D.

Now we have seen (page 8) that multiplying the log. of a number by 2 gives the log. of the square of the number. Hence, above any number on D we find its _square_ on A, or, conversely, below any number on A, we find its _square root_ on D. Thus, above 2 we find 4; under 49, we find 7 and so on. Obviously the same relation exists between the B and C scales.

THE CURSOR OR RUNNER.

All modern slide rules are now fitted with a _cursor_ or _runner_, which usually consists of a light metal frame moving under spring control in grooves in the edges of the stock of the rule. This frame carries a piece of glass, mica or transparent celluloid, about 1 in. square, across the centre of which a fine reference line is drawn exactly at right angles to the line of scales. To “set the cursor” to any value on the scales of the rule, the frame is taken between the thumb and forefinger and adjusted in position until the line falls exactly upon the graduation, or upon an estimated value, between a pair of graduations, as the case may be. Having fixed one number in this way, another value on either of the scales on the slide may be similarly adjusted in reference to the cursor line. The cursor will be found very convenient in making such settings, especially when either or both of the numbers are located by eye estimation. It also finds a very important use in referring the readings of the upper scale to those of the lower, or _vice versa_, while as an aid in continued multiplication and division and complex calculations generally, its value is inestimable.

_Multiple Line Cursors._—Cursors can be obtained with _two_ lines, the distance between them being that between 7·854 and 10 on the A scale. The use of this cursor is explained on page 57. Another multiple line cursor has short lines engraved on it, corresponding to the main graduations from 95 to 105 on the respective scales. This is useful for adding or deducting small percentages.

_The Broken Line Cursor._—To facilitate setting, broken line cursors are made, in which the hair-line is not continued across the scales, but has two gaps, as shown in Fig. 6.

_The Pointed Cursor_ has an index or pointer, extending over the bevelled edge of the rule, on which is a scale of inches. It is useful for summing the lengths of the ordinates of indicator diagrams, and also for plotting lengths representing the logarithms of numbers, sometimes required in graphic calculations.

_The Goulding Cursor._—It has been pointed out that in order to obtain the third or fourth figure of a reading on the 10 in. slide rule, it is frequently necessary to depend upon the operator’s ability to mentally subdivide the space within which the reading falls. This subdivision can be mechanically effected by the aid of the Goulding Cursor (Fig. 7), which consists of a frame fitting into the usual grooves in the rule, and carrying a metal plate faced with celluloid, upon which is engraved a triangular scale A B C. The portion carrying the chisel edges E is not fixed to the cursor proper, but slides on the latter, so that the index marks on the projecting prongs can be moved slightly along the scales of the rule, this movement being effected by the short end of the bent lever F working in the slot as shown. D is a pointer which can be moved along F under spring control. As illustrating the method of use, we will assume that 1 on C is placed to 155 on D, and that we require to read the value on D under 27 on C. This is seen to lie between 4150 and 4200, so setting the pointer D to the line B C—always the first operation—we move the whole along the rule until the index line on the lower prong agrees with 4200. We then move F across the scale until the index line agrees with 4100, set the pointer D to the line A C, and move the lever back until the index line agrees with 27 on the slide. It will then be found that the pointer D gives 85 on A B as the value of the supplementary figures, and hence the complete reading is 4185.

[Illustration: FIG. 6.]

[Illustration: FIG. 8.]

[Illustration: FIG. 7.]

[Illustration: FIG. 9.]

_Magnifying Cursors_ are of assistance in reading the scales, and in a good and direct light are very helpful. In one form an ordinary lens is carried by two light arms hinged to the upper and lower edges of the cursor, so that it can be folded down to the face of the rule when not in use. A more compact form, shown in Fig. 8, consists of a strip of plano-convex glass, on the under-side of which is the hair-line. In a cursor made by Nestler of Lahr, the plano-convex strip is fixed on the ordinary cursor. The magnifying power is about 2, so that a 5 in. rule, having the same number of graduations as a 10 in. rule, can be read with equal facility, by the aid of this cursor.

The Digit-registering Cursor, supplied by Mr. A. W. Faber, London, and shown in Fig. 9, has a semicircular scale running from 0 at the centre upward to −6 and downward to +6. A small finger enables the operator to register the number of digits to be added or subtracted at the end of a lengthy operation, as explained at page 28.

MULTIPLICATION.

In the preliminary notes it was shown that by mechanically adding two lengths representing the logarithms of two numbers, we can obtain the _product_ of these numbers; while by subtracting one log. length from another, the number represented by the latter is divided by the number represented by the former. Hence, using the C and D scales, we have the

RULE FOR MULTIPLICATION.—_Set the index of the C scale to one of the factors on D, and under the other factor on C, find the product on D._

[Illustration: FIG. 10.]

Thus, to find the product of 2 × 4, the slide is moved to the right until the left index (1) of C is brought over 2 on D, when under the other factor (4) on C, is found the required product (8) on D. Following along the slide, to the right, we find that beyond 5 on C (giving 10 on D), we have no scale below the projecting slide (Fig. 10). If we imagine the D scale prolonged to the right, we should have a repetition of the earlier portion, but, as with the two parts of the A scales, the repeated portion would be of tenfold value, and 10 on C would agree with 20 on the prolonged D scale. We turn this fact to account by moving the slide to the left until 10 on C agrees with 2 on D, and we can then read off such results as 2 × 6 = 12; 2 × 8 = 16, etc., remembering that as the scale is now of tenfold value, there will be two figures in the result. Hence, for those who prefer rules, we have the

RULE FOR THE NUMBER OF DIGITS IN A PRODUCT.—_If the product is read with the slide projecting to the_ LEFT, ADD THE NUMBER OF THE DIGITS IN THE TWO FACTORS; _if read with the slide to the_ RIGHT, _deduct 1 from this sum_.

EX.—25 × 70 = 1750.

The product is found with the slide projecting to the _left_, so the number of digits in the product = 2 + 2 = 4.

EX.—3·6 × 25 = 90.

The slide projects to the _right_, and the number of digits in the product is therefore 1 + 2 − 1 = 2.

EX.—0·025 × 0·7 = 0·0175.

The product is obtained with the slide projecting to the _left_, and the number of digits is therefore −1 + 0 = −1.

EX.—0·000184 × 0·005 = 0·00000092.

The sum of the number of digits in the two factors = −3 + (−2) = −5, but as the slide projects to the _right_, the number of digits will be −5 − 1 = −6.

From the last two examples it will be seen that when the first significant figure of a decimal factor does not immediately follow the decimal point, the minus sign is to be prefixed to the number of digits, counting as many digits _minus_ as there are 0’s following the decimal point. Thus, 0·03 has −1 digit, 0·0035 has −2 digits, and so on. Some little care is necessary to ensure these minus values being correctly taken into account in determining the number of digits in the answer. For this reason many prefer to treat decimal factors as whole numbers, and to locate the decimal point according to the usual rules for the multiplication of decimals. Thus, in the last example we take 184 × 5 = 920, but as by the usual rule the product must contain 6 + 3 = 9 decimal places, we prefix six cyphers, obtaining 0·00000092. When both factors consist of integers as well as decimals, the number of digits in the product, and therefore the position of the decimal point, will be determined by the usual rule for whole numbers.

Another method of determining the number of digits in a product deserves mention, which, not being dependent upon the position of the slide, is applicable to all calculating instruments.

