Multibody simulationDynamics of a multibody system (Newton-Euler formulation)

Dimitar Dimitrov

Orebro University

June 8, 2012

Main points covered

Newton-Euler formulation

forward dynamics

inverse dynamics

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Euler-Lagrange formulation

We derived the equations of motion for a multibody system by treatingthe multibody system as a whole and performing analysis using theLagrangian (the difference between the kinetic and potential energy) ofthe system.

Newton-Euler formulation

In this lecture, we describe an alternative formulation which treats eachlink in turn by forming the equations that govern its linear and angularmotion. The forces f and torques t acting on a particular link arecomputed by analyzing its interaction with its neighboring links.

Our starting point are the Newton-Euler equations for a single rigid body

f = mvc

t = ICω + ω × ICω

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Let us express the equations of motion for link i as

[f i

ti

]

=

[miI 0

0 Ii

] [vi

ωi

]

+

[0

ωi × Iiωi

]

vi denotes the linear velocity of the center of mass (CoM) of link i

ωi denotes the angular velocity of link i

Ii denotes the 3× 3 inertia matrix of link i, calculated about itsCoM and expressed in the world frame.

all vectors are expressed with respect to the world frame

We can write the above equations as

uei = M iξi + un

i ,

where

uei =

[f i

ti

]

, M i =

[miI 0

0 Ii

]

, ξi =

[vi

ωi

]

, uni =

[0

ωi × Iiωi

]

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For a system of n bodies (without constraints) we have

ue1

ue2

...uen

︸ ︷︷ ︸

ue

=

M1 0 . . . 0

0 M2 . . . 0

......

. . ....

0 0 . . . Mn

︸ ︷︷ ︸

M

ξ1ξ2...

ξn

︸ ︷︷ ︸

˙ξ

+

un1

un2

...unn

︸ ︷︷ ︸

un

or in short

ue = Mξ + un (1)

The fact that M is a block-diagonal matrix implies that there is nocoupling between the rigid bodies in the system. Or in other words, themotion of body i is independent of all other bodies.

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Constraints due to joints

When the bodies are interconnected using joints, we have to introduceconstraints on their motion. There are alternative ways to do so

form the constraint equations of each pair of bodies, and imposethem explicitly

since we are dealing only with open-loop systems, we could use thejoint variables q as generalized coordinates, and impose theconstraints implicitly by expressing ξ in (1) as a function of thegeneralized coordinates and their derivatives

We adopt the second approach

Recall that when using the Euler-Lagrange formulation, we expressed theLagrangian of the system as a function of the generalized coordinates,and then substituted it in the Euler-Lagrange equations. Because of that,no constraints appeared explicitly in the equations of motion.

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Implicit constraints

The relation between the velocityof link i and q can be expressed as

ξi =

[Jvi

Jωi

]

q = Jci q.

By using

Jc =

Jc1

Jc2

...Jcn

∈ R

6n×n

the velocities of all links can beexpressed as

ξ = Jcq

ξ = Jcq + Jcq (2)

Substitute (2) in ue = Mξ + un

ue = M(Jcq + Jcq) + un

= MJcq + uc, (3)

where uc = MJcq+un is the sum of allvelocity dependent inertial forces (andtorques) acting on the CoM of the links.

Multiply (3) by JTc from the left

JTc MJc

︸ ︷︷ ︸

H

q + JTc u

c

︸ ︷︷ ︸

c

= JTc u

e

︸ ︷︷ ︸

τ e

.

τ e are the generalized forces/torquesacting on the generalized coordinates as aresult of ue. Recall that τ e = JT

c ue is

the static relation.

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Computing c

c = JTc MJcq + JT

c un

The only term of the above equation that we have not computed yet isJc. Forming it explicitly is not usually done because we can computedirectly the product Jcq as follows.

If we set q = 0 in ξ = Jcq + Jcq, we obtain

Jcq = ξ0.

We use ξ0to denote the Cartesian accelerations of the links computed

with zero joint accelerations q. Hence, c can be computed using

c = JTc (Mξ

0+ un)

︸ ︷︷ ︸

uc

. (4)

Computing the vector ξ0∈ R

6n can be done in a recursive fashion (as weshow next).

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Link velocity ξ

ξ can be computed using ξ = Jcq, which corresponds to the followingrecursion for i = 1, . . . , n (we will assume fixed base v0 = 0, ω0 = 0).All vectors are expressed in the world frame.

