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33 2 BELTS.
then multiply the result by the following constants, according to
the arc of contact between the belt and the small pulley.
When the arc of contact between the belt and the small
pulley is 90° multiply by 1.60.
100° " " 1.45 140° multiply by 1.15
110° " " 1.35 150° " " 1.10
120° " " 1.25 160° ’• " 1.06
130° " " 1.20 170C - ’•’
1.04
Example.
The pulley on a dynamo is 15" in diameter, and it makes
1200 revolutions per minute. The driving pulley is so large that
the belt only lays around the dynamo pulley for a distance of
150 degrees. What is the necessary width of a light double belt,
weighing 21 ounces per square foot, when it takes 40 horse-power
to run the dynamo ?
Solution
:
If the arc of contact had been 180 degrees the belt would
40 X 50000 n . . , ,
be b = v>qo x 15 X 21
= inches wide, but as the arc
of contact is not 180 degrees, but only 150 degrees, this width
is multiplied by the constant 1.10. as given in the preceding
table. Thus, the width of the belt will be 5.3 X 1.1 = 5.83
inches or, practically, a belt six inches wide is required.
When belts are running in a horizontal direction,
and the driven pulley and the driver are of equal diameter
and finish, the belt will always, when overloaded, commence to
slip on the driver, and when pulleys are of unequal size it is
always more favorable for the belt when the driving pulley is
the larger than when vice versa.
To Find the Arc of Contact of Belts.
Make a scale drawing of the pulleys and the belt, and
measure the arc of contact from the drawing by means of a
protractor, or the arc of contact in degrees on the small
pulley for an open belt may be calculated by the formula
:
Cosine of half the angle = —
R = Radius of large pulley in inches.
r = Radius of small pulley in inches.
/ = Distance in inches between centers of the shafts.
Example.
The distance between centers of two shafts is 16 feet ; the
large pulley is 60 inches and the small pulley is 20 inches
in diameter. What is the arc of contact of the belt ?
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