ÿþ1
00:00:15,639 --> 00:00:22,639
The topic of today's lecture is number synthesis.
During this stage of kinematic synthesis called
2
00:00:24,520 --> 00:00:31,520
number synthesis, we determine the type and
number of different types of links and the
3
00:00:32,509 --> 00:00:39,509
number of simple pairs like revolute or prismatic
pairs that needed to yield a single degree
4
00:00:40,710 --> 00:00:47,710
of freedom planar linkage. It is needless
to say that all the single degree of freedom
5
00:00:48,120 --> 00:00:55,120
planar linkages will satisfy the Grubler's
criterions which are discussed earlier. However,
6
00:00:56,489 --> 00:01:03,489
before we get into the discussion or details
of number synthesis we shall first prove certain
7
00:01:03,629 --> 00:01:10,610
basic results, which are of vital important
for number synthesis. The first of these two
8
00:01:10,610 --> 00:01:17,610
questions is what is the minimum number of
binary links that such a linkage must posses?
9
00:01:19,380 --> 00:01:26,380
So, we determine the minimum number of binary
links in a single degree of a freedom planar
10
00:01:28,110 --> 00:01:35,110
linkage. Let n be the total number of links
in the linkage, n2 be the number of binary
11
00:01:39,810 --> 00:01:46,810
links, n3 be the number of ternary links and
n4 be the number of quaternary links and so
12
00:01:52,000 --> 00:01:59,000
on. Thus, we have the total number of links
in is equal to n2 plus n3 plus n4 upto ni
13
00:02:03,140 --> 00:02:10,140
where i denotes the highest order link that
is present in this linkage. Our first task
14
00:02:11,000 --> 00:02:18,000
is to determine the minimum value of n2. Towards
this goal let us consider this figure.
15
00:02:19,519 --> 00:02:26,519
In this figure, we see there is one link which
is connected to two other links through the
16
00:02:28,510 --> 00:02:35,510
revolute pairs here and here. To note that
at each of these revolute pairs we have two
17
00:02:37,650 --> 00:02:44,650
elements say 1 minus which is the spin which
goes into the hole which is denoted by one
18
00:02:47,349 --> 00:02:54,349
plus. So this 1 plus and 1 minus we shall
call elements, thus at each revolute pair
19
00:02:55,180 --> 00:03:02,180
we have two elements. Similarly the two elements
at these revolute pairs are these 2 plus and
20
00:03:02,890 --> 00:03:09,390
2 minus.
In this way, if we count the total number
21
00:03:09,390 --> 00:03:16,390
of elements that I can write e should be equal
to twice the number of joints or pairs say
22
00:03:20,480 --> 00:03:27,480
that is e equal to 2j. We can also count this
number of elements from this links. This is
23
00:03:28,819 --> 00:03:35,819
a binary link which has two elements because
it is connected to two other links, two revolute
24
00:03:39,129 --> 00:03:44,489
pairs. Similarly, a ternary link, we will
have three elements because it is connected
25
00:03:44,489 --> 00:03:50,060
to three other links and a quaternary link
we will have four such elements. So if we
26
00:03:50,060 --> 00:03:57,060
count the total number of elements from the
view point of links then I can write e equal
27
00:03:58,760 --> 00:04:05,760
to 2n2 plus 3n3 plus 4n4 plus in, where n2
is the number of binary links, n3 is the number
28
00:04:09,459 --> 00:04:16,459
of ternary links, n4 is the number of quaternary
links and ni is the number of ith order link.
29
00:04:26,650 --> 00:04:32,710
We have just now seen that the total number
of elements can be counted from two view points.
30
00:04:32,710 --> 00:04:38,839
If we count it from the view point of number
of pairs then I can write the total number
31
00:04:38,839 --> 00:04:45,839
of elements e equal to 2j. But, if we count
the number of elements from the view point
32
00:04:47,339 --> 00:04:53,339
of different links or of different orders
then we can write the total number of elements
33
00:04:53,339 --> 00:05:00,339
e equal to 2n2 plus 3n3 plus 4n4 plus ini.