GENERAL RULE FOR NUMBER OF DIGITS IN A PRODUCT.—_When the first significant figure in the product is smaller than in_ EITHER _of the factors, the number of digits in the product is equal to the_ SUM _of the digits in the two factors. When the contrary is the case, the number of digits is 1_ LESS _than the sum of the digits in the two factors. When the first figures are the same, those following must be compared._

_Estimation of the Figures in a Product._—We have given rules for those who prefer to decide the number of figures by this means, but experience will show that to make the best use of the instrument, the result, as read on the rule, should be regarded merely as the _significant figures of the answer_, the position of the decimal point, if not obvious, being decided by a very rough mental calculation. In very many instances, the magnitude of the result will be evident from the conditions of the problem—_e.g._, whether the answer should be 0·3 in., 3 in., or 30 in.; or 10 tons, 0·1 ton, 100 tons, etc. In those cases where the magnitude of the answer cannot be estimated, and the factors contain many figures, or have a number of 0’s following the decimal point, the use of notation by powers of 10 (page 8) is of considerable assistance; but more usually it will be found, that a very rough calculation will settle the point with comparatively little trouble. Considerable practice is needed to work rapidly and with certainty, when using rules. Moreover, the experience thus acquired is confined to slide-rule work. The same time spent in practising the “rough approximation” method will enable reliable results to be obtained rapidly, with the advantage that the method is applicable to calculations generally. However, the choice of methods is a matter of personal preference. Both methods will be given, but whichever plan is followed, the student is strongly advised to cultivate the habit of forming an idea of the magnitude of the result.

EX.—33·6 × 236 = 7930.

Setting 1 on C to 33·6 on D, we read under 236 on D and find 793 on D, as the significant figures of the answer. A rough calculation, as 30 × 200 = 6000, indicates that the result will consist of 4 figures, and is therefore to be read as 7930.

EX.—17,300 × 3780 = 65,400,000.

By factorising with powers of 10

1·73 × 10^4 × 3·78 × 10^3 = 1·73 × 3·78 × 10^7.

Setting 1 on C to 1·73 on D, we read, under 3·78 on C, the result of the simple multiplication, as 6·54. Multiplying by 10^7 moves the decimal point 7 places to the right, and the answer is 65,400,000.

If it is required to find a series of products of which one of the factors is _constant_, set 1 on C to the constant factor on D and read the several products on D, under the respective variable factors.

If the factors are required which will give a constant _product_ (really a case of division), set the cursor to the constant product on D. Then obviously, as the slide is moved along, any pair of factors found simultaneously under the cursor line on C, and on D under index of C, will give the product. A better method of working will be explained when we deal with the inversion of the slide.

It is sometimes useful to remember that although we usually set the slide to the rule, we can obtain the result equally well by setting the rule to the slide. Thus, bringing 1 (or 10) on D to 2 on C, we find on C, _over_ any other factor, _n_ on D, the product of 2 × _n_. But note that the slide and rule have now changed places, and if we use rules for the number of digits in the result, we must now deduct 1 from the sum of the digits in the factors, when the _rule projects_ to the _right of the slide_.

With the ordinary 10 in. rule it will be found in general that the extent to which the C and D scales are subdivided is such as to enable not more than three figures in either factor being dealt with. For the same reason it is impossible to directly read more than the first three figures of any product, although it is often possible—by mentally dividing the smallest space involved in the reading—to correctly determine the fourth figure of a product. Necessarily this method is only reliable when used in the earlier parts of the C and D scales. However, the last numeral of a three-figure, and in some cases the last of a four-figure, product can be readily ascertained by an inspection of the factors.

EX.—19 × 27 = 513. Placing the L.H. index of C to 19 on D, we find opposite 27 on C, the product, which lies between 510 and 515. A glance at the factors, however, is sufficient to decide that the third figure must be 3, since the product of 9 and 7 is 63, and the last figure of this product must be the last figure in the answer.

EX.—79 × 91 = 7189.

In this case the division line 91 on C indicates on D that the answer lies between 7180 and 7190. As the last figure must be 9, it is at once inferred that the last two figures are 89.

When there are more than three figures in either or both of the factors, the fourth and following figures to the right must be neglected. It is well to note, however, that if the first neglected figure is 5, or greater than 5, it will generally be advisable to increase by 1 the third figure of the factor employed. Generally it will suffice to make this increase in one of the two factors only, but it is obvious that in some cases greater accuracy will be obtained by increasing both factors in this way.

CONTINUED MULTIPLICATION.—To find the product of more than two factors, we make use of the cursor to mark the position of successive products (the value of which does not concern us) as the several factors are taken into the calculation. Setting the index of C to the 1st factor on D, we bring the line of the cursor to the 2nd factor on C, then the index of C to the cursor, the cursor to the 3rd factor, index of C to cursor, and so on, reading the final product on D under the last factor on C. (Note that the 1st factor and the result are read on D; all intermediate readings are taken on C.)

If the rule for the number of digits in a product is used, it is necessary to note the number of times multiplication is effected with the slide projecting to the right. This number, deducted from the sum of the digits of the several factors, gives the number of digits in the product. Ingenious devices have been adopted to record the number of times the slide projects to the right, but some of these are very inconvenient. The author’s method is to record each time the slide so projects, by a minus mark, thus −. These can be noted down in any convenient manner, and the sum of the marks so obtained deducted from the sum of the digits in the several factors, gives the number of digits in the product as before explained.

EX.—42 × 71 × 1·5 × 0·32 × 121 = 173,200.

The product given, which is that read on the rule, is obtained as follows:—Set R.H. index of C to 42 on D, and bring the cursor to 71 on C. Next bring the L.H. index of C to the cursor, and the latter to 1·5 on C. This multiplication is effected with the slide to the right, and a memorandum of this fact is kept by making a mark −. Bring the R.H. index of C to the cursor and the latter to 0·32 on C. Then set the L.H. index of C to the cursor and read the result, 1732, on D under 121 on C, while as a slide again projects to the right, a second − memo-mark is recorded. There are 2 + 2 + 1 + 0 + 3 = 8 digits in the factors, and as there were 2 − marks recorded during the operation, there will be 8 − 2 = 6 digits in the product, which will therefore read 173,200 (173,194·56).

For a very rough evaluation of the result, we note that 1·5 × 0·3 is about 0·5; hence, as a clue to the number of figures we have

40 × 70 × 60 = 3000 × 60 = 180,000.

DIVISION.

The instructions for multiplication having been given in some detail, a full discussion of the inverse process of division will be unnecessary.

RULE FOR DIVISION.—_Place the divisor on C, opposite the dividend on D, and read the quotient on D under the index of C._

EX.—225 ÷ 18 = 12·5.

Bringing 18 on C to 225 on D, we find 12·5 under the L.H. index of C.

As in multiplication, the factors are treated as whole numbers, and the position of the decimal point afterwards decided according to the following rule, which, as will be seen, is the reverse of that for multiplication:—

RULE FOR THE NUMBER OF DIGITS IN A QUOTIENT.—_If the quotient is read with the slide projecting to the_ LEFT, _subtract the number of digits in the divisor from those in the dividend; but if read with the slide to the_ RIGHT, ADD _1 to this difference_.[2]

In the above example the quotient is read off with the slide to the right, so the number of digits in the answer = 3 − 2 + 1 = 2.

EX.—0·000221 ÷ 0·017 = 0·013.

Here the number of digits in the dividend is −3, and in the divisor −1. The difference is −2; but as the result is obtained with the slide to the right, this result must be increased by 1, so that the number of digits in the quotient is −2 + 1 = −1, giving the answer as 0·013.

If preferred, the result can be obtained in the manner referred to when considering the multiplication of decimals. Thus, treating the above as whole numbers, we find that the result of dividing 221 by 17 = 13, since the difference in the number of digits in the factors, which is 1, is, owing to the position of the slide, increased by 1, giving 2 as the number of digits in the answer. Then by the rules for the division of decimals we know that the number of decimal places in the quotient is equal to 6 − 3 = 3, showing that a cypher is to be prefixed to the result read on the rule.

