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In an earlier lecture, we have already mentioned
that there are two types of problems in kinematics
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namely, kinematic analysis and kinematic synthesis.
In kinematic analysis, we determine the relative
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motion characteristics of a given mechanism.
From today's lecture, we shall start the discussion
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on this topic of kinematic analysis.
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Broadly, we can classify the kinematic analysis
problems into three headings namely, displacement
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analysis, velocity analysis and acceleration
analysis. For all these three types of problems,
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that is displacement analysis, velocity analysis
and acceleration analysis, we can use either
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a graphical method or an analytical method.
In today's lecture, we shall discuss only
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the graphical method and that too only of
displacement analysis. Later on, we shall
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take up velocity analysis and acceleration
analysis. Let us see, what do you mean by
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a displacement analysis of a mechanism?
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If we have given the kinematic dimensions
and the position or the movement of the input
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link, then we should be able to determine
the position or movement of all other links
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by this displacement analysis. In the graphical
method, first we draw the kinematic diagram
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of the mechanism to a suitable scale and then
the desired unknown quantities are determined
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through suitable geometrical constructions
and calculations. We shall demonstrate this
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graphical method through a series of examples.
However, before going into the details of
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each and every example, let me list the main
points that one should remember while using
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graphical method of displacement analysis.
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So, the main points are: First of all, we
must remember that the configuration of a
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rigid body in plane motion is completely defined
by the location of any two points on it. That
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means, if we know the location of any two
arbitrary points of a rigid body in plane
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motion, then we know the location of all other
points on that rigid body.
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The second point is as we will see when we
solve the examples that very often, we will
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need to draw two circles which are intersecting
or a line and a circle which are intersecting.
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We know that such intersection points can
be two that means two intersecting circles,
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in general intersect at two points. Similarly,
a line and a circle also intersect generally
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at two points. However, sometimes it may be
necessary to choose the correct point of intersection
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for the problem in hand. As I said, all these
points will be shown very clearly when we
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solve those particular examples.
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The third problem is sometimes we may use
a tracing paper as an overlay and we will
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see that this will very convenient, especially
if there are higher order links present in
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any particular mechanism and lastly, we must
know that the graphical method of displacement
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analysis cannot be used unless there are adequate
number of four-links closed loops in the particular
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mechanism. Unless, we have adequate number
of four-links closed loops, the graphical
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method of displacement analysis cannot proceed.
So, let me start with the first example, which
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is a slotted lever quick return mechanism
which is used in shipping machines. First,
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I will show the model, then the kinematic
diagram of the same mechanism and then post
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the question of displacement analysis.
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This is the model of the quick return mechanism
that is used in shipping machine. This is
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the cutting tool. This tool holder moves in
a slot in the frame of the machine. This is
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called the bull gear, which is rotated at
a constant speed. Here is a block, which is
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hinged to the bull gear and this block moves
up and down in this slotted lever, the slotted
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lever is hinged to the frame at this point
and the slotted lever is connected to the
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tool holder by an additional link. So, we
have a six-link mechanism where the continuous
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uniform rotation of the bull gear is converted
into the to and fro motion of the cutting
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tool. The cutting tool during the forward
motion is doing useful work and we have to
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maintain a proper cutting speed. However,
during the return stroke of the tool, this
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is not doing any useful work, so I would like
to make the return faster and that is why,
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it is called quick return mechanism. As we
see very clearly, if I rotate the bull gear
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at almost uniform speed, the forward motion
is slow but the return is faster.
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This is the kinematic diagram of the six-link
slotted lever mechanism, the model of which
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we have just seen. Here, this link O2A, that
is link number 2 represents the bull gear,
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link number 3 is the block which goes up and
down in the slotted lever, which is link number
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4, link number 5 connects the slotted lever
to the tool holder, which is represented by
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link 6. So, we have a six-link mechanism with
a revolute pair at O2, revolute pair at O4,
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revolute pair at A, revolute pair at B, revolute
pair at D and there are two prismatic pairs,
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one between 1 and 6 in this horizontal direction
and there is a prismatic pair between link
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3 and 4 along the slotted lever. The problem
is if link 2 rotates at a constant speed say,
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omega 2 then we have to find, what we call
Q.R.R- Quick Return Ratio.
