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As told at the end of the last lecture, we
shall continue our discussion on displacement
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analysis of planar mechanisms using graphical
method. Last time we had seen examples in
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which all were six link mechanisms with single
degree of freedom. Today, we shall discuss
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or take up a little more involved mechanism
with more number of links and may be with
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more number of degrees of freedom and so how
the graphical method can be used for displacement
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analysis. As the first example today, that
is the fourth example in the overall displacement
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analysis, we consider a nine link front end
loader.
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First, let us have a look at the model of
this loader. To start with, let us first discuss
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the model of this mechanism which is used
for a front end loader. As we see there are
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two hydraulic actuators between link number
2 and link number 3 and there is another hydraulic
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actuator between this red link and this link,
that is link 6 and link 7. By moving these
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two hydraulic actuators, we can move the front
end of this loader, that is this bin which
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is link number 9. We will first show, that
this particular mechanism has 2 degrees of
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freedom and with two inputs here and there,
so these two hydraulic actuators. Consequently,
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this output link cannot go anywhere with any
arbitrary orientation. It will have only two
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independent co-ordinates, which we can specify
either the xy co-ordinates of the point only
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or the orientation will be automatically fixed
up or only the x co-ordinate and the orientation
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of this bin and the y co-ordinate will be
automatically decided.
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Let us determine the displacement analysis
of such a mechanism by graphical method, that
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will be our next example. Let us now discuss
the problem of displacement analysis with
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reference to this front end loader, the model
of which we have just now seen. As we see
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there are two hydraulic actuators, one named
Z1and the other named Z2.
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Link number 1 is the body of the truck and
this hydraulic piston cylinder constitutes
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link 2 and link 3. There is obviously a prismatic
pair between link 2 and 3, which is the piston
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and the cylinder. Similarly, there is a prismatic
pair between link 6 and link 7 where there
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is a hydraulic actuator, again between the
piston and cylinder of this hydraulic actuator.
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Link 7 one end is connected to this body of
the truck, at the revolute pair at O7, Link
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4 one end is connected to a revolute pair
to the body of the truck O4 and link 2 one
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end is connected to the body of the truck
by a revolute pair at O2.
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Let us see this mechanism does not have all
binary links, as we have considered earlier.
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On the first example that we considered, they
had either binary link or ternary link. If
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we consider link number 4, as we see this
is connected to four different links, namely
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to link 1 at O4, link 3 at A, link 5 at B
and link 9 at C. So, link 4 which has four
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revolute pairs becomes a quaternary link and
link 5 is a ternary link connected to link
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8, link 4 and link 6. Link 9 is the front
end of the loader that is, the bin.
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In this mechanism, because we have a quaternary
link, that is a very higher order link, I
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will show how to conveniently use the tracing
paper as an over lay as we discussed in the
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previous lecture. Let me state the problem,
the problem is posed as follows: that if the
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point D of link number 9 comes to D prime.
This is the scale of the drawing, 25 cm is
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represented by this distance. When the point
D comes to D prime, can we locate all other
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revolute pairs? That means, where they move
in this consideration, when point D of link
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9 has moved to D prime.
Before that, let me show that this mechanism
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has 2 degrees of freedom. For that we count
the number of links n, which as we have seen
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1, 2, 3, 4, 5, 6, 7, 8, and 9, n is 9. Let
me count the revolute pairs and prismatic
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pairs, that is j number of joints. What we
see, we have 1, 2, 3, 4, 5, 6, 7, 8, and 9,
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9 revolute pairs plus 2 prismatic pairs, that
is at these two hydraulic actuators. So, total
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j is 9 + 2 that is 11. If we remember our
formula to calculate the degrees of freedom
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that is F is equal to 3 times n minus 1 minus
2j, which we get, 3 into 824 minus 2 into
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11, 22 that is F is equal to 2. Thus, it is
2 degrees of freedom mechanism because we
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have two independent inputs at these two hydraulic
actuators, it is a constant mechanism because
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it is 2 degrees of freedom, I can have 2 degrees
of freedom for this output bin, that is the
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point D can go to D prime but at that location
the orientation of bin will be automatically
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determined. I cannot take D to D prime and
then again change the orientation. So, we
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solved this problem, when D goes to D prime,
how do we determine the location of all other
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revolute pairs? That is the first part of
the problem.
