Part 6

Special graduations, marking the position of constant factors which frequently enter into engineering calculations, are found on most slide rules. Usually the values of π = 3·1416 and (π)/(4) = 0·7854—the “gauge points” for calculating the circumference and area of a circle—are marked on the upper scales. The first should be given on the lower scales also. Marks _c_ and _c_^1 are sometimes found on the lower scales at 1·128 = √((4)/(π)) and at 3·568 = √((40)/(π)). These are useful in calculating the contents of cylinders and are thus derived:—Cubic contents of cylinder of diameter _d_ and length _l_ = (π)/(4)_d_^2_l_; substituting for (π)/(4) its reciprocal (4)/(π), the formula becomes (_d_^2)/(1·273 × _l_), and by taking the square root of the fractional part we have (_d_)/(1·128)^2 × _l_. This is now in a very convenient form, since by setting the gauge point _c_ on C to _d_ on D, we can read over _l_ on B the cubic contents on A. This example indicates the principle to be followed in arranging gauge points. Successive multiplication is avoided by substituting the reciprocal of the constant, thus bringing the expression into the form (_a_ × _b_)/(_c_), which, as we know, can be resolved by one setting of the slide. The advantage of dividing _d_ before squaring is also evident. The mark _c_^1 = _c_ × √(10) is used if it is necessary to draw the slide more than one-half its length to the right.

A gauge point, M, at 31·83 = (100)/(π) is found on the upper scales of some rules. Setting this point on B to the diameter of a cylinder on A, the circumference is read over 1 or 100 on B or the area of the curved surface over the length on B.

As another example of establishing a gauge point, we will take the formula for the theoretical delivery of pumps. If _d_ is the diameter of the plunger in inches, _l_ the length of stroke in feet, and Q the delivery in gallons, we have

Q = _d_^2 × (π)/(4) × _l_ × (12)/(277). (N.B.—277 cubic inches = 1 gallon.)

Multiplying out the constant quantities and taking its reciprocal, we readily transform the statement into Q = (_d_^2_l_)/(29·4) or ((_d_)/(5·42))^2 × _l_. Hence set gauge point 5·42 on C to _d_ on D and over length of stroke in feet on B, read delivery in gallons per stroke on A; or over piston speed in feet per minute on B, read theoretical delivery in gallons per minute on A.

Several examples of gauge points will be found in the section on calculating the weights of metal (see pages 59 and 60). In most cases their derivation will be evident from what has been said above. In the case of the weight of spheres, we have Vol. = 0·5236_d_^3, and this multiplied by the weight of 1 cubic inch of the material will give the weight W in lb. Hence for cast-iron, W = 0·5236 × _d_^3 × 0·26, which is conveniently transformed into W = (_d_ × _d_^2)/(7·35) as in the example on page 60.

With these examples no difficulty should be experienced in establishing gauge points for any calculation in which constant factors recur.

_Marking Gauge Points._—The practice of marking gauge points by lines extending to the working edge of the scale is not to be recommended, as it confuses the ordinary reading of the scales. Generally speaking, gauge points are only required occasionally, and if they are placed clear of the scale to which they pertain, but near enough to show the connection, they can be brought readily into a calculation by means of the cursor. Usually there is sufficient margin above the A scale and below the D scale for various gauge points to be marked. Another plan consists in cutting two nicks in the upper and lower edges of the cursor near the centre and about ⅛ in. apart. These centre pieces, when bent out, form a tongue, which are in line with the cursor line and run nearly in contact with the square and bevelled edges of the rule respectively. A fine line in the tongue can then be set to gauge points marked on these two edge strips, the ordinary measuring graduations being removed, if desired, by a piece of fine sand-paper.

For gauge points marked on the face of the rule, the author prefers two fine lines drawn at 45°—thus, ✕—and crossing in the exact point which it is required to indicate. With the “cross” gauge point the meeting lines facilitate the placing of the cursor, and an exact setting is readily made.[7] All lines should be drawn in Indian ink with a very sharp drawing pen. For a more permanent marking the Indian ink may be rubbed up in glacial acetic acid or the special ink for celluloid may be used. If any difficulty is found in writing the distinguishing signs against the gauge point, the inscription may be formed by a succession of small dots made with a sharp pricker.

EXAMPLES IN TECHNICAL CALCULATIONS.

In order to illustrate the practical value of the slide rule, we now give a number of examples which will doubtless be sufficient to suggest the methods of working with other formulæ. A few of the rules give results which are approximate only, but in all cases the degree of accuracy obtained is well within the possible reading of the scales. In many cases the rules given may be modified, if desired, by varying the constants. In most of the examples the particular formula employed will be evident from the solution, but in a few of the more complicated cases a separate statement has been given.

