1
00:00:17,880 --> 00:00:24,900
In this lecture, we will continue our discussion
on displacement analysis of planar linkages
2
00:00:24,900 --> 00:00:31,989
by analytical method. Today, we shall start
our discussion with an example. Let us look
3
00:00:31,989 --> 00:00:37,870
at this figure which is a kinematic diagram
of a 10-link mechanism.
4
00:00:37,870 --> 00:00:48,510
This is the fixed link number 1, then O2A
is the link 2, ABC is this ternary link, link
5
00:00:48,510 --> 00:00:59,300
number 3, link number 4 is O4B, link number
5, link 6 is again a ternary link, link 7,
6
00:00:59,300 --> 00:01:07,909
link 8, link number 9, which is again a ternary
link, and link number 10. So we have ten links.
7
00:01:07,909 --> 00:01:15,900
Now let us look at the kinematic pairs.
As we see, at this point O2, three links are
8
00:01:15,900 --> 00:01:25,170
connected namely 1, 2 and 5 so this is a second
order hinge. Similarly, at O4 three links
9
00:01:25,170 --> 00:01:33,490
are connected namely 1, 4 and 10 so this is
again a second order hinge. And finally at
10
00:01:33,490 --> 00:01:41,009
C three ternary links namely 3 6 and 9 are
connected so this is again a second order
11
00:01:41,009 --> 00:01:46,460
hinge. Let us now calculate the degrees of
freedom of this 10-link mechanism.
12
00:01:46,460 --> 00:01:57,149
We have already seen the total number of links,
n is 10. The number of kinematic pairs j is
13
00:01:57,149 --> 00:02:08,009
equal to, we have seven simple hinges namely
at O, F, G, B, A, D, and E. So, j1 that is
14
00:02:08,009 --> 00:02:15,990
7 plus as already mentioned, there are three
second order hinges, one at O2, one at O4
15
00:02:15,990 --> 00:02:30,330
and the other at C. So, 2 into 3, that is,
j is 13. So, the degree of freedom F is 3
16
00:02:30,330 --> 00:02:44,500
times (n minus 1) minus 2j which is 3 into
9 is 27 minus 2 into 13 is 26 that is equal
17
00:02:44,500 --> 00:02:49,830
to 1.
So we see it is a single degree of freedom
18
00:02:49,830 --> 00:02:56,720
mechanism that means, whenever any link moves
all other links move in a unique fashion.
19
00:02:56,720 --> 00:03:05,900
The question is, as this mechanism moves let
us see how does the point O, that this point
20
00:03:05,900 --> 00:03:13,870
O moves. To find that we will use the links
as link-length vectors and try to find the
21
00:03:13,870 --> 00:03:15,460
vector O2O.
22
00:03:15,460 --> 00:03:41,430
Here as we see, the vector O2O we can write
as vector O2D plus vector DE plus vector EO.
23
00:03:41,430 --> 00:03:50,220
One has to note that in this particular mechanism
O2ACD this is a parallelogram. Similarly,
24
00:03:50,220 --> 00:03:59,090
O4BCG that is also a parallelogram and OECF
that is another parallelogram. That means,
25
00:03:59,090 --> 00:04:07,211
the link-length vector AC is same as O2D;
link-length vector EO is same as CF and so
26
00:04:07,211 --> 00:04:14,960
on. Not only that, the three ternary links
namely link number 3, 6 and 9 these consists
27
00:04:14,960 --> 00:04:21,269
of three similar triangles as indicated by
these three angles namely alpha, beta, gamma
28
00:04:21,269 --> 00:04:30,340
in each of these triangles. These three triangles
ABC, CGF and CED are three similar triangles.
29
00:04:30,340 --> 00:04:40,150
So let me now write the vector O2D as same
as the vector AC because they always remain
30
00:04:40,150 --> 00:05:01,620
parallel and of equal length plus DE
plus EO which is same as CF because CF and
31
00:05:01,620 --> 00:05:14,140
EO always remain parallel and of equal length.
Now, the vector AC can be written as AC by
32
00:05:14,140 --> 00:05:27,480
AB into vector AB that takes care of the magnitude
of AC. However, the vector AC is at an angle
33
00:05:27,480 --> 00:05:33,520
alpha in the counter clockwise direction from
the vector AB. So I write, e to the power
34
00:05:33,520 --> 00:05:44,270
of i alpha. So, the first term vector AC can
be represented as AC by AB into vector AB
35
00:05:44,270 --> 00:06:02,050
multiplied by e to the power i alpha. Now,
the vector DE can be written as DE by DC into
36
00:06:02,050 --> 00:06:08,669
vector DC that takes care of the magnitude
of DE. But the vector DE is again at an angle
37
00:06:08,669 --> 00:06:15,740
alpha from the vector DC so we multiply it
by e to the power i alpha.
38
00:06:15,740 --> 00:06:33,240
Then CF, CF I can write as CF by CG into vector
CG that takes care of the magnitude of CF
39
00:06:33,240 --> 00:06:40,640
and again to take care of the direction, I
multiply it by e to the power of i alpha.
40
00:06:40,640 --> 00:06:47,910
Now as we see, because these three triangles
are similar triangles; AC by AB is same as
41
00:06:47,910 --> 00:06:54,340
DE by DC it is also same as CF by CG, as all
these three ratios are same, so I can take
42
00:06:54,340 --> 00:07:01,930
any one of them say, AC by AB, that I can
take common from all these three expressions.
43
00:07:01,930 --> 00:07:08,330
Same is true for e to the power i alpha, so
that also I take common, that leaves me with
44
00:07:08,330 --> 00:07:19,040
vector AB from the first term; then vector
DC which is same as the vector O2A because
45
00:07:19,040 --> 00:07:23,710
DC is same length as O2A and they always remain
parallel.
46
00:07:23,710 --> 00:07:33,540
So here I write, instead of DC, I write same
vector O2A that leaves me with the vector
47
00:07:33,540 --> 00:07:39,930
CG which is same as BC because CG and BC are
of equal length and they remain parallel.
48
00:07:39,930 --> 00:07:54,820
So I write this as BC, which means the vector
O2O finally comes out as AC by AB e to the
49
00:07:54,820 --> 00:08:05,270
power of i alpha and summation of these three
vectors namely O2A plus AB plus BC which is
50
00:08:05,270 --> 00:08:18,699
nothing but the vector O2O4. So this vector
O2O which is vector O2O4 into AC by AB into
51
00:08:18,699 --> 00:08:23,280
e to the power of i alpha.
