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In this module, we will start our discussion with some advanced problems of dimensional
synthesis. We have already discussed three position synthesis of 4-link mechanisms namely:
motion generation, path generation and function generation.
We have already seen that using these three problems and a two stage synthesis technique
we could have designed, first a slider-crank function generator and then use that design
to generate an approximate straight line path generation of a 4-R linkage coupler curve
which could be coordinated with the input, that is the crank movement.
Today we will start our discussion with the design of a six-link fork lifter with only
revolute pairs. As we know these fork lifters are used in industries to shift article from
one place of the shop floor to another place but such fork lifters normally have two vertical
guides or two prismatic pairs. We all know that the prismatic pairs are difficult to
maintain and more costly to manufacture. There is more friction also in the prismatic pair
whereas, revolute pairs are easy to fabricate and there is virtually no maintenance and
because of a rollers that is the revolute pairs with the pin and hole joint there is
hardly any friction, because the pin diameters are normally not very large.
Let me first show you the model of this proposed six-link fork lifter with only R-pairs. This
is the model of that proposed six-link mechanism to be used as a fork lifter consisting of
only revolute pairs. As we see the body of the truck is the fixed link or link number
1 and there is a link number 2, 3, 4, 5 and 6. As this input link of the crank is moved
by a motor what we see that fork that is this link number 6 moves up and down almost vertically,
but this vertical movement is without any vertical guide. We will solve and try to design
this problem such that these two fixed pivots are located on the body of truck at convenient
positions and we have to come up with this linked lengths such that, the rotation of
the crank is converted into almost vertical movement of this fork. We will solve this
problem by what we call a multi-stage synthesis. Let me now pose the problem with the required
design specifications.
Let us now discuss the details of the design of this six-link fork lifter with only revolute
pairs, the model of which we have just now seen. We will achieve this design by a process
which we can call as multi-stage synthesis. That means the techniques that we have already
learnt for path generation, motion generation and function generation will be applied in
a number of stages to get to the final design. The design specifications, for example, in
this particular mechanism let us represent it through the following parameters:
First, the 60 degree rotation of the input link should cause 1.5 meter vertical travel
of the fork that is as the input link crank rotates through a 60 degree, the fork should
travel almost along a vertical line through a distance of 1.5 meter.
Not only that, the travel of the fork should be proportional to the input rotation. As
we know, we cannot have the vertical line exactly vertical at all configurations and
we go through what we call a precision point approach.
Then we decided to use three Chebyshev's accuracy points during this entire travel of 1.5 meter.
At this stage let me now go back to the sketch of this mechanism and define the other parameters.
Let me now refer to the proposed sketch of this fork lifter. This two fixed points O2
and O4 are to be placed on the body of the truck at convenient locations. These two points
B and D belong to the fork and the BD defines the fork completely and we want that 60 degree
rotation of this input link should cause 1.5 meter travel along this vertical line. So
we start the design assuming a convenient location of O2 and a vertical line in front
of the truck where I want the line BD to move. Let Bi represents the initial point of B and
Bf is the final point of B on this vertical line. Bi to BF, this distance is 1.5 meter
and the corresponding rotation of this input link should be 60 degree. Now, first step
is to determine the three Chebyshev's accuracy points in this range Bi to Bf which we shall
call it as B1, B2 and B3. The three Chebyshev's accuracy points in this interval from Bi to
Bf can be determined as we know both analytically and graphically. First of all let me determine
these three accuracy points B1, B2 and B3 analytically.
We have been given the delta s that is the total movement is 1.5 meter. We are going
for three Chebyshev's accuracy points that is n is equal 3. a, that is the distance of
the midpoint from Bi is delta s by 2, which is 0.75 meter with the origin at Bi. The half
of the interval h is also delta s by 2, so again h again comes out as 0.75meter.
Then, the three accuracy points, sj with j going from 1, 2 and 3 can be obtained from
the formula as 0.1005 meter, 0.75 meter and 1.3995 meter.
Let me now obtain the same three points B1, B2 and B3 graphically. To determine the three
accuracy points graphically, let me refer to this figure. This is the initial point
Bi which I have chosen on this chosen vertical line and this is the point Bf and to this
scale Bi to Bf this delta s is 1.5 meter. We have to locate three accuracy points in
this range from Bi to Bf, graphically to do that first we draw a semicircle with Bi, Bf as diameter. This
is a semicircle with Bi, Bf as diameter, because, n is equal to 3, now I draw a regular hexagon
with two of its sides perpendicular to this diameter and inscribed in this circle. The
half of this regular hexagon is shown, this hexagon is inscribed within this circle. So,
the projection of these three vertices 1, 2 and 3 on to this diameter Bi, Bf locates
three accuracy points namely, B1, B2 and B3. B1B2 is the travel of the fork we represent
by s 1 2. Similarly B2B3, the travel of the fork from the second to the third accuracy
points is what I call is s 2 3. So s 1 2, s 2 3 is now known I can either measure it
from this diagram to this scale or I have already calculated analytically.
