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Theoretical contents
I. Overview

II. Classification
III. Transmission Ratio
IV. Problem

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1. Overview

 The gears are matching together in series or parallel.
 Transmitting and distributing motion
 Increasing or decreasing rotation speed
 Large ratio and far centre distance

Planar Universal

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2. Classification.

Gear drives system

(Planar, universal)

Simple gear Epicyclic

Planetary Open Close
Differential differential

Combination

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2. Classification.
a. Simple gear drive system (hệ thường)

Centerlines of all gears are fixed

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2. Classification.
b. Epicyclic gear system (hệ ngoại luân)

 Epicyclic gear system has at least 1 gear’s centerline motion .

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2. Classification.
b. Epicyclic gear system

 Planetary gear (hành tinh) is at least 1 fixed center gear

Z2 Z’2

Z2 Z’2
C Z3
C

Z1

Z1 Z3

Degree of freedom:
W = 3𝑛 − 2𝑝 − 𝑝4 = 3.3 − 2.3 − 2 = 1

The gear that the centerline is in the main axis of the system is so-called center gear

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2. Classification.
b. Epicyclic gear system

 Open differential: (Vi sai hở) All center gears are moveable

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8 Z2

2. Classification. Z1 C
b. Epicyclic gear system Z3

 Open differential All center gears are moveable
𝒁𝒃

Z3
Z1

Z’2
Zb

Za

Z2 Z’2

𝒁𝒂 C C
Degree of freedom:
W = 3𝑛 − 2𝑝 − 𝑝4 = 3.4 − 2.4 − 2 = 2 Z1

Z3

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2. Classification.

b. Epicyclic gear system

 Close differential (vi sai kín) 2 in 3 kinematic links (C, Z1, Z3 ) are engaged

each other
Z2 Z’2 Z2

Z’3

Z’1 C Z’2
C

Z5

Z1 Z3 Z1 Z3 Z’3

Z4 Z’4

Z’4 Z4

Degree of freedom:
W = 3𝑛 − 2𝑝 − 𝑝4 = 3.5 − 2.5 − 4 = 1

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2. Classification.
c. Combination gear system


Simple gearing coupling with a epicyclic (planetary, open differential, close
differential)

Z2

Zb Z’2

C Z’3

Z3
Z1
Za Zd

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3. Transmission ratio
a. Simple gear system

 Ratio of 1 gearing

u12  ω1  n1   Z2 (+): Inside engage (𝜔 ↑↑ 𝜔 )
ω2 n2 Z1 (−): Outside engage (𝜔 ↑↓ 𝜔 )

 Ratio of simple gear system

u1n  ω1  n1  ()  Zbd
ωn nn  Zd


(−1)k Planar. where, k is number of outside engage

±
For Universal gear (bevel gear). Determining (+) (-) in reality

• If 𝑢 < 0 gear 1 and n gear are opposite direction.
• If 𝑢 > 0 gear 1 and n gear are same direction.

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3. Transmission ratio Z3
a. Simple gear system Z’3

 Exercise Z’2
Z1

Planary
Z2

Universal Z’2 Z2
C
Z3 Given:
Z’3 Z1 = 20 Z2 = 40 Z’2= 20
Z1
Z3 = 50 Z’3= 40 Z4 = 20
Calculating u13 ?
Z4


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3. Transmission ratio

b. Epicyclic system

the angular velocities of
the various gears in the
system is taken by
transforming the arm C
into stationary observer

• If arm C is stationary, angular velocity of each link will be −𝜔

𝐾ℎâ𝑢 1 𝑍1 𝜔 − 𝜔
𝐾ℎâ𝑢 2 (𝑍2 & 𝑍′2) 𝜔 − 𝜔
Kℎâ𝑢 3 (𝑍3) 𝜔 − 𝜔
𝐾ℎâ𝑢 4 (𝑪): 0

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3. Transmission ratio
b. Epicyclic system

 Planetary gear


Z2 Z’2 In reference frame C

u13/C  1  c  1  1
3  c c

C 0

In absolute system 𝑢 = −1 . =5

Z1 Z3 u1C  1  4
c
Given: Z1=20; Z2=60;
Z’2=30; Z3=50 (Arm C and Z1 are opposite direction)

𝑢 = =?

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3. Transmission ratio
b. Epicyclic system

 Open differential

Z2 Z’2

Given: Z1=20; Z2=60; Z’2=30; Z3=50

ω =8 rad/s ω =1 rad/s


C Z and C are the opposite direction

Calculating:

ω =? rad/s and direction Z3?

