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.