ÿþ1
00:00:05,140 --> 00:00:12,140
The
topic of today's lecture is mobility analysis.
By mobility analysis, we obtain the degrees
2
00:00:23,900 --> 00:00:30,900
of freedom of a given mechanism. This is accomplished
by the counting number of links and the number
3
00:00:32,669 --> 00:00:39,669
of different types of kinematic pairs those
are used to connect these links.
4
00:00:40,030 --> 00:00:47,030
Let me now elaborate, how we carry out this
mobility analysis for planar mechanisms. It
5
00:00:47,160 --> 00:00:54,160
is worthwhile to recall that in a planar mechanism
each link has 3 degrees of freedom 2 of which
6
00:00:56,170 --> 00:01:03,019
are translational in the plane of motion and
1 is rotational about an axis perpendicular
7
00:01:03,019 --> 00:01:05,100
to this plane of motion.
8
00:01:05,100 --> 00:01:11,390
Let there be n number of total links in a
mechanism, which includes the fixed link of
9
00:01:11,390 --> 00:01:18,390
the frame that means there are n minus 1 moving
links. When these links are not connected
10
00:01:19,479 --> 00:01:26,479
by any kinematic pair then the total degrees
of freedom is obviously 3 times n minus 1.
11
00:01:27,520 --> 00:01:31,899
For each of these n minus 1 links there are
3 degrees of freedom, so the total degree
12
00:01:31,899 --> 00:01:38,899
of freedom of the system is 3 times n minus
1. Let these n links be connected by j number
13
00:01:40,360 --> 00:01:47,360
of lower pairs. By lower pair in a planar
mechanism, we can mean either a revolute pair
14
00:01:48,079 --> 00:01:55,079
or a prismatic pair and each of these kinematic
pairs connects only 2 links. We also recall
15
00:01:58,079 --> 00:02:04,950
that whether it is a revolute pair or a lower
pair, at each of these pairs, 2 degrees of
16
00:02:04,950 --> 00:02:11,950
freedom is cuttled and only 1 out of 3 is
maintained. If there j number of total kinematic
17
00:02:14,819 --> 00:02:19,450
pairs 2 times j numbers of degrees of freedom
are cuttled.
18
00:02:19,450 --> 00:02:26,450
The effective degree of freedom of the mechanism
is reduced to f, which are the degrees of
19
00:02:27,260 --> 00:02:34,260
freedom of the mechanism is 3 times n minus
1 minus 2 times j. Let us consider a constant
20
00:02:37,019 --> 00:02:43,040
mechanism with a single degree of freedom;
that is, there exist a unique input-output
21
00:02:43,040 --> 00:02:50,040
relationship, where the degree of freedom
of the mechanism F is 1. Substituting F equal
22
00:02:50,290 --> 00:02:57,290
to 1 in the above equation, we get 2j minus
3n plus 4 is equal to 0.
23
00:03:01,110 --> 00:03:06,780
For a single degree of freedom mechanism,
maintaining a unique input-output relationship,
24
00:03:06,780 --> 00:03:12,700
the number of links and the number of lower
pairs must be related to this equation that
25
00:03:12,700 --> 00:03:19,700
is 2j minus 3n plus 4 equal to 0. This equation
is called Grubler's criterion for single degree
26
00:03:21,090 --> 00:03:28,090
of freedom mechanism. While deriving this
Grubler's criterion, we assume that each of
27
00:03:30,390 --> 00:03:37,390
these lower pairs is connecting only 2 links.
However, due to practical considerations some
28
00:03:38,140 --> 00:03:45,140
times more than 2 links can be connected at
a particular hinge. As an example of different
29
00:03:46,280 --> 00:03:52,299
types of kinematic pairs which, connects more
than 2 links let us consider this figure.
30
00:03:52,299 --> 00:03:59,299
Here 3 links namely 2, 3 and 4 are connected
by a single hinge at this location. Such hinges
31
00:04:04,769 --> 00:04:10,769
are called compound hinges or higher order
hinges. This particular compound hinge is
32
00:04:10,769 --> 00:04:17,440
equivalent to two simple hinges as explained
in the adjoining figure. For example, this
33
00:04:17,440 --> 00:04:24,440
particular hinge can be thought of as 2 hinges.
One connecting link number 2 and link number
34
00:04:24,700 --> 00:04:31,700
3, whereas another hinge connects link number
3 and link number 4.
35
00:04:32,540 --> 00:04:39,540
Thus, a hinge which connects 3 different links
is equivalent to 2 simple hinges. This way
36
00:04:42,250 --> 00:04:49,250
we can think of another type of hinge where
4 links are connected and such a hinge will
37
00:04:49,440 --> 00:04:56,220
obviously be equivalent to 3 simple hinges.
Maintaining this equivalent between higher
38
00:04:56,220 --> 00:05:02,320
order hinges and simple hinges, we would like
to modify the equation for calculating the
39
00:05:02,320 --> 00:05:08,350
degrees of freedom of a mechanism as follows.
When higher order hinges are present, the
40
00:05:08,350 --> 00:05:13,390
symbol j in the equation, we would like to
modify as follows.
