7.1 Introduction
CHAPTER 7
Transportation Route
Alignments
7.1 Introduction
The famous “Silk Road” has been one of the best known ancient trade routes. This route
consisted of many sections, links, and alternates. The “Silk Road” was built as a caravan route
connecting China and Western Europe. The route was more than 6,000 kilometers long. First
Egyptians, then Romans, and other westerners established trade relationships with China. Far East
products (primarily silk, paper, and spices) were transported along this route for many centuries.
Caravan “rest areas” were built at distances 30-40 km. from one another. In this way, it was possible
to travel between two caravan stations in eight to ten hours on foot.
The “Silk Road”, as well as many other ancient and modern roads have ben highly influenced
by terrain, and they followed the most convenient path, i.e. the “path of least resistance”. Clearly, the
ideal alignment would be a great circle from the point of origin to the point of destination. The most
convenient path is usually that one that follows the natural alignment of the countryside. A route
alignment represents a defined three-dimensional path. It is usual and convenient to present route
alignments in two dimensions: horizontal and vertical alignments. Vertical alignment (also called
profile grade line) is composed of a sequence of straight line profiles connected by vertical parabolic
curves. In some cases the profile grade increases from a flat alignment. In some other cases, profile
grade decreases from a flat alignment. A “plus grade” denotes the situation when profile grade is
increasing from a flat alignment (Figure 7.1). The opposite situation is known as a “minus grade”
(Figure 7.1).
7-1
7.1 Introduction
Tangent
Tangent
+ G1
-G
Curve
Curve
2
Figure 7.1 Definition of Vertical Profiles (Plus Grade G 1 and Minus Grade G 2 ).
Grades are expressed in [%] or in [m/m]. Horizontal alignment represents projection of the
alignment onto xy plane (Figure 7.2).
z
L
y
(0,0,0)
Horizontal alignment
x
L xy
Figure 7.2 Horizontal Alignment.
The position of a specific point on a highway is traditionally determined using concept of
stations. A datum point on a highway alignment is specified. This initial point is designated station 0
+ 000.000. The positions of all other points on the highway are calculated by measuring
corresponding distances on a horizontal plane along the highway from the initial point. Each station
contains one kilometer of highway alignment distance. For example, the point on a highway located
2345.6 m from the previously specified point, is designated station 2 + 345.600. The station 3 +
465.800 corresponds to the highway point located 3465.8 meters from the initial point.
7-2
7.2 Vertical Alignment
Let us create along horizontal alignment the surface orthogonal to the xy plane. Let us also
denote h the distance measured along the horizontal alignment L xy . When we stretch this surface, it
becomes flat. We call it hz plane. The vertical alignment L hz represents the projection of the
alignment L onto hz plane (Figure 7.3).
z
L hz
Vertical alignment
h
Figure 7.3 Vertical Alignment L hz .
The specification of a route alignment depends on demand for transportation services
expected between two points along the future route and should consider construction, operating and
maintenance cost as well as safety.
7.2 Vertical Alignment
Vertical alignment is composed of straight sections that are connected by vertical curves.
These straight sections are called grades or tangents. Once a vertical alignment is designed, the
elevations of all the points along the highway are established. Vertical curve are typically classified as
“crest curves” or “sag curves” (Figure 7.4).
The most important points on the vertical curve are: the Vertical Point of Curvature VPC , the
Vertical Point of Intersection VPI , and the Vertical Point of Tangency VPT (Figure 7.5). The VPC is
the initial point of the curve, while the Vertical Point of Tangency VPT is the final point of the
vertical curve. The vertical curve length L is indicated in the Figure 7.5. The grades of the initial and
final tangent segments are denoted by G 1 , and G 2 , respectively. The point VPI represents the point
of intersection of the initial and final grade.
In the most cases, the first half of the vertical curve length is located before the VPI , and the
second half after the VPI (Figure 7.5). Such curves are called equal tangent vertical curves.
7-3
7.2 Vertical Alignment
+G 2
Crest
Crest
-G 1
+G 1
Crest
- G2
+G 1
-G 2
+G 2
Sag
Sag
+G 2
+G 1
-G 1
-G 1
Sag
-G 2
Figure 7.4 Types of Vertical Curves (Crest and Sag Vertical Curves).
VPI
x
PVT
VPC
+G 1
parabolic curve
L/2
L/2
-G 2
L
Figure 7.5 Elements of a Vertical Curve.
