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The topic of today's lecture is number synthesis.
During this stage of kinematic synthesis called
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number synthesis, we determine the type and
number of different types of links and the
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number of simple pairs like revolute or prismatic
pairs that needed to yield a single degree
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of freedom planar linkage. It is needless
to say that all the single degree of freedom
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planar linkages will satisfy the Grubler's
criterions which are discussed earlier. However,
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before we get into the discussion or details
of number synthesis we shall first prove certain
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basic results, which are of vital important
for number synthesis. The first of these two
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questions is what is the minimum number of
binary links that such a linkage must posses?
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So, we determine the minimum number of binary
links in a single degree of a freedom planar
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linkage. Let n be the total number of links
in the linkage, n2 be the number of binary
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links, n3 be the number of ternary links and
n4 be the number of quaternary links and so
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on. Thus, we have the total number of links
in is equal to n2 plus n3 plus n4 upto ni
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where i denotes the highest order link that
is present in this linkage. Our first task
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is to determine the minimum value of n2. Towards
this goal let us consider this figure.
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In this figure, we see there is one link which
is connected to two other links through the
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revolute pairs here and here. To note that
at each of these revolute pairs we have two
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elements say 1 minus which is the spin which
goes into the hole which is denoted by one
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plus. So this 1 plus and 1 minus we shall
call elements, thus at each revolute pair
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we have two elements. Similarly the two elements
at these revolute pairs are these 2 plus and
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2 minus.
In this way, if we count the total number
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of elements that I can write e should be equal
to twice the number of joints or pairs say
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that is e equal to 2j. We can also count this
number of elements from this links. This is
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a binary link which has two elements because
it is connected to two other links, two revolute
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pairs. Similarly, a ternary link, we will
have three elements because it is connected
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to three other links and a quaternary link
we will have four such elements. So if we
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count the total number of elements from the
view point of links then I can write e equal
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to 2n2 plus 3n3 plus 4n4 plus in, where n2
is the number of binary links, n3 is the number
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of ternary links, n4 is the number of quaternary
links and ni is the number of ith order link.
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We have just now seen that the total number
of elements can be counted from two view points.
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If we count it from the view point of number
of pairs then I can write the total number
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of elements e equal to 2j. But, if we count
the number of elements from the view point
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of different links or of different orders
then we can write the total number of elements
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e equal to 2n2 plus 3n3 plus 4n4 plus ini.
We can equate these two numbers of elements
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counted from the view points the kinematic
pairs and from the view point of different
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order links we can write 2j equal to 2n2 plus
3n3 plus 4n4 plus ini.
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As I already told, that all these linkages
must satisfy the Grubler's criterion which
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is 2j minus 3n plus 4 equal to 0, where j
denotes the number of kinematic pairs and
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n denotes the number of total links. Substituting
e equal to 2j which we have just now saying
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to be given by 2n2 plus 3n3 plus 4n4 plus
ini minus, we replace this n by the number
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of counts of different links which is n2 plus
n3 plus n4 plus ni. Simplifying this equation
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we can see the 3 and 3 cancels and ultimately
we get n2 is equal to p minus 3 summed over
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all values of p starting from 4 up to i plus
4. That means n2 is given by p minus 3, p
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going from 4 to i plus 4. That is, this p
denotes the number of quaternary links and
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higher order links.
So we can easily see if the sum is 0, the
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minimum number of binary links (n2)min equal
to 4. This again convinces us what we have
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seen earlier, that this simplest linkage must
have 4 binary links what we call four bar
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linkage.
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Next we would like to add another question
that is what is the highest order link in
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an n link mechanism? That means, the total
number of links is n then in such a linkage
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what is the highest order link? We should
try to answer this question in a reverse manner.
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We will say, the highest order link, be ith
order that is we have some ni's. Then what
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is the minimum number of links that is needed
to produce the single degree of freedom planar
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linkage. Towards this goal let me consider
the following figure.