41
00:05:13,390 --> 00:05:20,390
j is equal to j1, which represents the number
of simple hinges, which connects only 2 links
42
00:05:20,570 --> 00:05:27,570
plus 2j2, where j2 is the number of hinges
to each one of which connects 3 links and
43
00:05:27,730 --> 00:05:34,730
so on; that is j3 represents the number of
hinges each one of which connects 3 plus 1,
44
00:05:35,180 --> 00:05:42,180
that is 4 links and so on up to ji. ji is
the number of compound hinges each of which
45
00:05:44,550 --> 00:05:51,550
connects i plus 1 number of links. In a mechanism,
there can be higher pair as well and as we
46
00:05:53,080 --> 00:05:59,040
recall, if there is a higher pair then at
each higher pair only 1 translational degree
47
00:05:59,040 --> 00:06:05,580
of freedom is cuttled that is along the common
normal to the point or line of contact. Two
48
00:06:05,580 --> 00:06:11,870
other degrees of freedom can be retained.
Consequently, at each higher pair only 1 degree
49
00:06:11,870 --> 00:06:18,660
of freedom is cuttled.
I would like to modify the equation, the degrees
50
00:06:18,660 --> 00:06:25,660
of freedom of a mechanism F is equal to 3
times n minus 1 minus 2j minus h, where h
51
00:06:27,490 --> 00:06:33,910
represents the number of higher pairs, j represents
the number of equivalent simple hinges and
52
00:06:33,910 --> 00:06:40,910
n represents the number of total links.
Sometimes there can be some redundant degree
53
00:06:41,940 --> 00:06:48,270
of freedom of a mechanism. What do we mean
by a redundant degree of freedom? Due to some
54
00:06:48,270 --> 00:06:54,350
typical kinematic pairs and their placement,
we may find that in a mechanism a particular
55
00:06:54,350 --> 00:07:01,350
link may be moved without transmitting any
motion to any other link. Such a degree of
56
00:07:01,380 --> 00:07:05,990
freedom is referred to as redundant degree
of freedom.
57
00:07:05,990 --> 00:07:12,000
Let me now explain some redundant degrees
of freedom and how to take care of that in
58
00:07:12,000 --> 00:07:19,000
the equation so that we get the effective
degrees of freedom. As an example of a redundant
59
00:07:19,330 --> 00:07:26,060
degree of freedom, let us look at this 4-link
mechanism, where we have link number 1 which
60
00:07:26,060 --> 00:07:31,990
is the fixed link; link number 2 which is
connected to link number 1 through this revolute
61
00:07:31,990 --> 00:07:38,990
pair at O2. There is link number 4, which
is connected to link number 1 through this
62
00:07:39,310 --> 00:07:46,310
revolute pair at O4. Link number 3, has 2
prismatic pairs connecting it to link number
63
00:07:49,370 --> 00:07:55,430
2 and link number 4.
The thing to be noted is that the direction
64
00:07:55,430 --> 00:08:02,430
of this revolute pair is same; both this prismatic
pair the direction is along this link 3. Consequently,
65
00:08:04,940 --> 00:08:11,940
link 3 can be dragged along this direction
without transferring any motion either to
66
00:08:12,150 --> 00:08:19,150
link 2 or to link 4. Consequently, this constitutes
a redundant degree of freedom. If we apply
67
00:08:20,520 --> 00:08:27,520
the formula bluntly, that is F is equal to
3 times n minus 1 minus 2j, we get, there
68
00:08:32,409 --> 00:08:39,409
are 4 links, so this is 3 into 4 minus 1 three
minus 2. There are 4 kinematic pairs - 2 revolute
69
00:08:41,880 --> 00:08:48,880
and 2 prismatic. So 2 into 4 is equal to 1.
It appears according to the formula that this
70
00:08:51,610 --> 00:08:58,300
is a single degree freedom mechanism implying
unique input-output relationship. However,
71
00:08:58,300 --> 00:09:05,300
this link 2 or link 4 cannot be moved at all.
This is permanently locked; so this acts like
72
00:09:07,180 --> 00:09:13,080
a structure. What is this degree of freedom
1? That is nothing but, this redundant degree
73
00:09:13,080 --> 00:09:20,080
of freedom of the link 3 along this direction
of the prismatic pairs.
74
00:09:20,550 --> 00:09:26,090
It may be interesting to see what happens
if the directions of the 2 prismatic pairs
75
00:09:26,090 --> 00:09:32,590
are different. Between 3 and 4, it is in this
direction; whereas, between 3 and 2 it is
76
00:09:32,590 --> 00:09:39,590
along a different direction. Consequently,
here the formula will work perfectly, because,
77
00:09:40,540 --> 00:09:46,260
there is no redundant degree of freedom. We
cannot move link 3 without transferring motions
78
00:09:46,260 --> 00:09:53,260
to links 2 and 4. So, here n is 4, j is 4
as we obtained earlier and F is equal to 3
79
00:09:57,970 --> 00:10:04,970
times n minus 1 minus 2j is again 3 into 3
minus into 2 into 4 which is equal to 1.