Z1 Z3

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3. Transmission ratio
b. Epicyclic system

 Open differential u13/C  1  c
In reference frame C 3  c

In absolute system u13  12 Z2  Z3  5

Z1  Z2'

where, 𝑢 / =𝑢

 3  1  4c
5

Assuming direction Z1 +  (8)  4(1)   4
Or C - 5 5


(Z3 and Z1 are the same direction)

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3. Transmission ratio
b. Epicyclic system

 Close differential

Z2 Z’2

In reference frame C 𝑢 / =

Z’3 ZZ
𝑢 = −1 . = 6
Z’1 C Z Z′

ω − ω = 6 (ω − ω ) =>ω =

Z1 Z3 Close differential: 𝑢 = = −1 Z’4 . Z’3=+4
Z’1 Z4

=>ω = + ;nên ω = + = + rad/s

Z’4 Z4

Given: Z1=20; Z2=60; Z’2=20; Z3=40 Z’1=20; Z’4=40; Z4=20; Z’3=40 ω =8 rad/s

Calculating ω =? rad/s

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3. Transmission ratio

c. Combination
Z1

𝑈 = 𝑈 .𝑈

𝑍
𝑈 =−

𝑍

𝑈= ω Z’2 Z4

ω ω −ω ω Z2

𝑈 / =ω −ω =1−ω

𝑈 = (−1) 𝑍 𝑍 −𝑍 Z3
=
𝑍′ 𝑍 𝑍′

Determining the sign ( + or -) Simple gear Planetary gear
following the ‘’k’’ indication


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Problem Solution Method

1. Determine kind of gear train
Planar: based on pair of external gearing k
Space: based on realistic schematic diagram

2. Epicyclic gear train
u13/C  ω1  ωC Planetary: ω = 0
ω3  ωC
  ' Z2 Z3 Open Given 2 of 3 speed C, Z1, Z3
Z1 Z2 differential Direction (the same or opposite)

Close
differential Close differential equation

3. Combination Solve the Epicyclic
Determine the remained variable in Simple train

Note: The notation must be based on the realistic schematic diagram

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Q 1. 𝐴𝑙𝑙 𝑔𝑒𝑎𝑟 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑚𝑜𝑑ule.

𝐻𝑒𝑟𝑒, Z = 2Z , Z = 20, Z = 40, Z′ = 35, Z = 70 teeth
Gear Z has n = 120 𝑟𝑝𝑚 , n = 70 𝑟𝑝𝑚 ,
𝑍 𝑎𝑛𝑑 𝑙𝑖𝑛𝑘 𝐶 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
D𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐞 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑛 𝑎𝑛𝑑 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑍 .

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Solution Q1. u13/C  n1  nC  na  3nC
n3  nC (1) & (2)  n3  2
Epicyclic gear
 (-1) ' 2 Z2 Z3   4 4
Simple gear Z1 Z2

Assumption C +  n3  n1  3nC (1)
or Za + 4

uab  na   Zb  2
nb Za

 n1  nb   na (2)
2

 (120)  3(70)
 n3  2
4   37,5 v / p

Conclusion
Z3 rotates at 37,5 rpm and the same direction Arm C


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C Zb 8
Problem Z3 Za

Q 2. Z1

Given: Z1=20; Z2=80; Z’2=40; Z3=40
Z = 2Z , n = 400 𝑟𝑝𝑚 , n = 200 𝑟𝑝𝑚
Z and C are opposite direction of rotation .
Calculate:

a. Degree of freedom W of gear?
b. Direction of gear?
c. Rotation speed n3 of gear Z3?
d. Rotation speed nb of gear Zb?

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Problem

Solution 2. u13/C  n1  nC   ' Z2 Z3   4
Z2 n3  nC Z1 Z2
Z’2
C Z3  n3  5nC  n1 (1)
Zb 4

Z1
uab  na   Zb  2
nb Za

 nc  nb   na (2)
2

 5na  n1
Za (1) & (2)  n3  2

4

Assume Z1 +  5(400)  (200)
 C-  Za +  n3  2

Conclusion: 4
  300 v / p

Z3 rotates at 300 rpm and opposite direction Z1 (or the same Arm C)

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Q 3. Z1= Z’2= Z’3= 50, Z2= Z3= 25,
Z4= 100, Z’4= 20, Z5= 40.
Rotation speed n1= 800 (rpm)
Calculating:
a. Degree of freedom W of gear?
b. Rotation speed n3 of gear Z3?

c. Rotation speed n5 of gear Z5?

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Q 4. Za= 25, Zb= 50, 𝑍 = 25, 𝑍 = 50, 𝑍′ = 15, 𝑍 = 30.
Gear Za and arm C turn same direction.
Rotation speed na= 200 (rpm), nc= 100 (rpm).
Determine the direction and rotation speed n3 of gear Z3?

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IV Problem

Q5. Z1= Z5= Z4= 30, Z2=60, Z3= 80, Z6=140.
n1= 960 rpm. Determine
a) DOF of the given gear train?

b) nC and rotating direction of Arm C

(compared to direction of Z1)?

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