parameters

ai or

Talking Specifics – Joint Parameters

i is an angle measured about the Zi-1 axis from Xi-1 to Xi and is a variable for a revolute joint – its fixed for a Prismatic Joint

di is a distance measured from the origin of Frame(i-1) to the intersection of Zi-1 and Xi and is a variable for a prismatic joint – its fixed for a Revolute Joint

Talking Specifics – Link Parameters

ai (or li) is the Link length and measures the distance from the intersection of Zi-1 to the origin of Framei measured along Xi

i is the Twist angle which measures the angle from Zi-1 to Zi as measured about Xi

Both of these parameters are fixed in value regardless of the joint type. A Further note: Fixed does not mean zero

degrees or zero length just that they don’t change

Very Important to note:

Two Design Axioms prevail in this modeling approach Axiom DH1: The Axis Xi must be designed to be

perpendicular to Zi-1

Axiom DH2: The Axis Xi must be designed to intersect Zi-1

Thus, within reason we can design the structure of the coordinate frames to simplify the math (they are under our control!)

Returning to the 4 ‘Frame-Pair’ Operators:

1st is which is an operation of pure rotation about Z or:

2nd is d which is a translation along Z or:

os 0 0

0 0

0 0 1 0

0 0 0 1

C Sin

Sin Cos

1 0 0 0

0 1 0 0

0 0 1

0 0 0 1

d

Returning to the 4 Frame Operators:

3rd is a Translation Along X or:

4th is which is a pure Rotation about X or:

1 0 0

0 1 0 0

0 0 1 0

0 0 0 1

a

1 0 0 0

0 0

0 0

0 0 0 1

Cos Sin

Sin Cos

The Overall Effect is:

os 0 0 1 0 0 0 1 0 0 1 0 0 0

0 0 0 1 0 0 0 1 0 0 0 0

0 0 1 0 0 0 1 0 0 1 0 0 0

0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

C Sin a

Sin Cos Cos Sin

d Sin Cos

Simplifying this Matrix Product:

0

0 0 0 1

i i i i i i i

i i i i i i i

i i i

C S C S S a C

S C C C S a S

S C d

This matrix is the general transformation relating each and every of the frame pairs

along a robot structure

So, Since We Can Control the Building of this Set Of Frames, What Are The Rules?

We will employ a method called the Denavit-Hartenberg Method

It is a Step-by-Step approach for modeling each of the frames from the initial (or 0) frame all the way to the n (or end) frame

This modeling technique makes each joint axis (either rotation or extension) the Z-axis of the appropriate frame (Z0 to Zn-1).

The Joint motion, thus, is taken W.R.T. the Zi-1 axis of the frame pair making up the specific transformation matrix under design

The D-H Modeling Rules:1) Locate & Label the Joint Axes: Z0 to Zn-1

2) Establish the Base Frame. Set Base Origin anywhere on the Z0 axis. Choose X0 and Y0 conveniently and to form a right hand frame.

3) Locate the origin Oi where the common normal to Zi-1 and Zi intersects Zi. If Zi intersects Zi-1 locate Oi at this intersection. If Zi-1 and Zi are parallel, locate Oi at Joint i+1

The D-H Modeling Rules:

4) Establish Xi along the common normal between Zi-1 and Zi through Oi, or in the direction Normal to the plane Zi-1 – Zi if these axes intersect

5) Establish Yi to form a right hand system Set i = i+1, if i<n loop back to step 3

(Repeat Steps 3 to 5 for I = 1 to I = n-1)

6) Establish the End-Effector (n) frame: OnXnYnZn. Assuming the n-th joint is revolute, set kn = a along the direction Zn-1. Establish the origin On conveniently along Zn, typically mounting point of gripper or tool. Set jn = o in the direction of gripper closure (opening) and set in = n such that n = oxa. Note if tool is not a simple gripper, set Xn and Yn conveniently to form a right hand frame.

The D-H Modeling Rules:

7) Create a table of “Link” parameters: 1) i as angle about Zi-1 between X’s

2) di as distance along Zi-1

3) i as angle about Xi between Z’s

4) ai as distance along Xi

8) Form HTM matrices A1, A2, … An by substituting , d, and a into the general model

9) Form T0n = A1*A2*…*An

Some Issues to remember:

If you have parallel Z axes, the X axis of the second frame runs perpendicularly between them

When working on a revolute joint, the model will be simpler if the two X directions are in alignment at “Kinematic Home” – ie. Our model’s starting point

To achieve this kinematic home, rotate the model about moveable axes (Zi-1’s) to align X’s

Kinematic Home is not particularly critical for prismatic joints

An ideal model will have n+1 frames However, additional frames may be necessary –

these are considered ‘Dummy’ frames since they won’t contain joint axes

Applying D-H to a General Case:

General Case: Considering Link i

Connects Frames: i-1 and I and includes Joint i

This information allowed us to ‘Build’ The L.P. (link parameter) Table as seen here

Frames Link Var d a S C S C

i -1 to i i R + 37 17.5 u 47.8 u 17.8 0.306 .952 S( + 37) C( + 37)

Leads to an Ai Matrix:

( 37 ) .952 ( 37 ) .306 ( 37 ) 47.8* ( 37 )

( 37 ) .952 ( 37 ) .306 ( 37 ) 47.8 ( 37 )

0 0.306 .952 17.5

0 0 0 1

C S S C

S C C S

Frame Skeleton for Prismatic Hand Robot

LP Table:

Frames Link Var d a S C S C

0 1 1 R 10 0 -90 -1 0 S1 C1

1 2 2 R 2-1 6 0 0 1 S2 C2

2 3 3 R 30.5 0 90 1 0 S3 C3

3 4 4 P 0 d4 + 4.25 0 0 0 1 S4 C4

4 n 5 R 51 0 0 0 1 S5 C5

Depends on Location of n(end)-frame!

