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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 show
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,
with 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 by the x, y coordinates of a point
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only and the orientation will be automatically
fixed up or only the x-coordinate and the
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orientation of this bin and the y-coordinate
will be automatically decided.
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So 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 at
Z1, the one named Z1 and 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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Now let us see this mechanism does not have
all binary links, as we have considered earlier.
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On the first three examples that we considered,
they had either binary link or ternary link.
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Here if we consider link number 4, as we see
it 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 and link 9 is the front
end of the loader, that is the bin.
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So 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. Now let me state the problem,
the problem is posed as follows: that if link
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number 9 the point D of link number 9 comes
to D prime. This is the scale of the drawing;
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25 cm is represented by this distance. Now
when the point D comes to D prime, can we
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locate all other revolute pairs? That means,
where they move in this configuration, when
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point D of link 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 plus 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 8, 24 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 constrained mechanism.
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Because it has 2 degrees of freedom, I can
have 2 degrees of freedom for this output
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bin, that is the point D can go to D prime
but at that location the orientation of bin
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will be automatically determined. I cannot
take D to D prime and then again change the
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orientation. So, we solve this problem, that
when D goes to D prime, how do we determine
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the location of all other 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 D1 and the objective is to find
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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
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link, that is number 9, that is the bin, this
distance also cannot change because I know
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the location of D prime, I put my centre of
a circle with DC as radius with D1 as the
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center.
If I draw this circular arc, C must lie on
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this circle. Thus, we get that C should lie
on this circle and also on this circle, that
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means the intersection of these two circles
determine the location of C, when D moves
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to D1, C moves to C1. Now to locate the positions
of the hinges, say A and B, which lie on link
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4, as we noted earlier link 4 is a rigid link
but it is a quaternary link having 4 hinges,
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one at O4, other at A, other at B and one
at C. When D comes to D1, C has moved to C1,
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O4 does not move, so, O4 remains there. To
locate the positions of A and B, corresponding
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to the second position, what we do, we take
a tracing paper and mark these points: O4,
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A, B and C. The relative positions of these
O4, A, B, C can never change because they
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lie on the same rigid body.
So what I do is I move this tracing paper
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now with O4 at O4 and C at C1. These two red
dots, which are the positions of A and B is
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automatically determined, A and B moves to
these new locations. So if we can now pierce
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this tracing paper here and there, I immediately
get the locations of A and B. This point will
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give me A1 and this point will give me the
location B1. Let me repeat it once more because
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O4, A, B, C are four points on the same rigid
link, I use this tracing paper as an over
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lay, mark the relative locations of O4, A,
B, and C because two points define the rigid
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body completely and in the second position
I know the location of O4 and the location
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of C1.
So I take these two red dots corresponding
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to O4 and C to this new location and where
ever these two dots come, they are the positions
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of A and B corresponding to the second configuration.
Now once we have obtained D1 and B1, it is
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very easy to determine the new location of
this revolute pair E, because the distance
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DE does not change. So with D1 as center,
I draw a circular arc with DE as the radius.
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So, E must lie on this circle and BE also
does not change. So, if B1 as center, I draw
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another circular arc with BE as the radius,
where ever these two circles intersect, that
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gives me the location of E, which I call E1.
Similarly, to determine F, I know the locations
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B, E and F. These are the three points on
the same rigid link 5 so, I can again use
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this tracing paper as an over lay and mark
B, F and E, their relative positions do not
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change and now I know where B and E has gone,
so I can replace this, coinciding this B with
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B1, this E with E1 and where ever F goes,
that becomes
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that becomes the location of F, which I call
F1.
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I have determined the location of all the
moving hinges namely: A, B, F, E, D as A1,
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B1, E1, F1 and D1 of course is given, and
C as C1. So we have solved this part of the
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problem that corresponding to the second position,
when D has moved to D1, where all other moving
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hinges go. Not only that, I can also find
what is the inclination of the bin from the
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first position. That is this line CD, that
was this initial orientation has now become
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this. So, angle between these two lines CD
and C1D1 determines the change in orientation
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of the bin. If I draw a parallel line to through
D1, let us say this is the line which is parallel
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to CD, then this amount of counter clock wise
rotation of the bin has taken place, when
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D has moved to D1. The problem that we have
just solved, let me now restate the problem
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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 D1. We have also found out what is
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the amount of the rotation of the bin corresponding
to this movement. Now 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. Now I will show you a figure
which will be a simple 3R-1P mechanism to
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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. So in this 3R-1P mechanism,
there are three revolute pairs at O2, A, and
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O4 and 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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O2O4 that is one link length, let me call
it 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
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call 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 oscillatory. 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 herring, which is there only to confuse
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you because it is the location of the revolute
pair O2, O4 and A, that I cannot change and
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prismatic pair has no location, it is only
a direction, that is marked by this line.
