1
00:00:04,880 --> 00:00:20,279
The
topic of today’s lecture is mobility analysis.
2
00:00:20,279 --> 00:00:27,480
By mobility analysis, we obtain the degrees
of freedom of a given mechanism.
3
00:00:27,480 --> 00:00:34,450
This is accomplished by the counting number
of links and the number of different types
4
00:00:34,450 --> 00:00:40,020
of kinematic pairs those are used to connect
these links.
5
00:00:40,020 --> 00:00:47,160
Let me now elaborate, how we carry out this
mobility analysis for planar mechanisms.
6
00:00:47,160 --> 00:00:55,120
It is worthwhile to recall that in a planar
mechanism each link has 3 degrees of freedom:
7
00:00:55,120 --> 00:01:02,079
2 of which are translational in the plane
of motion and 1 is rotational about an axis
8
00:01:02,079 --> 00:01:04,449
perpendicular to the plane of motion.
9
00:01:04,449 --> 00:01:11,219
Let there be ‘n’ number of total links
in a mechanism, which includes the fixed link
10
00:01:11,219 --> 00:01:17,329
of the frame that means there are n – 1
moving links.
11
00:01:17,329 --> 00:01:23,240
When these links are not connected by any
kinematic pair then the total degrees of freedom
12
00:01:23,240 --> 00:01:27,469
are obviously 3(n – 1).
13
00:01:27,469 --> 00:01:31,890
For each of these n – 1 links there are
3 degrees of freedom, so the total degree
14
00:01:31,890 --> 00:01:35,310
of freedom of the system is 3(n – 1).
15
00:01:35,310 --> 00:01:41,960
Let these ‘n’ links be connected by j
number of lower pairs.
16
00:01:41,960 --> 00:01:49,219
By lower pair in a planar mechanism, we can
mean either a revolute pair or a prismatic
17
00:01:49,219 --> 00:01:56,789
pair and each of these kinematic pairs connects
only 2 links.
18
00:01:56,789 --> 00:02:03,979
We also recall that whether it is a revolute
pair or a lower pair, at each of these pairs,
19
00:02:03,979 --> 00:02:09,550
2 degrees of freedom are curtailed and only
1 out of 3 is maintained.
20
00:02:09,550 --> 00:02:18,010
If there are ‘j’ number of total kinematic
pairs 2j numbers of degrees of freedom are
21
00:02:18,010 --> 00:02:19,010
curtailed.
22
00:02:19,010 --> 00:02:27,140
So, the effective degree of freedom of the
mechanism is reduced to F, which are the degrees
23
00:02:27,140 --> 00:02:32,240
of freedom of the mechanism is 3(n – 1)
– 2j.
24
00:02:32,240 --> 00:02:41,140
Let us consider a constrained mechanism with
a single degree of freedom; that is, there
25
00:02:41,140 --> 00:02:46,670
exist a unique input-output relationship,
where the degree of freedom of the mechanism
26
00:02:46,670 --> 00:02:49,030
F is 1.
27
00:02:49,030 --> 00:02:57,380
Substituting F = 1 in above equation, we get
2j – 3n + 4 = 0.
28
00:02:57,380 --> 00:03:06,770
For a single degree of freedom mechanism,
maintaining a unique input-output relationship,
29
00:03:06,770 --> 00:03:12,710
the number of links and the number of lower
pairs must be related to this equation that
30
00:03:12,710 --> 00:03:16,840
is: 2j – 3n + 4 equal to 0.
31
00:03:16,840 --> 00:03:26,000
This equation is called Grubler’s criterion
for single degree of freedom mechanism.
32
00:03:26,000 --> 00:03:32,510
While deriving this Grubler’s criterion,
we assume that each of these lower pairs is
33
00:03:32,510 --> 00:03:34,970
connecting only 2 links.
34
00:03:34,970 --> 00:03:41,150
However, due to practical considerations some
times more than 2 links can be connected at
35
00:03:41,150 --> 00:03:42,980
a particular hinge.
36
00:03:42,980 --> 00:03:50,300
As an example of a different types of kinematic
pairs which connects more than 2 links, let
37
00:03:50,300 --> 00:03:52,150
us consider this figure.
38
00:03:52,150 --> 00:04:02,650
Here 3 links namely 2, 3 and 4 are connected
by a single hinge at this location.
39
00:04:02,650 --> 00:04:08,379
Such hinges are called compound hinges or
higher order hinges.
40
00:04:08,379 --> 00:04:15,330
This particular compound hinge is equivalent
to two simple hinges as explained in the adjoining
41
00:04:15,330 --> 00:04:16,330
figure.
42
00:04:16,330 --> 00:04:21,590
For example, this particular hinge can be
thought of as 2 hinges.
43
00:04:21,590 --> 00:04:32,210
One connecting link 2 and link 3, whereas
another hinge connects link 3 and link 4.
44
00:04:32,210 --> 00:04:40,700
Thus, a hinge which connects 3 different links
is equivalent to 2 simple hinges.
45
00:04:40,700 --> 00:04:49,340
This way we can think of another type of hinge
where 4 links are connected and such a hinge
46
00:04:49,340 --> 00:04:54,620
will obviously be equivalent to 3 simple hinges.
47
00:04:54,620 --> 00:04:58,750
Maintaining this equivalent between higher
order hinges and simple hinges, we would like
48
00:04:58,750 --> 00:05:05,760
to modify the equation for calculating the
degrees of freedom of a mechanism as follows.
49
00:05:05,760 --> 00:05:10,960
When higher order hinges are present, the
symbol ‘j’ in the equation, we would like
50
00:05:10,960 --> 00:05:13,380
to modify as follows.
