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In our last lecture we have seen that an epicyclic gear train has 2 degrees of freedom.
Because it has 2 degrees of freedom it can create two output speeds from single input
speed, and that has lot of application. For example, very commonly used epicyclic gear
train for the same purpose, that it can create two output speeds from one input speed is
in automobile differential, we know that when an automobile goes in a straight path both
the rear wheels rotate at the same speed. If the car wants to take a turn, then the
input speed may not change, but the outer wheel must rotate faster than the inner wheel.
From the same input speed, I must now create two different output speeds when the car wants
to take a turn, and that is possible because the input shaft and the wheels are connected
by what we call an automobile differential which is nothing but an epicyclic gear train
having 2 degrees of freedom. Let me draw the differential sketch. These
are the two shafts of the wheels, say to the one of the rear wheels and this to the other.
This is a bevel gear connected to another bevel gear. But this V wheel gear is carried
by an arm. These are bevel gears connected to the two rear wheels. This is another bevel
gear which carries this arm carries this arm that means this bevel gear rotates about this
arm, there is a revolute pair here that is what is called a planetary gear. And this
bevel gear is driven by this bevel gear, this is the input.
The input comes to this gear which drives this bevel gear, this bevel gear carries this
arm, and this bevel gear is carried on that arm so there can be relative rotation between
this arm, and this bevel gear. This bevel gear connects these two bevel gears, which
are connected to the wheel. Let me say this gears we name 1, 2 and this is 3, this is
the arm which is same rigid body as this gear which is 4, and this is 5. So n5 is the input
speed, this small n I refer to the rpm we call it as input speed. This n5 decides the
rotation of 4, so n4 we can find that will be N5 by N4 into n5, where N5 and N4 are the
number of teeth on these two gears. Depending on the direction of this rotation, we can
find what is the direction of this rotation for 4? And 4 is nothing but the arm. To analyze
this epicyclic gear train, we follow the same tabular method which I said earlier we talk
of condition, and we are interested in gear number 1, gear number 2 and arm. This gear
1, gear 2 and arm or gear number 4 they are the same.
First, we say everything is locked to the arm all these elements this gear, this gear
everything is locked to the arm. All elements locked to arm, and then the arm is given some
y revolution, because 1, 2 and 3 everything is locked to the arm this whole thing rotates
like a rigid body, so 1 and 2 also get the same revolution y. We say the second condition.
Second condition is we lock the arm that means, we hold this fixed. Arm is locked so no revolutions
to the arm. Then these two gears are of same dimension they have the same number of teeth
only if the arm is locked. This gear is rotated, the arm is not moving so this is like a simple
gear train. This motion is transmitted to this gear through this bevel gear, and the
number of teeth on this gear is an important it was the intermediate gear, and because
these two gears are of same size only thing they rotate in opposite direction. If gear
1 is given, x revolution gear 2 gets minus x revolution. The resultant is x plus y, y
minus x and y. If the arm rotates, we can say speed of 4 which is y is nothing but n1 plus
n2 by 2, so whatever maybe the values of x and y, the average of n1 and n2 will be the
rotation of gear 4. When the car is moving on a straight path,
there is no relative rotation between this gear 3 and the gear arm, there is no epicyclic
action this acts like a teeth. So, n4 just drives n1 and n2 in the same speed and this
gear 3 does not rotate, this does not have any gearing action this just acts like a solid
connection then n1 is same as n2 is same as n4. So, straight track, if the car is moving
on a straight path there is no relative rotation of gear 3 with respect to the arm there is
any epicyclic action, gear 3 does not revolve about its own axis. Gear 3 acts as key, and
n4 is same as n1 is same as n2 whereas, if the car takes a turn then this gear starts
rotating about its own axis and average of n1 and n2 becomes n4. If it is x plus y and
y minus x, the speeds are different. One is y plus x, other is y minus x. This is like
how an automobile differential can create two different output speeds if necessary because
of this epicyclic gear train. We shall now solve 1 or 2 more problems to show this tabular
method for analyzing epicyclic gear train.