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In this figure, we start with a link with
i hinges. This is the link which has i hinges
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numbered as 1, 2, 3, 4 so on up to i. To produce
a at each of these hinges we connect another.
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At the 1st hinge we connect link number one,
at the second hinge we connect link number
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2, at the third hinge we connect link number
3 and so on this ith link at the hinge number
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i. To connect these two links 1 and 2, we
must have some motion transfer links, accordingly,
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(i plus 1), (i plus 2) and (i plus 3) so on
up to (2i minus 1). The thing to note that,
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the hinge number 2, 3, 4 up to (i minus 1),
we have ternary links. Because link number
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2 has three hinges here, here and here and
that is true for all other links connected
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at hinge number 4, hinge number 4 and so on.
Because, if we have binary link at 2 then
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this particular hinge will not remain because
three links namely (i plus 2) and link number
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2 will get connected at this higher order
hinge. Then hinge as a simple hinge all these
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links starting from number 2, 3, 4 and so
on up to (i minus 1) must ternary link. So
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we have produced a close chain minimum number
of links if we start from a link with of ith
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order that is with i hinge.
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Let me count total number of links n if we
have started with, I have already shown the
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number up to (2i minus 1) and we count this
starting link which is having i hinges so
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the total number links is 2i. Thus, we see
that this is an ith order hinge then minimum
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I need 2i number of links to produce a closed
chain. That means, total number of links is
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n then imax can go up to n by 2 and not more
than n by 2. I emphasize that is the possible
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value of imax, not necessarily imax has to
be n by 2, definitely it cannot be more than
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n by 2.
Next thing we have to prove, that this closed
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chain from this closed chain if I hold one
link fixed it must produce a single degree
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freedom mechanism that is this particular
closed chain must satisfy our old Grubler's
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criterion. For that we count n equal to 2i.
Let me count the maximum hinges j, we have
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started with i hinges on this initial link
so j equal to i plus there is one hinge here
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and there is another hinge here which is at
two plus on all other links two, three, four
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there are two external because these are all
ordinary links one of the hinge has been already
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count with this starting link. There are two
hinges extra hinges on each of it so that
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into i, how many such linkages? We have starting
from two to (i minus1) that is 2 times (i
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minus 3). So that is the number of hinge i
plus 2 plus2 times (i minus3) which will give
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us (3i minus 4).
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We see that in this closed chain, total number
of links n turns out to be 2i, where i denote
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the highest order link in this chain. The
total number of joins j turns out to be 3i
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minus2. If we write the Grubler's criterion
that is 2j minus 3n plus 4 we get 2 times
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(3i minus 2) minus 3 times (2i plus 4) equal
to 0. Thus the Grubler's criterion is satisfied
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by this closed finite chain and consequently
this can constitute a single degree of freedom
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planar linkage. So we concentrate on these
two results that we have just now derived.
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One is that the minimum number of binary links
in a linkage must be flow and the second is
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that is highest order link n link mechanism
that is imax is n by 2. Since, all the single
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degree freedom linkage must satisfy the Grubler's
criterion that 2j minus 3n plus 4 equal to
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0 that gives 3n equal to 2j plus 4. We note
that the right hand side 2j plus 4 is an even
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number and if 3n is equal to an even number
then the n must be even, which means all the
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planar linkages simple pairs and single degree
of freedom must have even number of links.
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We have already seen that the four-link mechanism
is the simplest mechanism. The next more complicated
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mechanism should be n equal to 6 that is a
six-link mechanism. If the kinematic requirements
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are little more complex, which cannot be satisfied
by a four-link mechanism then we have to try
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to use a six-link mechanism. Let me go into
this number synthesis of six-link mechanism.
