Part 13

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 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^2l; 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^2l)/(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. = 0·5236d^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. 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.