tte3810 geometric design part one_fall 2013.pdf
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TTE3810 CLASS
HIGHWAY GEOMETRIC DESIGN (PART-I-)
Prepared By
Dr. Haitham Al-Deek, P.E.
Professor of EngineeringDepartment of Civil and Environmental Engineering
University of Central Florida
Orlando, FL 32816-2450Phone: (407) 823-2988Fax: (407) 823-3315
E-Mail: [email protected]
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
Table of Contents
Page Number
I- Design Features 1
II- Elements of Highway Geometric Design 1
1. Design Speed 4
2. Horizontal Alignment 6
2.1 Simple Curve 6
2.2 Review of Curvilinear Motion and Derivation of 13 Superelevation Equation
2.3 Superelevation 16
2.4 Transition Curves (Spirals) 27
3. Vertical Alignment 32
4. Sight Distance 44
4.1 Types of Sight Distance 44
4.2 Sight Distance on Horizontal Curves 46
4.3 Sight Distance on Vertical Curves 51
Appendix A: Metric Tables for Sight Distance on Vertical Curves 55
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 1 -
I.
DESIGN FEATURES
Geometric Design
Highway Alignment X-Section Design
x
z
y
x - Latitude y - Departure (Positive y ⇒ N)
z - Elevation above mean sea level
pts. along route are called stations( the distance in feet from somereference point)
At right angle to direction of alignment
Describes widths, clearances and slopes...
The essential design features of a roadway are its location and its cross section. In the horizontal
plane, the locations of points are referenced to a coordinate system in which the positive y-axis is
north and the positive x-axis is east.
Points along the route are identified by stations. A station is defined as the distance in feet or
meter from some reference point, commonly the beginning point of the roadway. The location
of points in the vertical plane (or along z-axis) is given as the elevation above the mean sea level.
The cross section of a roadway is described by its dimensions at the right angle to the direction
of the alignment, including widths, clearances, slopes, and so on.
II. ELEMENTS OF HIGHWAY GEOMETRIC DESIGN
1. Design Speed
2. Horizontal Alignment
2.1. Simple Curve
2.2. Superelevation (e)
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 2 -
2.3. Transition Curves (Spirals)
3. Vertical Alignment
4. Sight Distance
4.1. Types of Sight Distance4.2. Sight Distance on Horizontal Curves
4.3. Sight Distance on Vertical Curves
5. Earthwork (covered in the “Surveying” course).
6. Cross Section Elements
7. Intersections and Interchanges
7.1. Intersection at Grade.
7.2. Grade Separations and Interchanges
The relationship of traffic to highway design
Traffic elements that influence highway design include the following:
• Average Daily Traffic (ADT)
• Design Hourly Volume (DHV)
• Directional Distribution (D)
• Percentage of Trucks (T)
• Design Speed
Selection of the Design Hourly Volume (DHV)
Average Annual Daily Traffic "AADT" (vehicles per day) is not appropriate to use in the
geometric design of highways. Traffic engineers use hourly traffic volumes for design. Which
hourly traffic volume should be used?
The maximum peak hourly traffic volume during the year? Or
The average hourly traffic volume during the year ?
Answer: None!!
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 3 -
5
10
15
20
30
25
H o u r l y V o l u m e a s p e r c e n t o f A A D T
Source: Minnesota Department of Transportation
100 300200 600500400 1000900800700 1100
EXHIBIT 8-8. RANKED HOURLY VOLUMES
Recreational Access Route MN 169
Main Rural Route I-35
Urban Circumferential Freeway I-494
Urban Radial Freeway I-35E
Hour Rank
Source: HCM2000 and AASHTO 2001
24
14
10
30 100
H o u r l y V o l u m e
Number of hours in one year with hourly volume
greater than that shown
AADT
The 30th highest hourly volume is used as the design hourly volume. This means that the
designer is willing to tolerate subjecting the public (travelers) to congested conditions for 29
hours during the year (this is only 0.3% of all hours during the year).
This Volume is referred to as DHV.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 4 -
DHV K ADT = ×
K does not vary much from year to year!
8 – 12% Urban highways
K= 10-15% Suburban highways
12-18% Rural highways
FDOT (1995) uses different values. See below:
Area Type K-FactorUrbanized 0.091
Urban 0.093Transitioning/Urban 0.093Rural Developed 0.095Rural Undeveloped 0.100
Note that DHV is the sum of hourly volume for both directions. To find the “Directional Design
Hourly Volume” or DDHV we need to multiply the DHV by a distribution factor called D to
split traffic volume between the two directions, or use ADT to calculate DDVH as follows:
DDHV K D ADT = × ×
Where,
D = Directional split factor, which splits traffic into the heavy direction. So, D might be as high
as 0.7 or 70% in the heavy direction in rural highways during design hour, and the off-peak
direction carries only 30% of the traffic. Normally, the peak/off-peak volume switches
directions between morning and evening peak hours. This means that one direction will be
heavy in the morning while the opposite direction will be heavy in the evening.
1. DESIGN SPEED
The elements of highway design are influenced by the design speed. The selection of the
appropriate design speed depends on:
Highway type (rural vs. urban, interstate, local)
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 5 -
Traffic characteristics (volume and composition).
Vehicle characteristics (trucks, passenger cars).
Cost of land.
Topography (level, rolling, mountainous).
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 6 -
2.
HORIZONTAL ALIGNMENT
Horizontal curves are curves in plan used to provide change in direction of the center line of the
road.
2.1 Simple Curve
A simple curve is a circular joining two tangents.
x
y
φ
Pl = arc
lengthPC to P
PC
R θ
θ
PC PT
PI
P
R R
R
100'
D
D = central angle
for 100' arc
∆
2∆
C
M
4∆
Back
Tangent Forward
Tangent
L =arc
length
PC to PT
∆
αT E
( ,R)
VARIABLES
PC = Point of curvature (Beginning of curve) PI = Point of intersection
PT = Point of tangency (End of curve) D = Degree of curvature (1)
PI = Point of intersection R = Radius of curve
∆ = Central angle (Intersection angle) E = External distance
L = Length of curve (PC to PT) M = Middle ordinate
l = Length of arc (PC to P) C = Chord Length
θ = Central angle for arc length l
T = Tangent length (PC to PI & PT to PI)
φ = Deflection angle at PC between tangent and chord for P
α = Deflection angle at PI between tangent and line from PI to P
x = Tangent distance from PC to P
y = Tangent offset P
Note: (1) This variable is used only for curve definition in traditional US units
Properties of a simple circular curve
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 8 -
Example 1:
A circular horizontal curve has an intersection angle ∆ = 23o 42' right and degree of curve
D = 1o 45′ (chord definition). Compute the tangent distance, curve length, and station of the
point of the tangency PT . The station of point of intersection PI = 27+85.50
Solution to Example 1:
Using the degree of curve (chord definition)
50 50R 3274.1 ft
1 45sin sin
2 2
D= = =
′°
23 42T = R * tan 3274.17* tan 686.99 ft
2 2
′∆ ° = =
2 R 2 *3274.17*23.7L 1354.34 ft
360 360
Π ∆ Π °= = =
PI station = 27+85.50 = 2785.50 ft
PC station = PI station – back tangent distance = 2785.50 - 686.99 = 2098.51 ft = 20+98.51
PT station = PC station + curve length = 2098.51 + 1354.34 = 3452.85 ft = 34+52.85
Example 2:
Two tangents are to be connected by a circular curve with D = 4°. The tangents intersect at
station (28+42.83) with ∆ = 52°30′. Find R, L, T, E, stations of PC and PT.
