Presentation is loading. Please wait.

Presentation is loading. Please wait.

Vertical Alignment CE 2710 Spring 2009 Chris McCahill www.Exchange3D.com.

Similar presentations


Presentation on theme: "Vertical Alignment CE 2710 Spring 2009 Chris McCahill www.Exchange3D.com."— Presentation transcript:

1 Vertical Alignment CE 2710 Spring 2009 Chris McCahill www.Exchange3D.com

2 Components of Highway Design Horizontal Alignment Vertical Alignment Plan View Profile View

3

4

5 Vertical Alignment & Topography Texas DOT

6 Today’s Class Properties of vertical curves (parabolic) Design of vertical curves

7 Crest Curve Sag Curve G1G1 G2G2 G3G3 Vertical Alignment - Overview

8 Parabolic Curves RadiusParabola Horizontal Curve Vertical Curve

9 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

10 Properties of Vertical Curves BVC EVC L G2G2 G1G1 Characterizing the curve: Rate of change of curvature: K = L / |A| Which is a gentler curve - small K or large K? L/2 PI

11 Properties of Vertical Curves BVC EVC L G2G2 G1G1 L/2 Characterizing the curve: 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 PI

12 Properties of Vertical Curves BVC EVC PI L G2G2 G1G1 Point elevation (meters or feet): y = y 0 + g 1 x + 1/2 rx 2 where, y 0 = elevation at the BVC (meters or feet) g = grade expressed as a ratio (positive /, negative \) x = horizontal distance from BVC (meters or feet) r = rate of change of grade expressed as ratio (+ sag, - crest) Point T Elevation = y

13 Properties of Vertical Curves Distance to turning point (high/low point) (x t ): Given y = y 0 + g 1 x + 1/2 rx 2 Slope: dy/dx = g 1 + rx At turning point, dy/dx = 0 0 = g 1 + rx t x t = -(g 1 /r) where, r is negative for crest, positive for sag

14 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 = 2500 m - 2400 m = 100 m Sta. BVC = Sta. PI - L/2 Sta. BVC = [24+00] - 100 m Sta. BVC = 23+00 L = 200 m

15 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

16 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

17 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

18 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

19 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

20 Design of Vertical Curves

21 Determine the minimum length (or minimum K) for a given design speed. –Sufficient sight distance –Driver comfort –Appearance

22 Design of Vertical Curves Crest Vertical Curve If sight distance requirements are satisfied then safety, comfort, and appearance will not be a problem. h 1 = height of driver’s eyes, in fth 2 = height of object, in ft

23 Design of Vertical Curves Crest Vertical Curve Sample Equation: From AASHTO: h 1 ≈ 3.5 ft h 2 ≈ 0.5 ft (stopping sight distance) h 3 ≈ 4.25 ft (passing sight distance)

24 Design of Vertical Curves Sag Vertical Curve Stopping sight distance not an issue. What are the criteria? –Headlight sight distance –Rider comfort –Drainage –Appearance

25 Sag Vertical Curve Check also: Comfort –Change in grade, A –Design Speed Appearance –Change in grade, A Design of Vertical Curves

26 Maximum Grade Context –Rural –Urban Design Speed Terrain –Level –Rolling –Mountainous 0.5% to 3.0%

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

28 Maximum Grade www.nebraskaweatherphotos.org

29 Maximum Grade Dee747 at picasaweb.google.com


Download ppt "Vertical Alignment CE 2710 Spring 2009 Chris McCahill www.Exchange3D.com."

Similar presentations


Ads by Google