Chapter 33 : Transmission of Power by Belts and Ropes 677
We know that the length of the belt,
L= Length of arc GJE + EF + Length of arc FKH + HG
= 2 (Length of arc JE + EF + Length of arc FK)
Substituting the values of length of arc JE from equation (ii), length of arc FK from equation
(iii) and EF from equation (iv) in this equation,
L=2
12
12(– )
2– –
222rr
rl r
l⎡⎤⎛⎞ππ⎛⎞
⎢⎥⎜⎟+α + + ⎜⎟α
⎣⎦⎝⎠ ⎝⎠=2
12
11 2 2(– )
2– –
222rr
rrl r r
l⎡⎤ππ
⎢⎥+α+ + α
⎣⎦=2
12
12 12(– )
2( ) (–) –
22rr
rr rr l
l⎡⎤π
⎢⎥++α +
⎣⎦=2
12
12 12(– )
()2(–)2–rr
rr rr l
lπ+ +α +Substituting the value of 12(– )rr
lα= from equation (i),L=2
12 12
12 12(– ) (– )
()2 (–)2–rr rr
rr rr l
llπ+ +× +=22
12 12
122( – ) ( – )
() 2–rr rr
rr l
llπ+ + +=2
12
12(– )
()2rr
rr l
lπ +++33.12.LENGTH OF A CROSS-BELT DRIVE
Fig. 33.7. Length of a cross belt drive.
In this case, both the pulleys rotate in the opposite directions as shown in Fig. 33·7.
Let O 1 and O 2 = Centre of the two pulleys,
r 1 and r 2 = Radii of the larger and smaller pulleys,
l= Distance between O 1 and O 2.