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With a six-link chain we have n equal to 6
that is imax is n by 2 that is 3. The highest
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possible order is a ternary link. So a six-link
mechanism constitutes a binary links and ternary
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links. So the total number of link n equal
to n2 plus n3 equal to 6 where n2 is the number
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of binary links, n3 is the number of ternary
links. For n equal to 6, we know to satisfy
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Grubler's criterion 2j must be equal to 3n
minus 4 equal to 3 times 6 minus 4 equal to14
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that is j equal to 7. We have got one equation,
numbered equation one number in terms of two
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unknown in terms of n2 and n3. We derived
another equation involving n2 and n3 by counting
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the number of elements.
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The number of elements e equal to 2j which
is also given by 2n2 plus 3n3. Thus 2n2 plus
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3n3 equal to 2j where j is equal to 7 this
comes out to be 14. This is the second equation
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involving these two unknowns namely n2 and
n3. Our previous equation was n2 plus n3 equal
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to 6 and the second equation is 2n2 plus 3n3
equal to 14.
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We can easily solve these two linear equations
in two unknown namely n2 and n3 as n2 equal
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to 4 and n3 equal to 2. Thus, a six-link mechanism
has four binary links and 2 ternary links.
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We shall see what are the possible combinations
of these binary and ternary links to generate
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different types of six-link mechanisms?
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This figure shows one possible six-link chain
with two ternary links and four binary links.
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As we see, the link number 1 is a ternary
link, link number 4 is another ternary link
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whereas link number 2, 3, 5 and 6 are all
binary links. The thing to note, that in this
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chain there are six-link and seven revolute
pairs, we can count at vertices of this hexagon
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and one inside the hexagon. Another thing
to note that here the two parallel links 1
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and 4 are directly connected by this revolute
pair and all the four binary links are connected
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to the ternary links, this chain is known
as Watt's-chain.
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So in a Watt's-chain, two ternary links are
directly connected to each other. In this
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Watt's-chain, we can see that the two ternary
links that is number 1 and 4 are equivalent.
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In the sense, both of them are connected to
a ternary link at one kinematic pair and two
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binary pair at the other two revolute pairs
like, 4 is connected to link number four by
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a revolute pair, to the binary link 4 is connected
to the another binary link 3 at this revolute
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pair and is connected to the ternary link
1 at this revolute pair. Exactly the same
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thing happened for the link number one, it
this connected to the ternary link 6 and this
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revolute pair binary link 2 to this revolute
pair and to ternary link 4 by this revolute
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pair.
Thus to topologically there is no difference
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between link number 1 and 4. The same is true
for all binary links namely 2, 3, 5 and 6
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each one of which is connected to a ternary
link at one end and to a binary link at the
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other end. For example, link number 2 is connected
to a ternary link at one end and to a binary
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link at the other end and the same is true
for all other binary links.
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Thus there are two types of links ternary
links and binary links but both the ternary
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links are equivalent and all the four binary
links are also equivalent. So from a Watt's-chain
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by kinematic inversion that is depending on
which link we hold fixed we can get two different
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types of Watt's mechanism. One type of Watt's
mechanism we can get by holding binary links
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fixed with 1, 2, 3 or 5 and 6, because all
of them are equivalent and the second type
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of Watt's mechanism we can get holding one
of the binary links that is either one or
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four are fixed.
We will now show a model of a six-link Watt's
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mechanism where we will find that one of the
binary links is held fixed.
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As an example of Watt's mechanism a with binary
link fixed let us go back to our old example
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of this parallel jaw player. We hold this
lower jaw that is this blue link fixed, this
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is a binary link because it has two revolute
pairs. Let us note that, this binary link
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is connected to another binary link at this
revolute pair and to this ternary link at
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this revolute pair. This lower jaw is a ternary
link because it has three revolute pairs and
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this small link is another ternary link which
has three revolute pair and these two ternary
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links are directly connected so it is one
type of Watt's chain where we know two ternary
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links must be directly connected. If we hold
this lower jaw fixed then we are holding this
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binary link fixed. We should also know that
this upper jaw is a binary link and this below
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link is another binary link. This binary link
is connecting this ternary link and this binary
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link. As a result of this we get a Watt's
mechanism by Watt's-chain. This is Watt's
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mechanism of one kind. Later on we will see
Watt's mechanism of another link where ternary
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link will be held fixed.