As in multiplication, so in division, we have a

GENERAL RULE FOR NUMBER OF DIGITS IN A QUOTIENT.—_When the first significant figure in the_ DIVISOR _is greater than that in the_ DIVIDEND_, the number of digits in the quotient is found by subtracting the digits in the divisor from those in the dividend. When the contrary is the case, 1_ IS TO BE ADDED _to this difference. When the first figures are the same, those following must be compared._

ESTIMATION OF THE FIGURES IN A QUOTIENT.—The method of roughly estimating the number of figures in a quotient needs little explanation.

EX.—3·95 ÷ 5340 = 0·00074.

Setting 534 on C to 3·95 on D we read under the (R.H.) index of C, the significant figures on D, which are 74. Then 3·9 ÷ 5 is about 0·8 and 0·8 ÷ 1000 gives 0·0008 as a rough estimate.

EX.—0·00000285 ÷ 0·000197 = 0·01446.

Regarding this as 2·85 × 10^{−6} ÷ 1·97 × 10^{−4} we divide 2·85 by 1·97 and obtain 1·446. Dividing the powers of 10 we have 10^{−6} ÷ 10^{−4} = 10^{−2}, so the decimal point is to be moved two places to the left and the answer is read as 0·01446.

Another method of dividing deserves mention as of special service when dividing a number of quantities by a _constant divisor_:—Set the index of C to the divisor on D and over any dividend on D, read the quotient on C.

For the division of a _constant dividend_ by a variable divisor, set the cursor to the dividend on D and bring the divisor on C successively to the cursor, reading the corresponding quotients on D under the index of C. Another method which avoids moving the slide is explained in the section on “Multiplication and Division with the Slide Inverted.”

CONTINUED DIVISION, if we can so call such an expression as

(3·14)/(785 × 0·00021 × 4·3 × 64·4) = 0·0688,

may be worked by repeating as follows:—Set 7·85 on C to 3·14 on D, bring cursor to index of C, 2·1 on C to cursor, cursor to index, 4·3 to cursor, cursor to index, 6·44 to cursor, and under index of C read 688 on D as the significant figures of the answer.

For the number of figures in the result, we deduct the sum of the number of digits in the several factors and add 1 for each time the slide projects to the right, which in this case occurs once. There are 3 + (−3) + 1 + 2 = 3 denominator digits, 1 numerator digit, and 1 is to be added to the difference. Therefore there are 1 − 3 + 1 = −1 digits in the answer, which is therefore 0·0688. The foregoing method of working may confuse the beginner, who is apt to fall into the process of continued multiplication. For this reason, until familiarity with combined methods has been acquired, the product of the several denominators should be first found by the continued multiplication process, and the figures in this product determined. Then divide the numerator by this product to obtain the result.

As the denominator product will be read on D, we may avoid resetting the slide by bringing the numerator on C to this product and reading the result on C _over_ the index of D. The slide and rule have here changed places; hence if rules are followed for the number of figures in the result, 1 must be added to the difference of digits, when the _rule projects_ to the _right of the slide_.

The author’s method of recording the number of times division is performed with the slide to the right is by vertical memorandum marks, thus |. The full significance of these memo-marks will appear in the following section.

For a rough calculation to fix the decimal point, in this example we move the decimal points in the factors, obtaining

(3)/(0·8 × 2 × 4 × 6) = (3)/(40) = 0·075.

THE USE OF THE UPPER SCALES FOR MULTIPLICATION AND DIVISION.

Many prefer to use the upper scales A and B, in preference to C and D. The disadvantage is that as the scales are only one-half the length of C or D, the graduation does not permit of the same degree of accuracy being obtained as when working with the lower scales. But the result can always be read directly from the rule without ever having to change the position of the slide after it has been initially set. Hence, it obviates the uncertainty as to the direction in which the slide is to be moved in making a setting.

When the A and B scales are employed, it is understood that the left-hand pair of scales are to be used in the same manner as C and D, and so far the rules relating to the latter are entirely applicable. But in this case the slide is always moved to the right, so that in multiplication the product is found either upon the left or right scales of A. If it is found on the left A scale, the rule for the number of digits in the product is found as for the C and D scales, and is equal to the _sum of the digits in the two factors, minus 1_; but if found on the right-hand A scale, the number of digits in the product is equal to the sum of the digits in the two factors.

In division, similar modifications are necessary. If when moving the slide to the right the division can be completely effected by using the L.H. scale of A, the quotient (read on A above the L.H. of index B) has a number of digits equal to the number in the dividend, less the number in the divisor, _plus 1_. But if the division necessitates the use of both the A scales, the number of digits in the quotient equals the number in the dividend, less the number in the divisor.

RECIPROCALS.

A special case of division to be considered is the determination of the _reciprocal_ of a number _n_, or (1)/(_n_). Following the ordinary rule for division, it is evident that setting _n_ on C to 1 on D, gives (1)/(_n_) on D under 1 on C. It is more important to observe that by inverting the operation—setting 1 (or 10) on C to _n_ on D—we can read (1)/(_n_) on C over 1 (or 10) on D. Hence whenever a result is read on D under an index of C, we can also read its reciprocal on C over whichever index of D is available.

_The Number of Digits in a Reciprocal_ is obvious when _n_ = 10, 100, or any power (_p_) of 10. Thus (1)/(10) = 0·1; (1)/(100) = 0·01; (1)/(10^{_p_}) = 1 preceded by _p_ − 1 cyphers. For all other cases we have the rule:—_Subtract from 1 the number of digits in the number._

EX.—(1)/(339) = 0·00295.

There are 3 digits in the number; hence, there are 1 − 3 = −2 digits in the answer.

EX.—(1)/(0·0000156) = 64,100.

There are −4 digits in the number; hence, there are 1 − (−4) = 5 digits in the result.

CONTINUED MULTIPLICATION AND DIVISION.

By combining the rules for multiplication and division, we can readily evaluate expressions of the form (_a_)/(_b_) × (_c_)/(_d_) × (_e_)/(_f_) × (_g_)/(_h_) = _x_. The simplest case, (_a_ × _c_)/(_b_) can be solved by one setting of the slide.[3] Take as an example, (14·45 × 60)/(8·5) = 102. Setting 8·5 on C to 14·45 on D, we can, if desired, read 1·7 on D under 1 on C, as the quotient. However, we are not concerned with this, but require its multiplication by 60, and the slide being already set for this operation, we at once read under 60 on C the result, 102, on D. The figures in the answer are obvious.

When there are more factors to take into account, we place the cursor over 102 on D, bring the next divisor on C to the cursor, move the cursor to the next multiplier on C, bring the next divisor on C to the cursor, and so on, until all the factors have been dealt with. Note that only the first factor and the result are read on D; also _that the cursor is moved for multiplying and the slide for dividing_.

_Number of Digits in Result in Combined Multiplication and Division._—For those who use rules the author’s method of determining the decimal point in combined multiplication and division may be used. Each time _multiplication_ is performed with the slide projecting to the _right_, make a − mark; each time _division_ is effected with the slide to the right, make a | mark; _but allow the_ | _marks to cancel the_ − _marks as far as they will_. Subtract the sum of the digits in the denominator from the sum of digits in the numerator, and to this difference _add_ any uncancelled memo-marks, if of | character, or _subtract_ them if of − character.

EX.—(43·5 × 29·4 × 51 × 32)/(27 × 3·83 × 10·5 × 1·31) = 1468.

[Sidenote: ⵜ ⵜ ⵏ ⵏ]

Set 27 on C to 43·5 on D, and as with this _division_ the slide is to the right, make the first ⵏ mark. Bring cursor to 29·4 on C, and as in this _multiplication_ the slide is to the right, make the first − mark, cancelling as shown. Setting 3·83 on C to the cursor, requires the second ⵏ mark, which, however, is cancelled in turn by the multiplication by 51. The division by 10·5 requires the third ⵏ mark, and after multiplying by 32 (requiring no mark) the final division by 1·31 requires the fourth ⵏ mark. Then, as there are 8 numerator digits, 6 denominator, and 2 uncancelled memo-marks (which, being 1, are additive) we have

Number of digits in result = 8 − 6 + 2 = 4.