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By Quick Return Ratio we mean the time that
the tool takes for the forward motion and
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ratio of the time taken for the forward motion
and the backward motion.
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If link 2 is rotating at a constant speed,
say omega 2, then the time it takes for the
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forward motion is thetaforward that is the
rotation of the link 2 during the forward
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motion and backward motion is say return,
thetar where thetar is the rotation of link
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2 during the return motion. So the problem
is during this mechanism, can we find the
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quick return ratio? Before I start solving
the problem, let me restate the problem for
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your benefit, which is what we are taking
as our example 1.
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Example 1 is determination of quick return
ratio of a slotted lever mechanism used in
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shapers. It is already been seen that the
quick return ratio which is defined as q.r.r
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is equal to thetaf divided by thetar, where
thetaf is the rotation of link 2 during the
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forward motion and thetar is the rotation
of same link 2, that is the bull gear during
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the return motion of the cutting tool. Of
course, we are assuming that the angular speed
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of link 2 remains constant.
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Let us now return to the kinematic diagram
of the same slotted lever quick return mechanism
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in which link 2, that is this line O2A represents
the bull gear of the input link and link 6,
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that is this block D represents the tool holder
of the cutting tool. As this bull gear rotates
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counterclockwise direction, this tool moves
from right to left and left to right. This
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motion from right to left is the cutting motion
and left to right is the return motion. Our
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objective is to determine the quick return
ratio of this particular mechanism and also
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to determine what the stroke length of the
tool is.
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To analyze the problem, let me first note
that this point A moves along this circle
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whose centre is at O2, the radius is O2A.
That is, this circle represents the path of
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the point A and I call it kA. To locate the
extreme positions of the slotted lever, that
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is this link 4, I draw two tangents from the
point O4 to this circle kA. This is the tangent
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to the right and similarly, this is the tangent
to the left. Consequently, the right most
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position of this slotted lever, that is link
4 is represented by this line and the left
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most extreme position of the same slotted
lever O4B is represented by this line. When
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the slotted lever along these two lines, the
point A this at this location let me call
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it AR and at this position, let me call AL.
Since the distance O4B does not change, I
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can also locate right most position of the
revolute pair at B by drawing this circular
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arc with O4 as centre and O4B as radius. So,
I gave this point which I call BR, denoting
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the right most position of B.
The same way on this line O4 AL, I locate
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the left most position of the revolute pair
BL. Now, let us notice that the point D is
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going along this horizontal straight line
and the distance BD does not change. So, when
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B occupies BR, let me locate where DR is.
To do that, I draw a circular arc with BR
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as centre and BD as radius intersects this
horizontal line. That is the line of the stroke
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of the cutting tool at this point which I
mark as DR. So DR indicates the extreme right
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most position of link 6.
The same way, I can look at DL on the same
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horizontal line. This is the line of movement
of D and D does not change, BL is here. So,
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I draw a circular arc with BL as centre and
BD as radius to locate the extreme position
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of the point D which I call DL. So this distance
DL, DR represents the stroke of the cutting
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tool according to the scale of the figure.
Now to find out the quick return ratio, we
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see that the link O2A rotates from O2AR to
O2AL during the forward movement. So, this
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is the angle through which link 2 is rotating
through its forward movement which I call
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thetaf.
Same way, during the return motion, the point
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A is going from AL to AR, that is link O2A
is going from O2AL to O2AR and rotating through
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an angle which I call thetar. So, we can easily
determine the quick return ratio, q.r.r as
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the ratio of the angle thetaf divided by thetar.