Revolute joint

ωi = ωi−1 + kiqi

vi = vi−1 + ωi−1 × hi + ωi × di

= vi−1 + ωi−1 × (ri − ri−1) + ki × diqi

Prismatic joint

ωi = ωi−1

vi = vi−1 + ωi−1 × hi + ωi × di + kiqi

= vi−1 + ωi−1 × (hi + di) + kiqi

= vi−1 + ωi−1 × (ri − ri−1) + kiqi

where di = di + kiqi

ri

ri−1

hi

diqi

ki

vi

vi−1

ωi

ωi−1

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Link acceleration ξ

The following computation for i = 1, . . . , n is called forward recursion. Afixed base is assumed, i.e., v0 = 0, v0 = 0, ω0 = 0, ω0 = 0.

revolute joint

ωi = ωi−1 + kiqi

ωi = ωi−1 + ωi−1 × kiqi + kiqi

vi = vi−1 + ωi−1 × (ri − ri−1) + ki × diqi

vi = vi−1 + ωi−1 × (ri − ri−1) + ωi−1 × (vi − vi−1) + ωi × (ki × di)qi

+ ki × diqi

prismatic joint

ωi = ωi−1

ωi = ωi−1

vi = vi−1 + ωi−1 × (ri − ri−1) + kiqi

vi = vi−1 + ωi−1 × (ri − ri−1) + ωi−1 × (vi − vi−1 + kiqi) + kiqi

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Some notes

d

dtki = ωi × ki = (ωi−1 + kiqi)× ki = ωi−1 × ki.

d

dt(ki × di) = (ωi−1 × ki)× di + ki × (ωi × di)

= (ωi−1 × ki)× di + ki × [(ωi−1 + kiqi)× di]

= (ωi−1 × ki)× di + ki × (ωi−1 × di) + ki × (ki × di)qi

= ωi × (ki × di).

The last equality is obtained by adding the following term to the equation

ωi−1 × (ki × di) + ki × (di × ωi−1) + di × (ωi−1 × ki) = 0.

Note that any three vectors a, b, c ∈ R3 satisfy the Jacobi identity

a× (b× c) + b× (c× a) + c× (a× b) = 0.

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Summary (Newton-Euler formulation - batch version)

Given

(q, q) - state of the system at time t

ue - external forces acting on the system (including gravity)

τ - torque delivered by joint motors

Do

form M , Jc, un

form the manipulator inertia matrix H = JTc MJc

compute link accelerations ξ0(forward recursion using q = 0)

form uc = Mξ0+ un

compute c = JTc u

c (backward recursion)

compute τ e = JTc u

e (backward recursion)

The above steps lead to the following equation of motion

Hq + c = τ e + τ

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Recursive computation of JTc (u

e − uc)

Define[

f eci

teci

]

= ueci = ue

i − uci

f eci ∈ R

3 is a force acting at the CoM of link i

teci ∈ R3 is a torque acting on link i.

see figure on next slide

The following computation for i = n, . . . , 1 is called backward recursion.It is computationally more efficient compared to using JT

c (ue − uc).

fJi= fec

i + fJi+1

tJi= teci + tJi+1

+ di × f eci + (di + hi+1)× fJi+1

fJi+1, tJi+1

- forces and torques acting on link i through joint i+ 1

fJi, tJi

- forces and torques acting on link i− 1 through joint i

fJn+1= 0, tJn+1

= 0 is assumed

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Figure (backward recursion)

fJi

tJi

fJi+1

tJi+1

feci

teci

di

hi+1

joint i

joint i+ 1

link i

backward recursion (i = n . . . , 1)

fJi= fec

i + fJi+1

tJi= teci + tJi+1

+ di × f eci + (di + hi+1)× fJi+1

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Project (fJi, tJi) on the ith axis of rotation/translation

Define

τ ec = τ e − c = JTc (u

e − uc).

τ ec is the generalized force/torque acting on the generalized coordinatesq as a result of uc and ue.

The final step of the backward recursion is to determine

the torque τeci around the axis of rotation ki, if joint i is revolute

the force τeci along the axis of translation ki, if joint i is prismatic

τeci =

{ki · tJi

if joint i is revoluteki · fJi

if joint i is prismatic

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Newton-Euler formulation (the big picture)

velo

citie

sand

accelerations (forward recursion)

forc

es

an

dto

rques (backward recursion)

the two recursions

forward recursion (1, . . . , n) for computing Cartesian velocities andaccelerations of the links

backward recursion (n, . . . , 1) for computing the generalized forcesτ ec as a result of the external and inertial forces acting on the links

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Forward dynamics ⇔ inverse dynamics

Forward dynamics

Solve the equations of motion for the accelerations q resulting from givengeneralized forces/torques

q = H−1(τ e + τ − c).

After solving for q, one can obtain q and q by using numericalintegration. Forward dynamics is useful for the purpose of simulation.

Inverse dynamics

Find joint forces/torques τ that should be applied by the joint motors, inorder to generate system motion specified by q, q, q, (possibly in thepresence of external forces). Solving the inverse dynamics problem isequivalent to performing forward and backward recursions. It is useful for

forming the equations of motion

manipulator trajectory planning

control algorithm implementation

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Recursive computation of the manipulator inertia matrix

We already know how to compute the manipulator inertia matrix H ,however, here we discuss an alternative (and more efficient) approach.

Step 0

set all external forces (including gravity) equal to zero

set q = 0.

This results in the following equation of motion (satisfied only by q = 0)

Hq = 0.

Recursive computation of H for i = 1, . . . , n

set qi = 1, and qj = 0 for all j 6= i

perform a forward and backward recursion

At the ith iteration, the result from the backward recursion would be avector of torques/forces that is equivalent to the ith column of H.

See file bMSd/examples/example_ID.m.17 / 17