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The topic of todayís 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 this single degree of freedom
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planar linkages will satisfy the Grublerís
criterions which were discussed earlier. However,
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before we get into the discussion or details
of number synthesis, we shall first prove
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certain basic results, which are of vital
important for number synthesis. The first
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of these two questions are, what is the minimum
number of binary links that such a linkage
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must possess?
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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 ëní be the total number of
links in the linkage, ën2í be the number
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of binary links, ën3í be the number of ternary
links and ën4í be the number of quaternary
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links and so on. Thus we have,
the total number of links, n = n2 + n3 + n4
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+ Ö. + ni
where i denotes the highest order link that
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is present in this linkage. So, our first
task is to determine the minimum value of
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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 1ñ which is the pin which goes
into the hole which is denoted by 1+. So this
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1+ and 1ñ, we shall call elements, thus at
each revolute pair we have two elements. Similarly,
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the two elements at this revolute pair are
2+ and 2ñ.
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In this way, if we connect count the total
number of elements that I can write ëeí
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should be equal to twice the number of joints
or pairs say that is e = 2j. We can also count
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this number of elements from this links. This
is a binary link which has two elements because
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it is connected to two other links, two revolute
pairs. Similarly, a ternary link, we have
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three elements because it is connected to
three other links and a quaternary link we
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will have four such elements. So if we count
the total number of elements from the view
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point of links then I can write, e = 2n2 (where
n2 is the number of binary links) + 3n3 (where
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n3 is the number of ternary links) + 4n4 (where
n4 is the number of quaternary links) + Ö.
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+ ini (where ni is the number of ith order
link). So we have just now seen that the total
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number of elements can be counted from two
viewpoints.
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If we count it from the viewpoint of number
of pairs then I can write the total number
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of elements e = 2j. But, if we count the number
of elements from the viewpoint of different
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links or of different orders then we can write
the total number of elements e = 2n2 + 3n3
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+ 4n4 + Ö. + ini. We can equate these two
numbers of elements counted from the viewpoints
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the kinematic pairs and from the viewpoint
of different order links, we can write,
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2j = 2n2 + 3n3 + 4n4 +Ö. + ini.
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As already told, that all these linkages must
satisfy the Grublerís criterion which is,
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2j ñ 3n + 4 = 0, where ëjí denotes the
number of kinematic pairs and ëní denotes
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the number of total links. Now substituting
e = 2j, which is just now saying to be given
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by,
2n2 + 3n3 + 4n4 + Ö. + ini ñ 3(n2 + n3 +
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n4 + Ö. + ni) + 4 = 0
Simplifying this equation, we get, n2 = summationip=4
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(p ñ 3) + 4
where ëpí denotes the number of quaternary
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links and higher order links.
So we can easily see if this sum is 0, the
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minimum number of binary links, (n2)min = 4.
This again convinces us what we have seen
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earlier, that this simplest linkage must have
4 binary links, what we call a four-bar 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 ëní link mechanism? That means, the total
number of links is ëní 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 niís. 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. Now
to produce a closed chain, at each of these
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hinges we connect another link. Like at the
1st hinge we connect link 1, at the second
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hinge we connect link 2, at the third hinge
we connect link 3 and so on this ith link
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at the hinge i. To connect these two links
1 and 2, we must have some motion transfer
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links, accordingly this, (i + 1), (i + 2)
and (i + 3) so on up to (2i ñ 1). The thing
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to note that, at the hinge number 2, 3, 4
up to (i ñ 1), we have ternary links. Because
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link 2 has three hinges and that is true for
all other links connected at hinge 3, hinge
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4 and so on. Because, if we have a binary
link at 2 then this particular hinge will
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not remain a simple hinge, because three links
namely (i + 2) and another link number 2 will
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get connected at this higher order hinge.
Then with all hinges as simple hinges all
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these links starting from number 2, 3, 4 and
so on up to (i ñ 1) must be ternary links.
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So, we have produced a close chain with minimum
number of links if we start from a link with
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of ith order that is with i hinges.