We have solved the problem. We have determined
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this given slotted lever quick return mechanism.
The quick return ratio is given by thetaf
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divided by thetar and the stroke of the tool
as DRDL for the particular given mechanism.
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We also note that if the stroke length DRDL
has to be decreased then I have to change
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the length O2A. As we decrease the length
O2A, the distance DRDL will change, that is
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the stroke length of the tool will be decreased.
However, as we decrease the radius O2A then
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this tangent from this O4 to this circle kA,
that is these points AR and AL move upward
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and the angle thetaf and thetar both tend
to value 180 degree. Consequently, q.r.r which
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is thetaf divided by thetar, when both of
them tend to the same value 180 degree, q.r.r
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reduces. That is, this mechanism is okay for
producing quick return effect as long the
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stroke of the tool is sufficiently large,
quick return effect continuously decreases
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as the stroke length of the tool decreases.
We have just now explained the slotted lever
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quick return mechanism used for shipping machines
is not good for smaller stroke length because
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the quick return ratio tends to become 1.
That is, the quick return ratio decreases
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with decreasing stroke length.
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However, for slotting machine where the stroke
lengths are normally short, we have to have
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different types of quick return mechanism
where the quick return ratio is the independent
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of the stroke length. Such a mechanism is
called Whitworth quick return mechanism. So
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we shall first see the model of such a Whitworth
quick return mechanism then, pose the same
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problem as example 2, determine the quick
return ratio of this Whitworth quick return
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mechanism and also see how it is independent
of the stroke length of the tool. Before showing
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the model and solving the problem, let me
restate the problem for your benefit.
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This is our example 2, that is determination
of quick return ratio of Whitworth quick return
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mechanism used in slotting machines. As I
said earlier, we will also show that in this
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mechanism the quick return ratio will be independent
of the stroke length.
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This is the model of Whitworth quick return
mechanism. Here, the cutting tool is this
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one which is moving in this horizontal slot.
So, if we rotate the input link at a constant
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speed, this is obvious that the forward and
return motion of the tool is taking at different
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time. It is going much faster in this direction
and coming back slower in this direction.
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So, if we consider this as the cutting tool,
which is cutting in that direction, then we
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should rotate it in the opposite direction.
It is cutting slowly and returning much faster.
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We shall see the kinematic diagram of this
Whitworth quick return mechanism and analyze
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its quick return ratio.
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This is the kinematic diagram of that Whitworth
quick return mechanism. As we see, this is
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also a six-link mechanism with five revolute
pairs and two prismatic pairs. This link 6
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is the tool holder and link 2 is the input
link which rotates at constant angular speed,
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link 3 is the block which goes along link
4 at this prismatic pair, link 4 is hinged
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to the fixed link at O4, link 2 is hinged
to the fixed link at O2, link 4 or 5 are connected
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by this revolute pair at C. Link 5 and 6 are
connected by this revolute pair and link 6
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has a prismatic pair with fixed link 1 along
this horizontal direction. For the given kinematic
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dimensions, our objective is to determine
the quick return ratio. If the link 2, that
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is input link rotates at constant angular
speed.
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First thing to note that the stroke of the
tool that is, this 6 is entirely decided by
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the length O4C. When D goes to the right most
position in this direction, C comes on this
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line, O4C and CD becomes collinear. Similarly,
for the left most position of this link 6,
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again O4C and CD becomes collinear but C comes
on this side on the line O4D and the stroke
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length of the tool is obviously equal to twice
of O4C because the maximum distance is O4C
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plus CD and the minimum distance is CD minus
O4C. So, the total movement of the tool is
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given by twice O4C, that is the stroke length
is changed by changing the length O4C, no
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change is made in the length O2A, whereas,
we shall see that the quick return ratio will
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be entirely decided by the link length O2A,
the position of O2 and the position of this
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horizontal line.
