Lecture 10 Reprise and Generalized Forces

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Lecture 10 Reprise and generalized forces

The Lagrangian

Holonomic constraints

Generalized coordinates

Nonholonomic constraints we haven’t done this, so let’s start with it Generalized forces

Euler‐Lagrange equaons

Hamilton’s equaons

1 Generalized forces are based on the principle of virtual work

i δW = F ⋅δr + τ ⋅δΦ = Qiδq convert to me derivaves € δW δW = δt ⇒ W˙ δt, Qδqi ⇒ Q q˙ iδt δt i i

˙ i ˙ i Wδ t = Qiq˙ δt ⇒ W = Qiq˙

€ ∂W˙ Q = i ∂q˙ i €

€ so what we need is W˙

2 € We need to look at the forces and torques applied to each link

W˙ = F ⋅ r˙ + τ ⋅ Φ˙

The translaonal part is obvious: F ⋅ r˙ = F ⋅ v € The rotaonal part € δΦ = δφk + δθI1 + δψK 3

˙ ˙ ˙ δΦ = φδ tk + θδ tI1 +ψ δtK 3 = ωδt = Ωδt € Φ˙ = ω € W˙ = F ⋅ v + τ ⋅ ω €

€ 3 So my total rate of work will be

K ˙ W = ∑(Fi ⋅ vi + τ i ⋅ ω i ) i=1

The velocies and rotaons must be consistent with the constraints and everything must be expressed in ineral coordinates €

I can believe that this looks mysterious, but it is actually simple, which we can see by looking at some examples

REMEMBER THAT THESE FORCES AND TORQUES DO NOT INCLUDE THE POTENTIAL FORCES AND TORQUES

4 This is the three link robot that we have seen before

It will have torques from the ground to link 1 from link 1 to link 2 from link 2 to link 3

5 The torque from the ground to link 1 τ 01k

€ The torque from link 1 to link 2 τ12I1

€ The torque from link 2 to link 3 τ 23I2

€ There is a force from the ground that balances the weight of the robot but it does no work because link 1 does not move up and down

6 The torque on link 1 τ 01k − τ12I1

€ The torque on link 2 τ12I1 − τ 23I2

€ The torque on link 3 τ 23I2

These will give us the generalized forces as we’ll see shortly €

7 There are a bunch of constraints aﬀecng the rotaon rates

φ1 = 0 = θ1

φ2 =ψ1 = φ3 € ψ2 = 0 =ψ3 € There are also enough connecvity constraints to reduce the system to three degrees of freedom € we’ll look at these shortly

8 All three I vectors are equal to

cos ψ   1  sinψ1     0 

The rotaon rates are €

 ˙   ˙   0  cosψ1θ 2 cosψ1θ 3    ˙   ˙  ω1 =  0 , ω 2 =  sinψ1θ 2 , ω 3 =  sinψ1θ 3   ˙   ˙   ˙  ψ 1   ψ1   ψ1 

€

9 Build the work terms for each link

˙ ω1 ⋅ (τ 01k − τ12I1) = τ 01ψ1

˙ ω2 ⋅ (τ12I1 − τ 23I2 ) = (τ12 − τ 23 )θ 2 € ˙ ω3 ⋅ τ 23I2 = τ 23θ 3 € so we have € ˙ ˙ ˙ ˙ W = τ 01ψ1 + (τ12 − τ 23)θ 2 + τ 23θ 3

€

10 The three generalized coordinates

ψ 1   ˙ q = θ 2  ⇒ W = τ 01q˙1 + (τ12 − τ 23)q˙ 2 + τ 23q˙ 3   θ 3 

Diﬀerenate to get the generalized forces €

Q1 = τ 01, Q2 = (τ12 − τ 23), Q3 = τ 23

€

11 We get a beer idea of the compability issues by looking further

Apply a horizontal force to the end of link 3 Let it be directed in the –I direcon

12 We have to take the connecvity constraints in to account to handle this

We see intuively that the only work is can do is to rotate the enre robot about k.

It will contribute a torque to Q1; can we get that using the rubric?

We have a force applied to link 3, and it is not at the center of mass, so there is an accompanying torque

1 1 1 τ = r K × F = r K × I F = r sinθ Fψ˙ 3 2 3 3 2 3 3 1 2 3 3 1

€ 13 The connecvity constraint

1 1 p = p + r K + r K ⇒ v = r K˙ + r K˙ 3 1 2 2 2 3 3 3 2 2 2 3 3

The force piece of the work term is €  1   1  W˙ = F ⋅ v = FI ⋅ r K˙ + r K˙  =  r sinθ + r sinθ  Fψ˙ F 3 1  2 2 2 3 3  2 2 2 3 3 1

combine that with the torque term and we have

€ ˙ ˙ W = (r2 sinθ2 + r3 sinθ3 )Fψ1

for an addional term in accord with intuion € Q1 = (r2 sinθ2 + r3 sinθ3 )F

14 € ??