As the mechanism moves, all the vectors changes
52
00:08:23,280 --> 00:08:28,320
but O2O4 never changes because O2 is a fixed
point; O4 is a fixed point, so the vector
53
00:08:28,320 --> 00:08:35,149
O2O4 is always on X-axis without changing
its length. Neither the length AC nor length
54
00:08:35,149 --> 00:08:40,060
AB changes, because those are the rigid link
lengths. Same is true for this angle alpha
55
00:08:40,060 --> 00:08:46,440
which is again the angle alpha of this ternary
link. Consequently, as the linkage moves the
56
00:08:46,440 --> 00:08:53,830
vector O2O never changes which means the point
O never moves in this mechanism. Though this
57
00:08:53,830 --> 00:08:58,850
is single degree freedom mechanism all other
points move as the mechanism moves but the
58
00:08:58,850 --> 00:09:04,360
point O never moves.
Just now we have seen that in the 10-link
59
00:09:04,360 --> 00:09:10,020
planar mechanism with specific dimensions
there is one point which was not moving though
60
00:09:10,020 --> 00:09:14,080
the overall mechanism had a single degree
of freedom.
61
00:09:14,080 --> 00:09:21,990
Consequently, if we fix that point O, the
point O which was not moving, let us fix it
62
00:09:21,990 --> 00:09:27,910
with the fixed link by putting a hinge thus
converting this hinge at O to a higher order
63
00:09:27,910 --> 00:09:34,810
hinge connecting three links namely 1, 7,
and 8. As a result, now we have an assembly
64
00:09:34,810 --> 00:09:42,990
of links where there are three 4 bar links:
one 4 bar link consisting of 1, 2, 3, and
65
00:09:42,990 --> 00:09:50,020
4 with the coupler point at C; there is a
second 4 bar linkage consisting of link 1,
66
00:09:50,020 --> 00:09:56,750
8, 9, and 10 with the same coupler point C;
and the third 4 bar linkage consisting of
67
00:09:56,750 --> 00:10:02,260
link number 1 that is the fixed link, link
7, link 6 and link 5.
68
00:10:02,260 --> 00:10:08,390
Now, all these three 4 bar linkages have the
same coupler point C and this assembly moves
69
00:10:08,390 --> 00:10:14,080
in a unique fashion. In other words, it means
there are three different 4 bar linkages we
70
00:10:14,080 --> 00:10:20,750
can generate the same coupler curve at the
point C. As a result, this gives a wider choice
71
00:10:20,750 --> 00:10:27,480
to the designer to choose one of these three
linkages to produce the same coupler curve.
72
00:10:27,480 --> 00:10:30,750
This is now what I will demonstrate with a
model.
73
00:10:30,750 --> 00:10:35,190
Let us now look at the model of the 10-link
mechanism which we have just discussed. There
74
00:10:35,190 --> 00:10:43,950
is a fixed link and there are three second
order hinges, all connected to the fixed link
75
00:10:43,950 --> 00:10:54,520
which we marked previously as O2, O4 and O.
There are three grey moving links, there are
76
00:10:54,520 --> 00:11:01,580
three red moving links and there are three
blue moving links. We also note that, this
77
00:11:01,580 --> 00:11:07,900
link length is equal to this link length and
this link length is equal to this link length.
78
00:11:07,900 --> 00:11:17,480
Thus, this point O2A and this was C and this
was D probably, these form a parallelogram.
79
00:11:17,480 --> 00:11:23,560
Similarly, we have a parallelogram here and
we have a parallelogram there. And these three
80
00:11:23,560 --> 00:11:28,260
ternary links, this red ternary link, the
grey ternary link and the blue ternary links,
81
00:11:28,260 --> 00:11:33,940
they are similar triangles. That means, this
angle is equal to this angle, this angle is
82
00:11:33,940 --> 00:11:37,810
equal to this angle, and this angle is equal
to this angle.
83
00:11:37,810 --> 00:11:43,760
We have seen as a consequence of these special
dimensions, this 10-link mechanism move in
84
00:11:43,760 --> 00:11:50,339
a unique fashion because it has single degree
of freedom. And this coupler point C can generate
85
00:11:50,339 --> 00:11:56,430
this coupler curve, whether I use only this
4-bar linkage or this 4-bar linkage or this
86
00:11:56,430 --> 00:12:12,500
4-bar linkage. All these three 4-bar linkages
generate the same coupler curve and this gives
87
00:12:12,500 --> 00:12:20,110
the designer a wider choice to choose a particular
one which may be convenient for the purpose.
88
00:12:20,110 --> 00:12:27,839
Now, we shall discuss some useful results
for 4R-linkage which are most commonly used
89
00:12:27,839 --> 00:12:30,959
and these results will be obtained analytically.
90
00:12:30,959 --> 00:12:43,540
For example, let us consider this 4R-linkage
namely, O2, A, B, and O4. As shown in this
91
00:12:43,540 --> 00:12:54,620
diagram, at this configuration O2A1B1O4, the
two links O2A1 and A1B1 are collinear.
92
00:12:54,620 --> 00:13:01,620
Consequently, this link O4B1 has taken one
of its extreme positions. It cannot go further
93
00:13:01,620 --> 00:13:09,350
to the left. As this link, the crank-rocker
mechanism, this crank O2A rotates. There is
94
00:13:09,350 --> 00:13:19,170
another configuration when O2A2 and A2B2 again
become collinear and the corresponding configuration
95
00:13:19,170 --> 00:13:25,149
of O4B that is, and this O4B2 is the other
extreme position of this follower link, which
96
00:13:25,149 --> 00:13:29,600
is link 4.
So we are considering a crank-rocker mechanism
97
00:13:29,600 --> 00:13:38,460
and we see that, as the follower goes from
O4B2 to O4B1, during this movement, the crank
98
00:13:38,460 --> 00:13:47,980
rotates from O2A2 to O2A1. That means, it
moves through an angle theta2 star. During
99
00:13:47,980 --> 00:13:55,230
the return from B1 to B2, the crank rotates
from A1 to A2. So it rotates to an angle 2
100
00:13:55,230 --> 00:14:02,180
pi minus theta2 star. If the crank rotates
at constant speed, then the time taken for
101
00:14:02,180 --> 00:14:08,360
the follower motion during the B2B1 and B1B2
are not same.
102
00:14:08,360 --> 00:14:14,070
Normally, we would like to have a quick return
that is returning from B1 to B2, it rotates
103
00:14:14,070 --> 00:14:20,950
through an angle 2 pi minus theta2 star and
the follower motion that is B2 to B1, it rotates
104
00:14:20,950 --> 00:14:26,620
through an angle theta2 star, which is more
than pi. The quick return ratio can be defined
105
00:14:26,620 --> 00:14:41,360
as theta2 star divided by, we can write q
r r, quick return ratio will be theta2 star
106
00:14:41,360 --> 00:14:52,529
divided by 2 pi minus theta2 star.