51
00:05:13,380 --> 00:05:20,570
j is equal to j1, which represents the number
of simple hinges, which connects only 2 links
52
00:05:20,570 --> 00:05:27,740
plus 2j2, where j2 is the number of hinges
to each one of which connects 3 links and
53
00:05:27,740 --> 00:05:35,169
so on; that is j3 represents the number of
hinges each one of which connects (3 + 1),
54
00:05:35,169 --> 00:05:43,740
that is 4 links and so on up to ji. that is
ji is the number of compound hinges, each
55
00:05:43,740 --> 00:05:49,090
of which connects (i + 1) number of links.
56
00:05:49,090 --> 00:05:54,380
In a mechanism, there can be higher pair as
well and as we recall, if there is a higher
57
00:05:54,380 --> 00:06:01,520
pair then at each higher pair only 1 translational
degree of freedom is curtailed that is along
58
00:06:01,520 --> 00:06:04,830
the common normal at the point or line of
contact.
59
00:06:04,830 --> 00:06:08,570
Two other degrees of freedom can be retained.
60
00:06:08,570 --> 00:06:13,180
Consequently, at each higher pair only 1 degree
of freedom is curtailed.
61
00:06:13,180 --> 00:06:21,010
I would like to modify the equation, the degrees
of freedom of a mechanism,
62
00:06:21,010 --> 00:06:29,900
F = 3(n – 1) – 2j – h,
where h represents the number of higher pairs,
63
00:06:29,900 --> 00:06:37,610
j represents the number of equivalent simple
hinges and n represents the number of total
64
00:06:37,610 --> 00:06:39,520
links.
65
00:06:39,520 --> 00:06:44,430
Sometimes there can be some redundant degree
of freedom of a mechanism.
66
00:06:44,430 --> 00:06:47,889
What we mean by a redundant degree of freedom?
67
00:06:47,889 --> 00:06:53,610
Due to some typical kinematic pairs and their
placement, we may find that in a mechanism
68
00:06:53,610 --> 00:07:00,830
a particular link may be moved without transmitting
any motion to any other link.
69
00:07:00,830 --> 00:07:05,990
Such a degree of freedom is referred to as
redundant degree of freedom.
70
00:07:05,990 --> 00:07:12,020
Let me now explain some redundant degrees
of freedom and how to take care of that in
71
00:07:12,020 --> 00:07:17,230
the equation so that we get the effective
degrees of freedom.
72
00:07:17,230 --> 00:07:24,290
As an example of a redundant degree of freedom,
let us look at this 4-link mechanism, where
73
00:07:24,290 --> 00:07:30,870
we have link number 1 which is the fixed link;
link 2 which is connected to link 1 through
74
00:07:30,870 --> 00:07:33,460
the revolute pair at O2.
75
00:07:33,460 --> 00:07:42,120
There is link 4, which is connected to link
1 through the revolute pair at O4.
76
00:07:42,120 --> 00:07:51,180
And link 3, has 2 prismatic pairs connecting
it to link 2 and link 4.
77
00:07:51,180 --> 00:07:59,050
The thing to note is that the direction of
this prismatic pair is same; both these prismatic
78
00:07:59,050 --> 00:08:04,030
pairs have the direction along the link 3.
79
00:08:04,030 --> 00:08:11,520
Consequently, link 3 can be dragged along
the direction without transferring any motion
80
00:08:11,520 --> 00:08:14,930
either to link 2 or to link 4.
81
00:08:14,930 --> 00:08:19,770
Consequently, this constitutes a redundant
degree of freedom.
82
00:08:19,770 --> 00:08:31,690
If we apply the formula bluntly, that is F
= 3(n – 1) – 2j, we get,
83
00:08:31,690 --> 00:08:43,890
n = 4, j = 4; Therefore, F = 3(4 – 1) – 24
= 1.
84
00:08:43,890 --> 00:08:51,601
i.e., F = 1
It appears according to the formula this is
85
00:08:51,601 --> 00:08:57,020
a single degree freedom mechanism implying
unique input-output relationship.
86
00:08:57,020 --> 00:09:03,760
However, the link 2 or link 4 cannot be moved
at all.
87
00:09:03,760 --> 00:09:08,700
This is permanently locked which acts like
a structure.
88
00:09:08,700 --> 00:09:10,899
What is this degree of freedom ‘1’?
89
00:09:10,899 --> 00:09:16,790
That is nothing but, this redundant degree
of freedom of the link 3 along this direction
90
00:09:16,790 --> 00:09:18,490
of the prismatic pairs.
91
00:09:18,490 --> 00:09:26,080
It may be interesting to see what happens
if the directions of the 2 prismatic pairs
92
00:09:26,080 --> 00:09:27,480
are different.
93
00:09:27,480 --> 00:09:33,560
Between 3 and 4, it is in this direction;
whereas, between 3 and 2 it is along a different
94
00:09:33,560 --> 00:09:34,560
direction.
95
00:09:34,560 --> 00:09:42,399
Consequently, here the formula will work perfectly,
because, there is no redundant degree of freedom.
96
00:09:42,399 --> 00:09:48,370
I cannot move link 3 without transferring
motions to links 2 and 4.
97
00:09:48,370 --> 00:10:06,119
So, here n is 4, j is 4 as we obtained earlier
and F = 3(n – 1) – 2j = 3(4 – 1) – 24
98
00:10:06,119 --> 00:10:07,589
= 1.
99
00:10:07,589 --> 00:10:12,000
i.e., F = 1.
100
00:10:12,000 --> 00:10:19,580
Actually, here link 2 can be moved to transmit
motion to link 4.
101
00:10:19,580 --> 00:10:27,690
A little thought would convince that the rotation
of link 2 and link 4 must be identical.
