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, we paisqa : By 1st method 300 X 300 Jest Volt Boilex 93 1500 150029 Of ad list and dai to tangent run from p.c. e, and 10000 DOG I 25 distance from other end to said tangent. 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 pas by simple curves. Thusto ko sa od ma bojonqs Jos Measure old chord, and you have the direction of the new. Having found the p. c. c. calculate the offsets from each 10007 chord separately. The above rule for ordinates, although not perfectly accurate, considering the divisor always=2R, while it is variable, is sufficiently near for centres to grade by, when the chord subtends not more 20° The levature. also apply placing centre points between stations. Thus zona de, bidha sala gd bonia an dicasts wil On a chord of 100 feet, radius 1000 feet, let sit be required to locate a point 30 feet from one end and 70 feet from the other Corpo Frodo to gran oil but os borimpot od al von sangqué Then we have =105 97100 "u a to dost 60€ of quibner | brodo lo dtgan asrig Joel 000 30 x 70 2000 38 0001 Let L= length of rail and R = length of radius. 97 Then paid 2 2 R Ꭱ spring in inches Let the rail be 20 feet long, and the radius 1200 feet. Then 24 × 20a 9600 &q août ant dregnar of To Hence the 8 1200 1200 gm bire of bing quilts plevitoogeen sbrodo tu ́abos rei ewenoM 1024 times are 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, edgga sanga of Svenong oras mentio eniyani 10) doctail oft eruccoon buk J2 noxqisagart ad Janlapins predio ult To find the length of chord for any rate of curvature (less than 8°) not specified in the Table of Chords (p 414.)o olquis qu 008 / 008 Let it be required to find the length of chord of chord corresponding to 800 feet of curve for a 7 10 curve. 7° curve gives 769.01 abtesque Inodo otannosh plotting 7° 16 curve gives 76679 of ordo GT live á eldener at Differences 2:2215) alt gaireblenes Then 15 10 :: 2.22 148, and 769 01 — 148 — 767-53 abis # 2:22; 74, and 76679 + 074—76753. 100 The result, as obtained by the table of sines, is 767 54, only of a foot difference.st 0000 aviber,4961 005 to Poiko s at That is, sine 28° 40′ x radius 800 x 2767 54. Jabg a gizool Suppose now it be required to find the length of chord corresponding to 950 feet of a 6° curve. 20 { 867.45 955-37 1822.82 911-41 Mean 950 Now sine 28° 30′ x radius 955 37 × 2 being only of a difference, so that length of chord = 91171, will be suffi cient for ordinary purposes. For common rates ofvature 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°, } BCE n = 6°, and the perpendicular distance DC=p=466} ft. Required the amount of curvature in the arc A B, its reversion BC, and the length of the common radius OB MB by which the arcs A B and B Care described Let mnat. vers. sine DA C, and n nat. vers. sine BCE Or curvature A B =m+r, and curvature BC= n + x.f 1 To find a we have, a m + n V. s. 18+ v. s. 6° 0048944 + 0005478" 2 =0.027211 2 nat. vers. sine 13° 23 48. 15° 23′ 48′′ – 31° 23′ 48′′ and BC =6° + Then by principles from which Proposi OBR, Therefore are A B = 18° + we have twice nat. vers. sine AB nat. vers. sine (m-n) R, Or nat. vers. sine ALG 13° 23' 48" 3° 21' If DC cannot be measured, measure A C and 1500 ft. we have 1500 × sine 18° 1500 x 0-30902 +DGE being equal to A OB, A OB~ m = ALG=CLE. gives 579 ft: arc BC" 3° 21' |