Vertical Alignment
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Vertical Alignment - Transportation Engineering - Lecture Slides, Slides of Transportation Engineering
Some concept of Transportation Engineering are Basic Transportation Model, Classification of Urban Streets, Example of Shock Wave, Geometric Design of Highways, Route Choice, Trip Assignment, Time-Distance Diagrams. Main points of this lecture are: Vertical Alignment, Components of Highway Design, Horizontal Alignment, Profile View, Plan View, Topography, Properties of Vertical Curves, Design of Vertical Curves, Parabolic Curves, Characterizing Curve
Typology: Slides
2012/2013
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Vertical Alignment
Components of Highway Design
Horizontal Alignment
Vertical Alignment
Plan View
Profile View
Today’s Class
- Properties of vertical curves (parabolic)
- Design of vertical curves
Crest Curve
Sag Curve
G 1
G 2 G 3
Vertical Alignment - Overview
Properties of Vertical Curves
BVC
EVC
L
G 2
G 1
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 / L /
PI
Properties of Vertical Curves
BVC
EVC
L
G 2
G 1
Characterizing the curve: Rate of change of curvature: K = L / | A | Which is a gentler curve - small K or large K?
L / L /
PI
Properties of Vertical Curves
BVC
EVC
PI
L
G 2
G 1
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
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
Properties of Vertical Curves
BVC
EVC
PI
G 2
G 1
Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+ Station of PI = 24+
r - value?
r = ( g 2 - g 1 )/ L r = (0.02 - [-0.01])/200 m r = 0.00015 / meter
Properties of Vertical Curves
BVC
EVC
PI
G 2
G 1
Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+ Station of PI = 24+
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+
Properties of Vertical Curves
BVC
EVC
PI
G 2
G 1
Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+ Station of PI = 24+
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
Properties of Vertical Curves
BVC
EVC
PI
G 2
G 1
Example: G 1 = -1% G 2 = +2% Elevation of PI = 125.00 m Station of EVC = 25+ Station of PI = 24+
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
Design of Vertical Curves
- Determine the minimum length (or minimum K ) for a
given design speed.
- Sufficient sight distance
- Driver comfort
- Appearance
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 ft h 2 = height of object, in ft