As I said, let me now solve this example of an epicyclic gear train. This is an epicyclic
gear train, where the input shaft 1 is connected to this compound gear 1 and 2, that is the
speed of this shaft is the same as that of speed of gear 1 and gear 2. Gear 1 is connected
to this planet gear, gear 3. Gear 3 is carried by this arm that means there is a revolute
pair hear and this arm is this gear 5 which is an internal gear. This 5 is an internal
gear which is this arm, for this planet gear 3, and the gear 3 is also connected to this
internal gear 4 which is fixed. This internal gear 4 is not rotating, this gear 3 is rotating
in space, moving in spaced axis of this gear as this gear rotates the axis of this gear
3 is carried in space. Gear 2 is connected to this planet gear 6, which is carried by
this arm there is a revolute pair between the arm and gear 6 here, and gear 6 is also
connected to this internal gear 5 which is rotating. Our objective is if the number of
teeth is specified, input speed is specified we have to find the output speed.
For example, let us say the input speed is given to be in these direction 1000 rpm, what
is the output speed? What is the second output speed? Because we are specifying one input
speed, it is an epicyclic gear train so we must specify two input speed the other input
speed is specified as this gear 4 is fixed that is speed of gear 4 is 0, so the other
is speed of gear 4 is 0. Now two input speeds are specified 0 and 1000, we have to find
out this output speed, what is nII? So that is the question. Obviously the number of teeth
on the various gears should be specified, let me say number of teeth on gear 1 is given
to be 20, number of tooth on gear 3 is 30, number of teeth on gear 2 is 24, and number
of teeth on gear 5 is 80. This is an intermediate gear the number of teeth is not given and
not necessary either. Here also number of teeth on gear 3 though it is given is not
necessary for any speed calculation, but we notice that the number of teeth on gear 4
which will be necessary, because that is not an intermediate gear, so that can be calculated
from this value. What is number of teeth on gear 4 that we need, for calculation of various
speeds? Here we can calculate the number of teeth
on gear 6 from this 80 and 24, but we will never need it. So, this is the problem the
epicyclic gear train is specified two speeds are specified 1000 and 0 we have to find out
this output speed. Let me now solve this problem, to solve this problem first of all let us
see that there are two epicyclic gear trains in this whole gear train. There are two epicyclic
parts because there are two arms, one arm for this, if this is the arm which is nothing
but gear 5 and for this portion, this is the arm which is carrying gear 6, and this arm
5 is carrying gear 3. Let me first consider one epicyclic gear train consisting of gear
1, 3, 4 and its arm which is nothing but gear 5. Number of teeth on 1 is 20; number of teeth
on 3 is 30, so we can find out what is the number of teeth on gear 4.
For that, we write pitch circle radius of gear 4 as we see this is the pitch circle
radius, pitch circle radius of gear 1 is rp1 plus pitch circle diameter of gear 3 that
is twice rp3 because all the gears have the same modules pitch circle radius is also proportional
to the number of teeth. From here, I can write number of teeth on gear 4 is number of teeth
on gear 1 plus twice the number of teeth on gear 3, n1 is 20 n3 is 30 so N4 is 80 so that
gives N4 is 80. To consider this gear train as before we make the tabular method we write
condition and revolutions of gear number 1, 3, 4 and the arm which is nothing but gear
5 so first conditions is lock all the elements to this arm. All elements locked to arm. After locking all the elements
to the arm we give y revolution to the arm because everything is locked so we get the
same number of revolutions in all the elements. We lock the arm, no revolution to the arm and give
any revolution x to say 1. If I give x to 1, then 3 rotates in the opposite direction
and minus x into N1 by N3 because this is an external gearing so that we are taking
opposite direction and inversely proportional to the number of teeth. Now, rotation of 3
will cause rotation of 4, but this is an internal gearing so they rotate in the same direction.