80
00:10:13,230 --> 00:10:20,230
Actually, here link number 2 can be moved
to transmit motion to link number 4. A little
81
00:10:21,490 --> 00:10:28,130
thought would convince that the rotation of
link 2 and link 4 must be identical. Let me
82
00:10:28,130 --> 00:10:35,130
explain why. As we see, link 2 and link 3
has a prismatic pair here, which means there
83
00:10:37,280 --> 00:10:44,280
is no relative notation between link number
2 and 3. Similarly, there is a prismatic pair
84
00:10:44,920 --> 00:10:50,930
here between link 3 and link 4. So there cannot
be any relative rotation between link number
85
00:10:50,930 --> 00:10:57,530
3 and link number 4. Consequently, there cannot
be any relative rotation between link number
86
00:10:57,530 --> 00:11:04,190
2 and link number 4, both of which are in
translation with respect to link number 3.
87
00:11:04,190 --> 00:11:09,760
What is the implication? That there is no
relative rotation between links 2 and 4. Both
88
00:11:09,760 --> 00:11:16,760
of them rotate but they rotate by the same
amount, so that, there is no relative rotation.
89
00:11:16,860 --> 00:11:23,430
Let me now take another example of a redundant
degree of freedom which is very commonly seen.
90
00:11:23,430 --> 00:11:29,390
In this figure, we see what is known as a
cam follower mechanism and we have a roller
91
00:11:29,390 --> 00:11:36,390
follower. Cam is this input link which is
number 2, which is hinged to link number 1
92
00:11:39,360 --> 00:11:46,360
the fixed link at this revolute joint at O2.
Follower that is link number 4 is hinged to
93
00:11:47,850 --> 00:11:54,850
roller at this revolute pair; roller is the
link number 3. It is intuitively pretty obvious
94
00:11:55,550 --> 00:12:02,550
that if we move link number 2, say I give
it a rotation then the follower will also
95
00:12:03,720 --> 00:12:10,720
have a rotation in this direction.
There exists a unique input-output relationship,
96
00:12:10,790 --> 00:12:16,910
unique rotations of link 2 causes unique rotation
of link 4. Let me calculate the degree of
97
00:12:16,910 --> 00:12:22,650
freedom. As we have seen, there is a unique
input-output relationship depending on the
98
00:12:22,650 --> 00:12:27,630
shape of the cam profile, so the degree of
freedom should turn out to be 1. But, let
99
00:12:27,630 --> 00:12:34,029
me do it by counting according to our formula.
We have already seen that there are 4 links;
100
00:12:34,029 --> 00:12:41,029
so n is 4. There are three revolute pairs
one between 1 and 2, one between 1 and 4 at
101
00:12:43,850 --> 00:12:50,850
O4 and one between 3 and 4, at the roller
centre; so j is 3.
102
00:12:54,230 --> 00:13:01,230
There is a higher pair between link number
2 and 3 at this point; so h is 1. If we calculate
103
00:13:06,670 --> 00:13:13,670
the degree of freedom F, which is 3 times
n minus 1 minus 2j minus h, which is 3 into
104
00:13:20,180 --> 00:13:27,180
3 is 9 minus 2 into j is 6, minus h, that
is 1, which gives 2. So, the degree of freedom
105
00:13:32,260 --> 00:13:37,350
according to the formula is standing out to
be true, because, there is a redundant degree
106
00:13:37,350 --> 00:13:44,350
of freedom and that is, roller 3 can be rotated
about this revolute pair without transferring
107
00:13:46,350 --> 00:13:52,720
any motion either to link 2 or link 4. So,
that is the redundant degree of freedom. So
108
00:13:52,720 --> 00:13:59,720
Fr if we call as the redundant degree of freedom,
Fr is 1. In view of this redundant degree
109
00:14:03,070 --> 00:14:07,410
of freedom, let us modify our equation which
we obtained earlier.
110
00:14:07,410 --> 00:14:12,779
Now that we have seen there can be some redundant
degrees of freedom, let us now modify the
111
00:14:12,779 --> 00:14:16,380
formula in view of this.
112
00:14:16,380 --> 00:14:23,380
Feff that is the really the input-output relationship
is governed by Feff is given by 3 times n
113
00:14:26,500 --> 00:14:33,500
minus 1 minus 2j minus h minus Fr, where Fr
is the total number of redundant degrees of
114
00:14:36,430 --> 00:14:43,430
freedom. Sometimes due to some other practical
considerations, a mechanism may have some
115
00:14:43,580 --> 00:14:50,580
redundant kinematic pairs, which means, those
kinematic pairs are not kinematically important,
116
00:14:51,740 --> 00:14:56,830
but they may be required due to some other
considerations. The simplest example is a
117
00:14:56,830 --> 00:15:03,830
shaft is normally mounted on 2 bearings, but
both the bearings act as 1 revolute pair permitting
118
00:15:04,930 --> 00:15:11,300
rotation about the same axis. By counting
we may call it 2, but kinematically, that
119
00:15:11,300 --> 00:15:17,860
is only 1 revolute pair. Let me show an example
of such redundant kinematic pair.
120
00:15:17,860 --> 00:15:24,860
Here, we consider a 6-mechanism. This is link
number 1 which is fixed, which is connected
121
00:15:25,839 --> 00:15:32,839
to link number 2 through a revolute pair at
O2. This is link number 3 having a revolute
122
00:15:33,720 --> 00:15:40,720
pair between 2 and 3 here. 4 is the next link
which is connected to 3 by this revolute pair.
