unit 2b power transmission by belts

UNIT 2 b – Power Transmission by belts

By

Dr. S. Murali

Professor & Head

Department of Mechanical Engineering

J k h N C ll f E i iJayaprakash Narayan College of Engineering

Dharmapur, Mahabubnagar-509001

Telangana IndiaTelangana, India

Introduction• Rotating elements which possess mechanical

h b ili d i d l benergy has to be utilized at required place by transmitting.– From prime mover to machine– From one shaft to anotherFrom one shaft to another

10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.2

Transmission system and Typesy yp• The system that is used to transmit power

f h i l l hfrom one mechanical element to another mechanical element

• TypesBelt drives– Belt drives

– Rope drives– Chain drives– Gear drives


Factors to select transmission systemy• Distance between driver and driven pulley

h fshaft.• Operational speed.p p• Power to be transmitted


Belt drives• Power is to be transmitted between the parallel

shaftshaft.• Consists of two pulleys over which a endless belt is

passed encircling the bothpassed encircling the both.• Rotary motion is transmitted from driving pulley to

driven pulleydriven pulley.


Terminology of a belt drive gy• Driver : in a transmission system the one which

drives or supplies power to other mechanicaldrives or supplies power to other mechanical element.

• Driven : in a transmission system the one which yfollows the driver or receives power from driver.

• Tight side : the portion of the belt in maximum g ptension. Denoted by T1 Newton.

• Slack side : the portion of the belt in minimum tension. Denoted by T2 Newton

• Arc / angle of contact : it is the portion of the belt h h h ll f dwhich is in contact with pulley surface. Denoted

by


Belt material• Rubber• Leather • CanvasCanvas • Cotton • Steel


Power transmission systemsyIn factories, the power or rotary motion, from

h f h id blone shaft to another at a considerabledistance is, usually, transmitted by means ofbelts, ropes and gears.


Belt Drives


Belt Drives

Drilling machine with speed cone pulleys A open belt drive in a jig-saw

machine Lathe machine withmachine Lathe machine with speed cones and timing belt


Types of Beltsyp

(a) Flat Belt (b) V Belt ( ) Ci l B lt(a) Flat Belt (b) V-Belt (c) Circular BeltModerate amount ofpower two pulleys

Great amount ofpower two pulleys

Great amount ofpower two pulleyspower two pulleys

are not more than10 m apart.

power two pulleysare very nearer toeach other.

power two pulleysnot more than 5 mapart.


10 m apart. each other. apart.

Types of Belt drivesyp

Open Belt Drive Cross Belt Drive

• two pulleys rotate in the samedirection

• Length of the belt is smaller

• pulleys rotate in the oppositedirections

• Length of the belt is largerLength of the belt is smaller• Angle of lap is different for driver

and driven pulley

Length of the belt is larger• Angle of lap is same for driver

and driven pulley


Velocity Ratio(VR)y ( )It is the ratio between the velocities of the driver and the driven (follower)

L d d DiVb= V1

Let d1,d2 -Diameters, md1

d2N1N2

N1, N2 - Speeds, rpm

V V Vb -Velocities m/min

Length of the belt length covered⎧ ⎫ ⎧ ⎫

For no slip Vb= V2

V1 ,V2 ,Vb -Velocities, m/min

Length of the belt length covered passing over driver pulley by a point on driver pulley

inone minute in one minute=

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

Length of the belt length covered ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎨ ⎬ ⎨ ⎬

1 1bV d Nπ=

2 2b

passing over driven pulley by a point on driven pulley inone minute in one minute

=

V d Nπ

⎪ ⎪ ⎪ ⎪⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

=


2 2bV d Nπ

Velocity Ratio(Contd..,)y ( )Length of the belt Length of the belt

passing over driver pulley passing over d= riven pulley⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪

1 1 2 2bV d N d NN dSpeed of Driven

inone minute inone minuteπ π

⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

= =

2 1

1

, N dSpeed of DrivenVelocity Ratio VRSpeed of Driver N

= = =2d

1N∴ ∝ Vb= V1Nd

∴ ∝

From the equation it is obvious that i i l b l d i if di f

d1d2N1

N2

b 1

in simple belt drive if diameter of the pulley decrease Speed will increases and vice versa.

