DEPARTMENT OF CIVIL ENGINEERING
HIGHWAY ENGINEERING I (CENG 3102)
CHAPTER 3: GEOMETRIC DESIGN OF HIGHWAY ALIGNMENT
3.1 Horizontal Alignment
3.2 Vertical Alignment
3.3 Phasing of Alignment
3.4 Road Furniture Designs
By Haile G.
After completion of this chapter,
students will be able to

Analyze and design highways
geometrically in accordance with
ERA(2013) with particularly
emphasis on

Horizontal alignment designs

Vertical alignment designs
March 2023
Introduction
2
Geometric design of highways refers to the design of the visible
dimensions of such features as horizontal and vertical alignments,
cross sections, intersections, and bicycle and pedestrian facilities.
Horizontal Alignment Design
3

It consists of a series of intersecting tangents & circular
curves with or without transition curves

Should be designed to the highest standard consistent with
the topography

Be chosen carefully to provide good drainage & minimize
earthworks

Should be designed to achieve a uniform operating speed
Horizontal Alignment Design
4

Considerations during horizontal alignment design:







Horizontal and vertical alignment should compliment each other
Alignment should be straight if possible within physical and
economic constraints
Alignment should be consistent
Avoid the use of minimum radii whenever possible
Avoid horizontal curves on bridges
Avoid abrupt reversals in reverse curves
Avoid broken back curves
Horizontal Alignment Design
5

Horizontal Alignment Design consists of design parameters:



Straight lines-tangents:
 Length of tangent section, so L=?
 Aligned direction angle i.e. bearing angle or azimuth =?
Horizontal Curves:
 Simple circular curves, so R=?
 Compound Curves, so R1 and R2=?
 Reverse Curves, so R1 and R2=?
 Transitional Spirals so Length of spiral, Ls=?
Superelevation:
 So the rate of superelevation (e)=?,
 Lengths of superelevation developments i.e.,
 Length of tangent runout (Lt)=? and
 length of superelevation runoff (Lr)=?.
Horizontal Alignment Design
6
Tangent Section:





Straight portion of roads
Provide better visibility & better appearance
Limited minimum length of tangent b/n curves is greater
than or equal to 6V (in meter), V is in Km/hr
Long straight sections encourage drivers to drive at speeds
in excess of the design speed therefore the maximum
length of tangent should be less than or equal to 20V (in
meter), where V in km/r.
Thus a safer alternative is obtained by a winding alignment
with tangents deflecting 5° to 10 ° alternately to the right
and to left. This scenario is to be followed when the terrain
is flat.
Horizontal Alignment Design
7
Straights (Tangents)

Short straights between curves in the same direction should not be used
because of the broken back effect

Where a reasonable tangent length is not attainable, the use of long
transitions or compound curvature should be considered.
Centerline Stationing


A station is a longitudinal measurement method used to provide location
along highway and other longitudinal projects.
Stationing typically begins on the bigger city or town (from Center to
outer side of the country line).

1 STA = 20 meter (100 ft)

STA. 45+060 = 45,060 meter from station 0+000
Horizontal Alignment Design
8
Curves





 Used to provide access
 Used to prevent demolishing of important places
 Used to make transition from tangent to tangent
Long tangent roadway segments joined by an isolated curve designed at
or near the minimum radius are unsafe.
Long straight sections encourage drivers to drive at speeds in excess of
the design speed, hence sudden and unexpected sharp curves are
dangerous.
Good design practice is to avoid the use of minimum standards in such
conditions.
For isolated curves, the minimum horizontal curve radius as shown in
Tables 2.6 through 2.16 of ERA(2013) should be increased by 50 %.
For small changes of direction it is often desirable to use a large radius of
curvature.
Horizontal Alignment Design
9
Types of Horizontal Curves:-
1. Simple Circular Curve

When a curve consists of a with a continuing
radius connecting the 2 tangents, it's said to
be a circular curve
2. Compound Curve


When a curve consists of two or more arcs
with different radii, it's called a compound
curve.
Such a curve lies on an equivalent side of a
standard tangent and centers of the various
arcs lie on an equivalent side of their
respective tangents.
Horizontal Alignment Design
10
Types of Horizontal Curves:-
3. Reverse Curve



A reverse curve consists of two arcs bending
in opposite directions.
Their centers lie on converse sides of the
curve.
Their radii could also be either equal or
different, and that they have one common
tangent
4. Transition Curve


A curve of variable radius is knoun as a
transition curve. it's also called a spiral
curve or easement curve.
Such a curve is provided on each side
of a circular curve to minimize super
elevation.
Horizontal Alignment Design
11
a) Simple Circular Curve
b) Circular Curve with transition curve (Spirals)
As per ERA(2013) manual:
R1<1.5*R2
c) Reverse Curves,
Broken-Back Curves,
and
Compound Curves
Horizontal Alignment Design
Degree of Curvature
 Arc Definition in meter
20 2R
1145.92

D
R
D 360
 Arc Definition in feet
100 𝑓𝑡
𝐷
5,729.6
=
≫𝐷=
𝑓𝑡
2𝜋𝑅
360
𝑅
 Chord Definition in meter
10
R
sin( D )
2
 Chord Definition in feet
𝑹=
𝟓𝟎
𝒇𝒕
𝑫
𝐬𝐢𝐧( )
𝟐
Horizontal Alignment Design
13
Simple Circular Curves - Terms
• PI=Point of Intersection
• PC= Point-of-Curvature
• PT=Point-of-Tangency
PC
PT







∆=deflection angle
L=Length of Curve
C=Chord Length
R=Radius of Curvature
M=Middle Ordinate
E=External Distance
T=Length of Tangent
Arc definition relations:
Stations of PC, PI, and PT:
PC = PI – T
P T = PC + Lc or
𝑇 = 𝑅 tan( Δ/2)
𝐶 = 2𝑅 sin( Δ/2)
𝐿 = Π𝑅Δ/180
𝐸 = 𝑅 sec( Δ/2) − 1
𝑀 = 𝑅 1 − cos( Δ/2)
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐶 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐼 − 𝐿𝑒𝑛𝑡ℎ 𝑜𝑓 𝑇𝑎𝑛𝑔𝑒𝑛𝑡(𝑇)
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝑇 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐶 + 𝐿𝑒𝑛𝑡ℎ 𝑜𝑓 𝐶𝑢𝑟𝑣𝑒(𝐿)
Horizontal Alignment Design
14
Example 1:

A curve has a deflection angle of ∆ = 23o18’ 02”, and a radius of 1432.6m. The Point of
Intersection (PI) is 5+053.87. Calculate the tangent distance (T), external distance (E),
Middle ordinate distance(M), curve length (L), Point of Curvature (PC), and Point of
Tangent (PT).
Solution:
Given ∆ = 23o18’ 02”, and R=1432.6m, PI= 5+053.87
𝑇 = 𝑅 tan( Δ/2) = 1432.6 ∗ tan(23.3/2) = 295.35m
𝐿 = Π𝑅Δ/180 = 3.14 ∗ 1432.6 ∗ 23.3/180 = 582.51𝑚
𝐸 = 𝑅 sec( Δ/2) − 1 = 1432.6 ∗ (1/(cos( 23.3/2)) − 1) = 30.13𝑚
𝑀 = 𝑅 1 − cos( Δ/2) = 1432.6 ∗ (1 − cos( 23.3/2)) = 29.51𝑚
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐶 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐼 − 𝐿𝑒𝑛𝑡ℎ 𝑜𝑓 𝑇𝑎𝑛𝑔𝑒𝑛𝑡(𝑇)
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝐶 = 5053.87 − 295.35 = 4758.49 𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 4 + 758.49
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝑇 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐶 + 𝐿𝑒𝑛𝑡ℎ 𝑜𝑓 𝐶𝑢𝑟𝑣𝑒(𝐿)
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑇 = 4758.49 + 582.51 = 5341 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 5 + 341
Horizontal Alignment Design
15
Example 2:

For design speed of 70km/hr, if it is required to use a curve radius of 300m for both curves
whose PI details is shown below. Determine the final length of the segment A to B.
Since deflection angle and circular curve radius is the same, thus L1=L2= 278.2m for both curves
and T1=T2 =150m for both curves. For a given curve,
Station PC = Station PI -Length of Tangent (T)
Station PT = Station PC + Length of Curve (L)
Horizontal Alignment Design
16
Solution:
Curve-1
 Station PC_1= Station PI_1 –T1 =10150 -150 =10,000 =≫ 10+000
 Station PT_1=Station PC_1 +L1 =10,000+278.2=10278.2 =≫ 10+278.2
Curve-2
Adjustments of PI_2 station due to Curve-1=Station of PT-1+ (distance b/n PI-2
and PI-1) - T1
Adjustments of station PI_2 = 10278.2+(10900-10150) -150=10878.2 =≫
10+878.2
 Station PC_2= Station PI_2 –T2 =10878.2 -150 = 10728.2 =≫ 10+728.2
 Station PT_2=Station PC_2 +L2 =10,728.2+278.2=11,006.4 =≫ 11+006.4
 Adjusted station of B or EP =Station PT_2+(Distance between EP and PI_2 )T2
Distance between EP and PI_2 =(11,700-10900)=800m
 Adjusted station of B, EP =11,006.4+800-150=11,656.4 =≫ 11+656.4
 The final length of the Segment A to B is Adjusted station of B – Station of A
=11656.4 – 8850= 2,806.4 m
Horizontal Alignment Design
17
Sight Distance on Horizontal Curves
 On the inside of horizontal curves, it may be
necessary to remove buildings, trees or other
sight obstructions or widen cuts on the insides
of curves to obtain the required sight distance
(see Figure).

