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35 2 PULLEYS.
Solution
:
Speed of counter-shaft will be:
N = *J F X S = V 360 X 40 = 120 revolutions per minute.
Diameter of smallest step on spindle will be :
18 X 40
120
6 inches.
Diameter of the intermediate step on spindle will be:
(18 + 6) X 120
D1 12 inches.
(120 + 120)
Thus, the sizes of each step, with regard to speed, should
be 6, 12 and 18 inches, but with regard to belt tension these
sizes have to be slightly altered.
To Correct the Diameter of Stepped Pulleys so that the
Belt will have the Same Tension on all the Steps.
At first thought, it may seem as if the belt would have
equal tension on each step when the sum of the diameters of
the largest and the smallest steps of the two pulleys are equal
to the sum of the diameters of the two middle steps ; but this is
only correct if a crossed belt is used on the pulleys. For a two-
step pulley it is also correct for either open or crossed belt, if
both pulleys are of the same size ; but if the pulleys are of dif-
ferent sizes, the diameter of the steps must be calculated for
two steps as well as if there were more.
It is evident from Fig. 2, that an open belt will be tighter
over the largest and the smallest pulleys than it would be over
the two middle pulleys, as the part a of the belt runs parallel to
the center line and will be as long as the distance between
centers, but the inclined line, b, will be as much larger as the
distance rtTto e. (See Fig. 2).
A convenient way to solve this is: First calculate pulleys
that will give the required speed, and of such sizes that the sum
of the diameters of the two steps which are to work together
Avill be equal, then calculate the length of the belt when laying
on the largest and smallest steps, with a given distance between
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