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.