Our objective is, to determine the relationship
107
00:14:52,529 --> 00:15:01,050
between the various link lengths namely, the
fixed length l1, the crank length l2, the
108
00:15:01,050 --> 00:15:08,550
coupler length l3, and the follower length
l4 so that we can determine whether there
109
00:15:08,550 --> 00:15:17,490
is any quick return effect or not. Towards
this end, we consider this figure, which has
110
00:15:17,490 --> 00:15:25,380
been drawn for a mechanism without any quick
return. One extreme position is O2, A1, B1,
111
00:15:25,380 --> 00:15:33,399
O4 and the other extreme position is O2, A2,
B2, O4. During these extreme positions, that
112
00:15:33,399 --> 00:15:40,160
the crank and the coupler are always collinear.
This is the outer dead center which is O2A1B1
113
00:15:40,160 --> 00:15:45,420
and this is called the inner dead center when
O2A2 and A2B2 are opposite to each other.
114
00:15:45,420 --> 00:15:50,570
The angle between them, once I can say 0 degree,
in the other case it is 180 degrees.
115
00:15:50,570 --> 00:15:59,959
So, for this configuration as we see, the
angle between O2A1 and O2A2 is pi. That means
116
00:15:59,959 --> 00:16:04,779
there is no quick return. It takes pi amount
of rotation of the crank for the forward motion
117
00:16:04,779 --> 00:16:11,700
and again another pi amount of rotation for
the return motion. This theta4 star gives
118
00:16:11,700 --> 00:16:18,139
you the swing angle of the rocker. Let us
now derive what is the relationship between
119
00:16:18,139 --> 00:16:30,850
various link lengths. If we consider the triangle
O2B1O4 what do we see? O2O4 is of length l1
120
00:16:30,850 --> 00:16:37,029
and O2B2 is of length l2 plus l3 and O4B1
is l4.
121
00:16:37,029 --> 00:16:48,230
And let us say, this angle is let me denote
it by chi. So considering the triangle O2O4B1,
122
00:16:48,230 --> 00:17:10,029
I can write, l1 squared is l4 squared, sorry
let me write l4 squared is l1 squared plus
123
00:17:10,029 --> 00:17:33,000
(l2 plus l3) whole squared minus twice l1
into (l2 plus l3) cosine of the angle chi.
124
00:17:33,000 --> 00:17:40,640
Now, to determine the angle cosine chi, this
value, let me draw a perpendicular from O4
125
00:17:40,640 --> 00:17:52,041
to the line B1B2. O2B1B2 is an isosceles triangle
because this length O4B1 is always equal to
126
00:17:52,041 --> 00:18:00,049
O4B2, so this perpendicular bisector meets
B1B2 at the mid point. Now, let me call this
127
00:18:00,049 --> 00:18:18,720
point say, D.
We can easily see that, B1B2 is O2B1 minus
128
00:18:18,720 --> 00:18:42,460
O2B2. Now, link length O2B1 is l2 plus l3
minus O2B2 is A2B2 is l3 and O2A2 is l2, so
129
00:18:42,460 --> 00:18:57,790
this is (l3 minus l2). So we get, 2l2, so
B1B2 is 2l2, so half of that B2D will be l2.
130
00:18:57,790 --> 00:19:17,450
So I can write B2D is l2. So O2D, which is
O2B2 plus B2D and O2B2 is this, so l3 minus
131
00:19:17,450 --> 00:19:29,440
l2 plus l2 which is l3. So cosine of this
angle chi is O2D divided by O2O4, so cosine
132
00:19:29,440 --> 00:19:45,040
chi is l3 that is, O2D divided by O2O4 that
is, l1. So, we substitute this l3 by l1 here
133
00:19:45,040 --> 00:19:54,330
and if we substitute this, we can easily show
that we will get l1 squared plus l2 squared
134
00:19:54,330 --> 00:20:07,160
will be same as l3 squared plus l4 squared.
Thus, for a 4R-linkage, to have no quick return
135
00:20:07,160 --> 00:20:13,270
effect that is, without any quick return effect,
the link lengths of a 4R-linkage must satisfy
136
00:20:13,270 --> 00:20:22,100
this relationship between its link lengths.
l1 is the fixed link length, l2 is the crank
137
00:20:22,100 --> 00:20:28,120
length, l3 is the coupler length and l4 is
the follower length. Not only that, we can
138
00:20:28,120 --> 00:20:36,240
also see that, B2D which we have got as l2
is nothing but l4 sine of theta4 star by 2.
139
00:20:36,240 --> 00:20:42,590
So, this angle is theta4 star by 2. So there
is another relationship for such a linkage
140
00:20:42,590 --> 00:20:54,669
without quick return that the crank length
must be follower length into sine theta4 star
141
00:20:54,669 --> 00:20:58,549
that is the swing angle of the rocker divided
by 2.
142
00:20:58,549 --> 00:21:05,900
So, these are the two very important relationships
which can be used while designing a mechanism.
143
00:21:05,900 --> 00:21:11,220
Let us now look at the model of this crank
rocker linkage without quick return that we
144
00:21:11,220 --> 00:21:12,860
have just discussed.
145
00:21:12,860 --> 00:21:22,240
This is a crank rocker linkage where l1 squared
plus l2 squared is l3 squared plus l4 squared.
146
00:21:22,240 --> 00:21:27,520
This is the one extreme position, where the
crank and the coupler has fallen in one line;
147
00:21:27,520 --> 00:21:39,490
this is one extreme position of the follower.
Now, as it rotates to 180 degree, again the
148
00:21:39,490 --> 00:21:45,080
crank and the coupler becomes collinear giving
rise to the other extreme position of the
149
00:21:45,080 --> 00:21:52,410
follower. As a result, this 4R-linkage, if
the crank rotates at uniform speed, the forward
150
00:21:52,410 --> 00:21:58,679
and return motion of the follower takes equal
time and there is no quick return.
151
00:21:58,679 --> 00:22:07,190
Now, let us look at the model of this crank
rocker linkage, where l1 squared plus l2 squared
152
00:22:07,190 --> 00:22:15,100
is not the same as l3 squared plus l4 squared.
As a result, the extreme position of the follower
153
00:22:15,100 --> 00:22:21,591
that it takes is due to the unequal rotation
of the crank. One extreme position is here,
154
00:22:21,591 --> 00:22:31,200
when the crank and the coupler has fallen
in one line. Now from here, it rotates through
155
00:22:31,200 --> 00:22:37,210
this angle, again the crank and the coupler
falls in one line, giving rise to extreme
156
00:22:37,210 --> 00:22:46,750
position of the follower.