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This is the figure of the same front end loader
which we are discussing right now. As I said
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earlier, the point D has moved to this point,
which I call D one and the objective is to
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find the location of all other movable revolute
pairs that is at A B E F and C. We should
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note that O4C being two revolute pairs on
the same rigid link 4, the distance O4C cannot
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change. So, C must move on a circular path
with O4 as centre and O4C as radius. This
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is the circle on which C must move, also D
and C being two points on the same rigid link,
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that is number 9, that is the bin, this distance
also cannot change because I know the location
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of D prime, I put my centre of a circle with
DC as radius with D one as the centre.
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If I draw this circular arc, C must lie on
this circle. Thus, we get that C should lie
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on this circle and also on this circle, that
means the intersection of these two circles
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determine the location of C, when D moves
to D one, C moves to C one. To locate the
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positions of the hinges, say A and B, which
lie on link 4, as we noted earlier link 4
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is a rigid link, but it is a quaternary link
having 4 hinges, one at O4, other at A, other
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at B and one at C. When D comes to D1, C has
moved to C one, O4 does not move, so, O4 remains
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there. To locate the positions of A and B,
corresponding to the second position, what
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we do, we take a tracing paper and mark these
points: O4, A, B and C. The relative positions
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of these O4, A, B, C can never change because
they lie on the same rigid body.
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What I do is I move this tracing paper now
with O4 at O4 and C at C one. These two red
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dots, which are the positions of A and B is
automatically determined, A and B moves to
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these new locations. If we can now pierce
this tracing paper here and there, I immediately
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get the locations of A and B. This point will
give me A one and this point will give me
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the location B one. Let me repeat it once
more because O4, A, B, C are four points on
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the same rigid link, I use this tracing paper
as an over lay, mark the relative locations
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of O4, A, B, and C because two points define
the rigid body completely and in the second
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position I know the location of O4 and the
location of C one.
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I take these two red dots corresponding to
O4 and C to this new location and where ever
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these two dots come, they are the positions
of A and B corresponding to the second configuration.
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Once, we have obtained D one and B one, it
is very easy to determine the new location
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of this revolute pair E, because the distance
DE does not change. With D one as centre,
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I draw a circular arc with DE as the radius.
So, E must lie on this circle and BE also
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does not change. So, if B one as centre, I
draw another circular arc with BE as the radius,
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where ever these two circles intersect, that
gives me the location of E, which I call E
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one. Similarly, to determine F, I know the
locations B, E and F. These are the three
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points on the same rigid link 5 so, I can
again use this tracing paper as an over lay
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and mark B, F and E, their relative positions
do not change and now, I know where B and
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E has gone. So, I can replace this, coinciding
this B with B one, this E with E one and where
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ever F goes, that
becomes the location of F, which I call F
one.
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I have determined the location of all the
moving hinges namely: A, B, F, E, D as A one,
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B one, E one, F one and D one of course is
given, and C as C1 .So we have solved this
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part of the problem that corresponding to
the second position, when D has moved to D
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one, where all other moving hinges go. Not
only that, I can also find what is the inclination
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of the bin from the first position, that is
this line CD, that was this initial orientation
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has now become this. So, angle between these
two lines CD and C one D one determines the
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change in orientation of the bin. If I draw
a parallel line to D one, let us say, this
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is the line which is parallel to CD, then
this amount of counter clock wise rotation
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of the bin has taken place, when D has moved
to D one. The problem that we have just solved,
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let me now restate the problem for your benefit,
this was our example 4.
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The first part of the problem that we have
just solved is to determine the location of
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all the revolute pairs, when the pair at D
moves to D one. We have also found out what
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is the amount of the rotation of the bin corresponding
to this movement. I will leave an assignment
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for you to do from the same figure.
Assume the actuator Z1 is held fixed, that
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is the distance O2A is not changing and the
actuator Z2 is activated to change the distance
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O7F in the figure to three fourths of it present
value, then what will be the corresponding
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rotation of the bin. You can follow exactly
the same logic and same technique of graphical
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analysis to solve the second part of the problem.
Let me now go to another example, which is
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our example 5. I will show you a figure which
will be simple 3R-1P mechanism to scale.
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First we will verify that this is a Non-Grashof
type linkage. That means, all the links can
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have only angular oscillation, no link can
rotate completely. The problem will be to
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determine the ranges of oscillation of these
links 2 and 4, in this particular Non-Grashof
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3R-1P mechanism. Let me now show the figure
of this problem and solve the problem for
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you.
