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In this lecture, we will continue our discussion
on displacement analysis of planar linkages
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by analytical method. Today, we shall start
our discussion with an example. Let us look
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at this figure which is a kinematic diagram
of a 10-link mechanism.
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This is the fixed link number 1, then O2A
is the link 2, ABC is this ternary link, link
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number 3, link number 4 is O4B, link number
5, link 6 is again a ternary link, link 7,
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link 8, link number 9, which is again a ternary
link, and link number 10 so we have ten links.
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now let us look at the kinematic pairs.
As we see, at this point O2, three links are
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connected namely 1, 2 and 5 so this is a second
order hinge. Similarly, at O4 three links
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are connected namely 1, 4 and 10 so this is
again a second order hinge. Finally, at C
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three ternary links namely 3 6 and 9 are connected
so this is again a second order hinge. Let
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us now calculate the degrees of freedom of
this 10-link mechanism.
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We have already seen the total number of links,
n is 10. The number of kinematic pairs j is
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equal to, we have seven simple hinges namely
at O, F, G, B, A, D, and E. So, j is 7 plus
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as already mentioned, there are three second
order hinges, one at O2, one at O4 and the
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other at C. So, 2 into 3, that is, j is 13.
So, the degree of freedom F is 3 times (n
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-- 1) -- 2j which is 3 into 9 is 27 -- 2 into
13 is 26 that is equal to 1.
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We see it is a single degree of freedom mechanism
that means, whenever any link moves all other
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links move in a unique fashion. The question
is, as this mechanism moves let us see how
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this point O moves. To find that we will use
the links as link lengths vectors and try
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to find the vector O2O.
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Here as we see, the vector O2O we can write
as vector O2D plus vector DE plus vector EO.
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One has to note that in this particular mechanism
O2ACD is a parallelogram. Similarly, O4BCG
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is also a parallelogram and OECF is another
parallelogram. That means, the link AC is
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same as the link O2D; link length EO is same
as the link length CF and so on. Not only
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that, the three ternary links namely link
number 3, 6 and 9 consists of three similar
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triangles as indicated by these three angles
namely alpha, beta, gamma in each of these
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triangles. These three triangles ABC, CGF
and CED are three similar triangles.
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Let me write the vector O2D as same as the
vector AC because they always remain parallel
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and of equal length plus DE plus EO which
is same as CF because CF and EO always remain
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parallel and of equal length. Now, the vector
AC can be written as AC by AB into vector
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AB that takes care of the magnitude of AC.
However, the vector AC is at an angle alpha
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in the counter clockwise direction from the
vector AB. So I write, e to the power of i
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alpha. So, the first term vector AC can be
represented as AC by AB into vector AB multiplied
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by e to the power i alpha. Now, the vector
DE can be written as DE by DC into vector
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DC that takes care of the magnitude of DE.
But the vector DE is again at an angle alpha
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from the vector DC so we multiply it by e
to the power i alpha.
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Then CF, I can write as CF by CG into vector
CG that takes care of the magnitude of CF
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and again to take care of the direction, I
multiply it by e to the power of i alpha.
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Now as we see, as these three triangles are
similar triangles; AC by AB is same as DE
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by DC it is also same as CF by CG, as all
these three ratios are same, so I can take
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any one of them say, AC by AB, that I can
take common from all these three expressions.
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Same is true for e to the power i alpha, so
that also I take common, that leaves me with
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vector AB from the first term; then vector
DC which is same as the vector O2A because
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DC is same length as O2A and they always remain
parallel.
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So here, instead of DC, I write the same vector
O2A that leaves me with the vector CG which
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is same as BC because CG and BC are of equal
length and they are even parallel. So I write
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this as BC which means the vector O2O finally
comes out as AC by AB e to the power of i
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alpha and summation of these three vectors
namely O2A plus AB plus BC which is nothing
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but the vector O2O4. So this vector O2O which
is vector O2O4 into AC by AB into e to the
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power of i alpha.
As the mechanism moves, all the vectors change
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but O2O4 never change because O2 is a fixed
point; O4 is a fixed point, so the vector
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O2 O4 is always on X-axis without changing
its length. Neither the length AC nor length
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AB changes, because those are the rigid link
lengths. Same is true for this angle alpha
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which is again the angle alpha of this ternary
link. Consequently, as the linkage moves the
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vector O2O never change which means the point
O never moves in this mechanism. Though this
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is single degree freedom mechanism all other
points move as the mechanism moves but the
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point O never moves.
Just now we have seen that in the 10 link
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planar mechanism with specific dimensions
there is one point which was not moving though
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the overall mechanism had a single degree
of freedom.
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Consequently, if we fix that point O which
was not moving, let us fix it with the fixed
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link by putting a hinge thus converting this
hinge at O to a higher order hinge connecting
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three links namely 1, 7, and 8. As a result,
now we have an assembly of links where there
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are three 4 bar links: one 4 bar link consisting
of 1, 2, 3, and 4 with the coupler point at
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C; there is a second 4 bar linkage consisting
of link 1, 8, 9, and 10 with the same coupler
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point C; and the third 4 bar linkage consisting
of link number 1 that is the fixed link, link
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7, link 6 and link 5.
Now, all these three 4 bar linkages have the
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same coupler point C and this assembly moves
in a unique fashion. In other words, it means
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there are three different 4 bar linkages we
can generate the same coupler curve at the
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point C. As a result, this gives a wider choice
to the designer to choose one of these three
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linkages to produce the same coupler curve.
