1

Lecture 6: Constraints II

Planar mechanisms (four bar linkage)

A three-link robot

A general hinge

I want to focus on constraintsstill holonomic — both simple and nonsimple

I can do this in the context of three mechanismsand I can put some of this into Mathematica

2

planar mechanisms fit into our rubricuse the x =0 plane

simple holonomic constraints

€

x i = 0, φi = 0 =ψ i

planar mechanisms

3

We have a choice of how to fit this into our existing process

We can preserve q or we can preserve the idea that the long axes are Ks

If we choose the former, then the long axes become Jsthe latter adds π/2 to q

I’m going to do the former for today

4

We have the planar picture

5

This tells us what the connectivity constraints are

€

y1 = b1 cosθ1, z1 = b1 sinθ1

y2 = y1 + b1 cosθ1 + b2 cosθ2, z2 = z1 + b1 sinθ1 + b2 sinθ2

y3 = y2 + b2 cosθ2 + b3 cosθ3, z3 = z2 + b2 sinθ2 + b3 sinθ3

The system shown (known as a kinematic chain) has three degrees of freedom

(The three link robot to come is related to this)

6

The Lagrangian for this simple three link chain is

€

L =1

2A1 + b1

2 m1 + 4m2 + 4m3( )( ) ˙ θ 12 +

1

2A2 + b2

2 m2 + 4m3( )( ) ˙ θ 22 +

1

2A3 + m3b2

2( ) ˙ θ 3

2

2m3b2b3 cos θ2 −θ3( ) ˙ θ 2 ˙ θ 3 + 2b1b2 cos θ1 −θ2( ) m2 + 2m3( ) ˙ θ 2 ˙ θ 1 + 2b1b3 cos θ1 −θ3( )m3˙ θ 3 ˙ θ 1

You can see that this will lead to some fairly complicated EL equations

Move on to a common planar mechanism, the four bar linkage

7

The four bar linkage

€

r1J1 + r2J2 + r3J3 + r4J4 = 0

crank

coupler

follower

ground link

loop closure equation

8

Several kinds:crank-rocker: crank can make a full rotationdouble rocker: neither crank nor follower can make a full rotationdrag link: both crank and follower can make full rotations

The picture on the previous slide is a double rocker.

9

The loop closure equation has two componentswe can find two variables

The text discusses finding two angles given all four lengthsand the crank angle (the J4 angle is always π)

If we are doing dynamics, we only need to do that once to give us an initial condition

Even kinematics can be converted to differential equations

The picture we have already seen is of a double rocker linkagefor which the crank cannot make a full circle

I’ll build a crank rocker mechanism for which crank can make a full circle

10

11

Differentiate the loop closure equationSolve for two of the rates of change of angle

integrate numerically

€

˙ θ 2 = −r1 ˙ θ 1 csc θ2 −θ3( )sin θ1 −θ3( )

r2

˙ θ 3 =r1 ˙ θ 1 csc θ2 −θ3( )sin θ1 −θ2( )

r3

The equations to be integrated

Specify

€

˙ θ 1 = π sin t

Integrate to get the other two angles

12

13

??

14

Three link robot

We’ll look at fancier robots later in the coursebut this is enough to locate the end of the robot

wherever you want it in the robots work space.

This one will be very simple, made up of three identical cylinders

15

16

How does this work?What can it do?

The red link can rotate about the vertical — y1

The blue link is hinged to the red link — f2 = y1

The green link is hinged to the blue link — f3 = y1

The free angles are y1, q2 and q3 — three degrees of freedom

17

There are simple orientation constraints

€

φ1 = 0 = θ1, ψ 2 = 0 =ψ 3, φ2 =ψ1 = φ3

€

x1 = 0 = y1, z1 = c1

The first link is attached to the groundalso a simple constraint

There are also two vector connectivity constraints(six altogether) which are nonsimple

€

r1 = r0 + c1K1 + c2K 2

r2 = r1 + c2K 2 + c3K 3

18

I put numbers into this one: m = 1, l = 1, and a = 1/20 with g = 1

€

L =6403

9600˙ θ 2

2 +1603

9600˙ θ 3

2 +1

2cos θ2 −θ3( ) ˙ θ 2 ˙ θ 3

+8030 − 6397cos 2θ2( ) −1597 cos2 θ3 − sin2 θ3( ) + 9600sinθ2 sinθ3

19200˙ ψ 1

2

1

22 + 3cosθ2 + cosθ3( )

We get a Lagrangian and we could go on and set up the differential equationsbut they are pretty awful

19

In this case we’ll have generalized forces (torques)

from the ground to link one

from link one to link two

from link two to link three

The torques react back on the link imposing them, so we’ll have

20

€

τ 01k, τ 12I1, τ 23I2

€

τ 01k − τ 12I1, τ 12I1 − τ 23I2, τ 23I2

€

Q1 = τ 01, Q2 = τ 12 − τ 23, Q3 = τ 23

The three external torques

The torques on the three links

The torques that do work when the variables change

21

The Euler-Lagrange equations

€

d

dt

∂L

∂ ˙ ψ 1

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂ψ1

= Q1 = τ 01

d

dt

∂L

∂ ˙ θ 2

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂θ2

= Q2 = τ 12 − τ 23

d

dt

∂L

∂ ˙ θ 3

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂θ3

= Q3 = τ 23

These are pretty messy, and we don’t know yet how to assign the Qs.

22

??

23

A more general hinge

We just looked at two hinges, and they were simple because the first link was anchored.

If no link is anchored, then we really need to exercise our understanding of rotationto figure out how the mechanism will work

I will look at a general hinge that keeps I1 = I2

24

I2 will be equal to I1 if all three Euler angles are equal for the two links.

That’s the trivial solutionbut it’s where we need to start

We can add a fourth rotation to model the hinge

25

Rotation of the inertial coordinates looks like

€

Rz ψ( )Rx θ( )Rz φ( )

We need to add a fourth rotation

€

Rx χ( )Rz ψ( )Rx θ( )Rz φ( )

Rotation of the body coordinates is the inverse of this

€

RzT φ( )Rx

T θ( )RzT ψ( )Rx

T χ( )

26

There are four rotation variables

There are three connectivity constraints

There are a total of seven degrees of freedom — seven generalized coordinates

27

??OK, let’s look at some of this in Mathematica