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This is the kinematic diagram of the same
Whitworth quick return mechanism which we
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have just seen. This link O2A is the input
link number 2 and this link 6 is the output
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link, that is the tool holder. Due to continuous
uniform rotation of this input link 2, this
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tool holder 6 oscillates along this horizontal
line.
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The point to note is this point A moves on
a circle with O2 as centre and O2A as the
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radius. This is the path of the point A, let
me call it kA. Again, we should note that
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the point A, O4 and C always lie on the same
straight line and O4 is never moving. So,
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when the point A comes here, that is the point
of intersection kA and the line of reciprocation
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of the tool, let me call this point of intersection
is A to the power L. Corresponding to this
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position of A, since A, O4 and C are always
on one line, C will also be on this horizontal
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line and at a distance O4C from O4, because
this is also a link length which is not changing.
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Corresponding position of C, as C moves on
this circle with O4 as centre and O4C as radius.
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So, I will call this point C to the power
L that is, the left most position of C. Correspondingly,
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D will move here such that C to the power
L, D to the power L is CD because this link
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length is also not changing. So, this is the
left most position of the tool holder. Let
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me call it D to the power L.
Exactly the same way, when the point A occupies
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this point, which is the intersection of kA
and the line of reciprocation of the tool
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through O4, let me call this point of intersection
as A to the power of R. Since C is moving
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on this circle and A, O4 and C must be on
line, if I draw this circle, that is the path
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of C, when it intersects this line of reciprocation,
I will call that the intersection as C to
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the power of R. Again, the distance CD is
unique. So, from C to the power of R, if I
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draw an arc with CD as the radius, I get the
right most position of D, which I call D to
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the power of R. Thus, this tool that is this
link 6 moves from D to the power L to D to
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the power R that is, the stroke length which
is exactly equal to two times O4C.
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However, during this movement from D to the
power L to D to the power of R, the point
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A goes from A to the power L to A to the power
of R, from D to the power of R to D to the
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power L, the point A goes from A to the power
of R to A to the power L. As we see the O2A,
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this link is rotating with constant input
speed. So from right to left, the rotation
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is this angle, which I call thetaf and from
left to right it rotates only through this
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angle which is 2 pi minus thetaf. That is,
during the return stroke thetar is 2 pi minus
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thetaf and the quick return ratio is thetaf
divided by thetar. As we see, the stroke length
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can be changed by changing the length O4C
which has no role to play so far as these
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two lines are concerned, that is O2A to the
power L and O2A to the power of R that is
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decided by the intersection of the circle
kA and the line of reciprocation of this point
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D. So, the quick return ratio remains same
even if we change the link length O4C which
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causes a change in the stroke length. So,
this is the quick return ratio of the Whitworth
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quick return mechanism which is independent
of the stroke length.
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As our next example, let us consider another
six-link mechanism which is shown in this
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figure. Here, as we see there is a four-bar
mechanism O2ABO4. Link 4 of this four-link
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mechanism, that is this link is connected
to another link 5 at this compound hinge B
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where 3 links are connected namely, 3, 4 and
5. Link 5 is connected to this slider which
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is link 6 and slider has a prismatic pair
with the fixed link such that the point D
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moves in the horizontal direction. The question
is if this link O2A, that is link number 2
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rotates completely, what is the stroke length
of the slider at D? The scale of the diagram
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has been shown here that this distance is
equal 5 cm.
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For this problem, first we have to see that
the four-link mechanism O2ABO4 happens to
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be a crank rocker because the maximum link
length O2A for which I may call l1 as lmax
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and l2 as lmin, lmax plus lmin is less than
l3plus l4, that is the other two link lengths.
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So, this is the Grashof linkage with O2A as
the shortest link. Consequently, this link
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O2A will rotate completely. We have to find
out what is the maximum right most position
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of this point D and what is the left most
position of this point D such that I can determine
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DL and DR which will give the stroke length
of the slider 6. Before we solve this problem,
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let me write out this problem for your benefit.