7-4
7.2 Vertical Alignment
Profile grade tangents are connected by the parabolic curve. the basic definition of a parabola
is,
2
y = a!x +b!x+c
(7.1)
where:
x - horizontal distance to point on curve measured from the VPC
y - elevation of point on curve located at distance x from VPC
By the definition x = 0 in the point of the vertical curve VPC . This means that the elevation
in VPC is:
2
y = a!0 +b!0+c = c
We see that the coefficient c represents elevation in PVC . Let us determine coefficients a
and b . The first derivative of Equation (7.1) is:
dy
----- = 2 ! a ! x + b
dx
(7.2)
At the initial point VPC , the slope equals G 1 , and x = 0 , i.e.:
dy
----- = 2 ! a ! x + b = 2 ! a ! 0 + b = b
dx
b = G1
(7.3)
We conclude that the coefficient b equals to the initial slope G 1 . The second derivative
equals:
2
dy
-------- = 2 ! a
2
dx
(7.4)
The second derivative of Equation 7.1 represents the rate of slope change, i.e.:
2
G2 – G1
dy
-------- = -----------------2
L
dx
(7.5)
7-5
7.2 Vertical Alignment
This implies,
G2 – G1
2 ! a = -----------------L
(7.6)
We calculate the coefficient a as:
G2 – G1
a = -----------------2!L
(7.7)
The grades in Equation 7.7 are expressed in [m/m]. Let A the difference in grades
( A = G 2 – G 1 ). The difference in grades A is positive for sag curves and negative for crest curves
(Equation 7.5). Table 7.1 shows the maximum recommended grades for urban and rural freeways.
Table 7.1 Maximum Grades [%] for Urban and Rural Freeways (Source AASHTO, 1990)
Design
speed
[m.p.h.]
Design
speed
[m.p.h.]
Design
speed
[m.p.h.]
Type of terrain
50
60
70
Level
4
3
3
Rolling
5
4
4
Mountainous
6
6
5
Example 7.1
The length of the equal tangent vertical curve (Table 7.6 ) equals 300 [m]. The initial and the
the final grades are known to be:
G 1 = 2.5 [%] ; G 2 = – 1.5 [%]
The grades intersect at the station 3 + 650 and at an elevation of 210.500 m. (a) Determine the
station and the elevation of the VPC and VPT . (b) Calculate the elevation of the point on the curve
100 meters from the VPC .(c) Determine the station and the elevation of the highest point on the
curve.
7-6
7.2 Vertical Alignment
VPI
x
VPT
VPC
g 1 = 2.5 [%]
L/2
L/2
g 2 = – 1.5 [%]
L = 300 [ m ] ]
Figure 7.6 Vertical Curve for Problem 1.
Solution:
(a)
The grades of the initial G 1 and final tangent G 2 are:
m
2.5
G 1 = --------- = 0.025 ---m
100
m
1.5
G 2 = – --------- = – 0.015 ---m
100
The station of the VPC and VPT . are:
L
300
VPCSta = VPISta – --- = 3 + 650 – --------- = 3 + 650 – 150 = 3 + 500
2
2
VPTSta = VPCSta + L = 3 + 500 + 300 = 3 + 800
The elevation of the VPC and VPT . are:
L
300
E VPC = E VPI – $ G 1 ! ---% = 210.500 – $ 0.025 ! ---------% = 210.500 – 3.75 = 206.75 [ m ]
"
#
"
2
2 #
7-7
7.2 Vertical Alignment
300
L
E VPT = E VPI – $ G 2 ! ---% = 210.500 – $ 0.015 ! ---------% = 210.500 – 2.25 = 208.25 [ m ]
"
"
#
2 #
2
(b)
The elevation y of any point on the curve located at distance x from VPC is described by the
parabola:
2
y = a!x +b!x+c
(7.8)
After substituting the known values for the coefficients, we get:
G2 – G1 2
- ! x + G 1 ! x + E VPC
y = -----------------2!L
(7.9)
– 0.015 – 0.025 2
y = ------------------------------------- ! x + 0.025 ! x + 206.75
2 ! 300
2
y = – 0.000067 ! x + 0.025 ! x + 206.75
For x = 100 we get:
2
y = – 0.000067 ! 100 + 0.025 ! 100 + 206.75 = 208.58
The elevation of the point equals 208.58 [m].
(c)
Grades are opposite in sign. This means that the highest point can be estimated by setting the
first derivative the equation of the parabola to zero, i.e.:
dy
----- = 2 ! a ! x + b = 0
dx
2 ! ( – 0.000067 ) ! x + 0.025 = 0
x = 185.57 [ m ]
The station of the highest point is:
7-8
7.2 Vertical Alignment
HighSta = VPCSta + 185.57 = 3 + 500 + 185.57 = 3 + 685.57
The elevation of the highest point is:
2
y = – 0.000067 ! x + 0.025 ! x + 206.75
For x = 185.57 we get:
2
y = – 0.000067 ! 185.57 + 0.025 ! 185.57 + 206.75 = 209.08 [ m ]
Example 7.2
Offset Y of curve from initial grade line (Figure 7.7) is especially important parameter in
vertical curve design. We consider again the curve from the previous example.
Offsets
VPI
x
Y
VPT
VPC
G 1 = 2.5 [%]
L/2
L/2
G 2 = – 1.5 [%]
L = 300 [ m ] ]
Figure 7.7 Vertical Curve for Problem 7.2.
