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Lecture 6: Constraints II

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1 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

2 planar mechanisms planar mechanisms fit into our rubric use the x =0 plane simple holonomic constraints

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 Js the latter adds π/2 to q I’m going to do the former for today

4 We have the planar picture
The angles are theta, and I am using a common convention in kinematics by which theta lies on (0, 2π) The arcs show the angles

5 This tells us what the connectivity constraints are
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
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 coupler follower crank ground link
The four bar linkage is connected to some frame, or the actual ground Now we have the loop closure equation, which reduces this to one degree of freedom ground link loop closure equation

8 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.

9 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


11 Differentiate the loop closure equation
Solve for two of the rates of change of angle integrate numerically The equations to be integrated Specify Integrate to get the other two angles


13 ??

14 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


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
The first link is attached to the ground also a simple constraint There are also two vector connectivity constraints (six altogether) which are nonsimple

18 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

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 The three external torques
The torques on the three links The torques that do work when the variables change I1 and I2 and I3 are equal I1 is perpendicular to k, so the reaction onto link one does not work and does not contribute to the generalized forces

21 The Euler-Lagrange equations
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 rotation to 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 solution but 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
We need to add a fourth rotation Rotation of the body coordinates is the inverse of this I leave it to you to multiply these out and see that you get Goldstein’s equations 4-46 and 4-47

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

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