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Let us now consider this 3R-1P mechanism,
as we see the link number 2 has a revolute
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pair with link number 1 at O2 and a revolute
pair at A with link number 3 and link number
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3 has a prismatic pair with link number 4
and link number 4 has a revolute pair at O4
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with link number 1. In this 3R-1P mechanism,
there are three revolute pairs at O2, A, and
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O4, there is a prismatic pair between link
3 and link 4, in this direction. First of
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all let us identify, what are the kinematic
dimensions of this 3R-1P mechanism.
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O2, O4 that is one link length, let me call
lmax and O2A is another link length which
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is l2, let me call it lmin. This 3R-1P mechanism
has only one more kinematic dimension, that
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is the offset. One should not be confused
by this dimension. This dimension has no kinematic
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significance because the location of this
revolute pair A, I cannot change but the prismatic
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pair between 3 and 4, which is along this
direction, I could have drawn it through A.
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This is the prismatic pair direction and this
perpendicular distance from O2 up to this
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direction of the prismatic pair, that is the
other kinematic dimension E, which we call
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offset.
If we measure, this lmin and e, we will find
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that, lmin plus e is greater than lmax. Consequently,
this will be a Non-Grashof 3R-1P chain and
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no link will be able to rotate completely,
all links will be oscillated. The question
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that is asked is, find the angle through which
the link number 2 and link number 4 can oscillate.
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Let me repeat, it is this distance which is
a red heading, which is there only to confuse
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you because it is the location of the revolute
pair O2, O4 and A, I cannot change and prismatic
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pair has no location, it is only a direction,
that is marked by this line. So, I draw a
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line through A, parallel to this direction
of prismatic pair and get the kinematic diagram
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for this particular linkage. Let me now solve
this problem to determine the angle through
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which links 2 and 4 can oscillate.
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Let us consider the kinematic diagram of the
3R-1P mechanism we are discussing. This O2O4,
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that is what we call lmax, O2A, that is what
we call lmin and perpendicular distance of
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O4 from the direction of sliding passing through
A, is what we call the offset e. By measurement,
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we can find that lmin plus e is more than
lmax. So, it is a 3R-1P Non-Grashof chain
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and we have to find the angle through which
the link O2A and link 4 that is this line
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can oscillate. To solve this problem, let
us imagine the equivalent 4R mechanism, that
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is the prismatic pair between link 3 and 4
in this direction. Let us replace it by a
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revolute pair at infinity in a direction perpendicular
to this direction of slide, that is the hinge
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B of the O2 A B O4 which is the equivalent
4R linkage, B is on this line at infinity.
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All parallel lines meet at infinity, so I
can take any line perpendicular to the direction
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of sliding and claim that B is at infinity
on any of these parallel lines because all
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these parallel lines meet at infinity. To
find the extreme position of link 2 or link
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4, what I know, the other two links, that
is say for extreme position of link 4, link
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2 and link 3, that is, O2A and AB must be
collinear. A must move on this circle which
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is centered at O2 and radius O2A. This is
the path of A, we call it kA and this offset
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e, this distance rotate with O4 as centre
and e as the radius, that is this corner moves
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on this circle.
Let me draw a common tangent to these two
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circles, one centered at O2 and the other
centered at O4. The common tangents between
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these two circles are these two lines. This
is one circle centered at O4, this is other
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circle centered at O2 with O2 as radius . So,
these two lines represent the common tangent
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to these two circles. Let me see what happens
when A occupies rather this point, if I call
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it B prime, B is at infinity, I call this
point B prime. When B prime comes here and
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if it oscillates in this direction up to this
point of tangency, let me call it as B2 prime.
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Let me find where is A. These are the extreme
positions of link 4 because as we will see
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O2A and AB has become collinear now because
A is here which I call A prime and A is here
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which I call A double prime, so O2A prime
because it is tangent, if perpendicular to
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this direction of sliding and B is also at
infinity to this direction of sliding. So,
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O2A prime and B at infinity has become collinear,
same is to for the other configuration, O2
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A double prime and B which is perpendicular
to this line direction of sliding has become
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collinear. Consequently, link 4 has taken
its extreme position. This is the angle through
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which link 4 oscillates, where O4B prime is
common tangent I have drawn and this is the
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point of tangency, this is the common tangent
to these two circles, this is the point of
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tangency. To find the extreme position of
link 2, I have to say that O4B prime and B
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prime B should become collinear.
These two circles intersect at say this I
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call A triple prime and this as A four prime.