Solution to Example 2:
5729.6R 1432.4 ft
4= =
PCPT
PI ∆
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 9 -
( ) ( )2*3.1415
L 1432.4 52.5 1312.5 ft360
= =
52.5T 1432.4* tan 706.4 ft
2
= =
52.5E 706.4* tan 164.7 ft
4
= =
PC station = (28+42.83) – (7+06.40) = (21+36.43)
PT station = (21+36.43)+(13+12.5) = (34+48.93)
Example 3:
The two tangents described below intersect 2000 ft beyond station 10+00, back tangent
N45°00′00′′ W, forward tangent N 15°00′00′′ E. It is desired to design a 3000 ft radius
horizontal curve between the two tangents. Determine degree of curve (arc. def.), the tangent
distance, length of curve, station of PC, and station of PT.
PC
PT
PI
∆
10+00
2000 ft
45°15°
N
Not to scale
R=3000 ft
N 15°00'00"E
N 45°00'00"W
Solution to Example 3:
∆ = 15o 00′00′′ + 45o
00′00′′ = 60o 00′00′′
R = 3000 ft
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 10 -
100*180D = 1.91
R = °
π------- (arc. def) or
5729.6D =
R
60°T = R*tan = 3000*tan = 1732.05 ft
2 2
∆
2 R 2 *3000*60°L = = = 3141.59 ft
360 360
Π ∆ Π
Station of PI = 1000 + 2000 = 3000 ft = 30+00
Station of PC = Station of PI - Tangent length
= 3000 - 1732.05 = 1267.95 ft = 12+67.95
Station of PT = Station of PC + Length of curve
= 1267.95 + 3141.59 = 4409.54 ft = 44+09.54
Example 4:
Given highway curve of 10°
PC PT
PI
Station 50+00
Δ= 200 34'
D = 100
Required
a) Find T, station of PC and PT, length of curve, length of long chord, and middle ordinate M.
b) Calculate the chord length between stations 49+50 and 50+50.
c) With forward tangent moved out 2 ft and PC held, what is the length of the new curve? See
figure below.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 12 -
Find R new ?
Toriginal = 103.95 ft
2X = = 5.69 ft
sin 20° 34′
Tnew = 103.95 + X = 103.95 + 5.69 = 109.64 ft
R original = 572.96’
new
109.64R = = 604.31 ft
20.5667tan
2
newL = 604.31* *20.5667 = 216.92 ft180
Π
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 13 -
2.2 Review of Curvilinear Motion and Derivation of Superelevation
Equation
Curvilinear Motion
an
at
V
R
C
Vehicle’s acceleration Horizontal (tangential) at
Normal an
2
, if = constant => 0t t
n
dva v a
dt
va
R
= =
=
Also, for a circular path R is fixed. Take a horizontal cross-section AA for a highway.
Direction of velocity is tangent to the
path.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 14 -
R
C
A A
t
n
A AF
W
R
Lateral effect
Side friction resists the tendency of vehicle to slide outward in the circular path. To minimize
this tendency, highway design provides for superelevation of the section of the roadway => the
cross-section is tilted by an angle θ.
F
W
R
θ
Superelevation is used to counteract the effect of the outward acting centrifugal force. The outer
edge of the road should be raised with respect to the inner edge producing a traverse slope
throughout the whole length of the horizontal curve.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 15 -
2
cosW vg R
θ
2W vg R
θ
θF
Fx
y
2
sinW vg R θ
W cos θ
W
W sin θ
The components of the vehicle’s weight along the tilted pavement surface (X-axis) resist the
sliding tendency of the vehicle. Friction + Weight component resist this tendency of sliding
outward.
f = friction factor F= f. N
( )
( )
2
2
cos sin 1
from X-axis => sin cos 2
w vF f w
g R
w vF w
g R
θ θ
θ θ
= +
+ =
From (1) and (2)
( )
( )
( )
2 2
2 2
222 2 2
2
cos sin sin cos 3
divide by cos
tan tan
Let tan be = e, f.e 01.47
32.16 14.88 15
15
Note: V is in mph, R i
wf v w vwf w
g R g R
w
v v f f
gR gR
V v V V e f
gR R R R
V e f
R
θ θ θ θ
θ
θ θ
θ
+ + =
+ + =
≈
+ = = = ≈
+ =
s in ft.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 16 -
2.3 Superelevation (e)
Superelevation (e) is defined as the ratio of the height of the outer edge to the horizontal width of
the road.
θ h
P.C.
P.T.
outer "0"
inner
"i"
h = e
L
L
2 cosW vg R θ
2W vg R
θ
θF
Fx
y
2
sinW vg R
θ
W cos θ
W
W sin θ
Analysis of the forces acting on the vehicle as it moves on a circular curve results the following
formula for calculating the minimum design radius :
R 15
Vf e
2
=+ (traditional U.S. units)
Where
e = superelevation rate
f = friction factor
V = design speed in mph
R = minimum radius, ft
R 127
Vf e
2
=+ (metric system)
Where
e = superelevation rate
f = friction factor
V = design speed in kph
R = minimum radius, m
This equation shows that the effect of centrifugal force is resisted partly by superelevation(e) and
partly by friction between tires and road surface(f). From this equation, it can be seen that as the
summation of (e+f) increases, the required radius R decreases. So, there is a minimum allowable
R and a maximum allowable (e and f).
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 17 -
emax : According to AASHTO "Green Book", the maximum allowable design value for "e" is
0.12 ft/ft. Present practice suggests emax = 0.10 ft/ft. In locations where snow and ice
conditions occur, emax = 0.08.
f max: The maximum design values for "f" depend on design speed. Value of "f" ranges
between 0.17 and 0.10 for design speed range of 20 to 70 mph.
