Lecture 10 Reprise and generalized forces

Transcription

1 Lecture 10 Reprise and generalized forces The Lagrangian Holonomic constraints Generalized coordinates Nonholonomic constraints Generalized forces we haven t done this, so let s start with it Euler Lagrange equa:ons Hamilton s equa:ons 1

2 Generalized forces are based on the principle of virtual work δw = F δr + τ δφ = Q i δq i convert to :me deriva:ves δw = δw δt δt W δt, Q i δq i Q i q i δt W δt = Q i q i δt W = Q i q i Q i = W q i so what we need is W 2

3 We need to look at the forces and torques applied to each link W = F r + τ Φ The transla:onal part is obvious: F r = F v The rota:onal part δφ = δφk + δθi 1 + δψk 3 δφ = φ δtk + θ δti 1 + ψ δtk 3 = ωδt = Ωδt Φ = ω W = F v + τ ω 3

4 So my total rate of work will be W = K i=1 ( ) F i v i + τ i ω i The veloci:es and rota:ons must be consistent with the constraints and everything must be expressed in iner:al 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

5 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

6 The torque from the ground to link 1 τ 01 k The torque from link 1 to link 2 τ 12 I 1 The torque from link 2 to link 3 τ 23 I 2 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

7 The torque on link 1 τ 01 k τ 12 I 1 The torque on link 2 τ 12 I 1 τ 23 I 2 The torque on link 3 τ 23 I 2 These will give us the generalized forces as we ll see shortly 7

8 There are a bunch of constraints affec:ng the rota:on rates φ 1 = 0 = θ 1 φ 2 =ψ 1 = φ 3 ψ 2 = 0 =ψ 3 There are also enough connec:vity constraints to reduce the system to three degrees of freedom we ll look at these shortly 8

9 All three I vectors are equal to cosψ 1 sinψ 1 0 The rota:on rates are ω 1 = 0 0, ω 2 = ψ 1 cosψ 1 θ 2 sinψ 1 θ 2, ω 3 = ψ 1 cosψ 1 θ 3 sinψ 1 θ 3 ψ 1 9

10 Build the work terms for each link ω 1 ( τ 01 k τ 12 I 1 ) = τ 01 ψ 1 ω 2 ( τ 12 I 1 τ 23 I 2 ) = ( τ 12 τ 23 ) θ 2 ω 3 τ 23 I 2 = τ 23 θ 3 so we have W = τ 01 ψ 1 + ( τ 12 τ 23 ) θ 2 + τ 23 θ 3 10

11 The three generalized coordinates q = ψ 1 θ 2 θ 3 W = τ 01 q 1 + ( τ 12 τ 23 ) q 2 + τ 23 q 3 Differen:ate to get the generalized forces Q 1 = τ 01, Q 2 = ( τ 12 τ 23 ), Q 3 = τ 23 11

12 We get a bewer idea of the compa:bility issues by looking further Apply a horizontal force to the end of link 3 Let it be directed in the I direc:on 12

13 We have to take the connec:vity constraints in to account to handle this We see intui:vely that the only work is can do is to rotate the en:re robot about k. It will contribute a torque to Q 1 ; 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 τ 3 = 1 2 r 3 K 3 F = 1 2 r 3 K 3 I 1 F = 1 2 r 3 sinθ 3 F ψ 1 13

14 The connec:vity constraint p 3 = p 1 + r 2 K r 3K 3 v 3 = r 2 K r 3 K 3 The force piece of the work term is W F = F v 3 = FI 1 r 2 K r 3 K 3 combine that with the torque term and we have W = ( r 2 sinθ 2 + r 3 sinθ 3 )Fψ 1 for an addi:onal term in accord with intui:on Q 1 = ( r 2 sinθ 2 + r 3 sinθ 3 )F = r sinθ r sinθ F 3 3 ψ 1 14

15 ?? We ll look at this in general once again 15

16 The Lagrangian in physical variables T = 1 2 m x 2 + y 2 + L = ( z 2 ) A K i=1 T i V i ( θ cosψ + φ sinθ sinψ) 2 ( ) C ( ψ + φ cosθ) B θ sinψ + φ sinθ cosψ We ve mostly been doing gravity poten:als 16

17 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

18 Nonholonomic constraints: linear in the deriva:ves of q Write them in matrix form C ji q j = 0 C q = 0 C is an M x N matrix number of constraints x number of generalized coordinates 18

19 Generalized forces Iden:fy the outside forces and torques on each link Form W Then Q i = W q i Choose between Euler Lagrange and Hamilton 19

20 Euler Lagrange d dt L q i = L + Q q i i + λ j C i j Convert to first order system the first set of equa:ons is really simple q i = u i We need to develop the symbolic representa:on of the second set 20

21 Here s the Lagrangian we want its contribu:on to the le` and right hand sides L = 1 2 q i M ij q j V ( q k ) right hand side L = 1 q i 2 q M m mn q i q n V q i subs:tute q m = u m, q n = u n right hand side L = 1 M mn q i 2 um q i u n V q i 21

22 le` hand side L = M q i ij q j = M ij u j d dt L q i = M ij u j + dm ij dt u j M ij u j + dm ij dt u j = M ij u j + M ij q k u j u k and we can combine the two and add back the constraint contribu:on and the generalized forces 22