Had the uncancelled marks been − in character, the number of digits would have been 8 − 6 − 2 = 0.

For quantities less than 0·1 the digit place numbers will be _negative_. The troublesome addition of these may be avoided by transferring them to the opposite side and treating them as positive.

_2_ _4_ 0·00356 × 27·1 × 0·08375 Thus:— ───────────────────────── = 288 0·1426 × 9·85 × 0·00002 _2_ _1_ _1_

The first numerator, 0·00356, has −2 digits. Note this by placing 2 _below the lower line_ as shown. 27·1 has 2 digits; place 2 over it. 0·08375 has −1 digit; hence place 1 _below the lower line_. The first denominator has no digits; the second, 9·85, has 1 digit; hence place 1 under it. 0·00002 has −4 digits; place 4 _above the upper line_. The sum of the top series is 2 + 4 = 6; of the bottom series 2 + 1 + 1 = 4. Subtracting the bottom from the top, we have 6 − 4 = 2 digits, to which 1 has to be added for an uncancelled memo-mark, and the result is read as 288.

Moving the decimal point often facilitates matters. Thus, (32·4 × 0·98 × 432 × 0·0217)/(4·71 × 0·175 × 0·00000621 × 412000) is much more conveniently dealt with when re-arranged as (32·4 × 9·8 × 432 × 2·17)/(4·71 × 17·5 × 6·21 × 4·12) = 141.

To determine the number of figures in the result by rough cancelling and mental calculation, we note that 4·71 enters 432 about 100 times; 9·8 enters 17·5 about 2; 6·21 into 32·4 about 5; and 2·17 into 4·12 about 2. This gives (500)/(4) = 125, showing that the result contains 3 digits. From the slide rule we read 141, which is therefore the result sought.

The occasional traversing of the slide through the rule, to interchange the indices—a contingency which the use of the C and D scales always involves—may often be avoided by a very simple expedient. Such an example as (6·19 × 31·2 × 422)/(1120 × 8·86 × 2.09) = 3·93 is sometimes cited as a particularly difficult case. Working through the expression as given, two traversings of the slide are necessary; but by taking the factors in the slightly different order, (6·19 × 31·2 × 422)/(8·86 × 2·09 × 1120), _so that the significant figures of each pair are more nearly alike_, we not only avoid any traversing the slide, but we also reduce the extent to which the slide is moved to effect the several divisions.

Such cases as (_a_ × _b_)/(_c_ × _d_ × _e_ × _f_ × _g_) or (_a_ × _b_ × _c_ × _d_ × _e_)/(_f_ × _g_) really resolve themselves into (_a_ × _b_ × 1 × 1 × 1)/(_c_ × _d_ × _e_ × _f_ × _g_) and (_a_ × _b_ × _c_ × _d_ × _e_)/(_f_ × _g_ × 1 × 1 × 1), but, of course, if rules are used to locate the decimal point, the 1’s so (mentally) introduced are not to be counted as additional figures in the factors.

MULTIPLICATION AND DIVISION WITH THE SLIDE INVERTED.

If the slide be inverted in the rule but with the same face uppermost, so that the Ɔ scale lies adjacent to the A scale, and the right and left indices of the slide and rule are placed in coincidence, we find the product of any number on D by the coincident number on Ɔ (readily referred to each other by the cursor) is always 10. Hence, by reading the numbers on Ɔ as decimals, we have over any unit number on D, its _reciprocal_ on Ɔ. Thus 2 on D is found opposite 0·5 on Ɔ; 3 on D opposite to 0·333; while opposite 8 on Ɔ is 0·125 on D, etc. The reason of this is that the sum of the lengths of the slide and rule corresponding to the factors, is always equal to the length corresponding to the product—in this case, 10.

It will be seen that if we attempt to apply the ordinary rule for multiplication, with the slide inverted, we shall actually be multiplying the one factor taken on D by the _reciprocal_ of the other taken on Ɔ. But multiplying by the _reciprocal of a number_ is equivalent to _dividing_ by that number, and _dividing_ a factor by the _reciprocal_ of a number is equivalent to _multiplying_ by that number. It follows that with the slide inverted the operations of multiplication and division are reversed, as are also the rules for the number of digits in the product and the position of the decimal point. Hence, in multiplying with the slide inverted, we place (by the aid of the cursor) one factor on Ɔ opposite the other factor on D, and read the result on D under either index of Ɔ. It follows that with the slide thus set, any pair of coinciding factors on Ɔ and D will give the same constant product found on D under the index of Ɔ. One useful application of this fact is found in selecting the scantlings of rectangular sections of given areas or in deciding upon the dimensions of rectangular sheets, plates, cisterns, etc. Thus by placing the index of Ɔ to 72 on D, it is readily seen that a plate having an area of 72 sq. ft. may have sides 8 by 9 ft., 6 by 12, 5 by 14·4, 4 by 18, 3 by 24, 2 by 36, with innumerable intermediate values. Many other useful applications of a similar character will suggest themselves.

PROPORTION.

With the slide in the ordinary position and with the indices of the C and D scales in exact agreement, the _ratio_ of the corresponding divisions of these scales is 1. If the slide is moved so that 1 on C agrees with 2 on D, we know that under any number _n_ on C is _n_ × 2 on D, so that if we read numerators on C and denominators on D we have

C 1 1·5 2 3 4 ───────────────────────────────────────── D1 2 3 4 6 8.

In other words, the numbers on D bear to the coinciding numbers on C a ratio of 2 to 1. Obviously the same condition will obtain no matter in what position the slide may be placed. The rule for proportion, which is apparent from the foregoing, may be expressed as follows:—

RULE FOR PROPORTION.—_Set the first term of a proportion on the C scale to the second term on the D scale, and opposite the third term on the C scale read the fourth term on the D scale._

EX.—Find the 4th term in the proportion of 20 ∶ 27 ∷ 70 ∶ _x_. Set 20 on C to 27 on D, and opposite 70 on C read 94·5 on D. Thus

C 20 70 ───────────────── D 27 94·5.

It will be evident that this is merely a case of combined multiplication and division of the form, (20 × 70)/(27) = 94·5. Hence, given any three terms of a proportion, we set the 1st to the 2nd, or the 3rd to the 4th, as the case may be, and opposite the other given term read the term required.[4]

Thus, in reducing vulgar fractions to decimals, the decimal equivalent of (3)/(16) is determined by placing 3 on C to 16 on D, when over the index or 1 of D we read 0·1875 on C. In this case the terms are 3 ∶ 16 ∷ _x_ ∶ 1. For the inverse operation—to find a vulgar fraction equivalent to a given decimal—the given decimal fraction on C is set to the index of D, and then opposite any denominator on D is the corresponding numerator of the fraction on C.

If the index of C be placed to agree with 3·1416 on D, it will be clear from what has been said that this ratio exists throughout between the numbers of the two scales. Therefore, against any _diameter_ of a circle on C will be found the corresponding _circumference_ on D. In the same way, by setting 1 on C to the appropriate conversion factor on D, we can convert a series of values in one denomination to their equivalents in another denomination. In this connection the following table of conversion factors will be found of service. If the A and B scales are used instead of the C and D scales, a complete set of conversions will be at once obtained. In this case, however, the left-hand A and B scales should be used for the initial setting, any values read on the right-hand A or B scales being read as of tenfold value. With the C and D scales a portion of the one scale will project beyond the other. To read this portion of the scale, the cursor or runner is brought to whichever index of the C scale falls within the rule, and the slide moved until the other index of the C scale coincides with the cursor, when the remainder of the equivalent values can then be read off. It must be remembered that if the slide is moved in the direction of notation (to the _right_), the values read thereon have a tenfold _greater_ value; if the slide is moved to the _left_, the readings thereon are _decreased_ in a tenfold degree. Although preferred by many, in the form given, the case is obviously one of multiplication, and is so treated in the Data Slips at the end of the book.

