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
• Types
Belt 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 drivegy
• 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 whichy
follows the driver or receives power from driver.
• Tight side : the portion of the belt in maximumg p
tension. 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 systemsy
In factories, the power or rotary motion, from
h f h id blone shaft to another at a considerable
distance is, usually, transmitted by means of
belts, 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 Belt
Moderate amount of
power two pulleys
Great amount of
power two pulleys
Great amount of
power two pulleyspower two pulleys
are not more than
10 m apart.
power two pulleys
are very nearer to
each other.
power two pulleys
not more than 5 m
apart.
10 m apart. each other. apart.

Types of Belt drivesyp
Open Belt Drive Cross Belt Drive
• two pulleys rotate in the same
direction
• Length of the belt is smaller
• pulleys rotate in the opposite
directions
• 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 Di
Vb= V1
Let d1,d2 -Diameters, m
d1
d2N1
N2
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 oneminute
=
⎧ ⎫ ⎧ ⎫
⎪ ⎪ ⎪ ⎪
⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪
⎩ ⎭ ⎩ ⎭
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 oneminute
=
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 N
N dSpeed of Driven
inone minute inone minute
π π
⎪ ⎪ ⎪ ⎪
⎩ ⎭ ⎩ ⎭
= =
2 1
1
,
N dSpeed of Driven
Velocity Ratio VR
Speed of Driver N
= = =
2d
1
N∴ ∝ Vb= V1N
d
∴ ∝
From the equation it is obvious that
i i l b l d i if di f
d1
d2N1
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 t
VR
N 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 and
running at 240 revolutions per minute. What sized pulley is required on
the 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 58
N d
= = =
N1 =240 rpm.
2 1 1440
6
N d
= = =Velocity Ratio, VR= 2 1
1 2
2.58
240N d
= = =
1
2
300d
d⇒ = =
2 1
1 2
6
240N d
= = =
1 300d
d⇒ = =2
3VR
2 100d mm= 2 50d mm=
2
6
d
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 beltsp
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 to

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 drivep
Let d1,d2, d3, d4, and N1,N2, N3, N4= diameters and speeds for pulleys 1, 2, 3 and 4.
N d
We know that velocity ratio of pulleys 1 and 2,
2 1
1
1 2
N d
VR
N d
= =
Similarly, velocity ratio of pulleys 3 and 4,
---------------- (1)
y, y p y ,
34
2
3 4
dN
VR
N d
= = ---------------- (2)
Multiplying the above equations
1 32 4
1 2
d dN N
VR VR
N N d d
××
× = =
× × ---------------- (3)
1 3 2 4N N d d× ×
Since N2=N3, therefore velocity ratio of compound belt drive
1 34
1 2
d dN
VR VR VR
N d d
×
= × = =
×
---------------- (4)
1 2 4N d d×
Speed of last follower product of diametersof drivers
VR
Speed of first driver product of diametersof followers
= =
Speed of first driver product of diametersof followers

Problem:
It is required to drive a shaft at 1440 revolutions per minute, by means of compound
belt 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.;4
Diameter 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 34
1 2
1 2 4
1440
6
240
d dN
VR VR VR
N d d
×
= × = = = =
×
1
2
1
300
150
2
d
d mm
VR
= = =
1 2
6 2 3 1
3 2 2
VR VR
VR Case
Case
×⎧
⎪
= × ←⎨
⎪ × ←⎩
3
4
2
300
100
3
d
d mm
VR
= = =
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
1
1
% 100bV V
V
S
−
= ×
d1
d2N
N2
2N1
V
Vb
V2
Slip at driven,S2,Expressed in % of belt speed
2
2 % 100bV V
V
S
−
= ×
2
bV