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Now let me count the total number of links
n if we have started with a link with i hinges,
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I have already shown the number up to (2i
ñ 1) and we count this starting link which
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is having i hinges so the total number links
is 2i. Thus, we see that if this is an ith
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order hinge then minimum I need 2i number
of links to produce a closed chain. That means,
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if the total number of links is n then imax
can go up to n/2 and not more than n/2. I
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emphasize that is the possible value of imax,
not necessarily imax has to be n/2, definitely
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it cannot be more than n/2.
Next thing we have to prove, that from this
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closed chain if I hold one link fixed it must
produce a single degree freedom mechanism
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that is, this particular closed chain must
satisfy our old Grublerís criterion. For
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that we count n = 2i. Let me count the maximum
hinges j, we have started with i hinges on
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this initial link so j is equal to i plus
there is one hinge here and there is another
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hinge here which are thus two plus on all
other links two, three, four there are two
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extra hinges because these are all ternary
links one of the hinge has been already counted
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with this starting link. There are two extra
hinges on each of it so that is 2.i, How many
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such links? We have starting from 2 to (i
ñ 1) that is 2(i ñ 3). So that is the number
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of hinges j = i + 2 + 2(i ñ 3) = (3i ñ 4).
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So we see that in this closed chain, the total
number of links n turns out to be 2i, where
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i denote the highest order link in this chain.
The total number of joints j = 3i ñ 2. If
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we write the Grublerís criterion that is,
2j ñ 3n + 4 = 2(3i ñ 2) ñ 3*2i + 4 = 0.
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Thus, the Grublerís criterion is satisfied
by this closed finite chain and consequently
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this can constitute a single degree of freedom
planar linkage. So we concentrate on these
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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 four and the second is
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that is highest order link in a n-link mechanism
that is imax is n/2. Since, all the single
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degree freedom linkage must satisfy the Grublerís
criterion that,
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2j ñ 3n + 4 = 0 => 3n = 2j + 4
Now we note that the right-hand side 2j +
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4 is an even number and if 3n is equal to
an even number then the n must be even, which
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means all the planar linkages simple pairs
and single degree of freedom must have even
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number of links. We have already seen that
the four-link mechanism is the simplest mechanism.
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So the next more complicated mechanism should
be n equal to 6 that is a six-link mechanism.
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If the kinematic requirements are little more
complex, which cannot be satisfied by a four-link
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mechanism then we have to try to use a six-link
mechanism. Now let me go into this number
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synthesis of six-link mechanism.
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With a six-link chain we have n = 6 that is,
imax = n/2 = 3. So the highest possible order
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is a ternary link. So a six-link mechanism
constitutes a binary links and ternary links.
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So the total number of link n equal to n2
(that is the number of binary links) plus
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n3 (that is the number of ternary links) is
equal to 6. For n = 6, we know to satisfy
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Grublerís criterion, 2j = 3n ñ 4 that is
3.6 ñ 4 = 14, that is j = 7.
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We have got one equation, numbered equation
one number in terms of two unknown in terms
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of n2 and n3. So we derived another equation
involving n2 and n3 by counting the number
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of elements.
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The number of elements e given by 2j which
is also given by 2n2 + 3n3. Thus, 2n2 + 3n3
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=2j, where j = 7 this turns out to be 14.
This is the second equation involving these
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two unknowns namely n2 and n3. Our previous
equation was n2 + n3 = 6 and the second equation
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is 2n2 + 3n3 = 14.
We can easily solve these two linear equations
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in two unknowns namely n2 and n3 as n2 = 4
and n3 = 2. Thus, a six-link mechanism has
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four binary links and 2 ternary links. Now
we shall see what are the possible combinations
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of these binary and ternary links to generate
different types of six-link mechanisms?
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Above 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, six revolute pairs we can count at
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vertices of this hexagon and one inside the
hexagon. Another thing to note that here the
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two ternary links 1 and 4 are directly connected
by revolute pair and all the four binary links
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are connected to the ternary links. This chain
is known as Wattís chain.
