Matakuliah
Tahun
: S0753 – Teknik Jalan Raya
: 2009
Geometric Design
Session 02-06
Contents
•Concepts
•Vertical Alignment
•Fundamentals
•Crest Vertical Curves
•Sag Vertical Curves
•Examples
•Horizontal Alignment
•Fundamentals
•Superelevation
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Introduction
• Alignment is a 3D problem broken
down into two 2D problems
– Horizontal Alignment (plan view)
– Vertical Alignment (profile view)
• Stationing
– Along horizontal alignment
Piilani Highway on Maui
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Introduction
Stationing
Horizontal Alignment
Vertical Alignment
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Introduction
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From Perteet Engineering
Geometric Design Elements
•
•
•
•
Sight Distances
Superelevation
Horizontal Alignment
Vertical Alignment
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Vertical Alignment
• Objective:
– Determine elevation to ensure
• Proper drainage
• Acceptable level of safety
• Primary challenge
– Transition between two grades
– Vertical curves
Sag Vertical Curve
G1
G2
G1
G2
Crest Vertical Curve
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Vertical Curve Fundamentals
• Parabolic function
– Constant rate of change of slope
– Implies equal curve tangents
y  ax  bx  c
2
• y is the roadway elevation x stations
(or feet) from the beginning of the curve
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Vertical Curve Fundamentals
G1
PVC
PVI
δ
G2
PVT
L/2
L
x
y  ax  bx  c
2
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Choose Either:
• G1, G2 in decimal form, L in feet
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• G1, G2 in percent, L in stations
Relationships
At the PVC : x  0 and Y  c
Choose Either:
• G1, G2 in decimal form, L in feet
• G1, G2 in percent, L in stations
dY
 b  G1
dx
At the PVC : x  0 and
d 2Y
G2  G1
G2  G1
Anywhere :
 2a 
a
2
dx
L
2L
G1
PVC
PVI
δ
G2
PVT
L/2
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L
x
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Example
A 400 ft. equal tangent crest vertical curve has a PVC station of
100+00 at 59 ft. elevation. The initial grade is 2.0 percent and the final
grade is -4.5 percent. Determine the elevation and stationing of PVI,
PVT, and the high point of the curve.
PVI
PVT
PVC: STA 100+00
EL 59 ft.
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PVI
PVT
PVC: STA 100+00
EL 59 ft.
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•G1, G2 in percent
•L in feet
Other Properties
G1
x
PVT
PVC
Y
Ym
A  G1  G2
A 2
Y
x
200 L
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G2
PVI
AL
Ym 
800
Yf
AL
Yf 
200
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Other Properties
• K-Value (defines vertical curvature)
– The number of horizontal feet needed for a 1% change in slope
L
K
A
high / low pt.  x  K G1
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Crest Vertical Curves
SSD
PVI
Line of Sight
PVC
G1
PVT
G2
h2
h1
L
For SSD < L
ASSD 
For SSD > L
2
L

100 2h1  2h2
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
2

200 h1  h2
L  2SSD  
A 16

2
Crest Vertical Curves
• Assumptions for design
– h1 = driver’s eye height = 3.5
ft.
– h2 = tail light height = 2.0 ft.
• Assuming L >
SSD…
2
SSD
K
2158
• Simplified Equations
For SSD < L
ASSD 
L
2158
2
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For SSD > L
2158
L  2SSD  
A
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Design Controls for Crest Vertical Curves
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
Design Controls for Crest Vertical Curves
Sag Vertical Curves
Light Beam Distance (SSD)
G1
headlight beam (diverging from LOS by β degrees)
PVT
PVC
h1
G2
PVI
h2=0
L
For SSD < L
ASSD 
L
200h1  S tan  
2
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For SSD > L
200h1  SSD  tan  
L  2SSD  
A
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Sag Vertical Curves
• Assumptions for design
– h1 = headlight height = 2.0 ft.
– β = 1 degree
• Simplified Equations
For SSD < L
ASSD 
L
400  3.5SSD 
2
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• Assuming L >
SSD…
2
SSD
K
400  3.5SSD
For SSD > L
 400  3.5SSD  
L  2SSD   

A


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Design Controls for Sag Vertical Curves
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
Design Controls for Sag Vertical Curves
Horizontal Alignment
• Objective:
– Geometry of directional transition to ensure:
• Safety
• Comfort
• Primary challenge
– Transition between two directions
– Horizontal curves
Δ
• Fundamentals
– Circular curves
– Superelevation
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Horizontal Curve Fundamentals
PI
Δ
T

