Part 7

To find the cylinder ratio in compound engines, invert the slide and set index of Ɔ to diameter of the low-pressure cylinder on D. Then over the diameter of the high-pressure cylinder on C, read cylinder ratio on A.

EX.—Diameter of high-pressure cylinder 7¾in., low-pressure 15 in. Find cylinder ratio.

Set index on Ɔ to 15 on D, and over 7·75 on Ɔ read 3·75, the required ratio, on A.

The cylinder ratios of triple or quadruple-expansion engines may be similarly determined.

EX.—In a quadruple-expansion engine, the cylinders are 18, 26, 37, and 54 inches in diameter. Find the respective ratios of the high, first intermediate, and second intermediate cylinders to the low-pressure.

Set (R.H.) index of Ɔ to 54 on D, and over 18, 26, and 37 on Ɔ read 9, 4·31, and 2·13, the required ratios, on A.

Given the mean effective pressures in lb. per square inch in each of the three cylinders of a triple-expansion engine, the I.H.P. to be developed in each cylinder, and the piston speed, to find the respective cylinder diameters.

Set 42,000 on B to piston speed on A; bring cursor to mean effective pressure in low-pressure cylinder on B, index of B to cursor, and under I.H.P. on A read low-pressure cylinder diameter on C. To find the diameters of the high-pressure and intermediate-pressure cylinders, invert the slide and place the mean pressure in the low-pressure cylinder on ᗺ to the diameter of that cylinder on D. Then under the respective mean pressures on ᗺ read corresponding cylinder diameters on D.

EX.—The mean effective pressures in the cylinders of a triple-expansion engine are:—L.P., 10·32; I.M.P., 27·5; and H.P., 77·5 lb. per square inch. The piston speed is 650 ft. per minute, and the I.H.P. developed in each cylinder, 750. Find the cylinder diameters.

Set 42,000 on B to 650 on A, and bring cursor to 10·32 on B. Bring index of B to cursor, and under 750 on A read 68·5 in. on C, the L.P. cylinder diameter. Invert the slide, and placing 10·32 on ᗺ to 68·5 on D, read, under 27·5 on ᗺ, the I.M.P. cylinder diameter = 42 in., on D; also under 77·5 on ᗺ read the H.P. cylinder diameter = 25 in., on D.

To compute brake or dynamometrical horse-power.

Set 525 on C to the total weight in lb. acting at the end of the lever (or pull of spring balance in lb.) on D; set cursor to length of lever in feet on C, bring 1 on C to cursor, and under number of revolutions per minute on C find brake horse-power on D.

Given cylinder diameter and piston speed in feet per minute, to find diameter of steam pipe, assuming the maximum velocity of the steam to be 6000 ft. per minute.

Set 6000 on B to cylinder diameter on D, and under piston speed on B read steam pipe diameter on D.

Given the number of revolutions per minute of a Watt governor, to find the vertical height in inches, from the plane of revolution of the balls to the point of suspension.

Set revolutions per minute on C to 35,200 on A, and over index of B read height on A.

Given the weight in lb. of the rim of a cast-iron fly-wheel, to find the sectional area of the rim in square inches.

Set the mean diameter of the wheel in feet on C to 0·102 on D, and under weight of rim on C find area on D.

Given the consumption of coal in tons per week of 56 hours, and the I.H.P., to find the coal consumed per I.H.P. per hour.

Set I.H.P. on C to 40 on D, and under weekly consumption on C read lb. of coal per I.H.P., per hour on D.

EX.—Find coal used per I.H.P. per hour, when 24 tons is the weekly consumption for 300 I.H.P.

Set 300 on C to 40 on D, and under 24 on C read 3·2 lb. per I.H.P. per hour on D.

(N.B.—For any other number of working hours per week divide 2240 by the number of working hours, and use the quotient in place of 40 as above.)

To find the tractive force of a locomotive.

Set diameter of driving wheel in inches on B to diameter of cylinder in inches on D, and over the stroke in inches on B read on A, tractive force in lb. for each lb. of effective pressure on the piston.

STEAM BOILERS.