TABLE OF CONVERSION FACTORS. ─────────────────────────────────────────────────────────────── GEOMETRICAL EQUIVALENTS. ──────────────────────────┬──────────────────────────┬───────── SCALE C. │ SCALE D. │If C = 1, │ │ D = ──────────────────────────┼──────────────────────────┼───────── Diameter of circle │Circumference of circle │3·1416 „ „ │Side of inscribed square │0·707 „ „ │„ equal square │0·886 „ „ │„ „ equilateral │ │ triangle │1·346 Circum. of circle │„ inscribed square │0·225 „ „ │„ equal square │0·282 Side of square │Diagonal of square │1·414 Square inch │Circular inch │1·273 Area of circle │Area of inscribed square │0·636 ──────────────────────────┴──────────────────────────┴───────── MEASURES OF LENGTH. ──────────────────────────┬──────────────────────────┬───────── Inches │Millimetres │25·40 „ │Centimetres │2·54 8ths of an inch │Millimetres │3·175 16ths „ „ │„ │1·587 32nds „ „ │„ │0·794 64ths „ „ │„ │0·397 Feet │Metres │0·3048 Yards │„ │0·9144 Chains │„ │20·116 Miles │Kilometres │1·609 ──────────────────────────┴──────────────────────────┴───────── MEASURES OF AREA. ──────────────────────────┬──────────────────────────┬───────── Square inches │Square centimetres │6·46 Circular „ │„ „ │5·067 Square feet │„ metres │0·0929 „ yards │„ „ │0·836 „ miles │„ kilometres │2·59 „ „ │Hectares │259·00 Acres │„ │0·4046 ──────────────────────────┴──────────────────────────┴───────── MEASURES OF CAPACITY. ──────────────────────────┬──────────────────────────┬───────── Cubic inches │Cubic centimetres │16·38 „ „ │Imperial gallons │0·00360 „ „ │U.S. gallons │0·00432 „ „ │Litres │0·01638 Cubic feet │Cubic metres │0·0283 „ „ │Imperial gallons │6·23 „ „ │U.S. gallons │7·48 „ „ │Litres │28·37 „ yards │Cubic metres │0·764 Imperial gallons │Litres │4·54 „ „ │U.S. gallons │1·200 Bushels │Cubic metres │0·0363 „ │„ feet │1·283 ──────────────────────────┴──────────────────────────┴───────── MEASURES OF WEIGHT. ──────────────────────────┬──────────────────────────┬───────── Grains │Grammes │0·0648 Ounces (Troy) │„ │31·103 „ (Avoird.) │„ │28·35 „ „ │Kilogrammes │0·02835 Pounds (Troy) │„ │0·3732 „ (Avoird.) │„ │0·4536 Hundredweights │„ │50·802 Tons │„ │1016·4 „ │Metric tonnes │1·016 ──────────────────────────┴──────────────────────────┴───────── COMPOUND FACTORS—VELOCITIES. ──────────────────────────┬──────────────────────────┬───────── Feet per second │Metres per second │0·3048 „ „ │„ minute │18·288 „ „ │Miles per hour │0.682 „ minute │Meters per second │0·00508 „ „ │„ minute │0·3048 „ „ │Miles per hour │0·01136 Yards per „ │„ „ │0·0341 Miles per hour │Metres per minute │26·82 Knots │„ „ │30·88 „ │Miles per hour │1·151 ──────────────────────────┴──────────────────────────┴───────── COMPOUND FACTORS—PRESSURES. ──────────────────────────┬──────────────────────────┬───────── Pounds per sq. inch │Grammes per sq. mm. │0·7031 „ „ │Kilos. per sq. centimetre │0·0703 „ „ │Atmospheres │0·068 „ „ │Head of water in inches │27·71 „ „ │„ „ feet │2·309 „ „ │„ „ metres │0·757 „ „ │Inches of Mercury │2·04 Inches of water │Pounds per square inch │0·0361 „ „ │Inches of mercury │0·0714 „ „ │Pounds per square foot │5·20 Inches of mercury │Atmospheres │0·0333 Atmospheres │Metres of water │10·34 „ │Kilos. per sq. cm. │1·033 Feet of water │Pounds per square foot │62·35 „ „ │Atmospheres │0·0294 „ „ │Inches of mercury │0·883 Pounds per sq. foot │„ „ │0·01417 „ „ │Kilos. per square metre │4·883 „ „ │Atmospheres │0·000472 Pounds per sq. yard │Kilos. per square metre │0·5425 Tons per sq. inch │„ square mm. │1·575 „ sq. foot │Tonnes per square metre │10·936 ──────────────────────────┴──────────────────────────┴───────── COMPOUND FACTORS—WEIGHTS, CAPACITIES, ETC. ──────────────────────────┬──────────────────────────┬───────── Pounds per lineal ft. │Kilos. per lineal metre │1·488 „ per lineal yd. │„ „ „ │0·496 „ per lineal mile │Kilos. per kilometre │0·2818 Tons „ „ │Tonnes „ │0·6313 Feet „ „ │Metres „ │1·894 Pounds per cubic in. │Grammes per cubic cm. │27·68 „ per cubic ft. │Kilos. per cubic metre │16·02 „ per cubic yd. │„ „ „ │0·593 Tons per cubic yard │Tonnes „ „ │1·329 Cubic yds. per pound │Cubic metres per kilo. │1·685 „ per ton │„ „ per tonne │0·7525 Cubic inch of water │Weight in pounds │0·03608 Cubic feet of water │„ „ │62·35 „ „ │„ kilos │28·23 „ „ │Imperial gallons │6·235 „ „ │U.S. gallons │7·48 Litre of water │Cubic inches │61·025 Gallons of water │Weight in kilos │4·54 Pounds of fresh water │Pounds of sea water │1·026 Grains per gallon │Grammes per litre │0·01426 Pounds per gallon │Kilos. per litre │0·0998 „ per U.S. gal. │„ „ │0·115 ──────────────────────────┴──────────────────────────┴───────── COMPOUND FACTORS—POWER UNITS, ETC. ──────────────────────────┬──────────────────────────┬───────── British Ther. Units. │Kilogrammetres. │108 „ „ │Joules │1058 „ „ │Calories (Fr. Ther. units)│0·252 „ „ per sq. ft. │„ per square metre │2·713 „ „ per pound │„ per kilogramme │0·555 Pounds per sq. ft. │Dynes, per sq. cm. │479 Foot-pounds │Kilogrammetres │0·1382 „ „ │Joules │1·356 „ „ │Thermal Units │0·00129 „ „ │Calorie │0·000324 Foot-tons │Tonne-metres │0·333 Horse-power │Force decheval (Fr.H.P.) │1·014 „ „ │Kilowatts │0·746 Pounds per H.P. │Kilos. per cheval │0·447 Square feet per H. P. │Square metres per cheval │0·0196 Cubic „ „ │Cubic „ „ │0·0279 Watts │Ther. Units per hour │3·44 „ │Foot-pounds per second │0·73 „ │„ per minute │44·24 Watt-hours │Kilogrammetres │367 „ „ │Joules │3600 Kilogrammetres │„ │9·806 ──────────────────────────┴──────────────────────────┴─────────

_Inverse Proportion._—If “more” requires “less,” or “less” requires “more,” the case is one of _inverse_ proportion, and although it will be seen that this form of proportion is quite readily dealt with by the preceding method, the working is simplified to some extent by inverting the slide so that the C scale is adjacent to the A scale. By the aid of the cursor, the values on the inverted C (or Ɔ) scale, and on the D scale, can be then read off. These will now constitute a series of inverse ratios. For example, in the proportion

─────────── Ɔ 8 4 ─────────── D 1·5 3

the 4 on the Ɔ scale is brought opposite 3 on D, when under 8 on Ɔ is found 1·5 on D.[5]

GENERAL HINTS ON THE ELEMENTARY USES OF THE SLIDE RULE.