123
00:15:42,100 --> 00:15:49,100
5 is this link which is connected to link
number 2 by this revolute pair. 5 is connected
124
00:15:50,570 --> 00:15:56,940
to 4 by this prismatic pair here. 6 is another
link which is connected to link number 5 through
125
00:15:56,940 --> 00:16:03,940
this revolute pair. 6 is connected to 4 by
this prismatic pair. 6 is connected to 1 by
126
00:16:04,450 --> 00:16:11,110
this prismatic pair and 4 is connected to
1 by this prismatic pair.
127
00:16:11,110 --> 00:16:16,740
Let me apply the formula and try to find the
degree of freedom of this mechanism. Here,
128
00:16:16,740 --> 00:16:23,740
we have n is 6. All these pairs are simple
pairs because they connect only 2 links. So
129
00:16:26,279 --> 00:16:33,279
j we count there are 1, 2, 3, 4, 5 revolute
pairs and 3 prismatic pairs, so j is 8. Consequently,
130
00:16:42,540 --> 00:16:49,540
the degree of freedom of the mechanism F is
3 times 6 minus 1 that is 5 minus twice of
131
00:16:51,520 --> 00:16:58,520
j that is 2 times 8 and we get 15 minus 16,
that is, minus 1. According to the formula,
132
00:17:03,870 --> 00:17:09,400
this mechanism is a structure rather a statically
indeterminate structure with negative degrees
133
00:17:09,400 --> 00:17:16,370
of freedom and no relative motion should be
possible between various links. However, this
134
00:17:16,370 --> 00:17:23,370
as we see shortly, has degree of freedom 1
and there is a unique input-output relationship
135
00:17:24,500 --> 00:17:31,500
that means, if I use link 2 as my input link,
I rotate it, link 4, which I may treat as
136
00:17:32,300 --> 00:17:39,300
output link will have some motion.
Why is this calculation failing? This is because
137
00:17:42,150 --> 00:17:49,150
if we notice these 3 revolute pairs, we should
note that all this 3 prismatic pairs are in
138
00:17:50,920 --> 00:17:57,920
the same direction. This prismatic pair is
allowing horizontal translation between link
139
00:17:58,929 --> 00:18:05,929
number 1 and link number 6 1758 min. This
prismatic pair here is allowing relative translation
140
00:18:07,570 --> 00:18:14,570
in the horizontal direction between link number
1 and link number 4. This prismatic pair which
141
00:18:16,620 --> 00:18:23,620
is there to ensure horizontal translation
between link 4 and link 6 may be redundant.
142
00:18:23,640 --> 00:18:30,640
Even we can replace, we can withdraw, any
of this 3 prismatic pairs because all of these
143
00:18:30,890 --> 00:18:37,890
are ensuring horizontal translation between
links 1, 4 and 6. Thus, j which we counted
144
00:18:42,050 --> 00:18:49,050
previously as 8 is actually j is 7 because,
kinematically, 1 of these 3 prismatic pairs
145
00:18:51,730 --> 00:18:57,720
is redundant. So, I can remove this as a redundant
pair and make j equal to 7, which will give
146
00:18:57,720 --> 00:19:04,720
me F is equal to 15 minus 2 into 7, 14, which
is 1. Now that we have seen there is a possibility
147
00:19:09,450 --> 00:19:16,070
in an actual mechanism to have some redundant
kinematic pairs, let us rewrite the formula
148
00:19:16,070 --> 00:19:21,790
in the light of such redundant kinematic pairs.
149
00:19:21,790 --> 00:19:28,790
If Feff implies the effective degree of freedom
of a mechanism that is given by 3 times n
150
00:19:28,980 --> 00:19:35,980
minus 1 minus 2 times j minus jr minus h minus
Fr, where Fr was the redundant degrees of
151
00:19:37,850 --> 00:19:44,850
freedom, h was the number of higher pairs,
jr is the number of redundant kinematic pairs,
152
00:19:47,110 --> 00:19:52,960
j is the total number of lower pairs and n
is the total number of link. Thus, we arrive
153
00:19:52,960 --> 00:19:59,390
at a formula by counting the number of links
and considering the different types of pairs
154
00:19:59,390 --> 00:20:05,590
and redundant degrees of freedom and redundant
kinematic pair, we are in a position to calculate
155
00:20:05,590 --> 00:20:12,590
the effective degrees of freedom of a planar
mechanism. At this stage, I would like to
156
00:20:12,890 --> 00:20:19,350
emphasize a very subtle difference between
this revolute pairs and prismatic pairs. So
157
00:20:19,350 --> 00:20:24,460
far this formula is concerned, we have not
made any distinction between a revolute pair
158
00:20:24,460 --> 00:20:30,830
and a prismatic pair because, both types of
pairs cuttled 2 degrees of freedom and allowed
159
00:20:30,830 --> 00:20:37,160
1 degree of freedom. Let me now point out
what is this subtle difference.
160
00:20:37,160 --> 00:20:44,160
Let us notice this 3-link closed mechanism
consisting of only 3 revolute pairs Link number
161
00:20:51,150 --> 00:20:58,150
1, link number 2 and link number 3 constitutes
a closed kinematic chain consisting of 3 revolute
162
00:21:00,520 --> 00:21:06,460
pairs. We are already familiar with this and
we have seen that this is not a mechanism.
