Contact Dynamics Modelling for Robots

Introduction 

Impulse-momentum contact 

Point contact 

Friction forces 

Contact in robot dynamics modelling 

Conclusions 

Contact Dynamics Modelling for Robots 

Mike Boos 

March 30, 2009 

Mike Boos 

Contact Dynamics Modelling for Robots

Outline 

Introduction 


Point contact 



Conclusions 

1 Introduction 

What is contact dynamics? 

Why contact modelling in robotics? 

Types of contact models 

2 Impulse-momentum contact 

3 Point contact 

Hunt-Crossley contact 

4 Friction forces 

5 Contact in robot dynamics modelling 

2-link planar manipulator 

Example simulation 

Mike Boos 


Outline 

Introduction 


Point contact 



Conclusions 















Mike Boos 


Introduction 


Point contact 



Conclusions 





Ff 

N 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Why model contact in robotics? 




Haptics 

Interactions with the 

environment 

Intentional (gripping, 

part placement, etc) 

Unintentional 

(collisions) 

Control and stability 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Examples of contact models 




f n 

kv 

Impulse 

Momentum 

Point 

Contact 

Volumetric 

Contact 

Finite Element 

Methods 

Hardwarein-the-loop 

Mike Boos 


Outline 

Introduction 


Point contact 



Conclusions 












Mike Boos 


Introduction 


Point contact 



Conclusions 

Impulse-momentum 

Assumptions 

vB 

vA 

x 

y - plane of 

impact 

Contact forces >> other 

forces 

Deformation and 

restitution is 

near-instantaneous 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Impulse-momentum 

vB 

y - plane of 

impact 

Conservation of Momentum 

m A (v Ay ) 1 = m A (v Ay ) 2 

m B (v By ) 1 = m B (v By ) 2 

m i (v ix ) 1 = ∑ m i (v ix ) 2 

i 

∑ 

i 

vA 

x 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Coefficient of restitution 

vB 

y - plane of 

impact 

Need a second equation to 

solve for x-velocities 

Energy is lost in impact 

vA 

x 

e = (v Ax −v Bx ) 2 

(v Bx −v Ax ) 1 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Coefficient of restitution 

For low impact velocities and most materials with a linear elastic 

range (e.g. metals): 

e ≈ 1 − α|(v Bx − v Ax ) 1 | 

= 1 − αv 1 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Limitations 

Not useful for prolonged contact 

Cannot solve for contact or reaction forces 

May not be very useful in robotic or other multibody systems 

While simple impulse-momentum models may not be useful, the 

concept of a coefficient of restitution may still be important in 

other models. 

Mike Boos 


Outline 

Introduction 


Point contact 



Conclusions 













Mike Boos 


Point contact 

Introduction 


Point contact 



Conclusions 


Common model: 

spring-damper system 

N = K(δ) + B( ˙δ) 

Often, the relationship between 

penetration and normal force is 

non-linear (even with 

linear-elastic solids) 

δ 

N 

Mike Boos 


Hertz theory 

Introduction 


Point contact 



Conclusions 


Assumptions: 

Non-conforming surfaces 

Contact patch is small relative to the geometries of the bodies 

Bodies are linear-elastic solids 

Mike Boos 


Hertz theory 

Introduction 


Point contact 



Conclusions 


N = kδ 3/2 

Two spheres: 

k = 

4 

3π(h m,1 +h m,2 ) ( R 1R 2 

R 1 +R 2 

) 1/2 

Sphere on plane: 

k = 

4 

3π(h m,1 +h m,2 ) R1/2 

Ri 

δ 

N 

δ 

h m,i = 1−ν2 i 

πE i 

Rj 

Mike Boos 


Introduction 


Point contact 



Conclusions 



Contact forces modelled with a spring in parallel with a non-linear 

damper: 

N = K(δ) + B( ˙δ, δ) 

= kδ 3/2 + (λδ 2/3 ) ˙δ 

Consistent with Hertzian contact in the static case (i.e. ˙δ = 0) 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Damping coefficient 