We can equate these two numbers of elements
34
00:05:03,649 --> 00:05:09,229
counted from the view points the kinematic
pairs and from the view point of different
35
00:05:09,229 --> 00:05:16,229
order links we can write 2j equal to 2n2 plus
3n3 plus 4n4 plus ini.
36
00:05:16,510 --> 00:05:23,510
As I already told, that all these linkages
must satisfy the Grubler's criterion which
37
00:05:25,960 --> 00:05:32,960
is 2j minus 3n plus 4 equal to 0, where j
denotes the number of kinematic pairs and
38
00:05:33,390 --> 00:05:40,390
n denotes the number of total links. Substituting
e equal to 2j which we have just now saying
39
00:05:42,960 --> 00:05:49,960
to be given by 2n2 plus 3n3 plus 4n4 plus
ini minus, we replace this n by the number
40
00:05:52,830 --> 00:05:59,830
of counts of different links which is n2 plus
n3 plus n4 plus ni. Simplifying this equation
41
00:06:05,899 --> 00:06:12,899
we can see the 3 and 3 cancels and ultimately
we get n2 is equal to p minus 3 summed over
42
00:06:15,589 --> 00:06:22,589
all values of p starting from 4 up to i plus
4. That means n2 is given by p minus 3, p
43
00:06:24,690 --> 00:06:31,690
going from 4 to i plus 4. That is, this p
denotes the number of quaternary links and
44
00:06:32,830 --> 00:06:37,969
higher order links.
So we can easily see if the sum is 0, the
45
00:06:37,969 --> 00:06:44,969
minimum number of binary links (n2)min equal
to 4. This again convinces us what we have
46
00:06:47,330 --> 00:06:54,330
seen earlier, that this simplest linkage must
have 4 binary links what we call four bar
47
00:06:54,990 --> 00:06:57,770
linkage.
48
00:06:57,770 --> 00:07:04,770
Next we would like to add another question
that is what is the highest order link in
49
00:07:06,740 --> 00:07:13,349
an n link mechanism? That means, the total
number of links is n then in such a linkage
50
00:07:13,349 --> 00:07:19,550
what is the highest order link? We should
try to answer this question in a reverse manner.
51
00:07:19,550 --> 00:07:26,270
We will say, the highest order link, be ith
order that is we have some ni's. Then what
52
00:07:26,270 --> 00:07:32,580
is the minimum number of links that is needed
to produce the single degree of freedom planar
53
00:07:32,580 --> 00:07:37,630
linkage. Towards this goal let me consider
the following figure.
54
00:07:37,630 --> 00:07:44,630
In this figure, we start with a link with
i hinges. This is the link which has i hinges
55
00:07:47,180 --> 00:07:54,180
numbered as 1, 2, 3, 4 so on up to i. To produce
a at each of these hinges we connect another.
56
00:08:00,149 --> 00:08:05,880
At the 1st hinge we connect link number one,
at the second hinge we connect link number
57
00:08:05,880 --> 00:08:12,880
2, at the third hinge we connect link number
3 and so on this ith link at the hinge number
58
00:08:12,969 --> 00:08:19,969
i. To connect these two links 1 and 2, we
must have some motion transfer links, accordingly,
59
00:08:22,130 --> 00:08:29,130
(i plus 1), (i plus 2) and (i plus 3) so on
up to (2i minus 1). The thing to note that,
60
00:08:33,680 --> 00:08:40,399
the hinge number 2, 3, 4 up to (i minus 1),
we have ternary links. Because link number
61
00:08:40,399 --> 00:08:47,269
2 has three hinges here, here and here and
that is true for all other links connected
62
00:08:47,269 --> 00:08:54,269
at hinge number 4, hinge number 4 and so on.
Because, if we have binary link at 2 then
63
00:08:54,660 --> 00:09:01,660
this particular hinge will not remain because
three links namely (i plus 2) and link number
64
00:09:02,900 --> 00:09:09,150
2 will get connected at this higher order
hinge. Then hinge as a simple hinge all these
65
00:09:09,150 --> 00:09:16,150
links starting from number 2, 3, 4 and so
on up to (i minus 1) must ternary link. So
66
00:09:16,560 --> 00:09:23,560
we have produced a close chain minimum number
of links if we start from a link with of ith
67
00:09:23,770 --> 00:09:26,270
order that is with i hinge.