Before the more complex operations of involution, evolution, etc., are considered, a few general hints on the use of the slide rule for elementary operations may be of service, especially as these will serve to enforce some of the more important points brought out in the preceding sections.

Always use the slide rule in as _direct_ a light as possible.

Study the manner in which the scales are divided. Follow the graduations of the C and D scales from 1 to 10, noting the values given by each successive graduation and how these values change as we follow along to the right. Do the same with the two halves of the A and B scales and note the difference in the value of the subdivisions, due to the shorter scale-lengths.

Practise reading values by setting 1 on C to some value on D and reading under 2, 3, 4, etc., on C, checking the readings by mental arithmetic. To the same end, find squares, square roots, etc., comparing the results with the actual values as given in tables. Practise setting both slide and cursor to values taken at random. Aim at accuracy; speed will come with practice.

When in doubt as to any method of working, verify by making a simple calculation of the same form.

Follow the orthodox methods of working until entirely confident in the use of the instrument, and even then do not readily make a change. If any altered procedure is adopted, first work a simple case and guard carefully against unconsciously lapsing into the usual method during the operation.

Unless the calculation is of a straightforward character, time taken in considering how best to attack it (rearranging the expression if desirable) is generally time well spent.

In setting two values together, set the cursor to one of them on the rule, and bring the other, on the slide, to the cursor line.

In multiplying factors, as 57 × 0·1256, take the fractional value first. It is easier to set 1 on C to 1256 on D and read under 57 on C, than to reverse the procedure. When both values are eye-estimated, set the cursor to the second factor on C and read the result on D, under the cursor line.

In continuous operations avoid moving the slide further than necessary, by taking the factors in that order which will keep the scale readings as close together as possible.

SQUARES AND SQUARE ROOTS.

We have seen that the relation which the upper scales bear to the lower set is such that over any number on D is its square on A, and, conversely, under any number on A is its square root on D, the same remarks applying to the C and B scales on the slide. Taking the values engraved on the rule, we have on D, numbers lying between 1 and 10, and on A the corresponding squares extending from 1 to 100. Hence the squares of numbers between 1 and 10, or the roots of numbers between 1 and 100, can be read off on the rule by the aid of the cursor. All other cases are brought within these ranges of values by factorising with powers of 10, as before explained.

The more practical rule is the following:—

_To Find the Square of a Number_, set the cursor to the number on D and read the required square on A under the cursor. The rule for

_The Number of Digits in a Square_ is easily deducible from the rule for multiplication. If the square is read on the _left_ scale of A, it will contain _twice_ the number of digits in the original number _less_ 1; if it is read on the _right_ scale of A, it will contain _twice_ the number of digits in the original number.

EX.—Find the square of 114.

Placing the cursor to 114 on D, it is seen that the coinciding number on A is 13. As the result is read off on the _left_ scale of A, the number of digits will be (3 × 2) − 1 = 5, and the answer is read as 13,000. The true result is 12,996.

EX.—Find the square of 0·0093.

The cursor being placed to 93 on D, the number on A is found to be 865. The result is read on the _right_ scale of A, so the number of digits = −2 × 2 = −4, and the answer is read as 0·0000865 [0·00008649].

_Square Root._—The foregoing rules suggest the method of procedure in the inverse operation of extracting the square root of a given number, which will be found on the D scale opposite the number on the A scale. It is necessary to observe, however, that if the number consists of an _odd_ number of digits, it is to be taken on the _left-hand_ portion of the A scale, and the number of digits in the root = (N + 1)/(2), N being the number of digits in the original number. When there is an even number of digits in the number, it is to be taken on the _right-hand_ portion of the A scale, and the root contains _one-half_ the number of digits in the original number.

EX.—Find the square root of 36,500.

As there is an _odd_ number of digits, placing the cursor to 365 on the L.H. A scale gives 191 on D. By the rule there are (N + 1)/(2) = (5 + 1)/(2) = 3 digits in the required root, which is therefore read as 191 [191·05].

EX.—Find √(0·0098.)

Placing the cursor to 98 on the right-hand scale of A (since −2 is an _even_ number of digits), it is seen that the coinciding number on D is 99. As the number of digits in the number is −2, the number of digits in the root will be (−2)/(2) = −1. It will therefore be read as 0·099 [0·09899+].

EX.—Find √(0·098).

The number of digits is −1, so under 98 on the left scale of A, we find 313 on D. By the rule the number in the root will be (−1 +1)/(2) = 0, and the root is therefore read as 0·313 [0·313049+].

EX.—Find √(0·149.)

As the number of digits (0) is _even_, the cursor is set to 149 on the right-hand scale of A, giving 386 on D. By the rule, the number of digits in the root will be (0)/(2) = 0, and the root will be read as 0·386 [0·38605+].

Another method of extracting the square root, by which more accurate readings may generally be obtained, is by using the C and D scales only, with the slide inverted. If there is an _odd_ number of digits in the number, the _right_ index, or if an even number of digits the _left_ index, of the inverted scale Ɔ is placed so as to coincide with the number on D of which the root is sought. Then with the cursor, the number is found on D which coincides with the same number on Ɔ, which number is the root sought.

EX.—Find √(22·2.)

Placing the left index of Ɔ to 222 on D, the two equal coinciding numbers on Ɔ and D are found to be 4·71.

Note that under the cursor line we have the original number, 22·2, on A, and from this the number of digits in the root is determined as before.

The plan of finding the square of a number by ordinary multiplication is often very convenient. The inverse process of finding a square root by trial division is not to be recommended.

To obtain a close value of a root or to verify one found in the usual way, the author has, on occasion, adopted the following plan:—Set 1 (or 10) on B to the number on the A scale (L.H. or R.H. as the case may require), and bring the cursor to the number on D. If the root found is correct, the readings on C under the cursor and on D under the index of C, will be in exact agreement.

If 1 on B is placed to a number _n_ on the L.H. A scale, the student will note that while root _n_ is read on D under 1 on C, the root of 10 _n_ is read on D under 10 on B. Hence, if preferred, the number can be taken always on the first scale of A and the root read under 1 or 10 on B, according to whether there is an odd or even number of digits in the number. Obviously the second root is the first multiplied by √(10).

CUBES AND CUBE ROOTS.

In raising a number to the third power, a combination of the preceding method and ordinary multiplication is employed.

TO FIND THE CUBE OF A NUMBER.—_Set the_ L.H. _or_ R.H. _index of C to the number on D, and opposite the number_ ON THE LEFT-HAND _scale of B read the cube on the_ L.H. _or_ R.H. _scale of A_.

By this rule four scales are brought into requisition. Of these, the D scale and the L.H. B scale are _always_ employed, and are to be read as of equal denomination. The values assigned to the L.H. and R.H. scales of A will be apparent from the following considerations.

Commencing with the indices of C and D coinciding, and moving the slide to the right, it will be seen that, working in accordance with the above rule, the cubes of numbers from 1 to 2·154 (= ∛(10)) will be found on the first or L.H. scale of A. Moving the slide still farther to the right, we obtain _on the_ R.H. _A scale_ cubes of numbers from 2·154 to 4·641 (or ∛(10) to ∛(100)). Had we a _third_ repetition of the L.H. A scale, the L.H. index of C could be still further traversed to the right, and the cubes of numbers from 4·641 to 10 read off on this prolongation of A. But the same end can be attained by making use of the R.H. index of C, when, traversing the slide to the right as before, the cubes of numbers from 4·641 to 10 on D can be read off _on the_ L.H. _A scale_ over the corresponding numbers on the L.H. B scale. Hence, using the L.H. index of C, the readings on the L.H. A scale may be regarded comparatively as units, those on the R.H. A scale as tens; while for the hundreds we again make use of the L.H. A scale in conjunction with the _right-hand_ index of C.