Slipp
1
1
1
100bV V
V
S
−
= ×
⎛ ⎞ N l ti 1 2S S×
1
1 1
100
bV V
S⎛ ⎞
= −⎜ ⎟
⎝ ⎠
---------------- (1)
V V
Neglecting
12 1 2
1
100
d
d
SN
N
S+⎛ ⎞
= −⎜ ⎟
⎝ ⎠
1 2
10000
S S×
---------------- (2)
2
2
2
2
% 100
1
b
b
b
V
S
V
V
V
S
V
−
= ×
⎛ ⎞
= −⎜ ⎟
1 2 100dN
⎜ ⎟
⎝ ⎠
Assuming S1+S2=S
N d S⎛ ⎞
(2)2 1
100
bV V ⎜ ⎟
⎝ ⎠
S b tit ti th l V i (2)
2 1
1 2
1
100
N d
N
S
d
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
Substituting the value Vb in eq (2)
1
2 1
2
1 1
100 100
S S
V V
⎛ ⎞⎛ ⎞
= − −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
If thickness(t) of the belt is considered
2 1
1 2
1
100
N d t
N d t
S+ ⎛ ⎞
= −⎜ ⎟
+ ⎝ ⎠
2 2 1 1
1 2 1 2
1
100 100 10000
S S
N d
S
N
S
dπ π
⎛ ⎞
=
×
− − +⎜ ⎟
⎝ ⎠

Length of an Open Belt Driveg p
From geometry O1M= O1E- ME= r1- r2
From right angled triangle O1MO2g g g 1 2
1 2
sin
r r
x
α
−
=
For small values of α,
sinα=α
1 2r r−
(1)1 2
x
α =
Arc length JE= 1
2
r
π
α
⎛ ⎞
+⎜ ⎟
⎝ ⎠
---------------- (1)
---------------- (2)g
2
⎜ ⎟
⎝ ⎠
(∵ l=rθ )
(2)
Arc length FK= 2
2
r
π
α
⎛ ⎞
−⎜ ⎟
⎝ ⎠
---------------- (3)

( )
22
2 1 2MO EF x r r= = − − (pythagoras theorem)
2
r r⎛ ⎞1 2
1
r r
x
x
−⎛ ⎞
= −⎜ ⎟
⎝ ⎠
2 4
1 2 1 21 1
1
r r r r
x
⎛ ⎞− −⎛ ⎞ ⎛ ⎞
= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
(Binomial expansion)
2
1 21
1
r r⎛ ⎞−⎛ ⎞
⎜ ⎟⎜ ⎟ (neglecting higher order terms)
1
2 8
x
x x
+⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
( p )
( )
2
1 2
2
1
2
r r
MO EF x
−
= = −
1 2
1
2
x
x
= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(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)

Substituting 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 ,
( )
2
1 2
1 2
1
2
2 2 2
r r
L r x r
x
π π
α α
⎡ ⎤⎛ ⎞−⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦
( )
( )
( )
2
1 2
2 2
r r
r r x r rπ α
−
+ + +( ) ( )1 2 1 22 2r r x r r
x
π α= + + − + −
( )
( )
( )
( )
2
1 2 1 2r r r r− −
( )
( )
2
1 2
2
r r
L
+
( )
( )
( )
( )1 2 1 2
1 2 1 22 2r r x r r
x x
π= + + − + −
( )
( )1 2
1 2 2L r r x
x
π= + + +

Length of an Cross Belt Driveg
From geometry O1M= O1E- ME= r1+r2
From right angled triangle O1MO2g g g 1 2
1 2
sin
r r
x
α
+
=
For small values of α, sinα=α
1 2r r
α
−
= ---------------- (1)
x
Arc length JE= 1
2
r
π
α
⎛ ⎞
+⎜ ⎟
⎝ ⎠
(1)
---------------- (2)2
⎜ ⎟
⎝ ⎠
(∵ l=rθ )
(2)
Arc length FK= 2
2
r
π
α
⎛ ⎞
+⎜ ⎟
⎝ ⎠
---------------- (3)

( )
22
2 1 2MO EF x r r= = − + (pythagoras theorem)
2
r r+⎛ ⎞1 2
1
r r
x
x
+⎛ ⎞
= −⎜ ⎟
⎝ ⎠
2 4
1 2 1 21 1
1
r r r r
x
⎛ ⎞+ +⎛ ⎞ ⎛ ⎞
= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
(Binomial expansion)
2
1 21
1
r r⎛ ⎞+⎛ ⎞
⎜ ⎟⎜ ⎟ (neglecting higher order terms)
1
2 8
x
x x
+⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
( p )
( )
2
1 2
2
1
2
r r
MO EF x
+
= = −
1 2
1
2
x
x
= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
(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)