68
00:09:26,270 --> 00:09:33,270
Let me count total number of links n if we
have started with, I have already shown the
69
00:09:36,020 --> 00:09:41,470
number up to (2i minus 1) and we count this
starting link which is having i hinges so
70
00:09:41,470 --> 00:09:48,470
the total number links is 2i. Thus, we see
that this is an ith order hinge then minimum
71
00:09:52,080 --> 00:09:59,080
I need 2i number of links to produce a closed
chain. That means, total number of links is
72
00:09:59,670 --> 00:10:06,670
n then imax can go up to n by 2 and not more
than n by 2. I emphasize that is the possible
73
00:10:10,860 --> 00:10:17,750
value of imax, not necessarily imax has to
be n by 2, definitely it cannot be more than
74
00:10:17,750 --> 00:10:22,120
n by 2.
Next thing we have to prove, that this closed
75
00:10:22,120 --> 00:10:28,340
chain from this closed chain if I hold one
link fixed it must produce a single degree
76
00:10:28,340 --> 00:10:34,510
freedom mechanism that is this particular
closed chain must satisfy our old Grubler's
77
00:10:34,510 --> 00:10:41,510
criterion. For that we count n equal to 2i.
Let me count the maximum hinges j, we have
78
00:10:45,130 --> 00:10:52,130
started with i hinges on this initial link
so j equal to i plus there is one hinge here
79
00:10:59,440 --> 00:11:06,440
and there is another hinge here which is at
two plus on all other links two, three, four
80
00:11:10,130 --> 00:11:15,240
there are two external because these are all
ordinary links one of the hinge has been already
81
00:11:15,240 --> 00:11:21,380
count with this starting link. There are two
hinges extra hinges on each of it so that
82
00:11:21,380 --> 00:11:28,380
into i, how many such linkages? We have starting
from two to (i minus1) that is 2 times (i
83
00:11:38,940 --> 00:11:45,940
minus 3). So that is the number of hinge i
plus 2 plus2 times (i minus3) which will give
84
00:11:51,100 --> 00:11:55,910
us (3i minus 4).
85
00:11:55,910 --> 00:12:02,910
We see that in this closed chain, total number
of links n turns out to be 2i, where i denote
86
00:12:05,090 --> 00:12:12,090
the highest order link in this chain. The
total number of joins j turns out to be 3i
87
00:12:14,430 --> 00:12:21,430
minus2. If we write the Grubler's criterion
that is 2j minus 3n plus 4 we get 2 times
88
00:12:30,220 --> 00:12:37,220
(3i minus 2) minus 3 times (2i plus 4) equal
to 0. Thus the Grubler's criterion is satisfied
89
00:12:48,090 --> 00:12:55,090
by this closed finite chain and consequently
this can constitute a single degree of freedom
90
00:12:55,320 --> 00:13:01,520
planar linkage. So we concentrate on these
two results that we have just now derived.
91
00:13:01,520 --> 00:13:08,440
One is that the minimum number of binary links
in a linkage must be flow and the second is
92
00:13:08,440 --> 00:13:15,440
that is highest order link n link mechanism
that is imax is n by 2. Since, all the single
93
00:13:19,970 --> 00:13:26,970
degree freedom linkage must satisfy the Grubler's
criterion that 2j minus 3n plus 4 equal to
94
00:13:27,080 --> 00:13:34,080
0 that gives 3n equal to 2j plus 4. We note
that the right hand side 2j plus 4 is an even
95
00:13:36,140 --> 00:13:42,970
number and if 3n is equal to an even number
then the n must be even, which means all the
96
00:13:42,970 --> 00:13:49,720
planar linkages simple pairs and single degree
of freedom must have even number of links.
