OTHER METHODS OF OBTAINING POWERS AND ROOTS.
A simple method of obtaining powers and roots, which may serve on occasion, is by scaling off proportional lengths on the D scale (or the A scale) of the ordinary rule. Thus, to determine the value of 1·251·67 we take the actual length 1–1·25 on D scale, and increase it by any convenient means in the proportion of 1 ∶ 1·67. Then with a pair of dividers we set off this new length from 1, and obtain 1·44 as the result. One convenient method of obtaining the desired ratio is by a pair of proportional compasses. Thus to obtain 1·52¹⁷⁄₁₆, the compasses would be set in the ratio of 16 to 17, and the smaller end opened out to include 1–1·52 on the D scale; the opening in the large end of the compasses will then be such that setting it off from 1 we obtain 1·56 on D as the result sought.
Fig. 11.
The converse procedure for
obtaining the nth root of a
number N will obviously resolve
itself into obtaining 1
nth
of the scale length 1-N, and
need not be further considered.
Simple geometrical constructions are also used for obtaining
scale lengths in the required ratio. A series of parallel lines ruled
on transparent celluloid or stout tracing paper may be placed in an
inclined position on the face of the rule and adjusted so as to divide
the scale as desired. When much work is to be done which requires
values to be raised to some constant but comparatively low power,
n, the author has found the following device of assistance:—On a
piece of thin transparent celluloid a line OC is drawn (Fig. 11) and
in this a point B is taken such that OC
OB is the desired ratio. It is
convenient to make OB = 1–10 on the A scale, so that assuming
we require a series of values of v1·35, OB would be 12·5 cm. and OC,
16·875 cm. On these lines semi-circles are drawn as shown, both
passing through the point O.
Applying this cursor to the upper scales so that the point O is on 1 and the semi-circle O M B passes through v on A, the larger semi-circle will give on A the value of vn. Thus for p vn = 39·5 × 4·91·35, set 1 on B to 39·5 on A (Fig. 12) and apply the cursor to the working edge of B, so that O agrees with 1 and O M B passes through 4·9 on B. The larger semi-circle then cuts the edge of the slide on a point, giving 337 on A as the result required.
Of course any number of semi-circles may be drawn, giving different ratios. If a number of evenly-spaced divisions are used as bases, the device affords a simple means of obtaining a succession of small powers or roots, while it also finds a use in determining a number of geometric means between two values as is required in arranging the speed gears of machine tools, etc. The converse operation of finding roots will be evident as will also many other uses for which the device is of service.
Fig. 12.
The lines should be drawn in Indian ink with a very sharp pen and on the under side of the celluloid so that the lines lie in close contact with the face of the rule.
The Radial Cursor, another device for the same purpose, is always used in conjunction with the upper scales. As will be seen from Fig. 13, the body of the cursor P carries a graduated bar S which can be removed in a direction transverse to the rule, and adjusted to any desired position. Pivoted to the lower end of S is a radial arm R of transparent celluloid on which a centre line is engraved.
A reference to the illustration will show that the principle involved is that of similar triangles, the width of the slide being used as one of the elements. Thus, to take a simple case, if 2 on S is set to the index on P, and 1 on B is brought to N on A, then by swinging the radial arm until its centre line agrees with 1 on C, we can read N2 on A. Evidently, since in the two similar triangles A O N2 and N t N2 the length of A O is twice that of N t, it results that A N2 = 2 A N. In general, then, to find the nth power of a number, we set the cursor to 1 or 10 on A, bring n on the cross bar S to the index on the cursor, and 1 on B to N on A. Then to 1 on C we set the line on the radial arm, and under the latter read Nn on A. The inverse proceeding for finding the nth root will be obvious.
Fig. 13.
An advantage offered by this and analogous methods of obtaining powers and roots is that the result is obtained on the ordinary scale of the rule, and hence it can be taken directly into any further calculation which may be necessary.
COMBINED OPERATIONS.