MENSURATION, ETC.

Given the chord _c_ of a circular arc, and the vertical height _h_, to find the diameter _d_ of the circle.

Set the height _h_ on B to half the chord on D, and over 1 on B read _x_ on A. Then _x_ + _h_ = _d_.

EX.—_c_ = 6; _h_ = 2; find _d_. Set 2 on B to 3 on D, and over 1 on B read 4·5 on A. Then 4·5 + 2 = 6·5 = _d_.

Given the radius of a circle _r_, and the number of degrees _n_ in an arc, to find the length _l_ of the arc.

Set _r_ on C to 57·3 on D, and over any number of degrees _n_ on D read the (approximate) length of the arc on C.

EX.—_r_ = 24; _n_ = 30; find _l_.

Set 24 on C to 57·3 on D, and over 30 on D read 12·56 = _l_ on C.

Given the diameter _d_ of a circle in _inches_, to find the circumference _c_ in _feet_.

Set 191 on C to 50 on D, and under any diameter in inches on C read circumference _c_, in feet on D.

EX.—Find the circumference in feet of a pulley 17 in. in diameter. Set 191 on C to 50 on D, and under 17 on C read 4·45 ft. on D.

Given the diameter of a circle, to find its area.

Set 0·7854 on B to 10 (centre index) on A and over any diameter on D read area on B.

When the rule has a special graduation line = 0·7854, on the right-hand scale of B, set this line to the R.H. index of A and read off as above. If only π is marked, set this special graduation on B to 4 on A.

On the C and D scales of some rules a gauge point marked _c_ will be found indicating √((4)/(π)) = 1·1286. In this case, therefore, set 1 on C to gauge point _c_ on D, and read area on A as above. If the gauge point _c_′ is used, divide the result by 10. Or set _c_ on C, to diameter on D, and over index of B read area on A. Cursors are supplied, having _two_ lines ruled on the glass, the interval between them being equal to (4)/(π) = 1·273 on the A scale. In this case, if the right hand of the two cursor lines be set to the diameter on D, the _area_ will be read on A under the _left_-hand cursor line. For diameters less than 1·11 it is necessary to set the middle index of B to the L.H. index of A, reading the areas on the L.H. B scale. The confusion which in general work is sometimes caused by the use of two cursor lines might be obviated by making the left-hand line in two short lengths, each only just covering the scales.

Given diameter of circle _d_ in _inches_, to find area _a_ in square _feet_.

Set 6 on B to 11 on A, and over diameter in inches on D read area in square feet on B.

To find the surface in square feet of boiler flues, condenser tubes, heating pipes, etc., having given the diameter in inches and length in feet.

Find the circumference in feet as above and multiply by the length in feet.

EX.—Find the heating surface afforded by 160 locomotive boiler tubes 1¾in. in diameter and 12 ft. long.

Set 191 on C to 50 on D; bring cursor 1·75 on C, L.H. index of C to cursor; cursor to 12 on C; 1 on C to cursor; and under 160 on C read 880 sq. ft. of heating surface on D.

If the dimensions are in the same denomination and the rule has a gauge point M at 31·83 (= (100)/(π)), set this mark on B to diameter of cylinder on A, and read cylindrical surface on A over length on B.

To find the side _s_ of a square, equal in area to a given rectangle of length _l_ and breadth _b_.

Set R.H. or L.H. index of B to _l_ on A, and under _b_ on B read _s_ on D.

EX.—Find the side of a square equal in area to a rectangle in which _l_ = 31 ft. and _b_ = 5 ft.

Set the (R.H.) index of B to 31 on A, and under 5 on B read 12·45 ft. on D.

To find various lengths _l_ and breadths _b_ of a rectangle, to give a constant area _a_.

Invert the slide and set the index of Ɔ to the given area on D. Then opposite any length _l_ on Ɔ find the corresponding breadth _b_ on D.

EX.—Find the corresponding breadths of rectangular sheets, 16, 18, 24, 36, and 60 ft. long, to give a constant area of 72 sq. ft.

Set the R.H. index of Ɔ to 72 on D, and opposite 16, 18, 24, 36, and 60 on Ɔ read 4·5, 4, 3, 2, and 1·2 ft., the corresponding breadths on D.

To find the contents in cubic feet of a cylinder of diameter _d_ in inches and length _l_ in feet.

Find area in feet as before, and multiply by the length.