Vb= V2

If thickness(t) of the belt is considered

2 1N d tVRN d t

+= =+


1 2N d t+

Prob: RS Khurmi/Example 33.1/page 672p p g• It is required to drive a shaft at 720 revolutions per minute, by means of a

belt from a parallel shaft, having a pulley A 300 mm diameter on it andrunning at 240 revolutions per minute. What sized pulley is required onthe shaft B ?

Given: Speed of the driver N = 620 rpm ; Case (2):Given: Speed of the driver, N2 = 620 rpm.; Diameter of pulley,d1 = 300 mm and speed of the pulley, N1 = 240 rpm.

Case (2):N2 = 1440 rpm.; d1 = 300 mm and N 240Find Diameter of the follower,d2

Solution:Velocity Ratio VR= 2 1 720 2 58N d= = =

N1 =240 rpm.

2 1 1440 6N d= = =Velocity Ratio, VR= 2 1

1 2

2.58240N d

= = =

12

300dd⇒ = =

2 1

1 2

6240N d

= = =

1 300dd⇒ = =2 3VR

2 100d mm= 2 50d mm=

2 6d

VR⇒


2 2

Limitations of simple belt drivep• Higher Velocity ratios ratios, lessen the arc of

i li d l fcontact, causing slippage and loss of power.• For maximum power transfer on the belts p

and pulleys, the pulley ratio should be 3 to 1 or lessor less

• To avoid excessive single-step ratios or undersize pulleys. – The central distance can be increased– Compound drive /two-step drive /counter-shaft)

can be used tocan be used to10-02-2015 Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCE 1.16

Compound belt drivep• power is transmitted, from one shaft to another, through a

number of pulleysnumber of pulleys• Consists of more number

of simple belt drivesp• Pulleys 2 and 3 are keyed

to the same shat• Pulley 1 and 2 forms one

belt drive and Pulley 3 and 4 another


Compound belt drivepLet d1,d2, d3, d4, and N1,N2, N3, N4= diameters and speeds for pulleys 1, 2, 3 and 4.

N dWe know that velocity ratio of pulleys 1 and 2,

2 11

1 2

N dVRN d

= =

Similarly, velocity ratio of pulleys 3 and 4,

---------------- (1)

y, y p y ,

342

3 4

dNVRN d

= = ---------------- (2)

Multiplying the above equations

1 32 41 2

d dN NVR VRN N d d

××× = =× × ---------------- (3)

1 3 2 4N N d d× ×Since N2=N3, therefore velocity ratio of compound belt drive

1 341 2

d dNVR VR VRN d d

×= × = =×

---------------- (4)1 2 4N d d×

Speed of last follower product of diameters of driversVRSpeed of first driver product of diameters of followers

= =


Speed of first driver product of diameters of followers

Problem:It is required to drive a shaft at 1440 revolutions per minute, by means of compoundbelt from a parallel shaft, having drivers 300 mm diameter and first driver is running

t 240 l ti i t Wh t i d ll i d f f ll ?at 240 revolutions per minute. What sized pulley are required for followers ?Given: Speed of the driver, N4 = 620 rpm.; 4Diameter of pulleys,d1 = d3 =300 mm and speed of the pulley, N1 = 240 rpm.Find Diameter of the follower d and dFind Diameter of the follower,d2 and d4

Solution:Velocity Ratio compound drive, For Case 1

1 341 2

1 2 4

1440 6240

d dNVR VR VRN d d

×= × = = = =×

12

1

300 1502

dd mmVR

= = =

1 2

6 2 3 13 2 2

VR VRVR Case

Case

×⎧⎪= × ←⎨⎪ × ←⎩

34

2

300 1003

dd mmVR

= = =


3 2 2Case⎪ × ←⎩

Slipp• If there is no firm frictional grip between the belts and the shafts. This may cause

forward motion of the driver pulley without carrying the belt with it or forward motion of the belt without carrying the driven pulley with it. This is called slip of the belt, and is generally expressed as a percentage and denoted by S.