To find the bursting pressure of a cylindrical boiler shell, having given the diameter of shell and the thickness and ultimate strength of the material.

Set the diameter of the shell in inches on C to twice the thickness of the plate on D, and under strength of material per square inch on C read bursting pressure in lb. per square inch on D.

EX.—Find the bursting pressure of a cylindrical boiler shell 7 ft. 6 in. in diameter, with plates ½in. thick, assuming an ultimate strength of 50,000 lb. per square inch.

Set 90 on C to 1·0 on D, and under 50,000 on C find 555 lb. on D.

To find working pressure for Fox’s corrugated furnaces by Board of Trade rule.

Set the least outside diameter in inches on C to 14,000 on D, and under thickness in inches on C read working pressure on D in lb. per square inch.

To find diameter _d_ in inches, of round steel for safety valve springs by Board of Trade rule.

Set 8000 on C to load on spring in lb. on D, and under the mean diameter of the spring in inches on C read _d_^3 on D. Then extract the cube root as per rule.

SPEED RATIOS OF PULLEYS, ETC.

Given the diameter of a pulley and its number of revolutions per minute, to find the circumferential velocity of the pulley or the speed of ropes, belts, etc., driven thereby.

Set diameter of pulley in inches on C to 3·82 on D, and over revolutions per minute on D read speed in feet per minute on C.

EX.—Find the speed of a belt driven by a pulley 53 in. in diameter and running at 180 revolutions per minute.

Set 53 on C to 3·82 on D, and over 180 on D read 2500 ft. per minute on C.

EX.—Find the speed of the pitch line of a spur wheel 3 ft. 6 in. in diameter running at 60 revolutions per minute.

Set 42 in. on C to 3·82 on D, and over 60 on D read 660 ft. per minute on C.

Given diameter and number of revolutions per minute of a driving pulley, and the diameter of the driven pulley, to find the number of revolutions of the latter.

Invert the slide and set diameter of driving pulley on Ɔ to given number of its revolutions on D; then opposite diameter of any driven pulley on Ɔ read its number of revolutions on D.

EX.—Diameter of driving pulley 10 ft.; revolutions per minute 55; diameter of driven pulley 2 ft. 9 in. Find number of revolutions per minute of latter.

Set 10 on Ɔ to 55 on D, and opposite 2·75 on Ɔ read 200 revolutions on 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 D.

To calculate simple interest.

Set 1 on C to rate per cent. on D; bring cursor to period on C and 1 on C to cursor. Then opposite any sum on C find simple interest on D.

For interest per annum.

Set R.H. index on C to rate on D, and opposite principal on C read interest on D.

EX.—Find the amount with simple interest of £250 at 8 per cent., and for a period of 1 year and 9 months.

Set 1 on C to 8 on D; bring cursor to 1·75 on C, and 1 on C to cursor; then opposite 250 on C read £35, the interest, on D. Then 250 + 35 = £285 = the amount.

To calculate compound interest.

Set the L.H. index of C to the amount of £1 at the given rate of interest on D, and find the logarithm of this by reading on the reverse side of the rule, as explained on page 46. Multiply the logarithm, so found, by the period, and set the result, on the scale of equal parts, to the index on the under-side of the rule; then opposite any sum on C read the amount (including compound interest) on D.

EX.—Find the amount of £500 at 5 per cent. for 6 years, with compound interest.

Set L.H. index of C to £1·05 on D, and read at the index on the scale of equal parts on the under-side of rule, 0·0212. Multiply by 6, we obtain 0·1272, which, on the scale of equal parts, is placed to the index in the notch at the end of the rule. Then opposite 500 on C read £670 on D, the amount required, including compound interest.

MISCELLANEOUS CALCULATIONS.

To calculate percentages of compositions.

Set weight (or volume) of sample on C, to weight (or volume) of substance considered, on D; then under index of C read required percentage on D.

EX.—A sample of coal weighing 1·25 grms. contains 0·04425 grm. of ash. Find the percentage of ash.

Set 1·25 on C to 0·04425 on D, and under index on C read 3·54, the required percentage of ash on D.

Given the steam pressure P and the diameter _d_ in millimetres, of the throat of an injector, to find the weight W, of water delivered in lb. per hour from W = (_d_^2√̅P)/(0·505).

