Transmission of Motion & Power3

8/17/2019 Transmission of Motion & Power3

1/67

& • ( ) & .

• , &

.• ;

1. (/)

2. 3.

.


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B .

, :

1.

, . .2.

.

, .. .


3/67

& B

The lower side of the belt will have more tension & is called theTight side.

The upper side of the belt will have less tension & is called theSlack side.

Belt Materials: Leather, rubber, canvas, balata (rubber withcotton)


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DriverDriven

Adjustable weight

Gravity Roller

Jockey pulleys are used to get proper arc of contact. It increases theangle of wrap and there by reduce the belt tensions required for a given

power. Normally the idler pulley is located near to the smaller diameterpulley.1. It is non reversible.2. The bending stress developed in the belt reduces the belt drive.

3. Requires endless belt.


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• A

, ,.Stepped Pulleys

Belt

Shaft

• 3 .

• B ,

.


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& • &

.

• A

Driver pulley

Belt

.

•

.

Belt shifter

Fast pulley Loose pulley


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&


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B B

ππππ++++2222αααα

αααα

αααα

A

−−−−

P

αααα

F

r

r

1

2

C

D

x

O O1 2


11/67

B B


αααα

αααα

A

C

ππππ−−−−2222ααααB

D

P

ααααE

F

r

r

1

2

OO

1 2

1 2

Consider an open belt drive as shown in fig.

Let r be the radius of the larger pulley & r be the radius of the smaller pulley

x=Center distance between the pulleys. From the fig, Length of open belt

L

{ }

1 2

1 2From triangle O ,

r r

O P

π α π α

open

open

= Arc ABC + 2 (AF )+ Arc DEF

L = ( + 2 )+ 2 (AF )+ ( - 2 )

1 11 1 2 2 2 2 2 2 2 2 2 22 2

2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )


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1 11 1


αααα

αααα

A

C


D

P

ααααE

F

x

r

r

1

2

O O1 2

{

{ }

( )

1 2

2

1 21 2

2

1 21 2

From triangle O ,

Expanding using binomial theorem & negelcting higher order terms,

1 1

2 2

O P

=

=

2 2 2 2 2 2 2 2 2 22 2

2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1

1 11 1

1 11 1 2 22 2 2 2 2 2 2 2 2 2 2 22 2

2 22 2

2 22 2

O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )

r - r r - r r - r r - r O P = (x )- (r - r ) x 1 - O P = (x )- (r - r ) x 1 - O P = (x )- (r - r ) x 1 - O P = (x )- (r - r ) x 1 -

r - r r - r r - r r - r O P = AF = x 1 - x - r - r O P = AF = x 1 - x - r - r O P = AF = x 1 - x - r - r O P = AF = x 1 - x - r - r

x

x x

2


13/67

1 2From fig, sinα α

≈ =

r - r r - r r - r r - r

x


αααα

αααα

A

C


D

P

ααααE

F

x

r

r

1

2

O O1 2

( )

( )

( )

( )

2

1 2

2

1 2

2 1 2

2

1 21 22

2

1 2

1

1

2

2

(

12

( ) 2

( ) 2

2r

π α π α

π α

π

π

⇒

− + −

− + −

∴ +

1 2 1 2 1 2 1 2

1 11 1

1 11 1

r x - r - r r r x - r - r r r x - r - r r r x - r - r r

r - r r - r r - r r - r r r (r r x r r (r r x r r (r r x r r (r r x

r - r r - r r - r r - r r - r r - r r - r r - r r r (r r x r r (r r x r r (r r x r r (r r x

r - r r - r r - r r - r

r x r x r x r x

x

x

x

x

x

open

open

op n

ope

e

n

L = ( + 2 )+ 2 + ( - 2

L = + )

L = + )+ 2

L = + )+ 2

+


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B B


αααα

A


D

P

r

r

1

2

αααα

C

αααα

F

x

O1 O2


15/67


αααα

A

C

ππππ++++2222ααααB

D

P

αααα

E

F

x

r

r

1

2

αααα

O1 O2

1 2

Consider an crossed belt drive as shown in fig.