102
00:10:27,690 --> 00:10:30,650
Let me explain why.
103
00:10:30,650 --> 00:10:38,070
As we see, link 2 and link 3 has a prismatic
pair here, which means there is no relative
104
00:10:38,070 --> 00:10:41,750
notation between link 2 and 3.
105
00:10:41,750 --> 00:10:48,070
Similarly, there is a prismatic pair here
between link 3 and link 4.
106
00:10:48,070 --> 00:10:52,640
So there cannot be any relative rotation between
link 3 and link 4.
107
00:10:52,640 --> 00:11:00,490
Consequently, there cannot be any relative
rotation between link 2 and link 4, both of
108
00:11:00,490 --> 00:11:03,610
which are in translation with respect to link
3.
109
00:11:03,610 --> 00:11:06,110
What is the implication?
110
00:11:06,110 --> 00:11:09,750
That there is no relative rotation between
links 2 and 4.
111
00:11:09,750 --> 00:11:14,270
Both of them rotate but they rotate by the
same amount, so that, there is no relative
112
00:11:14,270 --> 00:11:16,850
rotation.
113
00:11:16,850 --> 00:11:22,520
Let me now take another example of a redundant
degree of freedom which is very commonly seen.
114
00:11:22,520 --> 00:11:29,240
In the above figure, we see what is known
as a cam-follower mechanism and we have a
115
00:11:29,240 --> 00:11:32,410
roller follower.
116
00:11:32,410 --> 00:11:40,720
Cam is this input link which is link 2, which
is hinged to link 1, the fixed link at this
117
00:11:40,720 --> 00:11:45,060
revolute joint at O2.
118
00:11:45,060 --> 00:11:53,550
Follower is link 4 is hinged to roller at
this revolute pair; roller is the link 3.
119
00:11:53,550 --> 00:12:00,940
It is intuitively pretty obvious that if we
move link 2, say we give it a rotation then
120
00:12:00,940 --> 00:12:07,290
the follower will also have a rotation in
the given direction.
121
00:12:07,290 --> 00:12:13,610
There exists a unique input-output relationship,
unique rotations of link 2 causes unique rotation
122
00:12:13,610 --> 00:12:14,760
of link 4.
123
00:12:14,760 --> 00:12:18,630
Let me calculate the degree of freedom.
124
00:12:18,630 --> 00:12:23,260
As we have seen, there is a unique input-output
relationship depending on the shape of the
125
00:12:23,260 --> 00:12:26,910
cam profile, so the degree of freedom should
turn out to be 1.
126
00:12:26,910 --> 00:12:31,450
But, let me do it by counting according to
our formula.
127
00:12:31,450 --> 00:12:35,530
We have already seen that there are 4 links;
so n is 4.
128
00:12:35,530 --> 00:12:46,390
There are three revolute pairs: one between
1 and 2, one between 1 and 4 at O4 and one
129
00:12:46,390 --> 00:12:55,640
between 3 and 4, at the roller centre; so
j is 3.
130
00:12:55,640 --> 00:13:05,950
Now there is a higher pair between link number
2 and 3 at this point; so h is 1.
131
00:13:05,950 --> 00:13:21,800
If we calculate the degree of freedom F,
F = 3(n – 1) – 2j – h = 3(4 – 1) – 23
132
00:13:21,800 --> 00:13:29,580
– 1 = 2 i.e., F = 2.
133
00:13:29,580 --> 00:13:36,209
So, the degree of freedom according to the
formula is turning out to be two, because,
134
00:13:36,209 --> 00:13:44,140
there is a redundant degree of freedom and
that is, roller 3 can be rotated about this
135
00:13:44,140 --> 00:13:49,670
revolute pair without transferring any motion
either to link 2 or link 4.
136
00:13:49,670 --> 00:13:52,550
So, that is the redundant degree of freedom.
137
00:13:52,550 --> 00:14:01,640
So Fr if we call as the redundant degree of
freedom, Fr is 1.
138
00:14:01,640 --> 00:14:06,740
In view of this redundant degree of freedom,
let us modify our equation which we obtained
139
00:14:06,740 --> 00:14:07,740
earlier.
140
00:14:07,740 --> 00:14:12,800
Now that we have seen there can be some redundant
degrees of freedom, let us now modify the
141
00:14:12,800 --> 00:14:15,640
formula in view of this.
142
00:14:15,640 --> 00:14:23,019
Feff that is the really the input-output relationship
is governed by
143
00:14:23,019 --> 00:14:36,240
Feff = 3(n – 1) – 2j – h – Fr
where Fr is the total number of redundant
144
00:14:36,240 --> 00:14:37,980
degrees of freedom.
145
00:14:37,980 --> 00:14:44,870
Sometimes due to some other practical considerations,
a mechanism may have some redundant kinematic
146
00:14:44,870 --> 00:14:52,040
pairs, which means, those kinematic pairs
are not kinematically important, but they
147
00:14:52,040 --> 00:14:54,680
may be required due to some other considerations.
148
00:14:54,680 --> 00:15:02,010
The simplest example is a shaft is normally
mounted on 2 bearings, but both the bearings
149
00:15:02,010 --> 00:15:07,600
act as 1 revolute pair permitting rotation
about the same axis.
150
00:15:07,600 --> 00:15:13,820
By counting we may call it 2, but kinematically,
that is only 1 revolute pair.
151
00:15:13,820 --> 00:15:17,570
Let me show an example of such redundant kinematic
pair.
152
00:15:17,570 --> 00:15:21,060
Here, we consider a 6-link mechanism.
153
00:15:21,060 --> 00:15:31,950
Link 1 which is fixed, which is connected
to link 2 through a revolute pair at O2.