If dimensions are all in inches or feet, set the mark _c_ (= 1·128) on C to diameter on D and over length on B, read cubic contents on A.

To find the area of an ellipse.

Set 205 on C to 161 on D; bring cursor to length of major axis on C, 1 on C to cursor, and under length of minor axis on C read area on D.

EX.—Find the area of an ellipse the major and minor axes of which are 16 in. and 12 in. in length respectively.

Set 205 on C to 161 on D; bring cursor to 16 on C, 1 on C to cursor, and under 12 on C read 150·8 in. on D.

To find the surface of spheres.

Set 3·1416 on B to R.H. or L.H. index of A, and over diameter on D read by the aid of the cursor, the convex surface on B.

To find the cubic contents of spheres.

Set 1·91 on B to diameter on A, and over diameter on C read cubic contents on A.

WEIGHTS OF METALS.

To find the weight in lb. per lineal foot of square bars of metal.

Set index of B to weight of 12 cubic inches of the metal (_i.e._, one lineal foot, 1 square inch in section) on A, and over the side of the square in inches on C read weight in lb. on A.

EX.—Find the weight per foot length of 4½in. square wrought-iron bars.

Set middle index of B to 3·33 on A, and over 4½ on C read 67·5 lb. on A.

(N.B.—For other metals use the corresponding constant in column (2), below).

To find the weight in lb. per lineal foot of round bars.

Set R.H. or L.H. index of B to weight of 12 cylindrical inches of the metal on A (column (4), below), and opposite the diameter of the bar in inches on C, read weight in lb. per lineal foot on A.

EX.—Find the weight of 1 lineal foot of 2 in. round cast steel.

Set L.H. index of B to 2·68 on A, and over 2 on C read 10·7 lb. on A.

To find the weight of flat bars in lb. per lineal foot.

Set the breadth in inches on C to (1)/(weight of 12 cub. in.) of the metal (column (3), below) on D, and above the thickness on D read weight in lb. per lineal foot on C.

EX.—Find the weight per lineal foot of bar steel, 4½in. wide and ⅝in. thick.

Set 4·5 on C to 0·294 on D, and over 0·625 on D read 9·56 lb. per lineal foot on C.

To find the weight per square foot of sheet metal, set the weight per cubic foot of the metal (col. 1) on C to 12 on D, and

────────────┬──────────┬──────────┬──────────┬─────────── │ (1) │(2) │ (3) │ (4) │Weight in │Weight of │ (1)/(Wt. │ Weight of Metals. │ lb. per │12 cubic │of 12 cub.│ 12 │cubic ft. │in. │ in.) │cylindrical │ │ │ │ in. ────────────┼──────────┼──────────┼──────────┼─────────── Wrought iron│480 │3·33 │0·300 │2·62 Cast iron │450 │3·125 │0·320 │2·45 Cast steel │490 │3·40 │0·294 │2·68 Copper │550 │3·82 │0·262 │3·00 Aluminium │168 │1·166 │0·085 │0·915 Brass │520 │3·61 │0·277 │2·83 Lead │710 │4·93 │0·203 │3·87 Tin │462 │3·21 │0·312 │2·52 Zinc (cast) │430 │2·98 │0·335 │2·34 „ (sheet) │450 │3·125 │0·320 │2·45 ────────────┴──────────┴──────────┴──────────┴───────────

above the thickness of the plate in inches on D read weight in lb. per square foot on C.

EX.—Find the weight in lb. per square foot of aluminium sheet ⅜in. thick.

Set 168 on C to 12 on D, and over 0·375 on D read 5·25 lb. on C.

To find the weight of pipes in lb. per lineal foot.

Set mean diameter of the pipe in inches (_i.e._, internal diameter _plus_ the thickness, or external diameter _minus_ the thickness) on C to the constant given below on D, and over the thickness on D read weight in lb. per lineal foot on C.

┌────────────┬─────────────────────┬─────────────────────┐ │ Metals. │ Constant for Pipes. │Constant for Spheres.│ ├────────────┼─────────────────────┼─────────────────────┤ │Wrought iron│ 0·0955 │ 6·87 │ │Cast iron │ 0·1020 │ 7·35 │ │Steel │ 0·0936 │ 6·73 │ │Brass │ 0·0882 │ 6·35 │ │Copper │ 0·0834 │ 6·00 │ │Lead │ 0·0646 │ 4·65 │ └────────────┴─────────────────────┴─────────────────────┘

EX.—Find the weight per foot of cast-iron piping 4 in. internal diameter and ½in. thick.