So here, as we see the follower doesn’t
157
00:22:46,750 --> 00:22:55,630
take equal time during its forward and return
motion. There is some quick return effect
158
00:22:55,630 --> 00:23:02,490
depending on of course, whether I am rotating
clockwise or counter-clockwise. Here, theta2
159
00:23:02,490 --> 00:23:12,390
from here to there is more than here to there,
so if I rotate it clockwise, you can see the
160
00:23:12,390 --> 00:23:23,730
return is quicker.
Let us now look at another model where again
161
00:23:23,730 --> 00:23:31,570
it is a crank rocker linkage but l1 squared
plus l2 squared is much less than l3 squared
162
00:23:31,570 --> 00:23:38,070
plus l4 squared. Consequently, the quick return
effect will be much more predominant. For
163
00:23:38,070 --> 00:23:43,500
example, this is one extreme position and
the other extreme position is taken here.
164
00:23:43,500 --> 00:23:51,549
So, the angle that the crank rotates is more
than pi and the angle through which it returns
165
00:23:51,549 --> 00:24:01,350
is much less than pi. So if we move the crank
at uniform speed, the quick return effect
166
00:24:01,350 --> 00:24:07,400
is much more predominant. The follower is
taking much longer to come from right to left
167
00:24:07,400 --> 00:24:17,020
extreme and much less time to go back.
Now that we have discussed both graphical
168
00:24:17,020 --> 00:24:23,630
and analytical methods of displacement analysis,
let me show you through an example, how both
169
00:24:23,630 --> 00:24:28,240
these methods can even be combined while designing
a particular mechanism.
170
00:24:28,240 --> 00:24:38,660
As an example, let us consider this wind shield
wiper mechanism, which is a 6-link mechanism.
171
00:24:38,660 --> 00:24:48,600
This figure shows the mechanism to a particular
scale, for example, this distance is 50 mm,
172
00:24:48,600 --> 00:25:00,070
that is the scale of this diagram. Let us
now see this mechanism. This has O2ABO4, this
173
00:25:00,070 --> 00:25:08,470
part is a crank-rocker mechanism and then
this link 4 is extended beyond O4 and we have
174
00:25:08,470 --> 00:25:21,190
another 4R mechanism namely, O4, C, D and
O6 and the second 4R mechanism that is O4CDO6
175
00:25:21,190 --> 00:25:28,750
is in the form of a parallelogram. O4C is
same as length in O6D and length CD is same
176
00:25:28,750 --> 00:25:36,250
as the length O6O4. And this wiper blade is
an integral part of the coupler of this parallelogram
177
00:25:36,250 --> 00:25:44,500
linkage that is link number 5. The wiper blade
is same as link number 5. The first part of
178
00:25:44,500 --> 00:25:51,740
the question is what is the wiping field?
That means, as this crank O2A is driven by
179
00:25:51,740 --> 00:25:57,010
a motor, what is the field that is wiped by
this wiper blade?
180
00:25:57,010 --> 00:26:05,481
Now to solve this problem, what we do? First,
we use a little bit of graphical method. We
181
00:26:05,481 --> 00:26:14,000
study only this 4-bar linkage namely, O2ABO4
and determine the extreme positions of this
182
00:26:14,000 --> 00:26:23,809
link O4B that is, link number 4. For that,
we know as usual, B is going along a circle
183
00:26:23,809 --> 00:26:33,650
with O4B as radius. This is the circle with
O4 as center and O4B as radius, that we call
184
00:26:33,650 --> 00:26:41,830
the path of B say kB. The extreme positions
of B will be taken up when the links O2A and
185
00:26:41,830 --> 00:26:49,030
AB become collinear. So the farthest point
B can go is, when the distance of B from O2
186
00:26:49,030 --> 00:26:58,730
is O2A plus AB. So, I take O2 as center and
draw a circular arc with l3 plus l2 as radius
187
00:26:58,730 --> 00:27:06,780
and let that intersect kB at this point, this
I call B2.
188
00:27:06,780 --> 00:27:13,679
And other extreme position of B will be taken
up again, when AB and O2A will be collinear
189
00:27:13,679 --> 00:27:21,480
and the distance of B from O2 will be equal
to l3 minus l2. So I draw a circular arc with
190
00:27:21,480 --> 00:27:30,490
O2 as center and l3 minus l2 as radius and
let that intersect kB at this point B1. So
191
00:27:30,490 --> 00:27:37,000
O4B2 is one extreme position of link 4 and
O4B1 is the other extreme position of link
192
00:27:37,000 --> 00:27:47,460
4. Now as we see the link 4, this extension
O4C is not in line with O4B. There is an angle
193
00:27:47,460 --> 00:27:52,800
delta which has been prescribed as 16 degree.
194
00:27:52,800 --> 00:28:01,980
So we can draw the extreme positions of, we
have already determined B1 and B2 as explained
195
00:28:01,980 --> 00:28:10,250
earlier. When corresponding to B2, I draw
this line O4C2 at an angle delta which was
196
00:28:10,250 --> 00:28:17,490
16 degree and corresponding to B1. Again I
draw at an angle delta 16 degree to get C1.
197
00:28:17,490 --> 00:28:24,650
So, O4C2 and O4C1 are the two extreme positions
of the link 4.
198
00:28:24,650 --> 00:28:31,980
Now the question is, as we know, because the
second part of the mechanism was a parallelogram,
199
00:28:31,980 --> 00:28:37,799
so this line CD always remains horizontal
and the wiper blades always remain vertical,
200
00:28:37,799 --> 00:28:42,590
because there is no rotation of the coupler
of this parallelogram linkage.
201
00:28:42,590 --> 00:28:49,460
We have already determined the extreme position
C1 and C2, so I can draw the wiper blades
202
00:28:49,460 --> 00:29:01,500
as E2F2 and other extreme positions when C
is at C1 as E1F1. So these are the two extreme
203
00:29:01,500 --> 00:29:10,500
positions E2F2 and E1F1 for the wiper blade.
To determine the wiping field, we see that,
204
00:29:10,500 --> 00:29:15,590
because it is a parallelogram linkage, the
point C which goes in a circle with O4 as
205
00:29:15,590 --> 00:29:23,650
radius and O4C as oh sorry with O4 as center
and O4C as radius. Because it is a parallelogram
206
00:29:23,650 --> 00:29:28,040
linkage, all the points of the coupler move
in identical curves.