Let r be the radius of the larger pulley & r be the radius of the smaller pulley

x=Center distance between the pulleys. From the fig, Length of crossed

{ }

1 2

1 2

belt

From triangle O ,

r r

O P

π α π α

crossed L = Arc ABC + 2 (AF )+ Arc DEF

L = ( + 2 )+ 2 (AF )+ ( + 2 )1 11 1

2 2 2 2 2 2 2 2 2 22 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )

crossed


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αααα

A

C


D

P

αααα

E

F

x

r

r

1

2

αααα

O1 O2

{ }

{ }

( )

1 2

2

1 21 2

2

1 21 2

From triangle O ,

Expanding using binomial theorem & negelcting higher order terms,

1 1

2 2

O P

=

=

2 2 2 2 2 2 2 2 2 22 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1

1 11 1

1 11 1 2 22 2 2 2 2 2 2 2 2 2 2 22 2

2 22 2

2 22 2

O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )O P = (O O )- (O P )

r + r r + r r + r r + r O P = (x )- (r + r ) x 1 - O P = (x )- (r + r ) x 1 - O P = (x )- (r + r ) x 1 - O P = (x )- (r + r ) x 1 -

r + r r + r r + r r + r O P = AF = x 1 - x - r + r O P = AF = x 1 - x - r + r O P = AF = x 1 - x - r + r O P = AF = x 1 - x - r + r

x

x x

2


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1 2From fig, sinα α

≈ = r + r r + r r + r r + r


αααα

A

C


D

P

αααα

E

F

x

r

r

1

2

αααα

O1 O2

( )

( )

( )

( )2

1 2

1

2

1 2 2

2

1 2

2 1 2

2

1 2

2

2

1 22 1

1

2

( ) 2

( ) 2

( 2r

π α π α

π α

π

π ∴

⇒

+ + −

+ + −

+

1 11 1

1 11 1

1 11 1

r x - r + r r r x - r + r r r x - r + r r r x - r + r r

r + r r + r r + r r + r r r (r r x r r (r r x r r (r r x r r (r r x

r + r r + r r + r r + r r + r r + r r + r r + r r r (r r x r r (r r x r r (r r x r r (r r x

r + r r + r r + r r + r

r x r x r x r x

x

x

x x

crossed

crossed

cr

cross

o ed

ed

ss

L = ( + 2 )+ 2 + ( + 2

L =

L = + )

+ )+ 2

L = )

+

+ + 2

x


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A A B

dθ/2

T

TA2

µµµµ R

dθ/2

T

dθ

θ

dθ/2T

(T+δδδδT)

P

Q

B

O

1

R

Q

dθ/2

(T+δδδδT)


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dθ

θ

dθ/2

dθ/2T

T

(T+δδδδT)

T

P

Q

A

B

O

2

1

R

µµµµ R

dθ/2

T

P

Q

dθ/2

(T+δδδδT)

2

Consider a belt drive as shown in fig.

Let be the angle of contact & be the coefficient of friction.

Let & be the tensions on tight & slack sides of the belt

Consider an element PQ subtending an

θ µ

1 11 1 T TT T T

angle ' ' at the center of the

pulley such that the tensions at P & Q are & respectively.