Now B is moving on this circle, so B prime
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is also here, this point has moved to here
and O4 if I join this, the B goes to infinity
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in this direction, because B will be always
perpendicular to the direction of sliding
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at infinity. So what we see, that this link
4, O4B prime and link 3 AB prime have become
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collinear and consequently, link 2 has taken
its extreme position. Same will be true for
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this point of intersection and link 2 oscillates
through this large angle between this extreme
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positions.
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Just now we have seen very large angle of
oscillations of the input and output links
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of a 3R-1P mechanism which is of Non-Grashof
type. Here is a model, which is again a 3R-1P
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mechanism and this is the lmin and this is
the lmax and e is such that lmin plus e is
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more than lmax. Consequently, this becomes
a 3R-1P Non-Grashof type linkage. Let us now
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observe that angle of oscillations of this
link and this link. This is that common tangent
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that I was trying to draw between those two
circles, this line is parallel to that common
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tangent and consequently this has reached
its extreme position, it cannot be further
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inclined. However, this link can still go
and as we see from this extreme position,
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it is returning, that it is rotating counter
clock wise. Though this link is still going
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clock wise. This is the extreme position of
this link and now it cannot go any farther,
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it has to return.
Again, it has reached that extreme position
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and it cannot go any further, but this link
can go and the red link starts rotating in
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the opposite direction. This is the extreme
position of this link, whereas this is the
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extreme position of this link. From this,
this has to rotate in the opposite direction
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it cannot go any further. For this 3R-1P Non-Grashof
type linkage, as we see it is a very complicated
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movement with a very large angle of oscillation
for both link number 2 and link number 4.
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As a last example of displacement analysis,
let us now consider a really complicated problem
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as shown in this figure.
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This is the figure of arc movement machinery,
the model of which we have seen earlier. This
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00:29:52,499 --> 00:29:59,499
particular machine has three hydraulic actuators
one at Z1, other at Z2 and the third one at
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00:30:04,360 --> 00:30:11,360
Z3. This are three independent inputs, if
we count the number of links in this mechanism,
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I will get n = 12. This will be clear when
we draw the kinematic diagram of this particular
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machinery. If we count the number of pairs,
we will see there are 12 revolute pairs and
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3 prismatic pairs, so j is 15. Consequently,
if we calculate the degrees of freedom of
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this mechanism: F is equal to 3 of time's
n minus 1 of minus 2j, we get 33 minus 30,
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00:30:53,960 --> 00:30:58,039
that is 3.
So it is a 3 degrees of freedom mechanism
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00:30:58,039 --> 00:31:04,580
with 3 independent inputs. Consequently, it
is a constant mechanism and because of 3 degrees
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of freedom and with three inputs, this bin,
that is the output link which is defined completely
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00:31:10,639 --> 00:31:17,639
by these two hinges j and k can be taken anywhere
with any orientation. That means, I can independently
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00:31:20,549 --> 00:31:27,549
take j to say j2 and k to another position
k2, which will completely define the position
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00:31:29,129 --> 00:31:36,129
and orientation of this output link which
is the bin. Second thing to note is that,
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there is a very high order link, which is
connecting at C, D, E, H and K. This particular
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00:31:46,149 --> 00:31:52,690
link has 5 revolute pairs in it and it is
a link of fifth order. So, we will show how
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we can use the tracing paper very conveniently
to locate the revolute pairs in this kind
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00:31:58,460 --> 00:32:05,399
of higher order link. Let me now solve this
problem, that when j and k moves to j2 and
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00:32:05,399 --> 00:32:12,399
k2 respectively, how can we locate all other
moving revolute pairs, that is at B, C, D,
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00:32:13,789 --> 00:32:20,789
F, H and G and A. That is the problem that
we will solve right now.
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Let us consider the kinematic diagram of this
arc moving machinery which we have just seen.
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As we have said, there are 12 links and this
is the kinematic diagram, out of which link
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3 as we see have 4 hinges on it, O4, A, B,
D, so it is a quaternary link. Link 7 has
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00:32:46,239 --> 00:32:53,239
5 hinges on it - C, D, E, H and K, this is
a link of fifth order. Then question is, when
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this J and k move to J two and K two, can
we locate the corresponding positions of all
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00:33:04,080 --> 00:33:11,080
other revolute pairs? To solve this problem
first we note that the point D, because O4D
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00:33:15,559 --> 00:33:22,559
never changes and O4 is a fixed point. This
D point must move on this circle with O4 as
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00:33:24,360 --> 00:33:31,360
center and O4D as radius, it must move on
this circle. So, D two must lie on this line,
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00:33:36,259 --> 00:33:43,259
let me call it kD. Now k has gone to K two,
the distance KD also does not change, I measure
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00:33:50,950 --> 00:33:57,950
this KD and from K two,. I draw a circular
arc with kD as radius. So, D must move on
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00:33:59,519 --> 00:34:06,519
this circle and also the circle drawn earlier
through D with O4 as center. So, these two
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00:34:07,499 --> 00:34:14,499
circles one point of intersection is here
so I locate D two. Once I have located two
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00:34:16,889 --> 00:34:23,889
points, that is O4 and D two on this quaternary
link number 3, A and B I can easily determine
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00:34:24,859 --> 00:34:31,859
using the tracing paper. On a tracing paper,
I mark O4, A, B and D. These four points,
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their relative positions do not change because
they belong to the same rigid link 3.