163
00:21:06,460 --> 00:21:12,610
It is a structure; no relative motion between
various links is possible when all these pairs
164
00:21:12,610 --> 00:21:19,610
are revolute pairs. Let us see what happens
if all 3 becomes prismatic pairs in different
165
00:21:20,260 --> 00:21:22,340
directions.
166
00:21:22,340 --> 00:21:29,340
This is again; there are 3 links link 1, link
2 and link 3. This constitutes a closed kinematic
167
00:21:32,760 --> 00:21:38,830
chain and there are 3 prismatic pairs. One
in this horizontal direction between link
168
00:21:38,830 --> 00:21:45,580
1 and 2, one in the vertical direction between
link 1 and 3 and there is one in this inclined
169
00:21:45,580 --> 00:21:49,610
direction between links 2 and 3.
170
00:21:49,610 --> 00:21:56,610
The kinematic representation of this is as
follows There are a three links links 1, 2
171
00:21:57,090 --> 00:22:04,090
and 3 having 3 prismatic pairs in different
directions. It is obvious that here relative
172
00:22:04,960 --> 00:22:09,970
motion between various links is possible;
it is not a structure, the degree of freedom
173
00:22:09,970 --> 00:22:14,970
of this loop is not zero.
174
00:22:14,970 --> 00:22:21,040
As we can see link 2 can be moved in the horizontal
direction to produce a unique vertical movement
175
00:22:21,040 --> 00:22:28,040
for link 3. Thus, for this particular closed
loop mechanism, n is 3, j is also 3. So according
176
00:22:36,120 --> 00:22:43,120
to the formula, we should have had 3 into
n minus 1, that is 2 minus 2j that is 2 into
177
00:22:45,540 --> 00:22:52,540
3 is 0, which is true for the revolute pairs,
but not true for the prismatic pair. In light
178
00:22:55,120 --> 00:23:01,590
of this difference between revolute and prismatic
pair, let us modify our formula for calculating
179
00:23:01,590 --> 00:23:08,590
the degrees of freedom. In view of this single
degree of freedom, closed loop which is possible
180
00:23:10,510 --> 00:23:17,450
by 3 prismatic pairs connecting 3 links, let
us modify the formula for calculating the
181
00:23:17,450 --> 00:23:19,410
effective degrees of freedom.
182
00:23:19,410 --> 00:23:26,410
I would like to say Feff is equal to 3 times
n minus 1 into minus 2 times j minus jr minus
183
00:23:29,760 --> 00:23:36,760
h minus Fr plus PL, where PL is the number
of 3 link closed loops having 3 prismatic
184
00:23:40,010 --> 00:23:47,010
pairs in different directions. While deriving
this formula, we have not bothered with the
185
00:23:47,490 --> 00:23:54,059
kinematic dimensions of the mechanism. So,
this formula may have some exceptions for
186
00:23:54,059 --> 00:24:01,059
some very special kinematic dimensions, as
we shall see shortly through a number of examples.
187
00:24:01,360 --> 00:24:07,990
We have already said that due to some special
kinematic dimensions the formula that we derived
188
00:24:07,990 --> 00:24:09,870
may give wrong result.
189
00:24:09,870 --> 00:24:16,870
As an example, let us talk of this parallelogram
linkage. This is a 4-link mechanism with 4
190
00:24:16,880 --> 00:24:23,880
revolute pairs, but the opposite sides have
equal lengths. These 2 links are of same lengths
191
00:24:28,340 --> 00:24:35,070
and this coupler length is equal to the frame
length, that is, the distance between these
192
00:24:35,070 --> 00:24:42,070
2 fixed pivots. Obviously, this is a 4R mechanism,
which is degree of freedom 1 and it can transmit
193
00:24:46,460 --> 00:24:53,460
motion from this link to that link. During
this movement, the opposite sides always remain
194
00:24:53,620 --> 00:24:58,540
of same length; so a parallelogram remains
a parallelogram.
195
00:24:58,540 --> 00:25:05,070
In this parallelogram linkage, if we add an
extra coupler which is parallel to the original
196
00:25:05,070 --> 00:25:12,070
coupler then what happens? As we see now n
has become 5 and due to this extra coupler,
197
00:25:15,040 --> 00:25:22,040
we have introduced 2 revolute pairs at its
2 ends one there and one there. So, j has
198
00:25:22,400 --> 00:25:29,400
become 6. Consequently, from the formula,
we get F equal to 3 times 5 minus 1, that
199
00:25:32,309 --> 00:25:39,309
is 4 minus 2 times j, that is 2 times 6, which
is 0.
200
00:25:40,140 --> 00:25:46,600
So the formula tells us that, this is structure,
but intuitively we can realize that this extra
201
00:25:46,600 --> 00:25:52,040
coupler has not imposed any extra constant
and the mechanism still retains its single
202
00:25:52,040 --> 00:25:59,040
degree of freedom and this moves like a parallelogram
as before. Of course, this failure of the
203
00:25:59,150 --> 00:26:06,150
formula is only because these 2 couplers are
parallel and the original diagram was a parallelogram.
204
00:26:07,500 --> 00:26:12,870
If this coupler, extra coupler, I introduced
in an inclined fashion, say starting from
205
00:26:12,870 --> 00:26:19,530
this point to this point 2615 min, then the
formula will be correct and the assembly will
206
00:26:19,530 --> 00:26:26,530
become a structure. In fact, such an extra
coupler is normally used to drive a parallelogram
207
00:26:28,299 --> 00:26:31,940
mechanism.