Rmin: For a given design speed, R min and Dmax (max. safe degree of curvature) can be
calculated from the maximum rate of superelevation "e" and the side friction factor (f)
as follows:
)f e(15
VR
.max
2
.min +=
which is equivalent to:
2
max
.maxV
)f e(900,85D
+=
where,
R min.= minimum safe radius "R" of the horizontal curve.
Dmax= maximum safe degree of the horizontal curve.
Question: Why do we set maximum limits on superelevation e?
Answer:
There are three reasons:
1) To prevent slow moving vehicles from sliding to the inside of the curve,
2) In urban areas, to keep parking lanes relatively level; and
3) To keep the difference in slope between the roadway and any streets or driveways that
intersect it within reasonable bounds.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 18 -
Example 5:
• Calculate the maximum degree of curve and R min of a simple circular curve with an externalangle of 1000. The design speed is 50 mph, the corresponding value of f max is 0.14 and emax =0.10 ?
• Calculate edesign for a curve that has a radius of 800’ ? (given above f max)
Solution to Example 5:
( ) ( )
( )
( )
2 2
min
min
0
2
0
2 2
2
50694.4'
15 15 0.1 0.14
695'
85,900 0.1 0.148.2464
50,
5729.588.24
695
Note: External angle does not enter these calculations.
800'
15 15
50 0.14 0.0715 800
design
V R
e f
R
D
or
D
R
V V e f e f
R R
e
= = =+ +
≅
+= =
= =
= ⇒
+ = ⇒ = −
= − =
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 19 -
Example 6
Example of Minimum Raduis Data for Limiting Values of e and f :
Rural Highways and High-Speed Urban Streets
DesignSpeed
(km/hr)
Maximume
Maximumf
Total(e + f)
CalculatedRadius
(m)
RoundedRadius
(m)120 0.06 0.09 0.15 755.9 755
30 0.08 0.17 0.25 28.3 30
40 0.08 0.17 0.25 50.4 50
50 0.08 0.16 0.24 82.0 80
60 0.08 0.15 0.23 123.2 125
70 0.08 0.14 0.22 175.4 175
80 0.08 0.14 0.22 229.1 230
90 0.08 0.13 0.21 303.7 305
100 0.08 0.12 0.20 393.7 395
110 0.08 0.11 0.19 501.5 500
120 0.08 0.09 0.17 667.0 665
30 0.10 0.17 0.27 26.2 2540 0.10 0.17 0.27 46.7 45
50 0.10 0.16 0.26 75.7 75
60 0.10 0.15 0.25 113.4 115
70 0.10 0.14 0.24 160.8 160
80 0.10 0.14 0.24 210.0 210
90 0.10 0.13 0.23 277.3 275
100 0.10 0.12 0.22 357.9 360
110 0.10 0.11 0.21 453.7 455
120 0.10 0.09 0.19 596.8 595
30 0.12 0.17 0.29 24.4 25
40 0.12 0.17 0.29 43.4 45
50 0.12 0.16 0.28 70.3 7060 0.12 0.15 0.27 105.0 105
70 0.12 0.14 0.26 148.4 150
80 0.12 0.14 0.26 193.8 195
90 0.12 0.13 0.25 255.1 255
100 0.12 0.12 0.24 328.1 330
110 0.12 0.11 0.23 414.2 415
120 0.12 0.09 0.21 539.9 540Source: A Policy on Geometric Design of Highways and Streets , American Association of State Highway andTransportation Officials, Washington, DC, 1994.
Given the table above, calculate the superelevation rates for a roadway with a design speed of 90
km/hr that has a wide range of curve radii, that is, R has values of 585, 440, 350, 295, and 250 m.
(These values correspond to degrees of curve D of 3, 4, 5, 6, and 7.) Use emax = 0.10. Compare
the results with those obtained in the following figure.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 20 -
Solution to Example 6:
By the above table, use a maximum side friction factor of 0.13.
By the equation2
127
V e f
R+ = , the following values of e are calculated.
R (m) Computed value of e Recommended design e, fromfigure below
585 -0.021 0.070
440 +0.015 0.083
350 +0.052 0.094
295 +0.086 0.099
250 +0.125 Exceeds emax
Superelevation of Railway and Transit Guideway Curves
7000
5000
3000
20001500
1000
700
300
500
150
200
100
30
50
70
R a d i u s o f c u r v e ,
R ( m
)
e m a x
= 1 0 . 0
%
Superelevation rate, e (%)
0.0 10.09.08.07.06.05.04.03.02.01.0 11.0
30
40
50
100
80
70
60
90
110120
V, km/h
Design superelevation rates, emax
= 0.10. (Source: A Policy on Geometric Design of
Highways and Streets, copyright 1994, American Association of State Highway and Transportation
Officials, Washington, DC. Used by permission.)
Discussion:
For the sharpest curve (the 250-m curve), the combination of maximum superelevation rate and
the maximum side friction factor is insufficient to offset the centrifugal force. This curve is too
sharp for the given design speed and maximum superelevation rate and would be unsuitable forthe stated conditions.
At the other extreme, for the 585-m curve, a negative value of e was computed. Along this
curve, all of the centrifugal force could be offset without exceeding the recommended f max value
of 0.13, even with zero superelevation. The AASHTO favors a distribution of superelevation
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 21 -
that provides a logical relation between the side friction factor and applied superelevation rate
and recommends a positive amount of superelevation for the flattest curve.
The minimum radius of curve for a given design speed can be determined from the maximum
rate of superelevation and the side friction factor. The minimum comfortable radius R can becalculated from the following:
( )
2
min
max
VR
127 e f =
+ (metric system)
( )
2
min
max
VR
15 e f =
+ (traditional U.S. units)
The relationship between superelevation and minimum radius of a curve is shown in the previous
table for selected metric design speeds.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 22 -
Superelevation Design:
Types of X-Sections
P.T. P.C.
P
Normal Crown (nc)
P
Adverse Crown
Removed Cross Fall
efull
efull (design)
Crown runoff
Direction
of tilting
P
w
Superelevation runoff: Transition of the x-section from adverse crown removed on the tangent
to a fully superelevated pavement (edesign) on the curve.
Current Practice:
P
efull
Y
Y
P.T.
