Part 8

Owing to the rapidly diminishing differences of the values of the sines as the upper end of the scale is approached, the sines of angles between 60° and 90° cannot be accurately determined in the foregoing manner. It is therefore advisable to calculate the value of the sine by means of the formula:

Sine θ = 1 − 2 sin^2 (90 − θ)/(2).

To determine the value of sin^2 (90 − θ)/(2). With the slide in the normal position, set the value of (90 − θ)/(2). on S to the index on the under-side of the rule, and read off the value _x_ on B under the R.H. index of A. Without moving the slide find _x_ on A, and read under it on B the value required.

EX.—Find value of sine 79° 40′.

Sine 79° 40′ = 1 − 2sin^2 5° 10′.

But sine 5° 10′ = 0·0900, and under this value on A is 0·0081 on B. Therefore sine 79° 40′ = 1 − 0·0162 = 0·9838.

The sines of very small angles, being very nearly proportional to the angles themselves, are found by direct reading. To facilitate this, some rules are provided with two marks, one of which, a single accent (′), corresponds to the logarithm of (1)/(sine 1′) and is found at the number 3438. The other mark—a double accent (″)—corresponds to the logarithm of (1)/(sine 1″) and is found at the number 206,265. In some rules these marks are found on either the A or the B scales; sometimes they are on both. In either case the angle on the one scale is placed so as to coincide with the significant mark on the other, and the result read off on the first-named scale opposite the index of the second.

In sines of angles under 3″, the number of integers in the result is −5; while it is −4 for angles from 3″ to 21″; −3 from 21″ to 3′ 27″; and −2 from 3′ 27″ to 34′ 23″.

EX.—Find sine 6′.

Placing the significant mark for minutes coincident with 6, the value opposite the index is found to be 175, and by the rule above this is to be read 0·00175. For angles in seconds the other significant mark is used; while angles expressed in minutes and seconds are to be first reduced to seconds. Thus, 3′ 10″ = 190″.

_Tangents of Angles._—There remains to be considered the third scale found on the back of the slide, and usually distinguished from the others by being lettered T. In most of the more recent forms of rule this scale is placed near the lower edge of the slide, but in some arrangements it is found to be the centre scale of the three. Again, in some rules this scale is figured in the same direction as the scale of sines—viz., from left to right,—while in others the T scale is reversed. In both cases there is now usually an aperture formed in the back of the left extremity of the rule, with an index mark similar to that already referred to in connection with the scale of sines. Considering what has been referred to as the more general arrangement, the method of determining the tangents of angles may be thus explained:—

The tangent scale will be found to commence, in some rules, at about 34′, or, precisely, at the angle whose tangent is 0·01. More usually, however, the scale will be found to commence at about 5° 43′, or at the angle whose tangent is 0·1. The other extremity of the scale corresponds in all cases to 45°, or the angle whose tangent is 1. This explanation will suggest the method of using the scale, however it may be arranged. If the graduations commence with 34′, the T scale is to be used in conjunction with the right and left scales of A; while if they commence with 5° 43′ it is to be used in conjunction with the D scale.

In the former case the slide is to be placed in the rule so that the T scale is adjacent to the A scales, and, with the right and left indices coinciding, when opposite any angle on T will be found its tangent on A. From what has been said above, it follows that the tangents read on the L.H. scale of A have values extending from 0·01 to 0·1; while those read on the R.H. scale of A have values from 0·1 to 1·0. Otherwise expressed, to the values of any tangent read on the L.H. scale of A a cypher is to be prefixed; while if found on the R.H. scale, it is read directly as a decimal.

EX.—Find tan. 3° 50′.

Placing the slide as directed, the reading on A opposite 3° 50′ on T is found to be 67. As this is found on the L.H. scale of A, it is to be read as 0·067.

EX.—Find tan. 17° 45′.

Here the reading on A opposite 17° 45′ on T is 32, and as it is found on the R.H. scale of A it is read as 0·32.

As in the case of the scale of sines, the tangents may be found without reversing the slide, when a fixed index is provided in the back of the rule for the T scale.

We revert now to a consideration of those rules in which a single tangent scale is provided. It will be understood that in this case the slide is placed so that the scale T is adjacent to the D scale, and that when the indices of both are placed in agreement, the value of the tangent of any angle on T (from 5° 43′ to 45°) may be read off on D, the result so found being read as wholly decimal. Thus tan. 13° 20′ is read 0·237.

If a back index is provided, the slide is used in its normal position, when, setting the angle on the tangent scale to this index, the result can be read on C over the L.H. index of D.