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So, I draw a line through A parallel to this
direction of prismatic pair and get the kinematic
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diagram for this particular linkage. Let me
now solve this problem to determine the angle
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through which links 2 and 4 can oscillate.
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Let us consider now the kinematic diagram
of the 3R-1P mechanism we are discussing.
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This O2O4, that is what we call lmax, O2A,
that is what we call lmin and perpendicular
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distance of O4 from the direction of sliding
passing through A, is what we call the offset
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‘e’. By measurement, we can find that
lmin plus e is more than lmax. So, it is a
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3R-1P Non-Grashof chain and we have to find
the angle through which the link O2A and link
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4 that is this line can oscillate. To solve
this problem, let us imagine the equivalent
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4R mechanism, that is the prismatic pair between
link 3 and 4 in this direction. Let us replace
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it by a revolute pair at infinity in a direction
perpendicular to this direction of slide,
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that is the hinge B of the O2ABO4 which is
the equivalent 4R linkage, B is on this line
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at infinity.
All parallel lines meet at infinity, so I
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can take any line perpendicular to the direction
of sliding and claim that B is at infinity
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on any of these parallel lines because all
these parallel lines meet at infinity. Now
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to find the extreme position of link 2 or
link 4, what I know, the other two links,
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that is say for extreme position of link 4,
link 2 and link 3, that is, O2A and AB must
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be collinear. And ‘A’ must move on this
circle which is centered at O2 and radius
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O2A. This is the path of A, we call it kA
and this offset e, this distance rotate with
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O4 as centre and ‘e’ as the radius, that
is this corner moves on this circle.
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Now let me draw a common tangent to these
two circles, one centered at O2 and the other
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centered at O4. The common tangent between
these two circles are these two lines. This
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is one circle centered at O4, this is other
circle centered at O2, with O2 as radius.
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So, these two lines represent the common tangent
to these two circles. Let me see what happens
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when A occupies rather this point, if I call
it B prime, B is at infinity, I call this
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point B prime. When B prime comes here and
if it oscillates in this direction up to this
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point of tangency, let me call it as B two
prime.
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Let me find where is A now. These are the
extreme positions of link 4 because as we
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will see O2A and AB has become collinear now
because A is here which I call A prime and
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A is here which I call A double prime, so
O2A prime because it is tangent is perpendicular
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to this direction of sliding and B is also
at infinity to this direction of sliding.
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So, O2A prime and B at infinity has become
collinear, same is to for the other configuration,
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O2A 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. So this is the angle
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through which link 4 oscillates, where O4B
prime is common tangent I have drawn and this
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is the point of tangency, this is the common
tangent to these two circles, this is the
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point of tangency. To find the extreme position
of link 2, I have to say that O4B prime and
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B 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 that will yes this is the extreme
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position it cannot go any further and this
link can still go and the red link starts
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rotating in the opposite direction. This is
the extreme position of this link, whereas
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this is the extreme position of this link.
From this, this has to rotate in this opposite
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direction, it cannot go any further. And from
this it cannot go any further, it has to rotate
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in the opposite direction, from here.
So for this 3R-1P Non-Grashof type linkage,
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as we see it is a very complicated movement
with a very large angle of oscillation for
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both link number 2 and link number 4.
As a last example of displacement analysis,
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let us now consider a really complicated problem
as shown in this figure.
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This is the figure of an earth-moving machinery,
the model of which we have seen earlier. This
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00:29:52,500 --> 00:30:06,120
particular machine has three hydraulic actuators
one at Z1, other at Z2 and the third one at
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Z3. So these 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 equal to 3 of times n minus
1 of minus 2j, we get 33 minus 30, that is
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3.
So it is a 3 degrees of freedom mechanism
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with 3 independent inputs. Consequently, it
is a constrained mechanism and because of
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3 degrees of freedom and with three inputs,
this bin, that is the output link which is
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defined completely by these two hinges j and
k can be taken anywhere with any orientation.