R = curve radius measured to the centerline of the road in
meter

 = central angle of the curve in degrees,

L = length of curve in meter

PC = point of curve (the beginning point of the horizontal
curve), and

PT = point of tangent (the ending point of the horizontal
curve).
𝑀𝑠 = 𝑅𝑣 1 − cos(
90𝑆𝑆𝐷
)
Π𝑅𝑣
Horizontal Alignment Design
18
Sight Distance on Horizontal Curves
 SSD = stopping sight distance in meter,
 Rv = radius to the vehicle’s traveled path (usually
measured to the center of the innermost lane of the
road) in meter
 s = angle (in degrees) subtended by an arc equal
in length to the required stopping sight distance
(SSD),
 Ms = middle ordinate necessary to provide adequate
stopping sight distance (SSD) in meter.
Relationships
 If these equations are to be used for
passing sight distance(PSD), SSD in the
equations would be replaced by PSD.
Π𝑅𝑣Δ𝑠
180
180𝑆𝑆𝐷
Δ𝑠 =
Π𝑅𝑣
𝑆𝑆𝐷 =
90𝑆𝑆𝐷
)
Π𝑅𝑣
Π𝑅𝑣
𝑅𝑣 − 𝑀𝑠
𝑆𝑆𝐷 =
[sec(
)]
90
𝑅𝑣
𝑀𝑠 = 𝑅𝑣 1 − cos(
Horizontal Alignment Design
19
Example 3:
A horizontal curve on a two lane highway is designed with a 609.6 m radius,
3.6 m lanes, and a 96 km/hr design speed. Determine the distance that must be cleared from
the inside edge of the inside lane to provide a sufficient stopping sight distance.
(The coefficient of friction of the road surface f = 0.32 and PRT = 2.5 sec, g=0).
Solution:

The SSD for the design speed of 96 km/hr is 180 m

The radius of the middle of the inside lane (Rv=R-w/2)

Rv=609.6m-1.8m=607.8m
 6.65 m must be cleared as measured from the center of the inside lane or
4.85 m must be cleared as measured from the outside edge of inside lane.
Horizontal Alignment Design
Circular curve Design
Stability of a vehicle on flat surface
20








𝑁 = 𝑚𝑔
a=radial acceleration;
𝑚𝑉 2
m=mass of vehicle;

𝐹=
𝑅
V=speed of vehicle;

𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛,
R=Radius of curvature;
F=Frictional Resistance;

𝜇 = 𝐹/𝑁
N=Reaction (normal force)
When road has no camber and the vehicle is on the verge of
overturning

To avoid overturning
𝑚𝑉 2 ℎ/𝑅 < 𝑚𝑔𝑏 ⇒ 𝑉 2 ℎ/𝑅 < 𝑔𝑏

To avoid side slip
𝑚𝑉 2 /𝑅 < 𝜇𝑚𝑔 ⇒ 𝑉 2 /𝑅 < 𝜇𝑔
Horizontal Alignment Design
Stability of a vehicle on Super-elevated Surface Forces & Equilibrium
21
 Resolving the Forces // and |to the road
 (// to the road)
Wv 2
F  WSin 
Cos
gR
 (| to the road)
e
1
𝑾𝒗𝟐
𝑾𝑪𝒐𝒔𝜽 +
𝑺𝒊𝒏𝜽 = 𝑵
𝒈𝑹
 Frictional force, F=mN
𝐹 = 𝜇𝑁 =

2
𝑊𝑣 2
𝐶𝑜𝑠𝜃 − 𝑊𝑆𝑖𝑛𝜃
𝑔𝑅
𝑊𝑣 2
𝑊𝑣 2
𝜇
𝑆𝑖𝑛𝜃 + 𝑊𝐶𝑜𝑠𝜃 =
𝐶𝑜𝑠𝜃 − 𝑊𝑆𝑖𝑛𝜃
𝑔𝑅
𝑔𝑅
𝜇𝑣 2
𝑣2
𝑆𝑖𝑛𝜃
+ 1 = 𝐶𝑜𝑠𝜃
−𝜇
𝑔𝑅
𝑔𝑅
𝑣2
−𝜇
𝑔𝑅
𝑒 = 𝑡𝑎𝑛𝜃 = 2
𝜇𝑣
+1
𝑔𝑅
But the term 𝜇𝑣
has a very small value and could be ignored for all practical purposes.
𝑔𝑅
Check using typical values like V=50km/hr; m=0.16; and R=100m.
Horizontal Alignment Design
22
Continued…
𝑣2
𝑇ℎ𝑢𝑠, 𝑡𝑎𝑛𝜃 =
−𝜇 =𝑒
𝑔𝑅
𝑣2
𝑉 Τ3.6 2
⇒𝑒+𝜇 =
=
𝑔𝑅
9.81𝑅
Example 4:
Fundamental equation of designing curves
where
2
𝑉
• V is in Km/hr
𝑅=
127 𝑒 + 𝜇 min• R is in m
• e is in m/m or in % used in decimal
• µ is dimensionless
What is the minimum radius of curvature allowable for a roadway with a 100
km/h design speed, assuming that the maximum allowable superelevation rate is
10% and maximum allowable side friction is 0.11? What is the actual maximum
superelevation rate allowable under ERA recommended standards for a 100 km/h
design speed, if it is decided to use a radius of 500m? Round the answer down to
the nearest whole percent.
• Minimum radius of curvature for 100 km/h design speed:
𝑅=
𝑉2
127 𝑒+𝜇
=
1002
=375
127 0.1+0.11 min
m
Actual maximum superelevation rate allowable
1002
500 =
127 𝑒 + 0.11
𝑒 = 0.047, 𝑒 = 4.7% = 5%
Horizontal Alignment Design
23
ERA(2013) suggests the minimum radius value for a safe side friction and superelevation
rate of 4%,6%,8% and 10%. See the Tables below.

m is low for high speed design than for low speed design
Horizontal Alignment Design
24
Reading Assignment


Compound Curves
Reversed Curves
Horizontal Alignment Design
25
Compound Curves

A compound curve consists of two (or
more) circular curves between two main
tangents joined at point of compound
curve (PCC). Curve at PC is designated
as 1 (R1, L1, T1, etc) and curve at PT is
designated as 2 (R2, L2, T2, etc).
𝑰𝟏
𝟐
𝑰
𝑹𝟐 𝒕𝒂𝒏 𝟐
𝟐

𝑻𝟏 = 𝑹𝟏 𝒕𝒂𝒏

𝑻𝟐 =

𝑻𝑳 = 𝑻𝟏 + 𝒙,

𝑻𝒔 = 𝑻𝟐 + 𝒚

𝒙
𝒔𝒊𝒏𝑰𝟐
=
𝑻𝟏 +𝑻𝟐
𝒚
=
𝒔𝒊𝒏(𝟏𝟖𝟎−𝑰) 𝒔𝒊𝒏𝑰𝟏
Horizontal Alignment Design
26
Elements of compound curve
 ∆=I = angle of intersection = I1 + I2
Elements of compound curve
 Lc1 = length of first curve
 PC = point of curvature
 Lc2 = length of second curve
 PT = point of tangency
 L1 = length of first chord
 PI = point of intersection
 L2 = length of second chord
 PCC = point of compound curve
 L = length of long chord from PC to PT
 T1 + T2 = length of common tangent
 T1 = length of tangent of the first curve
measured from V1 to V2
 T2 = length of tangent of the second
 𝜃= 180° - I
curve
 x and y can be found from
 V1 = vertex of the first curve
triangle
V1-V2-PI.
 L can be found from triangle PC-PCC-PT
 V2 = vertex of the second curve
Finding the stationing of PT
 I1 = central angle of the first curve
Given the stationing of PC
 I2 = central angle of the second curve
Sta PT=Sta PC+Lc1+Lc2
 TL =Long Tangent
Given the stationing of PI
 TS =Short Tangent
Sta PT=Sta PI−x−T1+Lc1+Lc2
Compound Curves
Horizontal Alignment Design
27
Example 5:
 A compound curve ,consisting of
two simple circular curves of
radii 500m and 350m ,is to be
laid out between two straights.
 The angles of intersection
between the tangents and the
two straights are 25˚ and 55˚.
 The station PI is 10+500
 Calculate the various elements
of the compound curve. And
 Find the station of PC, PCC and
PT
Horizontal Alignment Design
28

Length of the common tangent,
250

𝑉1 𝑉2 = 𝑇1 + 𝑇2 = 𝑅1 𝑡𝑎𝑛

= 500 𝑡𝑎𝑛

I=∆ = 550+ 250= 800

From triangle V1-PI-V2, by sine rule
250
2
+ 350 𝑡𝑎𝑛
2
+ 𝑅2 𝑡𝑎𝑛
550
2
Finding the stationing of PT
Given the stationing of PC
Sta PT=Sta PC+Lc1+Lc2
550
2
=293.04 m

𝐿𝑐1 =

𝐿𝑐2 =
𝛱𝑅1 𝛥1
180
𝛱𝑅1 𝛥1
180
=
𝛱∗500∗25
180
𝛱∗350∗55
180
=218.17 m

𝑦
𝑠𝑖𝑛250

x = 293.04𝑠𝑖𝑛550 /𝑠𝑖𝑛1000 = 243.74 m

y = 293.04𝑠𝑖𝑛250 /𝑠𝑖𝑛1000 =125.754 m

Sta. of PC=PI-𝑇𝐿 =10500-354.59

Length of long tangent

Sta. of PC=10+145.41

𝑇𝐿 = 𝑇1 + 𝑥 = 𝑅1 𝑡𝑎𝑛 27.5 + 𝑥

Sta. of PCC=PC+𝐿𝑐1 =10145.41+218.17m

=500tan12.5+243.74= 354 .59 m

Sta. of PCC=10+363.58

𝑇𝑆 = 𝑇2 + 𝑦 = 𝑅2 𝑡𝑎𝑛 12.5 + y

Sta. of PT=PCC+𝐿𝑐2 =10363.58+335.98

=350tan27.5+125.754 =307.95 m

Sta. of PT=10+699.56
=
𝑇1 +𝑇2
𝑥
=
𝑠𝑖𝑛100 𝑠𝑖𝑛550
Given the stationing of PI
Sta PT=Sta PI−x−T1+Lc1+Lc2
=
= 335.98 𝑚
Horizontal Alignment Design
29
Reversed Curves
 Reversed curve, though pleasing to
the eye, would bring discomfort to
motorist running at design speed.
 The instant change in direction at
the PRC brought some safety
problems.
 Despite this fact, reversed curves
are being used with great success
on park roads, formal paths,
waterway channels, and the like.t
Horizontal Alignment Design
30
Elements of Reversed Curve














 Finding the stationing of PT
PC = point of curvature
 Given the stationing of PC
PT = point of tangency
Sta PT=Sta PC+Lc1+Lc2
PRC = point of reversed curvature
 Given the stationing of V1
T1 = length of tangent of the first curve
Sta PT=Sta V1−T1+Lc1+Lc2
T2 = length of tangent of the second curve
V1 = vertex of the first curve
V2 = vertex of the second curve
I1 = central angle of the first curve
I2 = central angle of the second curve
Lc1 = length of first curve
Lc2 = length of second curve
L1 = length of first chord
L2 = length of second chord
T1 + T2 = length of common tangent measured from V1 to V2
Horizontal Alignment Design
31
Elements of Reversed Curve














PC = point of curvature
PT = point of tangency
PRC = point of reversed curvature
T1 = length of tangent of the first curve
T2 = length of tangent of the second curve
V1 = vertex of the first curve
V2 = vertex of the second curve
I1 = central angle of the first curve
I2 = central angle of the second curve
Lc1 = length of first curve
Lc2 = length of second curve
L1 = length of first chord
L2 = length of second chord
T1 + T2 = length of common tangent measured from V1 to V2
Horizontal Alignment Design
32
Maximum Degree of Curvature



𝑉2
𝑅=
127 𝑒 + 𝜇
Minimum radius for safety (veh. stability)
min
Limiting value for a given design speed (given emax mmax)
The respective maximum Degree of Curvature(angle that subtends 20m arc
length) is: 20 𝑚 𝐷𝑚𝑎𝑥
20 ∗ 180
1145.92
2𝜋𝑅
=
360
,
100 𝑓𝑡 𝐷𝑚𝑎𝑥
=
,
2𝜋𝑅
360


𝐷𝑚𝑎𝑥 =
𝜋𝑅
≫ 𝐷𝑚𝑎𝑥 =
𝑅
𝑚
100 ∗ 180
5,729.6
≫ 𝐷𝑚𝑎𝑥 =
𝑓𝑡
𝜋𝑅
𝑅
Substituting R
𝐷𝑚𝑎𝑥 =

𝐷𝑚𝑎𝑥 =
1145.92
1145.92
143240 𝑒 + 𝜇
= 2
=
𝑅min
𝑉 Τ127 𝑒 + 𝜇
𝑉2
Sharper Curve might justify use of e>emax or a higher dependence on tyre
friction or both
What is the degree of curvature for 30 m chain (arc)?
Horizontal Alignment Design
33
Super-elevation rate, e
 Superelevation is the banking of the roadway along a horizontal curve,
so that the drivers can negotiate the curve at safe and comfortable speed.



As shown in the Figure below it is the raising of the outer edge of the road along a curve
in-order to counteract the effect of radial centrifugal force in combination with the
friction between the surface and tyres developed in the lateral direction
Maximum value is controlled by:

Climatic conditions: frequency & amount of snow/icing

Terrain condition: flat vs. mountainous

Area type: rural vs. urban

Frequency of very slow moving vehicles
Minimum super-elevation rate is determined by drainage requirements
Horizontal Alignment Design
34
Superelevation Transitions
 Consists of Runoff and
Tangent Runout sections
 Runout: length of
roadway needed to
accomplish a change in
outside lane cross slope
from normal rate to
zero.
 Runoff: length of
roadway needed to
accomplish a change in
outside lane cross slope
from zero to full
Horizontal Alignment Design
Methods of Achieving Superelevation (For Crowned Roadways)
35
Is done in two stages:
1.
Neutralizing the camber of
the road gradually, bringing
it in to a straight line slope
2. Then Increasing the slope
gradually until design
super-elevation is attained
• Slope, ed=(H/wT ),
H=ed*wT ;
Where:
• H=elevation difference b/n edges of
pavements
• wT=Total width of roadway pavement
(carriage way)
• ed=superelevation rate
Horizontal Alignment Design
36
Methods of Achieving Superelevation
(For Crowned Roadways with Simple Circular Curve)
•
The transition from a tangent section with a normal superelevation for drainage to a super
elevated horizontal curve occurs in two stages:
Tangent Runout:
• The outside lane of the curve must have a transition from the normal drainage
superelevation to a level or flat condition prior to being rotated to the full superelevation
for the horizontal curve.
• The length of this transition is called the tangent runout and is noted as Lt.
Superelevation Runoff:
 Once a flat cross-section is achieved
for the outside lane of the curve, it
must be rotated (with other lanes) to
the full superelevation rate of the
horizontal curve.
 The length of this transition is called
the superelevation runoff and is
noted as Lr.
Horizontal Alignment Design
37
Minimum length of superelevation runoff (Lr)



The widely used empirical expression, the superelevation runoff length is determined
as a function of the slope of the outside edge of the traveled way relative to the
centerline profile(AASHTO,2001).
Current practice is to limit the grade difference, referred to as the relative gradient, to a
maximum value of 0.50 percent or a longitudinal slope of 1:200 at 80 km/h [50 mph].
However longer runoff lengths are required at higher speeds and shorter lengths at lower
speeds. Experience indicates that relative gradients of 0.78 and 0.35 percent provide
acceptable runoff lengths for design speeds of 20 and 130 km/h, respectively.
 A comfortable and aesthetically pleasing runoff design
can be attained through the exclusive use of the
maximum relative gradient criterion(g). Table shown
presents maximum relative gradient values for each
design speed. It is taken from AASHTO(2001).
Table: Design speed Vs. max. rel. gradient(g )
Horizontal Alignment Design
38
Methods of Achieving Superelevation…
where:

Lr= minimum length of superelevation runoff, m;
 g = maximum relative gradient, percent;
 n1 = number of lanes rotated;
 bw = adjustment factor for number of lanes rotated, 1
for n1 = 1, 0.75 for n1 = 2, 0.67 for n1 = 3;
 w = width of one traffic lane, m (typically 3.6 m);
 ed = design superelevation rate, percent
• This Equation can be used to calculate Ls directly for undivided streets or highways
where the cross section is rotated about the highway centerline and n1 is equal to
one-half the number of lanes in the cross section.
• Where separate pavements on a divided highway are rotated around an edge, the
full number of lanes on the pavement would be used as n1.
Horizontal Alignment Design
39
Minimum length of tangent runout(Lt)
The length of tangent runout is determined by the amount of adverse cross slope to
be removed and the rate at which it is removed. To effect a smooth edge of
pavement profile, the rate of removal should equal the relative gradient used to
define the superelevation runoff length. Based on this rationale, the following
equation should be used to compute the minimum tangent runout length:
where:
Lt = minimum length of tangent runout, m;
𝑒𝑁𝐶
eNC = normal cross slope rate, percent;
𝐿𝑡 =
𝐿𝑟
𝑒𝑑
ed = design superelevation rate, percent;
Lr = minimum length of superelevation runoff, m
Example 6:

A four-lane highway with a superelevation rate of 8% achieved by rotating two 3.65mlanes around the centerline. The design speed of the highway is 80 km/hr and the normal
crown slope is 2.5%. What is the appropriate minimum length of superelevation runoff
and corresponding length of tangent runout?
Solution:
𝑤 ∗ 𝑛1 ∗ 𝑒𝑑 ∗ 𝑏𝑤 3.65 ∗ 2 ∗ 8% ∗ 0.75
𝐿𝑟 =
=
= 87.6𝑚
𝛥𝑔
0.5%
𝐿𝑡 =
𝑒𝑁𝐶
2.5%
𝐿𝑟 =
∗ 87.6 = 27.4𝑚
𝑒𝑑
8%
Horizontal Alignment Design
40
Application of Superelevation on circular curves with out spiral curves
• In the design of curves without spirals the super-elevation runoff is considered to be that
length beyond the tangent runout.
• Empirical methods are employed to locate the superelevation runoff length with respect to
the Point of curvature (PC).
• According to ERA(2013), the Current design practice is to place approximately two-thirds of
the runoff on the tangent approach and one-third on the curve, as shown in Figure 8.7.
• This procedure is reversed on leaving the curve.
2*Lr/3
Lt
Lr/3
Figure 8-7: Diagram of superelevation Transition on Circular Curve (no spiral curves)
Horizontal Alignment Design
41
Application of Superelevation on curve with transition curves (spirals)
• In alignment design with spirals, the super-elevation runoff is provided over the
whole of the transition curve.
• The length of runoff is the spiral length, with the tangent to spiral (TS) transition
point at the beginning and the spiral to curve (SC) transition point at the end.
• The change in cross slope begins by removing the adverse cross slope from the
lane or lanes on the outside of the curve on a length of tangent just ahead of TS
Lr
(the tangent runout, Lt).
L
t
The Figure shows how the full superelevation of
6% is attained on curve with spirals from a
tangent normal cross-slope of 2% .
Horizontal Alignment Design
42
Application of Superelevation on curve with transition curves (spirals)….
• Between the TS and SC (the super-elevation runoff) the travelled way is rotated to
reach the full super-elevation at the SC.
• This procedure is reversed on leaving the curve.
• By this design the whole of the circular curve has full super-elevation, as shown in
the Figure.
Lr
Lt
Figure 8-6:Diagram of Superelevation Transition with Spiral curves.
Horizontal Alignment Design
43
Example 7:
A two-lane highway (3.5 m lanes) with a design speed of 80
km/h has a 400 m radius horizontal curve. Determine the
superelevation rate. Use side friction=0.12
Solution:

Determine superelevation rate:
𝑉2
𝑅𝑢𝑠𝑒𝑑 =
,
127 𝑒 + 𝜇
1002
400 =
,
127 𝑒 + 0.12
𝑒 = 0.08, 𝑢𝑠𝑒 𝑒 = 8%
Horizontal Alignment Design
44
 Shoulder Super-elevation(for surfaced roads)
Note: For design classes
DC5 and lower the
shoulder may be sloped
with the carriageway,
but the shoulder should
then be surfaced on the
outside
of
the
curve(ERA,2013).
Horizontal Alignment Design
45
Transition Curves (Spirals)
Spirals: these are curves which provide a
gradual change in curvature from tangent to a
circular path
Advantages:
• Provides an easy-to-follow path so that
centrifugal force increases and decreases
gradually; lesser danger of overturning/ sideslipping
• Vehicle could keep to the middle of lane while
traversing a curve
• Is convenient for the application of superelevation
• Improved visual appearance, no “kinks”(sharp
curve)
Horizontal Alignment Design
46
• For large radius curves, the rate of change of lateral acceleration is small and
transition curves are not normally required.
Table 8-4 of ERA(2013) geometric design manual stated the condition when to
use Transition curves (Spirals).
• If a transition curve is required, the Euler spiral, which is also known as the clothoid, should be
used.
• The radius varies from infinity at that tangent end of the spiral to the radius of the circular arc at
the circular curve end.
• By definition, the radius at any point of the spiral varies inversely with the distance measured
along the spiral.
Horizontal Alignment Design
=∆
47
Transition Curves - Geometry
Elements of Spiral Curve
 TS = Tangent to spiral
 SC = Spiral to curve
 CS = Curve to spiral
 ST = Spiral to tangent
 LT = Long tangent
 ST = Short tangent
 R = Radius of simple curve
 Ts = Spiral tangent distance
 Tc = Circular curve tangent
 Ls = Length of spiral
 PI = Point of intersection
 ∆ =I = Angle of intersection
 L = Length of spiral from TS to any point along the spiral
 Ic = Angle of intersection of the simple curve
 p = Length of throw or the distance from tangent that the
circular curve has been offset
 X = Offset distance (right angle distance) from tangent to
any point on the spiral
 Xc = Offset distance (right angle distance) from tangent
to SC
 Y = Distance along tangent to any point on the spiral
 Yc = Distance along tangent from TS to point at right angle
to SC
 Es = External distance of the simple curve
∆=
Elements of Spiral Curve…
 θ = Spiral angle from tangent to any point on the spiral
 θs = Spiral angle from tangent to SC
 i = Deflection angle from TS to any point on the spiral, it
is proportional to the square of its distance
 is = Deflection angle from TS to SC
 D = Degree of spiral curve at any point
 Dc = Degree of simple curve
Horizontal Alignment Design
48
Transition Curves - Geometry
 Formulas for Spiral Curves

𝒀= 𝑳−
𝑳𝟓
𝟒𝟎𝑹𝟐 𝑳𝟐𝒔
At L=Ls , Y=Yc thus,
𝐿3𝑠
𝑌𝑐 = 𝐿𝑠 −
40𝑅2
 Offset distance from tangent to any point
on the spiral:

𝟔𝑹𝑳𝒔
1
𝐿2𝑠
4
24𝑅
𝑝 = 𝑋𝑐 =
 Spiral angle from tangent to any point on
the spiral (in radian)
𝑳𝟐
 θ= 𝟐𝑹𝑳
from TS to any point on
the spiral:
𝑳𝟑
𝑿=
At L = Ls, X = Xc, thus,
𝐿2𝑠
𝑋𝑐 =
6𝑅
 Length of throw:

 Deflection angle
𝒔
At L = Ls, θ = θs, thus,
𝐿𝑠
θ𝑠 =
2𝑅
 This angle is
proportional to the
square of its distance
Horizontal Alignment Design
49

Since the spiral is defined as the curve such that the reciprocal of the radius
varies linearly from zero at the TS to 1/Rc at the SC,
For a length of segment on spiral, L measured from TS the corresponding
reciprocal of the radius at L is
Y
Xs
Constant, A called flatness of spiral
𝐴=
𝐿𝑠 ∗ 𝑅𝑐
SC
Rc
Ls
Ys

The spiral angle θ is given by
In particular,
The coordinates X and Y of any point on the spiral may be expressed as
functions of L:
To calculate Ys and Xs, use L=Ls in the Equations above in X and Y
X
TS
𝑅=∞
Rc
Horizontal Alignment Design
50
 Other measurements of interest are

Determine stations of critical points:
TS station = P.I. station - (T`+ k)
SC station = TS station + Ls
CS station = SC station + Lc
ST station = CS station + Ls
Horizontal Alignment Design
51
Minimum length of spiral, Ls

According to AASHTO(2001) two criteria to be fulfilled in addition to superelevation runoff
length requirement. These are:
i. Driver comfort criterion: is intended to provide a spiral length that allows for a
comfortable increase in lateral acceleration as a vehicle enters a curve.
speed , v  constant
 rate of change of ar  1 rate of change of R
𝑚𝑣 2
𝑣2
𝑃=
⇒ 𝑎𝑟 =
𝑅
𝑅
 faster change in R  faster change in radial force
 greater passenger discomfort
Δ𝑎𝑟(𝑓𝑟𝑜𝑚𝑇𝑆−𝑆𝐶) = 𝑣 2 Τ𝑅
Δ𝑡(𝑓𝑟𝑜𝑚𝑇𝑆−𝑆𝐶) = 𝐿𝑠Τ𝑣
Δ𝑎𝑟 𝑣 2 Τ𝑅
𝑣3
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑟𝑎𝑑𝑖𝑎𝑙 𝑎𝑐𝑐. , 𝐶 =
=
=
Δ𝑡
𝐿𝑠Τ𝑣 𝐿𝑠𝑅
𝑣3
(0.278𝑉)3
(𝑉)3
(𝑉)3
𝐿𝑠 =
; 𝐿𝑠 =
=
= 0.0214
(𝑽)𝟑
(𝑽)𝟑
𝐶𝑅
𝐶𝑅
46.66𝐶𝑅
𝐶𝑅
𝑳𝒔, 𝒎𝒊𝒏 = 𝟎. 𝟎𝟐𝟏𝟒
= 𝟎. 𝟎𝟏𝟖
𝟏. 𝟐 ∗ 𝑹
𝑹
V = design speed, km/h; R= Circular curve radius, m
C = maximum rate of change in lateral acceleration (0.3 to 1.2 m/s3 is used)
In order to calculate the minimum required to satisfy this requirement AASHTO recommends C=1.2 m/s3
𝑎𝑛𝑑
Horizontal Alignment Design
52
Minimum length of spiral, Ls….
ii. lateral shift criterion: is intended to ensure that a spiral curve is
sufficiently long to provide a shift in a vehicle’s lateral position within its
lane that is consistent with that produced by the vehicle’s natural spiral
path.
𝐿S,min = 24 ∗ 𝑃 min ∗ 𝑅 ;

Ls,min = minimum length of spiral, m;

R = radius of circular curve, m;


pmin = minimum lateral offset between the tangent and circular curve
(0.20 m is recommended by AASHTO);
Substituting pmin =0.2 in the equation above gives
𝑳𝒔,𝒎𝒊𝒏 = 𝟐. 𝟏𝟗 𝑹
• Transition curve must, therefore, be long enough to ensure that the radius
can be changed at a slow rate
Horizontal Alignment Design
53
Maximum length of spiral, Ls,max

International experience indicates that there is a need to limit the length of spiral
transition curves.
 Safety problems have been found to occur on spiral curves that are long (relative to the
length of the circular curve).
 Such problems occur when the spiral is so long as to mislead the driver about the
sharpness of the approaching curve
𝐿𝑠,𝑚𝑎𝑥 =
24 ∗ 𝑃max ∗ 𝑅 ;
pmax = maximum lateral offset between the tangent and circular curve
(1.0 m is recommended by AASHTO);
Substituting p, max=1;
𝑳𝒔,𝒎𝒂𝒙 = 𝟐𝟒 ∗ 𝟏 ∗ 𝑹 = 𝟐𝟒𝑹;
Horizontal Alignment Design
54
Example 8:

For a two-lane highway with 3.65m lane width, If it is decided to use, design
speed=100km/h, e=0.10 m/m=10%, cross-slope=2.5%, and μ=0.11; determine the
length of the transition curve
Solution:
R=100*100/127(0.11+0.1)=375m<590m(as specified in Table 8-4 of ERA. Hence transition
curve is required.)
1) If it is decided to use a circular curve with R=375m , then Length of transition curve,
i.
𝐿𝑠, min
(𝑉)3
= 0.018 𝑅
= 0.018 ∗
(100)3
375
= 48𝑚
ii. 𝐿𝑠, min = 2.19 𝑅 = 2.19 375 = 42.4𝑚
iii. On the other hand, the spiral must be at least as long as the superelevation runoff
length, Ls:
𝒘 ∗ 𝒏𝟏 ∗ 𝒆𝒅 ∗ 𝒃𝒘 𝟑. 𝟔𝟓 ∗ 𝟏 ∗ 𝟏𝟎% ∗ 𝟏
𝑳=
=
= 𝟖𝟐. 𝟗𝟓 𝒎
𝜟𝒈
𝟎. 𝟒𝟒%
𝑪𝒉𝒆𝒄𝒌, 𝑳𝒔, 𝒎𝒊𝒏 = 𝟖𝟐. 𝟗𝟓 < 𝑳𝒔, 𝒎𝒂𝒙 = 𝟐𝟒 ∗ 𝟑𝟕𝟓 = 𝟗𝟒. 𝟖𝟔𝒎. . . . . 𝒐𝒌!
Hence, use Ls=90m, and tangent runout length, Lt
𝐿𝑡 =
𝑒𝑁𝐶
2.5%
𝐿𝑠 =
∗ 90 = 22.5𝑚
𝑒𝑑
10%
Horizontal Alignment Design
55
Example 9:
A 375 m radius horizontal curve with spiral curve length of 90m connecting tangents with
bearings of N75E° and S78E°. Determine the stations of the TS, SC, CS, and ST, given that the
temporary station of the P.I. is 15+000.