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Now O4 does not move, so it stays here. When
D goes to D two and O4 stays at O4, where
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00:34:54,049 --> 00:35:01,049
ever these two dots move, they define the
location of A two and B two. We have also
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00:35:04,059 --> 00:35:11,059
located the point K which has gone to K two
and the point D which has gone to D two. Now
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00:35:19,059 --> 00:35:26,059
link number 7 is a link of order 5 and it
has 5 hinges namely: K, H, E, D and C. These
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00:35:32,140 --> 00:35:37,390
five points do not change the relative position,
but K has gone to K two and D has gone to
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00:35:37,390 --> 00:35:44,390
D two. So, if I now take, this is K so K goes
to K two and let me mark this point, so that
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00:35:50,130 --> 00:35:55,930
will be easy to remember, this is K and this
is D
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00:35:55,930 --> 00:36:02,930
. K
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00:36:10,349 --> 00:36:16,809
has gone to K two and D has gone to D two
and these other mark points gives the location
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00:36:16,809 --> 00:36:23,809
of C two, E two and H two.
There are five points K, H, E, D and C. I
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00:36:32,490 --> 00:36:39,490
get K two, D two, C two, E two and H two.
Once I have determined all these points, rest
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00:36:41,599 --> 00:36:48,599
of the problem is trivial, because this distance
is GJ and HG do not change. If I know J, which
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00:36:49,359 --> 00:36:56,359
has gone to J two, H, I have located, so I
can easily locate G two by JG and HG these
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00:36:56,650 --> 00:37:02,660
two link lengths. Rest of the points, I leave
for the students to complete. What we have
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demonstrated, is that the use of tracing paper
is very convenient, when we have such higher
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00:37:08,849 --> 00:37:15,849
order links and the displacement analysis
can be completed if the problem is well posted.
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Let me now state the problem that we have
just solved which was our example number 6.
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In example 6, we had the figure which had
the kinematic diagram of an arc moving machinery
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consisting of 12 links and 3 inputs, which
had those three hydraulic actuators Z1, Z2,
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00:37:37,920 --> 00:37:44,680
Z3. Statement of the problem was to determine
the location of all the moving revolute pairs
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00:37:44,680 --> 00:37:51,680
corresponding to configuration 2, when the
hinges J and K occupy respectively the locations
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00:37:51,869 --> 00:37:58,869
J two and K two. I can extend this problem
for you to have some practice which are of
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00:38:00,740 --> 00:38:01,619
little tougher.
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The second part of the problem could be, the
range of movements of the actuators Z1 and
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Z2 are such that the maximum distance between
O2 and A, that is (O2Amax is say 1800 millimetre
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to the scale and (O2Amin is 1100 millimetre.
Similarly, the movement of the actuator Z2
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00:38:22,910 --> 00:38:29,910
is such, that the maximum distance between
C and B, (CBmax is 2400 millimetre and (CBmin
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00:38:31,240 --> 00:38:38,240
is 1500 millimetre. The question is to determine
the zone in which the hinge K lies when the
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00:38:39,099 --> 00:38:46,099
full ranges of movements of Z1 and Z2 are
utilized. The third part of the problem could
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be, determine the maximum tilt angle alpha
of the bin while keeping the lip of the bin
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00:38:53,339 --> 00:38:59,760
L at the location L2. As I said I will not
solve this problem in detail just leave you
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with a little bit of hints.
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If we look carefully at the original diagram,
what we can see, that this distance O4A never
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changes because they belong to the same rigid
body. Consequently A lies on a circle with
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O4 as center and O4A as its radius. The maximum
distance between O2 and A, which is given
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00:39:26,660 --> 00:39:31,990
to us, locates this point. Minimum distance
between O2 and A, which is again specified
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00:39:31,990 --> 00:39:38,990
in the problem locates this point.
