 # Practice English Speaking&Listening with: Module 8 Lecture 1 Kinematics Of Machines

Normal
(0)
Difficulty: 0

So far, we have discussed graphical method for kinematic synthesis.

Today, we shall start discussing on kinematic synthesis of planar linkages, but by analytical

methods. If we recall, we discussed four different types of problems when we had, the graphical

methods namely: motion generation, function generation, fast generation and generation

of prescribed dead-centre configurations. However, when we discuss analytical method,

we shall restrict ourselves only to one type of problem namely function generation. There

are various methods like analytical method for function generation. But we shall discuss

only one such method which is known as Freudenstein's method.

As an example, we take a 4R-linkage function generator and show how to get the link lengths

for generating a prescribed function. If you remember, when we discussed graphical methods,

we had function generator for three position synthesis and four position synthesis. However,

we shall restrict ourselves only to three positions synthesis so far as 4R-linkage function

generator is concerned. As we shall see later, for three position synthesis we can get away

by considering only linear algebraic equation which is very easy to solve.

Towards this end, let me first draw a 4R-linkage. These are the two fixed changes O4O2 which

defines the fixed link and we take our X-axis joining O4 and O2. O2, A, B, O4 is the 4R-linkage

AB is the coupler, which is link number 3; O4B- the output link is link number 4. The

orientation of the input and the output link is defined by the angle that they make with

X-axis namely, theta2 and theta4. We complete the coordinate axis by drawing the Y-axis

with O4 as the origin.

For three position synthesis, what we mean is to generate two pair of prescribed coordinated movements of the input and output link, that is, as

the input link goes from position one to two by changing theta2 by an amount theta2 1 2.

The output link should move from position 1 to 2 by a prescribed movement theta4 1 2.

Similarly, another pair of prescribed movement is theta2 2 3 and theta4 2 3. So, from position

1 to 2, change in angle theta2 is given by theta2 1 2 and the corresponding change in

the angle theta4 should be theta4 1 2. This is a pair of prescribed movement.

This is another pair of prescribed movement. As theta2 changes from second to third configuration

by an amount theta2 2 3, the corresponding change in theta4 should be theta4 2 3.

In Freudenstein's method for the three positions synthesis, we assume the values of theta2

corresponding to the first configuration and value of theta4 corresponding to the first

configuration. I would like to emphasize that we should note the difference in the assumed

parameters as far as the analytical method and the previous graphical method is concerned.

In the graphical method, what we assume is the location of the point A that is in the

corresponding to the first configuration, that is, the link length l2 and theta2 1 or

we could have assumed the point B corresponding to the first configuration, which means assume

l4 the fall wall length and theta4 1. Whereas here we are leaving the link lengths as unknown

parameters and we are assuming the value of the theta2 and theta4 corresponding to the

first configuration. This gives me theta2 2 which is theta2 1 plus theta2 1 2 and theta4

2 is theta4 1 plus theta4 1 2. These are already prescribed and we have assumed. So, we get

these two values namely, theta2 2 and theta4 two. Exactly the same way, I can get theta2

three, which is theta2 2 plus theta2 2 3 and theta4 three is theta4 2 plus theta4 2 3.

Once, we know theta2 2 and theta4 2 and we have given theta2 2 3 and theta4 2 3, I can

get the values of theta2 and theta4 corresponding to the third configuration namely, theta2

3 and theta4 3. So, the three position function generation has while down to determine the

link length such that these three pairs of values of theta2 and theta4 can take place

in this linkage. As a first step towards Freudenstein's methodwe obtain the displacement equation

in a particular form which is known as Freudenstein's equation.

First, we derive Freudenstein's equation, which is nothing but the displacement equation

for this 4R-linkage in a particular format. Towards this end, we note that the x-coordinate

of the point A, xA is given by l1 plus l2 cosine theta2, where l1 is the length of the

fixed link O4O2. This is the l1, O2A is l2, AB is the coupler length l3 and O4B is the

fall wall length l4; y-coordinate of the point A, yA is easily saying to be l2 sine theta2

and x and y coordinates of the point B namely, xB is l4 cosine theta4 and yB is l4 sine theta4.