The length of the equal tangent vertical curve (Figure 7.7) is 300 [m]. The initial and the final
grades are:
G 1 = 2.5 [%] ; G 2 = – 1.5 [%]
The grades intersect at station 3 + 650 and at an elevation of 210.500 m. (a) Determine the offset at
distance 100 [m] from the VPC . (b) Determine the offset at the end of vertical curve.
Solution:
7-9
7.2 Vertical Alignment
The coefficient c represents elevation in VPC (Figure 7.8). The equation of the initial tangent
is:
y = k!x+c
(7.10)
where k is the coefficient to be determined.
y
VPI
x
Y
VPT
VPC
c
x
Figure 7.8 Initial Tangent and Parabola.
The equation of the parabola is:
2
y = a!x +b!x+c
(7.11)
The first derivative is:
dy
----- = 2 ! a ! x + b
dx
(7.12)
The coordinates of the VPC are ( 0, c ) . The first derivative in this point is calculated to be:
2!a!0+b = b
We conclude that:
k = b
(7.13)
7-10
7.2 Vertical Alignment
The equation of the initial tangent is:
y = b!x+c
(7.14)
The offset Y of curve from initial grade line equals:
2
Y = (b ! x + c) – (a ! x + b ! x + c)
Y = –a ! x
2
G1 – G2 2
G2 – G1 2
- ! x = ------------------!x
Y = – -----------------2!L
2!L
(7.15)
(7.16)
(7.17)
m
where G 1 and G 2 are expressed in ---- .
m
When calculating offsets, grades are expressed in [%]. Equation 7.17 becomes:
G1 – G2 2
-!x
Y = -----------------200 ! L
(7.18)
The general offset formula for both sag and crest vertical curves are:
G1 – G2 2
-!x
Y = --------------------200 ! L
(7.19)
The offset at a distance 100 [m] from the VPC is:
2
2.5 – ( – 1.5 )
Y = -------------------------------- ! 100 = 0.67 [ m ]
200 ! 300
The offset at the end of vertical curve is calculated to be:
2
2.5 – ( – 1.5 )
Y ( 300 ) = -------------------------------- ! 300 = 6 [ m ]
200 ! 300
7-11
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
7.3
Stopping Sight Distance Considerations in Vertical Curve Design
Vertical curve design requires consideration of the average stopping sight distances. Drivers
should have a clear view of the road ahead to stop before and obstacle while driving on a vertical
curve. This visible length of a highway ahead is called sight distance. Let us assume for this
discussion that the vehicle speed is the design speed. The minimum stopping sight distance SSD is
the sum of two components: (a) the distance vehicle traveled from the time point the object is
observed to point in time when brakes are applied; and (b) the distance traveled by vehicle from the
point in time when the brakes are applied until the vehicle is fully stopped.
The American Association of State Highway and Transportation Officials (AASHTO) has
recommended the minimum sight distances. These distances are shown in Table 7.2.
Table 7.2 Stopping Sight Distances on Wet Pavements (Source AASHTO, 1994).
Coefficient
of friction
Braking
Distance
on Level
[m]
Computed
[m]
Rounded
for
Design
[m]
20.8-20.8
0.40
8.8-8.8
29.6-29.6
30-30
2.5
27.8-27.8
0.38
16.6-16.6
44.4-44.4
50-50
47-50
2.5
32.6-34.7
0.35
24.8-28.1
57.4-62.8
60-70
60
55-60
2.5
38.2-41.7
0.33
36.1-42.9
74.3-84.6
80-90
70
63-70
2.5
43.7-48.6
0.31
50.4-62.2
94.1-110.8
110-120
80
70-80
2.5
48.6-55.5
0.30
64.2-83.9
112.8-139.4
120-140
90
77-90
2.5
53.5-62.5
0.30
77.7-106.2
131.2-168.7
140-170
100
85-100
2.5
59.0-69.4
0.29
98.0-135.6
157.0-205.0
160-210
110
91-110
2.5
63.2-76.4
0.28
116.3-170.0
179.5-246.4
180-250
120
98-120
2.5
68.0-83.3
0.28
134.9-202.3
202.9-285.6
210-290
Design
Speed
[km/h]
Assumed
Speed for
Condition
[km/h]
Brake
Reaction
Time
[s]
Brake
Reaction
Distance
[m]
30
30-30
2.5
40
40-40
50
The distances shown in the table are calculated under the assumption of a “worse case
scenario”. For example, it is assumed that the driver’s perception/reaction time is 2.5 seconds. It is
also assumed that the brakes are applied in a situation when the pavement is wet.
7-12
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
7.4.1 Crest Vertical Curve Design
Let us note the crest vertical curve shown in Equation 7.9. Let S the available sight distance.
