12 Speed Gear Box Theory Notes by Prof. Sagar Dhotare

DESIGN OF GEAR BOXES For Machine Tool Application
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Contains Covered: Geometric progression , Standard step ratio, Ray diagram, kinematics layout -
Design of sliding mesh gear box and Constant mesh gear box. Design of multi speed gear box.
1. Requirements of a speed gear boxes
A speed gear box should have the following requirements:
 It should provide the designed series of spindle speeds
 It should transmit the required amount of power to the spindle
 It should provide smooth silent operation of the transmission
 It should have simple construction
 Mechanism of speed gear boxes should be easily accessible so that it is easier to
carry out preventive maintenance
2. The speeds in machine tool gear boxes are in geometric progression. Why?
The speeds in gear boxes can be arranged in arithmetic progression (A.P.),
geometric progression (G.P.), harmonic progression (H.P), and logarithmic progression
(L.P.). However, when the speeds are arranged in G.P., it has the following advantages
over the other progressions.
1. The speed loss is minimum
i.e., Speed loss = Desired optimum speed – Available speed
2. The number of gears to be employed is minimum
3. G.P. provides a more even range of spindle speeds at each step
4. The layout is comparatively very compact
5. Productivity of a machining operation, i.e., surface area of the metal removed in
unit time, is constant in the whole speed range
6. G.P. machine tool spindle speeds can be selected easily from preferred numbers.
Because preferred numbers are in geometric progression.
3. Methods for changing speed in gear boxes
The two important methods widely used are:
1. Sliding mesh gear box, and
2. Constant mesh gear box
4. Preferred Numbers
Preferred numbers are the conventionally rounded off values derived from
geometric series. There are five basic series, denoted as R 5, R 10, R 20, R 40 and R 80
series. Each series has its own step ratio i.e., series factor. The series factor for various
series are given in table.

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The series of preferred numbers is obtained by multiplying a step ratio with the
first number to get the second number. The third number is obtained by multiplying a
step ratio with the second number. Similarly the procedure is continued until the series is
completed.
5. Step ratio (or series ratio or progression ratio) (∅)
When the spindle speeds are arranged in geometric progression, then the ratio
between the two adjacent speeds is known as step ratio or progression ratio. It is
denoted by∅.
If N1, N2, N3, ………, Nn are the spindle speeds arranged in geometric
progression, then
N 2
N3
N1 N2
N
4
N3
......................
Nn
Nn
1
= constant = ∅
If ‘n’ is the number of steps of speed, then
𝑁𝑛
𝑁1
= ∅𝑛−1
𝑜𝑟 [
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
= ∅𝑛−1
] 𝑜𝑟 ∅ = [
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
]
1
𝑛−1
Note: Permissible deviation = ± 10 (∅ − 1 )%
No. of speeds to skip =
𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐿𝑜𝑔 𝑜𝑓 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 ∅
𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐿𝑜𝑔 𝑜𝑓 𝑠 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 ∅
=
𝑙𝑛(∅)
𝑙𝑛 (𝑠𝑡𝑑(∅))
Calculation of non standard speeds:
N1 = Nmin
N2 = Nmin x ∅1
N3 = Nmin x ∅2
.
.
Nn = Nmin x ∅
n-1

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 What is step ratio in a gear box?
When the spindle speeds are arranged in geometric progression, then the ratio between the
two adjacent speeds is known as step ratio.
 What does the ray-diagram indicate?
The ray diagram is a graphical representation of the drive arrangement in general form. It
serves to determine the specific values of all the transmission ratios and speeds of all the
shafts in the drive.
 Specify four types of gear boxes.
(i) Sliding mesh gear box
(ii) Constant mesh gear box
(iii)Synchromesh gear box
(iv)Planetary gear box
 What are the possible arrangements to achieve 12 speeds from a gear box
The possible arrangements to achieve 12 speeds from a gear box are
i. 3 2 2 scheme,
ii. 2 3 2 scheme and
iii. 2 3 3scheme.
 What are the points to be considered while designing a sliding-mesh type of multi-
speed gear box?
The basic rules to be followed while designing the gear boxes are as follows:
i. The transmission ratio (i) in a gear box is limited by
1
4
≤ 𝑖 ≤ 2.
ii. For stable operation, the speed ratio at any stage not be greater than 8.
In other words,
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
≤ 8.
iii. In all stages except in the first stage, 𝑁𝑚𝑎𝑥 ≥ 𝑁𝑖𝑛𝑝𝑢𝑡 > 𝑁𝑚𝑖𝑛 .
iv. The sum of teeth of mating gears in a given stage must be the same for same module
in a sliding gear set.
v. The minimum number of teeth on smallest gear in drives should be greater than or
equal to 17.

