Part 5

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.

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.