208
00:26:31,940 --> 00:26:37,870
As we shall see in a model that, when the
parallelogram moves, there is a configuration
209
00:26:37,870 --> 00:26:44,530
when all the links become collinear and that
mechanism loses its transmission quality.
210
00:26:44,530 --> 00:26:51,530
In fact, it can go into a non-parallelogram
or anti-parallelogram configuration. To ensure
211
00:26:54,360 --> 00:27:00,130
that a parallelogram always remains a parallelogram
such an extra coupler is necessary.
212
00:27:00,130 --> 00:27:07,130
In fact, to maintain the good transmission
quality at all configurations, these 2 extra
213
00:27:07,710 --> 00:27:14,710
couplers are connected to the input and output
link by making a 90 degree angle between the
214
00:27:18,059 --> 00:27:25,059
extensions of this input link and the output
link, such that, when this particular coupler
215
00:27:25,510 --> 00:27:32,510
is collinear with the line of frame, the other
coupler is parallel to the line of frame,
216
00:27:33,600 --> 00:27:40,120
this portion of the links become perpendicular
to the line of frame. This point will be much
217
00:27:40,120 --> 00:27:44,250
clearer when you demonstrate it through a
model.
218
00:27:44,250 --> 00:27:50,770
Let us now look at the model of this parallelogram
linkage. Here this red link and this blue
219
00:27:50,770 --> 00:27:57,150
link are of equal link length. This coupler
which is the yellow link has the same length
220
00:27:57,150 --> 00:28:03,700
as the fixed link or the distance between
the 2 fixed pivots. As we see this parallelogram
221
00:28:03,700 --> 00:28:10,700
linkage when it moves always remains a parallelogram.
However, when all the 4 links become collinear,
222
00:28:11,470 --> 00:28:17,210
there is a possibility that it flips into
anti-parallelogram configuration and it does
223
00:28:17,210 --> 00:28:24,210
not move as a parallelogram linkage. Again,
here, if sufficient care is taken, one may
224
00:28:24,870 --> 00:28:31,870
transfer it to a parallelogram linkage. To
get rid of this uncertainty configuration,
225
00:28:33,780 --> 00:28:39,460
it is better to have an extra coupler as explained
earlier and we shall demonstrate it through
226
00:28:39,460 --> 00:28:41,900
our next model.
227
00:28:41,900 --> 00:28:48,340
Let us now look at the model of this parallelogram
linkage with a redundant coupler. As we see
228
00:28:48,340 --> 00:28:55,340
these 2 links are extended at 90 degrees and
there are 2 parallel couplers. Consequently,
229
00:28:57,309 --> 00:29:04,049
here we shall be able to maintain the parallelogram
configuration throughout the cycle of motion.
230
00:29:04,049 --> 00:29:11,049
It can never flip back into anti-parallelogram
configuration.
231
00:29:12,410 --> 00:29:19,220
As we have just seen that for very special
kinematic dimensions, the formula for calculating
232
00:29:19,220 --> 00:29:24,950
the degrees of freedom may fail. In fact,
when the formula was telling that the degree
233
00:29:24,950 --> 00:29:31,950
of freedom is 0, we are getting single degree
of freedom mechanism. For special kinematic
234
00:29:32,330 --> 00:29:39,169
dimensions, when the degree of freedom calculation
fails, according to the formula, such linkages
235
00:29:39,169 --> 00:29:42,450
are called over closed linkages.
236
00:29:42,450 --> 00:29:49,450
As a further example of over close linkages,
let us look at this 10-link mechanism. Here,
237
00:29:49,669 --> 00:29:56,669
we have link number 1, which is the fixed
link; link 2 connected to link 3 connected
238
00:29:57,600 --> 00:30:04,429
to link 4 which in turn is again connected
to link 1. That means, we get a simple 4-bar
239
00:30:04,429 --> 00:30:11,429
mechanism. There is another 4-bar mechanism
link 8, link 9, link 10 and link 1. There
240
00:30:15,330 --> 00:30:22,330
is a third 4-bar mechanism consisting of link
7, link 6, link 5 and link 1. All these 4-bar
241
00:30:25,740 --> 00:30:29,590
mechanisms are connected at this revolute
pair C.
242
00:30:29,590 --> 00:30:36,590
So, in all we have 10 linked mechanisms and
let me also see, what typical revolute pairs
243
00:30:37,710 --> 00:30:44,710
are there. There is a revolute pair at O2
which connects 3 links namely 1, 2 and 5.
244
00:30:46,120 --> 00:30:53,120
There is a revolute pair at O4 which again
connects 3 links namely 1, 4 and 10. There
245
00:30:53,919 --> 00:31:00,919
is a revolute pair at O which again connects
3 links namely 7, 8 and 1. There is a revolute
246
00:31:01,970 --> 00:31:08,970
pair at C which connects 3 links namely 3,
6 and 9. These hinges are of j2 category and
247
00:31:12,460 --> 00:31:19,460
thus we have 4 such hinges of j2 category.
There are simple hinges at A, at B, at G,
248
00:31:22,820 --> 00:31:29,820
at F, at E and at D.