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 List the ways by which the number of intermediate steps may be arranged in a gear
box.
S.No. Number of speeds Preferred structural formula
1. 6 speeds
i. 3 (1) 2 (3)
ii. 2 (1) 3 (2)
2. 8 speeds
i. 2 (1) 2 (2) 2 (4)
ii. 4 (1) 2 (4)
3. 9 speeds i. 3 (1) 3 (3)
4. 12 speeds
i. 3 (1) 2 (3) 2 (6)
ii. 2 (1) 3 (2) 2 (6)
iii. 2 (1) 2 (2) 3 (4)
 Which type of gear is used in constant mesh gear box? Justify.
Helical gears are used in constant mesh type gear boxes tp provide quieter and smooth
operation.
 What is speed reducer?
Speed reducer is a gear mechanism with a constant speed ratio, to reduce the angular speed
of output shaft as compared with that of input shaft.
 What are preferred numbers?
Preferred number are the conventionally rounded off values derived from geometric series.
There are five basis series, denoted as R 5, R 10, R 20, R 40 and R 80 series.
 Name any two methods used for changing speeds in gear boxes.
The two methods used for changing speeds in gear boxes are
i. Silding mesh gear box and
ii. Constant mesh gear box.
 What is step ratio? Name the series in which speeds of multi-speed gear box are
arranged.

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When the spindle speeds are arranged in geometric progression, then the ratio between the
two adjacent speeds is known as step ratio or progression ratio. R20 and R40 series are used
in the design of multi-speed gear boxes.
 Distinguish between structural diagram and speed diagram.
The structural diagram is a kinematic layout that shows the arrangement of gears in a gear
box. The speed diagram, also known as ray diagram, is a graphical representation of the structural
formula.
 What are the methods of lubrication in speed reducers?
i. Splash or spray lubrication method
ii. Pressure lubrication method
 What is the function of spacers in a gear box?
The function of spacers is to provide the necessary distance between the gears and the
bearings.
 What are the possible arrangements to achieve 12 speeds from a gear box?
The possible arrangement are:
i. 3×2×2 scheme
ii. 2×3×2 scheme
iii. 2×2×3 scheme

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Problem.1: Find the progression ratio for a 12 speed gear box having speeds between 100 and
3.55 r.p.m. Also find the spindle speeds.
Given data: n = 12; Nmin = 100 r.p.m.; Nmax = 355 r.p.m.
To find Spindle speeds:
Progressive ratio, ∅ = [
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
]
1
𝑛−1
= [
355
100
]
1
12−1
= 1.122
Spindle speeds :
Since the calculated ∅ (=1.12) is a standard step ratio for R 20 series. Therefore the spindle
speed from R 20 series are
1 x 100 = 100 1.12 x 100 = 112 1.25 x 100 = 125 1.40 x 100 = 140
1.60 x 100 = 160 1.80 x 100 = 180 2.00 x 100 = 200 2.24 x 100 = 224
2.50 x 100 = 250 2.80 x 100 = 280 3.15 x 100 = 315 3.55 x 100 = 355

7 | P a g e
Problem.2: Selest the spindle speeds, 50 – 800 r.p.m., 12 speeds.
Given Data: Nmin = 50 r.p.m.; Nmax = 800 r.p.m.; n = 12.
To find: Spindle speeds.
Progressive ratio, ∅ = [
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
]
1
𝑛−1
= [
800
50
]
1
12−1
= 1.2866
We find ∅ = 1.2866 is not a standard ratio. So let us find out whether multiples of standard ratio
1.12 or 1.06 come close to 1.286.
Calculation of non standard speeds:
N1 = Nmin N1 = 50 rpm
N2 = Nmin x∅1
N2 = 50 x 1.28661
=64.3 rpm
N3 = Nmin x∅2
N3 = 50 x 1.28662
= 82.7 rpm
. .
. .
. .
Nn = Nminx ∅n-1
N12 = 50 x 1.286611
= 795.5 r.p.m.
STRUCTURAL FORMULA:
Let n = Number of speeds avaliable at thespindle.
p1, p2, p3 … = Stage numbers in the gear box, and
X1, X2, X3 ... = Characteristic of the stage.
Then, the structural formula is given as