Leading to 5 Ai Matrices

1

1 0 1 0

1 0 1 0

0 1 0 0

0 0 0 1

C S

S CA

3

3 0 3 0

3 0 3 0

0 1 0 .5

0 0 0 1

C S

S CA

2

2 2 0 6 2

2 2 0 6 2

0 0 1 1

0 0 0 1

C S C

S C SA

44

1 0 0 0

0 1 0 0

0 0 1 4.25

0 0 0 1

Ad

#5 is:

5

5 5 0 0

5 5 0 0

0 0 1 1

0 0 0 1

C S

S CA

Now, Lets Form the FKS: T0

n = A1*A2*A3*A4*A5

I’ll use a software: Mathematica

This value is called the Hand Span and depends on the Frame Skeleton we developed

Solving for FKS

Here we have a special case – two of the Joints are a “planer arm” revolute model – i.e. parallel, consecutive revolute joints

These are contained in the A2 and A3 Matrices These should be pre-multiplied using a

trigonometric tool that recognizes the sum of angle cases ((Full)Simplify in mathematica)

Basically then: T0n = A1*{A2A3}*A4*A5

Finalizing the FKS – perform a physical verification

Physical verification means to plug known angles into the variables and compute the Ai’s and FKS against the Frame Skeleton

Another? 6dof Articulating Arm – (The Figure Contains Frame Skelton)

LP Table

Frames

Link

Var d a S C S C

0 1 1 R 1 0 0 90 -1 0 S1 C1

1 2 2 R 2 0 a2 0 0 1 S2 C2

2 3 3 R 3 0 a3 0 0 1 S3 C3

3 4 4 R 4 0 a4 -90 -1 0 S4 C4

4 5 5 R 5 0 0 90 1 0 S5 C5

5 6 6 R 6* d6 0 0 0 1 S6 C6

* With End Frame in Better Kinematic Home. Here, as shown, it would be (6 - 90), which is a problem!

A Matrices

1

1 0 1 0

1 0 1 0

0 1 0 0

0 0 0 1

C S

S CA

2

22

2 2 0 2

2 2 0 2

0 0 1 0

0 0 0 1

C S a C

S C a SA

3

33

3 3 0 3

3 3 0 3

0 0 1 0

0 0 0 1

C S a C

S C a SA

4

44

4 0 4 4

4 0 4 4

0 1 0 0

0 0 0 1

C S a C

S C a SA

A Matrices, cont.

5

5 0 5 0

5 0 5 0

0 1 0 0

0 0 0 1

C S

S CA

66

6 6 0 0

6 6 0 0

0 0 1

0 0 0 1

C S

S CA

d

At Better Kinematic Home!

Leads To:

FKS of:

0 1 2 3 4 5 6nT A A A A A A

PreProcessg g g g g

After Preprocessing A2-A3-A4, with (Full)Simplify function, the FKS must be

reordered as A1-A234-A5-A6 and Simplified

Solving for FKS

Pre-process {A2*A3*A4} with Full Simplify

They are the “planer arm” issue as in the previous robot model

Then Form: A1* {A2*A3*A4}*A5*A6

Simplify for FKS!

Simplifies to(in the expected Tabular Form):

nx = C1·(C5·C6·C234 - S6·S234) - S1·S5·C6ny = C1·S5·C6 + S1·(C5·C6·C234 - S6·S234)nz = S6·C234 + C5·C6·S234

ox = S1·S5·S6 - C1·(C5·S6·C234 + C6·S234)oy = - C1·S5·S6 - S1·(C5·S6·C234 + C6·S234)oz = C6·C234 - C5·S6·S234

ax = C1·S5·C234 + S1·C5ay = S1·S5·C234 - C1·C5az = S5·S234

dx = C1·(C234·(d6·S5 + a4) + a3·C23 + a2·C2) + d6·S1·C5dy = S1·(C234·(d6·S5 + a4) + a3·C23 + a2·C2) - d6·C1·C5dz = S234·(d6·S5 + a4) + a3·S23 + a2·S2

Verify the Model

Again, substitute known’s (typically 0 or 0 units) for the variable kinematic variables

Check each individual A matrix against your robot frame skeleton sketch for physical agreement

Check the simplified FKS against the Frame skeleton for appropriateness

After Substitution: nx = C1·(C5·C6·C234 - S6·S234) - S1·S5·C6 = 1(1-0) – 0 = 1 ny = C1·S5·C6 + S1·(C5·C6·C234 - S6·S234) = 0+ 0(1 – 0) = 0 nz = S6·C234 + C5·C6·S234 = 0 + 0 = 0

ox = S1·S5·S6 - C1·(C5·S6·C234 + C6·S234) = 0 – 1(0 + 0) = 0 oy = - C1·S5·S6 - S1·(C5·S6·C234 + C6·S234) = -0 – 0(0 + 0) = 0 oz = C6·C234 - C5·S6·S234 = 1 – 0 = 1

ax = C1·S5·C234 + S1·C5 = 0 + 0 = 0 ay = S1·S5·C234 - C1·C5 = 0 – 1 = -1 az = S5·S234 = 0

dx = C1·(C234·(d6·S5 + a4) + a3·C23 + a2·C2) + d6·S1·C5= 1*(1(0 + a4) + a3 + a2) + 0 = a4 + a3 + a2

dy = S1·(C234·(d6·S5 + a4) + a3·C23 + a2·C2) - d6·C1·C5= 0(1(0 + a4) + a3 + a2) – d6 = -d6

dz = S234·(d6·S5 + a4) + a3·S23 + a2·S2= 0(0 + a4) + 0 + 0 = 0