So, the distance between these two points A and B, I can write, which is l3 as we see

the square of the distance l3square is equal to xA minus xB whole squared plus yA minus

yB whole squared. Substituting for xA, xB and yA, yB here, we get l1 plus l2 cos theta2

minus l4 cosine theta4 whole squared plus l2 sine theta2 minus l4 sine theta4 whole

squared. Expanding this, we can easily get l1 squared plus l2 squared plus l4 squared

plus twice l1l2 cosine theta2 minus twice l1l4 cosine theta4 minus twice l2l4 cosine

theta2 minus theta4.

We rewrite this equation: Cosine theta2 minus theta4 is divided by twice l2l4, l2 and l2

cancels, We get l1 by l4 cosine theta2 minus l1 by l2 cosine theta4 plus l1 squared plus

l2 squared minus l3 squared plus l4 squared divided by twice l2l4. We write l1 by l4 as

a design parameter, D1 cosine theta2; l1by l2 as another design parameter, D2 cosine

theta4, and we write this quantity as another design parameter, which is D3. This equation

is called the Freudenstein's equation, where three design parameters are D1 is equal to

l1 by l4, another design parameter D2 is equal to l1by l2 and the third design parameter

D3 is l1 squared plus l2 squared minus l3 squared plus l4 squared whole divided by twice

l2l4. I can substitute. This equation has to be satisfied by these

three states of theta2 and theta4 values. I can substitute theta2 1 theta4 one, then

this equation will be satisfied; If I substitute theta2 2 and theta4 2 for theta2 and theta4,

then also this equation will be satisfied; I can also substitute theta2 3 and theta4

3 in these equations. So this way, I get three equations which are linear in three design

parameters D1, D2 and D3. The values of theta2 and theta4 are known, the only unknown in

these three equations are D1, D2 and D3 and this equation is linear. So, I can easily

solve for three unknowns, namely, D1, D2 and D3 by substituting these three pairs of values

of theta2 and theta4 in this equation to generate three equations. This is known as Freudenstein's

method of function generation for three position synthesis by a 4R-linkage.

Things to notice: From these three design parameters once we have solved I can get the

link length ratios, namely l1 by l4 from D1, l1 by l2 from D2 and substituting for l2 and

l4 in this equation from D3, I can get l1 by l3. So, the three link length ratios, namely

l1 by l2, l1 by l4 and l1 by l3 These three link length ratios can be obtained by solving

a set of three linear algebraic equations involving D1, D2 and D3. At this recall that

after solving this state of linear equations,D1 turns out to be negative, so stage, it may

be worthwhile to is D2. D2 may be positive or negative; D1 may be positive or negative.

What does it imply? The negative value of D1 and D2 imply that

this l2 and l4 has to be interpreted in the vector sense. That means if l2 transferred

to be negative, this theta2 is theta2 plus 5 such that l2 becomes negative.

Similarly, l4 has to be interpreted in the vector sense that D1 is negative then l4 is

negative which means theta4 will be changed by theta4 plus 5. l1 is always taken as positive

and l3 also as positive by solving l1 by l3 whole squared from this equation and taking

the positive root of this l1 by l3.

So to summarize, I can say that six-link length and the coupler length is always positive.

If D1 turns out to be negative, that means, l4 can be negative and if D2 turns out to

be negative, then l2 is negative. So, l4 and l2, that is, the input link length and the

output link length may be positive or negative. The negative link length has to be interpreted

in the vector sense, that is, we have to think of this O2A in this direction, that is, theta2

changes by theta2 plus 5. Exactly the same is for the output length. If l4 is negative,

then O4B will be in this direction, that is, theta4 has to be replaced by theta4 plus 5.