Thus far the various operations have been separately considered,
and we now pass on to a consideration of the methods of working for
solving the various formulæ met with in technical calculations. We
propose to explain the methods of dealing with a few of the more
generally used expressions, as this will suffice to suggest the procedure
in dealing with other and more intricate calculations. In solving
the following problems, both the upper and lower scales are used,
and the relative value of the several scales must be observed
throughout. Thus, in solving such an expression as √74·5
15·8 = 6·86,
the division is first effected by setting 15·8 on B to 745 on A.
From the relation of the two parts of the upper scales (page 37)
we know that such values as 7·45, 745, etc., will be taken on the
left-hand A and B scales, while values as 15·8, 1580, etc., will be
taken on the right-hand A and B scales. Hence, 15·8 on the R.H.
B scale is set to 745 on the L.H. A scale, and the result read on D
under the index of C. Had both values been taken on the L.H. A and
B scales, or both on the R.H. A and B scales, the results would have
corresponded to x = √7·45
1·58 = 2·17, or to x =√74·5
15·8 = 2·17, i.e., to
6·86
√10. Hence if a wrong choice of scales has been made, we can
correct the result by multiplying or dividing by √10 as the case
may require. If the result is read on D, set to it the centre
index (10) of B and read the corrected result under the index of C.
To solve a × b2 = x. Set the index of C to b on D, and over a on B read x on A.
To solve a2
b = x. Set b on B to a on D by using the cursor, and over index of B read x on A.
To solve b
a2 = x. Set a on C to b on A, and over 1 on B read x on A.
To solve a × b2
c = x. Set c on B to b on D, and over a on B read x on A.
To solve (a × b)2 = x. Set 1 on C to a on D, and over b on C read x on A.
To solve (a
b)2 = x. Set b on C to a on D, and over 1 on C read x on A.
To solve √a × b = x. Set 1 on B to a on A, and under b on B read x on D.
To solve √a
b = x. Set b on B to a on A, and under 1 on C read x on D.
To solve a b
c2 = x. Set b on C to c on D and over a on B read x on A.
To solve c√a
b = x. Set b on B to a on A, and under c on C read x on D.
To solve √ ̅a
b = x. Set b on C to a on A, and under 1 on C read x on D.
To solve a
√ ̅b = x. Set b on B to a on D, and under 1 on C read x on D.
To solve b√ ̅a = x. Set 1 on C to b on D, and under a on B read x on D.
To solve √a3 = x. Treat as a√ ̅a.
To solve a√b3 = x. Treat as a√ ̅b × b.
To solve √ ̅a3
b = x. Treat as √ ̅a × a
b.
To solve √a3
b = x. Treat as √ ̅a × a
√ ̅b = √a
b × a.
To solve √a × b
c = x. Set c on B to a on A, and under b on B read x on D.
To solve a × b
√ ̅c = x. Set c on B to b an D, and under a on C read x on D.
To solve √a2 × b
c = x. Set c on B to a on D, and under b on B read x on D.
To solve a2 × b2
c = x. Set c on B to a on D, and over b on C read x on A.
To solve a√ ̅b
c = x. Set c on C to b on A, and under a on C read x on D.
To solve (a × √ ̅b
c)2 = x. Set c on C to a on D, and over b on B read x on A.
HINTS ON EVALUATING EXPRESSIONS.
As a general rule, the use of cubes and higher powers should be avoided whenever possible. Thus, in the foregoing section, we recommend treating an expression of the form a√b3 as a × b × √ ̅b; the magnitudes of the values thus met with are more easily appreciated by the beginner, and mistakes in estimating the large numbers involved in cubing are avoided.
Ex.—7·3 × √573 = 3140.
Set 1 on C to 57 on D; bring cursor to 57 on B (R.H., since 57 has an even number of digits); bring 1 on C to cursor, and under 7·3 on C read 3140 on D. As a rough estimate we have √57, about 8; 8 × 57, about 400; 400 × 7, gives 2800, showing the result consists of 4 figures.
