The L-E (Torque) Dynamical Model: Inertial Forces Coriolis & Centrifugal Forces Gravitational Forces Frictional Forces

Lets Apply the Technique -- Lets do it for a 2- Link “Manipulator” Link 1 has a Mass of m1; Link 2 a mass of m2

Before Starting lets define a L-E Algorithm: Step 1 Apply D-H Algorithm to build Ai matrices and find Fi the “link frame” Step 2Set T 0 0 =I; i=1; D(q)=0 Step 3 Find  c i the Homogenous coordinate of the center of mass of link I WRT Fi Step 4 Set Fc as the translation of Frame F1 to Cm of i Compute Inertia Tensor Di about Cm wrt Fc Step 5Compute: z i-1 (q); T 0 i ; c i bar (q); D i (q)

Before Starting lets define a L-E Algorithm: Step 6Compute Special Case of J i (q) Step 7 Partition J i and compute D(q) = D(q) + {A T m K A + B T D K B} Step 8 Set i = i+1 go to step 3 else (i=n+1) set i=1 & continue Step 9Compute C i (q); h i (q) and friction i Step 10Formulate Torque i equation Step 11Advance “i” go to step 9 until i>n

We Start with Ai’s Not Exactly D-H Legal (unless there is more to the robot than these 2 links!)

So Let’s find 0 T 2 0 T 2 = A 1 *A 2

I’ll Compute Similar Terms back – to – back rather than by the Algorithm

C 2(bar) Computation:

Finding D 1 Consider each link a thin cylinder These are Inertial Tensors with respect to a F c aligned with the link Frames at the C m

Continuing for Link 1

Simplifying:

Continuing for Link 2

Now lets compute the Jacobians

Finishing J 1 Note the 2 column is all zeros – even though Joint 2 is revolute – this is the special case!

Jumping into J 2 This is 4 th column of A 1

Continuing:

And Again:

Summarizing, J 2 is:

Developing the D(q) Contributions D(q) I = (A i ) T m i A i + (B i ) T D i B i A i is the “Upper half” of the J i matrix B i is the “Lower Half” of the J i matrix D i is the Inertial Tensor of Link i defined in the Base space

Building D 1 D(q) 1 = (A 1 ) T m 1 A 1 + (B 1 ) T D 1 B 1 Here:

Looking at the 1 st Term (Linear Velocity term)

Looking at the 2 nd Term (Angular Velocity term) Recall that D 1 is: Then:

Putting the 2 terms together, D(q) 1 is:

Building the Full Manipulator D(q) D(q) man = D(q) 1 + D(q) 2 Where – D(q) 2 = (A 2 ) T m 2 A 2 + (B 2 ) T D 2 B 2 And recalling (from J 2 ):

Building the 1 st D(q) 2 Term:

How about term (1,1) details!

Building the 2 rd D(q) 2 Term: Recall D 2 : Then:

Combining the 3 Terms to construct the Full D(q) term:

Simplifying then D(q) is: NOTE: D(q) man is Square in the number of Joints!

This Completes the Fundamental Steps: Now we compute the Velocity Coupling Matrix and Gravitation terms:

For the 1 st Link

Plugging ‘n Chugging From Earlier: THUS:

P & C cont:

Finding h 1 : Given: gravity vector points in –Y 0 direction (remembering the model!) g k =(0, -g 0, 0) T g 0 is gravitational constant In the ‘h’ model A ki j is extracted from the relevant Jacobian matrix Here:

Continuing: Note: Only k = 2 has a value for g k which is g 0 !

Stepping to Link 2

Computing h 2

Building “Torque” Models for each Link In General:

For Link 1: The 1 st terms: 2 nd Terms:

Writing the Complete Link 1 Model

And, Finally, For Link 2:

Ist 2 terms: 1st Terms: 2 nd Terms:

Finalizing Link 2 Torque Model: