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Sight Distance for horizontal curves

Civil Engineering

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Sight Distance for horizontal curves

  1. 1. Sight Distance forSight Distance for Horizontal CurvesHorizontal Curves 1
  2. 2. Provided Sight Distance • Potential sight obstructions – On horizontal curves: barriers, bridge-approach fill slopes, trees, back slopes of cut sections – On vertical curves: road surface at some point on a crest vertical curve, range of head lights on a sag curve
  3. 3. 3 S R M O TA Sight Obstruction on for Horizontal Curves
  4. 4. 4 Line of sight is the chord AT Horizontal distance traveled is arc AT, which is SD. SD is measured along the centre line of inside lane around the curve. See the relationship between radius of curve, the degree of curve, SSD and the middle ordinate S R M O TA
  5. 5. 5 Middle ordinate • Location of object along chord length that blocks line of sight around the curve • m = R(1 – cos [28.65 S]) R Where: m = line of sight S = stopping sight distance R = radius
  6. 6. 6 Middle ordinate • Angle subtended at centre of circle by arc AT is 2θ in degree then • S / πR = 2θ / 180 • S = 2R θπ / 180 • θ = S 180 / 2R π = 28.65 (S) / R • R-M/R = cos θ • M = [1 – cos 28.65 (S) / R ] θR M A T B T O θ
  7. 7. 7 Sight Distance Example A horizontal curve with R = 800 ft is part of a 2-lane highway with a posted speed limit of 35 mph. What is the minimum distance that a large billboard can be placed from the centerline of the inside lane of the curve without reducing required SSD? Assume p/r =2.5 and a = 11.2 ft/sec2 SSD = 1.47vt + _________v2 ____ 30(__a___ ± G) 32.2
  8. 8. 8 Sight Distance Example SSD = 1.47(35 mph)(2.5 sec) + _____(35 mph)2 ____ = 246 feet 30(__11.2___ ± 0) 32.2
  9. 9. 9 Sight Distance Example m = R(1 – cos [28.65 S]) R m = 800 (1 – cos [28.65 {246}])= 9.43’ 800 (in radians not degrees)
  10. 10. 10 SuperelevationSuperelevation DR ABDUL SAMI QURESHIDR ABDUL SAMI QURESHI M L N a E B
  11. 11. 11 Motion on Circular Curves dt dv at = R v an 2 = C.F Weight of Vehicle Friction force
  12. 12. 12 Definition • The transverse slope provided by raising outer edge w.r.t. inner edge • To counteract the effects of C. Force (overturning/skid laterally) L N a EB M
  13. 13. 13 •S.E. expressed in ratio of height of outerS.E. expressed in ratio of height of outer edge to the horizontal widthedge to the horizontal width e = NL / ML = tan= NL / ML = tan θ tantan θ = sin= sin θ,, θ is very smallis very small e =NL / MN = E / Be =NL / MN = E / B E = Total rise in outer edgeE = Total rise in outer edge B total width of pavementB total width of pavement M L N θ EB
  14. 14. 14 FgFW fp =+ α α C cos θ Cx Mx =M sin θ My =M cos a Ff Ff θ C M 1 ft e ≈ Rv 1. C.F 2. Weight of Vehicle 3. Friction force X-X Y-Y C N N C sin θ
  15. 15. 15 Theo- retical Consi- derati on
  16. 16. 16 Vehicle Stability on Curves where: gR v fe s 2 =+ (ft/s),speeddesign=v (-),tcoefficienfrictionside=sf ).ft/s(32onacceleratigravity 2 =g (-),tionsupereleva=e (ft),radius=R Assumed Desig n speed (mph) Maximum design fs max 20 0.17 70 0.10 Must not be too short (0.12)0.10-0.06max =e
  17. 17. 17 Selection of e and fs • Practical limits on super elevation (e) – Climate – Constructability – Adjacent land use • Side friction factor (fs) variations – Vehicle speed – Pavement texture – Tire condition • The maximum side friction factor is the point at which the tires begin to skid • Design values of fs are chosen somewhat below this maximum value so there is a margin of safety
  18. 18. 18
  19. 19. 19 Side Friction Factor from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004 New Graph
  20. 20. 20 Maximum Superelevation • Superelevation cannot be too large since an excessive mass component may push slowly moving vehicles down the cross slope. • Limiting values emax – 12 % for regions with no snow and ice conditions (higher values not allowed), – 10 % recommended value for regions without snow and ice conditions, – 8% for rural roads and high speed urban roads, – 4, 6% for urban and suburban areas.