δθ

(T + (T + (T + (T + δT )δT )δT )δT )T


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For equilibrum of the element, 0 & 0

Equating summaton of horizontal force to zero, we get

Tsin (T+ T)sin 0.2 2

Ignoring product of small quantities and putting

sin2

H V

Rδθ δθ

δ

δθ δθ

= =

+ − =

≈

∑ ∑

for small angles, we get2

dθ

θ

dθ/2

dθ/2T

T

(T+δδδδT)

T

P

Q

A

B

O

2

1

R

µµµµ R

T

0

Tcos (T+ T)cos 02 2

and putting cos 1 for small angles, we get2

V

Rδθ δθ

δ µ

δθ

δ δ µ

µ

=

⇒ − + =

≈

=

∑⋯⋯⋯

⋯⋯⋯

Similarly,Similarly,Similarly,Similarly,

T TT T T = R Or R (ii )T = R Or R (ii )T = R Or R (ii )T = R Or R (ii )

dθ/2

P

Q

dθ/2

(T+δδδδT)


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1

22 1

0

1

From (i) & (ii), we get

.

Integrating the above equation between the limits

T & T and 0 & , T T

T T

T

θ

δ µ µ

δ θ µ δθ

=

=

∫ ∫

θ θθ θ 1 11 1

T TT T δT δT δT δT T TT Tδθ = Or δθ δθ = Or δθ δθ = Or δθ δθ = Or δθ T TT T

T TT T

2

e

T

θ

=

2 22 2

Note : Note : Note : Note :

(1 )The an (1 )The an (1 )The an (1 )The an

gle mus gle mus gle mus gle mus

T TT T

t be t be t be t be

1 1 2

1 1 2

2sin

2sin

s

r r

x

r r

x

θ θ π

θ π

−

−

− = −

+ +

in radians in radians in radians in radians

(2 )For open belt drive, = (2 )For open belt drive, = (2 )For open belt drive, = (2 )For open belt drive, =

(3 )For crossed belt drive, = (3 )For crossed belt drive, = (3 )For crossed belt drive, = (3 )For crossed belt drive, =


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1 2

1 2

The net driving tension on the pulley =(T )

Torque on the pulley =(T )

where r=radius of the pulley.Power transmitted by the belt drive

T

T r

−

∴ − ×

Power Transmitted by Belt Drive :

1 2 P= 60000KW

−

1 2(T )

1000

2 where v=velocity of belt= m/sec

60 60

T vP KW

rn dnπ π

− ×⇒ =

=


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It is the ratio of speed of the driven pulley to the speed of the driving pulley.

If the driving pulley is suffixed as 1 & driven pulley as 2,

Velocity ratio Transmitted by

Velo

Belt Drive

city r

:

atio =

2 1

1 2

( neglecting slip & belt thickness)

n d

n d =

2 1

1 2

2 1

1 2

(neglecting slip &considering belt thickness)

1100

where S=Total % slip on driving & driven

n d t

n d S

n d

= +

= −

Velocity ratio =

Velocity ratio

pulleys


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& .

. .

2 1

1 2

1100

where S=Total % slip on driving & driven pulleys

n d S

n d

= −

Velocity ratio


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&

.

.

.


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• .

• , 1 & &

.

1 0 0 2

1 2 0

. . if =coefficient of expansion or contraction,

( ) ( )

( ) 2

i e

T T T T

T T T

α

α α − = −

⇒ + =

1 2 1 2 1 2 1 2 0 00 0

T +T T +T T +T T +T ∴T = ∴T = ∴T = ∴T =

2 22 2


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1. , , .

2.

.

3. B .

.

5. ,

.

1. .

2. .

3. .


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• & .

•

Grooved

Pulley

V-Belt

.

•

150 .

Tension fabric cord

Base rubber

Fabric cover


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1

, 3.

400 .

30 .

.


31/67

1

2

1

1

2

Velocity ratio = =3

Speed of driving pulley n 400 ,Diameter of driven pulley d 30

n

n

rpmcm

==

Data :

2 1 1

1 2

2 2

1

We know that 330

Also 3 3400

n d d

n d

n n

n

= ⇒ =

∴

= ⇒ =

∴

1 11 1 Diameter of driving pulley d = 90 cm Diameter of driving pulley d = 90 cm Diameter of driving pulley d = 90 cm Diameter of driving pulley d = 90 cm

Sp Sp Sp Sp

Solution :

2 22 2 eed of driven pulley n = 1200rpm eed of driven pulley n = 1200rpm eed of driven pulley n = 1200rpm eed of driven pulley n = 1200rpm


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2

1000

. 600 &

1800 ,

.


