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33° BELTS.
Example 1.
A double belt 10 inches wide, weighing 25 ounces per square
foot, runs over 50-inch pulleys, making 240 revolutions per min-
ute. How many horse-power will it properly transmit ?
Solution:
„ , .
r , ,
50 X 3.1416 X 240
velocity of belt = ^ = 3141.6 ft. per minute.
3141.6 X 10
= 500
= ^2 ’^ norse-Power-
Example 2.
One hundred horse-power is to be transmitted by a double
belt weighing 25 ounces per square foot. The pulleys are 66
inches in diameter and make 150 revolutions per minute. What
is the necessary width of belt ?
Solution :
Pulleys of 66 inches diameter, running 150 revolutions per
minute, will give a belt speed of
15° X 3’ 1416 X 66
= 2591.8;
say, 2592 feet per minute.
100 X 500 , , , , , ,
b — ^— = 19.3 inches; thus, a double belt 20
inches wide will do the work.
Example 3.
A light single belt 4 inches wide, weighing 13 ounces per
square foot, runs over pulleys of 36 inches diameter, making 100
revolutions per minute. How many horse-power may be trans-
mitted ?
Solution
:
TT , . P , , 36 X 3.1416 X 100
Velocity of belt = ^ — 942.48 ft. per minute.
The belt is a light single belt and its transmitting capacity
4 X 942.48
will be, H — vwS — 3.76992, about 3% horse-power.
To Calculate Size of Belt for Given Horse=Power when
Diameter of Pulley and Number of Revolutions
of Shaft Are Known.
The following formulas may be used for calculating belt
transmission, and will give results approximately consistent
with previously given rules, but they are more convenient for
use, as the velocity of the belt does not need to be first calculat-
ed, but the velocity of the belt must not exceed the practical
limit.
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