The length of the crest vertical curve is denoted by L . The height of the eye of the average driver is
denoted by h 1 . The h 2 denotes the height of the object perceived by the driver. The formulae for the
crest vertical curve length L are presented in Table 3. The formulae are derived using mathematical
properties of the parabola made up by two tangent segments.
S
VPI
h1
VPC
L
h2
PVT
Figure 7.9 Crest Vertical Curve: Stopping Sight Distance Analysis.
Table 7.3 Formulas to Estimate the Length of a Crest Vertical Curve.
Relation
between S
and L
S&L
S>L
Length of the Vertical Curve L
2
A !S
L = ---------------------------------------------2
200 ! ( h 1 + h 2 )
2
200 ! ( h 1 + h 2 )
L = 2 ! S – --------------------------------------------A
Older AASHTO design guidelines assumed h 1 = 1.08 [ m ] and h 2 = 0.15 [ m ] . Newer
design guidelines have modified the value of h2 to be 0.60 meters without apparent degradation in the
safety of the design (AASHTO 2004). This new design rationale produces more cost-effective
vertical designs without safety degradation.
7-13
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
In order to calculate the minimum curve length L min we assume that the available sight
distance S is the same as the stopping sight distance SSD . The minimum curve length equals:
2
2
A ! SSD
A ! SSD
L min = -------------------------------------------------------- = -----------------------2
404
200 ! ( 1.08 + 0.15 )
404
L min = 2 ! SSD – --------A
SSD < L
SSD > L
(7.20)
(7.21)
If the new AASHTO design criteria is used instead (with h2 = 0.60 meters), the resulting
equations are:
2
2
A ! SSD
A ! SSD
L min = -------------------------------------------------------- = -----------------------2
658
200 ! ( 1.08 + 0.60 )
658
L min = 2 ! SSD – --------A
SSD < L
SSD > L
Stopping sight distances are shown in column 4 of Table 7.4. For example, for a design speed
of 120 [km/h], the stopping sight distances range from[210-290] meters. The lower value (210
meters) is called minimum SSD. The higher value (290 meters) is known as desirable SSD.
The usual assumption made by the highway designers is that SSD < L . In this book we
exclusively use the following formula to calculate L min :
2
A ! SSD
L min = -----------------------404
(7.22)
Let K be the following expression:
2
SSD
K = ------------404
(7.23)
7-14
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
The quantity K is called the rate of vertical curvature. The K values are given in the Table 7.4.
Table 7.4 Design Controls for Crest Vertical Curves Based on Stopping Sight Distance (Source
AASHTO, 1994).
Rate of
Vertical
Curvature
K
[length[m]
per % of A]
Computed
Rate of
Vertical
Curvature
K
[length[m]
per % of A]
Rounded
for Design
Design
Speed
[km/h]
Assumed
Speed for
Condition
[km/h]
Coefficient
of friction
Stopping
Sight
Distance
Rounded
for
Design
[m]
30
30-30
0.40
30-30
2.17-2.17
3-3
40
40-40
0.38
50-50
4.88-4.88
5-5
50
47-50
0.35
60-70
8.16-9.76
9-10
60
55-60
0.33
80-90
13.66-17.72
14-18
70
63-70
0.31
110-120
21.92-30.39
22-31
80
70-80
0.30
120-140
31.49-48.10
32-49
90
77-90
0.30
140-170
42.61-70.44
43-71
100
85-100
0.29
160-210
61.01-104.02
62-105
110
91-110
0.28
180-250
79.75-150.28
80-151
120
98-120
0.28
210-290
101.90-201.90
102-202
The length L min can be calculated using the following equation:
L min = K ! A
(7.24)
The rate of vertical curvature K equals:
L min
K = ---------A
(7.25)
The units for L min are meters, while the units for A are % . This means that the units of K
are [length[m] per % of A].
7-15
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
Example 7.3
The initial and the final grade of an equal tangent vertical curve are:
G 1 = 2.5 [%] ; G 2 = – 1.5 [%]
The highway designers are considering design speeds ranging from 100 [km/h] to 120 [km/h].
Calculate corresponding minimum length of the vertical curve that satisfies the minimum stopping
sight distance.
Solution:
The grades are:
G 1 = 2.5 [%]
G 2 = – 1.5 [%]
The difference in grades A is:
A = G 2 – G 1 = – 1.5 – 2.5 = – 4
A = –4 = 4
The L min is:
L min = K ! A = 4 ! K
The minimum curve length that satisfy minimum stopping sight distance are:
For a design speed of 100 [km/h]:
L min = 4 ! K = 4 ! 62 = 248 [ m ]
For a design speed of 110 [km/h]:
L min = 4 ! K = 4 ! 80 = 320 [ m ]
7-16
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
For a design speed of 120 [km/h]:
L min = 4 ! K = 4 ! 102 = 408 [ m ]
In some situations there is also a need to design vertical curves allowing a vehicle to overtake
another one. Passing sight distance is especially important in the case of two-lane highways.