T  R tan
2
E
M
PC
L
Δ/2
PT

100
L
R 
180
D
 180 
100

  18,000

D

R
 R
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R
R
Δ/2 Δ/2
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Horizontal Curve Fundamentals
PI
Δ
T
E
M
PC
 1

E  R
 1
 cos  2 


M  R1  cos 
2

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L
Δ/2
R
PT
R
Δ/2 Δ/2
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W p  F f  Fcp
Rv
≈
Superelevation
Fc
e
W
1 ft
α

 WV 2
WV 2
W sin   f s W cos  
sin   
cos 
gRv

 gRv
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Superelevation

 WV 2
WV 2
W sin   f s W cos  
sin   
cos 
gRv

 gRv
V2
1  f s tan  
tan   f s 
gRv
V2
1  f s e
e  fs 
gRv
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V2
Rv 
g  f s  e
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Selection of e and fs
• Practical limits on superelevation (e)
– Climate
– Constructability
– Adjacent land use
• Side friction factor (fs) variations
– Vehicle speed
– Pavement texture
– Tire condition
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Side Friction Factor
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
Minimum Radius Tables
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WSDOT Design Side Friction Factors
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from the 2005 WSDOT Design Manual, M 22-01
For Open Highways and Ramps
WSDOT Design Side Friction Factors
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from the 2005 WSDOT Design Manual, M 22-01
For Low-Speed Urban Managed Access Highways
Design Superelevation Rates - AASHTO
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
Design Superelevation Rates - WSDOT
emax = 8%
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from the 2005 WSDOT Design Manual, M 22-01
Circular Curve Geometrics
•
•
•
•
•
PC = Point of Curvature
PT = Point of Tangency
PI = Point of Intercept
100/D = L/Δ, so,
L = 100 (Δ /D) where:
• L = arc length(measured in Stations (1 Sta =
100 ft)
• Δ = internal angle (deflection angle)
• D = 5729.58/R
• M = middle ordinate m=R [1 – cos(Δ /2) ]
M - is maximum distance from curve to long chord
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Circular Curve Geometrics
Degree of curvature: D = central angle which subtends
an arc of 100 feet
D=5729.58/R where R – radius of curve
For R=1000 ft. D = 5.73 degrees
Maximum degree of curve/min radius:
Dmax = 85,660 (e + f)/V2 or
Rmin = V2/[15 (e + f)]
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Horizontal Sight Distance
1) Sight line is a chord of the circular
curve
2) Applicable Minimum Stopping Sight
Distance (MSSD) measured along
centerline of inside lane
Criterion: no obstruction
within middle ordinate
Assume:
driver eye height = 3.5 ft
object height = 2.0 ft.
Note: results in line of sight obstruction height
at middle ordinate of 2.75 ft
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Horizontal Alignment
• Basic controlling expression:
e + f = V2/15R
• Example:
– A horizontal curve has the following characteristics: Δ = 45˚, L
= 1200 ft, e = 0.06 ft/ft. What coefficient of side friction would
be required by a vehicle traveling at 70 mph?
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Circular Curve Geometrics
•
•
•
•
•
PC = Point of Curvature
PT = Point of Tangency
PI = Point of Intercept
100/D = L/Δ, so,
L = 100 (Δ /D) where:
• L = arc length(measured in Stations (1 Sta =
100 ft)
• Δ = internal angle (deflection angle)
• D = 5729.58/R
• M = middle ordinate m=R [1 – cos(Δ /2) ]
M - is maximum distance from curve to long chord
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Stopping Sight Distance

100 s
SSD 
Rv  s 
180
D
180SSD 
s 
Rv

 90 SSD 

M s  Rv 1  cos
 Rv 

Rv 
 Rv  M s 

SSD 
cos 
90 
 Rv 
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1
SSD
Ms
Obstruction
Rv
Δs
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Cross Section
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Superelevation Transition
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from the 2001 Caltrans Highway Design Manual
Superelevation Transition
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
Spiral Curves
No Spiral
Spiral
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
Spiral Curves
•
•
•
•
Involve complex geometry
Require more surveying
Are somewhat empirical
If used, superelevation transition should occur entirely
within spiral
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Desirable Spiral Lengths
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from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001