207
00:29:28,040 --> 00:29:33,240
That means, the curve generated by the points
E or F that is, the end of the wiper blades
208
00:29:33,240 --> 00:29:40,850
also will be similar circles, exactly of same
radius as O4C. Only thing the center of the
209
00:29:40,850 --> 00:29:54,860
circle will be shifted from O4 to CE for the
point E and O4 to CF for the point F. That
210
00:29:54,860 --> 00:30:05,010
is, CE, CF is same as E2F2 and E1F1. So, with
center as CE and radius as O4C which is same
211
00:30:05,010 --> 00:30:10,510
as CEE1, I draw this circular arc. Similarly,
with center as CF, I draw this circular arc,
212
00:30:10,510 --> 00:30:16,070
and these are the two extreme positions and
the wiping field is what has been shown by
213
00:30:16,070 --> 00:30:21,490
these hatched lines. So we have obtained the
field of wiping for this particular mechanism.
214
00:30:21,490 --> 00:30:29,141
Let me repeat, first we said, determine the
path of B which is the circle with O4 as center
215
00:30:29,141 --> 00:30:37,870
and O4B as radius. On this circular path,
I locate B1 and B2 using the relation O2B1
216
00:30:37,870 --> 00:30:47,540
is AB minus O2A, O2B2 is AB plus O2A. Once
I got the extreme positions of link 4, I draw
217
00:30:47,540 --> 00:30:56,490
O4C2 and O4C1 corresponding to O4B2 and O4B1,
because link 4 is a rigid link, the same angle
218
00:30:56,490 --> 00:31:03,270
delta is maintained between the line O4B and
O4C. Once we get the extreme positions of
219
00:31:03,270 --> 00:31:09,279
the point C, I draw the wiper blades which
always remain vertical because of the parallelogram
220
00:31:09,279 --> 00:31:18,070
linkage as E2F2 and E1F1. Because of the parallelogram
linkage, all the coupler points generate same
221
00:31:18,070 --> 00:31:25,470
circular arc as O4C. Only thing, the center
of the circle is shifted in a symmetric fashion
222
00:31:25,470 --> 00:31:32,070
from C2 to E2 that is O4 to CE; C2 to F2 that
is O4 to CF; these are all on the same vertical
223
00:31:32,070 --> 00:31:38,440
line. As a result, we get the complete field
of wiping as generated by this particular
224
00:31:38,440 --> 00:31:40,540
wiping mechanism.
225
00:31:40,540 --> 00:31:47,570
We demonstrate the same wiper mechanism that
we have just now studied. This is the same
226
00:31:47,570 --> 00:31:55,149
wiper mechanism consisting of a crank rocker,
consisting of link 2, link 3, and link 4.
227
00:31:55,149 --> 00:32:02,529
O2 and O4 are the two fixed hinges. And there
is another parallelogram linkage starting
228
00:32:02,529 --> 00:32:09,659
from here link 4, link 5, and link 6. This
link length is same as this link length and
229
00:32:09,659 --> 00:32:14,600
this link length is same as this link length.
So it is a parallelogram, and we should note
230
00:32:14,600 --> 00:32:19,210
because it is a parallelogram linkage, the
coupler always remains parallel to itself,
231
00:32:19,210 --> 00:32:26,370
it never changes its orientation.
So, as the motor here rotates, the coupler
232
00:32:26,370 --> 00:32:38,200
wiper blade goes from left to right generating
a field of wiping, but the coupler blade always
233
00:32:38,200 --> 00:32:41,830
remains vertical.
234
00:32:41,830 --> 00:32:47,370
Now that we have obtained this field of wiping
for this particular given mechanism, let us
235
00:32:47,370 --> 00:32:58,059
observe that this wiping field that is E1E2F2F1E1.
This field of wiping is not symmetrical about
236
00:32:58,059 --> 00:33:04,960
the line vertical line passing through O4.
This O4V is the vertical line passing through
237
00:33:04,960 --> 00:33:11,960
O4. But the field of wiping is more on the
right and less on the left. So, as a designer,
238
00:33:11,960 --> 00:33:17,980
maybe we can make a very little change to
make this field of wiping symmetrical about
239
00:33:17,980 --> 00:33:25,870
this vertical line. So, the second part of
the problem is retaining all other link parameters
240
00:33:25,870 --> 00:33:32,470
same, change only the angle delta which was
given a 16 degree. Change only this angle
241
00:33:32,470 --> 00:33:42,049
delta to make the field of wiping symmetrical
about the line O4V. To solve this problem,
242
00:33:42,049 --> 00:33:49,299
what do we do? We find, what is the angle
of oscillation of this rocker link O4B? That
243
00:33:49,299 --> 00:34:01,279
is, from O4B1 to O4B2 that is the theta4 star
which we call the angle of swing. This theta4
244
00:34:01,279 --> 00:34:06,929
star is not symmetrical about the vertical
line, as a result the field of wiping is also
245
00:34:06,929 --> 00:34:09,480
not symmetrical about the vertical line.
246
00:34:09,480 --> 00:34:18,379
This theta4 star we measure. And then, theta4
star by 2 and theta4 star by 2 symmetrically
247
00:34:18,379 --> 00:34:25,660
about the vertical lines O4. We have already
seen that B1 which is the extreme position
248
00:34:25,660 --> 00:34:39,879
of the link O4B which we have obtained earlier,
here this B1. That is this B1. And now O4C1
249
00:34:39,879 --> 00:34:45,860
must be like this to make the field of wiping
symmetrical because now I have made O4C1 and
250
00:34:45,860 --> 00:34:52,210
O4C2 symmetrically placed about the vertical
line just at an angle theta4 star by 2 and
251
00:34:52,210 --> 00:34:58,719
theta4 star by 2 because theta4 star is entirely
decided by all the link lengths that I cannot
252
00:34:58,719 --> 00:35:06,839
change. But now, this is O4B1 and this is
O4C1. So the extension of O4B1 and this line
253
00:35:06,839 --> 00:35:14,759
O4C1, the angle is delta. This is the angle
delta, which should be provided rather than
254
00:35:14,759 --> 00:35:20,809
what we had earlier as 16 degree.
Now, the third part of the problem we can
255
00:35:20,809 --> 00:35:27,700
have some more specifications. For example,
we define what is the width of this wiping
256
00:35:27,700 --> 00:35:35,479
field? That is, this horizontal distance C1C2.
So, let me pose a problem that modifies this
257
00:35:35,479 --> 00:35:36,479
design.
258
00:35:36,479 --> 00:35:45,609
But you are allowed to change only the crank
length O2A, this coupler length AB and the
259
00:35:45,609 --> 00:35:52,589
angle delta to satisfy three requirements.