The tangents of angles above 45° are obtained by the formula: Tan. θ = (1)/(tan. (90 − θ)). For all angles from 45° to (90° − 5° 43′) we proceed as follows:—Place (90 − θ) on T to the R.H. index of D, and read tan. θ on D under the L.H. index of T. The first figure in the value thus obtained is to be read as an integer. Thus, to find tan. 71° 20′ we place 90° − 71° 20′ = 18° 40′ on T, to the R.H. index of D, and under the L.H. index of T read 2·96, the required tangent.

The tangents of angles less than 40′ are sensibly proportional to the angles themselves, and as they may therefore be considered as sines, their value is determined by the aid of the single and double accent marks on the sine scale, as previously explained. The rules for the number of integers are the same as for the sines.

Multiplication and division of tangents may be quite readily effected.

EX.—Tan. 21° 50′ × 15 = 6.

Set L.H. index of T to 15 on D, and under 21° 50′ on T read 6 on D.

EX.—Tan. 72° 40′ × 117 = 375.

Set (90° − 72° 40′) = 17° 20′ on T to 117 on D, and under R.H. index of T read 375 on D.

_Cosines of Angles._—The cosines of angles may be determined by placing the scale S with its indices coinciding with those of A, and when opposite (90 − θ) on S is read cos. θ on A. If the result is read on the L.H. scale of A, a cypher is to be prefixed to the value read; while if it is read on the R.H. scale of A, the value is read directly as a decimal. Thus, to determine cos. 86° 30′ we find opposite (90° − 86° 30′) = 3° 30′ on S, 61° on A, and as this is on the L.H. scale the result is read 0·061. Again, to find cos. 59° 20′ we read opposite (90° − 59° 20′) or 30° 40′ on S, 51 on A, and as this is found on the R.H. scale of A, it is read 0·51.

In finding the cosines of small angles it will be seen that direct reading on the rule becomes impossible for angles of less than 20°. It is advisable in such cases to adopt the method described for determining the _sines_ of the _large_ angles of which the complements are sought.

_Cotangents of Angles._—From the methods of finding the tangents of angles previously described, it will be apparent that the cotangents of angles may also be obtained with equal facility. For angles between 5° 45′ and 45°, the procedure is the same as that for finding tangents of angles greater than 45°. Thus, the angle on scale T is brought to the R.H. index of D, and the cotangent read off on D under the L.H. index of T. The first figure of the result so found is to be read as an integer.

If the angle (θ) lies between 45° and 84° 15′, the slide is placed so that the indices of T coincide with those of D, and the result is then read off on D opposite (90 − θ) on T. In this case the value is wholly decimal.

_Secants of Angles._—The secants of angles are readily found by bringing (90 − θ) on S to the R.H. index of A and reading the result on A over the L.H. index of S. If the value is found on the L.H. scale of A, the first figure is to be read as an integer; while if the result is read on the R.H. scale of A, the first _two_ figures are to be regarded as integers.

_Cosecants of Angles._—The cosecants of angles are found by placing the angle on S to the R.H. index of A, and reading the value found on A over the L.H. index of S. If the result is read on the L.H. scale of A, the first figure is to be read as an integer; while if the result is found on the R.H. scale of A, the first _two_ figures are to be read as integers.

It will be noted that some of the rules here given for determining the several trigonometrical functions of angles apply only to those forms of rules in which a single scale of tangents T is used, reading from left to right. For the other arrangements of the scale, previously referred to, some slight modification of the method of procedure in finding the tangents and cotangents of angles will be necessary; but as in each case the nature and extent of this modification is evident, no further directions are required.

THE SOLUTION OF RIGHT-ANGLED TRIANGLES.

From the foregoing explanation of the manner of determining the trigonometrical functions of angles, the methods of solving right-angled triangles will be readily perceived, and only a few examples need therefore be given.

Let _a_ and _b_ represent the sides and _c_ the hypothenuse of a right-angled triangle, and _a_° and _b_° the angles opposite to the sides. Then of the possible cases we will take

(1.) Given _c_ and _a_°, to find _a_, _b_, and _b_°.

The angle _b_° = 90 − _a_°, while _a_ = _c_ sin _a_° and _b_ = _c_ sin _b_°. To find _a_, therefore, the index of S is set to _c_ on A, and the value of _a_ read on A opposite _a_° on S. In the same manner the value of _b_ is obtained.

EX.—Given in a right-angled triangle _c_ = 9 ft. and _a_° = 30°. Find _a_, _b_, and _b_°.