Determine central (deflection) angle, :
=(90-75) +(90 -78) =27° rad =27*3.14/180=0.471 rad

Determine spiral angle and Length of circular curve, Lc
𝜃𝑠 =
𝐿𝑠
= 90/(2 ∗ 375) = 0.12𝑟𝑎𝑑
2𝑅𝑐
𝐿𝑐 = 𝑅𝑐 ∗ Δ𝑟𝑎𝑑 − 𝐿𝑠 = 375 ∗ 0.471 − 90 = 86.63𝑚

Then the flatness of spiral, A:-
𝐴=
𝐿𝑠 ∗ 𝑅𝑐 = 90 ∗ 375 = 183.71
Horizontal Alignment Design
56
Example…

Determine coordinates of SC point:
905
909
𝑋𝑠 = 90 −
+
= 89.87𝑚
40 ∗ 183.714 3456 ∗ 183.718
𝑌𝑠 =
903
907
90^11
−
+
= 3.6𝑚
6 ∗ 183.712 336 ∗ 183.71^6 42,240 ∗ 183.7110
 Determine p, k, T `:
𝑝 = 𝑌𝑠 − 𝑅𝑐 (1 − cos( 𝜃𝑠 )) = 3.6 − 375(1 − cos( 6.88°)) = 0.9
𝑘 = 𝑋𝑠 − 𝑅𝑐 ∗ sin( 𝜃𝑠 ) = 89.87 − 375 ∗ sin( 6.88°) = 44.95
𝑇 ∗ = (𝑅𝑐 + 𝑝) tan( Δ/2) = (375 + 0.9) ∗ tan( 27°/2) = 90.25
 Determine stations of critical points:
TS station = P.I. station - (T``+ k) =15000-(90.25+44.95)=14864.8 means 14+864.8
SC station = TS station + Ls=14864.8+90=14954.8 >> 14+954.8
CS station = SC station + Lc=14954.8+86.63=15,041.43 >> 15+041.43
ST station = CS station + Ls=15041.43+90=15131.43 >> 15+131.43
Horizontal Alignment Design
57
Setting out a Simple Horizontal Curve by deflection angles
To locate C

<BAC=dx from tangent line to point B

Align C with the theodolite

Measure AC from A(chord, Cx)
B (PI)

dx
A
PC
The deflection angle in radians dx to a point on
the curve at a distance x from the PC or SC is
given by
X
PT
Cx
2*dx

The corresponding chord length, cx to this point is given by
C
Horizontal Alignment Design
58
Example 10:
Setting out of a circular curve with spiral
Table below gives deflection angles and chords at stations of 20 m intervals for a 375
m radius curve with a central angle of 13.24 degree and a SC at station 14+954.8. It
is calculated using Excel sheet
Point
Station
x
dx,rad=x/(2R)
dx, deg=dx,rad*180/∏
C
SC
14+954.8
14+960
14+980
15+000
15+020
15+040
15+041.43
0
5.2
25.2
45.2
65.2
85.2
0.0000
0.0069
0.0336
0.0603
0.0869
0.1136
0.1155
0.000
0.397
1.925
3.453
4.981
6.509
6.618
0.00
5.20
25.20
45.17
65.12
85.02
86.44
CS
86.63
Horizontal Alignment Design
59
Setting out a Spiral Curve by deflection angles
The deflection angle d(in rad) is
and the corresponding chord length, c is
Setting out data for 1st Spiral curve of the above example
For distance measured on the spiral as measured from TS is L, the corresponding
coordinate (X,Y) on the spiral can be computed by using the equations discussed above.
The Table below is calculated using Excel sheet.
0.00000
0.00114
d(in deg)
=drad*180/∏
0
0.0654
0.00
15.20
35.20
0.00611
0.3503
35.20
0.83
55.19
0.00150
0.0862
55.20
5.748
2.10
75.15
0.02790
1.5994
75.18
6.879
3.59
89.87
0.03995
2.2903
89.94
Point
Station
L
θrad =L/(2R)
θdeg=θrad *180/∏
Y
X
d(in rad)
TS
14+864.8
14+880
0
15.2
0
0.0203
0.000
1.162
0.00
0.02
0.00
15.20
14+900
35.2
0.0469
2.690
0.22
14+920
55.2
0.0736
4.219
14+940
75.2
0.1003
14+954.8
90
0.1200
SC
C
Horizontal Alignment Design
60
Widening of Highway Curves
Need
 Rear wheels don’t follow front wheels,
 Trailers fitted on trucks, don’t follow path of trucks wheels
 To have adequate sight-distances
 Drivers tend to keep greater clearances with vehicles coming
from the opposite direction and might thus move out of a lane
when traversing a curve



Therefore widening of the carriageway is necessary where the
horizontal curve is tight.
This is to ensure that the rear wheels of the largest vehicles
remain on the road when negotiating the curve; and, on two lane
roads, to ensure that the front overhang of the vehicle does not
encroach on the opposite lane.
Widening is therefore also important for safety reasons.
Horizontal Alignment Design
61
Amount of Extra Widening required, f
Let





R1=radius of inner rear-wheel on a curved truck (m)
R2=radius of outer front-wheel (m)
B=width of vehicle
f=widening (m)
L=Length of vehicle (m)
Horizontal Alignment Design
62
Curve widening …


Widening on high embankments is recommended for design classes DC8
through to DC4. The steep drops from high embankments unnerve some
drivers and the widening is primarily for psychological comfort although it
also has a positive effect on safety.
In this case the height of fill is measured from the edge of the shoulder to the
toe of the slope. Widening for curvature and for high embankments should
be added where both cases apply(ERA,2013).
Horizontal Alignment Design
63
Widening - Methods





On a simple curve (i.e. with no spirals) widening should be applied
on the inside edge of a pavement only to match the tendency for
drivers to cut the inside edge of the travelled way.
For curves with spirals, widening could be applied on the inside
(only) or could be equally divided b/n the inside and outside
Widening should transition gradually on the approaches to the
curve so that the full additional width is available at the start of the
curve.
In practice, curve widening is thus applied over no more than the
length of the super-elevation runoff preceding the curve.
Widening is costly and very little is gained from a small amount of
widening.
Vertical Alignment Design
64



Vertical Alignment consists of straight sections of the highway
known as grades or tangents, connected by vertical curves.
The design involves the selection of suitable GRADES for the
tangent sections and the design of the VERTICAL CURVES.
The topography of the area through which the road traverses has
a significant impact on the design of the vertical alignment.
Vertical Alignment Design
65
GRADES






Effect of grade is more pronounced on Heavy Vehicles than on
Passenger Cars
Maximum Grade on a highway should be carefully selected
based on the design speed and design vehicle
Grades of 4 to 5 %  little or no effect on passenger cars,
except for those with high weight/horsepower ratios,
Grade > 5% speed of passenger cars decrease on upgrades
and increase on downgrades.
Grade has a greater impact on trucks than passenger cars.
Truck speeds may increase up to 5 percent on downgrades and
decrease by 7 percent on upgrades
Vertical Alignment Design
66
Maximum Grade
The recommended max. grades for different road standards is given in Table 94 of ERA(2013)
Vertical Alignment Design
67
Minimum Grade




Depend on the drainage conditions of the highway
zero-percent grades may be used on uncurbed pavements with
adequate cross slopes to laterally drain the surface water
For curbed pavements, however, a longitudinal grade should be
provided to facilitate the longitudinal flow of the surface water
A minimum grade of 0.5% is usually used; it may be reduced to
0.3% on high-type pavement constructed on suitably crowned,
firm ground.
Vertical Alignment Design
68
Effect of Grade
• Speed-distance curves for a typical heavy truck travelling at initial speed
of 120km/hr, the curves show its deceleration on upgrades of different
percent.
Figure: Speed-Distance Curves for a Typical Heavy Truck of 120 kg/kW for Deceleration on Upgrades (AASHTO,2001)
Vertical Alignment Design
69
Critical Length of Grade



It is the term that indicates the maximum length of a designated upgrade
on which a loaded truck can operate without an unreasonable reduction in
speed
For a given grade, lengths less than critical result in acceptable operation
in the desired range of speeds.
To maintain LOS on grades longer than critical

change in location to reduce grades

addition of extra lanes (climbing or crawler lanes): data for critical
lengths of grade are used with other pertinent considerations (such as
traffic volume in relation to capacity, % heavies) to determine where
added lanes are warranted.
Vertical Alignment Design
70
Critical Length of Grade

To establish design values for critical lengths of grade data or
assumptions are needed on the following:

Size and power of representative truck or truck combination to be
used as a design vehicle

