Gear Trains
 A combination of one or more gear pairs that are
interrelated is called a gear train.
 The velocity ratio for two meshing gears (2 and 3) is
only for parallel shaft
 Where the plus sign goes with an external gear meshing
with an internal gear, and the minus sign goes with two
external gears meshing.
At the pitch point, rolling occurs,
and the velocity at the pitch point
on both gears is the same.
n=3 (1, 2, 3)
j=3 (2 Revolute pairs, 1-2,
and 1-3; 1 Gear pairs, 2-3)
Pinion and rack
M=3(3-3-1)+(12+21)=1
13.1
[1]
Crossed Helical Gears
 Treat the pinion as a screw and the gear as fixed.
 Observe the motion of the pinion relative to the gear as the
pinion is rotated and the gear is viewed along the gear axis.
 If the pinion appears to advance toward the gear when the
pinion is rotated, in reality the gear would rotate
counterclockwise (Fig.13.2).
 If the pinion appears to withdraw from the gear when the
pinion is rotated, in reality the gear would rotate clockwise
(Fig.13.2).
Helix angle>45o
worm
13.2
[1]
Gear Trains
 Simple gear trains.
 Compound gear trains.
 Concentric Gear Trains.
 Planetary Gear Trains
Simple Gear Trains
 Simple gear trains have only one gear on each shaft.
These shafts rotate on bearings that are attached to the
same frame.
 Simple gear trains can be divided into two types
depending on whether idler gears are involved or not.
13.3
[1]
13.4
13.5
[1]
Idler Gears
 The idler gears in simple gear trains can serve two
purposes in design.
- One is to change the direction of motion of the output
gear.
- The second is to provide a spacer when two gears
cannot be directly meshed because of the shaft
locations.
13.6
[1]
The mechanism works well only when the gears are slowly moving or at
rest.
Each idler gear will have at least two mesh points.
13.3
[1]
&

=
The magnitude of the velocity ratio between the input and output
shafts is a function of the numbers of teeth on the input and output
gears only. The magnitude of the velocity ratio is independent of the
size and number of idler gears.
=
13.3
[1]
The sign of the train ratio for parallel-shaft gears does depend on the
number of idler gears.
At each mesh between external gears, the velocity ratio changes sign.
For internal gears, the velocity ratio remains the same sign.
If n is the number of meshes between external gears, the sign of the
velocity ratio is given by (-1)n.
Note that each idler gear will have at least two mesh points.
13.4
[1]
Compound Gear Trains
 In spur gears, the velocity ratio at any mesh should not
exceed 1:5. For larger reductions, compound gear trains
should be used.
 compound gear trains are characterized by the presence
of two or more gears attached to the same shaft.
 The shafts still rotate on bearings that are fixed to the
frame.
13.7
[1]
&
&

3=4

When parallel shaft
gearing is involved.
n=4 (1, 2, 3(4), 5)
j=5 (3 Revolute pairs, 1-2,
1-5 and 1-3(4); 2 Gear
pairs, 2-3 and 4-5)
M=3(4-5-1)+(13+22)=1
13.8
[1]
Compound Gear Trains
 The velocity ratio can be represented as the product of the
driven gear numbers divided by the product of the driver gear
numbers.
 If n is the number of gear meshes (including idlers that each
have two meshes), a general expression for the magnitude of the
velocity ratio can be written as
where m is the number of meshes involving external gears
EX. 13.1
13.9
13.9
13.1
[1]
Sol.
2  200 rpm
10  ?
v11  ?
9 N 2 N 4 N 6 N8