The angle _b_° = 90 − 30 = 60°. To find _a_, set R.H. index of S to 9 on A, and over 30° on S read _a_ = 4·5 ft. on A. Also, with the slide in the same position, read _b_ = 7·8 ft. [7·794] on A over 60° on S.

(2.) Given _a_ and _c_, to determine _a_°, _b_°, and _b_.

In this case advantage is taken of the fact that in every triangle the sides are proportional to the sines of the opposite angles. Therefore, as in this case the hypothenuse c subtends a right angle, of which the sine = 1, the R.H. index (or 90°) on S is set to the length of _c_ on A, when under _a_ on A is found _a_° on S. Hence _b_° and _b_ may be determined.

(3.) Given _a_ and _a_°, to find _b_, _c_, and _b_°.

Here _b_° = (90 − _a_°), and the solution is similar to the foregoing.

(4.) Given _a_ and _b_, to find _a_°, _b_°, and _c_.

To find _a_°, we have tan. _a_° = _a_/_b_, which in the above example will be (4·5)/(7·8) = 0·577. Therefore, placing the slide so that the indices of T coincide with those of D, we read opposite 0·577 on D the value of _a_° = 30°. The hypothenuse _c_ is readily obtained from _c_ = _a_/(sin _a_°).

THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

Using the same letters as before to designate the three sides and the subtending angles of oblique-angled triangles, we have the following cases:—

(1.) Given one side and two angles, as _a_, _a_°, and _b_°, to find _b_, _c_, and _c_°.

In the first place, _c_° = 180° − (_a_° + _b_°); also we note that, as the sides are proportional to the sines of the opposite angles, _b_ = (_a_ sine _b_°)/(sine _a_°) and _c_ = (_a_ sine _c_°)/(sine _a_°).

Taking as an example, _a_ = 45, _a_° = 57°, and _b_° = 63°, we have _c_° = 180 − (57 + 63) = 60°. To find _b_ and _c_, set _a_° on S to _a_ on A, and read off on A above 63° and 60° the values of _b_ (= 47·8) and _c_ (= 46·4) respectively.

(2.) Given _a_, _b_, and _a_°, to find _b_°, _c_°, and _c_.

In this case the angle _a_° on S is placed under the length of side _a_ on A and under _b_ on A is found the angle _b_° on S. The angle _c_° = 180 − (_a_° + _b_°), whence the length _c_ can be read off on A over _c_° on S.

(3.) Given the sides and the included angle, to find the other side and the remaining angles.

If, for example, there are given _a_ = 65, _b_ = 42, and the included angle _c_° = 55°, we have (_a_ + _b_) ∶ (_a_ − _b_) = tan. (_a_° + _b_°)/(2) ∶ tan. (_a_° − _b_°)/(2). Then, since _a_° + _b_° = 180° − 55° = 125°, it follows that (_a_° + _b_°)/(2) = (125°)/(2) = 62° 30′.

By the rule for tangents of angles greater than 45°, we find tan. 62° 30′ = 1·92. Inserting in the above proportion the values thus found, we have 107 ∶ 23 = 1·92 ∶ tan. (_a_° − _b_°)/(2). From this it is found that the value of the tangent is 0·412, and placing the slide with all indices coinciding, it is seen that this value on D corresponds to an angle of 22° 25′. Therefore, since (_a_° + _b_°)/(2) = 62° 30′, and (_a_° − _b_°)/(2) = 22° 25′, it follows that _a_° = 84° 55′, and _b_° = 40° 5′. Finally, to determine the side _c_, we have _c_ = (_a_ sin _c_°)/(sin _a_°) as before.

PRACTICAL TRIGONOMETRICAL APPLICATIONS.

A few examples illustrative of the application of the methods of determining the functions of angles, etc., described in the preceding section, will now be given.

To find the chord of an arc, having given the included angle and the radius.

With the slide placed in the rule with the C and D scales outward, bring one-half of the given angle on S to the index mark in the back of the rule, and read the chord on B under twice the radius on A.

EX.—Required the chord of an arc of 15°, the radius being 23 in.

Set 7° 30′ on S to the index mark in the back of the rule, and under 46 on A read 6 in., the required length of chord on B.

To find the area of a triangle, given two sides and the included angle.

Set the angle on S to the index mark on the back of the rule, and bring cursor to 2 on B. Then bring the length of one side on B to cursor, cursor to 1 on B, the length of the other side on B to cursor, and read area on B under index of A.