So, 3 has a minus sign, 4 will also have the minus sign and rotation of 3 N1 by N3 into
N3 by N4. As I said earlier, this intermediate gear number of teeth is not relevant so it
cancels. I am writing this line again x, this is minus N1 by N3, so it is 2 by 3 x minus
x , 20 by 80 is 4 and this is 0, this line I am just repeating. Let me talk of the resultant
motion, we add these two motions so here I get x + y here I do not need 3, because 3
is not given so I do not write it, 4 I get y minus x by 4 and arm I get y. Input speed
is given 1000 rpm which is the speed of one that is x plus y. I get x + y is 1000, I am
taking this direction as my positive direction as the direction which is shown which is counter
clockwise so this is the axis this is the positive direction, x + y is 1000 and n4 is
0 so y - x by 4 is 0, which if solve these two equations I can get x is equal to 800
and y is equal to 200 rpm and arm 5 which has the speed y that is speed of arm 5 is
200 rpm. When I analyze this part of the epicyclic
gear train I know the speed of arm 5 which is nothing but 200 rpm and gear 2 again the
speed will be known because that is same as shaft one gear 2 and gear 1 are one compound
gear integral to the shaft 1. In this epicyclic gear train I will know two speeds that was
speed of 2 is 1000 rpm and speed of 5 is 200 rpm and I should be able to solve for the
speed of the arm as I will show you just now.
Let me now analyze the second epicyclic gear train consisting of gear 2, gear 6, gear 5,
and this arm this arm carries gear 6. Remember this gear 5 was nothing but the arm of the
previous train which I have just now solved. We have already seen that n5 was obtained
as 200 rpm and n2 is given as 1000 rpm, and our objective is to find the rpm of this arm
which is the output shaft which is connected to the output shaft so gear train is 2, 6,
5 and the arm. As before we write condition and revolutions of 2, 6, 5 and arm. First
lock all elements to arm. All these elements are locked to the arm and give y revolution
to the arm, so all the elements get y revolution, then lock the arm, no revolution and let me
call it as y1 so that we do not confuse it x and y we used earlier let me call it as
y1 and lock the arm and give say x1 revolution to gear 2.
Gear 5 rotates in the opposite direction because this intermediate gear makes it opposite direction
6 and 5 rotate in the same direction so minus x1 into number of teeth on this gear is not
needed so 24 by 80. The resultant is x1 + y1, y1 and y1 - 3 by 10x1. The speeds given
at n2 and n5, n2 is x1 + y1 which is 1000 n2 and, n5 which is y1 - 3 by 10x1 and that
is equal to 200. We have to find what is the speed of the arm narm? That is the output
speed which is nothing but y1. If we solve for y1 from this what we get, we eliminate
x1, I multiply by 3 by 10 and add so I get 13y1 by 10. 3 by 10 I have multiplied and
adding so, x1 getting canceled 3 by 10 plus one gives me 13 by 10, and this is 3 by 10
that is 300, so 300 plus 200 is 500, y1 is 5000 by 13 rpm, because it is positive that
means it is in the same direction. We get the output speed as 5000 by 13 rpm in the
same direction. We shall solve one more problem involving epicyclic gear train.
Let me now solve one more problem of epicyclic gear train which is shown in this sketch.
This gear 1 is mounted on shaft 1 which is rotating at 2000 rpm in the direction shown.
The number of teeth on gear 1 is 20. Gear 1 is connected to gear 2 numbers of teeth
on which is 40. Gear 2 and gear 3 forms a compound gear that means they are the same
rigid body and number of teeth on gear 3 is 30. This shaft 2 which rotates at 350 rpm
carries this arm; this shaft is the arm as it rotates this takes these two bevel gears
with it. There is a revolute pair here between the arm and these two bevel gears. This bevel
gear 3 is connected to bevel gear 4, number of teeth on which is 64.
This gear 4 and 5 are compound gear again they have the same rigid body, and there is
a revolute pair between the arm and these two gears which is the same rigid body 4 and
5, and the number of teeth on 5 is 24. Gear 5 is connected to this bevel gear 6 number
of teeth on which is 18, and gear 6 is connected to the output shaft. These two input speeds
are specified that of the arm and that of gear 1 and we have to find out what is the
output speed in 3? The thing to note in this particular gear train is that, gear 1 decides
the speed of gear 2, because this 20 and 40, so if this rotates at 2000 rpm gear 2 rotates
at 1000 rpm opposite to this that is in this direction gear 2. n2 is known if I take this
as my positive direction, then what is given is narm I take this direction as positive
narm is 350 rpm, and n2 which is decided by n1 which is 2000 here so that is 1000 rpm
opposite to this that is the same direction as this.