Namely, the wiping field should be symmetrical
260
00:35:52,589 --> 00:35:59,309
about the vertical line O4B, the width of
the wiping field that is, the horizontal distance
261
00:35:59,309 --> 00:36:06,910
between these two extreme positions E1F1 and
E2F2 is say 450 mm to the same scale and there
262
00:36:06,910 --> 00:36:13,549
should be no quick return. That means, O2A1
and O2A2 these two extreme positions corresponding
263
00:36:13,549 --> 00:36:21,759
to these two extreme positions of the follower
the crank angle should be 180 degree.
264
00:36:21,759 --> 00:36:37,869
So we have no quick return for which we need
l1 squared plus l2 squared is same as l3 squared
265
00:36:37,869 --> 00:36:57,279
plus l4 squared. Now, the width C1C2 we can
easily see, it is 2 times O4C into sine of
266
00:36:57,279 --> 00:37:09,640
the angle theta4 star by 2. Because this angle
is theta4 star by 2, so this horizontal distance
267
00:37:09,640 --> 00:37:15,420
is O4C sine theta4 star by 2, twice of that
is the width of the field of wiping. This
268
00:37:15,420 --> 00:37:28,759
has been specified as 450 millimeter. The
length O4C has not been changed, so you had
269
00:37:28,759 --> 00:37:34,869
it already given in the design. So substituting
that value of O4C, I can find what is the
270
00:37:34,869 --> 00:37:45,660
theta4 star value, and we can find what is
the theta4 star. So we determine theta4 star,
271
00:37:45,660 --> 00:37:51,569
from this equation with the given value of
O4C. Now that we know theta4 star, if you
272
00:37:51,569 --> 00:38:05,869
remember for no quick return, we also had
a relationship that l2 must be l4 sine theta4
273
00:38:05,869 --> 00:38:13,849
star by 2.
Now l4 that was the link O4B was not allowed
274
00:38:13,849 --> 00:38:20,690
to be changed. So now that l4 is given, theta4
star we have obtained, so I can obtain the
275
00:38:20,690 --> 00:38:32,469
crank length O2A as l2. So I have obtained
l2. l1 has not been changed, l4 has not been
276
00:38:32,469 --> 00:38:39,550
changed and l2 we have obtained, so using
this relationship for no quick return, I can
277
00:38:39,550 --> 00:38:52,749
get the only remaining unknown that is l3.
So, we have designed the 4R-linkage O2ABO4.
278
00:38:52,749 --> 00:39:01,470
To obtain the required value of delta, we
just with the new lengths l2, l3, l1, l4 were
279
00:39:01,470 --> 00:39:12,369
unchanged, I again obtain what is the path
of B, which is this circle. And O2B1 as we
280
00:39:12,369 --> 00:39:23,339
know was l3 minus l2. From O2, I draw a circular
arc O2B1 as l3 minus l2. So I get the extreme
281
00:39:23,339 --> 00:39:27,359
position O4B1.
And the extreme position O4C1 is already known
282
00:39:27,359 --> 00:39:34,869
so the angle between the extension of O4B1
and O4C1 determines the required value of
283
00:39:34,869 --> 00:39:40,490
delta which as we see is much less than the
original value which was 16 degree. This delta,
284
00:39:40,490 --> 00:39:47,759
if you draw it correctly comes out around
5 degree. So we have modified the design to
285
00:39:47,759 --> 00:39:53,920
satisfy three requirements namely the field
of wiping has to be symmetrical has to be
286
00:39:53,920 --> 00:40:00,180
symmetrical about the vertical line O4B, has
to be of a particular width and also there
287
00:40:00,180 --> 00:40:05,670
should not be any quick return effect such
that the wiper blade takes equal time in the
288
00:40:05,670 --> 00:40:09,950
forward motion and return motion. It should
not go very fast in one direction and very
289
00:40:09,950 --> 00:40:16,519
slowly in the other direction which will definitely
disturb the driver.
290
00:40:16,519 --> 00:40:22,279
Now that we have completed our discussion
on displacement analysis both by graphical
291
00:40:22,279 --> 00:40:29,319
and analytical method let me start with a
very important index of a good mechanism.
292
00:40:29,319 --> 00:40:35,010
As you know, the mechanism has to satisfy
the geometric requirements, but satisfying
293
00:40:35,010 --> 00:40:41,809
the geometric requirement is not all. For
a real-life mechanism, it must move freely
294
00:40:41,809 --> 00:40:48,859
and this free running quality of a mechanism
is quantified by a parameter, which is called
295
00:40:48,859 --> 00:40:55,079
transmission angle. Let me now discuss the
concept of transmission angle and show how
296
00:40:55,079 --> 00:41:04,299
to calculate the transmission angle at least
for 4 link mechanisms.
297
00:41:04,299 --> 00:41:10,949
Let me repeat, for smooth running of a mechanism,
one requires that the output member receives
298
00:41:10,949 --> 00:41:16,799
a large component of the force or torque from
the member driving it along the direction
299
00:41:16,799 --> 00:41:23,890
of output movement. This will ensure that
the mechanism runs freely. Not only satisfies
300
00:41:23,890 --> 00:41:30,259
the geometric requirements or kinematic requirements,
it must have this quality of smooth or free
301
00:41:30,259 --> 00:41:37,380
running. However to ensure this smooth and
free running, one needs to have a complete
302
00:41:37,380 --> 00:41:41,109
dynamic analysis. That will be discussed much
later.
303
00:41:41,109 --> 00:41:49,869
However, even at this stage of kinematic design,
what we do?
304
00:41:49,869 --> 00:41:55,739
We neglect inertia, friction, and gravity
and treat all the binary links as two-force
305
00:41:55,739 --> 00:42:03,849
members that is, transmitting only axial force.
With this assumption, the free running quality
306
00:42:03,849 --> 00:42:11,030
of a mechanism can be expressed in terms of
what is called transmission angle. Let me
307
00:42:11,030 --> 00:42:18,599
explain this concept of transmission angle
for a 4R crank-rocker linkage.
308
00:42:18,599 --> 00:42:32,479
This diagram shows a crank rocker linkage
namely, O2, A, B and O4 and let O2A be the
309
00:42:32,479 --> 00:42:40,400
crank. As we said, if we assume that coupler
AB is a two-force member, then the entire
310
00:42:40,400 --> 00:42:47,420
force that AB exerts on the output member
O4B is along the line AB. So this is the direction
311
00:42:47,420 --> 00:42:53,910
of the coupler force. However, it is only
this component which is perpendicular to the
312
00:42:53,910 --> 00:43:02,319
follower, produces torque to drive the coupler.
So, you have to ensure that, this angle is
313
00:43:02,319 --> 00:43:09,119
as small as possible.