Crown
runoff -P
ww
Lc
o
L3
2L3
P
P
P'
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 23 -
w: width of section being rotated (ft), Lc to edge
L: Length of superelevation runoff (ft)
e: Superelevation rate (ft/ft)
r: slope ratio of outer edge of pavement in relation to Lc
Superelevation profile
Transition of the x-section from the normal crown on the tangent to fully superelevated
pavements (e design) on the curve.
crown
runoff
Y Y
23 L 1
3 L
Outer
edge
CLLevel
Inner
edge
on tangent on curve
PP
P'e
e fullCross
Fall
Normal
Crown
P . C . o r P . T .
E l e v a t i o n ( f t )
Adverse Crown
Removed
The Superelevation transition (runoff) for simple circular curves is located partly on the
tangent and partly on the curve. Standard practice assigns 2/3 of the length on tangent. This
rule will only apply to simple curves. A crown runoff is also provided before the
superelevation runoff.
r is defined as the relative change of slope between the edges of the section being rotated and
the centerline.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 24 -
r
ewL
⋅=
where :
w: width of section being rotated (ft)
L: length of superelevation runoff (ft)
e: superelevation rate (ft/ft)
r: slope ratio of outer edge of pavement in relation to centerline. Typical values of "r" are
1:200 (or 0.5%).
Example 7:
Draw a superelevation diagram (rotation around centerline) for a right-turn horizontal curve of a
2 lane 2 way rural highway. Given :
Lane width = 12.5 ft Normal crown slope (p) = 2% PC station = 101+00
Superelevation (e) = 6 % , "r" = 0.5%
The length of the horizontal curve = 200 ft
Solution to Example 7:
(see following two figures for superelevation profiles)
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Phone 407-823-2988 [email protected]
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
Example 7 Superelevation Profile
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Phone 407-823-2988 [email protected]
- 28 -
2.4 Transition Curves (Spirals)
Transition curves serve the purpose of providing a gradual change from the tangent section to the
circular curve and vice versa. A vehicle that enters a circular curve with transition travels
smoothly and naturally along a curve that gradually changes from straight line (infinite radius) to
a radius of some finite value, R, which is maintained throughout the length of the circular curve.
The most commonly used transition curve is the spiral. Spiral transitions normally are used only
on high-volume highways where the degree of curvature exceeds about 3°.
∆
∆c θ s
θ s
R c
S T
CSSC
L C
T S
K
L. T.
T S
PI
S. T.
θ s
E s
∆
PC
P φ c
PI - PT OF INTERSECTIONTS - TANGENT SPIRAL
SC - SPIRAL CURVE
CS - CURVE SPIRALST - SPIRAL TANGENT
RC - RADIUS OF CIRCULAR CURVE
L.T.- LONG TANGENT
S.T. SHORT TANGENT
LS - LENGTH OF SPIRAL
L - LENGTH TO ANY PT ON SPIRAL FROM TS
LC - CHORD DISTANCE - TS TO SC
S - CENTRAL ANGLE OF SPIRAL, "SPIRAL ANGLE"
- CENTRAL ANGLE OF ARC L
- DEFL. ANGLE FOR ANY PT. ON SPIRAL
FORMULAS3V
RCL = 1.6 , WHERE V = DESIGN SPEED
s
C - DEFL ANGLE FOR SC - INTERSECTION ANGLE,
TOTAL CENTRAL ANGLEC - CEN ANGLE OF CIRCULAR ARC
LC - LENGTH OF CIRCULAR ARCDC - DEGREE OF CURVE OF
SHIFTED CIRCLE
ES - EXTERNAL DIST FROM PI
TO CIRCULAR CURVE
P - OFFSET DIST OF CIRCULAR
CURVE PRODUCED
K - DIST FROM TS TO PC OF
CIRC CURVE PRODUCED
o
S CL D
S 200
C S
θ =
Δ =Δ 2θ−
( )
( )
( )( )
s
S
S
ΔS C 2
ΔS C 2
2O L3 L
100 Δ-2θ
C D
E = R +P exsec + P
T = R +P tan + k
=
L =
φ
Equations of Spirals
ss
c
Lθ = (radians)
2R or s c
s
L Dθ = (degrees)
200
C S2∆ = ∆ − θ
( )S
C
c
100 Δ 2θL =
D
−
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- 29 -
Where :
LS = length of spiral, ft θS = spiral angle
R c = Radius of circular curve Dc = degree of circular curve
Lc = length of circular curve, ft ∆C = circular angleθ = central angle of arc L Φ = deflection angle for any point on the spiral
ΦC = deflection angle for SC ∆ = intersection angle, total central angle
TS = tangent distance.
Length of transition curve
When spiral transitions are used, superelevation should be attained within the limits of the
transition, (i.e., T.S-S.C , C.S.- S.T.) segments. Note that the two third L rule does not apply
here as in the simple curve. The minimum length of the transition curve is given as
C
3
sR
V6.1L = (traditional U.S. units)
Where
V : design speed in mph
R c : radius in ft.
Ls : transition length in ft.
Example 8:
For a highway of design speed 44 mph, two tangents intersect at station (127+23.68) with
∆ = 54°. These tangents are to be connected by spiral curves in conjunction with a simple
circular curve of D = Dc = 4°. (P = 0.29, K = 50.00)
Find spiral length, spiral central angle, circular curve central angle, circular curve length, tangent
distance, and stations for the entire curve.
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
- 30 -
T S
P
SC CS
ES
∆C
θs
θs
S T
T S
K
PI (127+23.68)
∆ = 54°
L C
∆
R C
LS L
S
LC
Solution to Example 8:
( )33
s
1.58 441.58VL = 94 ft
5729.6R
4
= =
Round up to the nearest 50 feet Choose Ls = 100 ft. Question: Why do we round up?
s cs
L D 100*4 = = = 2
200 200θ
∆c + 2θs = ∆ = 54°, ∆c = 50°
cc
c
100 100*50L = = = 1250 ft
D 4
∆
sT = (R+P) tan + K = 780 ft2
∆
Station of TS = (127+23.68) – (7+80) = (119+43.68)
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Professor H. Al-Deek, Ph.D., P.E. Class TTE3810 Highway Geometric Design, Part-I
Phone 407-823-2988 [email protected]
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Station of SC = (119+43.68) + (1+00) = (120+43.68)
Station of CS = (120+43.68) + (12+50) = (132+93.68)
Station of ST = (132+93.68) + (1+00) = (133+93.68)
In the following page we show a superelevation highway figure that includes both simple
circular curve and transition curve.