The second requirement is that the travel of this fork should be proportional to the
rotation of this input link of the crank O2A. We calculate what is the corresponding movement
of this link 2, which I call it as theta2 1 2 and theta2 2 3. We have decided the three
accuracy points corresponding to the fourth point b namely at B1, B2 and B3 both analytically
and graphically.
We have got s12, that is s2 - s1 is equal to 0.6495 meter analytically and in the drawing
we got it to some scale and s23 is exactly the same, because the rotation of the crank
has to be proportional to the movement of the fork, the relationship is theta2 1 2 divided
by s12 is same as delta theta by deltas, where delta theta represents the total travel of
the crank and deltas represents the total travel of the fork. We are trying to design
O2 A, B as a slider crank, where B is the location of the slider and O2A is the crank.
Delta theta is given as 60 degree into deltas is given as 1.5 meter.
From this relation because s12 is known, we can find theta2 1 2 which turns out to be
26 degree counter-clockwise and theta2 2 3 again is 26 degree counter-clockwise. In the
first stage of the synthesis the problems boils down to designing a slider crank O2AB
such that, the sliding movement at the slider positioned at B is coordinated with the rotation
of the crank O2A according to this relation, that 26 degree counter- clockwise rotation
of the crank should produce 0.6495 meter vertical travel of the slider. Further 26 degree rotation
in the counter-clock wise direction of the crank will again cause further 0.6495 meter
vertical travel of the slider. Let me now at the first stage design this
slider cranks mechanism, which I now explain through the usual graphical method of which
we have learnt earlier.
At the first stage of this synthesis problem, let me start B1, B2 and B3, these are the
three accuracy points corresponding to the slider movement at B. We choose O2 arbitrarily
and the whole problem right now is a function generation problem as a slider crank such
that this downward movement from B1 to B2 which is 0.6495 meter should occur during
26 degree rotation of the crank in the counter-clock wise direction. Further 26 degree rotation
of the crank will make the point B2 to come up to B3. This we have already learnt how
to do this, we choose O2 at a convenient location on the body of the truck, we have chosen this
line conveniently at the front of the truck and I have located B1, B2 and B3 according
to Chebyshev's risk. We apply the method of inversion, we hold the crank fixed at its
first position and the crank pin at A1 has to be located. Here, we have already drawn
the solution but my objective is to locate the required point A1 such that the slider
crank O2AB acts as a function generator. If we hold the crank fixed, then we know this
O2B2 has to be rotated through 26 degree but in the clockwise direction. We rotate B2 with
O2 as center, this O2B2 is rotated by 26 degree and this point we call B2 inverted on the
first position or B2 1. Because, we are holding the first position
fixed, the inverted position of B3 I can get by rotating O2B3 by 52 degree. This line O2B3
is rotated through 52 degree and we get the inverted position of B3 which we call B3 1.
This is B3 1, this is B2 1 and this is B1 which is as same as B1 1. Then A1 has to be
at the center of the circle passing through B1 1, B2 1 and B3 1. To determine that, we
do the usual simple geometric procedure, we draw the mid normal of B2 1 and B3 1, which
is this line, this is the mid normal of B2 1, B3 1 and we also draw the mid normal of
B1 1 and B2 1. This is the line B1 1, B2 1 and we draw the perpendicular bisector of
this line and these two lines intersect at A1. Finally, at the end of the first stage,
we have designed a slider -crank mechanism namely O2A1B1, where A1 is the crank pin such
that I know the 26 degree counter-clockwise rotation of this crank will cause the slider
to come from B1 to B2, further 26 degree counter-clockwise rotation of the crank will cause the slider
to come from B2 to B3. In the next stage we will get to a four bar
linkage such that I can remove the slider and this coupler point B1 will automatically
pass through B1, B2 and B3 when the four bar linkage is moved and that is the second stage
of synthesis.
In the first stage we have obtained this O2, A1, and B. Now, I removed the slider at B1
and treat this link AB as the coupler of a four bar link namely, O2, A, C, O4 and B is
the coupler point of this 4-R link O2, A, C, O4. Our objective is how we determine the
correct location of O4 so that when this 4-R linkage is moved, this point B1 will automatically
pass through B1, B2 and B3 and maintaining the coordination with the input movement.
This problem we have solved earlier without elaborating.