This is now what I will demonstrate with a
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model.
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Let us now look at the model of the 10 link
mechanism which we have just discussed. There
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is a fixed link and there are three second
order hinges, all connected to the fixed link
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which we marked previously as O2, O4 and O.
There are three gray moving links, there are
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three red moving links, and there are three
blue moving links. We also note that, this
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link length is equal to this link length and
this link length is equal to this link length.
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Thus, this point O2A and this was C and this
was D probably, these form a parallelogram.
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Similarly, we have a parallelogram here and
we have a parallelogram there. These three
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ternary links, this red ternary link, the
gray ternary link and the blue ternary links,
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they are similar triangles. That means, this
angle is equal to this angle, this angle is
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equal to this angle, and this angle is equal
to this angle.
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We have seen as a consequence of these special
dimensions, this 10-link mechanism move in
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a unique fashion because it has single degree
of freedom and this coupler point C can generate
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this coupler curve. Whether I use only this
4 bar linkage or this 4 bar linkage or this
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4 bar linkage all these three 4 bar linkages
generate the same coupler curve and this gives
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the designer a wider choice to choose a particular
one which may be convenient for the purpose.
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Now, we shall discuss some useful results
for 4R-linkage which are most commonly used
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and these results will be obtained analytically.
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For example, let us consider this 4R-linkage
namely, O2, A, B, and O4. As shown in this
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diagram, at this configuration O2A1B1O4, the
two links O2A1 and A1B1 are collinear.
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Consequently, this link O4B1 has taken one
of its extreme positions. It cannot go further
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to the left. As this link, the crank rocker
mechanism, this crank O2A rotates, there is
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another configuration when O2A2 and A2B2 again
become collinear and the corresponding configuration
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of O4B that is, and this O4B2 is the other
extreme position of this follower link, which
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is link 4.
We are considering a crank rocker mechanism
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and we see that, as the follower goes from
O4B2 to O4B1, during this movement, the crank
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rotates from O2A2 to O2A1. That means, it
moves through an angle theta2 star. During
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the return from B1 to B2, the crank rotates
from A1 to A2. So it rotates to an angle 2
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pi minus theta2 star. If the crank rotates
at constant speed, then the time taken for
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the follower motion during the B2B1 and B1B2
are not same.
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Normally, we would like to have a quick return
that is returning from B1 to B2, it rotates
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through an angle 2 pi minus theta2 star and
the follower motion that is B2 to B1, it rotates
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through an angle theta2 star, which is more
than pi. The quick return ratio can be defined
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as theta2 star divided by, we can write q
r r, quick return ratio will be theta2 star
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divided by 2 pi minus theta2 star.
Our objective is, to determine the relationship
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between the various link lengths namely, the
fixed length l1, the crank length l2, the
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coupler length l3, and the follower length
l4 so that we can determine whether there
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is any quick return effect or not. Towards
this end, we consider this figure, which has
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been drawn for a mechanism without any quick
return. One extreme position is O2, A1, B1,
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O4 and the other extreme position is O2, A2,
B2, O4. During these extreme positions, that
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the crank and the coupler are always collinear.
This is the outer death center which is O2A1B1
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and this is called the inner death center
when O2A2 and A2B2 are opposite to each. The
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angle between them, once I can say 0 degree,
in the other case it is 180 degrees.
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So, for this configuration as we see, the
angle between O2A1 and O2A2 is pi. That means
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there is no quick return. It takes pi amount
of rotation of the crank for the forward motion
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and again another pi amount of rotation for
the return motion. This theta4 star gives
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you the swing angle of the rocker. Let us
now derive what is the relationship between
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various link lengths. If we consider the triangle
O2B1O4 what do we see? O2O4 is of length l1
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and O2B2 is of length l2 plus l3 and O4B1
is l4.
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Let us say, this angle is let me denote it
by chi. So considering the triangle O2O4B1,
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I can write, l4 square is l1 square plus (l2
+ l3) whole squared minus twice l1 into (l2
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+ l3) cosine of the angle chi. Now, to determine
the angle cosine chi, this value, let me draw
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a perpendicular from O4 to the line B1B2.
O2B1B2 is an isosceles triangle because this
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length O4B1 is always equal to O4B2, so this
perpendicular bisector meets B1B2 at the mid
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point. Now, let me call this point say, D.
We can easily see that, B1B2 is O2B1 -- O2B2.
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Now, link length O2B1 is l2 + l3 -- O2B2 is
A2B2 is l3 and O2A2 is l2, so this is (l3
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-- l2). So we get, 2 l2, so B1B2 is 2 l2,
so half of that B2D will be l2. So I can write
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B2 D is l2. So O2D, which is O2B2 + B2D and
O2B2 is this so, l3 -- l2 + l2 which is l3.
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So cosine of this angle chi is O2D divided
by O2O4, so cosine chi is l3 that is, O2D
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divided by O2O4 that is, l1. So, we substitute
this l3 by l1 here and if we substitute this,
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we can easily show that we will get l1 square
plus l2 square will be same as l3 square plus
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l4 square.
