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Spring 2015. Horizontal Alignment Geometric Elements of Horizontal Curves Superelevation Design Transition or Spiral Curves Sight Distance.

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Presentation on theme: "Spring 2015. Horizontal Alignment Geometric Elements of Horizontal Curves Superelevation Design Transition or Spiral Curves Sight Distance."— Presentation transcript:

1 Spring 2015

2 Horizontal Alignment Geometric Elements of Horizontal Curves Superelevation Design Transition or Spiral Curves Sight Distance

3 PC PT Circular Curve Tangent Point of Curvature Point of Tangency

4 SC ST Circular Curve Tangent Tangent to Spiral Spiral to Tangent Spiral TS Spiral to Curve CS Curve to Spiral

5 Design Elements of Horizontal Curves Deflection Angle Also known as Δ

6 Design Elements of Horizontal Curves Larger D = smaller Radius

7 Design Elements of Horizontal Curves E=External Distance M=Length of Middle Ordinate

8 Design Elements of Horizontal Curves LC=Length of Long Cord

9 Basic Formulas Where, e = superelevation f = side friction factor V = vehicle speed (mph) R = radius of curve (ft) Basic Formula that governs vehicle operation on a curve:

10 Basic Formulas Where, e = superelevation f = side friction factor V = vehicle speed (mph) R = radius of curve (ft) Minimum radius:

11 Minimum Radius with Limiting Values of “e” and “f”

12 Desirable superelevation: for R > R min Where, V= design speed in ft/s or m/s g = gravity (9.81 m/s 2 or 32.2 ft/s 2 ) R = radius in ft or m Various methods are available for determining the desirable superelevation, but the equation above offers a simple way to do it. The other methods are presented in the next few overheads.

13  Method 1: ◦ Superelevation and side friction are directly proportional to the inverse of the radius (straight relationship between 1/R=0 and 1/R =1/R min )  Method 2: ◦ Side friction is such that a vehicle traveling at the design speed has all the acceleration sustained by side friction on curves up to those requiring f max ◦ Superelevation is introduced only after the maximum side friction is used

14  Method 3: ◦ Superelevation is such that a vehicle traveling at the design speed has all the lateral acceleration sustained by superelevation on curves up to those required by e max ◦ No side friction is provided on flat curves ◦ May result in negative side friction  Method 4: ◦ Same approach as Method 3, but use average running speed rather than design speed ◦ Uses speeds lower than design speed ◦ Eliminate problems with negative side friction  Method 5: ◦ Superelevation and side friction are in a curvilinear relationship with the inverse of the radius of the curve, with values between those of methods 1 and 3 ◦ Represents a practical distribution for superelevation over the range of curvature ◦ This is the method used for computing values shown in Exhibits 3-25 to 3-29

15 e = 0 e max Reciprocal of Radius Side Friction Factor Five Methods f max M2 M1 M3 M5 M4 1/R f

16  Important considerations: ◦ Governed by four factors:  Climate conditions  Terrain (flat, rolling, mountainous)  Type of area (rural vs urban)  Frequency of slow-moving vehicles ◦ Design should be consistent with driver expectancy ◦ Max 8% for snow/ice conditions ◦ Max 12% low volume roads ◦ Recurrent congestion: suggest lower than 6%

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21  In overall sense, the method of rotation about the centerline (Method 1) is usually the most adaptable  Method 2 is usually used when drainage is a critical component in the design  In the end, an infinite number of profile arrangements are possible; they depend on drainage, aesthetic, topography among others

22 Median width Pivot points Example where pivot points are important Bad design Good design 15 ft to 60 ft

23  The superelevation transition consists of two components: ◦ The superelevation runoff: length needed to accomplish a change in outside-lane cross slope from zero (flat) to full superelevation ◦ The tangent runout: The length needed to accomplish a change in outside-lane cross slope rate to zero (flat)

24 Tangent Runout

25 Superelevation Runoff

26 Transition Design Control

27 http://techalive.mtu.edu/modules/module0003/Superelevation.htm

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29  = relative gradient in previous overhead

30 Values for n 1 and b w in equation

31 See Exhibit 3-32 for values of L t and L r

32 Location: 1/3 on curve Location: 2/3 on tangent

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34  All motor vehicles follow a transition path as it enters or leaves a circular horizontal curve (adjust for increases in lateral acceleration)  Drivers can create their own path or highway engineers can use spiral transitional curves  The radius of a spiral varies from infinity at the tangent end to the radius of the circular curve at the end that adjoins the curve

35 Need to verify for maximum and minimum lengths

36 Superelevation runoff should be accomplished on the entire length of the spiral curve transition Equation for tangent runout when Spirals are used:

37  The sight distance is measured from the centerline of the inside lane  Need to measure the middle-ordinate values (defined as M)  Values of M are given in Exhibit 3-53  Note: Now M is defined as HSO or Horizontal sightline offset.

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39 Included for your benefit

40 e = 0 e max (for the design speed) Reciprocal of Radius Side Friction Factor Selection of f design and e design (Method 5) f max (for the design speed) f design 1/R f

41 e = 0 e max Reciprocal of Radius Side Friction Factor Selection of f design and e design f max f design R f = V 2 /(gf max ) R o = V 2 /(ge max ) R 0 : f = 0, e = e max R min = V 2 /[g(f max + e max )] 1/R f

42 e = 0 e max (for the design speed) Reciprocal of Radius Side Friction Factor Selection of f design and e design f max (for the design speed) f design f design = α(1/R)+β(1/R) 2 α = f max R min [1-{R min /(R 0 -R min )}] β = f max R min 3 /(R 0 -R min ) 1/R f

43 Superelevation Design for High Speed Rural and Urban Highways

44 Example: Design Speed: 100 km/h f max = 0.128 e max = 0.06 Question? What should be the design friction factor and design superelevation for a curve with a radius of 600 m?

45 1. Compute R f, R 0, and R min : R f = V 2 /(gf max ) = 27.78 2 / (9.81 x 0.128) = 615 m R 0 = V 2 /(ge max ) = 27.78 2 / (9.81 x 0.06) = 1311 m R min = V 2 /[g(f max + e max )] = 27.78 2 / [9.81(0.128+0.06)] R min = 418 m

46 e = 0 e max = 0.06 Side Friction Factor Selection of f design and e design (example) f max = 0.128 f design 1 / 1311 1 / 615 1 / 418 1/R f

47 2. Compute α and β: α = 0.128 x 418 x [1 – 418 / (1311 – 418) ] = 28.45 m β = 0.128 x 418 3 / (1311 – 418) = 10502 m 2 3. Compute f design and e design : First, estimate the right-hand side of equation for designing superelevation e + f = V 2 /(gR) = 27.78 2 / (9.81 x 600) = 0.131 Then, f design = 28.45 / 600 + 10502 / 600 2 = 0.076 e design = 0.131 – 0.076 = 0.055 (< e max = 0.06)

48 e = 0 e max = 0.06 Side Friction Factor f max = 0.128 f design 1 / 1311 1 / 615 1 / 418 1 / 600 0.076 Selection of f design and e design (example) 1/R f

49 Selection of f design and e design (example) R=600 ft


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