Let me say, I choose C1 somewhere in the middle range of this AB not too close to A, not too
close to B, but somewhere in this middle region, so I have chosen C1 When this slider crank
goes from O2, A1, B1 to O2, A2, B2, C1 goes to C2. I draw A1 goes to A2, that is 26 degree,
B1 to B2 that is 0.6495 meter and I get O2, A2, B2 because this length is already fixed,
so from A2 I measure this length and get to C2. I locate the point C1. Similarly, in the
third configuration O2, A3, B3 I find the corresponding location of C and I call it
as C3. I can always draw a circle through these three points namely, C1, C2, C3 and
the center of this circle is at O4. The center can be easily determined by drawing the mid
normal of C1C2 which is this line and mid normal of C2C3, which is this line. These
two lines intersect at O4 determining the required location of the fixed hinge O4. The
point to note is that, because I choose C1 somewhat arbitrarily I have no control over
the location of O4. Here, it happens to be within the boundary of this truck body, but
with different choice of C1 this O4 could have been very much away from the body of
the truck and then we have to do more iteration such that with the proper choice of C1 only
I can get O4 in a convenient location. At this stage, that is at the end of the second
stage, I have got to a four bar linkage namely, O2, A, C, O4 such that when this 4-R-moves
this coupler point B1 has the desired movement and is also coordinated with the input movement.
At the second stage, I will do it little differently, because as I have seen that with an improper
choice of C1, O4 can go to an inconvenient position. The second method what we will do
I will choose that O4 conveniently at the body of the truck and try to locate the required
C1 on this coupler link AB.
In this alternative procedure we say that O2, A1, B1, O2A2B2 and O3, A3, B3 are already
known to us. Instead of choosing C1, I rather choose O4 conveniently on the body of the
truck say at this location which we call O4. To determine the point C1 on the coupler link,
I make a kinematic inversion holding the coupler fixed at its first configuration and obtain
the inverted position of O4 as we know, because this C belongs to the coupler and C, O4 does
not change. So, if we hold the coupler fixed at its first location, this point O4 will
move on a circle. To obtain the inverted position of O4 we follow the usual procedure, we mark
A2, B2 and O4. These are the relative positions of A2, B2 and O4 because the link AB is not
moving when I make a kinematic inversion, I will take this A2 and B2 to coincide with
A1 and B1 respectively so I move. If we remember this point represents B2, this point represents
A2 and this cross represents O4 and they define the relative position of A, B and O4 at the
second configuration. Kinematic inversion means this relative position should not change,
though I am not allowing AB to move from A1B1. I move this tracing paper A2 coincide with
A1, B2 coincide with B1 and wherever O4 goes that point I mark by piercing on this tracing
paper and mark this point as O4,2 1, because this is the second position of O4 inverted
on to the first position. Following the same logic, I also obtain the
inverted position of the third configuration that is on the tracing paper I mark A3, B3
and O4. These are the relative positions, but inverting on the first position A3, B3
coincides with A1 and B1 respectively and wherever O4 goes it has come here that point
I call as O4,3 1. So, this way we have obtained the three inverted position of O4 which is
as same as O4,1 1, O4,2 1 and O4,3 1. If I draw a circle passing through these three
inverted position, then the center of the circle will be C1. As usual I draw the perpendicular
bisector of this line and the perpendicular bisector of this line, these two bisectors
meet at the point C1 determining this revolute pair on the coupler link at C1. Now I have
got a different four bar linkage namely O2, A1, C1, O4 with B1 as the coupler point which
will pass through B1, B2, B3 maintaining the coordinated movement with the input link.
Here, the location of O4 is convenient, but as we see the coupler has become not a very
lean member but a heavier member, that is triangular in shape. Previously the coupler
was almost a straight link and now this coupler has become triangular in shape, but we have
controlled over the fixed hinge O4 that I can choose conveniently on the body of the
truck. At the end of this second stage either by following the previous procedure or all
this alternative procedure I can get through this four bar linkage with the coupler point
movement coordinated with the input link. In the third stage, I will convert this to
a six-link mechanism such that B, D which is not seen here B, D will be my fork of the
lifter.
So at the end of the second stage of the synthesis, we have determined A1, C1. We had already
chosen O2, O4 and this vertical line on which I had B1, B2 and B3. If this fork is hinged
at only one point this will be swinging like a pendulum. To guide this fork vertically
I choose another convenient point D1, this is chosen arbitrarily. B is at B1, D is at
D1. When at the second configuration, when B goes to B2, because this distance remains
same I can locate D2 and similarly D3 corresponding to the third position on the same vertical
line. The only task is to determine this revolute pair on this link at E1 such that D1 passes
through D1, D2 and D3. To determine this E1, I make again a kinematic inversion keeping
this link fixed, that is the follower link of this O2, A, C, O4. This four bar linkage,
the follower link I hold fixed at its first position and determine the inverted positions
of D1 generating the same relative movement, but holding this link four fixed.