Speed at entrance to critical length grade

Minimum speed on the grade below which interference to following
vehicles is considered unreasonable
Vertical Alignment Design
71
Critical Length of Grade
• It is recommended that a 15-km/h [10-mph] reduction criterion be
used as the general guide for determining critical lengths of grade.
• A 3% grade
cause a reduction
in speed of 15
KPH after 500 m
of ( if length of
grade exceeds
500 m)
Figure: Critical Lengths of Grade for Design, Assumed Typical Heavy Truck of 120 kg/kW [200 lb/hp], Entering Speed = 110 km/h [70
mph] (AASHTO,2001)
Vertical Alignment Design
72
Critical Length of Grade…
• Refer Table 9-5 of ERA(2013) which stated the criteria for providing climbing lane for
different design class of roads.,
Vertical Alignment Design
73
 When flatter grades cannot be
accommodated, consider climbing lane
when all 3 of the following criteria are
met (AASHTO):
1. Upgrade traffic flow rate in excess of
200 vehicles per hour.
2. Upgrade truck flow rate in excess of
20 vehicles per hour.
3. One of the following conditions exists:
• A 15 km/h or greater speed
reduction is expected for a typical
heavy truck.
• Level-of-service E or F exists on
the grade.
• A reduction of two or more levels
of service is experienced when
moving from the approach
segment to the grade.
Figure 9-4: Layout for Climbing Lane (ERA, 2013)
Vertical Alignment Design
74
Vertical Curves




Are parabolic curves used to provide a gradual change from one tangent
grade to another so that vehicles may run smoothly as they traverse the
highway.
• VPC: Vertical Point of Curvature
• VPI: Vertical Point of Intersection
Are of two types
• VPT: Vertical Point of Tangency
 Sag Vertical Curves
• G1, G2: Tangent grades in percent
 Crest Vertical Curves
• A: Algebraic difference in grades
• L: Length of vertical curve
Design Criteria for vertical curves
 Provision of minimum stopping sight distance
 Adequate drainage
 Comfortable in operation
 Pleasant appearance
The first criterion is the only criterion associated with crest vertical curves,
whereas all four criteria are associated with sag vertical curves.
Vertical Alignment Design
75
Types of Vertical Curves
 Crest



Stopping, or
Passing sight distance controls
Type I and Type II
 Sag
 Headlight/SSD distance,
 Comfort, drainage and
 Appearance control
 Type III and Type IV
 Vertical curves defined by
K = L/A
K= length of vertical
curve/difference in grades (in
percent)
K= length to change one percent in
grade
Vertical Alignment Design
76
K Values
Defintion:
 The horizontal distance in feet (meters) needed to make 1%
change in gradient.
𝐿
𝐿
𝐾=
=
𝑔2 − 𝑔1
𝐴
Application:
 To determine the minimum lengths of vertical curves
 To determine the horizontal distance from the VPC to the high
point of Type I or the low point of Type III
Vertical Alignment Design
77
Vertical curve Equation
• The rate of change of slope is given by
the second derivative,
D2y/dx2=2a, 2a is constant
• The rate of change of slope (2a) can
also be written as A/L
 The general equation of a parabola
y = ax2 + bx + c
Where:
• y = roadway elevation at distance x
• x = distance from beginning of
 d2y/dx2=2a=2(g2-g1)/2L=(g2-g1)/L
vertical curve
𝑔 –𝑔
𝐴
• 𝑎 = 2 1 = and
2𝐿
2𝐿
• b = g1
• c = elevation of BVC
 The slope of this curve at any point is
given by the first derivative,
• Slope=dy/dx=2ax+b,
 at x=0, dy/dx= b=g1
 At x=L, dy/dx= 2aL+g1=g2
>> a=(g2-g1)/2L=A/2L
Vertical Alignment Design
78
Vertical curve Equation
 The general equation of a parabola
y = ax2 + bx + c
• Finally equation of the curve is
𝑔2 − 𝑔1 2
𝑦=
𝑥 + 𝑔1 𝑥 + 𝐸𝑙𝑣. 𝐵𝑉𝐶
2𝐿
• Location of highest/lowest point on the
curve, 𝑥ℎ/𝑙
𝑑𝑦
𝑑𝑥
𝑔2 −𝑔1
𝑥 + 𝑔1 = 0,
𝐿
𝑔 𝐿
𝑔 𝐿
𝑥ℎ/𝑙 = − 1 = - 1 ,
𝑔2 −𝑔1
𝐴
=
2𝑎𝑥 + 𝑏 = 0
𝑔
𝑔
x= − 2𝑎1 = − 𝐴1L
• Elevation of highest/lower point,
𝑔 −𝑔
𝑦ℎ/𝑙 = 2 1 (𝑥ℎ/𝑙 )2 +𝑔1 𝑥ℎ/𝑙 + 𝐸𝑙𝑣. 𝐵𝑉𝐶
2𝐿
𝑔2 − 𝑔1
𝑔1 𝐿 2
𝑔1 𝐿
(−
) +𝑔1 (−
) + 𝐸𝑙𝑣. 𝐵𝑉𝐶
2𝐿
𝑔2 − 𝑔1
𝑔2 − 𝑔1
NB: BVC=PVC, EVC=PVT
𝑦ℎ/𝑙 =
Vertical Alignment Design
79
Vertical Curve Offsets


Offset – vertical distance from initial
tangent to the curve.
Relationships:
𝒚=
𝑨
𝟏
𝒙𝟐 =
𝒙𝟐
𝟐𝟎𝟎𝑳
𝟐𝟎𝟎𝑲
at L/2, y=E=M
E=M=
𝑨𝑳
𝟖𝟎𝟎
=
𝑳𝟐
𝟖𝟎𝟎𝑲
𝒙𝒉/𝒍 = K∗ 𝑮𝟏 ,Yf =
𝑨𝑳
𝟖𝟎𝟎
K=L/A
Where

G1=grade (in %)

y = vertical offset at any distance x
from the PVC

x = distance from PVC

𝑥ℎ/𝑙 = distance from the PVC to the
high or low point of the curve
Yf
Vertical Alignment Design
Example 11:
A +4.6% grade intersects a ‐3.0% grade at station 92+600 and at elevation of 1004.86 m. Given that a
360m curve is utilized, determine the station and elevation of the PVC and PVT. Calculate elevations at
every 20 m station and locate the station and elevation of the high point of the curve.
Solution:
Given: g1 = +4.6% (G1 = +0.046 m/m)
PVI sta. = 92+600 m,

g2= –3.0% (G2= –0.03 m/m)
PVI elev. = 1008.28 m and L = 360 m, L/2 = 180 m
Station locations for the PVC and PVT
PVC sta. = PVI sta. –L/2 = 92+600 –180 = 92+420
PVT sta. = PVC sta. + L = 92+420 + 360 = 92+780

Elevations for the PVC and PVT
Elev PVC = Elev PVI - G1(L/2) = 1004.86- 0046(180) = 996.5 m
Elev. PVT = Elev. PVI +G2(L/2) = 1004.86 – 0.03 (180) = 999.4 m
Location of high point @ xh= g1K = g1(L/A) = 4.6(360/7.6) = 217.89 m
High point sta. = PVC sta. + 217.89 m = 92+420 + 217.89 = 92+637.89
Elev. of high point (Elev. xh)= Elev. PVC + G1 xh+ A* (xh)2/200L
Elev.@ xh= 996.5 + 0.046(217.89) + (‐7.6* (217.89)2/(200*360) = 1001.51 m
Vertical Alignment Design
81
• For every x beginning from PVC point, corresponding elevation on the curve (y) can be
calculated by:
y0 =996.5 , g1 =4.6/100
r =(-3-4.6)/(360*100)
Calculation for point elevations at even 20m stations
Point
Station
X(m)
PVC
92+420
92+440
92+460
92+480
92+500
92+520
92+540
92+560
92+580
92+600
92+620
92+640
92+660
92+680
92+700
92+720
92+740
92+760
92+780
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
300
320
340
360
4.60%
PI
-3.00%
PVT
Elevation on initial Offset from initial
tangent
tangent
Elev.PVC +G1*X
off=r*X^2/2
996.5
997.42
998.34
999.26
1000.18
1001.1
1002.02
1002.94
1003.86
1004.78
1005.7
1006.62
1007.54
1008.46
1009.38
1010.3
1011.22
81
1012.14
1013.06
0.00
-0.04
-0.17
-0.38
-0.68
-1.06
-1.52
-2.07
-2.70
-3.42
-4.22
-5.11
-6.08
-7.14
-8.28
-9.50
-10.81
-12.20
-13.68
Final Elev. On curve, y
(Elev. On int. tangent+offset=Y)
996.50
997.38
998.17
998.88
999.50
1000.04
1000.50
1000.87
1001.16
1001.36
1001.48
1001.51
1001.46
1001.32
1001.10
1000.80
1000.41
999.94
999.38
Vertical Alignment Design
82
Exercise:
1) An equal tangent vertical curve connects an initial grade of -3% and a final grade of +1% and is
designed for 100 km/h. The PVI is at station 25+050 and elevation 2231m. What is the station and
elevation of the lowest point on the curve?
2) An equal-tangent crest curve connects a +1.0% and a -0.5% grade. The PVC is at station 1+653.24
and the PVI is at station 1734.92. Is this curve long enough to provide passing sight distance for a 100km/h design speed?
3) An equal‐tangent vertical curve is to be constructed between grades of ‐2.0% (initial) and +1.0%
(final). The PVI is at station 3+352.8 and at elevation 128.016 m. Due to a street crossing a roadway,
the elevation of the roadway at station 3+413.760 must be at 129.388 m. Design the curve.
4)
A vertical curve crosses a 1.219 m diameter pipe at right angles. The pipe is located at station
3+378.708 and its centerline is at elevation 332.720 m. The PVI of the vertical curve is at station
3+352.800 and elevation 334.792 m. The vertical curve is equal tangent, 182.880 m long, and
connects an initial grade of +1.20% and a final grade of ‐1.08%. Using offsets, determine the depth,
below the surface of the curve, of the top of the pipe and determine the station of the highest point of
the curve.
5)
A section of a two-lane highway (3.65 m lane width each) is designed for 120 km/h. At one point a
vertical curve connects a -2.5% and +1.5% grade. The PVT of this curve is at station 1 + 112.52. It is
known that a horizontal curve starts (has PC) 89.61m before the vertical curve’s PVC. If the superelevation
of the horizontal curve is 0.08 and the central angle is 38 degrees, what is the station of the PT?
Vertical Alignment Design
83
Vertical curve design using Offset (clearance) requirements
Example 12:
A vertical curve joins a -1.2% grade to a +0.8% grade. The P.I. of
the vertical curve is at station 75+000 and elevation 1500.90 m
above sea level. The centerline of the roadway must clear a pipe
located at station 75 +040 by 0.80 m. The elevation of the top of
the pipe is 1501.10 m above sea level. What is the minimum length
of the vertical curve that can be used?
Solution:
PVT
Vertical Alignment Design
Solution
 Elevation of PVC: Elv. of PVI+0.012*0.5L=1500.9+0.006L
 Exactly Above the pipe the road way elevation at least should be
1501.1+0.8=1501.9
 General equation of the sag curve
𝑦 = (0.8 − −1.2 ∗
𝑥2
200𝐿
−
1.2𝑥
100
+ 1500.9 + 0.006𝐿
But at 0.5L+40, y=1501.9m
1501.9= 2
(0.5𝐿+40)2
∗
200𝐿
−
1.2(0.5𝐿+40)
+
100
1500.9 + 0.006𝐿
 Rearrange and solve quadratically for L
L=417m
 Therefore, the minimum length of sag curve to satisfy the cover requirement
of the pipe by 0.8m is 417m
Vertical Alignment Design
Crest Vertical Curves