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2

2

1

1 2

1800Velocity ratio = = =3

600

Sum of diameters ( ) 1000

n

n

d d mm

n d d

+ =

Data :

1 2 2

1 2

1 2 2 2

2

e now t at

. . 3

Given ( ) 1000 (3 ) 1000

n d d

i e d d

d d d d

= ⇒ =

= ×

+ = ⇒

+ =∴Diameter of driven pulley d = 250 Diameter of driven pulley d = 250 Diameter of driven pulley d = 250 Diameter of driven pulley d = 250

o ut on :

1&

mm mm mm mm

Diameter of driving pulley d = 750 mm Diameter of driving pulley d = 750 mm Diameter of driving pulley d = 750 mm Diameter of driving pulley d = 750 mm


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3

A 200

300

. 500

.

8 & 4%.


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3

2

1

1

300Velocity ratio = = =1.5

200

Diameter of the driving pulley 500

Thickness of the belt 8 , Slip=4%

n

n

d mm

t mm

=

=

Data :

2 1

1 2

2

We know that 1100

500 8

1.5 18

n d t S n d t

d

+ = − +

+

⇒ = +

Solution :

2

4

100

−

∴Diameter of pulley on generator d = 317.12 Diameter of pulley on generator d = 317.12 Diameter of pulley on generator d = 317.12 Diameter of pulley on generator d = 317.12mm mm mm mm

4


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4

600

300

.

.

()

()

D t


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( )

1 1

2 2

Diameter of the larger pulley 600 0.3

Diameter of the smaller pulley 300 0.15

Center distance between pulleys 3

d mm r m

d mm r m

x m

= ⇒ =

= ⇒ =

=

Data :

2 22 2

1 2 1 2 1 2 1 2

open 1 2 open 1 2 open 1 2 open 1 2

Length of open belt drive Length of open belt drive Length of open belt drive Length of open belt drive

r - r r - r r - r r - r L = L = L = L = π (r + r )+ 2 x + π (r + r )+ 2 x + π (r + r )+ 2 x + π (r + r )+ 2 x + x xx x

::::

( )2 22 2

0.3 - 0.15 0.3 - 0.15 0.3 - 0.15 0.3 - 0.15 open open open open

( )

( )

∴

∴

open open open open

2 22 2

1 2 1 2 1 2 1 2

crossed 1 2 crossed 1 2 crossed 1 2 crossed 1 2

2 22 2

crossed crossed crossed crossed

crossed crossed crossed crossed

. .. .. .. .

3 33 3 L = 7.421m L = 7.421m L = 7.421m L = 7.421m

r + r r + r r + r r + r

L = L = L = L = π (r + r )+ 2 x + π (r + r )+ 2 x + π (r + r )+ 2 x + π (r + r )+ 2 x + x xx x

0.3 + 0.15 0.3 + 0.15 0.3 + 0.15 0.3 + 0.15 L = L = L = L = π (0.3 + π (0.3 + π (0.3 + π (0.3 +

Length of crossed beltLength of crossed beltLength of crossed beltLength of crossed belt

0.15 )+ (2 × 3 )0.15 )+ (2 × 3 )0.15 )+ (2 × 3 )0.15 )+ (2 × 3 )

drive :drive :drive :drive :

+ ++ + 3 33 3

L = 7.4812 m L = 7.4812 m L = 7.4812 m L = 7.4812 m

5


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5

A

1000 & 500

1500 .

0.3.

700 ,

400 .

