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  • 1. {Highway Curves1 20/01/2013
  • 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
  • 3. {Horizontal Curves3 20/01/2013
  • 4. Curves Reverse { Circular Curves Tangent Spiral Curve Spiral Curve4 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
  • 6. Curve Equations Tangent Distance T = R tan (θ/2)6 20/01/2013
  • 7. Length of Long Chord L.C. = 2R sin (θ/2)7 20/01/2013
  • 8. Middle Ordinate M = R[1 - cos (θ/2)]8 20/01/2013
  • 9. External Distance E = R[1/ cos (θ/2) -1] or E = R[sec (θ/2) -1]9 20/01/2013
  • 10. Length of the Curve L = (100 )/ For the chord definition of : L = (R  )/18010 20/01/2013
  • 11. Degree of Curvature and Radius of Curvature11 20/01/2013
  • 12. 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
  • 13. 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
  • 14. 100 ft R& 100 – ft arc  RSharp curve (large , small R) R  14 Flat curve (small , large R) 20/01/2013
  • 15. 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
  • 16. L   R R 360◦ c = 2RRelationship between the degree of curve () and the circle 16 20/01/2013
  • 17. From previous figure, L/2R = θ/360 L = 2R (θ/360)17 20/01/2013
  • 18. 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
  • 19.  and R From previous figure, /360 = 100/2R  = 5729.58/R19 20/01/2013
  • 20. Arc From previous figure, L/100 = θ/ L = 100 (θ/)20 20/01/2013
  • 21. ProblemFor a horizontal circular curve, these data aregiven; the P.I. is at station 64 + 32.2,  is 24 20and an  of 4 00 has been selected. Computethe necessary data and set up the field notes for50-ft stations.21 20/01/2013
  • 22. 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
  • 23. 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 1210)  32.17 ft 2 100 (100) (24.333333) L   608.33 ft  423 20/01/2013
  • 24. B.C. & E.C. P.I. = 64 + 32.2 - T = - (3 + 08.8) B.C. = 61 + 23.4 + L = 6 + 08.3 E.C. = 67 + 31.724 20/01/2013
  • 25. 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
  • 26. Field Notes Station Points Deflection angles Curve data67 + 31.7 E.C. 12 10 0067 + 00 11 31 55…………………. …………………….64 + 00 5 31 55 P.I. = 64 + 32.2 + 50 4 31 55  = 24 2063 + 00 3 31 55  = 4 00 + 50 2 31 55 R = 1432.4 ft62 + 00 1 31 55 T = 308.8 ft + 50 0 31 55 E = 32.9 ft + 23.4 B.C. 0 00 L = 608.3 ft61 + 0026 20/01/2013
  • 27. Deflection Angles  ECBC    1 2 Length of curve, chord basis27 20/01/2013
  • 28. Deflection Angles•   3 Use 100 - ft Chords.•  = 3 - 7 Use 50 - ft Chords.•  = 7 - 14 Use 25 - ft Chords.28 20/01/2013
  • 29. Arcs & Chords Comparison R = 5729.58/   = 2  = 3  = 6  = 10 Chord (ft) C = 2 R sin (deflec. angle) 99.995 99.989 99.954 99.873 Arc (ft) 100 100 100 100 For flat curves, say  = 2/ 3 chord  arc29 20/01/2013
  • 30. Deflection Angles [(25/100)  (3/2)] = 0 22.5 [0 22.5 + (3/2)] = 1 52.5 BCBack tangent  = 3 Staking out a horizontal curve30 20/01/2013
  • 31. Circular Curve Deflections PI 6 25 30 = /2 1 40 4 32 5 58 0 14 3 06 0 + 280 0 + 200 EC = 0 + 286.448 BC = 0 + 196.738 Field location for deflection angles Deflection angle = (arc/L)  (/2)31 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
  • 33. 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
  • 34. Metric Considerations If  = 6, the deflection angles would be as follows: Cross stations (m) Deflection angles 100  /2 = 300 50  /4 = 130 20  /5 = 036 10  /10 = 01834 20/01/2013
  • 35. 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
  • 36. 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
  • 37. 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
  • 38. S.P.I. S.C. Circular curve T.S. REnlargement of Spiral Curve 38 20/01/2013
  • 39. 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
  • 40. 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
  • 41. PI LS PI ¢ PI EC RS BC θ = 12 51 O41 Offset Curves 20/01/2013
  • 42. Problem of Offset Curves Station Computed Field Deflection angles Deflection angles EC 0 + 286.448 6 25 31 6 25 30 =  /2 (check) 0 + 280 5 57 49 5 58 0 + 260 4 31 52 4 32 0 + 240 3 05 55 3 06 0 + 220 1 39 58 1 40 0 + 200 0 14 01 0 14 BC 0 + 196.738 0 00 00 0 0042 20/01/2013
  • 43. Arc Distance Computations (o/s arc)/(¢ arc) = (o/s radius)/(¢ radius) Arithmetic Check: LS - ¢ = ¢ - RS43 20/01/2013
  • 44. Chord Distance Computations C = 2R (sin deflection angle) Arithmetic Check: LS chord - ¢ chord = ¢ chord – RS chord44 20/01/2013