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So in a Wattís chain, two ternary links are
directly connected to each other. In this
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Wattís-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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two binary pair at the other two revolute
pairs. Like 4 is connected to link number
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4 by a revolute pair, to the binary link 4
is connected to another binary link 3 at revolute
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pair and is connected to the ternary link
1 at revolute pair. Exactly same thing happened
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for the link number 1, it is connected to
the ternary link 6 to this revolute pair and
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binary link 2 to this revolute pair and to
ternary link 4 by this revolute pair.
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Thus topologically there is no difference
between link number 1 and 4. The same is true
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for all binary links namely 2, 3, 5 and 6
each one of which is connected to a ternary
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link at one end and to a binary link at the
other end. For example, link number 2 is connected
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to a ternary link at one end and to a binary
link at the other end and the same is true
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for all other binary links.
Thus, there are two types of links ternary
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links and binary links but both the ternary
links are equivalent and all the four binary
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links are also equivalent. So from a Wattís-chain
by kinematic inversion that is depending on
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which link we hold fixed we can get two different
types of Wattís mechanism. One type of Wattís
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mechanism we can get by holding binary links
fixed with 1, 2, 3 or 5 and 6, because all
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of them are equivalent and the second type
of Wattís mechanism we can get by holding
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one of the ternary links that is either one
or four are fixed.
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We will now show a model of a six-link Wattís
mechanism where we will find, that one of
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the binary links is held fixed.
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As an example of Wattís mechanism with a
binary link fixed let us go back to our old
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example of this parallel jaw plier. We hold
this lower jaw that is this blue link fixed,
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this is a binary link because it has two revolute
pairs. Now 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 pairs and these two ternary
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links are directly connected. So it is one
type of Wattís 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 is binary link, this upper jaw is a binary
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link and this below link is another binary
link. This binary link is connecting this
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ternary link and this binary link. As a result
of this we get a Wattís mechanism by Wattís-chain.
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This is Wattís mechanism of one kind. Later
on we will see Wattís mechanism of another
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link where ternary link will be held fixed.
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An example of another type of possible Wattís
mechanism let us consider above figure. Here
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as we started from a Wattís chain which has
two ternary links 1 and 4 and 4 binary links
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2, 3, 5 and 6. Here, is a binary link 1 which
is held fixed and this is known as Wattís
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walking beam engine. In this Wattís walking
beam engine, we must see that in the chain
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we have shown revolute pair between 1 and
6 which has been replaced by a prismatic pair
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between the cylinder and the piston. But in
our analysis, we always treated revolute pair
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and prismatic pair as equivalent. So here,
as we see that this great beam that is link
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4 is connected to ternary link directly by
this revolute pair. This is another type of
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Wattís mechanism which is possible to get
by kinematic inversion from a Wattís chain.
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Another example of a six-link mechanism with
seven hinges we can get the following figure.
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Here, as we see there are two ternary links
namely 1 and 4. However, unlike in a Wattís-chain
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these two ternary links are not directly connected
to each other rather they are connected via
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binary link number 6. Here, we have two ternary
links 1 and 4 which are connected by a binary
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link 6, binary link 5 and by two binary links
namely 2 and 3 and this particular chain where
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the two ternary links are not directly connected
is known as Stephensonís chain. So we see
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that in a Stephensonís chain two ternary
links are not directly connected. In a Stephensonís
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chain the ternary links 1 and 4 are equivalent
in a sense that both 1 and 4 are connected
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to three binary links at three revolute pairs.
For example, link 1 is connected to binary
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link 6, binary link 5 and binary link 2 at
these three revolute pairs and link number
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4, the other ternary links is also connected
to three binary links to link number 5 here,
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link number 6 here and number 3 here. Thus
both these ternary links are topologically
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equivalent because both of them are connected
to three binary links.