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This is our third example and example 3 is
the figure shows a six link mechanism. Determine
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the stroke-length of the output link, that
is the slider 6. Also, determine the quick
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return ratio assuming the constant angular
speed of link 2.
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Let us solve this example 3 and the kinematic
diagram of that mechanism is shown here again.
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The problem is first to find out the stroke
length of this link 6, that is the slider
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at D. To find the extreme right position of
the point D, first we have to find out what
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is the extreme right position of this point
B? Due to this length O4B that is link 4,
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B moves on a circle with O4 as centre and
O4B as radius. This circle represents the
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path of B. Let me call it kB. The extreme
right position of the point B will be taken
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up. As we have discussed earlier, when link
2, that is O2A and link 3, that is AB become
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collinear. So the extreme position of B let
me call it B to the power of R, when O2B to
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the power of R is equal to O2A plus AB. So,
I take O2A plus AB and from O2, I mark that
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distance at B arc.
Let me repeat. B moves on this circle and
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at the extreme position, O2A and AB become
collinear. So, O2B to the power of R becomes
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O2A plus AB. Similarly, for the extreme left
position, again link 2 and link 3 become collinear
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but O2A comes here. Let me call it A to the
power L. This A to the power L B to the power
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L is equal to AB and this point becomes B
to the power of L. That is, O2B to the power
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L becomes the difference of the link lengths
AB minus O2A. So, I take the difference of
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AB and O2A and take that from O2 and mark
it on kB. The link 4 oscillates from O2B to
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the power of R to O2B to the power L. So,
this is a crank rocker mechanism and this
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is the rocking movement of the link 4. BD
is of fixed length and B moves on this horizontal
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line. From this B to the power of R, if I
mark this is BD. From B to the power of R,
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I mark this circle and wherever it intersects
the horizontal line through D that determines
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the extreme position of D, I call it D to
the power of R.
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Similarly, from B to the power L, again taking
the same length BD, I draw a circular arc
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and wherever it intersects this horizontal
line through D that determines the extreme
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left position of D, that is D to the power
L. So, this distance D to the power of L D
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to the power of R determines the stroke length
of this slider 6. Now, to determine the quick
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return ratio that is the time taken from left
to right and right to left, I can find out
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assuming, of course that link 2 rotates at
uniform speed. For B to the power of R, the
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corresponding point of A and this is the circle
on which A moves with O2 as centre and O2A
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as radius which we call kA. For the
extreme right position A comes here, let me
call it A to the power of R and for the extreme
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left position, that is B to the power L, this
point which we have already mark as A to the
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power L. As the link rotates uniformly from
right to left, the rotation of O2A is given
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by this angle, that is the angle between O2A
to the power of R and O2A to the power L,
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from left to right, the rotation of same link
2 is given by this angle, that is 2 pi minus
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this. So if I call from right to left, that
is the forward motion is thetaf and return
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motion is thetar. Of course, thetar is nothing
but 2 pi minus thetaf and q.r.r-the quick
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return ratio of this mechanism, we obtain
as thetaf by thetar.
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Let me summarize, what we have learned today.
We have done the graphical method of displacement
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analysis and have discussed three different
examples of six link mechanism to show how
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graphically, we can determine the quick return
ratio or stroke length once the kinematic
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dimensions of those mechanisms are given.
We started with a slotted lever quick return
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mechanism used in shipper machines where we
saw that the quick return ratio depends on
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the stroke length.
Then we discussed the Whitworth quick return
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mechanism where the quick return effect is
independent of the stroke length and in the
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third example, we have seen how from a Grashof
Crank Rocker linkage. We can again get a quick
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return mechanism by using two more extra links
and converting it to a six link mechanism.
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We have also seen that whenever we need the
point of intersection of a straight line and
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a circle, we have to choose a correct point
of intersection.
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In our next lecture, we shall discuss a little
more involved and difficult problems on displacement
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analysis.
266