By keeping these points in view, the number of digits in the cube (N) of a given number (_n_) are readily deduced. Thus, if the units scale is used, N = 3_n_ − 2; if the tens scale, N = 3_n_ − 1; while if the hundreds scale be used, N = 3_n_. Placed in the form of rules:—

N = 3_n_ − 2 when the product is read on the L.H. scale of A with the slide to the _right_ (units scale).

N = 3_n_ − 1 when the product is read on the R.H. scale of A; slide to the _right_ (tens scale).

N = 3_n_ when the product is read on the L.H. scale of A with the slide to the _left_ (hundreds scale).

With decimals the same rule applies, but, as before, the number of digits must be read as −1, −2, etc., when one, two, etc., cyphers follow immediately after the decimal point.

EX.—Find the value of 1·4^3.

Placing the L.H. index of C to 1·4 on D, the reading on A opposite 1·4 on the L.H. scale of B is found to be about 2·745 [2·744].

EX.—Find the value of 26·4^3.

Placing the L.H. index of C to 26·4 on D, the reading on A opposite 26·4 on the L.H. scale of B is found to be about 18,400 [18,399·744].

EX.—Find the value of 7·3^3.

In this case it becomes necessary to use the R.H. index of C, which is set to 7·3 on D, when opposite 7·3 on the L.H. scale of B is read 389 [389·017] on A.

EX.—Find the value of 0·073^3.

From the setting as before it is seen that the number of digits in the number must be multiplied by 3. Hence, as there is −1 digit in 0·073, there will be −3 in the cube, which is therefore read 0·000389.

The last two examples serve to illustrate the principle of factorising with powers of 10. Thus

0·073 = 7·3 × 10^{−2}; 0·073^3 = 7·3^3 × (10^{−2})^3 = 389 × 10^{−6} = 0·000389.

_Cube Root_ (_Direct Method_).—One method of extracting the cube root of a number is by an inversion of the foregoing operation. Using the same scales, _the slide is moved either to the right or left until under the given number on A is found a number on the_ L.H. _B scale, identical with the number simultaneously found on D under the right or left index of C_. This number is the required cube root.

From what has already been said regarding the combined use of these scales in cubing, it will be evident that in extracting the cube root of a number, it is necessary, in order to decide which scales are to be used, to know the number of figures to be dealt with. We therefore (as in the arithmetical method of extraction) point off the given number into sections of three figures each, commencing at the decimal point, and proceeding to the left for numbers greater than unity, and to the right for numbers less than unity. Then if the first section of figures on the left consists of—

1 figure, the number will evidently require to be taken on what we have called the “units” scale—_i.e._, on the L.H. scale of A, using the L.H. index of C.

If of 2 figures, the number will be taken on the “tens” scale—_i.e._, on the R.H. scale of A, using the L.H. index of C.

If of 3 figures, the number will be taken on the “hundreds” scale—_i.e._, on the L.H. scale of A, using the R.H. index of C.

To determine the number of digits in cube roots it is only necessary to note that when the number is pointed off into sections as directed, there will be one figure in the root for every section into which the number is so divided, whether the _first_ section consists of 1, 2, or 3 digits.

Of numbers wholly decimal, the cube roots will be decimal, and for every group of _three_ 0s immediately following the decimal point, _one_ 0 will follow the decimal point in the root. If necessary, 0s must be added so as to make up complete multiples of 3 figures before proceeding to extract the root. Thus 0·8 is to be regarded as 0·800, and 0·00008 as 0·000080 in extracting cube roots.

EX.—Find ∛(14,000.)

Pointing the number off in the manner described, it is seen that there are _two_ figures in the first section—viz., 14. Setting the cursor to 14 on the R.H. scale of A, the slide is moved to the right until it is seen that 241 on the L.H. scale of B falls under the cursor, when 241 on D is under the L.H. index of C. Pointing 14,000 off into sections we have 14 000—that is, _two_ sections. Therefore, there are two digits in the root, which in consequence will be read 24·1 [24·1014+].

EX.—Find ∛(0·162.)

As the divisional section consists of _three_ figures, we use the “hundreds” scale. Setting the cursor to 0·162 on the L.H. A scale, and using the R.H. index of C, we move the slide to the left until under the cursor 0·545 is found on the L.H. B scale, while the R.H. index of C points to 0·545 on D, which is therefore the cube root of 0·162.

EX.—Find ∛(0·0002.)

To make even multiples of 3 figures requires the addition of 00; we have then 200, the cube root of which is found to be about 5·85. Then, since the first divisional group consists of 0s, one 0 will follow the decimal point, giving ∛(0·0002) = 0·0585 [0·05848].

_Cube Root (Inverted Slide Method)._—Another method of extracting the cube root involves the use of the inverted slide. Several methods are used, but the following is to be preferred:—_Set the_ L.H. _or_ R.H. _index of the slide to the number on A, and the number on ᗺ (i.e., B inverted), which coincides with the same number on D, is the required root._

Setting the slide as directed, and using first the L.H. index of the slide and then the R.H. index, it is always possible to find _three_ pairs of coincident values. To determine which of the three is the required result is best shown by an example.

EX.—Find ∛(5,) ∛(50,) and ∛(500.)

Setting the R.H. index of the slide to 5 on A, it is seen that 1·71 on D coincides with 1·71 on ᗺ. Then setting the L.H. index to 5 on A, further coincidences are found at 3·68 and at 7·93, the three values thus found being the required roots. Note that the first root was found on that portion of the D scale lying under 1 to 5 on A; the second root on that portion lying under 5 to 50 on A; and the third root on that portion of D lying under 50 to 100 on A. In this connection, therefore, scale A may always be considered to be divided into three sections—viz., 1 to _n_, _n_ to 10_n_, and 10_n_ to 100. For all numbers consisting of 1, 1 + 3, 1 + 6, 1 + 9—_i.e._, of 1, 4, 7, 10, or −2, −5, etc., figures—the coincidence under the first section is the one required. If the number has 2, 5, 8, or −1, −4, −7, etc., figures, the coincidence under the second section is correct, while if the number has 3, 6, 9, or 0, −3, etc., figures, the coincidence under the last section is that required. The number of digits in the root is determined by marking off the number into sections, as already explained.

_Cube Root (Pickworth’s Method)._—One of the principal objections to the two methods described is the difficulty of recollecting which scales are to be employed and with which index of the slide they are to be used. With the direct method another objection is that the readings to be compared are often some distance apart, the maximum distance intervening being _two-thirds_ of the length of the rule. To carry the eye from one to another is troublesome and time-taking. With the inverted scale method the reading of a scale reversed in direction and with the figures inverted is also objectionable.

With the author’s method these objections are entirely obviated. The _same scales and index are always used_, and are read in their normal position. The three roots of _n_, 10_n_ and 100_n_ (_n_ being less than 10 and not less than 1) are given with one setting and appear in their natural sequence, no traversing of the slide being needed. The readings to be compared are always close together, the maximum distance between them being _one-sixth_ of the length of the rule. The setting is always made in the earlier part of the scales where closer readings can be obtained, and finally, if desired, the result may be readily verified on the lower scales by successive multiplication.

For this method two gauge points are required on C. To conveniently locate these, set 53 on C to 246 on D; join 1 on D to 1 on A with a straight-edge and with a needle point draw a short fine line on C. Set 246 on C to 53 on D, and repeat the process at the other end of the rule. The gauge points thus obtained (dividing C into three equal parts) will be at 2·154 and 4·641, and should be marked ∛(10) and ∛(100) respectively.[6]

EX.—Find ∛(2·86,) ∛(28·6) and ∛(286).