97
00:13:49,720 --> 00:13:56,680
We have already seen that the four-link mechanism
is the simplest mechanism. The next more complicated
98
00:13:56,680 --> 00:14:03,680
mechanism should be n equal to 6 that is a
six-link mechanism. If the kinematic requirements
99
00:14:04,960 --> 00:14:11,600
are little more complex, which cannot be satisfied
by a four-link mechanism then we have to try
100
00:14:11,600 --> 00:14:18,600
to use a six-link mechanism. Let me go into
this number synthesis of six-link mechanism.
101
00:14:23,020 --> 00:14:30,020
With a six-link chain we have n equal to 6
that is imax is n by 2 that is 3. The highest
102
00:14:33,300 --> 00:14:40,300
possible order is a ternary link. So a six-link
mechanism constitutes a binary links and ternary
103
00:14:40,650 --> 00:14:45,210
links. So the total number of link n equal
to n2 plus n3 equal to 6 where n2 is the number
104
00:14:45,210 --> 00:14:52,210
of binary links, n3 is the number of ternary
links. For n equal to 6, we know to satisfy
105
00:14:55,420 --> 00:15:02,420
Grubler's criterion 2j must be equal to 3n
minus 4 equal to 3 times 6 minus 4 equal to14
106
00:15:03,730 --> 00:15:10,330
that is j equal to 7. We have got one equation,
numbered equation one number in terms of two
107
00:15:10,330 --> 00:15:17,330
unknown in terms of n2 and n3. We derived
another equation involving n2 and n3 by counting
108
00:15:18,610 --> 00:15:23,670
the number of elements.
109
00:15:23,670 --> 00:15:30,670
The number of elements e equal to 2j which
is also given by 2n2 plus 3n3. Thus 2n2 plus
110
00:15:35,150 --> 00:15:42,150
3n3 equal to 2j where j is equal to 7 this
comes out to be 14. This is the second equation
111
00:15:42,970 --> 00:15:49,970
involving these two unknowns namely n2 and
n3. Our previous equation was n2 plus n3 equal
112
00:15:50,720 --> 00:15:56,700
to 6 and the second equation is 2n2 plus 3n3
equal to 14.
113
00:15:56,700 --> 00:16:03,700
We can easily solve these two linear equations
in two unknown namely n2 and n3 as n2 equal
114
00:16:06,370 --> 00:16:13,370
to 4 and n3 equal to 2. Thus, a six-link mechanism
has four binary links and 2 ternary links.
115
00:16:16,589 --> 00:16:23,589
We shall see what are the possible combinations
of these binary and ternary links to generate
116
00:16:24,380 --> 00:16:28,750
different types of six-link mechanisms?
117
00:16:28,750 --> 00:16:35,750
This figure shows one possible six-link chain
with two ternary links and four binary links.
118
00:16:35,760 --> 00:16:42,760
As we see, the link number 1 is a ternary
link, link number 4 is another ternary link
119
00:16:43,700 --> 00:16:50,700
whereas link number 2, 3, 5 and 6 are all
binary links. The thing to note, that in this
120
00:16:52,890 --> 00:16:59,890
chain there are six-link and seven revolute
pairs, we can count at vertices of this hexagon
121
00:17:02,279 --> 00:17:09,279
and one inside the hexagon. Another thing
to note that here the two parallel links 1
122
00:17:09,529 --> 00:17:16,529
and 4 are directly connected by this revolute
pair and all the four binary links are connected
123
00:17:16,939 --> 00:17:23,939
to the ternary links, this chain is known
as Watt's-chain.
124
00:17:35,330 --> 00:17:42,330
So in a Watt's-chain, two ternary links are
directly connected to each other. In this
125
00:17:42,630 --> 00:17:49,160
Watt's-chain, we can see that the two ternary
links that is number 1 and 4 are equivalent.