V V

Vb

V1 Slip at driver,S1,Expressed in % of driver speed,1

11% 100bV VV

S −= ×

d1

d2NN2

2N1

V

Vb

V2Slip at driven,S2,Expressed in % of belt speed

22 % 100bV V

VS −= ×


2bV

Slipp

11

1 100bV VV

S −= ×

⎛ ⎞ N l ti 1 2S S×1

1 1100bV V S⎛ ⎞= −⎜ ⎟

⎝ ⎠---------------- (1)

V V

Neglecting

12 1 21100

dd

SNN

S+⎛ ⎞= −⎜ ⎟⎝ ⎠

1 2

10000S S×

---------------- (2)

2

2

2

2

% 100

1

b

b

b

VS VV

V SV

−= ×

⎛ ⎞= −⎜ ⎟

1 2 100dN ⎜ ⎟⎝ ⎠

Assuming S1+S2=S N d S⎛ ⎞(2)2 1

100bV V ⎜ ⎟⎝ ⎠

S b tit ti th l V i (2)

2 1

1 2

1100

N dN

Sd

⎛ ⎞= −⎜ ⎟⎝ ⎠

Substituting the value Vb in eq (2)

12 1

21 1100 100S SV V ⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

If thickness(t) of the belt is considered2 1

1 2

1100

N d tN d t

S+ ⎛ ⎞= −⎜ ⎟+ ⎝ ⎠

2 2 1 11 2 1 21

100 100 10000S SN d SN Sdπ π ⎛ ⎞= ×− − +⎜ ⎟

⎝ ⎠


Length of an Open Belt Driveg pFrom geometry O1M= O1E- ME= r1- r2

From right angled triangle O1MO2g g g 1 2

1 2sin r rx

α −=

For small values of α, sinα=α

1 2r r− (1)1 2

xα =

Arc length JE= 1 2r π α⎛ ⎞+⎜ ⎟⎝ ⎠

---------------- (1)

---------------- (2)g2⎜ ⎟

⎝ ⎠

(∵ l=rθ )

(2)

Arc length FK= 2 2r π α⎛ ⎞−⎜ ⎟⎝ ⎠

---------------- (3)


Length of an Open Belt Driveg p( )22

2 1 2MO EF x r r= = − − (pythagoras theorem)2r r⎛ ⎞1 21 r rx

x−⎛ ⎞= − ⎜ ⎟

⎝ ⎠2 4

1 2 1 21 11 r r r rx⎛ ⎞− −⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

(Binomial expansion)

21 211 r r⎛ ⎞−⎛ ⎞

⎜ ⎟⎜ ⎟ (neglecting higher order terms)

12 8

xx x

+⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠( p )

( )21 2

212

r rMO EF x

−= = −

1 212

xx

= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠(neglecting higher order terms)

---------------- (4)2 2 x

Length of the belt =Length of arc GJE + EF + Length of arc FKH + HG

L=2 (Length of arc JE + EF + Length of arc FK)


Length of an Open Belt Driveg pSubstituting the values of length of arc JE from equation (2), length of arc FK from equation (3) and EF from equation (4) in this equation,q ( ) q ( ) q ,

