Part 3

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.