126
00:17:49,160 --> 00:17:55,360
In the sense, both of them are connected to
a ternary link at one kinematic pair and two
127
00:17:55,360 --> 00:18:02,360
binary pair at the other two revolute pairs
like, 4 is connected to link number four by
128
00:18:02,530 --> 00:18:09,530
a revolute pair, to the binary link 4 is connected
to the another binary link 3 at this revolute
129
00:18:09,790 --> 00:18:16,100
pair and is connected to the ternary link
1 at this revolute pair. Exactly the same
130
00:18:16,100 --> 00:18:21,120
thing happened for the link number one, it
this connected to the ternary link 6 and this
131
00:18:21,120 --> 00:18:28,120
revolute pair binary link 2 to this revolute
pair and to ternary link 4 by this revolute
132
00:18:28,590 --> 00:18:31,900
pair.
Thus to topologically there is no difference
133
00:18:31,900 --> 00:18:38,900
between link number 1 and 4. The same is true
for all binary links namely 2, 3, 5 and 6
134
00:18:40,679 --> 00:18:46,559
each one of which is connected to a ternary
link at one end and to a binary link at the
135
00:18:46,559 --> 00:18:53,559
other end. For example, link number 2 is connected
to a ternary link at one end and to a binary
136
00:18:54,100 --> 00:18:58,940
link at the other end and the same is true
for all other binary links.
137
00:18:58,940 --> 00:19:05,940
Thus there are two types of links ternary
links and binary links but both the ternary
138
00:19:07,750 --> 00:19:14,500
links are equivalent and all the four binary
links are also equivalent. So from a Watt's-chain
139
00:19:14,500 --> 00:19:20,480
by kinematic inversion that is depending on
which link we hold fixed we can get two different
140
00:19:20,480 --> 00:19:27,380
types of Watt's mechanism. One type of Watt's
mechanism we can get by holding binary links
141
00:19:27,380 --> 00:19:34,350
fixed with 1, 2, 3 or 5 and 6, because all
of them are equivalent and the second type
142
00:19:34,350 --> 00:19:40,500
of Watt's mechanism we can get holding one
of the binary links that is either one or
143
00:19:40,500 --> 00:19:46,480
four are fixed.
We will now show a model of a six-link Watt's
144
00:19:46,480 --> 00:19:52,100
mechanism where we will find that one of the
binary links is held fixed.
145
00:19:52,100 --> 00:19:59,100
As an example of Watt's mechanism a with binary
link fixed let us go back to our old example
146
00:19:59,150 --> 00:20:06,059
of this parallel jaw player. We hold this
lower jaw that is this blue link fixed, this
147
00:20:06,059 --> 00:20:12,600
is a binary link because it has two revolute
pairs. Let us note that, this binary link
148
00:20:12,600 --> 00:20:18,610
is connected to another binary link at this
revolute pair and to this ternary link at
149
00:20:18,610 --> 00:20:25,610
this revolute pair. This lower jaw is a ternary
link because it has three revolute pairs and
150
00:20:25,740 --> 00:20:32,740
this small link is another ternary link which
has three revolute pair and these two ternary
151
00:20:34,170 --> 00:20:41,170
links are directly connected so it is one
type of Watt's chain where we know two ternary
152
00:20:42,260 --> 00:20:48,790
links must be directly connected. If we hold
this lower jaw fixed then we are holding this
153
00:20:48,790 --> 00:20:55,790
binary link fixed. We should also know that
this upper jaw is a binary link and this below
154
00:20:57,080 --> 00:21:04,080
link is another binary link. This binary link
is connecting this ternary link and this binary
155
00:21:04,460 --> 00:21:11,460
link. As a result of this we get a Watt's
mechanism by Watt's-chain. This is Watt's
156
00:21:14,610 --> 00:21:21,610
mechanism of one kind. Later on we will see
Watt's mechanism of another link where ternary
157
00:21:21,990 --> 00:21:24,040
link will be held fixed.