However for defining the transmission angle,
314
00:43:09,119 --> 00:43:21,569
it is defined as the acute angle between the
coupler that is AB and the follower O4B. This
315
00:43:21,569 --> 00:43:42,999
is what is shown as the angle mu. So, we define
the transmission angle is equal to mu, which
316
00:43:42,999 --> 00:43:47,609
is the acute angle between the coupler and
the follower.
317
00:43:47,609 --> 00:43:54,269
Obviously, the base possible value of mu is
90 degree. Then the entire coupler force is
318
00:43:54,269 --> 00:44:00,799
used to produce torque about O4, to drive
the follower. However, as the mechanism moves,
319
00:44:00,799 --> 00:44:07,171
this angle mu changes, but one is to ensure
that mu does not fall below a particular minimum
320
00:44:07,171 --> 00:44:15,479
value and normally, minimum value of mu prescribed
around 30 degree.
321
00:44:15,479 --> 00:44:24,930
Next, we will show, because we are only interested
in ensuring the minimum value of mu, can you
322
00:44:24,930 --> 00:44:33,190
find out for what crank position that is,
for what value of theta2 the minimum transmission
323
00:44:33,190 --> 00:44:38,609
angle offers? Because this angle mu keeps
on changing with the crank position, it depends
324
00:44:38,609 --> 00:44:46,410
on theta2. It can be easily shown that, if
it is a crank rocker linkage without any quick
325
00:44:46,410 --> 00:45:02,499
return that is l1 squared plus l2 squared
is l3 squared plus l4 squared, then mu attains
326
00:45:02,499 --> 00:45:12,299
it minimum value, that is mumin, when this
angle theta2 is either 0 or pi. That is the
327
00:45:12,299 --> 00:45:24,140
crank is along the line of frame O4O2. That
is theta2 is either 0 or pi. To get this result,
328
00:45:24,140 --> 00:45:30,269
that mu attains its minimum value for theta2
equal to 0 and pi, if there is no quick return,
329
00:45:30,269 --> 00:45:35,809
that is this l1 squared plus l2 squared is
equal to l3 squared plus l4 squared, this
330
00:45:35,809 --> 00:45:39,940
relationship holds good.
It is very easy to show, if we consider the
331
00:45:39,940 --> 00:45:52,229
length O4A. O4A, I can write in terms of l3,
l4 and mu. Considering the triangle O4AB,
332
00:45:52,229 --> 00:46:01,759
I can write O4A squared is equal to l4 squared
plus l3 squared minus twice l3 l4 cos mu.
333
00:46:01,759 --> 00:46:09,489
Same way, I consider again the triangle O4O2A
and write O4A squared as l1 squared plus l2
334
00:46:09,489 --> 00:46:19,459
squared minus twice l1l2 into cosine of pi
minus theta2, that is plus twice l1 l2 cos
335
00:46:19,459 --> 00:46:28,999
theta2. To obtain an expression for the transmission
angle mu, let us consider the triangle O4AB.
336
00:46:28,999 --> 00:46:51,390
Then we can write O4A squared is l3 squared
plus l4 squared minus twice l3l4 cosine mu.
337
00:46:51,390 --> 00:47:00,790
Same way, if we consider the triangle O4AO2,
then I can write again, O4A squared is l1
338
00:47:00,790 --> 00:47:16,729
squared plus l2 squared plus twice l1l2 cosine
theta2. If this crank rocker linkage has no
339
00:47:16,729 --> 00:47:22,809
quick return effect that means, l1 squared
plus l2 squared is same as l3 squared plus
340
00:47:22,809 --> 00:47:41,559
l4 squared, then I can get expression for
cosine mu as l1l2 divided by l3l4 into minus
341
00:47:41,559 --> 00:47:55,910
of cosine theta2. For mu to be maximum, we
see that the values of theta2 can be either
342
00:47:55,910 --> 00:48:02,630
0 or pi. Because we have to remember, if this
angle is more than 90 degree then I will take
343
00:48:02,630 --> 00:48:08,860
pi minus this angle between the coupler and
the follower as my transmission angle. Transmission
344
00:48:08,860 --> 00:48:13,279
angle is defined as the acute angle between
the coupler and the follower.
345
00:48:13,279 --> 00:48:19,689
Here in this diagram, angle happens to be
acute so it is cos mu, it is mu. But if when
346
00:48:19,689 --> 00:48:26,299
this angle becomes obtuse, then I have to
take 180 degree minus this angle as my transmission
347
00:48:26,299 --> 00:48:32,939
angle. So, transmission angle is minimized
when theta2 is either 0 or pi for this particular
348
00:48:32,939 --> 00:48:38,209
situation. That means, the crank falls in
line with the frame that is O4O2. This will
349
00:48:38,209 --> 00:48:46,329
show now through models. But if there is quick
return effect then the minimum transmission
350
00:48:46,329 --> 00:48:52,400
angle occurs either at theta2 is equal to
0 or at theta2 is equal to pi. Only when,
351
00:48:52,400 --> 00:48:59,069
there is no quick return effect then it will
be at both locations theta2 equal to 0 and
352
00:48:59,069 --> 00:49:03,859
pi.
This figure clearly shows that, when the crank
353
00:49:03,859 --> 00:49:14,489
O2A along the line of frame O4O2. The transmission
angle that is the angle between the coupler
354
00:49:14,489 --> 00:49:22,069
and the follower is at its minimum value,
this is what we call mumin, because it is
355
00:49:22,069 --> 00:49:23,140
already acute.
356
00:49:23,140 --> 00:49:34,109
In another situation, we see that the crank
O2A is again along the frame line O4O2 and
357
00:49:34,109 --> 00:49:43,289
the angle between the coupler and the follower
that is this angle, is more than 90 degree.
358
00:49:43,289 --> 00:49:50,619
So here, we will define pi minus mu as my
transmission angle. It will always take the
359
00:49:50,619 --> 00:50:01,140
acute angle and this angle is minimised when
O2A is along the line of frame. So there are
360
00:50:01,140 --> 00:50:07,589
two situations: either O2A is in this direction
when this angle between the coupler and the
361
00:50:07,589 --> 00:50:16,049
follower is more than 90 degree and the transmission
angle is pi minus mu; the other situation
362
00:50:16,049 --> 00:50:22,390
is, when O2A is again along the line of frame
and the coupler and the follower makes an
363
00:50:22,390 --> 00:50:28,249
acute angle and that itself is the minimum
transmission angle, mumin. This now, we will
364
00:50:28,249 --> 00:50:30,420
demonstrate through models.
365
00:50:30,420 --> 00:50:36,240
Consider this model of a crank rocker linkage
without any quick return effect. That is,
366
00:50:36,240 --> 00:50:43,799
l1 squared plus l2 squared is l3 squared plus
l4 squared. For such a linkage, as I told
367
00:50:43,799 --> 00:50:49,920
you, the minimum transmission angle occurs
when the crank is along the line of frame.