We’ll look at this in general once again

15 The Lagrangian in physical variables

L = ∑Ti −Vi i=1

1 1 2 T = m x˙ 2 + y˙ 2 + z˙ 2 + A θ˙ cosψ + φ˙ sinθ sinψ 2 ( ) 2 ( ) € 1 2 1 2 + B −θ˙ sinψ + φ˙ sinθ cosψ + C ψ˙ + φ˙ cosθ 2 ( ) 2 ( )

€ We’ve mostly been doing gravity potenals

16 Simple holonomic constraints

Nonsimple holonomic constraints

These remove some number of variables from the problem

At this point you can assign the remaining variables to a set of generalized coordinates

17 Nonholonomic constraints: linear in the derivaves of q

Write them in matrix form

i j C jq˙ = 0 ⇔ Cq˙ = 0

C is an M x N matrix — number of constraints € x number of generalized coordinates

18 Generalized forces

Idenfy the outside forces and torques on each link

Form W˙

∂W˙ Then Qi = € ∂q˙ i

€ Choose between Euler‐Lagrange and Hamilton

19 Euler‐Lagrange

d  ∂L  ∂L j  i  = i + Qi + λ jCi dt  ∂q˙  ∂q

Convert to ﬁrst order system € the ﬁrst set of equaons is really simple q˙ i = ui

We need to develop the symbolic representaon of the second set €

20 Here’s the Lagrangian we want its contribuon to the le and right hand sides 1 L = q˙ i M q˙ j −V qk 2 ij ( )

∂L 1 ∂M ∂V right hand side = q˙ m mn q˙ n − € ∂qi 2 ∂qi ∂qi substute q˙ m = um, q˙ n = un € right hand side ∂L 1 ∂M ∂V = um mn un − € ∂qi 2 ∂qi ∂qi

€ 21 le hand side

∂L j j d  ∂L  j dMij j i = Mijq˙ = Mij u ⇒  i  = Mij u˙ + u ∂q˙ dt  ∂q˙  dt

dMij ∂Mij € M u˙ j + u j = M u˙ j + u j uk ij dt ij ∂qk

and we can combine the two and add back € the constraint contribuon and the generalized forces

22 ∂Mij ∂L M u˙ j + u j uk = + λ C j + Q ij ∂qk ∂qi j 1 i

∂L 1 m ∂Mmn n ∂V i = u i u − i € ∂q 2 ∂q ∂q

€ ∂Mij 1 ∂M ∂V M u˙ j + u j uk = um mn un − + λ C j + Q ij ∂qk 2 ∂qi ∂qi j 1 i

€

23 ∂Mij 1 ∂M ∂V M u˙ j + u j uk = um mn un − + λ C j + Q ij ∂qk 2 ∂qi ∂qi j 1 i

1 ∂M ∂Mij ∂V M u˙ j = um mn un − u j uk − + λ C j + Q € ij 2 ∂qi ∂qk ∂qi j 1 i

€ q˙ i = ui

and we can (in principle) solve for the individual components of u €   j ki 1 m ∂Mmn n ∂Mij j k ∂V j u˙ = M  u i u − k u u − i + λ jC1 + Qi   2 ∂q ∂q ∂q 

€ 24 We did Hamilton the last me

I have the pair of sets of odes

j ji q˙ = M pi

∂L j p˙ i = + λ jC1 + Qi € ∂qi where all the q dots in the second set have been replaced by their expressions in terms of p €

25 We can write this out in detail, although it looks prey awful

1 L = q˙ i M q˙ j −V qk 2 ij ( )

∂L 1 ∂M ∂V right hand side = q˙ m mn q˙ n − € ∂qi 2 ∂qi ∂qi substute m mr n ns q˙ = M pr, q˙ = M ps €

∂L 1 ∂M ∂V right hand side = M mr p mn M ns p − € ∂qi 2 r ∂qi s ∂qi

€

26 ∂L p˙ = + λ C j + Q i ∂qi j 1 i

∂L 1 ∂M ∂V = M mr p mn M ns p − ∂qi 2 r ∂qi s ∂qi €

€ 1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i

€

i ij q˙ = M p j

27 € Both of these are of the same general form a form that we will see in the other methods we will address

i ij q˙ = M p j

1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i Euler‐Lagrange €

€ Hamilton q˙ i = ui

  j ki 1 m ∂Mmn n ∂Mij j k ∂V j u˙ = M  u i u − k u u − i + λ jC1 + Qi   2 ∂q ∂q ∂q  €