( )21 2

1 212

2 2 2r r

L r x rx

π πα α⎡ ⎤⎛ ⎞−⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦

( ) ( ) ( )2

1 22 2r r

r r x r rπ α−

+ + +( ) ( )1 2 1 22 2r r x r rx

π α= + + − + −

( ) ( ) ( ) ( )21 2 1 2r r r r− −

( ) ( ) 21 22r r

L+

( ) ( ) ( ) ( )1 2 1 21 2 1 22 2r r x r r

x xπ= + + − + −

( ) ( )1 21 2 2L r r x

xπ= + + +


Length of an Cross Belt DrivegFrom geometry O1M= O1E- ME= r1+r2

From right angled triangle O1MO2g g g 1 2

1 2sin r rx

α +=

For small values of α, sinα=α

1 2r rα −= ---------------- (1)x

Arc length JE= 1 2r π α⎛ ⎞+⎜ ⎟⎝ ⎠

(1)

---------------- (2)2⎜ ⎟⎝ ⎠

(∵ l=rθ )

(2)

Arc length FK= 2 2r π α⎛ ⎞+⎜ ⎟⎝ ⎠

---------------- (3)


Length of an Cross Belt Driveg( )22

2 1 2MO EF x r r= = − + (pythagoras theorem)2r r+⎛ ⎞1 21 r rx

x+⎛ ⎞= − ⎜ ⎟

⎝ ⎠2 4

1 2 1 21 11 r r r rx⎛ ⎞+ +⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

(Binomial expansion)

21 211 r r⎛ ⎞+⎛ ⎞

⎜ ⎟⎜ ⎟ (neglecting higher order terms)

12 8

xx x

+⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠( p )

( )21 2

212

r rMO EF x

+= = −

1 212

xx

= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠(neglecting higher order terms)

---------------- (4)2 2 x

L th f th b lt L th f GJE EFLength of the belt =Length of arc GJE + EF + Length of arc FKH + HG

L=2 (Length of arc JE + EF + Length of arc FK)


Length of an Cross Belt DrivegSubstituting the values of length of arc JE from equation (2), length of arc FK from equation (3) and EF from equation (4) in this equation,q ( ) q ( ) q ,

( )21 2

1 212

2 2 2r r

L r x rx

π πα α⎡ ⎤⎛ ⎞+⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦

( ) ( ) ( )2

1 22 2r r

r r x r rπ α+

+ + + +( ) ( )1 2 1 22 2r r x r rx

π α= + + − + +

( ) ( ) ( ) ( )21 2 1 2r r r r+ +

( ) ( ) 21 22r r

L+

( ) ( ) ( ) ( )1 2 1 21 2 1 22 2r r x r r

x xπ= + + − + +

( ) ( )1 21 2 2L r r x

xπ= + + +


Prob:RS Khurmi/Example 33.4/679 pFind the length of belt necessary to drive a pulley of 500 mm diameter running parallel at a distance of 12 meters from the driving pulley of diameter 1600 mm.

Given: Diameter of the driven pulley (d2) = 500 mm = 0·5 m or radius (r2) = 0·25 m;Diameter of the driving pulley (d ) 1600 mm 1 6 m or radius (r ) 0 8 mDiameter of the driving pulley (d1) = 1600 mm = 1·6 m or radius (r1) = 0·8 m.Distance between the centres of the two pulleys (x) = 12 m

Find length of the belt for 1. open belt drive 2. Cross belt drive Find length of the belt for . open belt drive . Cross belt driveSolution:1. Open Belt drive

( ) ( )2 20 8 0 25( ) ( ) ( ) ( )1 21 2

0.8 0.252 0.8 0.25 2 12 27.32

r rL r r x m

x xπ π

− −= + + + = + + × + =

2 C B lt d i2. Cross Belt drive

( ) ( ) ( ) ( )2 21 2

1 2

0.8 0.252 0.8 0.25 2 12 27.39

r rL r r x m

x xπ π

+ += + + + = + + × + =


x x

Open belt drive Vs Close belt driveItem Open Belt Drive Closed Belt Drive

Arrangement

Rotation of pulleys same direction opposite direction

useful alignment of shafts horizontal or inclined horizontal or inclined or verticalvertical

rubbing no rubbing point, the life of the belt is more

rubbing point, the life of the belt reduces.