EX.—The sides of a triangle are 5 and 6 ft. in length respectively, and they include an angle of 20°. Find the area.

Set 20 on S to index mark, bring cursor to 2 on B, 5 on B to cursor, cursor to 1 on B, 6 on B to cursor, and under 1 on A read the area = 5·13 sq. ft. on B.

To find the number of degrees in a gradient, given the rise per cent.

Place the slide with the indices of T coincident with those of D, and over the rate per cent. on D read number of degrees in the slope on T.

As the arrangement of rule we have chiefly considered has only a single T scale, it will be seen that only solutions of the above problem involving slopes between 10 and 100 per cent. can be directly read off. For smaller angles, one of the formulæ for the determination of the tangents of submultiple angles must be used.

In rules having a double T scale (which is used with the A scale) the value in degrees of any slope from 1 to 100 per cent. can be directly read off on A.

To find the number of degrees, when the gradient is expressed as 1 in _x_.

Place the index of T to _x_ on D, and over index of D read the required angle in degrees on T.

EX.—Find the number of degrees in a gradient of 1 in 3·8.

Set 1 on T to 3·8 on D, and over R.H. index of D read 14° 45′ on T.

Given the lap, the lead and the travel of an engine slide valve, to find the angle of advance.

Set (lap + lead) on B to half the travel of the valve on A, and read the angle of advance on S at the index mark on the back of the rule.

EX.—Valve travel 4½in., lap 1 in., lead ⁵⁄₁₆in. Find angle of advance.

Set 1⁵⁄₁₆ = 1·312 on B to 2·25 on A, and read 35° 40′ on S opposite the index on the back of the rule.

Given the angular advance θ, the lap and the travel of a slide valve, to find the cut-off in percentage of the stroke.

Place the lap on B to half the travel of valve on A, and read on S the angle (the supplement of the _angle of the eccentric_) found opposite the index in the back of the rule. To this angle, add the angle of advance and deduct the sum from 180°, thus obtaining the _angle of the crank_ at the point of cut-off. To the cosine of the supplement of this angle, add 1 and multiply the result by 50, obtaining the percentage of stroke completed when cut-off occurs.

EX.—Given the angular advance = 35° 40′, the valve travel = 4½in., and the lap = 1 in., find the angle of the crank at cut-off and the admission period expressed as a percentage of the stroke.

Set 1 on B to 2·25 on A, and read off on S opposite the index, the supplement of the angle of the eccentric = 26° 20′. Then 180° − (35° 40′ + 26° 20′) = 118° = the crank angle at the point of cut-off. Further, cos. 118° = cos. 62° = sin (90° − 62°) = sin 28°, and placing 28° on S to the back index, the cosine, read on B under R.H. index of A, is found to be 0·469. Adding 1 and placing the L.H. index of C to the result, 1·469, on D, we read off under 50 on C, the required period of admission = 73·4 per cent. on D.

The trigonometrical scales are useful for evaluating certain formulæ. Thus in the following expressions, if we find the angle _a_ such that sin. _a_ = _k_, we can write:—

(_k_)/(√1 − _k^2_) = tan. _a_; (√1 − _k^2_)/(_k_) = cot. _a_; √(1 − _k^2_) = cos. _a_; etc.

In the first expression, take _k_ = 0·298. Place the slide with the sine scale outward and with its indices agreeing with the indices of the rule. Set the cursor to 0·298 on the (R.H.) scale of A, and read 17° 20′ on the sine scale as the angle required. Then under 17° 20′ on the tangent scale, read 0·312 on D as the result.

SLIDE RULES WITH LOG.-LOG. SCALES.