2 N 3 N 5 N 7 N 9
9 
N 2 N 4 N 6 N8
2
N3 N5 N 7 N9
60 80 60 2
200
48 120 40 80
 6.25 rpm=10  CW 

[1]
13.10
13.1
The velocity of the rack will be equal to the linear velocity of the pitch point on gear 10.
N
65
d10  10 
 13"
Pd 10 5
 v11  10
d10 
2  13
  6.25 
  4.26 in/s   
2 
60  2
Concentric Gear Trains
 In a concentric gear train, the input and output shafts are
collinear. A concentric gear train with a two stage
reduction which principal requirement is r2+r3=r4+r5.
[1]
13.12
13.11
[1]
Reduction ratio
r2+r3=r4+r5
(2)
13.12
(1)
[1]
To mesh properly, gears 2 and 3 must have the same normal pitch, and
gears 4 and 5 must have the same normal pitch. If helical gears are
involved, we must select the helix angles.
(5)
(3)
=Pt2
=Pt4
(4)
(6)
There are 12 unknowns (r2, r3 , r4 , r5, N2, N3 , N4 , N5, Pn2, 2 , Pn4 , 4, )
and 6 equations. We can select six of the variables to solve the equations
subject to the constraints that the tooth numbers are integers.
Concentric Gear Design
 One design approach is to select first the tooth numbers,
which is typically the most difficult equation to satisfy.
This is equivalent to selecting three of the variables.
 When the machine function does not require an exact ratio,
it is usual to select tooth numbers for a meshing gear pair
that do not have common factor.
 If R=p/q, we would look for values of N3 and N5 such that
N3 N5=p and values of N2 and N4 such that N2 N4=q.
Concentric Gear Design
 After the tooth number are established, we
can select one of the normal diametral pitch,
for example, Pn2, and the corresponding
helix angle 2 . Then solve for r2 and r3.
 Solve for r4 and r5. Pick a standard value for
Pn4, and solve for the helix angle 4 .
EX. 13.2
r2+r3=r4+r5
R=(r3r5)/(r2r4)
12.1
Sol.
R  20 :1
n  200
Pn 2  8
 2  300
N2  ?
N3  ?
N4  ?
N5  ?
r2  ?
r3  ?
r4  ?
r5  ?
Pn 4  ?
4  ?
To avoid undercutting, we will
limit the tooth numbers for N2
and N4 to 12 teeth.
For the smallest possible gear
box, assume that N2 is 12, and
select that N4 to be 12 teeth
also.
N3 N5 N3 N5
 R  20 


N 2 N 4 12 12
When designing the two-stage gear reducer, it is generally desirable to
make the two gear reductions about the same. The factors for 2880, 48 and
60 will give reductions of 60/12=5 (First) and 48/12=4 (Second) for the
two stages.
R  20 
 N3  60
N3 N5 N3 N5


=5  4
N 2 N 4 12 12
N5  48
N
Pt  Pn cos 
2rp
N2
12
 r2 

2 Pn 2 cos 2 2  8cos 300
 0.866"
N3
60
r3 

2 Pn 2 cos 2 2  8cos 300
 4.33"
N 4 12 1 r4

 
N5 48 4 r5
 r5  4r4
r4  r5  5r4  r2  r3  0.866  4.33  5.196"
 r4  1.039"
r5  4r4  4.156"
N5
N4
12
Pn 4 cos 4 

 5.775 
2r4 2 1.039
2r5
select a normal diametral pitch of 7, then
5.775
 4  cos 1
 34.40
7
Planetary Gear Trains
 Both simple and compound gear trains have the
restriction that their gear shafts must rotation in bearing
fixed to the frame.
 If one or more shafts rotate around another shaft as well
as spinning about their own axes, the gear train is called
a planetary or epicyclic gear train.
Note that the carrier, ring
gear, and sun gear are all
rotate about concentric
axes.
The planetary gear train
has two degrees of
freedom.
n=5 (S, P, C, R, Frame)
13.15
[1]
j=6 (4 Revolute pairs, P-C and S-C-RFrame; 2 Gear pairs, S-P and P-R)
M=3(5-6-1)+(14+22)=2
Planetary gear trains are typically made up of the following:
S
P
C
R
13.13
[1]
13.14
[1]
Very high velocity reductions can be achieved with
compound planetary gear trains.
13.16
Ring gear to be
replaced by
another sun gear
[1]
Carrier can involve
several shafts
containing four or
more planetary
gears
Planetary gears
connect in series
13.17
[1]
Hypoid ring
gear and
pinion and
bevel gear
planets
13.18
[1]
Analysis of Planetary Gear Trains
Using Equations
 The procedure is to write relative angular velocity
equations (relative to frame) for each of the gears with
fixed rotation axes.
 Also, write relative velocity equations for the same
gears relative to the carrier.
 Typically, we could select counterclockwise (CCW) as
plus and clockwise (CW) as minus.
Example 13.3
13.15
4
3
3
2
2
4
1. There are two gears, (2
and 4) that rotate about
fixed axes in the system.
2  1c  c2
1
 c2  12  1c
4  1c  c4
1
[1]
13.15
 c4  14  1c
4
3
3
2
2
4
2. If we make the carrier
the reference link, the
gears will move as an
ordinary gear train in
which the planet gear
acts as an idler.
R
1 NS