158
00:21:24,040 --> 00:21:31,040
An example of another type of possible Watt's
mechanism let us consider this figure. Here
159
00:21:32,770 --> 00:21:39,770
we started from a Watt's chain which has two
planar links 1 and 4 and 4 binary links 2,
160
00:21:40,660 --> 00:21:47,660
3, 5 and 6. Here, is a binary link 1 which
is held fixed this is known as Watt's walking
161
00:21:57,960 --> 00:22:04,960
beam engine. In this Watt's walking beam engine
we must see that in the chain we have shown
162
00:22:07,720 --> 00:22:14,720
revolute pair between 1 and 6 which has been
replaced by a prismatic pair between the cylinder
163
00:22:14,870 --> 00:22:21,870
and the piston. But in our analysis, we always
treated revolute pair and prismatic pair as
164
00:22:22,110 --> 00:22:29,110
value equivalent. So here, as we see that
this great beam that is link 4 is connected
165
00:22:29,710 --> 00:22:36,710
to ternary link directly by the revolute pair.
This is another type of Watt's mechanism which
166
00:22:37,370 --> 00:22:42,360
is possible to get by kinematic inversion
from a Watt's chain.
167
00:22:42,360 --> 00:22:49,360
An another example of a six-link mechanism
with seven hinges we can get the following
168
00:22:49,990 --> 00:22:50,710
figure.
169
00:22:50,710 --> 00:22:57,710
Here, as we see two ternary links namely 1
and 4. However, unlike in a Watt's-chain these
170
00:23:02,340 --> 00:23:08,080
two ternary links are not directly connected
to each other rather they are connected via
171
00:23:08,080 --> 00:23:15,080
this binary link number 6. Here, we have two
ternary links 1 and 4 which are connected
172
00:23:15,610 --> 00:23:22,610
by a binary link 6, binary link 5 and by two
binary links namely 2 and 3 and these particular
173
00:23:25,390 --> 00:23:31,850
chain where the two ternary links are not
directly connected is known as Stephenson's
174
00:23:31,850 --> 00:23:38,850
chain. We see that in a Stephenson's chain
two ternary links are not directly connected.
175
00:23:43,470 --> 00:23:50,470
In a Stephenson's chains the ternary links
1 and 4 are equivalent in a sense that both
176
00:23:51,340 --> 00:23:58,340
1 and 4 are connected to three binary links
and three revolute pairs. For example, one
177
00:23:58,730 --> 00:24:05,730
is connected to binary link 6, binary link
5 and binary link 2 at these three revolute
178
00:24:08,760 --> 00:24:15,240
pairs and link number 4, the other ternary
links is also connected to three binary links
179
00:24:15,240 --> 00:24:22,240
to link number 5 here, link number 6 here
and number 3 here. Thus both these ternary
180
00:24:23,540 --> 00:24:29,820
links are topologically equivalent because
both of them are connected to three binary
181
00:24:29,820 --> 00:24:33,370
links.
However, so far the binary links are concerned
182
00:24:33,370 --> 00:24:40,370
there are two varieties, namely 5 and 6 and
2 and 3. We should note that both 5 and 6
183
00:24:46,900 --> 00:24:53,900
are connected to two ternary links at two
joints, 6 is also connected to two ternary
184
00:24:57,870 --> 00:25:04,870
links at two joins. Link number 2 and 3 at
one end is connected to a ternary link but
185
00:25:05,470 --> 00:25:10,830
at the other end is connected to a binary
link. So there are two types of binary links
186
00:25:10,830 --> 00:25:17,830
they can be grouped as (5, 6) and (2, 3).
By kinematic inversion we can get three different
187
00:25:21,590 --> 00:25:28,590
types of Stephenson's mechanism depending
on whether ternary links 1 or 4 is held fixed
188
00:25:29,480 --> 00:25:36,480
or one of the binary links in this group that
is either 5 or 6 is held fixed or one of this
189
00:25:37,240 --> 00:25:44,240
group namely 2 and 3 that is either 2 or 3
are held fixed. There are three different
190
00:25:45,120 --> 00:25:50,350
types of Stephenson's mechanism which can
be obtained by kinematic inversion from the
191
00:25:50,350 --> 00:25:57,350
same Stephenson's chain. We now see a model
of a Stephenson's chain to generate a Stephenson's
192
00:25:58,470 --> 00:26:02,110
linkage where a ternary link is held fixed.