Set cursor to 2·86 on A and drawing the slide to the right find 1·42 under 1 on C, when 1·42 on B is under the cursor. Then reading under 1, ∛(10) and ∛(100,) we have

∛(2·86) = 1·42; ∛(28·6) = 3·06 and ∛(286) = 6·59.

It will be seen that factorising with powers of 10, we multiply the initial root by ∛(10) and ∛(100). Obviously the three roots will always be found on D, in their natural order and at intervals of one-third the length of the rule. The number of digits in the roots of numbers which do not lie between 1 and 1000, is found as before explained.

In any method of extracting cube roots in which the slide has to be adjusted to give equal readings on B and D, the author has found it of advantage to adopt the following plan:—The cursor being set to, say, 4·8 on A, bring a near _main_ division line on B, as 1·7, to the cursor; then 1 on C is at 1·68 on D. The difference in the readings is two small divisions on D, and moving the slide forward by _one-third the space representing this difference_, we obtain 1·687 as the root required. With a little practice it is possible to obtain more accurate results by this method than by comparing the reading on D with that on the less finely-graded B scale.

MISCELLANEOUS POWERS AND ROOTS.

In addition to squares and cubes, certain other powers and roots may be readily obtained with the slide rule.

_Two-thirds Power._—The value of N^⅔ is found on A over ∛̅N on D. The number of digits is decided by the rule for squares, working from the number of digits in the cube root. It will often be found preferable to treat N^⅔ as N ÷ ∛̅N, as in this way the magnitude of the result is much more readily appreciated.

_Three-two Power._—N^{³⁄₂} can be obtained by cubing the square root, deciding the number of digits in each process. For the reason just given, it is preferable to regard N^{³⁄₂} as N × √̅N.

_Fourth Power._—For N^4 set the index of C to N on D and over N on C read N^4 on A; or find the square of the square of N, deciding the number of digits at each step.

_Fourth Root._—Similarly for ∜̅N, take the square root of the square root.

_Four-third Power._—N^{⁴⁄₃} = N^{1·33} (useful in gas-engine diagram calculations) is best treated as N × ∛̅N.

Other powers can be found by repeated multiplication. Thus setting 1 on B to N on A, we have on A, N^2 over N; N^3 over N^2; N^4 over N^3; N^5 over N^4, etc. In the same way, setting N on B to N on D, we can read such values as N^¾, N^⅞, etc.

POWERS AND ROOTS BY LOGARITHMS.

For powers or roots other than those of the simple forms already discussed, it is necessary to employ the usual logarithmic process. Thus to find _a^n_ = _x_, we multiply the logarithm of _a_ by _n_, and find the number _x_ corresponding to the logarithm so obtained. Similarly, to find _ⁿ√̅a_ = _x_ we divide the logarithm of _a_ by _n_, and find the number _x_ corresponding to the resulting logarithm.

_The Scale of Logarithms._—Upon the back of the slide of the Gravêt and similar slide rules there will be found three scales. One of these—usually the centre one—is divided equally throughout its entire length, and figured from right to left. It is sometimes marked L, indicating that it is a scale giving logarithms. The whole scale is divided primarily into ten equal parts, and each of these subdivided into 50 equal parts. In the recess or notch in the right-hand end of the rule is a reference mark, to which any of the divisions of this evenly-divided scale can be set.

As this decimally-divided scale is equal in length to the logarithmic scale D, and is figured in the reverse direction, it results that when the slide is drawn to the right so that the L.H. index of C coincides with any number on D, the reading on the equally-divided scale will give the decimal part of the logarithm of the number taken on D. Thus if the L.H. index of C is placed to agree with 2 on D, the reading of the back scale, taken at the reference mark, will be found to be 0·301, the logarithm of 2. It must be distinctly borne in mind that the number so obtained is the _decimal part_ or _mantissa_ of the logarithm of the number, and that to this the characteristic must be prefixed in accordance with the usual rule—viz., _The integral part, or characteristic of a logarithm is equal to the number of digits in the number, minus 1. If the number is wholly decimal, the characteristic is equal to the number of cyphers following the decimal point, plus 1._ In the latter case the characteristic is negative, and is so indicated by having the minus sign written _over_ it.

To obtain any given power or root of a number, the operation is as follows:—Set the L.H. index of C to the given number on D, and turning the rule over, read opposite the mark in the notch at the right-hand end of the rule, the decimal part of the logarithm of the number. Add the characteristic according to the above rule, and multiply by the exponent of the power, or divide by the exponent of the root. Place the _decimal part_ of the resultant reading, taken on the scale of equal parts, opposite the mark in the aperture of the rule, and read the answer on D under the L.H. index of C, pointing off the number of digits in the answer in accordance with the number of the characteristic of the resultant.

EX.—Evaluate 36^{1·414}.

Set 1 on C to 36 on D and read the decimal part of log. 36 on the scale of logarithms on the back of the slide. This value is found to be 0·556. As there are two digits in the number, the characteristic will be 1; hence log. 36 = 1·556. Multiply by 1·414, using the C and D scales, and obtain 2·2 as the log. of the result. Set the decimal part, 0·2, on the log. scale to the mark in the notch at the end of the rule and read 1585 on D under 1 on C. Since the log. of the result has a characteristic 2, there will be 3 digits in the result, which is therefore read as 158·5.

This example will suffice to show the method of obtaining the nth power or the _n_th root of _any_ number.

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 1·25^{1·67} 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 1·52^{¹⁷⁄₁₆}, 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.

[Illustration: FIG. 11.]

The converse procedure for obtaining the _n_th root of a number N will obviously resolve itself into obtaining (1)/(_n_)th 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 _v_^{1·35}, 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 _v^n_. Thus for _p_ _v^n_ = 39·5 × 4·9^{1·35}, 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.

[Illustration: 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 N^2 on A. Evidently, since in the two similar triangles A O N^2 and N _t_ N^2 the length of A O is twice that of N _t_, it results that A N^2 = 2 A N. In general, then, to find the _n_th 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 N^{_n_} on A. The inverse proceeding for finding the _n_th root will be obvious.

[Illustration: 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_ × _b_^2 = _x_. Set the index of C to _b_ on D, and over _a_ on B read _x_ on A.

To solve (_a_^2)/(_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_)/(_a_^2) = _x_. Set _a_ on C to _b_ on A, and over 1 on B read _x_ on A.

To solve (_a_ × _b_^2)/(_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_)/(_c_^2) = _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 √(_a_^3) = _x_. Treat as _a_√_̅a_.

To solve _a_√(_b_^3) = _x_. Treat as _a_√_̅b_ × _b_.

To solve (√_̅a_^3)/(_b_) = _x_. Treat as (√_̅a_ × _a_)/(_b_).

To solve √((_a_^3)/(_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 √((_a_^2 × _b_)/(_c_)) = _x_. Set _c_ on B to _a_ on D, and under _b_ on B read _x_ on D.

To solve (_a_^2 × _b_^2)/(_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_√(_b_^3) 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 × √(57^3) = 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_∛(_b_^2), or _a_ _b_^⅔, is better dealt with by rearranging as _a_ × (_b_)/(∛_b_).

EX.—3·64∛(4·32^2) = 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_ × _b_^{⁴⁄₃} are best dealt with as _a_ × _b_ × ∛_b_.

Factorising expressions sometimes simplifies matters, as, for instance, in _x_^4 − _y_^4 = (_x_^2 + _y_^2)(_x_^2 − _y_^2). 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, D_{1} = ∛((D^4 − _d_^4)/(D)) giving the diameter D_{1} of a solid shaft equal in torsional strength to a hollow shaft whose external and internal diameters are D and _d_ respectively. Rearranging as D_{1} = ∛(((D^2 + _d_^2)(D^2 − _d_^2))/(D)) and taking, as an example, D = 15 in. and _d_ = 7 in., we have D^2 + _d_^2 = 274 and D^2 − _d_^2 = 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 _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