Thus, for a 4R-linkage, to have no quick return
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effect that is, without any quick return effect,
the link lengths of a 4R-linkage must satisfy
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this relationship between its link lengths.
l1 is the fixed link length, l2 is the crank
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length, l3 is the coupler length and l4 is
the follower length. Not only that, we can
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also see that, B2D which we have got as, l2
is nothing but l4 sine of theta4 star by 2.
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So, this angle is theta4 star by 2. So there
is another relationship for such a linkage
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without quick return that the crank length
must be follower length into sine theta4 star
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that is the swing angle of the rocker divided
by 2.
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So, these are the two very important relationships
which can be used while designing a mechanism.
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Let us now look at the model of this crank
rocker linkage without quick return that is
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we have just discussed.
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This is a crank rocker linkage where l1 square
+ l2 square is l3 square + l4 square. This
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is the one extreme position, where the crank
and the coupler have fallen in one line; this
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is one extreme position of the follower. Now,
as it rotates to 180 degree, again the crank
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and the coupler becomes collinear giving rise
to the other extreme position of the follower.
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As a result, this 4R-linkage, if the crank
rotates at uniform speed the forward and return
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motion of the follower takes equal time and
there is no quick return.
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Now, let us look at the model of this crank
rocker linkage, where l1 square + l2 square
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is not the same as l3 square + l4 square As
a result, the extreme position of the follower
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that it takes is due to the unequal rotation
of the crank. One extreme position is here,
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when the crank and the coupler has fallen
in one line. Now from here, it rotates through
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this angle, again the crank and the coupler
falls in one line, giving rise to extreme
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position of the follower.
So here, as we see the follower does not take
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equal time during its forward and return motion.
There is some quick return effect depending
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on of course, whether I am clockwise or counter-clockwise.
Here, theta2 from here to there is more than
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here to there, so if I rotate it clockwise,
you can see the return is quicker.
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Let us now look at another model where again
it is a crank rocker linkage but l1 square
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+ l2 square is much less than l3 square +
l4 square. Consequently, the quick return
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effect will be much more predominant. For
example, this is one extreme position and
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the other extreme position is taken here.
So, the angle that the crank rotates is more
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than pi and the angle through which it returns
is much less than pi. So if we move the crank
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at uniform speed, the quick return effect
is much more predominant. The follower is
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taking much longer to come from right to left
extreme and much less time to go back.
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Now, that we have discussed both graphical
and analytical methods of displacement analysis.
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Let me take you through an example on how
both these methods can even be combined while
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designing a particular mechanism.
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As an example, let us consider this wind shield
wiper mechanism, which is a 6-link mechanism.
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This figure shows the mechanism to a particular
scale, for example, this distance is 50 mm,
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that is the scale of this diagram. Let us
now see this mechanism. This has O2ABO4, this
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part is a crank rocker mechanism and then
this link 4 is extended beyond O4 and we have
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another 4R mechanism namely, O4, C, D and
O6 and the second 4R mechanism that is O4CDO6
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is in the form of a parallelogram. O4C is
same as length in O6D and length CD is same
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as the length O6O4. This wiper blade is an
integral part of the coupler of this parallelogram
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linkage that is link number 5. The wiper blade
is same as link number 5. The first part of
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the question is what is the wiping field?
That means, as this crank O2A is driven by
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a motor, what is the field that is wiped by
this wiper blade?
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Now to solve this problem, what we do? First,
we use a little bit of graphical method. We
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study only this 4 bar linkage namely, O2ABO4
and determine the extreme positions of this
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link O4B that is, link number 4. For that,
we know as usual, B is going along a circle
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with O4B as radius. This is the circle with
O4 as center and O4B as radius that we call
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the path of B say kB. The extreme positions
of B will be taken up when the links O2A and
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00:26:41,780 --> 00:26:48,780
AB become collinear. So the farthest point
B can go is, when the distance of B from O2
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00:26:49,010 --> 00:26:56,010
is O2A plus AB. So, I take O2 as center and
draw a circular arc with l3 plus l2 as radius
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00:26:58,580 --> 00:27:05,580
and let that intersect kB at this point, this
I call B2.
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00:27:07,740 --> 00:27:13,679
Other extreme position of B will be taken
up again, when AB and O2A will be collinear
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00:27:13,679 --> 00:27:20,679
and the distance of B from O2 will be equal
to l3 -- l2. So I draw a circular arc with
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00:27:21,419 --> 00:27:28,419
O2 as center and l3 minus l2 as radius and
let that intersect kB at this point B1. So
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00:27:31,370 --> 00:27:37,020
O4B2 is one extreme position of link 4 and
O4B1 is the other extreme position of link
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00:27:37,020 --> 00:27:44,020
4. As we see, the link 4, this extension O4C
is not in line with O4B. There is an angle
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00:27:47,400 --> 00:27:52,809
delta which has been prescribed as 16 degree.
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00:27:52,809 --> 00:27:59,809
So we can draw the extreme positions of, we
have already determined B1 and B2 as explained
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00:28:01,990 --> 00:28:08,990
earlier. When corresponding to B2, I draw
this line O4C2 at an angle delta which was
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00:28:10,059 --> 00:28:17,059
16 degree and corresponding to B1. Again I
draw at an angle delta 16 degree to get C1.
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00:28:17,500 --> 00:28:24,500
So, O4C2 and O4C1 are the two extreme positions
of the link 4.