  21. 21. 21 Example • A section of road is being designed as a high-speed highway. The design speed is 70 mph. Using AASHTO standards, what is the maximum super elevation rate for existing curve radius of 2500 ft and 300 ft for safe vehicle operation? • Assume the maximum super elevation rate for the given region is 8%. • max e = ? • For 70 mph, f = 0.10 • 1. 2500 = V2/15(fs+e) = (70 )2/(0.10 + e) = 0.0306 • e = 3% • 2. 300 = V2/(fs+e) = (70 x 1.47)2/32.2(0.10 + e) = 0.988 • e = 9.8%
  22. 22. • 300 = V2/g(fs+e) = (70)2/15(fs + 0.8) • f = 1.008 > 0.10 • 300 = V2/15(fs+e) = (V )2/15(0.10+ 0.8) • V = 28.46 mph 22
  23. 23. 23 Crown / super elevation runoff length
  24. 24. 24 Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
  25. 25. 25 Attainment of Superelevation - General 1. Tangent to superelevation 2. Must be done gradually over a distance without appreciable reduction in speed or safety and with comfort 3. Change in pavement slope should be consistent over a distance 4. Methods (Exhibit 3-37 p. 186) a. Rotate pavement about centerline b. Rotate about inner edge of pavement c. Rotate about outside edge of pavement
  26. 26. 26 Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
  27. 27. 27 Common methods of developing the transition to super elevation • At (2)the out side edge is far below the centre line as the inside edge • At (3)the out side edge has reached the level of the centre line • At point (4) the out side edge is located as far above as the inside edge is below the centerline. • Finally , at point (5) the cross section is fully super elevated and remain through out the circular curve • The reverse of these profiles is found at the other end of circular curve.
  28. 28. 28 Common methods of developing the transition to super elevation • Location of inside edge, centre line, and out side edge are shown relative to elevation of centerline • The difference in elevation being equal to the normal crown times the pavement width. • At A the out side edge is far below the centre line as the inside edge • At B the out side edge has reached the level of the centre line • At point C the out side edge is located as far above as the inside edge is below the centerline. • Finally , at point E the cross section is fully super elevated • The reverse of these profiles is found at the other end of circular curve.
  29. 29. 29 Attaining Superelevation (1) •Location of inside edge, centre line, and out side edge are shown relative to elevation of centerline
  30. 30. 30 Attaining Superelevation (2)
  31. 31. 31 Attaining Superelevation (3)
  32. 32. 32 Superelevation Transition Section • Tangent Runout (Crown Runoff) Section + Superelevation Runoff Section. • Tangent runout = the length of highway needed to change the normal cross section to the cross section with the adverse crown removed.
  33. 33. Super elevation runoff • Super elevation runoff = the length of highway needed to change the cross section with the adverse crown removed to the cross section fully super elevated. 33
  34. 34. 34 Superelevation Runoff and Tangent Run out (Crown Runoff) Normal cross section Fully superelevated cross section Cross section with the adverse crown removed
  35. 35. 35 Location of Runout and Runoff
  36. 36. 36 Tangent Runout Section • Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero For rotation about centerline
  37. 37. 37 Superelevation Runoff Section • Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa • For undivided highways with cross- section rotated about centerline
  38. 38. 38 Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
  39. 39. 39 Minimum Length of Tangent Runout Lt = eNC x Lr ed where • eNC = normal cross slope rate (%) • ed = design superelevation rate • Lr = minimum length of superelevation runoff (ft) (Result is the edge slope is same as for Runoff segment)
  40. 40. 40 Length of Superelevation Runoff α = multilane adjustment factor Adjusts for total width r
  41. 41. 41 Minimum Length of Runoff for curve • Lr based on drainage and aesthetics and design speed. • Relative gradient is the rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road)
  42. 42. 42 Design Requirements for Runoffs Maximum Relative Gradient Relative gradient is the rate of transition of edge line from NC to full super elevation Relative gradient
  43. 43. 43 Relative Gradient (G) • Maximum longitudinal slope • Depends on design speed, higher speed = gentler slope. For example: • For 15 mph, G = 0.78% • For 80 mph, G = 0.35% • See table, next page
  44. 44. 44 Maximum Relative Gradient (G) Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.
