D.

BELTS AND ROPES.

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.

SPUR WHEELS.

Given diameter and pitch of a spur wheel, to find number of teeth.

Set pitch on C to π (3·1416) on D, and under any diameter on C read number of teeth on D.

Given diameter and number of teeth in a spur wheel, to find the pitch.

Set diameter on C to number of teeth on D, and read pitch on C opposite 3·1416 on D.

Given the distance between the centres of a pair of spur wheels and the number of revolutions of each, to determine their diameters.

To twice the distance between the centres on D, set the sum of the number of revolutions on C, and under the revolutions of each wheel on C find the respective wheel diameters on D.

EX.—The distance between the centres of two spur wheels is 37·5 in., and they are required to make 21 and 24 revolutions in the same time. Find their respective diameters.

Set 21 + 24 = 45 on C to 75 (or 37·5 × 2) on D, and under 21 and 24 on C find 35 and 40 in. on D as the respective diameters.

To find the power transmitted by toothed wheels, given the pitch diameter _d_ in inches, the number of revolutions per minute _n_, and the pitch _p_ in inches, by the rule, H.P. = (_n_ _d_ _p_^2)/(400).

Set 400 on B to pitch in inches on D; set cursor to d on B, 1 on B to cursor, and over any number of revolutions n on B read power transmitted on A.

EX.—Find the horse-power capable of being transmitted by a spur wheel 7 ft. in diameter, 3 in. pitch, and running at 90 revolutions per minute.

Set 400 on B to 3 on D; bring cursor to 84 in. on B, 1 on B to cursor, and over 90 revolutions on B read 170, the horse-power transmitted, on A.

SCREW-CUTTING.

Given the number of threads per inch in the guide screw, to find the wheels to cut a screw of given pitch.

Set threads per inch in guide screw on C, to the number of threads per inch to be cut on D. Then opposite any number of teeth in the wheel on the mandrel on C, is the number of teeth in the wheel to be placed on the guide screw on D.

STRENGTH OF SHAFTING.

Given the diameter _d_ of a steel shaft, and the number of revolutions per minute _n_, to find the horse-power from:—

H.P. = _d_^3 × _n_ × 0·02.

Set 1 on C to _d_ on D, and bring cursor to _d_ on B. Bring 50 on B to cursor, and over number of revolutions on B read H.P. on A.

EX.—Find horse-power transmitted by a 3 in. steel shaft at 110 revolutions per minute.

Set 1 on C to 3 on D, and bring cursor to 3 on B. Bring 50 on B to cursor, and over 110 on B read 59·4 horse-power on A.

Given the horse-power to be transmitted and the number of revolutions of a steel shaft, to find the diameter.

Set revolutions on B to horse-power on A, and bring cursor to 50 on B. Then move the slide until the same number is found on B under the cursor that is simultaneously found on D under the index of C. This number is the diameter required.

To find the deflection _k_ in inches, of a round steel shaft of diameter _d_, under a uniformly distributed load in lb. _w_, and supported by bearings, the centres of which are _l_ feet apart (_k_ = (_w_ _l_^3)/(78,000_d_^4)).

Modifying the form of this expression slightly, we proceed as follows:—Set _d_ on C to _l_ on D, and bring the cursor to the same number on B that is found on D under the index of C. Bring _d_ on B to cursor, cursor to _w_ on B, 78,800 on B to cursor, and read deflection on A over index of B.

EX.—Find the deflection in inches of a round steel shaft 3½in. diameter, carrying a uniformly distributed load of 3200 lb., the distance apart of the centres of support being 9 ft.

Set 3·5 on C to 9 on D, and read 2·57 on D, under the L.H. index of C. Set cursor to 2·57 on B, and bring 3·5 on B to cursor, cursor to 3200 on B, 78,000 on B to cursor, and over L.H. index of B read 0·199 in., the required deflection on A.

To find the diameter of a shaft subject to twisting only, given the twisting moment in inch-lb. and the allowable stress in lb. per square inch.

Set the stress in lb. per square inch on B to the twisting moment in inch-lb. on A, and bring cursor to 5·1 on B. Then move the slide until the same number is found on B under the cursor that is simultaneously found on D under the index of C.

EX.—A steel shaft is subjected to a twisting moment of 2,700,000 inch-lb. Determine the diameter if the allowable stress is taken at 9000 lb. per square inch.

