Part 26

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:—