  45. 45. 45 Multilane Adjustment factor • Runout and runoff must be adjusted for multilane rotation.
  46. 46. 46 Length of Superelevation Runoff Example For a 4-lane divided highway with cross- section rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft) Lr = 12eα G •
  47. 47. 47 Lr = 12eα = (12) (0.04) (1.5) G 0.5 Lr = 144 feet
  48. 48. 48 Tangent runout length Example continued • Lt = (eNC / ed ) x Lr as defined previously, if NC = 2% Tangent runout for the example is: LT = 2% / 4% * 144’ = 72 feet
  49. 49. 49 From previous example, speed = 50 mph, e = 4% From chart runoff = 144 feet, same as from calculation Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.
  50. 50. 50 Transition curve
  51. 51. 51 Spiral CurveSpiral Curve TransitionsTransitions
  52. 52. 52 Source: Iowa DOT Design Manual SPIRAL TERMINOLOGY
  53. 53. 53 Transition Curves – Spirals (Safety) • Provided between tangents and circular curves or between two circular curves • It provides the path where radial force gradually increased or decreased while entering or leaving the circular curves
  54. 54. 54 Tangent-to-Curve Transition (Appearance) • Improves appearance of the highways and streets • Provides natural path for drivers
  55. 55. 55 Tangent-to-Curve Transition (Superelvation or curve widening requirement) • Accommodates distance needed to attain super elevation • Accommodates gradual roadway widening
  56. 56. 56 Ideal shape of transition curve • When rate of introduction of C.F. is consistent • When rate of change C. Acceleration is consistent • When radius of transition curve consistently change from infinity to radius of circular curve
  57. 57. 57 Shape of transition curves • Spiral (Clotoid)= mostly used • Lemniscates (rate of change of radius not constant) • Cubic parabola
  58. 58. 58 Transition Curves - Spirals The Euler spiral (clothoid) is used. The radius at any point of the spiral varies inversely with the distance.
  59. 59. 59 Minimum Length of Spiral Possible Equations: When consistent C.F is considered Larger of (1) L = 3.15 V3 RC Where: L = minimum length of spiral (ft) V = speed (mph) R = curve radius (ft) C = rate of increase in centripetal acceleration (ft/s3 ) use 1-3 ft/s3 for highway)
  60. 60. 60 • V= 50 mph • C = 3 ft p c. sec • R= 929 • Ls = 141 ft
  61. 61. 61 Minimum Length of Spiral When appearance of the highways is considered 1.Minimum- 2.Maximum Length of Spiral Or L = (24pminR)1/2 Where: L = minimum length of spiral (ft) = 121.1 ft R = curve radius (ft) = 930 pmin = minimum lateral offset between the tangent and circular curve (0.66 feet)
  62. 62. 62 Maximum Length of Spiral • Safety problems may occur when spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)
  63. 63. 63 Maximum Length of Spiral L = (24pmaxR)1/2 Where: L = maximum length of spiral (ft) = 271 R = curve radius (ft) pmax = maximum lateral offset between the tangent and circular curve (3.3 feet)
  64. 64. 64 Length of Spiral o AASHTO also provides recommended spiral lengths based on driver behavior rather than a specific equation. o Super elevation runoff length is set equal to the spiral curve length when spirals are used. o Design Note: For construction purposes, round your designs to a reasonable values; e.g. Ls = 141 feet, round it to Ls = 150 feet.