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00:28:24,650 --> 00:28:31,650
Now the question is, as we know, because the
second part of the mechanism was a parallelogram,
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this line CD always remains horizontal and
the wiper blades always remain vertical, because
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there is no rotation of the coupler of this
parallelogram linkage.
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We have already determined the extreme position
C1 and C2, so I can draw the wiper blade sets
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00:28:49,590 --> 00:28:56,590
E2F2 and other extreme positions when C is
at C1 as E1F1. So these are the two extreme
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00:29:01,500 --> 00:29:08,500
positions E2F2 and E1F1 for the wiper blade.
To determine the wiping field, we see that,
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00:29:10,090 --> 00:29:15,559
because it is a parallelogram linkage, the
point C which goes in a circle with O4 as
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00:29:15,559 --> 00:29:22,559
center and O4C as radius. Because it is a
parallelogram linkage, all the points of the
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00:29:25,200 --> 00:29:30,130
coupler move in identical curves.
That means, the curve generated by the points
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00:29:30,130 --> 00:29:36,679
E or F that is, the end of the wiper blades
also will be similar circles, exactly of same
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00:29:36,679 --> 00:29:43,679
radius as O4C. Only thing the center of the
circle will be shifted from O4 to CE for the
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00:29:47,909 --> 00:29:54,909
point E and O4 to CF for the point F. That
is, CECF is same as E2 F2 and E1 F1. So, with
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00:30:01,090 --> 00:30:08,090
center as CE and radius as O4C which is same
as CEE1, I draw this circular arc. Similarly,
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00:30:08,090 --> 00:30:13,270
with center as CF, I draw this circular arc,
and these are the two extreme positions and
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00:30:13,270 --> 00:30:18,840
the wiping field is what has been shown by
this hashed lines. So we have obtained the
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00:30:18,840 --> 00:30:25,840
field of wiping for this particular mechanism.
Let me repeat, first, we said determine the
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00:30:26,450 --> 00:30:33,279
path of B which is the circle with O4 as center
and O4B as radius. On this circular path,
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00:30:33,279 --> 00:30:40,279
I locate B1 and B2 using the relation O2B1
is AB -- O2A, O2B2 is AB plus O2A. Once I
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00:30:44,580 --> 00:30:51,580
got the extreme positions of link 4, I draw
O4C2 and O4C1 corresponding to O4B2 and O4B1,
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00:30:54,289 --> 00:31:00,909
because link 4 is a rigid link, the same angle
delta is maintained between the line O4B and
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00:31:00,909 --> 00:31:06,900
O4C. Once we get the extreme positions of
the point C, I draw the wiper blades which
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00:31:06,900 --> 00:31:13,900
always remain vertical because of the parallelogram
linkage as E2F2 and E1F1. Because of the parallelogram
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00:31:14,440 --> 00:31:21,289
linkage, all the coupler points generate same
circular arc as O4 C. Only thing, the center
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00:31:21,289 --> 00:31:28,289
of the circle is shifted in a symmetric fashion
from C2 to E2 that is O4 to CE; C2 to F2 that
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00:31:29,539 --> 00:31:35,309
is O4 to CF; these are all on the same vertical
line. As a result, we get the complete field
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00:31:35,309 --> 00:31:40,549
of wiping as generated by this particular
wiping mechanism.
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00:31:40,549 --> 00:31:47,549
We demonstrate the same wiper mechanism that
we have just now studied. This is the same
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00:31:47,570 --> 00:31:54,570
wiper mechanism consisting of a crank rocker,
consisting of link 2, link 3, and link 4.
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00:31:55,149 --> 00:32:02,149
O2 and O4 are the two fixed hinges. There
is another parallelogram linkage starting
225
00:32:02,529 --> 00:32:09,529
from here link 4, link 5, and link 6. This
link length is same as this link length and
226
00:32:09,669 --> 00:32:14,490
this link length is same as this link length.
So it is a parallelogram, and we should note
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00:32:14,490 --> 00:32:18,809
that because it is a parallelogram linkage,
the coupler always remains parallel to itself
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00:32:18,809 --> 00:32:25,809
and it never changes its orientation.
So, as the motor here rotates, the wiper blade
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00:32:27,179 --> 00:32:34,179
goes from left to right generating a field
of wiping, but the coupler blade always remains
230
00:32:40,020 --> 00:32:41,830
vertical.
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00:32:41,830 --> 00:32:47,340
Now that we have obtained this field of wiping
for this particular given mechanism, let us
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00:32:47,340 --> 00:32:54,340
observe that this wiping field that is E1E2F2F1E1.
This field of wiping is not symmetrical about
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00:32:58,070 --> 00:33:05,070
the vertical line passing through O4. This
O4V is the vertical line passing through O4.
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00:33:05,760 --> 00:33:12,760
But the field of wiping is more on the right
and less on the left. So, as a designer, maybe
235
00:33:12,880 --> 00:33:18,570
we can make a very little change to make this
field of wiping symmetrical about this vertical
236
00:33:18,570 --> 00:33:25,570
line. So, the second part of the problem is
retaining all other link parameters same,
237
00:33:26,580 --> 00:33:33,529
change only the angle delta which was given
a 16 degree. Change only this angle delta
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00:33:33,529 --> 00:33:40,529
to make the field of wiping symmetrical about
the line O4V. To solve this problem what do
239
00:33:43,399 --> 00:33:50,399
we do? We find, what is the angle of oscillation
of this rocker link O4B? That is, from O4B1
240
00:33:50,590 --> 00:33:57,590
to O4B2 that is the theta4 star which we call
the angle of swing. This theta4 star is not
241
00:34:02,520 --> 00:34:08,049
symmetrical about the vertical line as a result
the field of wiping is also not symmetrical
242
00:34:08,049 --> 00:34:09,730
about the vertical line.