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Phone 407-823-2988 [email protected]
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Phone 407-823-2988 [email protected]
- 35 -
Vertical Curve Derivations
( )
2
2
2
2
1 2
0
1
2 11 2
To make life easier use constant = r
. . constant
To have a smooth ride and
1
What is r ?
x x L
x L
d y
dx
dy d y
dx r dx rxdx dx
dy dyg g
dx dx
dyg rx
dx
dy g gg rL g r dx L
→ →
=
⊕ → =
= = = +
⊕ → → →
⇒ = +
−= + = ⇒ =
∫ ∫
( )
( )
2 11
2
1 1
2
Combine (1) and (2)
grade at any point x
Elevation constant2
dy g gg x
dx L
dy
dx
dy rx y dx g rx dx g x
dx
−= +
=
= = + = + +
↑
∫ ∫
( )
( ) ( )
0
2
0 1
1 2
= = Elevation at BVC
32
Units: Any consistent set of units ft, meters...etc
g ,g are dimens
y
rx y x y g x= + +
• ionless
r ? must have compatable units as x and y.
1r units are reciprocal feet i.e. if x and y are in ft.
ft Practicing engineers use % for grades
e.g., g=0.03 3% rise of 3 ft in one stat
• →
•
→ → ( )ion or 100'
r=0.6% per station instead of 0.00006/ft, therefore Eq. (3) :becomes
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Phone 407-823-2988 [email protected]
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( )
( )
0
0
1 2 1 2
1 2 1 2
G (G G )
100 200L
G (G G )Elev. of Point P on the Curve Elev. of BVC 4
100 200L
, :Where
y x y x x
x x
y is the BVC and this equation can be stated as
−= + +
−= + +
Equations of vertical curves
VPC or BVC VPT or EVC
Sta. of BVC = Sta. PVI -2
L
Elev. of BVC = Elev. of PVI -1G L
100 2
Sta. of EVC = Sta. PVI +2
L = Sta. VPC + L
Elev. of EVC = Elev. of PVI2G L
100 2
For any point "p" at a distance x from "BVC":
Sta. of P = Sta. of BVC + x
1 2 1 2G (G G )Elev. of P Elev. of BVC
100 200L x x
−= + +
Grade at P = G1 + AL
x
For any two points:
2
2 2 21
1 2 E
L
2
y y
x x
′ ′= =
High or Low Point on the Curve
The high (or low) point on the vertical curve is located at a distance XHP in case of crest curve
and XLP in case of sag curve. For simplicity we refer to this as X. The high (or low) point on the
vertical curve is located at a distance X.
1
1 2
LGX=
G -G
Where : L is the length of vertical curve in ft., X in ft.
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- 37 -
BVC EVC
PI
L/2
x
High point
Crest
BVC EVC
PI
L/2
x
Low point
Sag
1
1 2
is in stations x is in stations isin ft x is in ft
LG x
G G
If L If L
=−
⇒⇒
Hints on Vertical Curves
( )2 1
The distance from the P.I to the middle of the parabolic curve:
800
where - , difference in grade in percent e.g., 0.03=3%
AL E
A G G
=
=
[ ]
Note that A is +ve if "sag" A is -ve if "crest"
+ve for "sag"E and L are in ft. and E is
-ve for "crest"
E is the distance that connects the midpoint of the curve above or below the P.I
to the P.I
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E
E
Sag (E is below the
curve and above PI
Crest (E is above the curve
and below PI)
Offsets (Two Types)
2
1) Offset: is between curve and the gradient passing through the BVC. This is given by:
' 4
E, x, and L are all in ft and y' is in ft.
x y E
L
=
2
22 1
or, offset ' 4 800 200
x and L are in ft.
AL x G G
y x L L
−
= =
BVC
EVC
PI
Offset y'
( )
2
22 1
If x, L are in stations and E is in ft. Use same equation ' 4
If x, L are in stations and ' is calculated as:
' y' will be in stations not ft2
x y E
L
y
G G y x
L
⇒ =
−= ⇒
2) Offset: distance between curve and both gradients that pass through BVC and EVC.
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BVC
EVC
PIy-
y'
L/2 L/2
( )
( )
2
2
4 ,2
'
4 - - ,2 2100
x L E x L
y x x A L L E x x L
≤
= ≥
Example 9:
A 600 ft. sag vertical curve has a -4% grade meeting a +4% grade. How long must a new vertical
curve be in order to retain the same grade rates and to raise the elevation at the center of the
curve by 2 ft.
Solution to Example 9:
( ) ( )( )(4-(-4)) 600 8 600AL (G2-G1)LE= = = = = 6
800 800 800 800′
new E = 6+2 = 8
( )8 L8
800=
Lnew = 800 ft
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Example 10:
An ascending 5% grade and a descending 3% grade are joined by an 800 ft VC. Elevation at Sta
24+00 on the 5% grade is 1776.39 ft and elevation at Sta 36+00 on the 3% grade is 1780.39 ft.
Find the station of the BVC.
Solution to Example 10:
PVI
- 3%
EVC400'400'
d2
Sta 36+00
Elev 1780.39
Sta 24+00
Elev 1776.39
d1
5%BVC
d 1 + d 2 = 3600 - 2400
d 1 + d 2 = 1200 (1)
1776.39 + 0.05(d 1) - 0.03(d 2) = 1780.39
0.05(d 1) - 0.03(d 2) = 4 (2) substitute (1) into (2)
0.05(1200 – d 2) - 0.03(d 2) = 4
60 - 0.08 d 2 = 4 d 2 = 700 ft d 1 = 500 ftSta of P.I. = 2400 + d 1 = 2400 + 500 = 2900 ft = 29+00
Sta of BVC = 2900 - 400 = 2500 ft = 25+00
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- 41 -
Example 11:
Two tangent highway alignments intersect in plan view as shown in the sketch below. Each
alignment consists of a vertical curve in the vicinity of the intersection point. The two vertical
curves data are given in the table below. Find the vertical clearance between the highwayalignments at point 'P'.
A l i g n m e n t # 1
A l i g
n m e n t #
2
Point 'P'
Sta 80+25 on Alignment #1
Sta 130+50 on Alignment #2
Data Alignment #1 Alignment #2
PVI Station 84+00 130+00
PVI Elevation (ft) 175.00 145.00
Length of Vertical Curve 1200 ft 800 ft
Tangent grades G1
G2
+2 %
-3 %
-1 %
+2 %
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AL (G2-G1)L (2-(-1)(800) (3)(800)E = = = = = 3
800 800 800 800′
22
' 3
(13400 13050) 800
2
y=
−
y′ at Sta. 130+50 = 2.297′
Elev of 130+50 on tangent = 145 + (0.02) (13050-13000) = 146′
Elev of 130+50 on curve = 146 +2.297 = 148.297′
Clearance is the difference between Elevation as in Alignment#1 and elevation as in
Alignment#2 . Clearance = 166.445 - 148.297 = 18.148′
Example 12:
The grade line is being developed for a six-lane freeway with a design speed of 70 mph. A rising
2.0% grade meets a falling grade of 1.5% at station 100+00. The elevation of the P.I. of the two
grades is 98.00 ft.
a. Using vertical curve length of 1400 ft, compute the middle ordinate of the vertical curve.
b. For each station ,(i.e., 93+00, 94+00, etc…), calculate VC elevation.