1


C
S
NR
C
[1]
13.15
C
R R  C
1 N

  1 S
S S  C
NR
C
簡單齒輪鏈
行星齒輪減速比分析
R  C
1 N
  1 S  
S  C
NR
0  C
N
  S =   ，  S  C   C
S  C
NR
C 1     S ， 
0<<1
S
1
 1
C

N

R  0
1
  S =   ， S  
S  0
NR
R

0  C
N


  S =   ， C 
S  C
NR
S 1  
R  C
N

1
  S =   ， C 
0  C
NR
R 1  
[2]
37
行星齒輪減速比分析
R  C
1 N
  1 S =  
S  C
NR
0<<1
N
R  0

  S =   ， R  
S  0
NR
S
R  C
N

  S =   ， R  1  
0  C
NR
C
  C
N
  S =   ，C  
  C
NR
[2]
行星齒輪具三元件，即便
指定一輸入件之轉速，僅
配合一轉速方程式，無法
求解剩餘二元件之轉速
38
四檔自動變速箱
辛普森(Simpson)行
星齒輪組。
前太陽輪和後太
陽輪連結，前行星
架和後環齒輪(亦
為輸出軸)連結
Over Drive
超速傳動
[2]
Toyota A40
行星齒輪組
前行星齒輪組
(1=42/79)
後行星齒輪組
(2=33/79)
OD行星齒輪組
(2=33/79)
齒數
檔位
減速比
太陽輪
42
一
2.804
行星小齒輪
19
二
1.532
環齒輪
79
三
1
太陽輪
33
四(OD)
0.705
行星小齒輪
23
倒
2.394
環齒輪
79
太陽輪
33
行星小齒輪
23
環齒輪
79
=太陽輪齒數/環齒輪齒數
D檔位-1檔
R  C
N
=  S  
S  C
NR
R +S  1    C  0
前行星齒輪組
R1 +1S1  1  1  C1  0
後行星齒輪組
R 2 +2S 2  0
 i +1  1 2  o  1  1  o  0
i
1
42 42 79
 1  1   1  
o
2
79 33 79
 2.804
2 - 41
S1 =S 2 =S
C1 =R 2  o
R1  i
i +1S  1  1  o  0
o +2S  0
[1]
Example 13.4
13.22
Sol.
2
2 N3 N5
& c   1
N2 N4
5
c
c  150 rpm
1
5  50 rpm
1
2  ?
1
2  1c N3 N5
1

1
5  c N 2 N 4
1
2   150 
1
30 20


50   150  18 28
 12  30.95 rpm
 30.95 rpm (CW)
[1]
13.19
There are two gears, (2
and 5) that rotate about
fixed axes in the system.
13.4
2  1c  c2
1
5  1c  c5
1
 c2  12  1c  c5  15  1c
Example 13.5
13.20
2 and 16 are known,
1
Sol.
2  60 rpm
6  0
N3 N 6
2 12  1c
c 1
  1
1
N2 N4
6
6  c
5  ?
60  1c
30 76