5


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5

1

2

2

Diameter of the larger pulley 1000

Diameter of the smaller pulley 500

Center distance 1500Speed of the smaller pulley 400

d mm

d mm

x mmn rpm

=

=

==

Data :

1

Coefficient of f

1 11 2

riction =0.3

Angle of contact for open belt1000 500

= 2sin 2sin2 2 1500

d d

x

µ

θ π π − −− −

− = − ×

∴θ = 2.8067 rad ∴θ = 2.8067 rad ∴θ = 2.8067 rad ∴θ = 2.8067 rad

Solution :

5


40/67

5

1

2

. .i e

=

= ⇒ =

μθ 0.3 μθ 0.3 μθ 0.3 μθ 0.3 × 2.8067 × 2.8067 × 2.8067 × 2.8067

2 22 2

2 22 2

Ratio of tensions in the belt : Ratio of tensions in the belt : Ratio of tensions in the belt : Ratio of tensions in the belt : T TT T

= e e = 2.321 = e e = 2.321 = e e = 2.321 = e e = 2.321 T TT T

700 700 700 700 2.321 T 301.6N 2.321 T 301.6N 2.321 T 301.6N 2.321 T 301.6N

T TT T

2 2

1 2

0.5 400Velocity of the belt V=60 60

( - ) (700 -301.6) 10.4

d n V

V

P T T V

π π × ×⇒ =

∴ =

= × = ×

Power transmit Power transmit Power transmit Power transmit

1 11 1

ted by the beltted by the beltted by the beltted by the belt

0.472 0.472 0.472 0.472

: :: :

m / sec m / sec m / sec m / sec

72 4172 Watt

P

=

∴ = 4.172 KW= 4.172 KW= 4.172 KW= 4.172 KW

6


41/67

6

,

1600

& 0.28.

5 , .

Data :


42/67

0

1

Angle of contact =160 160 2.794180

Width of the belt =200 mm

Maximum tension in the belt =5N per mm width of belt

i.e. T 5 200 1000

Coefficient of friction =0.28

rad

N

π

θ

µ

= × =

= × =

Data :

1

2

1 2

. .

1000 457.24

2 2

i e

=

= ⇒ =

+ + = =

μθ 0.28 μθ 0.28 μθ 0.28 μθ 0.28 × 2.794 × 2.794 × 2.794 × 2.794

2 22 2

2 22 2

0 00 0

0 00 0

Initi Initi Initi Initi

T TT T = e e = 2.187 = e e = 2.187 = e e = 2.187 = e e = 2.187 T TT T

1000 1000 1000 1000 2.187 T 45 2.187 T 45 2.187 T 45 2.187 T 45

al tension in the al tension in the al tension in the al tension in the

7.24N 7.24N 7.24N 7.24N

T TT T

T T T T T T T T T = 7 T = 7 T = 7 T = 7

belt T belt T belt T belt T

28.62N 28.62N 28.62N 28.62N


43/67

•

.• ,

.

• ( & ) .

• & , , , .

• 2 1

1 2

1

2

1

2

, where

Speed of driving pulleySpeed of driven pulley

No of teeth on driver sprocket

No of teeth on driven sprocket

n z

n z

nn

z

z

=

==

=

=


44/67

Roller Chains


45/67

Inverted tooth Chains

Silent Chains

(So named due to quietoperation even underheavy loads)


46/67

.

:()

,

,

() ()

/ 1 /

1 /

B


47/67

()

&

&

()

()

B


48/67

•

. & .• .• B ,

.

• , , .

Driving member(Pinion)

BDriven member(Gear)

Gear

Pinion


49/67

Spur gear pair(External &

Internal)

B

Spur gearanimation

Pitch cylinderswith pure rollingfriction


50/67

• .

• , (10 0 ).

• .

• . ( ) .

B

ββββ

Driving member(Pinion)

Driven member(Gear)

Helixangle

A


51/67

A


52/67

• B

.• . .

• .

• , .

B

θθθθ


53/67

• ,

.• A .

• 60:1.

