Lecture 6: Constraints II
I want to focus on constraints
still holonomic — both simple and nonsimple
I can do this in the context of three mechanisms
and I can put some of this into Mathematica
Planar mechanisms (four bar linkage)
A three-link robot
A general hinge
1
planar mechanisms
planar mechanisms fit into our rubric
use the x =0 plane
simple holonomic constraints
x i  0,  i  0   i

2
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 Js
the latter adds π/2 to q
I’m going to do the former for today
3
We have the planar picture
4
This tells us what the connectivity constraints are
y1  b1 cosq1, z1  b1 sin q1
y 2  y1  b1 cosq1  b2 cosq 2, z2  z1  b1 sin q1  b2 sin q 2
y 3  y 2  b2 cosq 2  b3 cosq 3, z3  z2  b2 sin q2  b3 sin q 3
The system shown (known as a kinematic chain) has three degrees of freedom

(The three link robot to come is related to this)
5
The Lagrangian for this simple three link chain is
1
1
1
2
2
2
2
Ý
Ý
Ý2
A

b
m

4m

4m
q

A

b
m

4m
q

A3  m3b22 q







1
1
1
2
3  1
2
2
2
3  2
3
2
2
2
Ýq
Ý
Ý Ý  2b b cosq  q m q
ÝÝ
2m3b2b3 cosq 2  q 3 q
2 3  2b1b2 cosq1  q 2 m 2  2m 3 q2q
1
1 3
1
3
3 3q
1
L
You can see that this will lead to some fairly complicated EL equations
Move on to a common planar mechanism, the four bar linkage
6
The four bar linkage
coupler
follower
crank
ground link
loop closure equation

r1J1  r2 J 2  r3 J 3  r4 J 4  0
7
Several kinds:
crank-rocker: crank can make a full rotation
double rocker: neither crank nor follower can make a full rotation
drag link: both crank and follower can make full rotations
The picture on the previous slide is a double rocker.
8
The loop closure equation has two components
we can find two variables
The text discusses finding two angles given all four lengths
and 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 linkage
for which the crank cannot make a full circle
I’ll build a crank rocker mechanism for which crank can make a full circle
9
10
Differentiate the loop closure equation
Solve for two of the rates of change of angle
integrate numerically
The equations to be integrated
Ýcscq  q sinq  q 
r
q
1
2
3
1
3
Ý 1
q
2
r2
Ýcscq  q sinq  q 
r1q
2
3
1
2
Ý
q3  1
r3
Specify

Ý   sin t
q
1
Integrate to get the other
 two angles
11
12
??
13
Three link robot
We’ll look at fancier robots later in the course
but 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
14
15
How does this work?
What can it do?
The red link can rotate about the vertical — 1
The blue link is hinged to the red link — 2 = 1
The green link is hinged to the blue link — 3 = 1
The free angles are 1, q2 and q3 — three degrees of freedom
16
There are simple orientation constraints
1  0  q1,  2  0  3 , 2  1  3

The first link is attached to the ground
also a simple constraint
x1  0  y1, z1  c1

There are also two vector connectivity constraints
(six altogether) which are nonsimple
r1  r0  c1K1  c 2K 2
r2  r1  c 2K 2  c 3K 3
17
I put numbers into this one: m = 1, l = 1, and a = 1/20 with g = 1
We get a Lagrangian and we could go on and set up the differential equations
but they are pretty awful
6403 Ý2 1603 Ý2 1
Ýq
Ý
q2 
q3  cosq 2  q 3 q
2 3
9600
9600
2
8030 6397cos2q2  1597cos2 q 3  sin 2 q 3  9600sin q 2 sinq 3
L

19200
Ý12

1
2  3cosq2  cosq3 
2

18
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
19
The three external torques
 01k, 12I1,  23I2
The torques on the three links

 01k  12I1, 12I1   23I2 ,  23I2
The torques that do work when the variables change

Q1   01, Q2  12   23 , Q3   23

20
The Euler-Lagrange equations
d  L  L
 Q1   01
 Ý 
dt 1  1
d  L  L
 Q2  12   23
 Ý 
dt q2  q 2
d  L  L
 Q3   23
 Ý 
dt q3  q 3
These are pretty messy, and we don’t know yet how to assign the Qs.

21
??
22
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 rotation
to figure out how the mechanism will work
I will look at a general hinge that keeps I1 = I2
23
I2 will be equal to I1 if all three Euler angles are equal for the two links.
That’s the trivial solution
but it’s where we need to start
We can add a fourth rotation to model the hinge
24
Rotation of the inertial coordinates looks like
Rz Rx qRz 
We need to add a fourth rotation

Rx  Rz Rx q Rz 

Rotation of the body coordinates is the inverse of this
RzT RxT qRzT RxT  

25
There are four rotation variables
There are three connectivity constraints
There are a total of seven degrees of freedom — seven generalized coordinates
26
??
OK, let’s look at some of this in Mathematica
27