243
00:34:09,730 --> 00:34:16,659
This theta4 star we measure.
And then, theta4 star by 2 and theta4 star
244
00:34:16,659 --> 00:34:23,659
by 2 are symmetrical about the vertical lines
O4. We have already seen that B1 which is
245
00:34:24,740 --> 00:34:31,740
the extreme position of the link O4 B which
we have obtained earlier, here this B1,. Now
246
00:34:38,929 --> 00:34:44,649
O4 C1 must be like this to make the field
of wiping symmetrical because now I have made
247
00:34:44,649 --> 00:34:51,599
O4C1 and O4C2 symmetrically placed about the
vertical line just at an angle theta4 star
248
00:34:51,599 --> 00:34:57,579
by 2 and theta4 star by 2 because theta4 star
is entirely decided by all the link lengths
249
00:34:57,579 --> 00:35:04,579
that I cannot change. But now, this is O4
B1 and this is O4 C1. The extension of O4
250
00:35:05,469 --> 00:35:12,469
B1 and this line O4 C1, the angle is delta.
This is the angle delta, which should be provided
251
00:35:14,029 --> 00:35:20,819
rather than what we had earlier at 16 degree.
Now, the third part of the problem we can
252
00:35:20,819 --> 00:35:27,819
have some more specifications. For example,
we define the width of this wiping field.
253
00:35:27,959 --> 00:35:34,959
That is, this horizontal distance C1 C2. So,
let me post the problem that modifies this
254
00:35:35,479 --> 00:35:36,180
design.
255
00:35:36,180 --> 00:35:43,180
But you are allowed to change only the crank
length O2 A, this coupler length AB and the
256
00:35:45,609 --> 00:35:52,599
angle delta to satisfy three requirements.
Namely, the wiping field should be symmetrical
257
00:35:52,599 --> 00:35:59,309
about the vertical line O4B, the width of
the wiping field that is, the horizontal distance
258
00:35:59,309 --> 00:36:06,309
between these two extreme positions E1F1 and
E2F2 is say 450 mm to the same scale and there
259
00:36:06,920 --> 00:36:13,920
should be no quick return. That means, O2A1
and O2A2 corresponding to these two extreme
260
00:36:14,979 --> 00:36:21,759
positions of the follower the crank angle
should be 180 degree.
261
00:36:21,759 --> 00:36:28,759
So we have no quick return for which we need
l1 square + l2 square is same as l3 square
262
00:36:37,819 --> 00:36:44,819
+ l4 square. Now, the width C1C2 we can easily
see, it is 2 times O4C into sine of the angle
263
00:36:57,839 --> 00:37:04,839
theta4 star by 2. Because this angle is theta4
start by 2, so this horizontal distance is
264
00:37:09,999 --> 00:37:15,880
O4C sine theta4 star by 2, twice of that is
the width of the field of wiping. This has
265
00:37:15,880 --> 00:37:22,880
been specified as 450 millimeter. The length
O4C has not been changed, so you have already
266
00:37:29,700 --> 00:37:36,700
given in the design. Substituting that value
of O4C, I can find the theta4 star value,
267
00:37:43,369 --> 00:37:50,369
so we determine theta4 star from this equation
with the given value of O4C. Now that we know
268
00:37:50,709 --> 00:37:57,709
theta4 star, if you remember for no quick
return, we also had a relationship that l2
269
00:37:59,430 --> 00:38:06,430
must be l4 sine theta4 star by 2.
Now l4 that was the link O4 B is not allowed
270
00:38:13,880 --> 00:38:20,759
to be changed. So now that l4 is given, theta4
star we have obtained, so I can obtain the
271
00:38:20,759 --> 00:38:27,759
crank length O2A as l2. So I have obtained
l2. l1 has not been changed, l4 has not been
272
00:38:32,519 --> 00:38:39,519
changed and l2 we have obtained, so using
this relationship for no quick return, I can
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00:38:39,549 --> 00:38:46,549
get the only remaining unknown that is l3.
So, we have designed the 4R-linkage O2ABO4.
274
00:38:51,140 --> 00:38:58,140
To obtain the required value of delta, we
just with the new lengths l2, l3, l1, l4 were
275
00:39:01,630 --> 00:39:08,630
unchanged. I again obtain the path of B, which
is this circle. O2B1 as we know was l3 -- l2.
276
00:39:15,359 --> 00:39:22,359
From O2, I draw a circular arc O2B1 as l3
-- l2.So I get the extreme position O4B1.
277
00:39:24,759 --> 00:39:30,509
The extreme position O4C1 is already known
so the angle between the extension of O4B1
278
00:39:30,509 --> 00:39:37,509
and O4C1 determines the required value of
delta which as we see is much less than the
279
00:39:38,559 --> 00:39:43,509
original value which was 16 degree. This delta,
if you draw it correctly comes out around
280
00:39:43,509 --> 00:39:50,509
5 degree. We have modified the design to satisfy
three requirements namely the field of wiping
281
00:39:51,329 --> 00:39:58,329
has to be symmetrical about the vertical line
O4B has to be of a particular width and also
282
00:39:59,930 --> 00:40:05,299
there should not be any quick return effect
such that the wiper blade takes equal time
283
00:40:05,299 --> 00:40:09,759
in the forward motion and return motion. It
should not go very fast in one direction and
284
00:40:09,759 --> 00:40:16,519
very slowly in the other direction which will
definitely disturb the driver.