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N = p1 (X1) . p2 (X2) . p3 (X3) . p4 (X4)
1st stage 2nd stage 3rd stage 4th stage
Where ,
X1 = 1;
X2 = p1;
X3 = p1 x p2 ;
X4 = p1 x p2 x p3
Preferred structural formulas:
1. Six speed gear box
i) Divide the given six speed by 2
Here p1 = 2, p2 = 3
2 6 =3
X1 = 1, X2 = p1 then structural formula is 2 (1) . 3(2)
ii) Divide the given six speed by 3 3 6 =2
Here
p1 = 3, p2 = 2
X1 = 1, X2 = p1 then structural formula is 3 (1) . 2 (3)
2. Eights speed gear box
i) Divide the given eight speed by 2 we get 2 x 2 x 2 =8
Here
p1 = 2, X1 = 1
p2 = 2, X2 = 2
p3 = 2 X3 = 2 x 2 = 4 then structural formula is 2 (1) . 2 (2) . 2 (4)
3. Twelve speed gear box
i) Divide the given Twelve speed by 2 and 3 we get 2 x 2 x 3 = 12
ii) Divide the given Twelve speed by 2 and 3 we get 3 x 2 x 2 = 12
iii) Divide the given Twelve speed by 2 and 3 we get 2 x 3 x 2 = 12
Here
Structural formula is i) 2 (1) . 2 (2) . 3 (4)
ii) 3 (1) . 2 (3) . 2 (6)
iii) 2 (1) . 3 (2) . 2 (6)

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KINEMATIC LAYOUT (OR KINEMATIC ARRANGEMENT)
The kinematic arrangement of a multi – speed gear box is shown in figure.
From the figure, it is clear that the kinematic layout shows the arrangement of gears in a
gear box. The kinematic layout provides the following informations required for gear box design.
 The number of speeds available at the spindle, i.e., at the driven shaft.
 The number of stages used to achieve the required spindle speeds.
 The number of simple gear trains required to obtain the required spindle
speeds and their arrangement.
 The overall working principle of the gear box.
 The information required for structural formula and ray diagram.
Illustration: in figure, the power is transmitted from driving shaft to driven shaft through a
intermediate shaft. In this conventional gear box, speed changing is obtained using sliding gear
mechanism. It can be seen that the number of speeds from driving shaft to intermediate shaft is 3,
and that from intermediate shaft is 3. Then the number of spindle speeds is equal to 3 x 3 = 9.

Prof, Sagar A. Dhotare, ViMEET, Khalapur 9 | P a g e
The structural formula for the kinematic arrangement of gear box shown in figure, is
given by
n = p1 (X1) . p2 (X2)
Where p1 = 3 (i.e., in stage 1, there are 3 speeds available)
p2 = 3 (i.e., in stage 2, there are 3 speeds available)
X1 = 1; and X2 = p1 = 3
structural formula, z = 3 (1) . 3 (3).
Where n = Number of speeds available at the driven shaft
= p1 . p2 = 3 x 3 = 9
Calculation of No. of shaft to draw kinetic arrangement:
No. of shafts = No. of stages + 1 that is
If 3 stages are there in gear box design, then No. of shaft will be (3+1 = 4 shafts)
Calculation of No. of gears to draw kinematic arrangement:
Shaft 1 = p1 gears
Shaft 2 = p1 and p2 gears
Shaft 3 = p2 and p3 gears
Shaft n = pn and pn+1 gears
Example: Draw kinematic arrangement for 12 speed gear box
Case.1 z = 2 (1) . 2 (2) . 3(4)
Calculation of No. of shaft to draw kineticarrangement:
No. of shafts = No. stages + 1
= 3 + 1 = 4
Shaft 1 = p1 gears shaft 1 = 2 gears
Shaft 2 = p1+ p2 gears shaft 2 = 2 + 2 = 4 gears
Shaft 3 = p2 + p3 gears shaft 3 = 2 + 3 = 5 gears

.
. .
. .
Shaft n = pn shaft4 = 3 gears
Total No. of gear = 14 gears
Case.2 z = 2 (1) . 3 (2) . 2(6)
= 3 + 1 = 4
. .
. .

. .
Case.3 z = 3 (1) . 2 (3) . 2(6)
= 3 + 1 = 4
. .
. .
. .

RAY DIAGRAM (OR SPEED DIAGRAM)
The ray diagram is graphical representation of the drive arrangement in general from. In
other words, the ray diagram is a graphical representation of the structural formula, as shown in
figure.
It provides the following data on the drive:
 The number of stages (a stage is a set of gear trains arranged on two consecutive
shafts)
 The number of speeds in each stage.
 The order of kinematic arrangement of the stages.
 The specific values of all the transmission ratios in the drive
 The total number of speeds available at the spindle.