I can locate position of A, for the extreme
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00:39:39,210 --> 00:39:46,210
movements of this actuator Z1. O4, A, B and
D, all four belong to the same rigid body,
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so using the tracing paper technique that
we have just now seen, I can locate the corresponding
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position of D as D prime and D2 prime. For
the extreme positions, one can see that link
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4 and link 7, get rigidly connected and that
distance DC also does not change. Corresponding
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00:40:10,720 --> 00:40:15,809
to the maximum position of DC and minimum
position of DC, depending on the movement
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00:40:15,809 --> 00:40:22,809
of the actuator Z2 corresponding to D two
prime, I get two locations for C. Same way
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00:40:24,500 --> 00:40:31,500
corresponding to D prime, I get two locations
of C, corresponding to the maximum and minimum
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00:40:32,190 --> 00:40:39,190
distance of BC. Now D, C, K, they belong to
the same rigid link, so corresponding to the
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00:40:40,809 --> 00:40:47,809
D, C, I can locate corresponding positions
of K here and here. Same way, corresponding
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00:40:50,470 --> 00:40:57,010
to D prime, I get the extreme positions of
K as this one and this one.
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Now the question is, what is the range in
which the hinge K lies when the maximum movements
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of these two actuators, between O2and A, and
between B and C is utilized. This is the maximum
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distance BC, this is the minimum distance
BC. One can easily see that when the two actuators
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00:41:18,700 --> 00:41:24,789
have taken the extreme positions, the distance
DK does not change. Consequently, this is
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a circular arc with D double prime as the
center and this as the radius, same way this
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00:41:32,030 --> 00:41:39,030
as the radius. So I get, these two circular
arcs. Same way one can say, the distance from
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00:41:39,799 --> 00:41:46,799
O4 does not change again because if the actuators
are at maximum locations, link number 4 and
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00:41:47,650 --> 00:41:53,530
7 get rigidly connected and K moves on a circle
with O4 as center.
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00:41:53,530 --> 00:41:59,450
This is another circular arc, joining K1 double
prime and K1 prime with O4 as center and another
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00:41:59,450 --> 00:42:06,450
circular arc with O4 as center and radius
O4K2 prime. So these four circular arcs, determine
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00:42:08,319 --> 00:42:15,319
the region in which the hinge K can lie when
the maximum and minimum distance between O2
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00:42:18,730 --> 00:42:25,619
and A, and B and C are utilized by the hydraulic
actuators. In the full range of operation,
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00:42:25,619 --> 00:42:31,059
the hinge K lies within this region. With
this hinge we should be able to complete the
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00:42:31,059 --> 00:42:31,680
problem.
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To answer the third part of the problem, that
is the tilt angle of the bin when the lip
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00:42:39,160 --> 00:42:46,160
is kept at this location L two, we solve it
as follows: We have already determined that
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the region in which the hinge K lies when
the total range of movements actuators are
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00:42:55,210 --> 00:43:01,740
used. The hinge K can never go outside these
regions consisting of four circular arcs.
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00:43:01,740 --> 00:43:07,480
But L and K are two points on the same rigid
body, so their distance cannot change. If
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00:43:07,480 --> 00:43:14,480
the lip is kept fixed at L then I rotate this
line LK two to reach the boundary of this
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00:43:14,930 --> 00:43:21,930
region, that is K three and this angle gives
the angle of rotation that the bin undergoes
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00:43:23,220 --> 00:43:29,109
keeping the lip at L. It cannot rotate any
further because then K will go outside this
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00:43:29,109 --> 00:43:36,109
range which is not possible for the range
of movement of actuators which are given.
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Let me now summarize what we have learned
today. Today, we have solved three rather
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00:43:43,859 --> 00:43:49,359
intricate or I would say more advance problems
of displacement analysis of planar mechanism
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00:43:49,359 --> 00:43:56,220
through examples having more than 1 degree
of freedom, that is the front end loader and
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00:43:56,220 --> 00:44:02,880
the arc moving machinery and the other is
a 3R-1P Non-Grashof linkage. What we have
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00:44:02,880 --> 00:44:09,450
seen, that this graphical method is very powerful
and clever use of the tracing paper as an
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00:44:09,450 --> 00:44:15,200
over lay can become very handy to solve even
more complicated problems of displacement
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00:44:15,200 --> 00:44:22,200
analysis. In our next lecture, we shall discuss
the analytical method of displacement analysis.
310