Set 0·505 on C to P on A; bring cursor to _d_ on C and index of C to cursor. Then under _d_ on C read delivery of water on D.

To find the pressure of wind per square foot, due to a given velocity in miles per hour.

Set 1 on B to 2 on A, and over the velocity in miles per hour on D read pressure in lb. per square foot on B.

To find the kinetic energy of a moving body.

Set 64·4 on B to velocity in feet per second on D, and over weight of body in lb. on B read kinetic energy or accumulated work in foot-lb. on A.

TRIGONOMETRICAL APPLICATIONS

_Scales._—Not the least important feature of the modern slide rule is the provision of the special scales on the under-side of the slide, and by the use of which, in conjunction with the ordinary scales on the rule, a large variety of trigonometrical computations may be readily performed.

Three scales will be found on the reverse or under-side of the slide of the ordinary Gravêt or Mannheim rule. One of these is the evenly-divided scale or scale of equal parts referred to in previous sections, and by which, as explained, the decimal parts or mantissæ of logarithms of numbers may be obtained. Usually this scale is the centre one of the three, but in some rules it will be found occupying the lowest position, in which case some little modification of the following instructions will be necessary. The requisite transpositions will, however, be evident when the purposes of the scales are understood. The upper of the three scales, usually distinguished by the letter S, is a scale giving the logarithms of the sines of angles, and is used to determine the natural sines of angles of from 35 minutes to 90 degrees. The notation of this scale will be evident on inspection. The main divisions 1, 2, 3, etc., represent the degrees of angles; but the values of the subdivisions differ according to their position on the scale. Thus, if any primary space is subdivided into 12 parts, each of the latter will be read as 5 minutes (5′), since 1° = 60′.

_Sines of Angles._—To find the sine of an angle the slide is placed in the groove, with the under-side uppermost, and the end division lines or indices on the slide, coinciding with the right and left indices of the A scale. Then over the given angle on S is read the value of the sine of the angle on A. If the result is found on the left scale of A (1 to 10), the logarithmic characteristic is −2; if it is found on the right-hand side (10 to 100), it is −1. In other words, results on the right-hand scale are prefixed by the decimal point only, while those on the left-hand scale are to be preceded by a cypher also. Thus:—

Sine 2° 40′ = 0·0465; sine 15° 40′ = 0·270.

Multiplication and division of the sines of angles are performed in the same manner as ordinary calculations, excepting that the slide has its under-face placed uppermost, as just explained. Thus to multiply sine 15° 40′ by 15, the R.H. index of S is brought to 15 on A, and opposite 15° 40′ on S is found 4·05 on A. Again, to divide 142 by sine 16° 30′, we place 16° 30′ on S to 142 on A, and over R.H. index of S read 500 on A.

The rules for the number of integers in the results are thus determined: Let N be the number of integers in the multiplier M or in the dividend D. Then the number of integers P, in the product or Q, in the quotient are as follows:—

When the result is found to the right of M or D, │P = N − 2│Q = N and in the same scale │ │ When the result is found to the right of M or D, │P = N − 1│Q = N + 1 and in the other scale │ │ When the result is found to the left of M or D, and│P = N − 1│Q = N + 1 in the other scale │ │ When the result is found to the left of M or D, and│P = N │Q = N + 2 in the same scale │ │

If the division is of the form (20° 30′)/(50), the result cannot be read off directly on the face of the rule. Thus, if in the above example 20° 30′ on S, is placed to agree with 50 on the right-hand scale of A, the result found on S under the R.H. index of A is 44° 30′. The required numerical value can then be found: (1) By placing the slide with all indices coincident when opposite 44° 30′ on S will be found 0·007 on A; or (2) In the ordinary form of rule, by reading off on the scale B opposite the index mark in the opening on the under-side of the rule. The above rules for the number of integers in the quotient do not apply in this case.

If it is required to find the sine of an angle simply, this may be done with the slide in its ordinary position, with scale B under A. The given angle on scale S is then set to the index on the under-side of the rule, and the value of the sine is read off on B under the right index of A.