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That means, I can independently take j to
say j2 and k to another position k2, which
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will completely define the position and orientation
of this output link which is the bin. Second
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00:31:35,201 --> 00:31:38,419
thing to note is that, there is a
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very high order link, which is connecting
at C, D, E, H and K. This particular link
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00:31:46,899 --> 00:31:53,200
has 5 revolute pairs in it and it is a link
of fifth order. So, we will show how we can
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use the tracing paper very conveniently to
locate the revolute pairs in this kind of
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higher order link. Let me now solve this problem,
that when j and k moves to j2 and k2 respectively,
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how can we locate all other moving revolute
pairs, that is at B, C, D, F, H and G and
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A. That is the problem that we will solve
right now.
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Let us consider the kinematic diagram of this
earth-moving machinery which we have just
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seen. As we have said, there are 12 links
and this is the kinematic diagram, out of
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which link 3 as we see have 4 hinges on it
– O4, A, B, D, so it is a quaternary link.
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Link 7 has 5 hinges on it – C, D, E, H and
K, this is a link of fifth order. Then question
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00:32:56,360 --> 00:33:04,090
is, when this J and k move to J2 and K2, can
we locate the corresponding positions of all
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00:33:04,090 --> 00:33:13,020
other revolute pairs?
To solve this problem first we note that the
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point D, because O4D never changes and O4
is a fixed point, this D point must move on
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this circle with O4 as center and O4D as radius,
it must move on this circle. So, D2 must lie
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on this line, let me call it kD. Now K has
gone to K2, the distance KD also does not
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change, so I measure this KD and from K2,
I draw a circular arc with KD as radius. So,
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D must move on this circle and also the circle
drawn earlier through D with O4 as center.
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00:34:05,900 --> 00:34:16,810
So, these two circles one point of intersection
is here so I locate D2. Once I have located
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two points, that is O4 and D2 on this quaternary
link number 3, A and B I can easily determine
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using the tracing paper. So 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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00:34:46,869 --> 00:34:54,379
Now O4 does not move, so it stays here. When
D goes to D2 and O4 stays at O4, where ever
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00:34:54,379 --> 00:35:06,440
these two dots move, they define the location
of A2 and B2. We have also located the point
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K which has gone to K2 and the point D which
has gone to D2. Now link number 7 is a link
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of order 5 and it has 5 hinges namely: K,
H, E, D and C. These five points do not change
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00:35:33,569 --> 00:35:44,700
the relative position, but K has gone to K2
and D has gone to D2. So, if I now take, this
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is K so K goes to K2 and let me mark this
point, so that will be easy to remember, this
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00:35:52,910 --> 00:36:13,280
is K and this is D. K goes to K2 and D goes
to D2. K has gone to K2 and D has gone to
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00:36:13,280 --> 00:36:26,170
D2 and these other mark points gives the location
of C2, E2 and H2 and Ok
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Those are five points K, H, E, D and C. I
get K2, D2, C2, E2 and H2. Once I have determined
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all these points, rest of the problem is trivial,
because this distance is GJ and HG do not
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00:36:46,710 --> 00:36:52,760
change. If I know J, which has gone to J2,
H, I have located, so I can easily locate
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G2 by JG and HG these two link lengths. Rest
of the points, I leave for the students to
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complete. What we have demonstrated that the
use of tracing paper is very convenient, when
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we have such higher order links and the displacement
analysis can be completed if the problem is
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00:37:13,829 --> 00:37:20,380
well posed. Let me now state the problem that
we have just solved which was our example
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00:37:20,380 --> 00:37:23,809
number 6.
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00:37:23,809 --> 00:37:30,260
In example-6, we had the figure which had
the kinematic diagram of an earth-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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00:37:37,740 --> 00:37:44,680
Z3. And statement of the problem was, determine
the location of all the moving revolute pairs
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corresponding to configuration 2, when the
hinges J and K occupy respectively the locations
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00:37:51,710 --> 00:38:01,180
J2 and K2. I can extend this problem for you
to have some practice which are of little
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00:38:01,180 --> 00:38:02,180
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 (O2A)max is say 1800 millimeter
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to the scale and (O2A)min is 1100 millimeter.
