2.
Highway and railroad routes are chosen after studying all possiblelocations. This involves;Use of aerial imagery, satellite imagery and ground surveys as well asanalysis of existing plans and maps. Design requirement with minimalsocial, environmental and financial impact should be considered. { 2 20/01/2013
5.
PI (θ) (E) A (L) (M) B (θ/2) BC (θ/2) EC (L.C.) 2 1 (θ)5 O 20/01/2013
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Curve Equations Tangent Distance T = R tan (θ/2)6 20/01/2013
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Length of Long Chord L.C. = 2R sin (θ/2)7 20/01/2013
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Middle Ordinate M = R[1 - cos (θ/2)]8 20/01/2013
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External Distance E = R[1/ cos (θ/2) -1] or E = R[sec (θ/2) -1]9 20/01/2013
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Length of the Curve L = (100 )/ For the chord definition of : L = (R )/18010 20/01/2013
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Degree of Curvature and Radius of Curvature11 20/01/2013
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3 Methods of Defining Degree of Curvature1. Radius of curvature; (often in highways) R isselected as a multiple of 100 ft. The smaller R, thesharper the curve.2. Degree of curvature on the chord basis is thecentral angle subtended by a chord of 100 ft.3. Degree of curvature on the arc basis is the centralangle of a circle subtended by an arc of 100 ft.12 20/01/2013
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R& Chord basis 100 ft arc 100 ft chord 50 ft 50 ft R R Arc basis /2 50 5729.58 R R sin13 2 20/01/2013
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100 ft R& 100 – ft arc RSharp curve (large , small R) R 14 Flat curve (small , large R) 20/01/2013
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R& Degree of Curvature & Radius of Curvature R, Chord basis R, Arc basis = 1 5729.65 ft 5729.58 ft = 5 1146.28 ft 1145.92 ft = 10 573.69 ft 572.96 ft15 20/01/2013
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L R R 360◦ c = 2RRelationship between the degree of curve () and the circle 16 20/01/2013
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From previous figure, L/2R = θ/360 L = 2R (θ/360)17 20/01/2013
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The degree of curve () is used as a term to define the sharpness of the curve. ( ) {is that central angle subtended by 100 ft of arc.18 20/01/2013
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and R From previous figure, /360 = 100/2R = 5729.58/R19 20/01/2013
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Arc From previous figure, L/100 = θ/ L = 100 (θ/)20 20/01/2013
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ProblemFor a horizontal circular curve, these data aregiven; the P.I. is at station 64 + 32.2, is 24 20and an of 4 00 has been selected. Computethe necessary data and set up the field notes for50-ft stations.21 20/01/2013
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Computation of Necessary Data 5729.58 5729.58 R 1432.39 ft 4 T R tan (1432.39) (0.21560) 308.82 ft 2 L.C. 2 R sin (2)(1432.39) (0.21076) 603.78 ft 222 20/01/2013
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Computation of Necessary Data 1 1 E R[ 1] (1432.39) ( 1) 32.91 ft cos 12 10 cos 2 M R (1 cos ) (1432.39) (1 cos 1210) 32.17 ft 2 100 (100) (24.333333) L 608.33 ft 423 20/01/2013
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As the degree of curvature is between 3 and 7, the curve will be staked out with 50-ft chords. The distance from B.C. (61 + 23.4) to the first 50-ft station (61 + 50) is 26.4 ft and the deflection angle to be used for that point is (26.6/100) (/2) = (26.6/100) (4/2) = 0 31 55. For each of subsequent 50-ft stations from 61 + 50 to 67 + 00, the deflection angles will increase by /4 or 1 00 00. Finally, for the E.C. at station 67 + 31.7, the deflection angle will be 11 32 + (31.7/100) (4/2) = 12 10 00 = /2 as it should.25 20/01/2013
32.
Chord Calculations C = 2R (sin deflection angle) R = 400.000 m First chord C = 2 400 (sin 0 14 01) = 3.2618 m = 3.262 m (at three decimals, chord = arc) Even station chord C = 2 400 (sin 1 25 57) = 19.998 m Last chord C = 2 400 (sin 0 27 42) = 6.448 m32 20/01/2013
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Metric Considerations The full station is 1 km (e.g. 1 + 000), and the stakes are at 50-m, 20-m and 10-m intervals. In field work, the use of (as opposed to R) allows quick determination of the deflection angle for even stations. In metric system, would be the central angle subtended by 100 m of arc and the deflection angles would be similarly computed as have done for ft units.33 20/01/2013
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Metric Considerations If = 6, the deflection angles would be as follows: Cross stations (m) Deflection angles 100 /2 = 300 50 /4 = 130 20 /5 = 036 10 /10 = 01834 20/01/2013
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Field Procedures to Lay Out A Curve • Extend the two straight lines 1 and 2, to determine the P.I. point of intersection • Measure (Intersection angle). • Select . • Two or more of these elements should be known or assumed , R, T, E or . • Calculate the stations B.C. = P.I. – T & E.C. = B.C. + L35 20/01/2013
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Horizontal Curves Passing Through Certain Points T T-x x P.I. y 35 10 C A T RR-y /2 /2 R 36 B 20/01/2013
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P.I. TS TS Circular curve T.S. S.T. C.S. S.C. LS LS R RSpiral Curves 1st spiral curve 2nd spiral curve37 20/01/2013
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S.P.I. S.C. Circular curve T.S. REnlargement of Spiral Curve 38 20/01/2013
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Moving up on PI θ = 12 51 the Curve 0 + 240 3 06 EC Tree BC 6 12 θ = 12 51 39 20/01/2013
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Problem of Offset Curves Given Data: = 12 51, R = 400.000 m and PI at 0 + 241.782 Calculated Data: T = 45.044 m, L = 89.710 m, BC at 0 + 196.738 and EC at 0 + 286.448 Required: Curbs to be laid out on 6-m offsets at 20-m stations.40 20/01/2013
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PI LS PI ¢ PI EC RS BC θ = 12 51 O41 Offset Curves 20/01/2013
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