1. 1 www.youtube.com

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3. 3 D=0 D= X D=0

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5. Spiral Curve Spiral Curve is a transition curve is sometimes used in horizontal alignment design It is used to – provide a gradual change in curvature from tangent to circular curve sections. – Provides a better transition for Superelevation. – Improves driver safety, comfort, and the road appearance. 5

6. Different types of transition curve may be used but the most common is the Euler Spiral Properties of Euler Spiral (reference: Surveying: Principles and Applications, Kavanagh and Bird, Prentice Hall] Spiral Curve

7. No Spiral With Spiral

8. Circle Curve 8 ΔcΔc ΔsΔs ΔsΔs Spiral Curve Ls Lc

9. Degree of Curvature of a spiral at any point is proportional to its length at that point The spiral curve is defined by ‘k’ the rate of increase in degree of curvature per station (100 ft) In other words, k = 100 D/ Ls (1) Characteristics of Euler Spiral

10. As with circular curve the central angle is also important for spiral Recall for circular curve Δ c = L c D / 100 (2) But for spiral Δ s = L s D / 200 (3) Central (or Deflection) Angle of Euler Spiral The total deflection angle for a spiral/circular curve system is Δ = Δ c + 2 Δ s (4)

11. Length of Euler Spiral Note: The total length of curve (circular plus spirals) is longer than the original circular curve by one spiral leg L total = L C + 2 L s =L circle + L s (5)

12. Example 4) Calculation – Spiral and Circular Curve The central angle for a curve is 24 degrees - the radius of the circular curve selected for the location is 1000 ft If a spiral with angle of 4 degrees is to be used instead of the simple circular curve, determine the i)length of each spiral leg, ii)k for the spiral, iii)total length of curve

13. 13 Part iii) L circle = 100 Δ / D or L circle = 24*100/5.73 = 418.8 ft Part i) Δ s = 4 degrees Δ s = L s D / 200  4 = L s * 5.73/200  L s = 139.6 ft Total Length of curve = length with no spiral + L s = 418.8+139.76 = 558.4 feet Solution Part ii) k = 100 D/ Ls  k= 100*5.73/139.6  k=4.1

14. Example 5 In a spiral curve the Δc is 33 degree. If the degree of curvature is 6 and the rate of increase in degree of curvature is 4. determine the i)length of each spiral leg, ii)the intersection angel, iii) Radius iv)total length of curve 14

15. Solution 15 Given: Δc =33, D=6, k=4 i) Ls ? k = 100 D/ Ls,  4=100*6/Ls  Ls= 150 ft ii) Δ ? Δ = Δc + 2.Δs Δs = Ls D / 200  Δs = 150*6 / 200  Δs = 4..5 Δ = 33+ 4.5*2= 42

16. 16 iii) Radius R=5730/D  R =5730/6= 955 ft iv) total length of curve L total = L C + 2 L s = L circle + L s Method a) L total =L circle + L s L = 100(Δ)/D = 100*(42)/6 = 700 ft. L total =700 + 150 =850 ft Method b) L total = L C + 2 L s Δc = Lc D / 100  Lc = Δc *100 / D =33*100/6=550 L total =550 + 2*150 =850 ft

17. 17 http://techalive.mtu.edu/modules/module000 3/Superelevation.htm

18. 18 http://www.istockphoto.com/

21. 21 Saddle Road the road between the east and west coasts of Hawaii http://www.both-ears.com/

22. Vertical Alignment Vertical alignment provide a means to smoothly shift from one tangent to another. Vertical curve provide a transition between two grades 22

23. 23 http://urbangeographies.tumblr.com/

24. Vertical Alignment Grade & Topography Texas DOT

25. Maximum Grade Harlech, Gwynedd, UK (G = 34%) www.geograph.org.uk

26. 26 montecarloforum.com Passo dello Stelvio in Alps, Italy Average Grade 7.5%

27. Maximum Grade www.nebraskaweatherphotos.org

28. Maximum Grade Dee747 at picasaweb.google.com

29. Maximum and Minimum Grade What is the maximum and minimum allowable grade for tangents? The minimum grade used is typically 0.5% The maximum grade is generally a function of the Design Speed Terrain (Level, Rolling, Mountainous)  On high speed facilities such as freeways the maximum grade is generally kept to 5% where the terrain allows (3% is desirable since anything larger starts to affect the operations of trucks)  At 30 mph design speed the acceptable maximum is in the range of 7 to 12 %

30. Crest Curve Sag Curve G1G1 G2G2 G3G3 Vertical Alignment Tangents and Curves The vertical curve is a parabolic curve rather than a circular or spiral curve. Implies Constant rate of change of slope

31. Parabolic vs Circle Curve 31

32. Types of Vertical Curves Typical Crest Curves Typical Sag Curves

33. Properties of Vertical Curves BVC EVC L G2G2 G1G1 BVC: Beginning of Vertical Curve EVC: End of Vertical Curve PI: Point of Intersection G 1 and G 2 : Slope of tangents L/2 PI

34. Properties of Vertical Curves BVC EVC L G2G2 G1G1 Change in grade: A = G 2 - G 1 where G is expressed as % (positive /, negative \) For a crest curve, A is negative For a sag curve, A is positive L/2 PI

35. Properties of Vertical Curves BVC EVC L G2G2 G1G1 K : defines vertical curvature Rate of change of curvature: K = L / |A| Which is a gentler curve - small K or large K? Minimum K and Design Speed L/2 PI

36. Properties of Vertical Curves BVC EVC L G2G2 G1G1 L/2 Rate of change of grade: r = (g 2 - g 1 ) / L where, g is expressed as a ratio (positive /, negative \) L is expressed in feet or meters Note – K and r are both measuring the same characteristic of the curve but in different ways PI

37. Properties of Vertical Curves BVC EVC PI L G2G2 G1G1 Equation for determining the elevation at any point on the curve y = y 0 + g 1 x + 1/2 rx 2 where, y 0 = elevation at the BVC g = grade expressed as a ratio x = horizontal distance from BVC r = rate of change of grade expressed as ratio Elevation = y x

38. Properties of Vertical Curves Distance BVC to the turning point (high/low point on curve) x t = -(g 1 /r) This can be derived as follows y = y 0 + g 1 x + 1/2 rx 2 dy/dx = g 1 + rx At the turning point, dy/dx = 0 0 = g 1 + rx t Therefore, x t = -(g 1 /r) Low Point xtxt

39. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Length of curve? L/2 = Sta. EVC – Sta. PI L/2 = 2500 m - 2400 m = 100 m L = 200 m

40. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 r - value? r = (g 2 - g 1 )/L r = (0.02 - [-0.01])/200 m r = 0.00015 / meter

41. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Station of low point? x = -(g 1 /r) x = -([-0.01] / [0.00015/m]) x = 66.67 m Station = [23+00] + 67.67 m Station 23+67

42. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Elevation at low point? y = y 0 + g 1 x + 1/2 rx 2 y 0 = Elev. BVC Elev. BVC = Elev. PI - g 1 L/2 Elev. BVC = 125 m - [-0.01][100 m] Elev. BVC = 126 m

43. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Elevation at low point? y = y 0 + g 1 x + 1/2 rx 2 y = 126 m + [-0.01][66.67 m] + 1/2 [0.00015/m][66.67 m] 2 y = 125.67 m

44. Properties of Vertical Curves BVC EVC PI G2G2 G1G1 Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+00 Station of PI = 24+00 Elevation at station 23+50? y = 126 m + [-0.01][50 m] + 1/2 [0.00015/m][50 m] 2 y = 125.69 m Elevation at station 24+50? y = 126 m + [-0.01][150 m] + 1/2 [0.00015/m][150 m] 2 y = 126.19 m

45. Example 6 The length of a tangent vertical curve equal to 500 m. The initial and final grades are 2.5% and -1.5% respectively. The grades intersect at the station 10+400 and at an elevation of 210.00 m Determine: a)the station and the elevation of the BVC and EVC points b) the elevation of the point on the curve 100 and 300 meters from the BVC point c) the station and the elevation of the highest point on the curve 45

46. 46 BVC EVC PI 2.5% -1.5%

47. Part a) the station and the elevation of the BVC and EVC points – Station-BVC= 10400-250=10150~10+150 – Station-EVC=10400+250=10650~10+650 – Elevation-BVC= 210-0.025*250= 203.75 m – Elevation-EVC= 210-0.015*250= 206.25 m 47 Solution

48. Part b) the elevation of the point on the curve 100 and 300 meters from the BVC point. y = y 0 + g 1.x + 1/2.r.x^2, y 0 = 203.75,g 1 = 0.025 – r=(g 2 -g 1 )/L – r=(-0.015-(0.025))/500=-0.04/500 r=-0.00008 y = 203.75 + 0.025.x - 0.00004.x^2,  y(100)=203.75+0.025*100-0.00004*100^2  y(100) = 205.85  y(300)= 207.65 48

49. Part c) the station and the elevation of the highest point on the curve – The highest point can be estimated by setting the first derivative of the parabola as zero. Set dy/dx=0,  dy/dx= 0.025-0.0008*x=0  X=0.025/0.00008= 312.5  y(312.5) = 203.75+0.025*312.5-0.00004*(312.5)^2  y(312.5)= 207.66 49

50. Example 7 For a vertical curve we know that G1=-4%, G2=-1%, PI: Station 20+00, Elevation: 200’, L=300’ Determine: i)K and r ii) station of BVC and EVC iii) elevation of point at a distance, L/4, from BVC iv) station of turning point v) elevation of turning point vi) elevation of mid-point of the curve vii) grade at the mid-point of the curve 50

51. i) K and r – K = L / |A| – A=G2-G1, A=-4-(1)=-5 K=300/5=60 – r=(g2-g1)/L – r=(-0.01-(-0.04))/300=0.03/300 r=0.0001 51 Solution -4% -1%

52. ii) station of BVC and EVC – L=300’, L/2=150’ – BVC= 2000-150=1850’, ~ 18+50 – EVC= 2000+150=2150’, ~ 21+50 iii) elevation of point at a distance, L/4, from BVC – y = y0 + g1x + 1/2 rx2, – r= -0.0001, g1=0.04 – y0=200+150*0.04=206 y=206-0.04x+0.00005.x^2 L/4~ x=75  Y(75)=203.28’ 52

53. iv) station of turning point – at turning point: dy/dx=0 – dy=dx=- 0.04+0.0001x=0  x(turn)=400’ v) elevation of turning point – x=400’ – Y (400) = 206-0.04*(150)+0.00005.(150)^2  Y(400)=198’  This point is after vertical curve (Turning point is not in the curve) 53

54. vi) elevation of mid-point – x=150’ – Y(150)= 206-0.04*(150)+0.00005.(150)^2  y (150)=201.12’ vii) grade at the mid-point of the curve – Grade at every points: dy/dx= =- 0.04+0.0001x – if x= 150  Grade (150)=-0.04+0.0001*150  grade(150)= -0.025 54

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56. 56 Photo :caraccidenttodayer.blogspot.com

57. Driver sees an object Driver decides to brake Driver applies brakes Car Stops 57 2-3 Seconds (average: 2.5 ) 2-3 Seconds (average: 2.5 ) Perception Reaction Braking

58. Factors affecting Perception and Reaction Driver Factors: – Physical ability – Mental ability Driving speed (Design Speed) Contextual Factors: – Weather and visibility – Time of the day Braking Driving speed Breaks condition and technology Road Surface Friction Grade of the road 58

59. Breaking Distance 59 W F = m.a F f= f f. N N Using Newton’s law we can calculate the acceleration (a) of the vehicle 1) ∑ f y = 0 2) ∑ f x = m.a y x α  a = g.cos(α) * ( tan(α)+f f ) If we set G= tan (α) and [ G is the road slope] g.cos(α) = g [α~0]  a = g (f f + G)

60. 60 An equation from Dynamic law: d b = -(u 1 ^2- u 0 ^2) / 2.a  d b = -(0- u 0 ^2) / 2.a = u 0 ^2 / 2.a a = g.( G+ f f ) Breaking Distance  d b = u 0 ^2 /[ 2 g.(f f + G) ]

61. Stopping Distance = Perception reaction distance+ Breaking Distance  SD= u 0.t+u 0 ^2 /[ 2g.(f f +- G) ] Positive G = upgrade Negative G = downgrade 61

62. Example 8 A vehicle initially traveling at 50 mph reached to a stop on a 5% downgrade, where the coefficient of friction is equal to 0.2. How far does the vehicle travel before coming to a stop? 62

63. Solution u 0 = 50 mph, f f = 0.2, G = 0.05 d b = u 0 ^2 /[ 2 g.( f f + G) ]  d b = (40*1.47) ^2 /[ 2 *32.2*( 0.2 -0.05 ) ]  d b = 357 ft 63

64. Example 9 Calculate total braking distance (including perception-reaction time) for a vehicle with a highway with +3% grade. The vehicles traveling speed is 30 mph. Compare findings for both wet and dry surface conditions. Dry weather: f f = 0.35 Wet weather: f f = 0.22 64

65. Solution G= 3%, V= 30 mph = 44.1 fps,  Stopping Distance = perception & reaction distance+ breaking distance Dry condition: Ds= 2.5*44.1+ 44.1^2/[2*32.2(0.35+0.03)] = 110.25+ 79.47 =189.72 ft Wet condition: Ds= 2.5*44.1+ 44.1^2/[2*32.2(0.22+0.03)] = 110.25+ 120.79 =231.05 ft 65