L th f th b ltLength of the belt(same centre distance, pulley diameters.)

less length Require more length

A l fAngle of contact(big and small pullys) different Same


Power transmitted by a belt• the effective turning (i.e. driving) force at the circumference of the

follower isF T T

y

F=T1 – T2.• Work done per second

= Force × Distance = T1 – T2 × v( ) /Power = (T1 – T2)v J/s

2 11 1TP T T⎛ ⎞

⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟

21 1

112

1 1

1

P T v T vTTT

⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎜ ⎟⎝ ⎠

⎛ ⎞1

11P T veμθ

⎛ ⎞= −⎜ ⎟⎝ ⎠

Notes:1. The torque exerted on the driving pulley =(T1 – T2) r12. Similarly, the torque exerted on the follower = (T1 – T2) r2

where r1 and r2 are in metres.


1 2

Centrifugal tensiong• The belt continuously runs over both the pulleys.

It carries some centrifugal force in the belt atIt carries some centrifugal force in the belt, at both the pulleys,

• whose effect is to increase the tension on both• whose effect is to increase the tension on both, tight as well as the slack sides. Th t i d b th t if l f i• The tension caused by the centrifugal force is called centrifugal tension.

• At lower speeds, the centrifugal tension is very small and may be neglected.

• But at higher speeds, its effect is considerable, and thus should be taken into account.


Centrifugal tensiongm = Mass of the belt per unit length,v = Linear velocity of the belt,y ,r = Radius of the pulley over which the belt runs,TC = Centrifugal tension acting tangentially at P and Q anddθ = Angle subtended by the belt AB at the centre of the pulley.

Length of belt AB = r dθl f h b l dθtotal mass of the belt M = mr dθ

2 2

CMv mrd vP

r rθ= =We know that centrifugal force of the belt AB,

2

Now resolving the forces (i.e., centrifugal force and centrifugal tension) horizontally and equating the same

2CP md vθ=

and equating the same,

22 sin2C

dT md vθ θ=


Centrifugal tensiongFor small values of ,sin

2 2 2d d dθ θ θ=

22 sin2C

dT md vθ θ=

2TCT mv=

Notes: 1. When the centrifugal tension is taken into account,

the total tension in the tight side = T1 + TC and total tension in the slack side = T2 + TC

2 Th t if l t i th b lt h ff t th2. The centrifugal tension on the belt has no effect on the power transmitted by it. The reason for the same is that while calculating the power transmitted, we have to use the values :

Total tension in tight side Total tension in the slack side= Total tension in tight side – Total tension in the slack side= (T1 + TC) – (T2 + TC) = (T1 – T2).


Maximum tension in the belt• Let σ = Maximum safe stress in the belt, N/mm2

b = Width of the belt in mm andb = Width of the belt in mm, andt = Thickness of the belt in mm.

• We know that maximum tension in the belt• We know that maximum tension in the belt,T = Maximum stress × Cross-sectional area of

beltbelt= σ bt

Wh t if l t i i l t d th• When centrifugal tension is neglected, then maximum tension, T = T1 and

• when centrifugal tension is considered then• when centrifugal tension is considered, then maximum tension, T = T1 + TC


Condition for transmission of maximum power

• that the power transmitted by a belt,

111

1

P T veμθ

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞1

1assuming 1P T Cv Ceμθ

⎛ ⎞⎛ ⎞= = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

B t T T T and T m 2But T1=T-Tc and Tc=mv2

∴P= (Tv-mv3)CDifferentiating w.r.t v and equating to zero

( )3dP d Tv mv= − 3T T=( )2

2

0 3

Tv mvdv dv

T mv= −

3 cT T=

It shows that when the power transmitted is maximum


23T mv= 1/3 rd of the maximum tension is absorbed as centrifugal tension

Belt speed for maximum powerp pfor maximum power transmission, 23T mv=

Speed of the belt for maximum transmission of power

3Tvm

=3m

Note: Maximum tension in the belt is equal to sum of tensions in tight side (T1) and centrifugal tension (TC).


