Gear Train

Main types:

• Simple gear train: one gear per

shaft

• Compound gear train: two or

more gears per shaft, but for the

first, last

• Reverted gear train: input and

output gears are coaxial.

• Planetary/epicyclic gear train:

one with a relative motion of

the axes.

A combination of gears used to transmit motion from one shaft to another.

Why combination?: to obtain large speed reduction within a small space.

Simple Gear Train

Gear Ratio

+ : Internal gearing

- : External gearing

Simple Gear Train

- -

Intermediate gears

Odd Even

Input :Output motion Alike Opposite

Intermediate gears are called idlers in SGT as they do not affect the speed ratio of the

gear train. They are primarily used to;

(a) Connect gears with large center distance

(b) To obtain the desired direction for the driven

Compound Gear Train

The compound gears are rigidly fixed to the shaft,

hence have the same speed.

- - -

-

(-1)pairs

Reverted Gear Train

Epicyclic Gear TrainThe axes of the shafts over which the gears are mounted may

move relative to the fixed axis.

Gear A and the arm C have a common axis at O1 about which

they can rotate.

The gear B meshes with gear A, and has its axis on the arm

at O2 about which it can rotate.

If the arm is fixed: Simple gear train

If the gear A is fixed and the arm is rotated about O1 then the gear B is forced to

rotate upon and around gear A. Such a motion is called epicyclic (epi: upon; cyclic:around)

EGTs are useful for transmitting high speed ratios with gears of moderate size………

….backgear of lathe; Differential gears of automobiles, wrist watches…

Epicyclic Gear Train

Planetory Gear Train

Sun (Center)

Planet (Around Sun

Annulus (Internal T)

Epicyclic Gear TrainSpeed ratio: Algebraic method; Tabular method

The motion of each element of the EGT relative to the arm is written in the form of equations

These equations are solved using two conditions:

(i) Some element is fixed, implying zero velocity

(ii) Motion of some other element is specified.

One caution: You’ll need to associate

+/- signs to clockwise and anticlockwise directions

Anticlockwise: +

Clockwise: -

Epicyclic Gear Train: Problem-1

- No. of teeth are as shown in the diagram

-A is fixed

-B and C are carried on an arm which revolves at 100 rpm about the axis of A or D.

# Find the number of teeth of C and the speed & sense of rotation of C?

Epicyclic Gear Train: Problem-1tA= 150; tB=25; tD=40; NA=0; Narm =100 CL =-100

NC=?

Geometry:

dA/2 = dB + dC + dD/2

tA/2 = tB + tC + tD/2

75=25+tC+20 tC=30

Engagement of AB: Internal implies +

(NB-Narm) / (NA-Narm) = tA/tB

NB- (-100) = (150/25) * (0-(-100)) NB=500

Engagement of BC: External implies -

(NB-Narm) / (NC-Narm) = -tC/tB

NC- (-100) = (-25/30) * (500-(-100)) NC=-600 = 600 rpm clockwise

Epicyclic Gear Train: Problem-2

# Find the speed and direction of wheel D when wheel A is fixed and arm F makes 200rpm

clockwise.

Epicyclic Gear Train: Problem-2tA= 80; tC=72; tD=48

NA=0; Narm =200 CL =-200

Note: NB=NC due to compound shaft; ND?

Geometry:

rA + rB = rC + rD

tA + tB = tC + tD

tB = 40

Engagement of AB: External implies -

(NB-Narm) / (NA-Narm) = -tA/tB

NB- (-200) = (-80/40) * (0-(-200)) NB=-600 600 rpm CL

Engagement of CD: External implies -

(ND-Narm) / (NC-Narm) = -tC/tD

Note NC= NB

ND = 400 = 400 rpm CCL

Torques in Epicyclic Gear Train

When a geared system transmits power, torques are transmitted from one element

to another.

Torques in Epicyclic Gear Train

When the rotating parts of an epicyclic

gear train have uniform speeds, i.e., zero

angular acceleration, then the gear train

is kept in equilibrium by three externally

applied torques. The net (external)

torque applied must be zero.