If we reintroduce the concept of a coefficient of restitution, where 

e ≈ 1 − αv 1 , it can be shown that: 

λ ≈ 3 2 αk 

Mike Boos 


Introduction 


Point contact 



Conclusions 



N = kδ 3/2 + (λδ 2/3 ) ˙δ 

where 

k = 

4 

3π(h m,1 + h m,2 ) ( R 1R 2 

R 1 + R 2 

) 1/2 

λ ≈ 3 2 αk 

Ri 

δ 

Rj 

e ≈ 1 − αv 1 

Mike Boos 


Outline 

Introduction 


Point contact 



Conclusions 












Mike Boos 


Introduction 


Point contact 



Conclusions 

Coulomb friction 

Static 

F f ≤ −µ S Nû 

where û is in the direction of 

other forces in the plane of 

contact. 

c 

vcn 

Dynamic 

F f = −µ k Nˆv ct 

Ff 

N 

vct 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Coulomb friction approximation 

μ 

μs 

μc 

vct 

vs 

vd 

Mike Boos 


Outline 

Introduction 


Point contact 



Conclusions 














Mike Boos 


Introduction 


Point contact 



Conclusions 




y 

q2 

Dynamic equations 

τ 1 = D 1 (q)¨q+C 1 (q, ˙q) ˙q+φ 1 (q) 

τ 2 = D 2 (q)¨q+C 2 (q, ˙q) ˙q+φ 2 (q) 

q1 

x 

Mike Boos 


Introduction 


Point contact 



Conclusions 




What happens to the dynamic equations with contact? 

y 

x 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Including contact forces 



F 

y 

Contact forces 

F = N + F f 

Cannot include contact forces 

in Lagrangian, since they are 

non-conservative (damping and 

friction). 

x 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Modified dynamic equations 



The torques caused by the contact forces must be included with 

the driver torques on the left-hand side of the dynamic equations: 


τ 1 + T c1 = D 1 (q)¨q + C 1 (q, ˙q) ˙q + φ 1 (q) 

τ 2 + T c2 = D 2 (q)¨q + C 2 (q, ˙q) ˙q + φ 2 (q) 

What are T c1 and T c2 ? 

Mike Boos 


Link 2 

Introduction 


Point contact 



Conclusions 



r2 

F 

Consider link 2 and the contact 

forces independently from the 

original system and its forces. 

There is an additional reaction 

at P from link 1. 

-F 

P 

T c2 = r 2 × F 

Mike Boos 


Link 1 

Introduction 


Point contact 



Conclusions 



-F 

O 

r1 

F 

Tc2 = r2 x F 

Now consider link 1. 

We have the reaction at P 

from link 2, as well as the 

torque from link 2. 

T c1 = r 1 × F + T c2 

= (r 1 + r 2 ) × F 

Mike Boos 


Introduction 


Point contact 



Conclusions 

Putting it all together 



F 

y 

r1 

r2 


τ + T c = D(q)¨q + C(q, ˙q) ˙q + φ(q) 

where[ ] 

((r 

T c = 1 + r 2 ) × F ) · ˆk 

(r 2 × F ) · ˆk 

x 

Mike Boos 


Introduction 


Point contact 



Conclusions 




Setup 

y 

45˚ 

1 m 

x 

Slender links of mass 1 kg, 

length 1 m 

Initially at rest 

End effector radius: 5 cm 

Contact properties of end 

effector and wall similar to 

that of steel for 

Hunt-Crossley 

τ 1 = −10Nm, τ 2 = −2Nm 

Mike Boos 


Joint angles 

Introduction 


Point contact 



Conclusions 



Mike Boos 


Introduction 


Point contact 



Conclusions 

End effector position 



Mike Boos 


Summary 

Introduction 


Point contact 



Conclusions 












Mike Boos 


Introduction 


Point contact 



Conclusions 

Questions? 

Mike Boos

Contact Dynamics Modelling for Robots

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