For occasional requirements, the method described on page 45 of determining powers and roots other than the square and cube, is quite satisfactory. When, however, a number of such calculations are to be made, the process may be simplified considerably by the use of what are known as _log.-log._, _logo-log._, or _logometric_ scales, in conjunction with the ordinary scales of the rule. The principle involved will be understood from a consideration of those rules for logarithmic computation (page 8) which refer to powers and roots. From these it is seen that while for the multiplication and division of numbers we _add_ their logarithms, for involution and evolution we require to _multiply_ or _divide_ the logarithms of the numbers by the exponent of the power or root as the case may be. Thus to find 3^{2.3}, we have (log. 3) × 2·3 = log. _x_, and by the ordinary method described on page 45 we should determine log. 3 by the aid of the scale L on the back of the slide, multiply this by 2·3 by using the C and D scales in the usual manner, transfer the result to scale L, and read the value of _x_ on D under 1 on C. By the simpler method, first proposed by Dr. P. M. Roget,[8] the multiplication of log. 3 by 2·3 is effected in the same way as with any two ordinary factors—_i.e._, by adding their logarithms and finding the number corresponding to the resulting logarithm. In this case we have log. (log. 3) + log. 2·3 = log. (log. _x_). The first of the three terms is obviously the _logarithm of the logarithm_ of 3, the second is the simple logarithm of 2·3, and the third the _logarithm of the logarithm of_ the answer. Hence, if we have a scale so graduated that the distances from the point of origin represent the logarithms of the logarithms (the log.-logs.) of the numbers engraved upon it, then by using this in conjunction with the ordinary scale of logarithms, we can effect the required multiplication in a manner which is both expeditious and convenient. Slightly varying arrangements of the log.-log. scale, sometimes referred to as the “P line,” have been introduced from time to time, but latterly the increasing use of exponential formulæ in thermodynamic, electrical, and physical calculations has led to a revival of interest in Dr. Roget’s invention, and various arrangements of rules with log.-log. scales are now available.

_The Davis Log.-Log. Rule._—In the rule introduced by Messrs. John Davis & Son Limited, Derby, the log.-log. scales are placed upon a separate slide—a plan which has the advantage of leaving the rule intact for all ordinary purposes, while providing a length of 40 in. for the log.-log. scales.

In the 10 in. Davis rule one face of the slide, marked E, has two log.-log. scales for numbers greater than unity, the lower extending from 1·07 to 2, and the upper continuing the graduations from 2 to 1000. On the reverse face of the slide, marked -E, are two log.-log. scales for numbers less than unity, the upper extending from 0·001 to 0·5, and the lower continuing the graduations from 0·5 to 0·933. Both sets of scales are used in conjunction _with the lower or D scale of the rule_, which is to be primarily regarded as running from 1 to 10, and constitutes a scale of exponents. In the 20 in. rule the log.-log. scales are more extensive, and are used in conjunction with the upper or A scale of the rule (1 to 100); in what follows, however, the 10 in. rule is more particularly referred to.

It has been explained that on the log.-log. scale the distance of any numbered graduation from the point of origin represents the log.-log. of the number. The point of origin will obviously be that graduation whose log.-log. = 0. This is seen to be 10, since log. (log. 10) = log. 1 = 0. Hence, confining attention to the E scale, to locate the graduation 20, we have log. (log. 20) = log. 1·301 = 0·11397, so that if the scale D is 25 cm. long, the distance between 10 and 20 on the corresponding log.-log. scale would be 113·97 ÷ 4 = 28·49 mm. For numbers less than 10 the resulting log.-logs. will be negative, and the distances will be spaced off from the point of origin in a negative direction—_i.e._, from right to left. Thus, to locate the graduation 5, we have

log. (log. 5) = log. 0·699 = ̅1·844; _i.e._, −1 + 0·844 or −0·156;

so that the graduation marked 5 would be placed 156 ÷ 4 = 39 mm. distant from 10 in a _negative_ direction, and proceeding in a similar manner, the scale may be extended in either direction. In the -E scale, the notation runs in the reverse direction to that of the E scale, but in all other respects it is precisely analogous, the distance from the point of origin (0·1 in this case) to any graduation _x_ representing log. [-log. _x_.]. It follows that of the similarly situated graduations on the two scales, those on the -E scale are the _reciprocals_ of those on the E scale. This may be readily verified by setting, say, 10 on E to (R.H.) 1 on D, when turning to the back of the rule we find 0·1 on -E agreeing with the index mark in the aperture at the right-hand extremity of the rule.

In using the log.-log. scales it is important to observe (1) that the values engraved on the scale are definite and unalterable (_e.g._, 1·2 can only be read as 1·2 and not as 120, 0·0012, etc., as with the ordinary scales); (2) that the upper portion of each scale should be regarded as forming a prolongation to the right of the lower portion; and (3) that immediately above any value on the lower portion of the scale is found the 10th power of that value on the upper portion of the scale. Keeping these points in view, if we set 1·1 on E to 1 on D we find over 2 on D the value of 1·1^2 = 1·21 on E. Similarly, over 3 we find 1·1^3 = 1·331, and so on. Then, reading across the slide, we have, over 2, the value of 1·1^{2 × 10} = 1·1^{20} = 6·73, and over 3 we have 1·1^{3 × 10} = 1·1^{30} = 17·4. Hence the rule:—_To find the value of x^n, set x on E to 1 on D, and over n on D read x^n on E._