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An example of another type of possible Watt's
mechanism let us consider this figure. Here
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we started from a Watt's chain which has two
planar links 1 and 4 and 4 binary links 2,
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3, 5 and 6. Here, is a binary link 1 which
is held fixed this is known as Watt's walking
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beam engine. In this Watt's walking beam engine
we must see that in the chain we have shown
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revolute pair between 1 and 6 which has been
replaced by a prismatic pair between the cylinder
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and the piston. But in our analysis, we always
treated revolute pair and prismatic pair as
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value equivalent. So here, as we see that
this great beam that is link 4 is connected
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to ternary link directly by the revolute pair.
This is another type of Watt's mechanism which
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is possible to get by kinematic inversion
from a Watt's chain.
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An another example of a six-link mechanism
with seven hinges we can get the following
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figure.
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Here, as we see two ternary links namely 1
and 4. However, unlike in a Watt's-chain these
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two ternary links are not directly connected
to each other rather they are connected via
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this binary link number 6. Here, we have two
ternary links 1 and 4 which are connected
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by a binary link 6, binary link 5 and by two
binary links namely 2 and 3 and these particular
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chain where the two ternary links are not
directly connected is known as Stephenson's
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chain. We see that in a Stephenson's chain
two ternary links are not directly connected.
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In a Stephenson's chains the ternary links
1 and 4 are equivalent in a sense that both
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1 and 4 are connected to three binary links
and three revolute pairs. For example, one
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is connected to binary link 6, binary link
5 and binary link 2 at these three revolute
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pairs and link number 4, the other ternary
links is also connected to three binary links
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to link number 5 here, link number 6 here
and number 3 here. Thus both these ternary
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links are topologically equivalent because
both of them are connected to three binary
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links.
However, so far the binary links are concerned
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there are two varieties, namely 5 and 6 and
2 and 3. We should note that both 5 and 6
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are connected to two ternary links at two
joints, 6 is also connected to two ternary
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links at two joins. Link number 2 and 3 at
one end is connected to a ternary link but
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at the other end is connected to a binary
link. So there are two types of binary links
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they can be grouped as (5, 6) and (2, 3).
By kinematic inversion we can get three different
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types of Stephenson's mechanism depending
on whether ternary links 1 or 4 is held fixed
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or one of the binary links in this group that
is either 5 or 6 is held fixed or one of this
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group namely 2 and 3 that is either 2 or 3
are held fixed. There are three different
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types of Stephenson's mechanism which can
be obtained by kinematic inversion from the
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same Stephenson's chain. We now see a model
of a Stephenson's chain to generate a Stephenson's
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linkage where a ternary link is held fixed.
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Let us now look at the model of this Stephenson's
mechanism where one of the ternary links is
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held fixed. Here we have this fixed link as
the ternary link which has three hinges one
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here, one there and another there and this
is the other ternary link which is connected
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to the fixed link by two binary links. The
two ternary links are not directly connected,
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they are connected via binary links at these
two points and by two binary links at this
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point, this is a binary link this is a binary
link. This binary links are equivalent because
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at one end this binary link is connected to
a ternary link, at this end this binary link
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is connected to another binary link. Similarly,
this binary link is connected to a ternary
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link at this end and at to a binary link at
this end. These two binary links are of same
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nature. Similarly these two binary links are
also of same nature because they are connected
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at both ends to ternary links. One of the
ternary links is held fixed and we get one
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variety of a Stephenson's linkage.
Another example of Stephenson's linkage let
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us consider the same Stephenson's chain and
consider one of the binary links to be fixed.
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As shown in this figure, this is the Stephenson's
chain that we have considered earlier and
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in this chain if we hold a binary link say
link number 2 fixed then we get this mechanism.
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As we see, 2 is connected to ternary link
1 and link 2 is the fixed link which is connected
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to a binary link 3 and to a ternary link 1.
