Suppose the tangent DC produced 100 feet to E, measure CX= 100 feet, measure E X. Now suppose it is 21 feet. Now the deflection of 1° for 100 feet is 1.75, and vature. 21 6 Suppose it is required to divide the curve into 6 stations. Then 35, the deflection for 2° in 100 feet. Hence it is a 2° curve. Or 12° divided by 6 stations gives a 2° curve also. The deflection being 1:75 from tangent to curve. Between two fixed points to supply the intermediate points by ordinates from the chord. FIG. 22. By what has been previously demonstrated, the middle ordinate 4 to 4 will be expressed by 4 x 4 At 3 the deflection from tan denominator, its relative value may be expressed by 4 x 4. At points 3 and 5 on chord the distance will be (4 × 4)-(1 × 1) 3 x 515. At 6 and 2 (4· × 4) — (2 × 2) = 2 × 6 = 12. ` At Then we observe that the sum of the two factors is equal, namely the length of chord. Hence the following rule : Multiply together the two segments of the chord or distance, divide by twice the radius, and the result is the distance from chord Had it been a 1° curve of 5730 feet radius, the ordinates would have been: so in proportion to any other rate of curvature in degrees. Hence when the rate of curvature is in degrees and no minutes, we have the following rule: Multiply together the distances in stations each side of the point, and the rate of curvature, deduct from this product of itself, the remainder will be the ordinate required. *The departure in 100 ft. of a 1° curve from tangent being 75= } of a foot. CASE 2D. Suppose that between the points 0 and 8 there occurs a point of c. c., for instance at 3 or 5, the curves compound from a 5000 feet radius to a 4000 feet radius. Measure from ends of chords respectively 11.25 and 25 feet; on this line, at a distance 300 feet from 11.25 offset, and 500 feet from 25 feet offset, would be the point of compound curvature sought. Or imagine either curve produced to a point opposite the end of the other; calculate by Proposition XI., and measure the distance. between the two curves, then on the new chord find the p. c. c. as by simple curves. Thus: Measure 2.25 from the old chord, and you have the direction of the new. Having found the p. c. c. calculate the offsets from each chord separately. The above rule for ordinates, although not perfectly accurate, considering the divisor always 2 R, while it is variable, is sufficiently near for centres to grade by, when the chord subtends not more than 20° curvature. This rule will also apply to placing centre points between stations. Thus On a chord of 100 feet, radius 1000 feet, let it be required to locate a point 30 feet from one end and 70 feet from the other. 30 × 70 Then we have = 1·05. 2000 FOR SPRINGING RAILS. Let L= length of rail and R = length of radius. Then: L2 ( 241 ) = spring in sixteenths of an inch. R EXAMPLE. Let the rail be 20 feet long, and the radius 1200 feet. Then 24 times the square of the length of rail in feet divided by length of radius in feet, will give the spring in middle in sixteenths of an inch. To find the length of chord for any rate of curvature (less than 8°) not specified in the Table of Chords (p 414.) EXAMPLE. Let it be required to find the length of chord corresponding to 800 feet of curve for a 7° 10' curve. Then 15: 10 :: 2·22: 1·48, and 769 01 — 1·48 — 767·53 ; or 15: 5 :: 2·22: 74, and 766.79 ± 0·74 = 767·53. The result, as obtained by the table of sines, is 767 54, only Too of a foot difference. 1 100 That is, sine 28° 40′ x radius 800 × 2767 54. Suppose now it be required to find the length of chord corresponding to 950 feet of a 6° curve. Now sine 28° 30′ × radius 955·37 × 2 3 length of chord 91171, being only of a foot difference, so that this table will be suffi cient for ordinary purposes. For common rates of curvature for a less distance, say 650 feet, the variations from the true length would be scarcely perceptible. PROBLEM.-Let A and C be two fixed tangent points, the positions of whose tangents are determined by the angles DAC: = m = 18°, B C E = n = 6o, and the perpendicular distance D C p = 4633 ft.* Required the amount of curvature in the arc A B, its reversion B C, and the length of the common radius 0 B = MB by which the arcs A B and B C are described. A B F Let m Let nat. vers. sine DA C, and n = nat. vers. sine BCE. n). Or curvature A Bm + x, and curvature BC= n + x.† m + n v. s. 18+ v. s. 6° To find x we have, a 0.048944 + 0:005478 2 Therefore arc A B 18° +15° 23′48′′ = 31° 23' 48" and BC 6°+ 13° 23′ 48′′ — 19° 23' 48". Then by principles from which Proposition XII. is derived, to find OB=R, * If DC cannot be measured, measure AC and calculate D C. 1500 ft. we have 1500 x sine 18° - 1500 × 0·30902 = 463·53. DGE being equal to A OB, A OB-m=ALG = CLG. being_equal nat. vers. sine A LG 13° 23′ 48′′, Thus if A C = Therefore x = |