Minimum length of the vertical curve (L) is determined by sight distance
(SD) requirements
That length is generally are satisfactory from the standpoint of safety,
comfort, and appearance.
Derivation is done for the two cases of:
 SD > L
 SD < L
SD > L
General
𝐿 = 2𝑆 −
H1=1.07
H2=0.15
SD < L
200
𝐻1 +
𝐴
404
𝐿 = 2𝑆 −
𝐴
𝐻2
2
𝐴𝑆 2
𝐿 =
200
𝐻1 +
85
𝐴𝑆 2
𝐿 =
404
𝐻2
2
Vertical Alignment Design
Crest Vertical Curves
Minimum length when S>L





Vehicle on the grade at C
H1 height of the driver's eye at C
H2 height of an object at D
PN is line of sight, and
S is the sight distance
Note: that the line of sight is not
necessarily horizontal, but in
calculating the sight distance, the
horizontal projection is
considered
Vertical Alignment Design
Crest Vertical Curves
Minimum length when S>L
 From the properties of the parabola,
X3 = L/2
 The sight distance S is then given as
S = X1 + L/2+ X2
X1 and X2 can be found in terms of the grades G1 and G2 and their algebraic difference A.
The minimum length of the vertical curve for the required sight distance is obtained as
where,
L = length of vertical curve, m
𝐿 = 2𝑆 −
S = Sight distance, m
A = algebraic difference in grades, %
H1 = height of eye above roadway surface, m
H2 = height of object above roadway surface, m
200
𝐻1 +
𝐴
𝐻2
2
Vertical Alignment Design
Crest Vertical Curves
minimum length when S>L
 When the height of eye and the height of object are 1070 mm
and 150 mm, respectively, as used for stopping sight distance,
the length of the vertical curve is,
𝐿 = 2𝑆 −
200
1.07 +
𝐴
404
𝐿 = 2𝑆 −
𝐴
0.15
2
Vertical Alignment Design
Crest Vertical Curves
minimum length when S<L
Similarly:
𝐴𝑆 2
𝐿 =
200
𝐻1 +
𝐻2
2
AASHTO recommends H1=1070mm and H2=150mm
Substituting 1070 mm for H1 and 150 mm for H2 gives
Summary for S=SSD:
 Case i: S>L ,use
𝐿 = 2𝑆 −
404
𝐴
 Case ii: S<L ,use
𝐴𝑆 2
𝐿 =
404
𝐴𝑆 2
𝐿 =
404
Vertical Alignment Design
90
Example 13:
Determine the minimum length of a crest vertical curve between a +0.5% grade and a 1.0% grade for a road with a 100-km/h design speed. The vertical curve must provide
190-m stopping sight distance.
Solution:
Stopping sight distance criterion:
 Assume S<L thus use
Since A=(0.5-(-1)=1.5 and S=190
𝐴𝑆 2
1.5 ∗ 1902
𝐿 =
,
𝐿 =
= 134𝑚
404
404
Check your assumption if it is true. In this case S=190m < L=134m …Not Ok!
 Use the other equation for case S>L use
𝐿 = 2𝑆 −
404
,
𝐴
𝐿 = 2 ∗ 190 −
404
1.5
= 110.7𝑚
Check your assumption if it is true. S=190m > L=110.7m ………Ok!
Thus the minimum length of crest curve to fulfil SSD on this curve is 110.7m
Vertical Alignment Design
91
Design for Passing sight distance(PSD)

Differ from those for stopping sight distance because of the
different height criterion (i.e. 1300 mm height of object) results in
the following specific formulas with the same terms as above:
PSD > L
When
General
𝐿 = 2𝑆 −
H1=1.07
H2=1.3
Where S=PSD
PSD < L
200
𝐻1 +
𝐴
946
𝐿 = 2𝑆 −
𝐴
𝐻2
2
𝐴𝑆 2
𝐿 =
200
𝐻1 +
𝐴𝑆 2
𝐿 =
946
𝐻2
2
Vertical Alignment Design
Example 14:
Determine the minimum length of a crest vertical curve between a +0.5% grade and a
-1.0% grade for a road with a 100-km/h design speed. The vertical curve must
provide 675-m passing sight distance.
Solution:
Passing sight distance criterion: A=(0.5-(-1))=1.5 and PSD=S=675 m
Assume S<L thus use;
𝐴𝑆 2
𝐿 =
946
1.5 ∗ 6752
𝐿 =
= 722.5𝑚
946
92
Check your assumption if it is true. In this case S=675m < L= 722.5....
Ok!
Thus a minimum length of 722.5m of crest curve to fulfil PSD requirement on this curve.
Vertical Alignment Design
93
Design of Crest curve using K-values
• The horizontal distance required to effect a 1% change in the slope is called K-value.
For crest curve, Table 9-1 of ERA(2013) presented the recommended minimum
values of K-values for SSD and PSD requirements of different design speed.
𝐿𝑆𝑆𝐷
𝐴 ∗ 𝑆𝑆𝐷2
=
404
𝐾𝑆𝑆𝐷
𝑆𝑆𝐷2
=
404
𝑳𝑷𝑺𝑫
𝑨 ∗ 𝑷𝑺𝑫𝟐
=
𝟗𝟒𝟔
𝑲𝑷𝑺𝑫
𝑷𝑺𝑫𝟐
=
𝟗𝟒𝟔
Vertical Alignment Design
94
Design of Crest curve using K-values…
Lmin = Kmin*A
Where,
Lmin=the minimum length of crest curve
Kmin = the horizontal distance required to effect a 1% change in the slope
A = absolute value of the difference in grades |G1 ‐ G2| , expressed as a %.
Example 15:
Determine the minimum length of a crest vertical curve between a +3.5% grade and a -5.0%
grade for a road with a 70-km/h design speed. The vertical curve must provide adequate
stopping sight distance, hence the minimum K-value for this speed for SSD requirement is 30.
Solution:
A=(3.5-(-5)=8.5
Lmin = Kmin*A=30*8.5=255m
Vertical Alignment Design
Sag Vertical Curves
Design Criteria:
1. Headlight sight distance
2. Driver Comfort
3. Aesthetics (rule of thumb)
4. Drainage Control
Fig. Sight distance at undercrossing
95
Vertical Alignment Design
Headlight Sight Distance, S

Height of the headlight, h1= 600mm

Upward divergence of the light beam,β = 1o

(The upward spread of the light beam provides some additional
visible length, but that is generally ignored.)
Vertical Alignment Design
97
Length of curve with adequate SD
When
SD > L
SD < L
General
𝟐𝟎𝟎(𝑯𝟏 + 𝑺 𝒕𝒂𝒏 𝜷)
𝑳 = 𝟐𝑺 −
𝑨
𝑨𝑺𝟐
𝑳 =
𝟐𝟎𝟎(𝑯𝟏 + 𝑺 𝒕𝒂𝒏 𝜷)
𝜷=1.0
H1=0.6
𝟏𝟐𝟎 + 𝟑. 𝟓𝑺
𝐿 = 𝟐𝑺 −
𝑨
𝑨𝑺𝟐
𝐿 =
𝟏𝟐𝟎 + 𝟑. 𝟓𝑺
Where,



L=length of curve (m),
A=algebraic difference in
grade (%), and
S=Sight distance (m)
Vertical Alignment Design
Length of Curve for comfort



Considers that both the gravitational and centrifugal forces act in combination,
resulting in a greater effect than on a crest vertical curve
Comfort is affected by:

weight carried, body suspension of the vehicle, and tire flexibility

Measuring Comfort = Difficult!

Indicator = radial acceleration is not greater than 0.3 m/s2
The general expression for such a criterion is:
𝑨𝑽𝟐
𝑳 =
𝟑𝟗𝟓
V is the design speed, km/h.

