Chapter C: N. P (2)

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.

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.

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.

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.