I would also like to say that sometimes, may be, the input link is on the left-hand side

and the output length is on the right-hand side. That means, the 4R-linkage

may be in this configuration: this is input link; this is coupler; this is the output

link; this is the fixed link. It is easy to see that in such a situation, I take X-axis

again along the fixed that is along O2O4 and Y-axis perpendicular to that. This is theta2

and this angle is theta4. So, we get Freudenstein's equation, corresponding to this 4R-linkage

exactly from this equation, only by replacing theta2 by theta4 and l4 by l2 and theta4 by

theta2 and l2 by l4. Here, of course, l2, l4 interchanging is not

changing this at all; only D1 and D2 will be changed; D3 remains the same. So, this

equation, cosine theta2 minus theta4 will be l1 by l2 cosine theta4 minus l1 by l4 cosine

theta2 plus D3. So we define, l1 by l2 is one design parameters and l1 by l4 as another

design parameter like the situation we faced here. So, that is the Freudenstein's equation

and this is the Freudenstein's method.

As explained just now, for three position synthesis, we write Freudenstein's equation

in this form: cos theta2 minus theta4 is D1 cos theta2 minus D2 cos theta4 plus D3; theta2

and theta4 takes three pair of coordinated values, namely, theta2 1, theta4 1, theta2

1, theta4 1, or theta2 2, theta4 2, theta2 2, theta4 2 and theta2 3, theta4 3, theta2

3 and theta4 3 that gives three sets of linear equations in D1, D2 and D3 to solve for. Once

we solve for D1, D2 and D3, we can get the three link length ratios. It may point it

out; it is the link length ratios that matter because, if the whole figure is killed up,

we get the same relative movement between various links. That is the linkage remains

same, it is only of the scale of the drawing. So l1, I can choose arbitrarily or conveniently

and then we can get the other three link lengths l2, l3 and l4. This equation can also be used

for generation of instantaneous kinematic relationship, not just position, but velocity

and acceleration as well. Instead of generating just position relationship between the input

and output link, we can use this equation for generation of instantaneous other kind

of kinematic relationship between the input and output links namely, the angular velocity

As an example, let us use this Freudenstein's equation to design a 4R-linkage, such that:

theta2 is 90 degree; theta4 is 60 degree; at this configuration, omega2 is 2 radians

per second and at this configuration, omega4 is 3 radians per second. At the same configuration,

that is theta2 equal to 90 degree and theta4 equal to 60 degree angular acceleration of

link 2 is 0, but angular acceleration of the output link is, say, minus1 radian per second

to satisfy these three conditions when theta2 is 90 degree, theta4 should be 60 degree and

this same configuration omega2 and omega4 are prescribed, alpha2 and alpha4 are also

prescribed? We have to determine the link length. Towards

this end, I start from the Freudenstein's equation for j is equal to 1, that is, cosine

theta2 minus theta4 is D1 cosine theta2 minus D2 cosine theta4 plus D3. Since, this equation

is valid for all instances what we can do, we can differentiate this equation with respect

to time and that gives me sine theta2 minus theta4 into omega2 minus omega4, where omega2

is theta2 dot, theta2 to D2 is the velocity of the input link, the omega4 is the angular

velocity of the output link and that is equal to D1 sine theta2 omega2 minus D2 sine theta4

omega4. Differentiating this equation once more with

respect to time we get sine theta2 minus theta4 into alpha2 minus alpha4 plus cosine theta2

minus theta4 into omega2 minus omega4 whole squared is equal to, if I differentiate the

right hand side with respect to time, we get D1 sine theta2 alpha2 plus cosine theta2 omega2

squared minus D2 sine theta4 alpha4 plus cosine theta4 omega4 squared. So I get three equations,

namely, 1, 2 and 3. In these three equations, I substitute the desired conditions which

are given here.