Solution to Example 12:
PVI
Sta 100+00
Elev 98.00
EVC
Sta 107+00
-1.5%+2%BVC
Sta 93+00
S t a
9 4 + 0 0
S t a
9 5 + 0 0
S t a
9 6 + 0 0
S t a
9 7 + 0 0
S t a
9 8 + 0 0
S t a
9 9 + 0 0
S t a
1 0 0 + 0 C
S t a
1 0 1 + 0 C
S t a
1 0 2 + 0 C
S t a
1 0 3 + 0 C
S t a
1 0 4 + 0 C
S t a
1 0 5 + 0 C
S t a
1 0 6 + 0 C
E
L = 1400'
AL (G2-G1)L (-1.5-2)(1400) (-3.5)(1400)E = = = = = -6.125
800 800 800 800′
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( )2 2
E offset=
( )L
2
x Offset =
( )
2 22
2 2
E 6.1250.0000125
(700)1400
2
x x x
−= = −
Station Tangent Elevation VC Offset VC ElevationBVC 93+00 84.00 0.000 84.0094+00 86.00 - 0.125 85.87595+00 88.00 - 0.500 87.50096+00 90.00 - 1.125 88.87597+00 92.00 - 2.000 90.00098+00 94.00 - 3.125 90.87599+00 96.00 - 4.500 91.500P.I. 100+00 98.00 - 6.125 91.875101+00 96.50 - 4.500 92.000102+00 95.00 - 3.125 91.875103+00 93.50 - 2.000 91.500
104+00 92.00 - 1.125 90.875105+00 90.50 - 0.500 90.000106+00 89.00 - 0.125 88.875EVC 107+00 87.50 - 0.000 87.500
Example 13:
A 600 ft VC connects a +4% grade to a -2% grade at station 25+60.55 and elevation 648.64 ft.
Calculate the location and elevation of BVC, the high point, EVC, and the curve elevation at
stations 24+00 and 27+00 ?
Solution to Example 13:
PVI
Sta 25+60.55
Elev 648.64
BVC
Sta 22+60.55
EVC
Sta 28+60.55
+ 4% - 2% E
High Point
300' 300'
XHP
Sta. of BVC = Sta. of PVI -2
L = 2560.55 –
600
2
= 2260.55 = (22+60.55)
Elev. of BVC = Elev. of PVI -G1 L
100 2
= 648.64 -4 600
100 2
= 648.64 - 12 = 636.64'
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- 45 -
AL (G2-G1)L (-2-4)(600) (-6)(600)E= = = = =-4.5'
800 800 800 800
Elevation for Station 24+00
( ) 2G2-G1G1Elev. of Sta. 24 + 00 = Elev. of BVC + +
100 200L
x x
where x = Sta. 24+00 - Sta. BVC = 2400 - 2260.55 = 139.45′
24 ( 2 4)Elev. Sta. 24 00 636.64 (139.45) 139.45 641.25'
100 (200)(600)
− − + = + + =
or2
G1Elev. of Sta. 24 00 Elev. of BVC 4E
100 L
x x
+ = + +
24 139.45
Elev. of Sta. 24 00 636.64 (139.45) 4( 4.5)100 600
+ = + + −
= 641.25
The High Point : '4006
4*600
)2(4
4*600
2G1G
1LGHPX ==−−=−=
Point x (ft) Tangent Elevation (ft)(This is the tangent that passes through BVC)
VC Offset*(ft)
2
y 4EL
′ =
x
VC Elevation (ft)
BVC 22+60.55 0 636.64 0.00 636.6424+00.00 139.45 642.22 - 0.97 641.25
HP 26+60.55 400 652.64 - 8.00 644.6427+00.00 439.45 654.22 - 9.66 644.56
EVC 28+60.55 600 660.64 - 18.00 642.64* Note that the VC Offset was calculated with respect to gradient G1 only in this table.
An alternative method is to calculate the “VC Offset” with respect to gradients +G1 and G2, sowe can get the following table instead:Point x (ft) x* (ft) Tangent
Elevation (ft)(This tangent iseither +G1 or
G2)
VC Offset (ft)2
y 4EL
′ =
x
VC Elevation (ft)
BVC 22+60.55 0 0 636.64 0.00 636.64
24+00.00 139.45 139.45 642.22 - 0.97 641.25HP 26+60.55 400 200 646.64 - 2.00* 644.6427+00.00 439.45 160.55 645.85 - 1.29* 644.56
EVC 28+60.55 600 0 642.64 0.00* 642.64*It is important to note that x* is measured from right to left (i.e., from EVC to left). The term x* is used instead of x in the “VC Offset” formula above only when x is larger than L/2, that is300 ft in this example. Hence, in this example, x*= 600- x for x larger than L/2 and x*= x when x is smaller than L/2.
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4. SIGHT DISTANCE
4.1 Types of Sight Distance
Sight distance is defined as the distance at which objects become visible to the driver. The
design concept of Sight Distance is that drivers traveling at design speed should be able to stop
their vehicles (Stopping Sight Distance SSD) or overtake slower vehicles safely (Passing Sight
Distance PSD).
Minimum SSD
The minimum distance sufficient to enable a driver traveling at design speed to stop his/her
vehicle before striking an unexpected object on the highway.
Minimum SSD = 1.47 V t +)gf (30
V2
±
Where,
SSD is in ft.
V = Design speed, mph
t = perception and reaction time. ( ~ 2.5 seconds)
f = friction factor, depends on tire, surface and speed
g = grade as decimal not as percent (e.g., 6%, g = 0.06). +ve for upgrade and -ve for
downgrade.
+ -
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SSD is provided in tables for different pavement conditions
(AASHTO 2001, Tables III-1 and III-2).