Set 4·5 on C to 0·102 on D, and over 0·5 on D read 22·1 lb. on C, the required weight.

To find the weight in lb. of spheres or balls, given the diameter in inches. (W = 0·5236_d_^3 × wt. of 1 cub. in. of material).

Set the constant for spheres (given above) on B to diameter in inches on A, and over diameter on C read weight in lb. on A.

EX.—Find the weight of a cast-iron ball 7½in. in diameter.

Set 7·35 on B to 7·5 on A, and over 7·5 on C read 57·7 lb. on A.

To find diameter in inches of a sphere of given weight.

Set the cursor to the given weight in lb. on A, and move the slide until the same number is found on C under the cursor that is simultaneously found on A over the constant for the sphere on B.

EX.—Find diameter in inches of a sphere of cast-iron to weigh 7½lb.

Setting the cursor to 7·5 on A, and moving the slide, it is found that when 3·8 on C falls under the cursor, 3·8 on A is simultaneously found over 7·35 on B. The required diameter is therefore 3·8 in.

The rules for cubes and cube roots (page 40) should be kept in view in solving the last two examples.

FALLING BODIES.

To find velocity in feet per second of a falling body, given the time of fall in seconds.

Set index on C to time of fall on D, and under 32·2 on C read velocity in feet per second on D.

To find velocity in feet per second, given distance fallen through in feet.

Set 1 on C to distance fallen through on A, and under 64·4 on B read velocity in feet per second on D.

EX.—Find velocity acquired by falling through 14 ft.

Set (R.H.) index of C to 14 on A, and under 64·4 on B read 30 ft. per second on D.

To find distance fallen through in feet in a given time.

Set index of C to time in seconds on D, and over 16·1 on B read distance fallen through in feet on A.

CENTRIFUGAL FORCE.

To find the centrifugal force of a revolving mass in lb.

Set 2940 on B to revolutions per minute on D; bring cursor to weight in lb. on B; index of B to cursor, and over radius in feet on B read centrifugal force in lb. on A.

To find the centrifugal stress in lb. per square inch, in rims of revolving wheels of cast iron.

Set 61·3 on C to the mean diameter of the wheel in feet on D, and over revolutions per minute on C read stress per square inch on A.

EX.—Find the stress per square inch in a cast-iron fly-wheel rim 8 ft. in diameter and running at 120 revolutions per minute.

Set 61·3 on C to 8 on D, and over 120 on C read 245 lb. per square inch on A.

THE STEAM ENGINE.

Given the stroke and number of revolutions per minute, to find the piston speed.

Set stroke in inches on C to 6 on D, and over number of revolutions on D read piston speed in feet per minute on C.

To find cubic feet of steam in a cylinder at cut-off, given diameter of cylinder and period of admission in inches.

Set 2200 on B to cylinder diameter on D, and over period of admission on B read cubic feet of steam on A.

EX.—Cylinder diameter 26 in., stroke 40 in., cut-off at ⅝ of stroke. Find cubic feet of steam used (theoretically) per stroke.

Set 2200 on B to 26 on D, and over 40 × ⅝ or 25 in. on B, read 7·68 cub. ft. on A, as the number of cubic feet of steam used per stroke.

Given the diameter of a cylinder in inches, and the pressure in lb. per square inch, to find the load on the piston in tons.

Set pressure in lb. per square inch on B to 2852 on A, and over cylinder diameter in inches on D read load on piston in tons on B.

EX.—Steam pressure 180 lb. per square inch; cylinder diameter, 42 in. Find load in tons on piston.

Set 180 on B to 2852 on A, and over 42 on D read 111 tons, the gross load, on B.

Given admission period and absolute initial pressure of steam in a cylinder, to find the pressure at various points in the expansion period (isothermal expansion).

Invert the slide and set the admission period, in inches, on Ɔ to the initial pressure on D; then under any point in the expansion stroke on Ɔ find the corresponding pressure on D.

EX.—Admission period 12 in., stroke 42 in., initial pressure 80 lb. per square inch. Find pressure at successive fifths of the expansion period.

Set 12 on Ɔ to 80 on D, and opposite 18, 24, 30, 36 and 42 in. of the whole stroke on Ɔ find the corresponding pressures on D:—53·3, 40, 32, 26·6 and 22·8 lb. per square inch.

To find the mean pressure constant for isothermally expanding steam, given the cut-off as a fraction of the stroke.