23 M ij u j + M ij q k u j u k = L + λ q i j C j 1 + Q i L = 1 M mn q i 2 um q i u n V q i M ij u j + M ij q k u j u k = 1 2 um M mn q i u n V + λ q i j C j 1 + Q i 23

24 M ij u j + M ij q k u j u k = 1 2 um M mn q i u n V + λ q i j C j 1 + Q i M ij u j = 1 M mn 2 um q i u n M ij q k u j u k V + λ q i j C j 1 + Q i q i = u i and we can (in principle) solve for the individual components of u 1 M u j = M ki mn 2 um q i u n M ij q k u j u k V + λ q i j C j 1 + Q i 24

25 We did Hamilton the last :me I have the pair of sets of odes q j = M ji p i p i = L + λ q i j C j 1 + Q i where all the q dots in the second set have been replaced by their expressions in terms of p 25

26 We can write this out in detail, although it looks prewy awful L = 1 2 q i M ij q j V ( q k ) right hand side L = 1 q i 2 q M m mn q i q n V q i subs:tute q m = M mr p r, q n = M ns p s right hand side L = 1 q i 2 M M mr p mn r q i M ns p s V q i 26

27 p i = L + λ q i j C j 1 + Q i L = 1 q i 2 M M mr p mn r q i M ns p s V q i p i = 1 2 M M mr p mn r q i M ns p s V + λ q i j C j 1 + Q i q i = M ij p j 27

28 Both of these are of the same general form a form that we will see in the other methods we will address q i = M ij p j Euler Lagrange p i = 1 2 M M mr p mn r q i M ns p s V + λ q i j C j 1 + Q i Hamilton 1 M u j = M ki mn 2 um q i q i = u i u n M ij q k u j u k V + λ q i j C j 1 + Q i 28

29 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 mul:pliers Find the rate of work and the generalized forces 29

30 For Hamilton s equa:ons Define the momentum and use it to eliminate q i = M ij p j (These are half the differen:al equa:ons) The other half are the momentum equa:ons p i = 1 2 M M mr p mn r q i M ns p s V + λ q i j C j 1 + Q i 30

31 Solve some of the momentum equa:ons for the Lagrange mul:pliers Combine the remaining momentum equa:ons with the differen:ated constraints These are the new momentum equa:ons Solve the new momentum equa:ons simultaneously with q i = M ij p j 31

32 An example link 1, the frame link 2, the fork link 3, the rear wheel link 4, the front wheel link 2 link 1 link 4 link 3 32

33 The (physical) Lagrangian There are four iden:cal 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

34 Simple holonomic constraints I want the system to be erect Let the frame and fork rotate only about the ver:cal φ 1 = 0 = φ 2, θ 1 = 0 = θ 2 minus 4 Define the difference between their k = K rota:ons ψ 2 =ψ 1 + α no change 34

35 Simple holonomic constraints The heights of all the links are fixed z 1 = h 1, z 2 = h 2, z 3 = r W = z 4 minus 4 35

36 Simple holonomic constraints I want the wheels to be upright θ 3 = π 2 = θ 4 minus 2 The rear wheel points in the same direc:on as the frame the front wheel in the same direc:on as the fork φ 3 =ψ 1, φ 4 =ψ 1 + α minus 2 The horizontal loca:ons of the centers of mass are connected the front connec:on is simple x 4 = x 3, y 4 = y 3 minus 2 36

37 Nonsimple holonomic constraints The other horizontal loca:ons of the centers of mass are also connected x 3 = l 1 cosψ 1, y 3 = l 1 sinψ 1 x 2 = l 2 cosψ 1, y 2 = l 2 sinψ 2 minus 4 At this point we have introduced eighteen constraints, leaving six variables 37

38 Assign the generalized coordinates x 1 y 1 ψ q = 1 α ψ 3 ψ 4 We s:ll have rolling constraints 38

39 Nonholonomic constraints rolling constraints v 3 = ω 3 r 3, v 4 = ω 4 r 4 r 3 = k = r 4 ver:cal components are already in place, so there are but four of these constraint matrix x 3 r W ψ 3 cosφ 3 = 0 = y 3 r W ψ 3 sinφ 3 x 4 r W ψ 4 cosφ 4 = 0 = y 4 r W ψ 4 sinφ l 2 sinq 3 0 r W cosq l 2 cosq 3 0 r W sinq 3 0 C = 1 0 l 1 sinq r W cos q 3 + q l 1 cosq r W sin q 3 + q 4 ( ) ( ) 39

40 Nonholonomic constraints in physical coordinates 1 0 l 2 sinψ 1 0 r W cosψ l 2 cosψ 1 0 r W sinψ 1 0 C = 1 0 l 1 sinψ r W cos( ψ 1 + α) 0 1 l 1 cosψ r W sin( ψ 1 + α) There will be four Lagrange mul:pliers We ll add λ j C i j to the momentum equa:ons 40

41 Generalized forces There is torque from the frame to the fork τ 12 k and from the frame to the rear wheel τ 13 K 3 The rate of work will be W = ω 1 ( τ 12 k τ 13 K 3 ) + ω 2 τ 12 k + ω 3 τ 13 K 3 from which we can find Q = { τ 12 τ 13 0} 41

42 At this point we have the structure of the problem It s :me to go to Mathema:ca to see how to do this in prac:ce And to see how to convert this to numerical form for integra:on And some simple examples 42