Table 16.2. Coefficients of Skidding Friction
Reference: See page 16-5 of the Civil Engineering
Reference Manual, 6th
Edition, by Michael Lindeburg
Exhibit 3-1 and Exhibit 3-7 of AASHTO 2001
Minimum Sight Distances
AASHTO 2001, pages 112 and 124.BC: bituminous concrete, dry
SA: sand asphalt, dry
RA: rock asphalt, dry
CC: portland cement concrete, dry
wet: all wet pavements
stopping sight distance
wet pavements
passing sight
distances
design
speed
mph
Calculatedsight
distance
ft
Designstopping
sightdistance
ft
Assumed passingspeed,mph
Distance2-lane
highway,ft
condition BC SA RA CC wet
new tires
11 mph
20
30
40
5060
70
.74
.76
.79
.75
.75
.75
.79
.75
.78
.76
.74
.74
.76
.73
.78
.76
.40
.36
.33
.31
.30
.29
20
30
40
50
60
70
111.9
196.7
300.6
423.8
566.0
727.6
115
200
305
425
570
730
28
36
44
51
57
64
710
1090
1470
1835
2135
24801 t p is also known as the PIEV time. This name is an acronym for the various
elements of reaction time, including perception, identification, emotion, andvolition
badly worn tires
11 mph20
30
40
50
60
70
.61
.60
.57
.48
.66
.57
.48
.39
.73
.65
.59
.50
.68
.50
.47
.33
.40
.36
.33
.31
.30
.29
2 For the purpose of determining minimum stopping sight distances, t p is takenas 2.5 seconds. For determining passing sight distances, t p is taken as 3.5 to 4.5seconds3 It is assumed that the driver’s eyes are 3.5 feet above the surface of the
roadway. The object being viewed (e.g., an oncoming car) is assumed to be at aheight of 3.5 feet.
Minimum PSD
The minimum distance sufficient to enable a driver traveling at design speed to pass (overtake)
slower vehicles. Passing sight distance is calculated only in case of two lanes of traffic moving
in opposite directions.
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4.2 Sight Distance on Horizontal Curves
For S < L
200
*cos1
DS
M R
−=
)(cos200 1
R
M R
DS
−= − , this is also the same equation in figure
Where;
L = Length of horizontal curve, ft.M = distance from centerline of inside lane to edge of obstruction, ftS = stopping distance along the center of the inside lane, ftR = radius of centerline curvature of inside lane, ftD = degree of curvature
Figure III-24(A), p. 220, AASHTO 1994.
Sight Distance (S)Highway
Inside Lane
Line of Sight
M
R
D
Sight
Obstruction
LC LC
V = 1 1 0 k m / h S = 2 4 6 .4 m
10
20
40
60
80
100
200
400
600
800
1000
2000
4000
6000
V = 1 2 0 k m / h S = 2 8 5 .6 m V = 1 0 0 k m / h S = 2 0 5 .0 m
V = 9 0 k m / h S = 1 6 8 .7 m
V = 8 0 k m / h S = 1 3 9 .4 m
V = 7 0 k m / h S = 1 1 0 .8 m
V = 6 0 k m
/ h S = 8 4 . 5 m
V = 5 0 k m / h S = 6 2 . 8 m
V = 4 0 k m
/ h S = 4 4 . 4 m
V = 3 0 0 k
m / h S = 2 9 . 6 m
( M e a s u r e d a l o n g C .L . I n s i d e L a n e ) R a d i u s ,
R ,
C e n t e r l i n e o f I n s i d e L a n e
( m )
Line of sight
S i g h t Dis tance ( S )
H i g h w a y
I n s i d e L
ane
Sightobstruction
Min. R when
a = 10%
Min. R when
a = 10%
28.65 SM =R 1- cos
Rwhere
S=Stopping Sight
Distance
M=Middle OrdinateR=Radius
R
M
Min. R when
a = 10%
18161412108642 Middle Ordinate, MCenterline of Inside Lane to Sight Obstruction (M) in meters
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Derivation of Horizontal Sight Distance Formula
M = )]2
cos(1[R )2
cos(*R R oo θ−=θ−
S =180
R o π
θ
θ o =R *
S*180
π=
R
S*30.57
get M = )]
R
S65.28cos(1[R −
S
θθ
θ
Example 14:
The corner of a building is situated next to a horizontal curve with a radius of 132 ft on a rural
highway. The inside lane is 10 ft wide and the inside edge of the road is 6 ft from the corner of a
building. Determine what speed limit should be imposed on this section of the highway.
Assume shoulder width = 4 feet, t =2.5 sec, f = 0.40 (wet conditions), g = 0 (level grade).
Solution to Example 14:
R = 132 –10
2
= 127 ft
M = 6 + 4 + 5 = 15 ft
S = 1127 127 15cos 125 ft
28.65 127
− − =
Note:S is calculated using the same equation shown inside AASHTO Figure III-24(A) on page
26.
S = 125 ft = 1.47 V t +)gf (30
V 2
±
125 = 1.47 V (2.5) +)4.0(30
V2
solve for V = 22.5 mph speed limit should be 22.5 mph (~ 20 mph)
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Example 15:
A 100 ft right-of-way (R/W) highway with 1250 ft proposed minimum centerline curvature is
being located through the Redwoods. A particularly fine tree stands 80.1 ft from the P-line and
321.3 ft from the PI of tangents which intersect at 135o.1. What is the largest radius curve joining the tangents such that the tree will not encroach on
the right-of-way (R/W)?
2. What is the available stopping-sight distance if the roadway section consists of a 16 ft
median and two 12 ft lanes each way?
See figure below.
P-line
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Solution to Example 15:
1. O.PC.PI (big triangle)
tan 22.5 =R
3.321X +
X = 0.41421 R - 321.3 => (1)
2. O.C.T (small triangle)
OT2 = CT2 + OC2
(R-50)2 = X2 + (R-80.1)2
R 2 - 100R + 2500 = X2 + R 2 -160.2 R +6416.01
60.2 R - 3916.01 = X2
=> (2) substitute (1) into (2)
60.2 R - 3916.01 = (0.41421 R - 321.3) 2
R = 1480.4 ft
--Note: This is not the R that we use in sight distance equation!
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3. Cross section at the tree location
M = 50 - (8+12+6) = 24 ft
28.65M 1 cos
S R
R
= −
R to be used in sight distance formula = 1480.4 - (8+12+6) = 1480.4 - 26 = 1454.4 ft
Note that this is the R that we need to use in the sight distance equation.
28.6524 1454.4 1 cos
1454.4
S = −
S = 529.16 ft
S ~ 161.3 meters
AASHTO Exhibit 3-1 or Table III-3, or Figure III-24(A) on page 48 (after conversion to
proper units). Enter figure with M and R in metric units and with interpolation
See where the red arrows meet on this figure.