There are no losses in power transmission,

i.e., there are no internal friction losses at

the bearings and at the contact surfaces. In

other words, the net energy dissipated by the

gear train is zero.

Torques in Epicyclic Gear Train: Problem 2Shown is a compound epicyclic

gear train in which two gears S1

and S2 are integral with the

Input shaft B. The arm A2 is

integral with the output shaft, C.

The planet gear P2 revolves on a

pin attached to the arm A2, and

meshes with the Sun gear S2 and

internal gear I2 that is co-axial

with the input shaft.

The planet gear P1 meshes with

the fixed internal gear I1 and sun

gear S1.

The planet gear P1 revolves on a pin fixed to internal gear I2.

Arm

A1

The number of teeth are as stated in the figure.

The input power on the shaft B is 10.47 kW

at 1000 rpm (CCL). Find:

•The speed and torque at the output shaft

•The torque required to hold the internal gear I1.

I1 P1 A1

I2 P2S1

S2

Torques in Epicyclic Gear Train: Problem 2

Arm

A1

The input power on the shaft B

is 10.47 kW at 1000 rpm (CCL).

Find:

•The speed and torque at the

output shaft

•The torque required to hold the

internal gear I1.

Inp:B S1 P1 I1 A1 I2 P2 S2 A2/C

Teeth

(T)

26 88 83 31

Speed

(N)

1000

CCL

1000

CCL

0 1000

CCL

Inp:B S1 P1 I1 A1 I2 P2 S2 A2/C

Teeth

(T)

26 31 88 83 26 31

Speed

(N)

1000

CCL

1000

CCL

0 1000

CCL

Rad(S1) + Dia (P1) = Rad(I1) Rad(S2) + Dia (P2) = Rad(I2)

(TS1)/2 + TP1 = (TI1)/2 (TS2)/2 + TP2 = (TI2)/2

Torques in Epicyclic Gear Train: Problem 2

The input power on the shaft B is 10.47

kW at 1000 rpm (CCL).

Find:

•The speed and torque at the output

shaft

•The torque required to hold the

internal gear I1.

I1 P1 A1

S1

Torques in Epicyclic Gear Train: Problem 2

The input power on the shaft B is 10.47

kW at 1000 rpm (CCL).

Find:

•The speed and torque at the output

shaft

•The torque required to hold the

internal gear I1.

I2 P2 S2

Torques in Epicyclic Gear Train: Problem 2

The input power on the shaft B is 10.47

kW at 1000 rpm (CCL).

Find:

•The speed and torque at the output

shaft

•The torque required to hold the

internal gear I1.

Epicyclic Gear Train with Bevel Gear

Bevel gears are used for high speed reduction with fewer gears.

The method applicable for EGT with spur gears is also applicable for EGT with bevel gears

with some change:

For those gears whose axes are inclined to the main axis, the terms clockwise and

anticlockwise are not applicable, because direction of rotation for different gears ought to

be with respect to the same axis.

Epicyclic Gear Train with Bevel Gear: Problem 1

The figure shows an epicyclic bevel GT (Humpage’s reduction gear) in which the gear B is

connected to the input shaft, and gear F is connected to the output shaft.

The Arm (A), carrying the compound wheels D & E, turns freely on the output shaft.

If the input speed is 1000 rpm CCL (seen from the right), determine the speed of the

output shaft, when:

I: gear C is fixed

II: gear C is rotated at 10 rpm CCL.

Epicyclic Gear Train with Bevel Gear: Problem 1

NB = 1000 rpm CCL, determine NF, when:

I: gear C is fixed

II: gear C is rotated at 10 rpm CCL.

C D B

F E

3

2 1

To find NF, find NE

to find which, find ND

Epicyclic Gear Train with Bevel Gear: Problem 1

NB = 1000 rpm CCL, determine NF, when:

I: gear C is fixed

II: gear C is rotated at 10 rpm CCL.

C D B

F E

3

2 1