154
00:15:31,950 --> 00:15:35,540
Link 3 having a revolute pair between 2 and
3.
155
00:15:35,540 --> 00:15:45,029
Link 4 is next link which is connected to
link 3 by this revolute pair.
156
00:15:45,029 --> 00:15:48,620
Link 5 is connected to link 2 by this revolute
pair.
157
00:15:48,620 --> 00:15:52,920
Link 5 is connected to link 4 by this prismatic
pair here.
158
00:15:52,920 --> 00:15:58,980
Link 6 is another link which is connected
to link 5 through this revolute pair.
159
00:15:58,980 --> 00:16:02,079
And link 6 is connected to link 4 by this
prismatic pair.
160
00:16:02,079 --> 00:16:09,250
Link 6 is connected to link 1 by this prismatic
pair and link 4 is connected to link 1 by
161
00:16:09,250 --> 00:16:11,100
this prismatic pair.
162
00:16:11,100 --> 00:16:15,740
Let me apply the formula and try to find the
degree of freedom of this mechanism.
163
00:16:15,740 --> 00:16:21,880
Here, we have n is 6.
164
00:16:21,880 --> 00:16:26,019
All these pairs are simple pairs because they
connect only 2 links.
165
00:16:26,019 --> 00:16:40,670
So j, we count there are 1, 2, 3, 4, 5 revolute
pairs and 3 prismatic pairs, so j is 8.
166
00:16:40,670 --> 00:16:49,180
Consequently, the degree of freedom of the
mechanism F is given by
167
00:16:49,180 --> 00:17:01,769
F = 3(6 – 1) – 28 = 15 – 16 = –1 i.e.,
F = –1
168
00:17:01,769 --> 00:17:06,949
That means according to the formula, this
mechanism is a structure rather a statically
169
00:17:06,949 --> 00:17:11,790
indeterminate structure with negative degrees
of freedom and no relative motion should be
170
00:17:11,790 --> 00:17:14,860
possible between various links.
171
00:17:14,860 --> 00:17:23,449
However, as we see shortly, this has degree
of freedom 1 and there is a unique input-output
172
00:17:23,449 --> 00:17:31,270
relationship that means, if we use link 2
as my input link and rotate it, link 4, which
173
00:17:31,270 --> 00:17:37,320
we may treat as output link will have some
motion.
174
00:17:37,320 --> 00:17:40,700
Now why is this calculation failing?
175
00:17:40,700 --> 00:17:49,440
This is because if we notice these 3 prismatic
pairs, we should note that all these 3 prismatic
176
00:17:49,440 --> 00:17:53,520
pairs are in the same direction.
177
00:17:53,520 --> 00:18:02,720
This prismatic pair is allowing horizontal
translation between link 1 and link 6.
178
00:18:02,720 --> 00:18:09,770
This prismatic pair here is allowing relative
translation in the horizontal direction between
179
00:18:09,770 --> 00:18:14,429
link 1 and link 4.
180
00:18:14,429 --> 00:18:21,260
This prismatic pair which is there to ensure
horizontal translation between link 4 and
181
00:18:21,260 --> 00:18:24,100
link 6 may be redundant.
182
00:18:24,100 --> 00:18:30,299
Even we can replace, we can withdraw, any
of these 3 prismatic pairs because all of
183
00:18:30,299 --> 00:18:39,110
these are ensuring horizontal translation
between links 1, 4 and 6.
184
00:18:39,110 --> 00:18:49,650
Thus, j which we counted previously as 8 is
actually j is 7 because, kinematically, 1
185
00:18:49,650 --> 00:18:52,380
of these 3 prismatic pairs is redundant.
186
00:18:52,380 --> 00:18:57,860
So, I can remove this as a redundant pair
and make j equal to 7, which will give me
187
00:18:57,860 --> 00:19:06,820
F = 1.
188
00:19:06,820 --> 00:19:12,440
Now that we have seen there is a possibility
in an actual mechanism to have some redundant
189
00:19:12,440 --> 00:19:21,790
kinematic pairs, let us rewrite the formula
in the light of such redundant kinematic pairs.
190
00:19:21,790 --> 00:19:28,150
If Feff implies the effective degree of freedom
of a mechanism that is given by,
191
00:19:28,150 --> 00:19:38,309
Feff = 3(n – 1) – 2(j – jr) – h – Fr
where Fr was the redundant degrees of freedom,
192
00:19:38,309 --> 00:19:47,530
h is the number of higher pairs, jr is the
number of redundant kinematic pairs, j is
193
00:19:47,530 --> 00:19:51,510
the total number of lower pairs and n is the
total number of link.
194
00:19:51,510 --> 00:19:58,240
Thus, we arrive at a formula by counting the
number of links and considering the different
195
00:19:58,240 --> 00:20:03,980
types of pairs and redundant degrees of freedom
and redundant kinematic pair, we are in a
196
00:20:03,980 --> 00:20:10,490
position to calculate the effective degrees
of freedom of a planar mechanism.
197
00:20:10,490 --> 00:20:16,890
At this stage, we would like to emphasize
a very subtle difference between these revolute
198
00:20:16,890 --> 00:20:19,260
pairs and prismatic pairs.
199
00:20:19,260 --> 00:20:24,200
So far as this formula is concerned, we have
not made any distinction between a revolute
200
00:20:24,200 --> 00:20:30,600
pair and a prismatic pair, because both types
of pairs curtailing 2 degrees of freedom allowed
201
00:20:30,600 --> 00:20:32,900
1 degree of freedom.
202
00:20:32,900 --> 00:20:37,160
Let me now point out what is this subtle difference.