193
00:26:02,110 --> 00:26:08,040
Let us now look at the model of this Stephenson's
mechanism where one of the ternary links is
194
00:26:08,040 --> 00:26:14,370
held fixed. Here we have this fixed link as
the ternary link which has three hinges one
195
00:26:14,370 --> 00:26:21,360
here, one there and another there and this
is the other ternary link which is connected
196
00:26:21,360 --> 00:26:28,360
to the fixed link by two binary links. The
two ternary links are not directly connected,
197
00:26:28,700 --> 00:26:35,020
they are connected via binary links at these
two points and by two binary links at this
198
00:26:35,020 --> 00:26:41,919
point, this is a binary link this is a binary
link. This binary links are equivalent because
199
00:26:41,919 --> 00:26:47,730
at one end this binary link is connected to
a ternary link, at this end this binary link
200
00:26:47,730 --> 00:26:53,730
is connected to another binary link. Similarly,
this binary link is connected to a ternary
201
00:26:53,730 --> 00:27:00,730
link at this end and at to a binary link at
this end. These two binary links are of same
202
00:27:00,850 --> 00:27:06,250
nature. Similarly these two binary links are
also of same nature because they are connected
203
00:27:06,250 --> 00:27:13,250
at both ends to ternary links. One of the
ternary links is held fixed and we get one
204
00:27:14,040 --> 00:27:21,040
variety of a Stephenson's linkage.
Another example of Stephenson's linkage let
205
00:27:21,919 --> 00:27:28,620
us consider the same Stephenson's chain and
consider one of the binary links to be fixed.
206
00:27:28,620 --> 00:27:34,370
As shown in this figure, this is the Stephenson's
chain that we have considered earlier and
207
00:27:34,370 --> 00:27:41,370
in this chain if we hold a binary link say
link number 2 fixed then we get this mechanism.
208
00:27:42,460 --> 00:27:49,460
As we see, 2 is connected to ternary link
1 and link 2 is the fixed link which is connected
209
00:27:53,610 --> 00:28:00,610
to a binary link 3 and to a ternary link 1.
Link number 4 is the ternary link which is
210
00:28:03,200 --> 00:28:10,200
connected to link 3 here, link 6 here and
link 5 here. This is known as Stephenson's
211
00:28:18,360 --> 00:28:25,360
valve gear mechanism which is used in a steam
engine. We have seen, a six-link chains consists
212
00:28:31,390 --> 00:28:37,710
of four binary links two ternary links and
various combinations which are possible as
213
00:28:37,710 --> 00:28:41,020
Watt's linkage or Stephenson's linkage.
214
00:28:41,020 --> 00:28:48,020
Let us consider the next higher order link
possible to give you the flavor
of number synthesis. The next most complicated
215
00:28:56,650 --> 00:29:03,650
mechanism we consider with n event is an eight-link
mechanism that is n equal to 8. Consequently
216
00:29:06,880 --> 00:29:13,270
highest order link possible in an eight-link
mechanism that is imax equal to n by 2 equal
217
00:29:13,270 --> 00:29:20,270
to 4. An eight-link mechanism we will have
binary link it is possible to have ternary
218
00:29:25,650 --> 00:29:30,840
link and it is possible to have quaternary
link.
219
00:29:30,840 --> 00:29:37,480
If the number of binary links is n2, the number
of ternary links is n3 and the number of quaternary
220
00:29:37,480 --> 00:29:44,480
links is n4 then the total number of links
n which is equal to n2 plus n3 plus n4 equal
221
00:29:45,610 --> 00:29:52,610
to 8. This is our first equation to determine
n2, n3 and three and n4. We also know that
222
00:29:54,880 --> 00:30:01,880
the number of elements e equal to 2j equal
to 3n minus 4 so that Grubler's criterion
223
00:30:06,880 --> 00:30:13,880
is satisfied, which means 3 times 8 minus
4 equal to 20. Counting the number of elements
224
00:30:19,419 --> 00:30:26,419
from the view point of links we can write
2n2 plus 3n3 plus 4n4 equal to 20 because
225
00:30:36,600 --> 00:30:43,600
this is also equal to the number of elements
so this is our second equation two.