Link number 4 is the ternary link which is
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connected to link 3 here, link 6 here and
link 5 here. This is known as Stephenson's
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valve gear mechanism which is used in a steam
engine. We have seen, a six-link chains consists
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of four binary links two ternary links and
various combinations which are possible as
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Watt's linkage or Stephenson's linkage.
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Let us consider the next higher order link
possible to give you the flavor
of number synthesis. The next most complicated
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mechanism we consider with n event is an eight-link
mechanism that is n equal to 8. Consequently
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highest order link possible in an eight-link
mechanism that is imax equal to n by 2 equal
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to 4. An eight-link mechanism we will have
binary link it is possible to have ternary
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link and it is possible to have quaternary
link.
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If the number of binary links is n2, the number
of ternary links is n3 and the number of quaternary
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links is n4 then the total number of links
n which is equal to n2 plus n3 plus n4 equal
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to 8. This is our first equation to determine
n2, n3 and three and n4. We also know that
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the number of elements e equal to 2j equal
to 3n minus 4 so that Grubler's criterion
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is satisfied, which means 3 times 8 minus
4 equal to 20. Counting the number of elements
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from the view point of links we can write
2n2 plus 3n3 plus 4n4 equal to 20 because
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this is also equal to the number of elements
so this is our second equation two.
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It may now appear that we have three unknowns,
n2, n3 and n4 to determine but we have only
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two equations namely 1 and 2. Thus there may
be infinite solutions a little thought would
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convince us that is not the situation we still
have finite number of solutions because we
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should remember all these numbers n2, n3 and
n4 are integers not only that the minimum
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value of n2 is also 4. Number n2 can start
from 4 then can go up to 5, 6 and so on.
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Let us see what are the various solutions
possible to these two equations that under
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such restrictions that all these numbers n2,
n3 and n4 must be positive integers there
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is no point having a negative number for the
number of links and also that minimum values
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of n2 is 4.
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For an eight-link mechanism, we have got two
equations namely n2 plus n3 plus n4 equal
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to 8 and 2n2 plus 3n3 plus 4n4 equal to 20.
If we assume that the values of n2 is 4 then
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from these two equations we get n3 plus n4
equal to 8 minus n2 equal to 8 minus 4 equal
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to 4 and from the second equation we get 3n3
plus 4n4 equal to 20 minus 2n2 that is 20
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minus 2 times 4 equal to 12. We get these
two equations to solve for n3 and n4 and the
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obvious solution is n3 equal to 4 and n4 equal
to 0. That means we can get an eight-link
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mechanism consisting of four binary links
and four ternary links. There is no necessity
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that we must have a quaternary link.
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If we take n2 equal to 5 then we get n3 plus
n4 equal to 8 minus 5 equal to 3 and 3n3 plus
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4 n4 equal to 20 minus 2 times 5 equal to
10. Solving these two equations we get n3
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equal to 2 and n4 equal to 1. Thus, we can
also have an eight-link mechanism
with 5 binary links with 2 ternary links and
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1 quaternary links. Similarly, if we take
n2 equal to 6, one can easily find that will
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get n3 equal to 0 and n4 equal to 2. That
means, we can have an eight-link mechanism
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with 6 binary links and 2 quaternary links.
From these three different types of eight
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link chains by kinematic inversions one can
get a very large number of different mechanisms.
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In conclusion, let me now repeat the foremost
important points that we have learnt today
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during this discussion of number synthesis
of planar linkages. The first point is that,
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the minimum number of binary links in any
such linkage must be four: that means, we
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must have at least four binary links. The
second point is that the highest order link
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in an nth-link mechanism is n by 2, that is
in a six-link mechanism the highest order
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is ternary in an eight-link mechanism the
highest order is quaternary. Third thing we
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have seen that, the total number of links
must be even and the last point is that with
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increase in the number of total links the
possible types of various mechanisms that
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we can have some such chains by kinematic
inversion increases drastically.
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