The classical AASHTO design guidelines assumed h 1 = 1.08 [ m ] and h 2 = 1.3 [ m ] (The
height of the vehicle is assumed to be 1.3 meters). Recent trends in vehicle technology has dictated a
reduction in the object height used for passing sight distance (PSD). The new AASHTO standard
recommends h 2 = 1.08 [ m ] instead of the original 1.3 meters recognizing that smaller vehicles
populate the market today.
In order to calculate the minimum curve length L min we assume that the available sight
distance S equals the passing sight distance PSD . The minimum curve length is then:
L min
2
2
A ! PSD
A ! PSD
= ----------------------------------------------------- = ------------------------2
946
200 ! ( 1.08 + 1.3 )
946
L min = 2 ! PSD – --------A
PSD < L
PSD > L
(7.26)
(7.27)
If new AASHTO standards are used the following expressions are derived:
L min
2
2
A ! PSD
A ! PSD
= -------------------------------------------------------- = ------------------------2
864
200 ! ( 1.08 + 1.08 )
864
L min = 2 ! PSD – --------A
PSD < L
PSD > L
Passing sight distances recommended by the AASHTO are shown in the Table 7.5.
7-17
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
Table 7.5 Design Controls for Crest Vertical Curves Based on Passing Sight Distance.
Design Speed
[km/h]
Minimum Passing
Sight Distance for
Design
[m]
Rate of Vertical Curvature
K
[length[m] per % of A]
Rounded for Design
30
217
50
40
285
90
50
345
130
60
407
180
70
482
250
80
541
310
90
605
390
100
670
480
110
728
570
120
792
670
7.5.2 Sag Vertical Curve Design
Sight distance on a sag vertical curve is not limited during a day.
S
H
'
VPC
PVT
VPI
L
Figure 7.10 Sag Vertical Curve: Stopping Sight Distance Analysis
7-18
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
Driving conditions at night are the key factor that influences sag vertical curve design. The
highway length lighted by the vehicle headlights depends on headlight height above the highway H ,
and on the angle ' between the headlight beam and the car horizontal plane (Equation 7.10).
It has been shown that the minimum curve length could be calculated using the relations
shown in the Table 7.6 9 (AASHTO, 2004).
Table 7.6 Length of the Sag Vertical Curves.
Relation
between S
and L
S&L
S>L
Length of the Vertical Curve L
2
A !S
L = ---------------------------------------------200 ! ( h + S ! tan ' )
200 ! ( h + S ! tan ' )
L = 2 ! S – ---------------------------------------------A
AASHTO design guidelines assumes h = 0.6 [ m ] and
. In order to calculate
' = 1[°]
the minimum curve length L min we assume that the available sight distance S equals the stopping
sight distance SSD . The minimum curve length equals:
2
2
2
A ! SSD
A ! SSD
A ! SSD
L = ---------------------------------------------- = ------------------------------------------------------------- = ----------------------------------------200 ! ( h + S ! tan ' )
200 ! ( 0.6 + SSD ! tan 1° )
120 + 3.49 ! SSD
SSD & L
(7.28)
200 ! ( h + SSD ! tan ' )
120 + 3.49 ! SSD
L = 2 ! SSD – ------------------------------------------------------ = 2 ! SSD – ----------------------------------------A
A
SSD > L
(7.29)
The usual assumption made by the highway designers is that SSD < L . In this book we use
the following formula to calculate L min :
7-19
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
L min
2
A ! SSD
= ----------------------------------------120 + 3.49 ! SSD
(7.30)
Let K be:
2
SSD
K = ----------------------------------------120 + 3.49 ! SSD
(7.31)
The K values are given in the Table 7.7.
Table 7.7 Design Controls for Sag Vertical Curves Based on Stopping Sight Distance (Source
AASHTO, 1994).
Stopping
Sight
Distance
Rounded
for
Design
[m]
Rate of
Vertical
Curvature
[length[m]
per % of A]
Computed
Rate of
Vertical
Curvature
[length[m]
per % of A]
Rounded
for Design
Design
Speed
[km/h]
Assumed
Speed for
Condition
[km/h]
Coefficient
of friction
Stopping
Sight
Distance for
Design
[m]
30
30-30
0.40
29.6-29.6
30-30
3.88-3.88
4-4
40
40-40
0.38
44.4-44.4
50-50
7.11-7.11
8-8
50
47-50
0.35
57.4-62.8
60-70
10.20-11.54
11-12
60
55-60
0.33
74.3-84.6
80-90
14.45-17.12
15-18
70
63-70
0.31
94.1-110.8
110-120
19.62-24.08
20-25
80
70-80
0.30
112.8-139.4
120-140
24.62-31.86
25-32
90
77-90
0.30
131.2-168.7
140-170
29.62-39.95
30-40
100
85-100
0.29
157.0-205.0
160-210
36.71-50.06
37-51
110
91-110
0.28
179.5-246.4
180-250
42.95-61.68
43-62
120
98-120
0.28
202.9-285.6
210-290
49.47-72.72
50-73
Example 7.4
The highway alignment should follow a crest vertical curve, then a constant grade section, and
finally a sag vertical curve (Figure 7.11). The starting and ending alignment grades are 0 [%]. The
elevation and the stationing at point A are:
7-20
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
E A = 225 [ m ] ; ASta = 3 + 120.000
The constant grade is G = – 1.5 [%]. The length of the constant grade section is planned to be
150 [m]. The design speed is 110 [km/h]. Determine the stationing and the elevations at points B, C,
and D in Figure 7.11.
