The document summarizes the kinematic analysis of an epicyclic gear train using a tabular method. It shows three conditions of the arm movement and calculates the speed of the gears and arm. In condition 1, the arm is fixed as gear A rotates 1 turn, giving gear B a speed of -TA/TB. In condition 2, gear A rotates x turns with arm fixed, giving gear B a speed of -x*TA/TB. Condition 3 adds a +y rotation of the arm. The tabular method is used to calculate the speeds by adding the individual steps.

1. Sanjivani Rural Education Society’s
Sanjivani College of Engineering, Kopargaon-423603
( An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune)
NAAC ‘A’ Grade Accredited, ISO 9001:2015 Certified
Subject :- Theory of Machines II
T.E. Mechanical (302043)
Unit 3
3.2 Kinematic Analysis of Epicyclic Gear Train
By
Prof. K. N. Wakchaure(Asst Professor)
Department of Mechanical Engineering
Sanjivani College of Engineering
(An Autonomous Institute)
Kopargaon, Maharashtra
Email: wakchaurekiranmech@Sanjivani.org.in Mobile:- +91-7588025393

2. Sun Gear= Rotates about own axis
Arm= Rotates about the axis of sun
gear at one end and another end
consist of Planet gear/s
• Mounted on another shaft
Planet Gear= Rotates about the
own axis as we all over the
periphery of Sun Gear (Epicyclic
action)
EPICYCLIC GEAR TRAIN
Sun Gear
Planet Gear
Arm

3. Kinematic Analysis (Tabular Method)
Gear A
Gear B
Arm C
Condition 1: ARM C is fixed and gear A competes +1 rotation (Clockwise)
𝑵 𝑪= 0
𝑵 𝑨= + 1
𝑵 𝑩
𝑵 𝑨
=
𝐓 𝐀
𝐓 𝐁
𝑵 𝑩 =
𝐓 𝐀
𝐓 𝐁
• Gear A and Gear B are having External Meshing.
• hence Gear A and Gear B will be rotate in opposite direction.
• As gear A rotation in Clockwise direction, gear B will move in
Anticlockwise direction.
𝑯𝒆𝒏𝒄𝒆 𝑵 𝑩 = −
𝐓 𝐀
𝐓 𝐁

4. Kinematic Analysis (Tabular Method)
Gear A
Gear B
Arm C
Condition 2: ARM C is fixed and gear A competes +𝒙 rotation
(Clockwise)
𝑵 𝑪= 0
𝑵 𝑨= + 𝒙
𝑵 𝑩
𝑵 𝑨
=
𝐓 𝐀
𝐓 𝐁
𝑵 𝑩 =
𝐓 𝐀
𝐓 𝐁
∗ 𝒙
• Gear A and Gear B are having External Meshing.
• hence Gear A and Gear B will be rotate in opposite direction.
• As gear A rotation in Clockwise direction, gear B will move in
Anticlockwise direction.
𝑵 𝑩 = −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙

5. Kinematic Analysis (Tabular Method)
Gear A
Gear B
Arm C
Condition 3: ARM C moving with +y rotation
𝑵 𝑪= +y
• As arm C rotates which add +y rotations in the speed of Gear
A and Gear B

6. Kinematic Analysis (Tabular Method)
Gear A
Gear B
Arm C
STEP OPERATION
SPEED OF THE ELEMENTS
ARM C GEAR A GEAR B
1 ARM C is fixed and gear A competes +1
rotation
0 +1
-
𝐓 𝐀
𝐓 𝐁
2 ARM C is fixed and gear A competes +x
rotation
0 +x
- 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
3 Add +y rotation of ARM C +y +y +y
4 Add Step 2 +Step 3 𝐲 x+y
y - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙
𝑵 𝑪 = +𝒚
𝑵 𝑨 = 𝒙 + 𝒚

7. Ref. https://nptel.ac.in/courses/116/102/116102012/
Kinematic Analysis (Tabular Method)
STEP OPERATION
SPEED OF THE ELEMENTS
ARM C GEAR A GEAR B
1
ARM C is fixed and gear A competes
+1 rotation
0 +1 -
𝐓 𝐀
𝐓 𝐁
2
ARM C is fixed and gear A competes
+x rotation
0 +x - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
3 Add +y rotation of ARM C +y +y +y
4 Add Step 2 +Step 3 𝐲 x+y y - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝐓 𝑩 = 𝟑𝟐
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚
𝑵 𝑪 = +𝟏 𝑵 𝑨 = 𝟎
𝒙 + 𝒚 = 𝟎𝑵 𝑪 = 𝒚 = 𝟏 𝒙 = −𝒚 𝒙 = −𝟏
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙 𝑵 𝑩 = 𝟏 −
𝟑𝟐
𝟑𝟐
∗ (−𝟏) 𝑵 𝑩=2
𝑵 𝑪 = +𝟏 𝑵 𝑨 = 𝟎

8. Ref. https://nptel.ac.in/courses/116/102/116102012/
Kinematic Analysis (Tabular Method)
STEP OPERATION
SPEED OF THE ELEMENTS
ARM C GEAR A GEAR B
1
ARM C is fixed and gear A competes
+1 rotation
0 +1 -
𝐓 𝐀
𝐓 𝐁
2
ARM C is fixed and gear A competes
+x rotation
0 +x - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
3 Add +y rotation of ARM C +y +y +y
4 Add Step 2 +Step 3 𝐲 x+y y - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝐓 𝑩 = 𝟑𝟐
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚
𝑵 𝑪 = +𝟏 𝑵 𝑨 = 𝟏
𝒙 + 𝒚 = 𝟏𝑵 𝑪 = 𝒚 = 𝟏 𝒙 = 𝟏 − 𝒚 𝒙 = 𝟎
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙 𝑵 𝑩 = 𝟏 −
𝟑𝟐
𝟑𝟐
∗ (𝟎) 𝑵 𝑩=1
𝑵 𝑨 = 𝟏 𝑵 𝑪 = +𝟏

9. Ref. https://nptel.ac.in/courses/116/102/116102012/
Kinematic Analysis (Tabular Method)
STEP OPERATION
SPEED OF THE ELEMENTS
ARM C GEAR A GEAR B
1
ARM C is fixed and gear A competes
+1 rotation
0 +1 -
𝐓 𝐀
𝐓 𝐁
2
ARM C is fixed and gear A competes
+x rotation
0 +x - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
3 Add +y rotation of ARM C +y +y +y
4 Add Step 2 +Step 3 𝐲 x+y y - 𝐱 ∗
𝐓 𝐀
𝐓 𝐁
𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝐓 𝑩 = 𝟑𝟐
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚
𝑵 𝑪 = +𝟏 𝑵 𝑨 = −𝟏
𝒙 + 𝒚 = −𝟏𝑵 𝑪 = 𝒚 = 𝟏 𝒙 = −𝟏 − 𝒚 𝒙 = −𝟐
𝑵 𝑩 = 𝒚 −
𝐓 𝐀
𝐓 𝐁
∗ 𝒙 𝑵 𝑩 = 𝟏 −
𝟑𝟐
𝟑𝟐
∗ (−𝟐) 𝑵 𝑩=3
𝑵 𝑨 = −𝟏 𝑵 𝑪 = +𝟏

10. •Thank You…