285
00:40:16,519 --> 00:40:22,279
Now that we have completed our discussion
on displacement analysis both by graphical
286
00:40:22,279 --> 00:40:29,279
and analytical method let me start with a
very important index of a good mechanism.
287
00:40:29,699 --> 00:40:35,019
As you know, the mechanism has to satisfy
the geometric requirements, but satisfying
288
00:40:35,019 --> 00:40:41,819
the geometric requirement is not all. For
a real life mechanism, it must move freely
289
00:40:41,819 --> 00:40:48,819
and this free running quality of a mechanism
is quantified by a parameter, which is called
290
00:40:48,869 --> 00:40:55,089
transmission angle. Let me now discuss the
concept of transmission angle and show how
291
00:40:55,089 --> 00:41:02,089
to calculate the transmission angle at least
for 4 link mechanisms.
292
00:41:04,309 --> 00:41:10,949
Let me repeat, for smooth running of a mechanism,
one requires that the output member receives
293
00:41:10,949 --> 00:41:16,799
a large component of the force or torque from
the member driving it along the direction
294
00:41:16,799 --> 00:41:23,269
of output movement. This will ensure that
the mechanism runs freely. Not only satisfies
295
00:41:23,269 --> 00:41:30,229
the geometric requirements or kinematic requirements,
it must have this quality of smooth or free
296
00:41:30,229 --> 00:41:37,229
running. To ensure this smooth and free running,
one needs to have a complete dynamic analysis
297
00:41:39,329 --> 00:41:41,519
that will be discussed much later.
298
00:41:41,519 --> 00:41:48,519
However, even at this stage of kinematic design,
what we do?
299
00:41:49,880 --> 00:41:55,739
We neglect inertia, friction, and gravity
and treat all the binary links as two-force
300
00:41:55,739 --> 00:42:02,739
members that is, transmitting only axial force.
With this assumption, the free running quality
301
00:42:03,859 --> 00:42:10,859
of a mechanism can be expressed in terms of
what is called transmission angle. Let me
302
00:42:11,029 --> 00:42:18,029
explain this concept of transmission angle
for a 4R crank rocker linkage.
303
00:42:18,609 --> 00:42:25,609
This diagram shows a crank rocker linkage
namely, O2, A, B and O4 and let O2A be the
304
00:42:32,329 --> 00:42:39,329
crank. As we said, if we assume that coupler
AB is a two-force member, then the entire
305
00:42:40,309 --> 00:42:47,309
force that AB exerts on the output member
O4B is along the line AB. So this is the direction
306
00:42:47,430 --> 00:42:54,279
of the coupler force. However, it is only
this component which is perpendicular to the
307
00:42:54,279 --> 00:43:01,279
follower, produces torque to drive the coupler.
So, you have to ensure that, this angle is
308
00:43:02,319 --> 00:43:09,319
as small as possible.
For defining the transmission angle, it is
309
00:43:09,699 --> 00:43:16,699
defined as the acute angle between the coupler
that is AB and the follower O4 B. This is
310
00:43:23,599 --> 00:43:30,599
what is shown as the angle mu. So, we define
the transmission angle is equal to mu which
311
00:43:42,999 --> 00:43:47,619
is the acute angle between the coupler and
the follower.
312
00:43:47,619 --> 00:43:54,269
Obviously, the base possible value of mu is
90 degree. Then the entire coupler force is
313
00:43:54,269 --> 00:44:00,969
used to produce torque about O4, to drive
the follower. However, as the mechanism moves,
314
00:44:00,969 --> 00:44:06,890
this angle mu changes, but one is to ensure
that mu does not fall below a particular minimum
315
00:44:06,890 --> 00:44:13,890
value and normally, minimum value of mu prescribed
around 30 degree.
316
00:44:15,489 --> 00:44:22,489
Next, I will show, because we are only interested
in ensuring the minimum value of mu, can you
317
00:44:24,940 --> 00:44:31,940
find out for what crank position that is,
for what value of theta2 the minimum transmission
318
00:44:33,199 --> 00:44:38,369
angle offers? Because this angle mu keeps
on changing with the crank position it depends
319
00:44:38,369 --> 00:44:45,369
on theta2. It can be easily shown that, if
it is a crank rocker linkage without any quick
320
00:44:46,420 --> 00:44:53,420
return that is l1 square + l2 square is l3
square + l4 square, then mu attains it minimum
321
00:45:03,009 --> 00:45:10,009
value that is, mu min, when this angle theta2
is either 0 or pi. That is the crank is along
322
00:45:12,839 --> 00:45:19,839
the line of frame O4O2. That is theta2 is
either 0 or pi. To get this result, that mu
323
00:45:24,969 --> 00:45:30,249
attains its minimum value for theta2 equal
to 0 and pi, if there is no quick return,
324
00:45:30,249 --> 00:45:36,279
that is this l1 square plus l2 square is equal
to l3 square plus l4 square, this relationship
325
00:45:36,279 --> 00:45:39,890
holds good.