 4.52
1
18 28
0  c
1
1
1
c
 1c  10.86 rpm (CCW)
5 15  1c
2 N2 N4



1
 
c
1
1
N3 N5
2
2  c
c
There are three
 c2  12  1c
1
gears, (2, 5, and
c
1
1
5  5  c  5  10.86  18 28  0.84
6) that can rotate
60  10.86 30 20
c
1
1
6  6  c
about fixed axes
 15  52.14 rpm (CCW)
in the system.
[1]
13.20
13.5
Example 13.6
13.21
Sol.
input: gear 2
output: carrier 7
4  0
1
2
?
1
7
1
[1]
13.21
13.6
There are three gears, (2, 4, and 6) that can rotate about fixed axes in the system.
We must separate the two stages of the planetary drivers when we write the equations.
second stage:
6  16  17
7
4  14  17
7
6
N5 N 4
N4
& 7 

N 6 N5
N6
4
7
first stage:
[1]
6  17
N4
1

1
N6
4  7
1
2  2  6
6
1
1
4  14  16
6
The first stage includes
gears 2, 3, and 4 and the
carrier is member 6.
The second stage includes
gears 6, 5, and 4 and
carrier 7.
Two stages can be
analyzed independently.
N N

N
& 6 2  3 4  4
N 2 N3
N2
4
6
 7 
1
2 1   N 4 N 2  

1
2  16  
 6 
1
N4
4  16
N2

2
1
1   N4 N2 
1   N4 N6 
1
2  6
N4
1

1
N2
4  6
1
6
1
1   N4 N6 

2  N 4  N 4 
 1  1 
1 

N
N
7 
6 
2 
1
 58  58 
 1  1    68.06
8 
8 

Example 13.7
13.22
Sol.
input: gear 2
output: gear 6
2  100 rpm
1
7  0
1
6  ?
1
[1]
13.22
13.7
The planet gears
rotate about the arm
axis, which is skewed
relative to the axis of
the other gears.
Therefore, the
angular velocity of
the planets (direction)
is not obtained by a
simple algebraic
addition of the values
from steps 1 and 2.
There are three gears, (2, 6, and 7) that can rotate about fixed axes in the system.
There are three gears, (2, 6, and 7) that
can rotate about fixed axes in the system.
 32  12  13
6  16  13
3
7  17  13
3
[1]
2 and 17 are known,
1
N7
2 12  13
3  1

1
N2
7
6  3
3
100  13
76


 3.8
1
20
0  3
6 16  13
N 2 N5 The overall velocity ratio
1

3
1
N 4 N 6 for the gear box is:
2
2  3
1
1
2 100
6  20.83
20 24

 69.2


1
6 1.44
100  20.83
56 35
3
 0.24
1

6  1.44 rpm (CCW)
 3  20.83 rpm (CCW)
1
Analysis of Planetary Gear Trains
Using Tabular
 The absolute angular velocity of any gear x that rotates
about an axis fixed to the frame can be written as
1 =1 +c . where 1 is the absolute angular velocity
x
c
x
c
of the carrier, and cx is the angular velocity of the gear
relative to the carrier.
1  = 1  +5 
2
5
2
1  = 1  +5 
3
5
3
1  = 1  +5 
4
5
4
13.23
[1]
Example 13.8
13.24
3
N2


