To find the ratio of tensions in the two sides of a belt, given the
coefficient of friction between belt and pulley μ and the number of
degrees θ in the arc of contact (log. R = (μθ)/(132)).
Set 132 on C to the coefficient of friction on D, and read off the value
found on D under the number of degrees in the arc of contact on C. Place
this value on the scale of equal parts on the back of the slide, to the
index mark in the aperture, and read the required ratio on D under the
L.H. index of C.
EX.—Find the tension ratio in a belt, assuming a coefficient of
friction of 0·3 and an arc of contact of 120 degrees.
Set 132 on C to 0·3 on D, and under 120 on C read 0·273. Place this on
the scale to the index on the back of the rule, and under the L.H.
index C read 1·875 on D, the required ratio.
Given belt velocity and horse-power to be transmitted, to find the
requisite width of belt, taking the effective tension at 50 lb. per inch
of width.
Set 660 on C to velocity in feet per minute on D, and opposite
horse-power on D find width of belt in inches on C.
Given velocity and width of belt, to find horse-power transmitted.
Set 660 on C to velocity on D, and under width on C find horse-power
transmitted on D.
(N.B.—For any other effective tension, instead of 660 use as a gauge
point:—33,000 ÷ tension.)
Given speed and diameter of a cotton driving rope, to find power
transmitted, disregarding centrifugal action, and assuming an effective
working tension of 200 lb. per square inch of rope.
Set 210 on B to 1·75 on D, and over speed in feet per minute on B read
horse-power on A.
EX.—Find the power transmitted by a 1¾in. rope running at 4000 ft. per
minute.
Set 210 on B to 1·75 on D, and over 4000 on B read 58·3 horse-power
on A.
Find the “centrifugal tension” in the previous example, taking the
weight per foot of the rope as = 0·27_d_^2.
Set 655 on C to the diameter, 1·75 in., on D, and over the speed, 4000
ft. on C, read centrifugal tension = 114 lb. on A.