203
00:20:37,160 --> 00:20:50,651
Let us notice this 3-link closed mechanism
consisting of only 3 revolute pairs: Link
204
00:20:50,651 --> 00:21:01,330
1, link 2 and link 3 constitutes a closed
kinematic chain consisting of 3 revolute pairs.
205
00:21:01,330 --> 00:21:05,910
We are already familiar with this and we have
seen that it is not a mechanism.
206
00:21:05,910 --> 00:21:12,549
It is a structure; no relative motion between
various links is possible when all these pairs
207
00:21:12,549 --> 00:21:13,580
are revolute pairs.
208
00:21:13,580 --> 00:21:21,330
Let us see what happens if all 3 becomes prismatic
pairs in different directions.
209
00:21:21,330 --> 00:21:30,860
Again there are 3 links: link 1, link 2 and
link 3.
210
00:21:30,860 --> 00:21:36,050
This constitutes a closed kinematic chain
and there are 3 prismatic pairs.
211
00:21:36,050 --> 00:21:42,400
One in this horizontal direction between link
1 and 2, one in the vertical direction between
212
00:21:42,400 --> 00:21:49,620
link 1 and 3 and there is one in this inclined
direction between links 2 and 3.
213
00:21:49,620 --> 00:21:56,110
The kinematic representation of this is as
follows: There are a three links: links 1,
214
00:21:56,110 --> 00:22:02,520
2 and 3 having 3 prismatic pairs in different
directions.
215
00:22:02,520 --> 00:22:08,669
It is obvious that here relative motion between
various links is possible; it is not a structure,
216
00:22:08,669 --> 00:22:14,960
the degree of freedom of this loop is not
zero.
217
00:22:14,960 --> 00:22:21,039
As we can see link 2 can be moved in the horizontal
direction to produce a unique vertical movement
218
00:22:21,039 --> 00:22:22,900
for link 3.
219
00:22:22,900 --> 00:22:35,520
Thus, for this particular closed loop mechanism,
n is 3, j is also 3.
220
00:22:35,520 --> 00:22:46,950
So according to the formula, we should have
{3(n – 1) – 2j = 32 – 23 = 0, i.e.,
221
00:22:46,950 --> 00:22:54,830
F = 0}, which is true for the revolute pairs,
but not true for the prismatic pair.
222
00:22:54,830 --> 00:23:00,660
In light of this difference between revolute
and prismatic pair, let us modify our formula
223
00:23:00,660 --> 00:23:04,490
for calculating the degrees of freedom.
224
00:23:04,490 --> 00:23:12,679
In view of this single degree of freedom closed
loop, which is possible by 3 prismatic pairs
225
00:23:12,679 --> 00:23:19,820
connecting 3 links, let us modify the formula
for calculating the effective degrees of freedom.
226
00:23:19,820 --> 00:23:29,870
Therefore,
Feff = 3(n – 1) – 2(j – jr) – h – Fr
227
00:23:29,870 --> 00:23:38,380
+ PL,
where PL is the number of 3 link closed loops
228
00:23:38,380 --> 00:23:42,370
having 3 prismatic pairs in different directions.
229
00:23:42,370 --> 00:23:50,620
While deriving this formula, we have not bothered
with the kinematic dimensions of the mechanism.
230
00:23:50,620 --> 00:23:56,510
So, this formula may have some exceptions
for some very special kinematic dimensions,
231
00:23:56,510 --> 00:24:01,360
which we shall see shortly through a number
of examples.
232
00:24:01,360 --> 00:24:08,071
We have already said that due to some special
kinematic dimensions the formula that we derived
233
00:24:08,071 --> 00:24:09,870
may give wrong result.
234
00:24:09,870 --> 00:24:13,880
As an example, let us talk of this parallelogram
linkage.
235
00:24:13,880 --> 00:24:26,020
It is a 4-link mechanism with 4 revolute pairs,
but the opposite sides have equal lengths.
236
00:24:26,020 --> 00:24:32,080
These 2 links are of same length and this
coupler length is equal to the frame length,
237
00:24:32,080 --> 00:24:36,500
that is, the distance between these 2 fixed
pivots.
238
00:24:36,500 --> 00:24:46,950
Obviously, this is a 4R mechanism, which is
degree of freedom 1 and it can transmit motion
239
00:24:46,950 --> 00:24:49,340
from this link to that link.
240
00:24:49,340 --> 00:24:55,880
During this movement, the opposite sides always
remain of same length; so a parallelogram
241
00:24:55,880 --> 00:24:58,539
remains a parallelogram.
242
00:24:58,539 --> 00:25:04,670
In this parallelogram linkage, if we add an
extra coupler which is parallel to the original
243
00:25:04,670 --> 00:25:08,080
coupler then what happens?
244
00:25:08,080 --> 00:25:16,159
As we see now ‘n’ has become 5 and due
to this extra coupler, we have introduced
245
00:25:16,159 --> 00:25:20,360
2 revolute pairs at its 2 ends.
246
00:25:20,360 --> 00:25:24,840
So, j has become 6.
247
00:25:24,840 --> 00:25:30,850
Consequently, from the formula, we get, F
= 3(5 – 1) – 26 = 34 – 26 = 0, i.e.,
248
00:25:30,850 --> 00:25:31,850
F = 0.
249
00:25:31,850 --> 00:25:43,049
So
the formula tells us that, this is structure,
250
00:25:43,049 --> 00:25:49,659
but intuitively we can realize that this extra
coupler has not imposed any extra constant
251
00:25:49,659 --> 00:25:55,409
and the mechanism still retains its single
degree of freedom and this moves like a parallelogram
252
00:25:55,409 --> 00:25:57,330
as before.