Let us try to calculate the degrees of freedom
249
00:31:31,690 --> 00:31:38,690
of this particular mechanism. We have already
seen n, which is the total number of links
250
00:31:40,059 --> 00:31:47,059
are 10. j1 - that is the number of simple
hinges which are at A, B, G, F, E and D that
251
00:31:52,390 --> 00:31:59,390
is j1 is 6; number of compound hinges each
one of which connects 3 links, that is j2
252
00:32:02,260 --> 00:32:09,260
is at O2, O4, O and C that is j2 is equal
to 4. Degree of freedom of this mechanism
253
00:32:11,980 --> 00:32:18,980
according to the formula is F equal to 3 times
n minus 1 that is 10 minus 1, 9, minus 2 times,
254
00:32:23,150 --> 00:32:30,150
that is j1, that is 6 plus 2 times j2 that
is 2 into 4 is equal to 8, that is 27 minus
255
00:32:32,480 --> 00:32:39,480
14 into 2 equals to 28, which is minus 1.
So, without any special dimensions this assembly
256
00:32:43,360 --> 00:32:49,929
is a structure with degree of freedom minus
1. However, if we look at this figure what
257
00:32:49,929 --> 00:32:56,929
we see that O2 A C D is a parallelogram; O4
G C B that is another parallelogram and O
258
00:33:02,250 --> 00:33:09,250
F C E is another parallelogram. Not only that
this ternary links that is number 3, number
259
00:33:10,929 --> 00:33:17,929
9 and number 6, all these 3 ternary links
are similar triangles as indicated by the
260
00:33:18,090 --> 00:33:25,090
angles alpha, beta and gamma. Due to these
special dimensions, we will find that the
261
00:33:25,690 --> 00:33:32,690
degree of freedom of this assembly will become
equal to 1. This is another example of an
262
00:33:33,520 --> 00:33:39,600
over closed linkage, where some of the constants
may be redundant, but this will not be highlighted
263
00:33:39,600 --> 00:33:45,600
in this lecture. We will just show you the
model of this particular mechanism.
264
00:33:45,600 --> 00:33:51,169
Let us now look at the model of this 10-link
mechanism which has just been discussed. As
265
00:33:51,169 --> 00:33:55,620
we have seen according to the calculation
the degree of freedom should have been minus
266
00:33:55,620 --> 00:34:02,620
1, but notice that these 4 hinges constitutes
a parallelogram; so does these 4. These 4
267
00:34:05,070 --> 00:34:11,119
hinges also constitute another parallelogram.
These 3 triangles, the ternary links are similar
268
00:34:11,119 --> 00:34:18,119
to each other. Consequently, this constitutes
a single degree freedom mechanism, which is
269
00:34:19,049 --> 00:34:26,049
an over closed linkage which has mobility;
it is not a structure.
270
00:34:28,149 --> 00:34:34,940
As the further example of an over closed linkage,
let us consider this 8-link mechanism which
271
00:34:34,940 --> 00:34:38,899
is known as Kempe Burmeister focal mechanism.
272
00:34:38,899 --> 00:34:45,899
As we see, there are 8 links link 1, link
2, link 3, 4, 5, 6, 7 and 8. These 8 links
273
00:34:57,049 --> 00:35:04,049
are connected by revolute pairs one at O2,
at S, A, P, B, Q, O4 and R. There are 8 simple
274
00:35:14,239 --> 00:35:21,239
hinges and there is a higher order hinge at
this point T where 4 links namely 5, 6, 7
275
00:35:23,359 --> 00:35:30,359
and 8 are connected. So, if we calculate the
degree of freedom, we see n is 8, j1 is 8,
276
00:35:34,869 --> 00:35:41,869
j2 is 0, but there is a j3 at T, where 4 links
are connected so j3 is 1. The degree of freedom
277
00:35:49,049 --> 00:35:56,049
F is 3 into n minus 1, that is 7 minus twice
j1 which is 8 plus 3 times j3 which is 3 into
278
00:36:01,299 --> 00:36:08,299
1 is equal to 3. We get 21 minus 2 into 11
that is 22 which is minus 1. According to
279
00:36:18,839 --> 00:36:25,349
the formula, this should be a structure. However,
for very special dimensions, as indicated
280
00:36:25,349 --> 00:36:32,349
by these similar triangles BTQ with O2TS,
this angle is equal to this angle and this
281
00:36:37,989 --> 00:36:43,549
angle is equal to this angle 3634 min. Similarly,
there are other similar triangles in this
282
00:36:43,549 --> 00:36:49,440
figure. For such special dimensions as we
see in our model the degree of freedom will
283
00:36:49,440 --> 00:36:56,440
turn out to be 1. F will be 1that means it
will be a constant mechanism with single degree
284
00:37:00,119 --> 00:37:02,279
of freedom.
285
00:37:02,279 --> 00:37:07,950
Let us now consider the model of this Kempe
Burmeister focal mechanism, which we have
286
00:37:07,950 --> 00:37:14,950
just discussed. As we see including this fixed
link, we have 8 links 2, 3, 4, 5, 6, 7, 8
287
00:37:18,979 --> 00:37:25,979
and this is a hinge where 4 links are connected
and all other hinges are simple hinges. Accordingly,
288
00:37:26,950 --> 00:37:32,960
the formula said the degree of freedom should
be minus 1. But however, as we see this mechanism
289
00:37:32,960 --> 00:37:39,960
can be moved very easily and there is a unique
input-output relationship. That means, the
290
00:37:40,210 --> 00:37:46,930
effective degree of freedom of this mechanism
is 1; that is only because of the special
291
00:37:46,930 --> 00:37:52,569
dimension. If we change any of these points
a little bit this will really become a structure
292
00:37:52,569 --> 00:37:59,569
and no relative movement would be possible.