S= 161.3 m ⇒ Speed = 90 kph ~ 56.25 mph.
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4.3
Sight Distance on Vertical Curves
Design of Vertical Curves Based on Sight Distance
1. CREST:
2. SAG:
Design of Vertical Curves
Crest Sag
S < L
( )2
21
2
hh200
SAL
+=
( )β+=
tanSH200
SAL
2
S > L ( )A
hh200S2L
2
21 +−=
L = 2S- ( )
A
tanSH200 β⋅+
Where S = Sight distance, feet (SSD or PSD)
L = Length of vertical curve, feet
,GGA 12 −= in percent
=1h height of driver's eye, in feet
(used for both SSD and PSD)
=2h For SSD, height of object in feet
For PSD, height of oncoming vehicle
S = Sight distance, feet SSD
L = Length of vertical curve, feet
,GGA 12 −= in percent
H = headlight height.
β = beam angle.
Default
Values
=1h 3.5 ft
=2h 2 ft (SSD) or 3.5 ft (PSD)
H = 2 ft.
β = 1°
L
L
S < L
S > L
β
LS < L
β
LS > L
h2
h1
h1 h2
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Minimum length is that which meets all required criteria.
Table III-3 Stopping Sight Distance on Wet Pavements(AASHTO 2001, Exhibit 3-1, p. 112) –See Appendix A for Metric Units.
DesignSpeed(mph)
Brake ReactionDistance
(ft)
BrakingDistanceon Levela
Stopping Sight Distance
Calculated(ft)
Rounded forDesign
(ft)15202530354045505560
65707580
55.173.591.9
110.3128.6147.0165.4183.8202.1220.5
238.9257.3275.6294.0
21.638.460.086.4117.6153.6194.4240.0290.3345.5
405.5470.3539.9614.3
76.7111.9151.9196.7246.2300.6359.8423.8492.4566.0
644.4727.6815.5908.3
80115155200250305360425495570
645730820910
Note: Brake reaction distance predicted on a time of 2.5 s; deceleration rate of 3.4 m/s [11.2 ft/s ] used todetermine calculated sight distance.
Table III-4 Design Controls for Stopping Sight Distance and for Crest Vertical Curves(AASHTO 2001, Exhibit 3-76, p. 274)-- See Appendix A for Metric Units
Designspeed(mph)
Stopping SightDistance
(ft)
Rate of Vertical Curvature, K a
[length (ft) per percent of A]
Calculated Rounded
for Design
15202530354045505560657075
80
80115155200250305360425495570645730820
910
3.06.111.118.529.043.160.183.7
113.5150.6192.8246.9311.6
383.7
37
121929446184114151193247312
384
aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).
LK=
A.
Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Associationof State Highway and Transportation Officials, Washington, DC.
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Table III-5 Design Controls for Sag Vertical Curves(AASHTO 2001, Exhibit 3-79, p. 280) --See Appendix A for Metric Units
Designspeed(mph)
Stopping SightDistance
(ft)
Rate of Vertical Curvature, K a
[length (ft) per percent of A]
Calculated Rounded
for Design15202530354045505560657075
80
80115155200250305360425495570645730820
910
9.416.525.536.449.063.478.195.7
114.9135.7156.5180.3205.6
231.0
1017263649637896115136157180206
231aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).
LK=
A.
Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Association ofState Highway and Transportation Officials, Washington, DC.
Table III-6 Design Controls for Crest Vertical Curves Based on Passing Sight Distance(AASHTO 2001, Exhibit 3-77, p. 276) --See Appendix A for Metric Units
Design Speed
(mph)
Passing Sight
Distance
(ft)
Rate of Vertical Curvature, K ,a
Rounded for Design
[length (ft) per percent of A]
20253035404550556065
707580
710900
10901280147016251835198521352285
248025802680
1802894245857729431203140716281865
219723772565
aRate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A).
LK=
A.
Source : A Policy on Geometric Design of Highway and Streets, copyright 2001 by the American Association of StateHighway and Transportation Officials, Washington, DC.
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Example 16
A car is travelling at 40 mph up a hill with a +1.25% grade. The descending grade is –2.75%.
What is the required length of curve for proper stopping sight distance?
Solution to Example 16:
The design speed (assumed travel speed) = 40 mph
See either Table III-3 or III-4 for SSD ⇒ SSD = 305 ft
Use table of crest vertical curves ⇒ K= 44 ft/percent of A
|A| = |-2.75-1.25| = 4%
Lmin = K|A| = (44) (4) = 176 ft
Check formula :
S = 305 > L = 176
L computed =
2
'200 3.5 2
2(305) 704
+ − = (305 > 70) Initial Assumption is O.K.
Recommended value for Lmin is 176 ft.
Note:
Sight distance tables in AASHTO meet additional criteria in addition to SSD and PSD minimum
length requirement. These tables meet the comfort and drainage criteria, so the length rounded
from these tables can be longer than what the equation calculations show. Unless you are asked
to use the equations for sight distance on vertical curves, it is recommended to use these tables
instead of the equations.
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APPENDIX –A-
METRIC TABLES FOR SIGHT DISTANCE ON VERTICAL CURVES
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Table 3: Design Controls for Sag Vertical Curves
Design
Speed
(km/hr)
Assumed
Speed for
Condition
(km/hr)
Coefficient of
Friction, f
Stopping
Sight Distance for
Design (m)
Rate of Vertical Curvature, K
[length (m) per percent of A]
Computed Rounded for
Design
30 30-30 0.40 29.6-29.6 3.88-3.88 4-4
40 40-40 0.38 44.4-44.4 7.11-7.11 8-8
50 47-50 0.35 57.4-62.8 10.20-11.54 11-12
60 55-60 0.33 74.3-84.6 14.45-17.12 15-18
70 63-70 0.31 94.1-110.8 19.62-24.08 20-25
80 70-80 0.30 112.8-139.4 24.62-31.86 25-32
90 77-90 0.30 131.2-168.7 29.62-39.95 30-40100 85-100 0.29 157.0-205.0 36.71-50.06 37-51
110 91-110 0.28 179.5-246.4 42.95-61.68 43-62
120 98-120 0.28 202.9-285.6 49.47-72.72 50-73
Table 4: Design Controls for Crest Vertical Curves Based on Passing Sight Distance
Design Speed(km/hr)
Minimum Passing
Sight Distancefor Design
(m)
Rate of Vertical Curvature, K ,a
Rounded for Design
[length (m) per percent of A]
30 217 50
40 285 90
50 345 130
60 407 180
70 482 250
80 541 310
90 605 390
100 670 480