A
B
Constant grade
section
Lc
L
C
D
Ls
Figure 7.11 Vertical Alignment Geometry for Problem 7.4.
Solution:
The grades are:
Constant grade:
G = – 1.5 [%]
Crest vertical curve (index c):
c
G1 = 0
c
G 2 = G = – 1.5 [%]
c
Ac = G2 – G1
c
= – 1.5 – 0 = 1.5 [%]
7-21
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
Sag vertical curve (index s):
s
G 1 = G = – 1.5 [%]
s
G2 = 0
s
As = G2 – G1
s
= 0 – ( – 1.5 ) = 1.5 [%]
L min is:
L min = K ! A
The rates of the vertical curvature are shown in Table 7.7 For a design speed of 110 [km/h] we
know:
c
s
K = 80 ; K = 43
The minimum curve length that satisfies the minimum stopping sight distances are:
c
c
s
s
L min = A c ! K = 1.5 ! 80 = 120 [ m ]
L min = A c ! K = 1.5 ! 43 = 64.5 [ m ]
In other words, the length of the vertical curves are:
L c = 120 [ m ]
L s = 64.5 [ m ]
The stationing at point B is :
BSta = ASta + L c = 3 + 120.000 + 120 = 3 + 240.000
The offset Y of curve from initial grade line is:
7-22
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
G1 – G2 2
-!x
Y = --------------------200 ! L
(7.32)
In the case of the crest vertical curve, the offset in the point B :
G 1c – G 2c
G 1c – G 2c
2
1.5
- ! L c = --------- ! 120 = 0.9 [ m ]
- ! L c = -------------------------Y B = -------------------------200 ! L c
200
200
The elevation E B at point B is:
E B = E A – Y B = 225 – 0.9 = 224.1 [ m ]
The stationing of point C is:
CSta = BSta + L = 3 + 240.000 + 150 = 3 + 390.000
The constant grade G :
EB – EC
------------------ = G
L
(7.33)
224.1 – E C
m
-------------------------- = 0.015 ---m
150
The elevation of the point C:
E C = 221.85 [ m ]
The stationing at point D is:
DSta = CSta + L s = 3 + 390.000 + 64.5 = 3 + 454.500
In the case of the sag vertical curve, the offset in the point C:
G 1s – G 2s
G 1s – G 2s
2
1.5
- ! L s = --------- ! 64.5 = 0.48375 [ m ]
- ! L s = ------------------------Y C = ------------------------200 ! L s
200
200
The elevation at point D:
7-23
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
E D = E C – Y C = 221.85 – 0.48375 = 221.36625 [ m ]
Example 7.5
The highway alignment should follow crest vertical curve, constant grade section, and finally
a sag vertical curve (Figure 7.12). The starting and ending alignment grades are 0 [%]. The elevation
and the stationing of the points A and D are known to be:
E A = 240 [ m ] ; ASta = 3 + 120.000
E D = 225 [ m ] ; DSta = 3 + 620.000
Points should be connected with crest and sag equal tangent vertical curves with a constant
grade section in between (Figure 7.12). The design speed equals 70 [km/h]. Determine the grade of
the constant section, and the lengths of the crest vertical curve, sag vertical curve, and the constant
grade section.
YB
A
B
Y
EA – ED
C
YC
Lc
L
Ls
D
Figure 7.12 Geometry for Problem 7.5.