253
00:25:57,330 --> 00:26:03,850
Of course, this failure of the formula is
only because these 2 couplers are parallel
254
00:26:03,850 --> 00:26:07,490
and the original diagram was a parallelogram.
255
00:26:07,490 --> 00:26:12,870
If this extra coupler which is introduced
in an inclined fashion, say starting from
256
00:26:12,870 --> 00:26:19,900
this point to this point, then the formula
will be correct and the assembly will become
257
00:26:19,900 --> 00:26:21,250
a structure.
258
00:26:21,250 --> 00:26:31,919
In fact, such an extra coupler is normally
used to drive a parallelogram mechanism.
259
00:26:31,919 --> 00:26:37,860
As we shall see in a model that, when the
parallelogram moves, there is a configuration
260
00:26:37,860 --> 00:26:44,520
when all the links become collinear and that
mechanism loses its transmission quality.
261
00:26:44,520 --> 00:26:53,350
In fact, it can go into a non-parallelogram
or anti-parallelogram configuration.
262
00:26:53,350 --> 00:26:59,990
To ensure that a parallelogram always remains
a parallelogram such an extra coupler is necessary.
263
00:26:59,990 --> 00:27:07,700
In fact, to maintain the good transmission
quality at all configurations, these 2 extra
264
00:27:07,700 --> 00:27:18,730
couplers are connected to the input and output
link by making a 900 angle between the extensions
265
00:27:18,730 --> 00:27:26,360
of the input link and the output link, such
that, when this particular coupler is collinear
266
00:27:26,360 --> 00:27:34,720
with the line of frame, the other coupler
is parallel to the line of frame, this portion
267
00:27:34,720 --> 00:27:38,500
of the links become perpendicular to the line
of frame.
268
00:27:38,500 --> 00:27:44,240
This point will be much clearer when you demonstrate
it through a model.
269
00:27:44,240 --> 00:27:48,090
Let us now look at the model of this parallelogram
linkage.
270
00:27:48,090 --> 00:27:53,150
Here red link and blue link are of equal link
length.
271
00:27:53,150 --> 00:27:59,580
The coupler which is yellow link has the same
length as the fixed link or the distance between
272
00:27:59,580 --> 00:28:02,059
the 2 fixed pivots.
273
00:28:02,059 --> 00:28:07,760
As we see, this parallelogram linkage when
it moves always remains a parallelogram.
274
00:28:07,760 --> 00:28:14,429
However, when all the 4 links become collinear,
there is a possibility that it flips into
275
00:28:14,429 --> 00:28:19,990
anti-parallelogram configuration and it does
not move as a parallelogram linkage.
276
00:28:19,990 --> 00:28:27,900
Again, here, if sufficient care is taken,
one may transfer it to a parallelogram linkage.
277
00:28:27,900 --> 00:28:37,159
To get rid of this uncertainty configuration,
it is better to have an extra coupler as explained
278
00:28:37,159 --> 00:28:41,900
earlier and we shall demonstrate it through
our next model.
279
00:28:41,900 --> 00:28:47,170
Let us now look at the model of above parallelogram
linkage with a redundant coupler.
280
00:28:47,170 --> 00:28:55,940
As we see these 2 links are extended at 900
and there are 2 parallel couplers.
281
00:28:55,940 --> 00:29:02,370
Consequently, here we shall be able to maintain
the parallelogram configuration throughout
282
00:29:02,370 --> 00:29:04,039
the cycle of motion.
283
00:29:04,039 --> 00:29:12,400
It can never flip back into anti-parallelogram
configuration.
284
00:29:12,400 --> 00:29:19,210
As we have just seen that for very special
kinematic dimensions, the formula for calculating
285
00:29:19,210 --> 00:29:21,559
the degrees of freedom may fail.
286
00:29:21,559 --> 00:29:27,510
In fact, when the formula was telling that
the degree of freedom is 0, we are getting
287
00:29:27,510 --> 00:29:31,220
single degree of freedom mechanism.
288
00:29:31,220 --> 00:29:36,851
For special kinematic dimensions, when the
degree of freedom calculation fails, according
289
00:29:36,851 --> 00:29:42,600
to the formula, such linkages are called over
closed linkages.
290
00:29:42,600 --> 00:29:48,700
As a further example of over close linkages,
let us look at this 10-link mechanism.
291
00:29:48,700 --> 00:29:58,400
Here, we have link 1, which is the fixed link;
link 2 connected to link 3 connected to link
292
00:29:58,400 --> 00:30:02,299
4 which in turn is again connected to link
1.
293
00:30:02,299 --> 00:30:05,220
That means, we get a simple 4-bar mechanism.
294
00:30:05,220 --> 00:30:14,780
There is another 4-bar mechanism: link 8,
link 9, link 10 and link 1.
295
00:30:14,780 --> 00:30:24,190
There is a third 4-bar mechanism consisting
of: link 7, link 6, link 5 and link 1.
296
00:30:24,190 --> 00:30:29,570
All these 4-bar mechanisms are connected at
this revolute pair C.
297
00:30:29,570 --> 00:30:37,559
So, in all we have 10 linked mechanisms and
let me also see, what typical revolute pairs
298
00:30:37,559 --> 00:30:39,039
are there.
299
00:30:39,039 --> 00:30:46,120
There is a revolute pair at O2 which connects
3 links namely 1, 2 and 5.
300
00:30:46,120 --> 00:30:53,900
There is a revolute pair at O4 which again
connects 3 links namely 1, 4 and 10.
301
00:30:53,900 --> 00:31:01,230
There is a revolute pair at O which again
connects 3 links namely 7, 8 and 1.
302
00:31:01,230 --> 00:31:07,980
There is a revolute pair at C which connects
3 links namely 3, 6 and 9.