Similarly, the movement of the actuator Z2
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is such, that the maximum distance between
C and B, (CB)max is 2400 millimeter and (CB)min
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is 1500 millimeter. The question is to determine
the zone in which the hinge K lies when the
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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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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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00:39:18,950 --> 00:39:26,660
O4 as center and O4A as radius. The maximum
distance between O2 and A, which is given
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to us, locates this point. Minimum distance
between O2 and A, which is again specified
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in the problem locates this point.
So I can locate position of A, for the extreme
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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 we have seen, I can locate
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the corresponding position of D as D prime
and D two prime. For the extreme positions,
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one can see that link 4 and link 7, get rigidly
connected and that distance DC also does not
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00:40:07,380 --> 00:40:15,170
change. So corresponding to the maximum position
of DC and minimum position of DC, depending
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00:40:15,170 --> 00:40:23,859
on the movement of the actuator Z2 corresponding
to D two prime, I get two locations for C.
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Same way corresponding to D prime, I get two
locations of C, corresponding to the maximum
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and minimum distance of BC. Now D, C, K, they
belong to the same rigid link. So corresponding
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00:40:40,270 --> 00:40:50,470
to the 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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00:40:57,010 --> 00:41:03,930
Now the question is, what is the range in
which the hinge K lies when the maximum movements
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00:41:03,930 --> 00:41:11,420
of these two actuators, between O2 and A,
and between B and C is utilized. This is the
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00:41:11,420 --> 00:41:18,700
maximum 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 --> 00:41:24,790
have taken the extreme positions, the distance
DK does not change. Consequently, this is
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00:41:24,790 --> 00:41:32,650
a circular arc with D double prime as the
center and this as the radius, same way this
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00:41:32,650 --> 00:41:39,799
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 --> 00:41:47,651
O4 does not change again because if the actuators
are maximum locations, link number 4 and 7
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00:41:47,651 --> 00:41:52,849
get rigidly connected and K moves on a circle
with O4 as center.
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00:41:52,849 --> 00:41:58,890
So this is another circular arc, joining K1
double prime and K1 prime with O4 as center
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00:41:58,890 --> 00:42:06,530
and another circular arc with O4 as center
and radius O4K2 prime. So these four circular
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00:42:06,530 --> 00:42:18,130
arcs, determine the region in which the hinge
K can lie when the maximum and minimum distance
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00:42:18,130 --> 00:42:24,569
between O2 and A, and B and C are utilized
by the hydraulic actuators. So in the full
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00:42:24,569 --> 00:42:29,770
range of operation, the hinge K lies within
this region. With this hint you should be
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00:42:29,770 --> 00:42:31,900
able to complete the problem.
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00:42:31,900 --> 00:42:39,160
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 --> 00:42:47,450
is kept at this location L2, we solve it as
follows: We have already determined that the
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00:42:47,450 --> 00:42:55,920
region in which the hinge K lies when the
total range of movements actuators are used.
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The hinge K can never go outside these regions
consisting of four circular arcs. But L and
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00:43:02,559 --> 00:43:08,640
K are two points on the same rigid body, so
their distance cannot change. If the lip is
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00:43:08,640 --> 00:43:15,790
kept fixed at L then I rotate this line LK2
to reach the boundary of this region, that
305
00:43:15,790 --> 00:43:24,339
is K3 and this angle gives the angle of rotation
that the bin undergoes keeping the lip at
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00:43:24,339 --> 00:43:29,859
L. It cannot rotate any further because then
K will go outside this range which is not
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00:43:29,859 --> 00:43:36,650
possible for the range of movement of actuators
which are given.
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00:43:36,650 --> 00:43:43,859
Let me now summarize what we have learned
today. Today, we have solved three rather
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00:43:43,859 --> 00:43:49,830
intricate or I would say more advance problems
of displacement analysis of planar mechanism
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00:43:49,830 --> 00:43:56,010
through examples, two, having more than 1
degree of freedom, that is the front-end loader
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00:43:56,010 --> 00:44:02,870
and the earth-moving machinery and the other
is a 3R-1P Non-Grashof linkage. What we have
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00:44:02,870 --> 00:44:09,849
seen, that this graphical method is very powerful
and clever use of the tracing paper as an
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00:44:09,849 --> 00:44:15,240
over lay can become very handy to solve even
more complicated problems of displacement
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00:44:15,240 --> 00:44:36,599
analysis. In our next lecture, we shall discuss
the analytical method of displacement analysis.
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1