As a last example of an over closed linkage,
293
00:38:00,569 --> 00:38:06,130
let us look at this 5-link mechanism which
is known as cross slider trammel.
294
00:38:06,130 --> 00:38:13,130
Here we have link 1 which is the fixed link.
Link 3 which is connected to link 1 and link
295
00:38:13,650 --> 00:38:20,650
3 is connected to link 4 and link 2. Link
number 4 and 2 are having prismatic pairs
296
00:38:23,319 --> 00:38:29,799
with link number 5. Link 2 has a prismatic
pair in the horizontal direction with link
297
00:38:29,799 --> 00:38:36,799
number 5. Link 4 has a prismatic pair in the
vertical direction with link number 5. Thus,
298
00:38:38,469 --> 00:38:45,469
we have n equal to 5; we have 4 revolute pairs
here and here and here and here 3846 min.
299
00:38:54,989 --> 00:39:01,989
So, j is 4 revolute pairs plus 2 prismatic
pairs. Thus, F turns out to be according to
300
00:39:05,839 --> 00:39:12,839
the formula 3 times 5 minus 1 which is 4,
minus 2 times j that is 6.
301
00:39:16,390 --> 00:39:23,390
So, the effective degree of freedom of this
mechanism according to the formula F is 3
302
00:39:23,469 --> 00:39:30,469
times n minus 1 which is 4 minus 2 times j
which is 6; that is F equal to 12 minus 12
303
00:39:35,119 --> 00:39:41,859
which is 0. Without any special dimension,
this will be a structure; there should not
304
00:39:41,859 --> 00:39:48,859
be any mobility any relative movement. However,
for these special dimensions when I make OC
305
00:39:49,440 --> 00:39:56,440
is same as BC is same as AC then, we will
see that there will be an effective degree
306
00:40:02,869 --> 00:40:09,869
of freedom of this mechanism will turn out
to be just 1. In fact, as we see the angular
307
00:40:12,739 --> 00:40:19,739
velocity of link number 3 to that of link
number 5 which are both rotating with respect
308
00:40:20,180 --> 00:40:27,180
to fixed link 1 will be exactly half for these
special dimensions. This is known as cross
309
00:40:28,469 --> 00:40:35,469
slider trammel and I would like to encourage
the student to show that why it moves by starting
310
00:40:36,279 --> 00:40:41,259
from the elliptic trammel that we have discussed
in an earlier lecture.
311
00:40:41,259 --> 00:40:47,390
Now, I shall demonstrate this cross slider
trammel through a model. Let us now look at
312
00:40:47,390 --> 00:40:53,809
the model of this cross slider trammel. This
is that link number 3 4052 min, which has
313
00:40:53,809 --> 00:40:59,670
a revolute pair with fixed link here. This
is that link number 5 4058 min, which has
314
00:40:59,670 --> 00:41:06,670
a revolute pair with fixed link here. There
are 2 sliders - 2 and 4 which are hinged to
315
00:41:07,959 --> 00:41:14,959
link number 3 here and here 4108 min and these
2 sliders move in these 2 perpendicular slots.
316
00:41:15,819 --> 00:41:21,680
For the special dimensions, as we see this
has degree of freedom 1 and rotation of link
317
00:41:21,680 --> 00:41:28,680
3 produces unique rotation of link 5. In fact,
we can see that 2 revolutions of link 3 produces
318
00:41:31,190 --> 00:41:38,190
1 revolution of link 5. That is, one can show
that omega3 by omega5 at all instance remain
319
00:41:39,479 --> 00:41:44,599
half.
Let me now summarize, what has been covered
320
00:41:44,599 --> 00:41:51,599
in today's lecture. What we have seen how
we can calculate the degrees of freedom of
321
00:41:51,739 --> 00:41:57,979
a planar mechanism by counting the number
of links and different times of kinematic
322
00:41:57,979 --> 00:42:04,979
pairs. Attention has been also drawn to the
fact that there is a possibility of some redundant
323
00:42:06,339 --> 00:42:10,479
degrees of freedom that has to be accounted
for.
324
00:42:10,479 --> 00:42:16,670
We have also seen there may be some kinematic
pairs which are redundant in the sense they
325
00:42:16,670 --> 00:42:22,269
do not serve any purpose so far kinematics
is concerned, but they must be there due to
326
00:42:22,269 --> 00:42:28,229
some other practical considerations.
At the end we have seen, that these formulas
327
00:42:28,229 --> 00:42:35,209
which are derived only from the count without
any consideration of any kinematic dimensions
328
00:42:35,209 --> 00:42:42,209
may fail when there are some special kinematic
dimensions. We have also seen some such over
329
00:42:42,819 --> 00:42:48,119
closed linkages through the models, how they
move, though the formula says they should
330
00:42:48,119 --> 00:42:55,119
be structures.
331