303
00:31:07,980 --> 00:31:16,620
Thus, we have 4 such hinges of j2 category.
304
00:31:16,620 --> 00:31:29,279
There are simple hinges at A, at B, at G,
at F, at E and at D.
305
00:31:29,279 --> 00:31:36,010
Let us try to calculate the degrees of freedom
of this particular mechanism.
306
00:31:36,010 --> 00:31:43,169
We have already seen ‘n’, which is the
total number of links are 10.
307
00:31:43,169 --> 00:31:56,880
j1 - that is the number of simple hinges which
are at A, B, G, F, E and D that is j1 is 6;
308
00:31:56,880 --> 00:32:04,690
number of compound hinges each one of which
connects 3 links, that is j2 is at O2, O4,
309
00:32:04,690 --> 00:32:10,110
O and C, that is j2 is equal to 4.
310
00:32:10,110 --> 00:32:13,980
Degree of freedom of this mechanism according
to the formula is
311
00:32:13,980 --> 00:32:38,460
F = 3(10 – 1) – 2(6 + 24) = 27 – 214
= –1, i.e., F = –1
312
00:32:38,460 --> 00:32:47,029
So, without any special dimensions this assembly
is a structure with degree of freedom –1.
313
00:32:47,029 --> 00:32:59,429
However, if we look at this figure what we
see that O2ACD is a parallelogram; O4GCB that
314
00:32:59,429 --> 00:33:06,240
is another parallelogram and OFCE is another
parallelogram.
315
00:33:06,240 --> 00:33:16,149
Not only that this ternary links that is 3,
9 and 6, all these 3 ternary links are similar
316
00:33:16,149 --> 00:33:22,919
triangles as indicated by the angles alpha,
beta and gamma.
317
00:33:22,919 --> 00:33:27,840
Due to these special dimensions, we will find
that the degree of freedom of this assembly
318
00:33:27,840 --> 00:33:31,140
will become equal to 1.
319
00:33:31,140 --> 00:33:36,340
That means, this is another example of an
over closed linkage, where some of the constraints
320
00:33:36,340 --> 00:33:40,669
may be redundant, but this will not be highlighted
in this lecture.
321
00:33:40,669 --> 00:33:45,590
We will just show you the model of this particular
mechanism.
322
00:33:45,590 --> 00:33:51,169
Let us now look at the model of this 10-link
mechanism which has just been discussed.
323
00:33:51,169 --> 00:33:56,559
As we have seen according to the calculation
the degree of freedom should have been –1,
324
00:33:56,559 --> 00:34:04,380
but notice that these 4 hinges constitute
a parallelogram; so does these 4 and these
325
00:34:04,380 --> 00:34:08,179
4 hinges also constitutes another parallelogram.
326
00:34:08,179 --> 00:34:13,679
These 3 triangles, the ternary links are similar
to each other.
327
00:34:13,679 --> 00:34:20,349
Consequently, this constitutes a single degree
freedom mechanism, which is an over closed
328
00:34:20,349 --> 00:34:28,139
linkage which has mobility; it is not a structure.
329
00:34:28,139 --> 00:34:34,940
As the further example of an over closed linkage,
let us consider this 8-link mechanism which
330
00:34:34,940 --> 00:34:38,899
is known as Kempe Burmeister focal mechanism.
331
00:34:38,899 --> 00:34:53,709
As we see, there are 8 links: link 1, link
2, link 3, 4, 5, 6, 7 and 8.
332
00:34:53,709 --> 00:35:11,890
These 8 links are connected by revolute pairs
one at O2, at S, A, P, B, Q, O4 and R. There
333
00:35:11,890 --> 00:35:20,789
are 8 simple hinges and there is a higher
order hinge at this point ‘T’ where 4
334
00:35:20,789 --> 00:35:25,579
links namely 5, 6, 7 and 8 are connected.
335
00:35:25,579 --> 00:35:41,519
So, if we calculate the degree of freedom,
we see n is 8, j1 is 8, j2 is 0, but there
336
00:35:41,519 --> 00:35:47,799
is a j3 at T, where 4 links are connected
so j3 is 1.
337
00:35:47,799 --> 00:36:12,849
The degree of freedom F is calculated below:
F = 3(8 – 1) – 2(8 + 31) = 21 – 22 = –1,
338
00:36:12,849 --> 00:36:18,299
i.e., F = –1.
339
00:36:18,299 --> 00:36:21,440
According to the formula, it should be a structure.
340
00:36:21,440 --> 00:36:34,569
However, for very special dimensions, as indicated
by these similar triangles BTQ with O2TS,
341
00:36:34,569 --> 00:36:39,999
this angle is equal to this angle and this
angle is equal to this angle.
342
00:36:39,999 --> 00:36:44,190
Similarly, there are other similar triangles
in this figure.
343
00:36:44,190 --> 00:36:49,880
For such special dimensions as we see in our
model the degree of freedom will turn out
344
00:36:49,880 --> 00:36:51,789
to be 1.
345
00:36:51,789 --> 00:37:02,279
F will be 1, that means it will be a constrained
mechanism with single degree of freedom.
346
00:37:02,279 --> 00:37:07,969
Let us now consider the model of this Kempe
Burmeister focal mechanism, which we have
347
00:37:07,969 --> 00:37:09,440
just discussed.
348
00:37:09,440 --> 00:37:20,289
As we see including this fixed link, we have
8 links: 2, 3, 4, 5, 6, 7, 8 and this is a
349
00:37:20,289 --> 00:37:25,859
hinge where 4 links are connected and all
other hinges are simple hinges.
350
00:37:25,859 --> 00:37:30,269
Accordingly, the formula said the degree of
freedom should be minus 1.