At the second stage of our synthesis we have determined this linkage O2, A1, C1, O4 out
of which O2 and O4 had been chosen conveniently on the body of the truck. The line of travel
of the fork was also conveniently chosen in front of the truck and we have come up to
this design O2, A1, C1, B1 and O4. As we see this fork, if it is hinged only at one point
it will swing like a pendulum. To guide the fork along this vertical path, I will choose
another point which I call it as D1 which I choose conveniently on the fork and now
this point D1 with a revolute pair has to be connected to this link O4, which is O4,
C1 such that D1 goes through D2 and D3. This distance BD remains the same, so from B2 I
can locate D2 on this vertical line and D3 also on this vertical line. I have got D1,
D2 and D3. To locate the position of E1, that is this revolute pair on link number 4, again
I apply the method of inversion holding this link that is link number four O4C fixed at
its first configuration and determine where is D2 1 and where is D3 1. To do this we follow
the same method of kinematic inversion. This is D2, this is O4 and this is C2. So,
I mark O4, C2 and D2. Now, I am holding this link O4, C fixed at its first position. So,
What do I do? I make C2 coincide with C1 and O4 does not move and wherever D1 goes that
becomes my D2 1. Let me repeat, to get the inverted position corresponding to the second
configuration inverted on the first position, I mark O4, C2 and D2. Since O4, C is not moving
from the first configuration, I make O4C2 coinciding with O4, C1, that is C2 coincides
with C1, O4 coincides with O4 and wherever this D2 moves that I call it as D2 1.
Similarly, to obtain D3 1, I mark O4, C3 and D3. Holding the link 4 fixed at its first
position, I make O4 at O4, C3 at C1 and wherever D3 moves that I call it as D3 1. Since link
4 is fixed at its first position, this link length is of constant length so E1 can be
located at the center of the circle passing through D1 which is same as D1 1, D2 1 and
D3 1. Again following the usual technique, I find the perpendicular bisector of these
two inverted position D2 1 and D3 1 and the perpendicular bisector of D1 1 and D2 1 which
is given by this line. These two lines intersect at E1 determining the location of this revolute
pair on link 4 and I connect E1 and D1 by a rigid link.
If we do not allow this link to move then obviously D1 goes on a circle with E1 as center.
So, E1 is determined at the center of the circle passing through D1 1, D2 1 and D3 1.
At this stage I have finished the design of a six-link mechanism which gives me a fork
lifter, where the fork goes almost vertically, at least it passes through B1, B2, B3 and
the point B passes through D1, D2, D3 lying on this vertical line. Of course there will
be deviation a little bit as it goes from B1 to B2 from this vertical configuration,
but the deviation is not much and it acts as a very good fork lifter particularly the
load is not heavy, where the movement of the fork has been coordinated with the rotation
of the crank and the fork goes almost vertically and we end up getting a fork lifter without
any vertical guide.
Let me now summarize this multi- stage synthesis of guideless fork lifter as we have completed
in today's lecture. We have chosen first O2 this is a fixed pivot at a convenient location
of the body of the truck. We have also chosen a vertical line in front of the truck which
is convenient for the movement of the fork. First, we design O2, A, B as a slider crank
such that the slider movement at B is coordinated with the input motion of the crank. 60 degree
counter-clock wise rotation of the crank has caused 1.5 meter vertical travel of this point
B using three Chebyshev's accuracy points in this interval.
Next, we chose O4 conveniently on the body of the truck. Then remove the slider and got
coupler point here at C such that O2, A, C, O4 this 4-R-link has a coupler point B which
has an approximate straight line path and the movement along that straight line path
is coordinated with the motion of the input crank. In the third stage, we have chosen
this point D again conveniently on the body of the fork such that a second point is guided
along the same vertical path. After choosing this point D, we found the inverted positions
of the D corresponding to the second and third configuration holding this link, the link
number 4 fixed at its first configuration and the center of the circle passing through
these three inverted positions of D namely, D1 1, D2 1 and D3 1 I found the center of
that circle at this point which I call it as E.
Finally, we get this six-link mechanism where I have a fork lifter without any vertical
guide. The techniques that we have learnt for function generation, motion generation
and path generation can be synthesized together to have a multi-stage synthesis problem for
even more complicated problem. In today's lecture, we have designed a six-link
fork lifter using only revolute pairs by multi- stage synthesis processes which are basically
the combination of the three position synthesis for function motion and path generation of
four-link mechanisms which we had discussed earlier. In our next lecture, we will try
to design another six-link mechanism for a rail-less garage door. A garage door which
will be opening inside the garage rather than coming out of the garage and in the open position
it will become horizontal to the roof of the garage using again only revolute pairs and
not using any prismatic pair.