Find the logarithm of the ratio of the expansion _r_, by the method previously explained (page 46). Prefix the characteristic and to the number thus obtained, on D, set 1 on C. Then under 2·302 on C read _x_ on D. To _x_ + 1 on D set _r_ on C, and under index of C read mean pressure constant on D. The latter, multiplied by the initial pressure, gives the mean forward pressure throughout the stroke. (N.B.—Common log. × 2·302 = hyperbolic log.)

EX.—Find the mean pressure constant for a cut-off of ¼th, or a ratio of expansion of 4.

Set (L.H.) index of C to 4 on D, and on the reverse side of the slide read 0·602 on the logarithmic scale. The characteristic = 0; hence to 0·602 on D set (R.H.) index of C, and under 2·302 on C read 1·384 on D. Add 1, and to 2·384 thus obtained on D set _r_ (= 4) on C, and under 1 on C read 0·596, the mean pressure constant required.

Mean pressure constants for the most usual degrees of cut-off are given below:—

Cut-off in fractions of stroke Mean pressure constant ¾ 0·968 ⁷⁄₁₀ 0·952 ⅔ 0·934 ⅝ 0·919 ⅗ 0·913 ½ 0·846 ⅖ 0·766 ⅜ 0·750 ⅓ 0·699 ³⁄₁₀ 0·664 ¼ 0·596 ⅕ 0·522 ⅙ 0·465 ⅐ 0·421 ⅛ 0·385 ⅑ 0·355 ⅒ 0·330 ¹⁄₁₁ 0·309 ¹⁄₁₂ 0·290 ¹⁄₁₃ 0·274 ¹⁄₁₄ 0·260 ¹⁄₁₅ 0·247 ¹⁄₁₆ 0·236

To find mean pressure:—Set 1 on C to constant on D, and under initial pressure on C read mean pressure on D.

Given the absolute initial pressure, length of stroke, and admission period, to find the absolute pressure at any point in the expansion period, it being assumed that the steam expands adiabatically. (P_{2} = (P_{1})/(R^{¹⁰⁄₉}) in which P_{1} = initial pressure and P_{2} the pressure corresponding to a ratio of expansion R.)

Set L.H. index of C to ratio of expansion on D, and read on the back of the slide the decimal of the logarithm. Add the characteristic, and to the number thus obtained on D set 9 on C, and read off the value found on D under the index of C. Set this number on the logarithmic scale to the index mark, in the opening on the back of the rule, and under L.H. index of C read the value of R^{¹⁰⁄₉} on D. The initial pressure divided by this value gives the corresponding pressure due to the expansion.

EX.—Absolute initial pressure 120 lb. per square inch; stroke, 4 ft.; cut-off ¼. Find the respective pressures when ½ and ¾ths of the stroke have been completed.

In the first case R = 2. Therefore setting the L.H. index of C to 2 on D, we find the decimal of the logarithm on the back of the slide to be 0·301. The characteristic is 0, so placing 9 on C to 0·301 on D, we read 0·334 as the value under the R.H. index of C. (N.B.—In locating the decimal point it is to be observed that the log. of R has been multiplied by 10, in accordance with the terms of the above expression.) Setting this number on the logarithmic scale to the back index, the value of R^{¹⁰⁄₉} is found on D, under the L.H. index of C, to be 2·16. Setting 120 on C to this value, it is found that the pressure at ½ stroke, read on C over the R.H. index of D, is 55·5 lb. per square inch. In a similar manner, the pressure when ¾ths of the stroke is completed is found to be 35·4 lb. per square inch.

For other conditions of expanding steam, or for gas or air, the method of procedure is similar to the above.

To find the horse-power of an engine, having given the mean _effective_ pressure, the cylinder diameter, stroke, and number of revolutions per minute.

To cylinder diameter on D set 145 on C; bring cursor to stroke in feet on B, 1 on B to cursor, cursor to number of revolutions on B, 1 on B to cursor, and over mean effective pressure on B find horse-power on A.

(N.B.—If stroke is in inches, use 502 in place of 145 given above.)

EX.—Find the indicated horse-power, given cylinder diameter 27 in., mean effective pressure 38 lb. per square inch, stroke 32 in., revolutions 57 per minute.

Set 502 on C to 27 on D, bring cursor to 32 on B, 1 on B to cursor, cursor to 57 on B, 1 on B to cursor, and over 38 on B read 200 I.H.P. on A.

To determine the horse-power of a compound engine, invert the slide and set the diameter of the _high_-pressure cylinder on Ɔ to the cut-off in that cylinder on A. Use the number then found on A over the diameter of the _low_-pressure cylinder on Ɔ as the cut-off in that cylinder, working with the same pressure and piston speed, and calculate the horse-power as for a single cylinder.