It is very easy to show, if we consider the
326
00:45:39,890 --> 00:45:46,890
length O4A. O4A. I can write in terms of l3,
l4 and mu. Considering the triangle O4AB,
327
00:45:52,229 --> 00:45:59,229
I can write O4A square is equal to l4 square
plus l3 square minus twice l3 l4 cos mu. Same
328
00:46:02,059 --> 00:46:09,059
way, I consider again the triangle O4O2A and
write O4A square as l1 square plus l2 square
329
00:46:10,049 --> 00:46:17,049
minus twice l1 l2 into cosine of pi minus
theta2, that is plus twice l1 l2 cos theta2.
330
00:46:20,890 --> 00:46:27,890
To obtain an expression for the transmission
angle mu, let us consider the triangle O4AB.
331
00:46:29,009 --> 00:46:36,009
Then we can write O4A square is equal to l3
square plus l4 square minus twice l3 l4 cosine
332
00:46:49,709 --> 00:46:56,709
mu. Same way, if we consider the triangle
O4AO2, then I can write again, O4A square
333
00:46:58,959 --> 00:47:05,959
is l1 square plus l2 square plus twice l1
l2 cosine theta2. If this crank rocker linkage
334
00:47:16,430 --> 00:47:22,209
has no quick return effect that means, l1
square plus l2 square is same as l3 square
335
00:47:22,209 --> 00:47:29,209
plus l4 square, then I can get expression
for cosine mu as l1 l2 divided by l3 l4 into
336
00:47:40,119 --> 00:47:47,119
minus of cosine theta2. For mu to be maximum,
we see that the values of theta2 can be either
337
00:47:56,619 --> 00:48:02,640
0 or pi. Because we have to remember, if this
angle is more than 90 degree then I will take
338
00:48:02,640 --> 00:48:08,859
pi minus this angle between the coupler and
the follower as my transmission angle. Transmission
339
00:48:08,859 --> 00:48:13,279
angle is defined as the acute angle between
the coupler and the follower.
340
00:48:13,279 --> 00:48:20,199
Here in this diagram, angle happens to be
acute so it is mu, but if when this angle
341
00:48:20,199 --> 00:48:26,719
becomes obtuse, then I have to take 180 degree
minus this angle as my transmission angle.
342
00:48:26,719 --> 00:48:33,719
So, transmission angle is minimized when theta2
is either 0 or pi for this particular situation.
343
00:48:33,839 --> 00:48:40,839
That means, the crank falls in line with the
frame that is O4O2. This will show now two
344
00:48:42,039 --> 00:48:47,420
models, but if there is quick return effect
then the minimum transmission angle occurs
345
00:48:47,420 --> 00:48:53,660
either at theta2 is equal to 0 or at theta2
is equal to pi. Only when, there is no quick
346
00:48:53,660 --> 00:49:00,309
return effect then it will be at both locations
theta2 equal to 0 and pi.
347
00:49:00,309 --> 00:49:07,309
This figure clearly shows that, when the crank
O2A along the line of frame O4O2. The transmission
348
00:49:09,079 --> 00:49:16,079
angle that is the angle between the coupler
and the follower is at its minimum value,
349
00:49:18,039 --> 00:49:23,140
this is what we call mu min, because it is
already acute.
350
00:49:23,140 --> 00:49:30,140
In another situation, we see that the crank
O2A is again along the frame line O4O2 and
351
00:49:34,119 --> 00:49:41,119
the angle between the coupler and the follower
that is, this angle is more than 90 degree.
352
00:49:43,299 --> 00:49:50,299
So here, we will define pi minus mu as my
transmission angle. It will always take the
353
00:49:50,609 --> 00:49:57,609
acute angle and this angle is minimised when
O2A is along the line of frame. So there are
354
00:50:01,209 --> 00:50:07,599
two situations: either O2 A is in this direction
when this angle between the coupler and the
355
00:50:07,599 --> 00:50:14,599
follower is more than 90 degree and the transmission
angle is pi minus mu; the other situation
356
00:50:16,059 --> 00:50:22,749
is, when O2A is along the line of frame and
the coupler and the follower makes an acute
357
00:50:22,749 --> 00:50:29,709
angle and that itself is the minimum transmission
angle, mu min. This now, we will demonstrate
358
00:50:29,709 --> 00:50:30,579
through models.
359
00:50:30,579 --> 00:50:36,079
We consider this model of a crank rocker linkage
without any quick return effect. That is,
360
00:50:36,079 --> 00:50:43,079
l1 square + l2 square is l3 square + l4 square.
For such a linkage, as I told you, the minimum
361
00:50:45,209 --> 00:50:50,959
transmission angle occurs when the crank is
along the line of frame. This angle between
362
00:50:50,959 --> 00:50:57,959
the coupler and the follower is its minimum
value. However, again when the crank falls
363
00:51:01,150 --> 00:51:08,150
along the line of frame, this angle is maximised,
that is the transmission angle which is defined
364
00:51:08,380 --> 00:51:12,910
as the acute angle between the coupler and
the follower that I can see by the extension
365
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of the coupler and the follower, that angle
is minimised. The minimum transmission angle
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occurs at two configurations: one is this
and the other is this.