If I substitute in the first equation, theta2 equal to 90 degree and theta4 equal to 60

degree, we get 90 minus 60 that is cos 30 degree that is root 3 divided by 2 is equal

to D1 cos theta2 is zero minus D2 cos 60 degree is half minus D2 by 2 plus D3. This is what

I get from the first equation, after substituting theta2 and theta4. If I substitute omega2,

omega4, theta2, theta4 in the second equation, we get sine 30 degree that is 1/2 into omega2

minus omega4 that is minus1 should be equal to D1, sine theta2 is 1 and omega2 is 2, so

2D1 minus D2 (sine theta4 is root 3 by 2 and omega4 is 3) so, 3 root 3 divided by 2. This

is equation one. This is equation two. In the third equation, I substitute all these

values to get sine 30 degree is half, alpha2 minus alpha4 is 1 plus cos 30 degree is root

3 by 2 into omega2 minus omega4 is minus 1 squared is 1 that is, 1 is equal to alpha2

is 0, so that term is zero; cos theta2 is 0, so this term is also 0; alpha2 is 0 and

cos theta2 two is 0,so this term is 0. So, we get minus D2 sin theta4 is root 3 divided

by 2 and alpha4 is minus 1 and that gives me minus root 3 divided by 2; cos theta4 is

1 by 2 and omega4 square is 9 so 9 by 2. This is equation three. From the third equation,

I can solve for D2. After I solve for D2, from the second equation, I can solve for

D1 and if D2 is known, I can solve for D3 from the first equation.

If we do the algebra, we will get for this particular set, D1 is minus 0.738, D2 is minus

0.376 and D3 is 0.678. As we said earlier, that D1, D2 can come out to be negative and

this particular problem D1 and D2 are turning out to be negative, which means l1 by l4 is

minus 0.738, l1 by l2 is minus 0.376 and from D3 if I substitute these values of l4 and

l2 in terms of l1, we can find out l1 by l3, which turns out to be equal to 0.446. So,

we get all three link lengths ratio and as we said, l1l3 are always positive. For this

particular problem, l2 and l4 are turning out to be negative. So, if we choose the length

l1, we can solve for l2, l4 and l3. If we draw this mechanism at this particular

configuration we see that this is O4 and this is O2 theta4 was 60 degree, but l4 has turned

out to be negative. So, I have to draw it at 240 degree. This is l4 which is negative;

l2 is longer than l4, but l2 is also negative and theta2 is 90 degree that is this way;

so, I draw it this way. All the prescribed instance relationship will be valued for this

4R-linkage where, this is l2; this is l4; this is l3; for a given value of l1, which

I can solve once I know these three values. This is what we mean by Freudenstein's method

We discussed the function generation problem with reference to 4R-linkage function generator

for three position synthesis but we emphasize that we got linear equations in the design

variables namely, D1, D2 and D3, only because we assumed theta2 one and theta4 one. If we

have to extend this method to four position syntheses, then beyond D1, D2 and D3, we have

to leave one of these parameters also as an unknown to be decided or to be determined.

That means D1, D2, D3 and either theta2 one or theta4 one have to be left undetermined

and then we can plug-in the four sets values of theta2 and theta4 to solve for these four

unknowns D1, D2, D3 and whichever is left unknown, but the trouble with that is that

the equations will not be linear because of this angular term coming into the equation.

We have cosine and things like that non-linear functions of these angles.

Consequently, the algebraic difficulty increases. Though it can be done, but we are keeping

it outside the scope of this course. So, we have discussed Freudenstein's method, for

function generation with reference to three position synthesis or two pair of coordinated

moments. Things to remember is that, after solution input or the output length l2 and

l4 may turn out to be negative. They have to be interpreted in the vector sense because

we are assuming this theta2 one and theta4 4 one arbitrarily. There cannot be any solution

with those values if l2 and l4 turn out to be negative, but the solution will be available

if we had pi to these terms. That means l2 instead of O2. This is what I call theta2

one; l2 negative means theta2 one, I have to take this angle, which is pi plus theta2

one. In our next lecture, we shall discuss how

the same methodology can be used for function generation by a slider-crank, that is, the

input-output correlation and the slider displacement to rotation of the crank for a slider-crank

mechanics.

The Description of Module 8 Lecture 1 Kinematics Of Machines

### Channels for learning English

• English Lessons with Adam - Learn English [engVid]
• EnglishAnyone
• VideoSparkNotes
• Miracle English Language & Literature Institute
• Flocabulary
• Cool School
• TheGrammarheads
• English Singsing