5
N3
2
5
Sol.
2  100 rpm
1
5  200 rpm
 53  
3  ? 4  ?

1
1
1
1  = 1  +5 
2
5
2
1  = 1  +5 
3
5
3
1  = 1  +5 
4
5
4
N2
N3

5
2
40
 100   200
20
4
N2


5
N4
2
5
[1]
13.24
13.8
13.1
N2
 4  
N4
5
13.8
200 15
0
200 15
200 15
-100 52
100 12
200 15
200 53
400 13

40


 100 
200 15
80
50 54
 50
1
250 4
5
2
Example 13.8
13.24
Sol.
3 13  15
N2



5
N3
2 12  15
5
2  100 rpm
1
5  200 rpm
1
3  200
1
40
 2
100  200
20
 13  400 rpm (CCW)

3  ?
1
(using equation)
[1]
13.24
13.8

N7
2
N4 N7


3
N
N
N2
7
2
4
3
Example 13.9
Sol.
1 =100
2
 32  
rpm
1 =0
7
1  = 1  +3 
2
3
2
1  = 1  +3 
7
3
7

7   
3
6 N5 N 7

3
7 N 6 N 4
3
1 /1 =?
2
6
1  = 1  +3 
6
3
6
N7
N2
x  3.8 x  100
 x  20.83 rpm
[1]

13.22
13.7
 36 
1
x
0
1
3
x 13
13.9
x
3.8x
1
3
x
3
2
100
1
2
N5 N 7
N6 N4

3
7
24 76
  x   0.9306 x
35 56
1
13.2
76
  x   3.8 x
20
2
4.8 x

 69.2
6 0.0694 x
-0.9306x
1
3
3
6
x 13
-x 37
0.0694x
1
6
0 17
Example 13.10
13.25
100, 101, and 99
35
3
N2
2 N5 N 2


1



6
N3 N5 N3
2
6
Sol.
1 =100
6
rpm
 63 
1 =0
2
1  =?
3

1 =?
4
N2
N3

2  
6
100
 100 
101
100
1.01
N2
N4

6
2
100
100
 100   
99
0.99
5
N2
N2 6
6






2 

5
6
N5
N5
2
6
1  = 1  +6 
3
6
3
1  = 1  +6 
5
6
5
4 
6

1 =?
5
1  = 1  +6 
4
6
4
4 N 2

6
2 N 4
6
[1]
13.25
13.10
13.3
100
0
100

100
100
 100  
35
0.35
13.10
100
-100
0
100
-100/1.01
1/1.01 1
100
-100/0.99
-1/0.99 -1
100
100/0.35
135/0.35
Harmonic Drive Speed Reduction
 Very high speed reduction can be achieved by connecting
several planetary drives in series. However, the resulting system
will require a relatively large number of gears and will be
relatively heavy.
 The harmonic drive is a simple solution for mechanical systems
requiring a large speed reduction in a light-weight space. Speed
reduction from 30 to 130 is available in a single stage.
 The function of harmonic drive is based on the principle of
strain-gearing which requires a flexible element.
Harmonic Drive Speed Reduction
 The input shaft is connected to an ellipse-shaped wave generator. The
output shaft is connected to a flexible sleeve with machined spline
teeth around one end (flexspline). The wave generator distorts the
splined end of the flexspline so that it engages with a rigid circular
splined member.
Harmonic Drive Speed Reduction
 The reduction ratio for a harmonic drive is given
by
Nf=number of teeth on the flexible flex spine
Nc=number of teeth on the rigid circular spine (Nc>Nf)
 For the largest reduction, we want the difference between Nc and Nf
to be as small as possible, which also minimize the distortion and the
stresses in the flexspline. However, the difference needs to be large
enough that the two splines touch at only the ends of the major axis
of the ellipse. Typical the minimum difference required is 2.
Harmonic Drive Speed Reduction
 In addition to a very high-speed reduction ratio in a single stage,
harmonic drives tend to have a number of other advantages over
rigid gear drives. They have very low backlash and very precise
positioning accuracy. Harmonic drives also tend to be very small
and lightweight compared to conventional gear reducers for
given torque and power requirements. In addition, their high
efficiency allows the drives to have a high torque to weight ratio.
 The main disadvantage of harmonic drives is relatively high cost.
They are a specialty item that is manufactured by a limited
number of company, mainly in Japan and Germany.
Reference
[1] K. J. Waldron, G. L. Kinzel, and S. K. Agrawal,
Kinematics, Dynamics, and Design of Machinery, 3rd
Edition, John Wiley & Sons, 2016.
[2] 賴瑞海編著, 汽車學II(底盤篇), 2011，全華圖書。