Procedure
 In this diagram, shafts are shown by vertical equidistant and parallel lines.
 The speeds are plotted vertical on a logarithmic scale with log ∅ as aunit.
 Transmission engaged at definite speeds of the driving and driven shafts are shown
on the diagram by rays connecting the points on the shaft lines representing these
speeds.
 Figure shows the ray diagram for a 9 speed gear box, having the structural formula,
z = 3 (1) . 3 (3)
BASIC RULES FOR OPTIMUM GEAR BOX DESIGN
The basic rules to be followed while designing the gear boxes are as follows:
1. The transmission ratio (i) in a gear box is limited by
1
≤ i ≤ 2 refer the figure
4
In other words, imin = N min ≥ 1 and
Ninput 4
Imax =
N max ≤ 2 …
Ninput
2. For stable operation, the speed ratio at any stage should not be greater than 8. In other words
𝑁𝑚𝑎𝑥
𝑁𝑚𝑖𝑛
≤ 8
3. In all stages except in the first stage, Nmax ≥ Ninput ≥ Nmin
4. The sum of teeth of mating gears in a given stage must be the same for same module
in a sliding gear set.
5. The minimum number of teeth on smallest gear in drives should be greater than or
equal to 17.
6. The minimum difference between the number of teeth of adjacent gears must be 4.
7. Gear box should be of minimum possible size. Both radial as well as axialdimensions
should be as small as possible.

Problem 3: A gear box is to be designed to provide 12 output speeds ranging from 160 to 2000
r.p.m. The input speed of motor is 1600 r.p.m. Choosing a standard speed ratio, construct the
speed diagram and the kinematic arrangement.
Given data: n = 12; Nmin = 160 r.p.m.; Nmax = 2000 r.p.m.; Ninput = 1600 r.p.m.
To find: Construction of the speed diagram and the kinematic arrangement.
☺Solution:
Selection of spindle speeds:
We know that,
Nmax
Nmin
= ϕ n-1
or
2000
= ϕ12-1
or ∅=1.258
160
We can write, 1.12 x 1.12 = 1.254 …
So ∅ =1.12 satisfies the requirement. Therefore the spindle speeds from R 20 series,
skipping one speed, are given by
160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600 and 2000 r.p.m.
Structural formula: For 12 speeds, the preferred structural formula
= 3 (1) 2 (3) 2 (6)
1st
stage 2nd
stage 3rd
stage
Speed diagram (or Ray diagram):
Procedure:
 Since there are 4shafts, draw 4 vertical equidistant lines to represent shafts.
 Since there are 12 spindle speeds, draw 12 horizontal equidistant lines.
 From the structural formula, it is clear that there are stages. In the third stage, i.e., in 2
(6), 2 represents the number of speeds available in that stage and (6) represents the steps
or intervals between these two speeds.
 Locate the first point A on the lowest speed i.e., at 160 r.p.m. on the last shaft. After 6
steps above, locate the second point B at 630 r.p.m. These are the two output speeds.
 Locate the input speed at any point on the preceding shaft (i.e., shaft 2), meeting the ratio
requirements. We find, the input speed 400 r.p.m. at point C satisfies the ratio
requirements.

 In the second stage, there are two speeds. Lowest is at C, which is already located. Now
locate point D on the 3rd
shaft, above point C, in a three step interval. For these two
output speeds in the second stage, the input should be from shaft 2. We find, the input
speed 630 r.p.m. at point E on shaft 2 satisfies the ratio requirements.
 In the first stage, there speeds. Lowest speed is at E, which is already located. Now locate
points F and G on the shaft 2, above point E, in a single step interval.
 Input speed can be located anywhere on shaft 1 meeting the ratio requirements. But in
this problem, given that, input speed is at 1600 r.p.m.
 In stage 2, we find input speed at E gives two output speeds at C and D. Similarly, input
speeds at F and G, should give two output speeds. This can be achieved by drawing lines
parallel to EC and ED, from points F and G, as shown in figure
 Now for stage 3, to get the output speeds to all the input speeds in shaft 3, draw lines
parallel to CA and CB. Thus we have located all the input and the output speeds. The
completed ray diagram is shown in figure
Stage 3: Nmin =
160
= 0.4 >
1
; and
N max 400 4
Stage 2:
Nmax
Ninput
Nmin
=
630
= 1.57 < 2
400
=
400
= 0.63 >
1
; and
Ninput 630 4
Nmax
Ninput
=
800
= 1.27 < 2
630
Stage 1:
N min
Ninput
=
630
1600
= 0.39 >
1
; and
4
Nmax
Ninput
=
1000
= 0.625 < 2 .
1600
∴ Ratio requirements are satisfied.

Kinematic arrangement: The kinematic arrangement for 12 speed gear box is drawn, as shown
in figure.

12 Speed Gear Box Theory Notes by Prof. Sagar Dhotare

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12 Speed Gear Box Theory Notes by Prof. Sagar Dhotare