Initial tension in the belt• the motion of the belt (from the driver) and the follower

(from the belt) is governed by a firm grip due to friction(from the belt) is governed by a firm grip due to friction between the belt and the pulleys, therefore the belt is tightened up, in order to keep a proper grip of the belt over the pulleys. Initially, even when the pulleys are stationary the belt is subject to some tension, called initial tension.

Let T0 = Initial tension in the belt,T1 = Tension in the tight side of the belt,T2 = Tension in the slack side of the belt, andn = Coefficient of increase of the belt length per unit force.


Initial tension in the beltthat increase of tension in the tight side = T1 – T0

increase in the length of the belt on the tight sideincrease in the length of the belt on the tight side= n (T1 – T0)Similarly, decrease in tension in the slack side = T0 – T2Similarly, decrease in tension in the slack side T0 T2

decrease in the length of the belt on the slack side= n (T0 – T2)( 0 2)For constant length of the belt, when it is at rest or in motion, The increase in length on the tight side = equal to decrease in the length on the slack sideTherefore, equating

(T T ) (T T ) Note: If centrifugal tension is takenn (T1 – T0) = n (T0 – T2)∴ T=(T1+T2)/2

Note: If centrifugal tension is taken into consideration, then T=(T1+T2)/2+Tc


References1. R.S. Khurmi & J.K . gupta,2005, “a text book

f hi d i ” fi l i l di iof machine design”, first multicolor edition, eurasia publishing house (pvt.) Ltd.

2. K.L. Narayana, P. Kannaiah & K. Venkatreddy 2006 ”machine drawing” 3rd editionreddy,2006, machine drawing , 3 edition, new age international (p) limited, publishers

3. wikipedia.org


Contact:[email protected]


THE BINOMIAL THEOREM(a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3

n+ nC4 an-4 x4 + ... + nCn xn = kknn

kk

n xaC −

=∑

0 nk 0(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ... + nCn xn =

kn

kk

n xC∑=0!n n

( )

( )

!! !

! ! 0! 1 1

nk

n

nCn k k

n nC

=−

= = = =∵( ) ( )

( )( )

( )1

0! 1 10 !0! ! 1

1 !!1 !1! 1 !

o

n

Cn n

n nnC n

= = = =− ×

× −= = =

∵

( ) ( )

( )( ) ( )( )

( )

1

1

1 !1! 1 !

1 2 ! 1!2 !1! 2 ! 2! 2!

n

n n

n n n n nnCn n

− −

× − × − × −= = =

− − ×( ) ( )2 !1! 2 ! 2! 2!n n ×

1010--0202--20152015 1.1.4141Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCEDr. S. Muali, Prof & Head, Department of Mechanical, JPNCE

Pythagoras theorem (Sulba Sutra)"In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs"

relationship can be stated as:

Pythagoras of SamosBAUDHAYANA(800BCE) Pythagoras of Samos(Greek: . 570 BC – c. 495BC)

Ionian Greek philosopher and mathematician

U (800 C )

“dīrghasyākṣaṇayā rajjuH pārśvamānī, tiryaDaM mānī, cha yatpṛthagbhUteg y ṣ ṇ y jj p , y , y pṛ gkurutastadubhayāṅ karoti.”There are certain sets of numbers that have a very special property. Not only do thesenumbers satisfy the Pythagorean Theorem, but any multiples of these numbers also satisfyfy y g , y p f fythe Pythagorean Theorem (numbers 3, 4, and 5)

1010--0202--20152015 1.1.4242Dr. S. Muali, Prof & Head, Department of Mechanical, JPNCEDr. S. Muali, Prof & Head, Department of Mechanical, JPNCE

unit 2b power transmission by belts

Engineering

belt drives power

belt drives10

endless belt

prof head

belt drivesdrilling

b power transmission

belt h h h ll f dwhich

belt drive gy driver