Set 9000 on B to 2,700,000 on A, and bring the cursor to 5·1 on B. Moving the slide to the left, it is found that when 11·51 on the R.H. scale of B is under the cursor, the L.H. index of C is opposite 11·51 on D. This, then, is the required diameter of the shaft.

(N.B.—The rules for the scales to be used in finding the cube root (page 42) must be carefully observed in working these examples.)

MOMENTS OF INERTIA.

To find the moment of inertia of a square section about an axis formed by one of its diagonals (I = (_s_^4)/(12)).

Set index of C to the length of the side of square _s_ on D; bring cursor to _s_ on C, 12 on B to cursor, and over index of B read moment of inertia on A.

To find the moment of inertia of a rectangular section about an axis parallel to one side and perpendicular to the plane of bending.

Set index of C to the height or depth _h_ of the section, and bring cursor to _h_ on B. Set 12 on B to cursor, and over breadth _b_ of the section on B read moment of inertia on A.

EX.—Find the moment of inertia of a rectangular section of which _h_ = 14 in. and _b_ = 7 in.

Set index of C to 14 on D, and cursor to 14 on B. Bring 12 on B to cursor, and over 7 on B read 1600 on A.

DISCHARGE FROM PUMPS, PIPES, ETC.

To find the theoretical delivery of pumps, in gallons per stroke.

Set 29·4 on B to the diameter of the plunger in inches on D, and over length of stroke in feet on B read theoretical delivery in gallons per stroke on A.

(N.B.—A deduction of from 20 to 40 per cent. should be made to allow for slip.)

To find loss of head of water in feet due to friction in pipes (Prony’s rule).

Set diameter of pipe in feet on B to velocity of water in feet per second on D and bring cursor to 2·25 on B; bring 1 on B to cursor, and over length of pipe in miles on B, read loss of head of water in feet, on A.

To find velocity in feet per second, of water in pipes (Blackwell’s rule).

Set 2·3 on B to diameter of pipe in feet on A, and under inclination of pipe in feet per mile on B read velocity in feet per second on D.

To find the discharge over weirs in cubic feet per minute and per foot of width. (Discharge = 214√(_h_^3))

Set 0·00467 on C to the head in feet _h_ on D, and under _h_ on B read discharge on D.

To find the theoretical velocity of water flowing under a given head in feet.

Set index of B to head in feet on A, and under 64·4 on B read theoretical velocity in feet per second on D.

HORSE-POWER OF WATER WHEELS.

To find the effective horse-power of a Poncelet water wheel.

Set 880 on C to cubic feet of flow of water per minute on D, and under height of fall in feet on C, read effective horse-power on D.

For breast water wheels use 960, and for overshot wheels 775, in place of 880 as above.

ELECTRICAL ENGINEERING.

To find the resistance per mile, in ohms, of copper wire of high conductivity, at 60° F. the diameter being given in mils. (1 mil. = 0·001 in.).

Set diameter of wire in mils. on C to 54,900 on A, and over R.H. or L.H. index of B read resistance in ohms on A.

EX.—Find the resistance per mile of a copper wire 64 mils. in diameter.

Set 64 on C to 54,900 on A, and over R.H. index of B read 13·4 ohms on A.

To find the weight of copper wire in lb. per mile.

Set 7·91 on C to diameter of wire in mils. on D, and over index of B read weight per mile on A.

Given electromotive force and current, to find electrical horse-power.

Set 746 on C to electromotive force in volts on D, and under current in ampères on C read electrical horse-power on D.

Given the resistance of a circuit in ohms and current in ampères, to find the energy absorbed in horse-power.

Set 746 on B to current on D, and over resistance on B read energy absorbed in H.P. on A.

EX.—Find the H.P. expended in sending a current of 15 ampères through a circuit of 220 ohms resistance.

Set 746 on B to 15 on D, and over 220 on B read 66·3 H.P. on A.

COMMERCIAL.

To add on percentages.

Set 100 on C to 100 + given percentage on D, and under original number on C read result on D.

To deduct percentages.

Set R.H. index of C to 100 − the given percentage on D, and under original number on C read result on D.

EX.—From £16 deduct 7½ per cent.

Set 10 on C to 92·5 on D and under 16 on C, read 14·8 = £14, 16s. on