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When the transmission characteristics are
very bad, because most of the tool of the
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00:51:29,949 --> 00:51:35,989
coupler is not going to drive the follower,
this perpendicular component of this actual
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00:51:35,989 --> 00:51:42,059
force is minimum. Here, it is very good. When
it is 90 degree, all the coupler force is
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00:51:42,059 --> 00:51:48,969
trying to drag the follower. Here, the transmission
is very good; here the transmission is worst;
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and here the transmission is worst.
Now, let us look at this another model, where
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l1 square plus l2 square is not the same as
l3 square plus l4 square. Here, the minimum
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00:52:02,739 --> 00:52:09,739
transmission angle occurs only for this configuration,
when the crank and the frame are along the
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00:52:11,229 --> 00:52:17,439
same line. The angle between them is maximised
so the transmission angle is minimum. Here,
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00:52:17,439 --> 00:52:21,949
the transmission quality is poor and here,
the transmission quality is very good, when
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00:52:21,949 --> 00:52:28,949
the angle is close to 90 degree. Let us look
at this model again where, l1 square + l2
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square is not the same as l3 square + l4 square.
So here, the minimum transmission angle occurs
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00:52:37,979 --> 00:52:42,910
only for this configuration when the crank
is along the line of frame and the angle between
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00:52:42,910 --> 00:52:49,009
the follower and the coupler is very small
giving rise to very poor transmission characteristics.
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00:52:49,009 --> 00:52:55,229
However, in this configuration, when again
the crank is along the line of frame, the
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00:52:55,229 --> 00:52:58,660
angle is quite large and this is not the minimum
transmission angle.
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00:52:58,660 --> 00:53:05,269
So, if there is no quick return, then the
minimum transmission angle occurs either here
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00:53:05,269 --> 00:53:11,739
or when this angle is 180 degree, but if there
is no quick return, then it happens both at
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00:53:11,739 --> 00:53:18,739
this position and at the 180 degree position
as shown earlier. We now explain the concept
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00:53:20,849 --> 00:53:25,529
of transmission angle with reference to a
slider-crank mechanism.
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00:53:25,529 --> 00:53:32,529
This figure shows a slider-crank mechanism,
where the slider is at B undergoing horizontal
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00:53:32,900 --> 00:53:39,900
translation and O2A is the crank. It is obvious
that, if the connecting rod AB is horizontal
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00:53:43,859 --> 00:53:49,599
then the entire connecting rod force is in
the direction of movement of the slider.
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00:53:49,599 --> 00:53:56,319
Consequently, we define the transmission angle
as mu that is the angle between the connecting
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00:53:56,319 --> 00:54:03,319
rod and a direction perpendicular to the line
of movement. The most desired value of mu
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00:54:04,160 --> 00:54:11,160
is 90 degree, but to ensure smooth free running
of the mechanism, mu min, minimum value of
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00:54:14,380 --> 00:54:21,380
mu should not fall below say, around 30 degree.
Now we can find out, for what value of crank
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00:54:24,499 --> 00:54:31,499
angle that is theta2 the minimum transmission
angle occurs. For that, we see this vertical
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00:54:34,910 --> 00:54:41,910
distance is l2 sine theta2, which is also
same as l3, this is 90 degree minus mu, so
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00:54:51,599 --> 00:54:58,599
that is l3 cos mu plus e. As theta2 changes,
l3 l2 and e are constants so, mu changes.
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00:55:05,049 --> 00:55:12,049
The minimum value of mu will take place depending
on whether e is this way or suppose the offset
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00:55:13,579 --> 00:55:18,410
was below this line, the direction of sliding
is like this, then this distance would have
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00:55:18,410 --> 00:55:25,410
been called e depending on whether e is upward
or downward, one can easily find that mu min
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00:55:30,229 --> 00:55:37,229
will be cos inverse l2 plus e by l3. That
occurs either at theta2 equal to pi by 2 or
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00:55:46,309 --> 00:55:53,309
3 pi by 2, depending on the direction of e.
This I leave for the students to decide for
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00:55:55,749 --> 00:56:02,319
themselves and that will add to the understanding.
One can also see, this slider-crank mechanism
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00:56:02,319 --> 00:56:09,319
add a 4Rlinkage with O4B the hinge O4 at infinity.
Then you see, O2ABO4 is the equivalent 4R-linkage
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00:56:15,869 --> 00:56:22,179
and mu is nothing but the angle between the
coupler AB and the follower O4B. This concept
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00:56:22,179 --> 00:56:28,429
of transmission angle we have used is same
for both the 4R-linkage, crank rocker linkage,
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00:56:28,429 --> 00:56:35,429
and this slider-crank mechanism.
Let me now summarize what we have learnt today.
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00:56:35,900 --> 00:56:42,049
We continued our discussion on displacement
analysis of planar mechanisms by analytical
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00:56:42,049 --> 00:56:49,049
methods. Then we obtained certain important
results so far as 4R crank rocker linkages
408
00:56:51,809 --> 00:56:57,579
are concerned, with reference to its quick
return effect and transmission angle and also
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00:56:57,579 --> 00:57:02,709
where the minimum transmission angle occurs,
which has to be ensured for a free running
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00:57:02,709 --> 00:57:09,709
of a design mechanism. We have also seen through
an example, that how we can combine both graphical
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00:57:10,259 --> 00:57:17,259
and analytical methods to improve the design
or modify an existing design.
412