• , .. .

B

Worm wheel

Worm

shaft

&


54/67

• , & .

• .

Pinion

B

Rack & pinion Animation

Rack


55/67

•

.• ,

, & .

• , ,

, .

B

Focus center of ellipse


56/67

• &

.

• &

•

.

•

.


57/67

B

Generation of involute Profile


58/67

B

A famous curve that was named by Galileo in 1599 is called acycloid. A cycloid is the path traced out by a point on thecircumference of a circle as the circle rolls (without slipping) along

a straight line.


59/67

1. .

2. , , , ,

.3. .

1. .

2. .

3. ,

.

A A


60/67

Velocity ratio in Gear Drives : Velocity ratio in Gear Drives : Velocity ratio in Gear Drives : Velocity ratio in Gear Drives :


61/67

2 1 11 2

1 2 2

1 2

1

, where Speed of driving pulley, Speed of driven pulley

Pitch circle diameter (PCD) of driver gear, PCD of of driven gear

No of teeth on driv

n d z

n nn d z

d d

z

= = = =

= =

= 2er gear, No of teeth on driven gear z =

A

& .

. .

, .


62/67

A ;

1.

2.

.

A A • A simple gear train is one in whicheach shaft carries only one gear


63/67

A B C

each shaft carries only one gear.

• From the fig, gear A is the drivinggear and gear D is the driven gear.B & C are the intermediate gears orIdler gears.

• The idler gears do not affect thevelocity ratio but simply bridge thegap between the driver & the drivengears.

A

Velocity ratio C A

A C

N z

N z

=

• Also if odd number of intermediategears are used, the driver & thedriven gears rotate in the same direction .

• If even number of intermediate

gears are used, the driver & thedriven gears rotate in the opposite directions .

B

D

C

Velocity ratio D A

A D

N z

N z

=


64/67

From the fig, and also

The speed of gear C,

C B A B

A B B C

C B A B

A B B C

N N z z

N z N z

N N z z

N N z z

= =

∴ × = × ⇒

= ⇒ =

AAAAC CC C

C CC C

AAAA

AAAA

A C A C A C A C

C CC C

C CC C

z zz z 30 30 30 30 N × N N × 90 = 180N × N N × 90 = 180N × N N × 90 = 180N × N N × 90 = 180

z 15 z 15 z 15 z 15

N NN N z zz z = == = N z N z N z N z

rpm.rpm.rpm.rpm.

A A • In a compound gear train the intermediate shaft carries two


65/67

intermediate shaft carries two

or more gears which are keyedto it.

• Compound gears are used whena high velocity ratio is required ina limited space.

• The intermediate gears will havean effect on the overall velocityratio.

A

B

C

D

From the fig, and also

As gears B & C are on same shaft, N

Speed of l. .

C B A D

A B C D

C B D A

A C B D

B C

z N z N N z N z

z N N z

N N z z

N

i e

= =

×

∴ × = ×

=

×

⇒ A C A C A C A C D DD D

A B D A B D A B D A B D

z z z z z z z z N NN N = == =

N z z N z z N z z N z z

ast driven shaft Product of no of teeth on driver

Speed of the first driving shaft Product of no of teeth on driven

=

• A reverted gear train is acompound gear train in


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compound gear train in

which the first & the lastgears are on the same axis.

• Hence, in a reverted gear

train, the center distancesfor the two gear pairs mustbe same.

AD

• everte gear tra ns are

used in automotivetransmissions, lathe backgears, and in clocks.

C

ddAs ,2 2

But d=mz, and the module 'm' is same for all gears,

z

C D A B

A B C D

d d

z z z

++ =

∴ + = +

• An epicyclic gear train isone in which the axis of one


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one in which the axis of one

or more gears movesrelative to the frame.

• Large speed reductions are

obtained with an epicyclictrain.

• They are compact in sizean automo e erent a .

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Transmission of Motion & Power3