We can forget about gear1, and we have an epicyclic gear train that means this gear
1 is not a part of this epicyclic gear train which is obvious because, when we lock all
the gears to the arm I cannot lock this gear because then this gear cannot rotate. This
is mounted here in the gearing, but if I remove this gear from this epicyclic gear train rest
of it is a simple epicyclic gear train, only thing speed of two is decided by that of speed
of one that is the only difficult point in this problem. Once we have done this stage
this epicyclic gear train is simple. What it consist of? We can write the same table
revolutions of gears 2 and 3 which is same then 4 and 5 we are not interested that is
intermediate gear so we write gear 6. Gear 6 is rotating about the same axis like the
gear 2 and 3 and also the arm which is rotating about this axis so we write arm.
We write condition, first condition all elements locked to the arm, and if arm is given y revolution
then gear 6 gets y revolution, 2 and 3 also gets y revolution. We are not writing these
two bevel gears because as we see the rotation of this gears allow with the arm and their
rotations above the arm are about different axis. These two gears rotate with respect
to the arm about this axis, whereas the arm is rotating about the horizontal axis, so
we cannot add those kinds of things, and we are not interested, because these are intermediate
gears so we do not write it here. Here, we write lock the arm, arm is given no revolution
and if I give x revolution to 3, then 4 rotates due to gearing action which is 30 by 64 into
x. If it rotates in this direction then it goes in this direction so this gear will rotate
in opposite direction. These two gears due to this bevel action rotates in opposite direction,
and their rotation is decided by the number of teeth on this and this which is minus x
into 30 by 64 and these two rotate in the same direction, then ratio here is... rotation
of this I get [as minus x not minus because their rotation is different direction so I
am not talking of that] 6 is rotating due to this 24 by18. This has 24 teeth and this
has 18 teeth, so it is not 24 by 18 this will rotate more, so it is alright 24 by 18. This
is rotating 30 by 64 this is rotating 24 by 18.
We get the resultant so it is x + y and this is y, and this is y minus. If I cancel 18
and 35 and if I cancel 8, 3, 8, 3 and 3 cancels, so y minus 5x by 8 out of which we are given
narm y is 350 and 2 is 1000, x + y is 1000. We get x = 650 and we have to find the revolution
of 6 which is n3, so n3 which is same as n6 which is y - 5x by 8 which is y is 350 - 5
by 8 into 650. We should calculate this, this is 325 into this is 4 so 4 into 350 is 1400
and 5 into 325 is minus 625 so it is minus 225 by 4. If this rotates in this direction
this is minus so it is in this direction, this is this axis, and the value is 225 by
4, which is 56.25 rpm. The output speed is 56.25 rpm opposite to this direction. That
is how we solve this epicyclic gear train, only thing to note that this gear does not
belong to the epicyclic gear train and if we have bevel gears, then we should not write
it in this table because then we cannot add simply like this because all this rotations
are about this axis whereas bevel gear with the arm rotates about this axis, but there
gearing action make them rotate about a different axis.
We cannot add rotations about two different axes algebraically, so this is how we solve
the problems of epicyclic gear trains. Now that we have explained how to analyze these
epicyclic gear trains so for the kinematics is concerned, that is how to obtain the speeds
of different gears and what is the difference between gear train and gear box. ?
Gear box as we know, there is one input speed, but I may require to get various different
output speeds and accordingly, we change the gears within the gear box the gear engagement
we change, and using different set of gears within the same gear box I can create different
output speed. As we do in automobile from the same input speed I can go to first gear,
second gear, third gear as we want to increase the speed of the car, but there are two kinds
of gear boxes, one is where the gears are changed that is output speed is changed after
bringing the entire system to rest. We change the gear engagement within gear box by sliding
or something, and then get switch on the machine and get a different output as we do in the
case of machine tools. The motor of the lathe machine rotates at
a constant speed, but I want to get different output speed at the head stock or I want to
rotate the job of the machine at different speed depending on the diameter of the job.
In machine tools we use a kind of gear box, where we bring the entire system to rest change
the gear engagement and then again switch on the machine. Whereas, in automobile we
need to change the gears while the car is in motion or the input motion is on. We cannot
switch of the car and then change the gears. We have to change the gear while it is in
motion for that this gear boxes are more complicated, and we have to use what is known as synchromesh.