226
00:30:45,210 --> 00:30:51,910
It may now appear that we have three unknowns,
n2, n3 and n4 to determine but we have only
227
00:30:51,910 --> 00:30:58,910
two equations namely 1 and 2. Thus there may
be infinite solutions a little thought would
228
00:30:58,960 --> 00:31:05,120
convince us that is not the situation we still
have finite number of solutions because we
229
00:31:05,120 --> 00:31:12,120
should remember all these numbers n2, n3 and
n4 are integers not only that the minimum
230
00:31:12,419 --> 00:31:19,419
value of n2 is also 4. Number n2 can start
from 4 then can go up to 5, 6 and so on.
231
00:31:28,380 --> 00:31:33,679
Let us see what are the various solutions
possible to these two equations that under
232
00:31:33,679 --> 00:31:40,309
such restrictions that all these numbers n2,
n3 and n4 must be positive integers there
233
00:31:40,309 --> 00:31:45,150
is no point having a negative number for the
number of links and also that minimum values
234
00:31:45,150 --> 00:31:46,890
of n2 is 4.
235
00:31:46,890 --> 00:31:53,890
For an eight-link mechanism, we have got two
equations namely n2 plus n3 plus n4 equal
236
00:32:01,240 --> 00:32:08,240
to 8 and 2n2 plus 3n3 plus 4n4 equal to 20.
If we assume that the values of n2 is 4 then
237
00:32:18,080 --> 00:32:25,080
from these two equations we get n3 plus n4
equal to 8 minus n2 equal to 8 minus 4 equal
238
00:32:30,380 --> 00:32:37,380
to 4 and from the second equation we get 3n3
plus 4n4 equal to 20 minus 2n2 that is 20
239
00:32:46,240 --> 00:32:53,240
minus 2 times 4 equal to 12. We get these
two equations to solve for n3 and n4 and the
240
00:32:56,179 --> 00:33:03,179
obvious solution is n3 equal to 4 and n4 equal
to 0. That means we can get an eight-link
241
00:33:07,490 --> 00:33:14,490
mechanism consisting of four binary links
and four ternary links. There is no necessity
242
00:33:14,539 --> 00:33:18,049
that we must have a quaternary link.
243
00:33:18,049 --> 00:33:25,049
If we take n2 equal to 5 then we get n3 plus
n4 equal to 8 minus 5 equal to 3 and 3n3 plus
244
00:33:37,210 --> 00:33:44,210
4 n4 equal to 20 minus 2 times 5 equal to
10. Solving these two equations we get n3
245
00:33:53,450 --> 00:34:00,450
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
246
00:34:14,049 --> 00:34:20,089
1 quaternary links. Similarly, if we take
n2 equal to 6, one can easily find that will
247
00:34:20,089 --> 00:34:27,089
get n3 equal to 0 and n4 equal to 2. That
means, we can have an eight-link mechanism
248
00:34:29,839 --> 00:34:36,559
with 6 binary links and 2 quaternary links.
From these three different types of eight
249
00:34:36,559 --> 00:34:43,559
link chains by kinematic inversions one can
get a very large number of different mechanisms.
250
00:34:44,409 --> 00:34:50,509
In conclusion, let me now repeat the foremost
important points that we have learnt today
251
00:34:50,509 --> 00:34:56,710
during this discussion of number synthesis
of planar linkages. The first point is that,
252
00:34:56,710 --> 00:35:02,789
the minimum number of binary links in any
such linkage must be four: that means, we
253
00:35:02,789 --> 00:35:09,339
must have at least four binary links. The
second point is that the highest order link
254
00:35:09,339 --> 00:35:16,200
in an nth-link mechanism is n by 2, that is
in a six-link mechanism the highest order
255
00:35:16,200 --> 00:35:22,839
is ternary in an eight-link mechanism the
highest order is quaternary. Third thing we
256
00:35:22,839 --> 00:35:29,759
have seen that, the total number of links
must be even and the last point is that with
257
00:35:29,759 --> 00:35:35,609
increase in the number of total links the
possible types of various mechanisms that
258
00:35:35,609 --> 00:35:42,609
we can have some such chains by kinematic
inversion increases drastically.
259