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However, so far the binary links are concerned
there are two varieties, namely (5 and 6)
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and (2 and 3). We should note that both 5
and 6 are connected to two ternary links at
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two joints, link 6 is also connected to two
ternary links at two joints. However link
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number 2 and 3 at one end is connected to
a ternary link but at the other end it is
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connected to a binary link. So there are two
types of binary links they can be grouped
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as (5, 6) and (2, 3). So by kinematic inversion
we can get three different types of Stephensonís
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mechanism depending on whether the ternary
link 1 or 4 is held fixed or one of the binary
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links in this group that is either 5 or 6
is held fixed or one of the other group namely
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2 and 3 that is either 2 or 3 is held fixed.
So, there are three different types of Stephensonís
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mechanism which can be obtained by kinematic
inversion from the same Stephensonís chain.
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We now see a model of a Stephensonís chain
to generate a Stephensonís linkage where
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a ternary link is held fixed.
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Let us now look at the model of this Stephensonís
mechanism where one of the ternary link 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. So
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. These 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. So these two binary links are of
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same 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. So one of the
ternary links is held fixed and we get one
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variety of a Stephensonís linkage.
Another example of Stephensonís linkage let
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us consider the same Stephensonís chain and
consider one of the binary links to be fixed.
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As shown in above figure, this is the Stephensonís
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 other ternary link which
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is connected to link 3 here, link 6 here and
link 5 here. This is known as Stephensonís
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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 and two ternary links
and various combinations which are possible
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as Wattís linkage or Stephensonís linkage.
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Let us consider the next higher order link
possible to give you the flavour of number
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synthesis. Obviously the next most complicated
mechanism we consider with ëní even is an
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eight-link mechanism that is n = 8. Consequently,
highest order link possible in an eight-link
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mechanism that is imax = n/2 = 4. So an eight-link
mechanism we will have binary links, it is
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possible to have ternary 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 = n2 + n3 + n4 = 8. This is our first equation
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to determine n2, n3 and n4. We also know that
the number of elements, e = 2j = 3n ñ 4,
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so that Grublerís criterion is satisfied,
which means 3*8 ñ 4 = 20. Counting the number
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of elements from the viewpoint of links, we
can write 2n2 + 3n3 + 4n4 = 20, because this
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is also equal to the number of elements. So,
this is our second equation 2.
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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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So for an eight-link mechanism, we have got
two equations namely,
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n2 + n3 + n4 = 8 and 2n2 + 3n3 + 4n4 = 20.
If we assume that the value of n2 = 4 then
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from these two equations we get, n3 + n4 = 4
and from the second equation we get 3n3 +
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4n4 = 12. Now we get these two equations to
solve for n3 and n4 and the obvious solution
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is n3 = 4 and n4 = 0. That means we can get
an eight-link mechanism consisting of 4 binary
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links and 4 ternary links. There is no necessity
that we must have a quaternary link.
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However if we consider n2 = 5 then we get
n3 + n4 = 3 and 3n3 + 4n4 = 10. Solving these
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two equations, we get n3 = 2 and n4 = 1. Thus,
we can also have an eight-link mechanism
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with 5 binary links with 2 ternary links and
1 quaternary link. Similarly, if we take n2
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= 6, one can easily find that we will get
n3 = 0 and n4 = 2. That means, we can have
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an eight-link mechanism with 6 binary links
and 2 quaternary links. From these three different
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types of eight link chains by kinematic inversions
one can get a very large number of different
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mechanisms.
In conclusion, let me now repeat the foremost
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important points that we have learnt today
during this discussion of number synthesis
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of planar linkages. The first point is that
the minimum number of binary links in any
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such linkage must be four: that means, we
must have at least four binary links. The
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second point is that the highest order link
in an nth-link mechanism is n/2, that is in
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a six-link mechanism the highest order is
ternary link and in an eight-link mechanism
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the highest order is quaternary link. Third
thing we have seen that, the total number
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of links must be even and the last point is
that with increase in the number of total
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links the possible types of various mechanisms
that we can have from some such chains by
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kinematic inversion increases drastically.
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8