This is a complicated system where a clutch action is needed such that gears which are
going to engage are speeded up, and get into the same speed or proper speed before they
get into engagement. Whereas, this machine tool gear boxes which are easier to design
and easier to analyze uses, what to be known as sliding clusters. We shall end this course
by having a little discussion one this sliding cluster gear boxes, which are used in machine
tools that is from one input speed I want to get several output speeds by changing the
gear engagement, I will show you through figures now such a sliding gear box which are used
in machine tools.
This picture shows what I just now said is a sliding cluster gear boxes, which are used
in machine tool application. This is the say the input shaft and this is the output shaft.
For the same speed of the input shaft omegai, I may like to get three different output speed
omegao, to do that I have a compound gear, this is the same rigid body and this whole
cluster can be moved along this shaft, this shaft is a splined shaft and on this splined
shaft this whole cluster can be moved, right now this gear b is in engagement with this
gear e so it transmits a particular speed ratio. If I want a different speed ratio,
I stop the entire machine then slide this cluster such that this gear c comes in engagement
with this gear f and depending on the number of teeth, they will transmit a different output.
Again if I want to have another third different speed I will stop the machine and move this
entire cluster such that this gear comes in contact with this engagement with this, and
depending on their number of teeth they will transmit a particular speed. This is a three
speed gear box using 6 gears and this is the sliding cluster, this is one rigid body but
number of teeth is different on these three gear bodies a, b, and c.
The thing to note is that before this gear comes in engagement with this gear, the gear
which was already in engagement must be totally clear. It must leave this engagement before
its starts engagement. Similarly, when I move in this way before this gear comes in engagement
with this gear, this engagement must be fully clear. Suppose, if the gear face widths of
all these gears are same and I call it as B. If every gear the width is B then this
total distance as we see is B plus B, because if it has to clear that movement is B so before
it comes into engagement so B plus B plus B plus B plus B again this gap must be B and
B, so this whole distance becomes at least 7B. It should be more than 7B, one more thing
to note if that the number of teeth difference, the difference in the number of teeth between
these two gears must be at least 4. Otherwise, while moving in this direction this gear will
foul with this gear. As we see pitch circle radius of this gear
b plus pitch circle radius of this gear re that is the centre distance but when this
gear is sliding over this, this must be greater than the outer radius sum of the outer radius
of a and b. Otherwise, a will interfere with e the outer radius of a plus outer radius
of e, the center distance should be more than that, such that this should not interfere
with this when I am sliding it to the right. If we remember the outer radius is pitch circle
radius plus the module, so outer radius is pitch circle radius plus the module so if
I cancel pitch circle radius of this so pitch circle radius of b must be greater than pitch
circle radius of a plus 2m, and if we remember the number of teeth module is nothing but
number of teeth pitch circle diameter by number of teeth this is the module. So rb I can write,
module into number of teeth on gear b divided by 2 should be greater than module into number
of teeth on gear a divided by 2 plus 2 m, where m is the module. If we cancel m I get
Nb should be greater than Na plus 4, so for designing such sliding cluster gear boxes
one has to ensure that the minimum number of teeth between these two adjacent gears
on a cluster is at least 4, should be more than 4 and this width should be sufficient
such that total disengagement takes place be from this pair before the other pair this
way or that way comes into engagement.
Using the same thing I can get a nine speed gear box. If this is the input shaft and this
is the output shaft, and this intermediate shaft I make it splined and here I keep two
sliding clusters. Depending on where the connection is taking place, I can get from the same input
speed nine different speeds. For example, right now this and this is in engagement and
this and this is in engagement so depending on the number of teeth on these 4 gears I
get a particular output speed. If I change, slide it this comes out of engagement and
this becomes in engagement and here it is the same so depending on the number of teeth
on these two gears I get a different speed. I can get three here and three here and by
combining them three into three I can get nine different output speeds from the same
input speed. This is what is called sliding clusters gear box which are used in machine
tools. This is a nine speed gear box from the same input speed I can get nine different
output speeds. The thing to note that because splined shaft
are much more costly we have put both the sliding clusters on the same shaft, input
shaft, intermediate shaft which is a splined shaft and this is the output shaft, input
and output shaft have fixed gears. Whereas, this intermediate shaft which is a splined
shaft has sliding gears so this is a case of a nine speed sliding cluster gear box,
but as I said for when the gear changes to take place while the whole system is running
that is much more difficult to design and we need what I call synchromesh and that is
beyond the scope of this course.