Solution:
The grades are:
Constant grade:
G [%]
7-24
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
Crest vertical curve (index c):
c
G1 = 0
c
G 2 = G [%]
c
c
A c = G 2 – G 1 = G – 0 = G [%]
Sag vertical curve (index s):
s
G 1 = G [%]
s
G2 = 0
s
As = G2 – G1
s
= 0 – ( G ) = G [%]
The difference in the elevation E A – E D between points A and D (Figure 7.12) are:
YB + Y + YC = EA – ED
(7.34)
Since:
Y
--- = G [%]
L
i.e.:
G m
Y
--- = --------- ---100 m
L
and:
Ac
Ac
2
G
- ! L c = -------- ! L = --------- ! L c
Y B = -----------------200
200 ! L c
200 c
(7.35)
As
As
2
G
- ! L s = --------- ! L s
Y C = -----------------! L s = -------200
200 ! L s
200
(7.36)
we get:
7-25
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
G
G
G ! L ---------------- ! L c + -------------- + - ! Ls = EA – ED
200
200
100
(7.37)
Since:
L c = K c ! A c and L s = K s ! A s , we have:
G
G
G ! L ---------------- ! K c ! A c + -------------- + - ! Ks ! As = EA – ED
200
200
100
(7.38)
G
G
G ! L ---------------- ! K c ! G + -------------- + - ! Ks ! G = EA – ED
200
200
100
(7.39)
The rates of the vertical curvature (Tables 7.6 and 7.7) for a design speed of 100[km/h] are:
K c = 22 ; K s = 20
Our equation reads:
2
2
22
20
G ! L ---------------- ! G + -------------- + - ! G = 240 – 225
200
200
100
2
22 ! G + 2 ! G ! L + 20 ! G
2
= 3000
2
42 ! G + 2 ! G ! L – 3000 = 0
The total length of the curve is:
L c + L + L s = DSta – ASta
L = ( DSta – ASta ) – L c – L s = ( DSta – ASta ) – ( K c ! G ) – ( K s ! G )
L = [ ( 3 + 620.000 ) – ( 3 + 120.000 ) ] – ( 22 ! G ) – ( 20 ! G )
L = 500 – 42 ! G
Further substitution yields:
7-26
7.3 Stopping Sight Distance Considerations in Vertical Curve Design
2
42 ! G + 2 ! G ! ( 500 – 42 ! G ) – 3000 = 0
A quadratic equation is:
2
– 42 ! G + 1000 ! G – 3000 = 0
The solutions are:
G 1 = 3.52 and G 2 = 20.29
We must reject the solution G 2 = 20.29 . If we accepted this solution the length of the
curves would be:
L c = K c ! G = 22 ! 20.29 = 446.38 [ m ]
L s = K s ! G = 20 ! 20.29 = 405.8 [ m ]
The total length of the vertical curves would be greater than the difference in stationing
between points D and A, i.e.:
L c + L s = 446.38 + 405.8 = 852.18 > 500 = DSta – ASta
The final solution is:
G = G 1 = 3.52 [%]
The length of the curves are:
L c = K c ! G = 22 ! 3.52 = 77.44 [ m ]
L s = K s ! G = 20 ! 3.52 = 70.4 [ m ]
L = 500 – 77.44 – 70.4 = 352.16 [ m ]
7-27
7.6 Horizontal Alignment and Horizontal Curves
7.6 Horizontal Alignment and Horizontal Curves
Consider the simple horizontal curve shown in Figure 7.13. We denote PC as the point of the
curve. This is the point where the horizontal curve starts. The curve ends at point PT that is called
point of tangency. The central angle of the curve ( is equal to the deflection angle between the curve
tangents (Figure 7.13). Point PI is called point of tangent intersection. Let R the radius of the
horizontal curve, L the length of curve (from PC to PT ), T tangent length, E the external distance,
and M the middle ordinate.
PI
T
E
M
PC
(
--2
Arc Length L
(
R
(
--2
PT
(
--2
R
Figure 7.13 Simple horizontal Alignment Circular Curve.
The following circular curve equations are obtained (Figure 7.13):
T
(
--- = tan --R
2
(7.40)
i.e., the tangent length is calculated to be:
(
T = R ! tan --2
(7.41)
The length of curve L is:
7-28
7.6 Horizontal Alignment and Horizontal Curves
2!R!)!(
R!)!(
L = --------------------------- = ------------------360
180
(7.42)
The quantities E and M can be calculated as:
$
%
R
1
*
E = ------------ – R = R ! ------------ – 1+
*
+
(
(
cos --" cos --#
2
2
(7.43)
(
(
M = R – R ! cos --- = R ! $ 1 – cos ---%
"
2
2#
(7.44)
Example 7.6
The stationing of PC is 5 + 210.000 . The curve radius is 650 [m]. The central angle is 30 [ ° ] .
Determine the stationing of PI and PT .
PI
Arc Length L
T
PC
R = 650
PT
(
--- = 15
2
Figure 7.14 Horizontal curve with the radius R = 650 and the central angle ( = 30
Solution:
The tangent length is:
7-29
7.6 Horizontal Alignment and Horizontal Curves
(
30
T = R ! tan --- = 650 ! tan ------ = 174.075
2
2
The stationing of PI is:
PISta = PCSta + T = 5 + 210.000 + 0 + 174.075 = 5 + 384.075
The length of curve L is:
650 ! 3.14
650 ! ) ! 30
R!)!(
L = ------------------- = --------------------------- = ------------------------ = 340.167
6
180
180
The stationing of PT is:
PTSta = PCSta + L = 5 + 210.000 + 0 + 340.167 = 5 + 550.167
Different forces act on the vehicle as vehicle travels around a horizontal curve of constant
radius. Basic analysis of the forces acting in a vehicle negotiating a horizontal curve shows the
following relationship:
2
v
R = ----------------------g ! (e + f)
(7.45)
where:
R - curve radius
v - vehicle speed [m/s]
g - gravitational constant
e - rate of superelevation (the vertical rise - in meters - for every 100 meters of horizontal distance)
f - coeeficient of side friction
AASHTO recommendations for minimum radii for limiting values of e and f are given in the
Table 7.8.