An expression of the form a∛b2, or a b⅔, is better dealt with by
rearranging as a × b
∛b.
Ex.—3·64∛4·322 = 9·65.
Set cursor to 4·32 on A, and move the slide until 1·63 is found simultaneously under the cursor on B and on D under 1 on C; bring cursor to 1 on C; 4·32 on C to cursor, and over 3·64 on D read 9·65 on C. (Note that in this case it is convenient to read the answer on the slide; see page 22). From the slide rule we know ∛4·32 = about 1·6; this into 4·32 is roughly 3; 3·64 × 3 is about 10, showing the answer to be 9·65.
Similarly products of the form a × b⁴⁄₃ are best dealt with as a × b × ∛b.
Factorising expressions sometimes simplifies matters, as, for instance,
in x4 − y4 = (x2 + y2)(x2 − y2). Here, working with the fourth
powers involves large numbers and the troublesome determination
of the number of digits in each factor; but squares are read on
the rule at once, the number of digits is obvious, and, in general,
the method should give a more accurate result. Take the expression,
D1 = ∛D4 − d4
D giving the diameter D1 of a solid shaft
equal in torsional strength to a hollow shaft whose external and
internal diameters are D and d respectively. Rearranging as
D1 = ∛(D2 + d2)(D2 − d2)
D and taking, as an example, D = 15 in.
and d = 7 in., we have D2 + d2 = 274 and D2 − d2 = 176; hence
D_1 = ∛274 × 176
15 = ∛3210 = 14·75 in.
Reversed Scale Notation.—With expressions of the form 1 − x, or 100 − x, it is often convenient to regard the scales as having their notation reversed, i.e., to read the scale backwards. When this is done the D scale is read as shown on the lower line—
| Direct Notation | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| D Scale | ||||||||||
| Reversed Notation | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
The new reading can be found by subtracting the ordinary reading from 1, 10, 100, etc., according to the value assigned to the R.H. index, but actually it is unnecessary to make this calculation, as with a little practice it is quite an easy matter to read both the main and subdivisions in the reversed order. Applications are found in plotting curves, trigonometrical formulæ, etc.
Ex.—Find the per cent. of slip of a screw propeller from
100 − S = 10133V
PR
taking the speed, V, as 15 knots, the pitch of the propeller, P, as 27 ft. 6 in., and the revolutions per minute, R, as 60.
Set 27·5 on B to 10133 on A (N.B.—Take the setting near the centre index of A); bring the cursor to 15 on B and 60 on B to cursor. Reading the L.H. A scale backwards, the slip, S, = 8 per cent. is found on A over 10 on B.
Percentage Calculations.—To increase a quantity by x per cent. we multiply by 100 + x; to diminish a quantity by x per cent. we multiply by 100 − x. Hence, to add x per cent., set 100 + x on C to 1 on D and read new values on D under original values on C. To deduct x per cent. read the D scale backwards from 10 and set R.H. index of C to x per cent. so read. Then read as before.
GAUGE POINTS.
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 c1 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
4d2l; substituting for π
4 its reciprocal 4
π, the formula becomes
d2
1·273 × l, and by taking the square root of the fractional part we
have d
1·1282 × 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 c1 = 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 = d2 × 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 = d2l
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. =
0·5236d3, 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 × d3 × 0·26, which is conveniently transformed into W = d × d2
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
| Metals. | (1) Weight in lb. per cubic ft. |
(2) Weight of 12 cubic in. |
(3) 1 Wt. of 12 cub. in. |
(4) Weight of 12 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 = 0·5236d3 × 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. (P2 = P1
R¹⁰⁄₉ in which P1 = initial pressure and P2
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 R¹⁰⁄₉ 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 R¹⁰⁄₉ 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.
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 d3 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 = 0·27d2.
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 p2
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. = d3 × 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 l3
78,000d4).
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.