  65. 65. 65 Location of Runouts and Runoffs • Tangent runout proceeds a spiral • Superelevation runoff = Spiral curve
  66. 66. 66
  67. 67. 67Source: Iowa DOT Design Manual
  68. 68. 68Source: Iowa DOT Design Manual
  69. 69. 69Source: Iowa DOT Design Manual
  70. 70. 70 Source: Iowa DOT Design Manual SPIRAL TERMINOLOGY
  71. 71. 71 Attainment of superelevation on spiral curves See sketches that follow: Normal Crown (DOT – pt A) 1. Tangent Runout (sometimes known as crown runoff): removal of adverse crown (DOT – A to B) B = TS 2. Point of reversal of crown (DOT – C) note A to B = B to C 3. Length of Runoff: length from adverse crown removed to full superelevated (DOT – B to D), D = SC 4. Fully superelevate remainder of curve and then reverse the process at the CS.
  72. 72. 72 Source: Iowa DOT Standard Road Plans RP-2 With Spirals Same as point E of GB
  73. 73. 73 With Spirals Tangent runout (A to B)
  74. 74. 74 With Spirals Removal of crown
  75. 75. 75 With Spirals Transition of superelevation Full superelevation
  76. 76. 76
  77. 77. 77 Transition Example Given: • PI @ station 245+74.24 • D = 4º (R = 1,432.4 ft) ∀∆ = 55.417º • L = 1385.42 ft
  78. 78. 78 With no spiral … • T = 752.30 ft • PC = PI – T = 238 +21.94
  79. 79. 79 For: • Design Speed = 50 mph • superelevation = 0.04 • normal crown = 0.02 Runoff length was found to be 144’ Tangent runout length = 0.02/ 0.04 * 144 = 72 ft.
  80. 80. 80 Where to start transition for superelevation? Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff: Distance before PC = Lt + 2/3 Lr =72 +2/3 (144) = 168 Start removing crown at: PC station – 168’ = 238+21.94 - 168.00 = Station = 236+ 53.94
  81. 81. 81 Location Example – with spiral • Speed, e and NC as before and ∀∆ = 55.417º • PI @ Station 245+74.24 • R = 1,432.4’ • Lr was 144’, so set Ls = 150’
  82. 82. 82 Location Example – with spiral See Iowa DOT design manual for more equations: http://www.dot.state.ia.us/design/00_toc.htm#Ch • Spiral angle Θs = Ls * D /200 = 3 degrees • P = 0.65 (calculated) • Ts = (R + p ) tan (delta /2) + k = 827.63 ft
  83. 83. 83 • TS station = PI – Ts = 245+74.24 – 8 + 27.63 = 237+46.61 Runoff length = length of spiral Tangent runout length = Lt = (eNC / ed ) x Lr = 2% / 4% * 150’ = 75’ Therefore: Transition from Normal crown begins at (237+46.61) – (0+75.00) = 236+71.61 Location Example – with spiral
  84. 84. 84 With spirals, the central angle for the circular curve is reduced by 2 * Θs Lc = ((delta – 2 * Θs) / D) * 100 Lc = (55.417-2*3)/4)*100 = 1235.42 ft Total length of curves = Lc +2 * Ls = 1535.42 Verify that this is exactly 1 spiral length longer than when spirals are not used (extra credit for who can tell me why, provide a one-page memo by Monday) Location Example – with spiral
  85. 85. 85 Also note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately ½ of the spiral length used. (good check – but why???) Location Example – with spiral
  86. 86. 86 Notes – Iowa DOT
  87. 87. 87 Quiz Answers What can be done to improve the safety of a horizontal curve?  Make it less sharp  Widen lanes and shoulders on curve  Add spiral transitions  Increase superelevation
  88. 88. 88 Quiz Answers 5. Increase clear zone 6. Improve horizontal and vertical alignment 7. Assure adequate surface drainage 8. Increase skid resistance on downgrade curves
  89. 89. 89 Some of Your Answers  Decrease posted speed  Add rumble strips  Bigger or better signs  Guardrail  Better lane markers  Sight distance  Decrease radius

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