368
00:50:49,920 --> 00:50:56,849
This angle between the coupler and the follower
reaches its minimum value. However, again
369
00:50:56,849 --> 00:51:06,630
when the crank falls along the line of frame,
this angle is maximised, that is the transmission
370
00:51:06,630 --> 00:51:11,519
angle which is defined as the acute angle
between the coupler and the follower that
371
00:51:11,519 --> 00:51:16,210
I can see by the extension of the coupler
and the follower, that angle is minimised.
372
00:51:16,210 --> 00:51:20,680
So the minimum transmission angle occurs at
the two orientations two configurations: one
373
00:51:20,680 --> 00:51:26,499
is this and the other is this.
When the transmission characteristics are
374
00:51:26,499 --> 00:51:33,529
very bad, because most of the tool of the
coupler is not going to drive the follower,
375
00:51:33,529 --> 00:51:39,599
this perpendicular component of this axial
force is minimum. Here, it is very good. When
376
00:51:39,599 --> 00:51:44,809
it is 90 degree, all the coupler force is
trying to drag the follower. Here, the transmission
377
00:51:44,809 --> 00:51:50,999
is very good; here the transmission is worst;
and here the transmission is worst.
378
00:51:50,999 --> 00:51:58,819
Now, let us look at this another model, where
l1 squared plus l2 squared is not the same
379
00:51:58,819 --> 00:52:06,949
as l3 squared plus l4 squared. Here, the minimum
transmission angle occurs only for this configuration,
380
00:52:06,949 --> 00:52:14,449
when the crank and the frame are along the
same line. And the angle between them is maximised
381
00:52:14,449 --> 00:52:19,920
so the transmission angle is minimum. Here,
the transmission quality is poor and here
382
00:52:19,920 --> 00:52:27,730
the transmission quality is very good, when
the angle is close to 90 degree.
383
00:52:27,730 --> 00:52:32,790
Let us look at this model again where, l1
squared plus l2 squared is not the same as
384
00:52:32,790 --> 00:52:39,719
l3 squared plus l4 squared. So here, the minimum
transmission angle occurs only for this configuration
385
00:52:39,719 --> 00:52:44,680
when the crank is along the line of frame
and the angle between the follower and the
386
00:52:44,680 --> 00:52:50,170
coupler is very small giving rise to very
poor transmission characteristics. However,
387
00:52:50,170 --> 00:52:55,950
in this configuration, when again the crank
is along the line of frame, the angle is quite
388
00:52:55,950 --> 00:52:58,650
large and this is not the minimum transmission
angle.
389
00:52:58,650 --> 00:53:05,259
So, if there is no quick return, then the
minimum transmission angle occurs either here
390
00:53:05,259 --> 00:53:11,709
or when this angle is 180 degree, but if there
is no quick return, then it happens both at
391
00:53:11,709 --> 00:53:20,839
this position and at the 180 degree position
as shown earlier. We now explain the concept
392
00:53:20,839 --> 00:53:25,519
of transmission angle with reference to a
slider-crank mechanism.
393
00:53:25,519 --> 00:53:32,900
This figure shows a slider-crank mechanism,
where the slider is at B undergoing horizontal
394
00:53:32,900 --> 00:53:43,859
translation and O2A is the crank. It is obvious
that, if the connecting rod AB is horizontal
395
00:53:43,859 --> 00:53:50,150
then the entire connecting rod force is in
the direction of movement of the slider.
396
00:53:50,150 --> 00:53:56,309
Consequently, we define the transmission angle
as mu that is the angle between the connecting
397
00:53:56,309 --> 00:54:04,150
rod and a direction perpendicular to the line
of movement. The most desired value of mu
398
00:54:04,150 --> 00:54:14,380
is 90 degree, but to ensure smooth free running
of the mechanism, mumin, minimum value of
399
00:54:14,380 --> 00:54:24,490
mu should not fall below say, around 30 degree.
Now we can find out, for what value of crank
400
00:54:24,490 --> 00:54:34,859
angle, that is theta2, the minimum transmission
angle occurs. For that, we see this vertical
401
00:54:34,859 --> 00:54:51,609
distance is l2 sine theta2, which is also
same as l3, this is 90 degree minus mu, so
402
00:54:51,609 --> 00:55:04,660
that is l3 cos mu plus e. As theta2 changes,
l3, l2 and ‘e’ are constants so, mu changes.
403
00:55:04,660 --> 00:55:12,380
And the minimum value of mu will take place
depending on whether ‘e’ is this way or
404
00:55:12,380 --> 00:55:17,969
suppose the offset was below this line, the
direction of sliding is like this, then this
405
00:55:17,969 --> 00:55:23,339
distance would have been called ‘e’ depending
on whether ‘e’ is upward or downward,
406
00:55:23,339 --> 00:55:41,459
one can easily find that mumin will be cos
inverse l2 plus e by l3. That occurs either
407
00:55:41,459 --> 00:55:52,749
at theta2 equal to pi by 2 or 3 pi by 2, depending
on the direction of ‘e’.
408
00:55:52,749 --> 00:55:59,319
This I leave for the students to decide for
themselves and that will add to the understanding.
409
00:55:59,319 --> 00:56:08,619
One can also see, this slider-crank mechanism
had a 4Rlinkage with O4B the hinge O4 at infinity.
410
00:56:08,619 --> 00:56:17,819
Then you see, O2ABO4 is the equivalent 4R-linkage
and mu is nothing but the angle between the
411
00:56:17,819 --> 00:56:25,349
coupler AB and the follower O4B. This concept
of transmission angle we have used is same
412
00:56:25,349 --> 00:56:32,069
for both the 4R-linkage, crank-rocker linkage,
and this slider-crank mechanism.
413
00:56:32,069 --> 00:56:38,719
Let me now summarize what we have learnt today.
We continued our discussion on displacement
414
00:56:38,719 --> 00:56:47,939
analysis of planar mechanisms by analytical
method. Then we obtained certain important
415
00:56:47,939 --> 00:56:54,069
results so far as 4R crank rocker linkages
are concerned, with reference to its quick
416
00:56:54,069 --> 00:57:00,240
return effect and transmission angle and also
where the minimum transmission angle occurs,
417
00:57:00,240 --> 00:57:06,130
which has to be ensured for a free running
of a design mechanism. And we have also seen
418
00:57:06,130 --> 00:57:13,039
through an example, that how we can combine
both graphical and analytical methods to improve
419
00:57:13,039 --> 00:57:20,210
the design or modify an existing design.
420
00:57:20,210 --> 00:57:24,579
9