98
Usually this length is about 50 percent of that required to satisfy the headlight sight
distance at various design speeds (for normal conditions).
Vertical Alignment Design
Minimum Length for Aesthetics
 Rule of thumb
𝐿min=30A
• Longer curves are
necessary for high
type of highways to
improve appearance
99
Vertical Alignment Design
Max. Length of Curve for drainage
Here the drainage criteria sets a limit on the MAXIMUM length of
curve!
 Long curves would have a relatively flat portion near the bottom
of curve. A minimum grade of 0.3% should be provided with in
15m of the level point of the curve
 Max length (drainage) is usually greater than min. length for
other criteria up to 100km/h and nearly equal to min length for
other criteria up to 120km/h
Vertical Alignment Design
Example 16:
Determine the minimum length of a sag vertical curve between a -0.7% grade and a +0.5% grade for a
road with a 110 km/h design speed. The vertical curve must provide 220 m stopping sight distance and
meet appearance criteria and comfort standard criteria. Round your answer to next 20m interval.
Solution
1) Stopping sight distance criterion:
i.
𝐴𝑆 2
1.2 ∗ 2202
𝐿 =
=
= 65.3 𝑚
120 + 3.5 ∗ 𝑆 120 + 3.5 ∗ 220
Assume S<L
Check your assumption if it is true. In this case S=220m<L=65.3 ….. Not Ok!
ii.
Use the other case equation S>L
𝐿 = 2𝑆 −
120 + 3.5𝑆
120 + 3.5 ∗ 220
= 2 ∗ 220 −
= −301𝑚 ≅ 0 𝑚
𝐴
1.2
2) 𝑳𝒎𝒊𝒏 for comfort
𝐿min
3) 𝑳𝒎𝒊𝒏 for Aesthetics
𝐴𝑉 2 1.2 ∗ 1102
=
=
= 36.8 = 36.8 ≈ 40 𝑚
395
395
𝐿min = 30A=30∗1.2 = 36.0 ≈ 40 𝑚
Hence use L=40m
Vertical Alignment Design
Design of Sag curves with K-values
𝑳𝒎𝒊𝒏 = 𝑲𝒎𝒊𝒏 ∗ 𝑨
Table 9.3 Minimum values of sag curves (ERA, 2013)
Where,
Lmin=the minimum length of sag curve
Kmin = the horizontal distance required to effe
ct a 1% change in the slope
A = absolute value of the difference in grades
𝑮𝟏 − 𝑮𝟐 , expressed as a %.
Design Speed (km/h)
20
25
30
40
50
60
70
80
85
90
100
110
120
K for driver comfort
1
1.5
2.5
4
6.5
9
12
16
18
20
25
30
36
Vertical Alignment Design
Underpass Sight Distance and Sag Vertical Curve Design
AASHTO recommends,
 H1 = 2.40 m for a truck driver
 H2= 0.60 m for object height (taillights of a vehicle)
Vertical Alignment Design
Underpass Sight Distance and Sag Vertical Curve Design …
When
SD > L
General
𝑳 = 𝟐𝑺𝑺𝑫 −
H1=2.4
H1=0.6
SD < L
𝟖𝟎𝟎(𝑯𝑪
𝑳 = 𝟐𝑺𝑺𝑫 −
𝑯 +𝑯
− 𝟏 𝟐 𝟐)
𝑨
𝟖𝟎𝟎(𝑯𝑪 − 𝟏. 𝟓)
𝑨
𝑨 ∗ 𝑺𝑺𝑫𝟐
𝑳 =
𝑯 + 𝑯𝟐
𝟖𝟎𝟎(𝑯𝑪 − 𝟏
)
𝟐
𝑨 ∗ 𝑺𝑺𝑫𝟐
𝑳 =
𝟖𝟎𝟎(𝑯𝑪 − 𝟏. 𝟓)
Where,
L=minimum length of sag curve required to fulfil required sight distance, m
Hc = clearance height of overpass structure above roadway in meter
SSD=Stopping sight distance required for the design speed considered
104
A= 𝑮𝟏 − 𝑮𝟐 in %
Vertical Alignment Design
Example 17:
An equal tangent sag curve has an initial grade of ‐4.0%, a final grade of +3.0
%, and a length of 387.096 m. an overhead guide sign is being placed directly
over the PVI of this curve.
The stopping sight distance required for the design speed is 285.9m.
At what height above the roadway should the bottom of this sign be placed.
Solution:
The necessary clearance must be provided based on SSD
Since S=285.9<L=387.96
𝐴 ∗ 𝑆𝑆𝐷2
𝐿=
800(𝐻𝐶 − 1.5)
7 ∗ 285. 62
387.96 =
800(𝐻𝐶 − 1.5)
Clearance Hc = 7(285.9)2/800(387.096) + 1.5 = 3.35 m

The bottom of this sign be placed at 3.35m above the roadway in order not obstruct
truck driver at height of 2.4m. (usually 4-5m is used)
Vertical Alignment Design
106





Especially for trunk and link roads, where the algebraic
difference between successive gradients is often small,
the intervening minimum vertical curve, obtained by
applying the above formulae, becomes very short.
This can create the impression of a kink in the grade
line.
If the vertical alignment is allowed to contain many
curves of short length, the result can be a ‘hidden dip’
profile, and/or a ‘roller coaster’ type profile, as
indicated in Figure 9.3.
For this reason, where the algebraic difference in
gradient is less than 0.5 percent, a minimum curve
length is recommended for purely aesthetic reasons.
The minimum length should not be less than twice the
design speed in km/h and, for preference, should be
400 metres or longer, except in mountainous or
Figure 9-3: Hidden Dip and Roller Coaster Profiles
escarpment terrain.
Vertical Alignment Design
107
• Where a crest curve and a succeeding sag curve have a common
beginning and end, the visual effect created is that the road has suddenly
dropped away.
• In the reverse case, the illusion of a hump is created.
Either effect is removed by inserting
a short length of straight grade
between the two curves.
Typically, 60 m to 100 m is adequate
for this purpose
Figure :Typical roller coaster alignment.
Phasing of Alignment
Combination of Horizontal and Vertical Alignments
 Horizontal and Vertical Alignments should not be designed independently and should be
considered together
 Correcting alignment deficiencies is extremely difficult and costly!
 Phasing of the vertical and horizontal curves of a road implies their coordination so that
the line of the road appears to a driver to flow smoothly, avoiding the creation of hazards
and visual defects.
 It is particularly important in the design of high-speed roads on which a driver must be
able to anticipate changes in both horizontal and vertical alignment well within the safe
stopping distance.
 It becomes more important with small radius curves than with large.
 Defects may arise if an alignment is mis-phased.
 Defects may be purely visual and do no more than present the driver with an aesthetically
displeasing impression of the road.
 Such defects often occur on sag curves.
 When these defects are severe, they may create a psychological obstacle and cause some
drivers to reduce speed unnecessarily.
 In other cases, the defects may endanger the safety of the user by concealing hazards on
the road ahead.
 A sharp bend hidden by a crest curve is an example of this kind of defect.
Phasing of Alignment
Possible solutions to Mis-phasing
 This refers to the coordination of HA & VA so that the line of the
road appears to a driver to flow smoothly, avoiding the creation of
hazards and visual defects.
 Is particularly important in the design of high-speed roads on
which a driver must be able to anticipate changes in both HA & VA
well with in the SSD and on curves with small radius.
 When the horizontal and vertical curves are adequately separated or
when they are coincident, no phasing problem occurs and no
corrective action is required.
 Where defects occur, phasing may be achieved either by separating
the curves or by adjusting their lengths such that vertical and
horizontal curves begin at a common station and end at a common
station.
 In some cases, depending on the curvature, it is sufficient if only
one end of each of the curves is at a common station.
Phasing of Alignment
Types of Mis-phasing and Corresponding Corrective Action
 Vertical curve overlaps one end of the horizontal curve


Insufficient Separation between the Curves


Corrective action consists of increasing the separation between the curves,
or making the curves concurrent
Both ends of the Vertical curve lie on the Horizontal curve


The defect may be corrected in both cases by completely separating the
curves
The corrective action is to make both ends of the curves coincident, or to
separate them.
Vertical crest curve overlaps both ends of the Horizontal Curve

The corrective action is to make both ends of the curves coincident.
Phasing of Alignment
111
Phasing of Alignment
112
Phasing of Alignment
113
GOOD
POOR
Phasing of Alignment
114
Figure 10-1: Phasing of Horizontal and Vertical Curves
Phasing of Alignment
The Economic Penalty Due to Phasing
The correct phasing of vertical curves restricts the designer in fitting
the road to the topography at the lowest cost.
Therefore, phasing is usually bought at the cost of extra earthworks
and the designer must decide at what point it becomes uneconomic.
He will normally accept curves that have to be phased for reasons of
safety.
In cases when the advantage due to phasing is aesthetic, the designer
will have to balance the costs of trial alignments against their
elegance.
Road Furniture Design
116



Road furniture means a fixture or structure on the road or
within the road reserve intended to provide information or
safety to a road user and includes a traffic light, sign post,
traffic sign, guard rail, fence, marker post, reflector or
center-line pad.
You have learned traffic controls play a major role for the
safe and efficient operations and communications of the
traffic in your Transport Engineering course.
Traffic controls such as lane markings, signs and posts,
signals, safety barriers and rumble strips (humps) might
need to be applied at different locations on the highway.
Road Furniture Design
117




Now it is the time to select the appropriate traffic control on each
highway segment or spot. For example if you are forced to design a
sharp curve with a circular radius less than the minimum required
for the design speed.
What controls would you suggest that need to be located at spots a
head of this curve?
Posting Warning sign would be adequate?
However, it depends on the sharpness of the curve, you may opt for
rumple strips or humps to effectively enforce speed change (as a
speed brake) and hence safe movement on the curve.
Road Furniture Design
118



Starting from traffic lane marking details to complex controls such
as signal design for junction need to be carried out.
At the segment of high fill, you should specify the type of barrier
to be implemented;
Obviously, regulatory, warning and information posts also required
to be posted frequently such as speed limit, land use signs, KM a
head & Direction to Towns, KM posts, clearance posts, Axle load
limits. (Reading Assignment: Read ERA(2013 Manual on this
topic)
Road Furniture Design
119
Road Furniture Design
References
1.
Fred L. (2013), Principles of Highway Engineering and Traffic
Analysis, 5th edition
2.
ERA (2013), Geometric Design Manual
3.
Daniel J. (2016), Highway Engineering: Planning, Design, and
Operations, 1st edition
Thank You !
120