7-30
7.6 Horizontal Alignment and Horizontal Curves
Table 7.8 Minimum Radius for Limiting Values of e and f (Source AASHTO, 1994)
Design
Speed
[km/h]
Maximum
e
Maximum
f
Total
(e + f)
Calculated
Radius
[m]
Rounded
Radius
[m]
30
0.04
0.17
0.21
33.7
35
40
0.04
0.17
0.21
60.0
60
50
0.04
0.16
0.20
98.4
100
60
0.04
0.15
0.19
149.2
150
70
0.04
0.14
0.18
214.3
215
80
0.04
0.14
0.18
280.0
280
90
0.04
0.13
0.17
375.2
375
100
0.04
0.12
0.16
492.1
490
110
0.04
0.11
0.15
635.2
635
120
0.04
0.09
0.13
872.2
870
30
0.06
0.17
0.23
30.8
30
40
0.06
0.17
0.23
54.8
55
50
0.06
0.16
0.22
89.5
90
60
0.06
0.15
0.21
135.0
135
70
0.06
0.14
0.20
192.9
195
80
0.06
0.14
0.20
252.0
250
90
0.06
0.13
0.19
335.7
335
100
0.06
0.12
0.18
437.4
435
110
0.06
0.11
0.17
560.4
560
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
7-31
7.6 Horizontal Alignment and Horizontal Curves
Table 7.8 Minimum Radius for Limiting Values of e and f (Source AASHTO, 1994)
Design
Speed
[km/h]
Maximum
e
Maximum
f
Total
(e + f)
Calculated
Radius
[m]
Rounded
Radius
[m]
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
25
40
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
Example 7.7
Calculate minimum radius for a design speed of 120 [km/h]. The superelevation rate is limited to 0.06
due to icing effects in winter.
Solution:
The minimum radius formula is:
2
v
R = ----------------------g ! (e + f)
The speed expressed in [m/s] is:
7-32
7.6 Horizontal Alignment and Horizontal Curves
m
1000 [ m ]
km
120 ------- = 120 ! --------------------- = 33.33 ---s
3600 [ s ]
h
At 120 km/hr AASHTO recommends using a side friction coefficient of 0.09. This value is read from
Table 7.8. Therefore, the minimum radius is:
2
( 33.33 )
R = ----------------------------------------------- = 755.1 [ m ]
9.81 ! ( 0.06 + 0.09 )
Depending on a terrain, the length of horizontal curves need to be adjusted to allow drivers be
able to see ahead and potentially avoid obstacles. In the context of horizontal curves, sight distance
depends on the presence of obstruction (Figure 7.15). The stopping sight distance SSD is measured
along the horizontal curve.
SSD
Object
Ms
Eye
Highway centerline
(s
Line of sight
R
Rv
Figure 7.15 Horizontal Curve: Analysis of the Stopping Sight Distance and Middle Ordinate.
The radius of the vehicle traveled path is denoted by R v . Using formulae for the length of the
curve, we can express the stopping sight distance SSD as:
7-33
7.6 Horizontal Alignment and Horizontal Curves
Rv ! ) ! (s
SSD = ----------------------180
(7.46)
where ( s is the angle that corresponds to the arc that is equal to the SSD length (Figure 7.15).
(7.47)
This angle is:
180 ! SSD
( s = -----------------------Rv ! )
(7.48)
The value of the midle ordinate M s that can satisfy SSD is:
(
M s = R v ! $ 1 – cos ---%
"
2#
(7.49)
After substitution from equation (7.48) into equation (7.49), we get:
180 ! SSD%
$
-----------------------*
Rv ! ) +
(
M s = R v ! $ 1 – cos ---% = R v ! * 1 – cos -----------------------+
"
2
2#
*
+
"
#
(7.50)
90 ! SSD
M s = R v ! $ 1 – cos ---------------------%
"
Rv ! ) #
(7.51)
M
90 ! SSD
------s = 1 – cos $ ---------------------%
"
Rv ! ) #
Rv
(7.52)
M
90 ! SSD
cos $ ---------------------% = 1 – ------s
" R !) #
R
(7.53)
M
90 ! SSD
--------------------- = acos $ 1 – ------s%
"
Rv #
Rv ! )
(7.54)
v
v
7-34
7.6 Horizontal Alignment and Horizontal Curves
The stopping sight distance SSD is then,
Rv ! )
M
- ! acos $ 1 – ------s%
SSD = -----------"
90
Rv #
(7.55)
7-35
7.6 Horizontal Alignment and Horizontal Curves
7-36