351
00:37:30,269 --> 00:37:36,430
But however, as we see this mechanism can
be moved very easily and there is a unique
352
00:37:36,430 --> 00:37:38,819
input-output relationship.
353
00:37:38,819 --> 00:37:46,229
That means, the effective degree of freedom
of this mechanism is 1; that is only because
354
00:37:46,229 --> 00:37:48,030
of the special dimension.
355
00:37:48,030 --> 00:37:52,759
If we change any of these points a little
bit, this will really become a structure and
356
00:37:52,759 --> 00:37:57,130
no relative movement would be possible.
357
00:37:57,130 --> 00:38:03,430
As a last example of an over closed linkage,
let us look at this 5-link mechanism which
358
00:38:03,430 --> 00:38:06,160
is known as cross slider trammel.
359
00:38:06,160 --> 00:38:09,989
Here we have link 1 which is the fixed link.
360
00:38:09,989 --> 00:38:19,299
Link 3 which is connected to link 1 and link
3 is connected to link 4 and link 2.
361
00:38:19,299 --> 00:38:25,309
Link 4 and 2 are having prismatic pairs with
link 5.
362
00:38:25,309 --> 00:38:31,150
Link 2 has a prismatic pair in the horizontal
direction with link 5.
363
00:38:31,150 --> 00:38:37,009
Link 4 has a prismatic pair in the vertical
direction with link 5.
364
00:38:37,009 --> 00:38:55,019
Thus, we have n equal to 5; we have 4 revolute
pairs here and here and here and here.
365
00:38:55,019 --> 00:39:02,239
So, j is 4 revolute pairs plus 2 prismatic
pairs.
366
00:39:02,239 --> 00:39:16,599
Thus, F turns out to be according to the formula,
{3(5 – 1) – 26 = 34 – 12 = 0}.
367
00:39:16,599 --> 00:39:31,109
So, the effective degree of freedom of this
mechanism according to the formula; Feff is
368
00:39:31,109 --> 00:39:38,440
0.
369
00:39:38,440 --> 00:39:43,039
Without any special dimension, it will be
a structure and there should not be any mobility
370
00:39:43,039 --> 00:39:44,459
any relative movement.
371
00:39:44,459 --> 00:39:57,609
However, for these special dimensions when
we make OC is same as BC as same as AC, then
372
00:39:57,609 --> 00:40:05,190
we will see that there will be an effective
degree of freedom of this mechanism will turn
373
00:40:05,190 --> 00:40:08,449
out to be just 1.
374
00:40:08,449 --> 00:40:17,660
In fact, as we see the angular velocity of
link number 3 to that of link number 5 which
375
00:40:17,660 --> 00:40:27,390
are both rotating with respect to fixed link
1 will be exactly half for these special dimensions.
376
00:40:27,390 --> 00:40:32,819
This is known as cross slider trammel and
we would like to encourage the students to
377
00:40:32,819 --> 00:40:40,180
show that why it moves by starting from the
elliptic trammel that we have discussed in
378
00:40:40,180 --> 00:40:41,400
an earlier lecture.
379
00:40:41,400 --> 00:40:46,459
Now, we shall demonstrate cross-slider trammel
through a model.
380
00:40:46,459 --> 00:40:51,549
Let us now look at the model of cross-slider
trammel.
381
00:40:51,549 --> 00:40:57,719
This is that link 3, which has a revolute
pair with fixed link here.
382
00:40:57,719 --> 00:41:05,119
This is that link 5, which has a revolute
pair with fixed link here and there are 2
383
00:41:05,119 --> 00:41:11,930
sliders: 2 and 4 which are hinged to link
3, here and here and these 2 sliders move
384
00:41:11,930 --> 00:41:14,519
in these 2 perpendicular slots.
385
00:41:14,519 --> 00:41:21,150
For the special dimensions, as we see this
has degree of freedom is 1 and rotation of
386
00:41:21,150 --> 00:41:25,259
link 3 produces unique rotation of link 5.
387
00:41:25,259 --> 00:41:33,619
In fact, we can see that 2 revolutions of
link 3 produces 1 revolution of link 5.
388
00:41:33,619 --> 00:41:42,099
That is, one can show that Omega3/Omega5 at
all instance remain half.
389
00:41:42,099 --> 00:41:46,599
Let me now summarize, what has been covered
in today’s lecture.
390
00:41:46,599 --> 00:41:53,369
What we have seen how we can calculate the
degrees of freedom of a planar mechanism by
391
00:41:53,369 --> 00:41:59,219
counting the number of links and different
times of kinematic pairs.
392
00:41:59,219 --> 00:42:06,339
Attention has been also drawn to the fact
that there is a possibility of some redundant
393
00:42:06,339 --> 00:42:09,640
degrees of freedom that has to be accounted
for.
394
00:42:09,640 --> 00:42:15,990
Then, we have also seen there may be some
kinematic pairs which are redundant in the
395
00:42:15,990 --> 00:42:22,009
sense they do not serve any purpose so far
kinematics is concerned, but they maybe there
396
00:42:22,009 --> 00:42:25,230
due to some other practical considerations.
397
00:42:25,230 --> 00:42:31,609
At the end we have seen, that these formulas
which are derived only from the count without
398
00:42:31,609 --> 00:42:38,359
any consideration of any kinematic dimensions
may fail when there are some special kinematic
399
00:42:38,359 --> 00:42:39,589
dimensions.
400
00:42:39,589 --> 00:42:46,349
We have also seen some such over closed linkages
through the models, how they move, though
401
00:42:46,349 --> 00:43:03,640
the formula says they should be structures.
402
00:43:03,640 --> 00:43:18,880
1