Substituting 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 ,
( )
2
1 2
1 2
1
2
2 2 2
r r
L r x r
x
π π
α α
⎡ ⎤⎛ ⎞+⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟= + + − + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦
( )
( )
( )
2
1 2
2 2
r r
r r x r rπ α
+
+ + + +( ) ( )1 2 1 22 2r r x r r
x
π α= + + − + +
( )
( )
( )
( )
2
1 2 1 2r r r r+ +
( )
( )
2
1 2
2
r r
L
+
( )
( )
( )
( )1 2 1 2
1 2 1 22 2r r x r r
x x
π= + + − + +
( )
( )1 2
1 2 2L r r x
x
π= + + +

Prob:RS Khurmi/Example 33.4/679p
Find 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 driveFind length of the belt for . open belt drive . Cross belt drive
Solution:
1. Open Belt drive
( ) ( )
2 2
0 8 0 25
( )
( )
( )
( )1 2
1 2
0.8 0.25
2 0.8 0.25 2 12 27.32
r r
L r r x m
x x
π π
− −
= + + + = + + × + =
2 C B lt d i2. Cross Belt drive
( )
( )
( )
( )
2 2
1 2
1 2
0.8 0.25
2 0.8 0.25 2 12 27.39
r r
L r r x m
x x
π π
+ +
= + + + = + + × + =
x x

Open belt drive Vs Close belt drive
Item 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 is
F T T
y
F=T1 – T2.
• Work done per second
= Force × Distance = T1 – T2 × v
( ) /Power = (T1 – T2)v J/s
2 1
1 1
T
P T T
⎛ ⎞
⎛ ⎞ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
2
1 1
11
2
1 1
1
P T v T v
TT
T
⎜ ⎟
= − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎜ ⎟
⎝ ⎠
⎛ ⎞
1
1
1P T v
eμθ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
Notes:
1. The torque exerted on the driving pulley =(T1 – T2) r1
2. 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.

m = 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 and
dθ = Angle subtended by the belt AB at the centre of the pulley.
Length of belt AB = r dθ
l f h b l dθtotal mass of the belt M = mr dθ
2 2
C
Mv mrd v
P
r r
θ
= =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
2
CP md vθ=
and equating the same,
2
2 sin
2
C
d
T md v
θ
θ=

For small values of ,sin
2 2 2
d d dθ θ θ
=
2
2 sin
2
C
d
T md v
θ
θ=
2
TCT 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, and
t = 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,
1
1
1
1
P T v
eμθ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
⎛ ⎞⎛ ⎞
1
1
assuming 1P TCv C
eμθ
⎛ ⎞⎛ ⎞
= = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
B t T T T and T m 2But T1=T-Tc and Tc=mv2
∴P= (Tv-mv3)C
Differentiating w.r.t v and equating to zero
( )3dP d
Tv mv= − 3T T=( )
2
2
0 3
Tv mv
dv dv
T mv= −
3 cT T=
It shows that when the power transmitted is maximum
2
3T mv= 1/3 rd of the maximum tension is absorbed as centrifugal tension

Belt speed for maximum powerp p
for maximum power transmission,
2
3T mv=
Speed of the belt for maximum transmission of power
3
T
v
m
=
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, and
n = Coefficient of increase of the belt length per unit force.

Initial tension in the belt
that 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 side
Therefore, 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

References
1. 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. Venkat
reddy 2006 ”machine drawing” 3rd editionreddy,2006, machine drawing , 3 edition,
new age international (p) limited, publishers
3. wikipedia.org

Contact:
smuralichinna@gmail.com

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 = kkn
n
k
k
n
xaC −
=
∑
0 nk 0
(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ... + nCn xn =
k
n
k
k
n
xC∑
=0!n n
( )
( )
!
! !
! !
0! 1 1
n
k
n
n
C
n k k
n n
C
=
−
= = = =∵
( )
( )
( )
( )
( )1
0! 1 1
0 !0! ! 1
1 !!
1 !1! 1 !
o
n
C
n n
n nn
C n
= = = =
− ×
× −
= = =
∵
( ) ( )
( )
( ) ( )
( )
( )
1
1
1 !1! 1 !
1 2 ! 1!
2 !1! 2 ! 2! 2!
n
n n
n n n n nn
C
n 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 Samos
BAUDHAYANA(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 g
kurutastadubhayāṅ karoti.”
There are certain sets of numbers that have a very special property. Not only do these
numbers satisfy the Pythagorean Theorem, but any multiples of these numbers also satisfyfy y g , y p f fy
the 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

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Unit 2b Power Transmission by Belts