€ 28 What you have to do in general — whether you use Euler‐Lagrange or Hamilton

Set up the physical Lagrangian

Put in the simple holonomic constraints

Put in the nonsimple holonomic constraints

Deﬁne the generalized coordinates

Find the nonholonomic constraints and the constraint matrix

Deﬁne the Lagrange mulpliers

Find the rate of work and the generalized forces

29 For Hamilton’s equaons

i ij Deﬁne the momentum and use it to eliminate q˙ = M p j

(These are half the diﬀerenal equaons) € The other half are the momentum equaons 1 ∂M ∂V p˙ = M mr p mn M ns p − + λ C j + Q i 2 r ∂qi s ∂qi j 1 i

€

30 Solve some of the momentum equaons for the Lagrange mulpliers

Combine the remaining momentum equaons with the diﬀerenated constraints

These are the new momentum equaons

i ij Solve the new momentum equaons simultaneously with q˙ = M p j

€

31 An example

link 1, the frame link 2, the fork link 2 link 3, the rear wheel link 4, the front wheel

link 1 link 4

link 3

32 The (physical) Lagrangian

There are four idencal Lagrangians — we’ve seen them before I’m not going to write them out

I am going to hold this system erect, so that gravity does nothing I can set g = 0

33 Simple holonomic constraints

I want the system to be erect Let the frame and fork rotate only about the vercal

φ1 = 0 = φ2, θ1 = 0 = θ 2 minus 4

Deﬁne the diﬀerence between their k = K rotaons €

ψ2 =ψ1 + α no change

€

34 Simple holonomic constraints

The heights of all the links are ﬁxed

z1 = h1, z2 = h2, z3 = rW = z4 minus 4

€

35 Simple holonomic constraints

I want the wheels to be upright

π θ = − = θ minus 2 3 2 4

The rear wheel points in the same direcon as the frame the front wheel in the same direcon as the fork €

φ3 =ψ1, φ4 =ψ1 + α minus 2

The horizontal locaons of the centers of mass are connected the front connecon is simple €

x4 = x3, y4 = y3 minus 2

36 € Nonsimple holonomic constraints

The other horizontal locaons of the centers of mass are also connected

x3 = −l1 cosψ1, y3 = −l1 sinψ1 minus 4 x2 = l2 cosψ1, y2 = l2 sinψ2

At this point we have introduced eighteen constraints, leaving six variables €

37 Assign the generalized coordinates

 x1    y  1  ψ  q =  1   α  ψ   3  ψ 4 

€ We sll have rolling constraints

38 Nonholonomic constraints

rolling constraints

v3 = ω 3 × r3, v4 = ω 4 × r4

r3 = k = r4

vercal components are already in place, so there are but four of these €

x˙ 3 −€r Wψ˙ 3 cosφ3 = 0 = y˙ 3 − rWψ˙ 3 sinφ3

x˙ 4 − rWψ˙ 4 cosφ4 = 0 = y˙ 4 − rWψ˙ 4 sinφ4

constraint matrix

€  3 3  1 0 l2 sinq 0 −rW cosq 0  3 3  0 1 −l2 cosq 0 −rW sinq 0  C =  3 3 4  1 0 −l1 sinq 0 0 −rW cos(q + q )  3 3 4  0 1 l1 cosq 0 0 −rW sin(q + q )

€ Nonholonomic constraints

in physical coordinates

1 0 l2 sinψ1 0 −rW cosψ1 0    0 1 −l cosψ 0 −r sinψ 0  C =  2 1 W 1  1 0 −l1 sinψ1 0 0 −rW cos(ψ1 + α)   0 1 l1 cosψ1 0 0 −rW sin(ψ1 + α)

There will be four Lagrange mulpliers €

j We’ll add λ jCi to the momentum equaons

€

40 Generalized forces

There is torque from the frame to the fork τ12k

and from the frame to the rear wheel τ13K 3 €

The rate of work will be € ˙ W = ω1 ⋅ (−τ12k − τ13K 3) + ω 2 ⋅ τ12k + ω 3 ⋅ τ13K 3

from which we can ﬁnd €

Q = {0 0 0 τ12 τ13 0}

€ 41 At this point we have the structure of the problem

It’s me to go to Mathemaca to see how to do this in pracce

And to see how to convert this to numerical form for integraon

And some simple examples