Geometric Design of Highways
Highway Alignment is a three-dimensional
problem
– Design & Construction would be difficult in 3-D so
highway design is split into 2-D problems
– Horizontal alignment and vertical alignment
Geometric Design of Highways
Geometric Design of Highways
Geometric Design of Highways
Components of Highway Design
• Horizontal Alignment - Plan View
• Vertical Alignment - Profile View
Horizontal Alignment
Tangents
Curves
Tangents and Curves
Tangents and Curves
Tangent
Curve
Tangent to
Circular Curve
Tangent to Spiral
Curve
Tangents and Curves
Tangents and Curves
Types of Curves
What is a Horizontal Curve?
• Provides a transition
between two tangent
lengths of roadway.
• Necessary for
gradual change in
direction when a
direct point of
intersection is not
feasible.
Curves
•The purpose of the curves is to deflect a vehicle
traveling along one of the straights safely and
comfortably through a deflection angle θ to
enable it to continue its journey along the other
straight.
Guidelines to Horizontal Curves
•Horizontal Alignment
Considerations
– Radius
– Design Speed
– Side Friction Factor
– Superelevation
Considerations for Horizontal Curves
•Safety
•Economically Practical
Types of Circular Curve
Simple Curve
Types of Circular Curve
Compound Curve
Types of Circular Curve
Reverse Curve
Layout of a Simple Curve
Elements of Simple Curve
R = Radius of Curve
PC = T1 = BC = Beginning of Curve (PC = Point of Curvature)
PT = T2 = EC = End of Curve (PT = Point of Tangency)
PI = V = Point of Intersection
T = Tangent Length (T = PI – BC = EC - PI)
Lc = Length of Curvature (Lc = EC – BC)
M = Middle Ordinate
E = External Distance
C = Chord Length (L)
Δ or I = Intersection or Central Angle
Definitions
1. Back tangent or First Tangent - (AT1) Previous to the curve
2. Forward Tangent or Second tangent - (BT2)- Following the
curve.
3. Point of Intersection ( PI) or Vertex (v) - If the tangents AT1
and BT2 are produced they will meet at a point called the
point of Intersection
4. Point of curve ( PC) – Beginning Point T1 of a curve.
Alignment changes from a tangent to curve.
5. Point of Tangency (PT) – End point of curve ( T2) is called..
6. Intersection Angle (I) - The angle between the back tangent
and forward tangent is called...
7. Deflection Angle (Δ) - The angle at P.I. between back
tangent and forward tangent is called..
8. Tangent Distance – It is the distance between P.C. and P.I.
Definitions
9. External Distance (E) - The distance from the mid point of
the curve to PI. It is also called the apex distance.
10. Length of curve (Lc) - It is the total length of curve from PC
to PT
11. Long chord (L)(C) – It is the chord joining PC to PT, T1, T2 is
a long chord.
12. Normal Chord - A chord between two successive regular
station on a curve is called normal chord. Normally , the
length of normal chord is 1 chain ( 20m).
13. Mid Ordinate (M) - The distance between mid point of long
chord and the apex, is called ...
14. Right hand curve - If the curve deflects to the right of the
direction of the progress of survey.
15. Left hand curve - If the curve deflects to the left of the
direction of the progress of survey.
Designation of Curve
The sharpness of the curve is designated by two
ways.
a. Radius ( R) - Curve is known by the length of its
radius
b. Degree of Curvature ( D ) - an angle subtended
by an arc whose length is 1 station.
c. Station - a linear distance of 20 m or 100 ft.
along some described alignment.
d. Deflection angle (α or ϴ) - half of the internal
angle
Designation of Curve
• Degree of Curvature (D)
– Chord definition (Chord Basis) - the angle
subtended at the centre of curve by a chord of
20m/30m is called the degree of curvature. If an
angle subtended at the centre of curve by a chord
of 20m is 5o, the curve is called 5o curve.
– Arc definition (Arc Basis) - the angle subtended
at the centre of curve by an arc of 20m/30m length
is called the degree of curve. Used in America,
Canada, India etc.
Relation between Radius and degree of curve
•By chord definition
Relation between Radius and degree of curve
• By Arc Definition
Elements of a Simple Curve
Length of curve (L)
•If curve is designated by Radius:
Lc = R(I) (When I is in Radian)
Lc = R(I)/180 (When I is in degree)
•If curve is designated by degree:
Length of arc = 20 mt.
Lc = 20(I)/D
Formulae for Horizontal Circular Curves
I
T  R tan
2
1
ER(
 1)
I
cos
2
I
M  R (1  Cos )
2
I
C  2 R Sin
2
I
L C  R ( I ) ; L C  2 R (
)
360
Setting out of a Simple Curve
• Normal chord and Sub chord:
–
–
–
–
–
–
For alignment pegs are driven.
The distance between two pegs is normally 20m
Peg station are called main stations.
The chord joining the tangents point T1 and the first
Main peg station is called First sub chord.
All the chord joining adjacent peg stations are called full
chord or normal chord.
– The length of normal chord is 20 m.
– The point joining last main peg station and tangent T2 is
called last sub chord.
Methods of Curve Setting Out
There are a number of different methods by which a centerline
can be set out, all of which can be summarized in two categories:
•Traditional methods: which involve working along the
centerline itself using the straights, intersection points and
tangent points for reference.
– The equipment used for these methods include, tapes and
theodolites or total stations.
•Coordinate methods: which use control networks as reference.
These networks take the form of control points located on site
some distance away from the centerline.
– For this method, theodolites, totals stations or GPS
receivers can be used.
Methods of Curve Setting Out
The methods for setting out curves may be divided into 2 classes
according to the instrument employed .
1.Linear or Chain & Tape Method
2.Angular or Instrumental Method
Peg Interval:
Usual Practice - Fix pegs at equal interval on the curve
20 m to 30 m ( 100 feet or one chain)
66 feet ( Gunter’s Chain)
Strictly speaking this interval must be measured as the Arc intercept
b/w them, however it is necessarily measured along the chord. The
curve consist of a series of chords rather than arcs. Along the arc it is
practically not possible that is why it is measured along the chord.
Methods of Setting out of Simple Curve
1. Linear Methods
a. By offsets or ordinate from the long chord.
b. By successive bisection of arcs or chords.
c. By offsets from the tangents.
d. By offsets from the chord produced.
Methods of Setting out of Simple Curve
2. Angular Methods
a. Rankine method of tangential angles (Tape and
Theodolite method
b. Two theodolite method
c. Tacheometric method
SAMPLE PROBLEM 1
Simple Curve
1. A simple curve has a radius of 286.48 m. It's
distance from PC to PT along the curve is equal to
240 m.
a. Compute the central angle of the curve. Use arc basis.
Lc
20

I
D
1145 . 916
1145 . 916
D
; D
; D  4o
R
286 . 48
240 20
 ; I  48o
I
4
SAMPLE PROBLEM 1
Simple Curve
b. Compute the distance from the mid-point of the long chord to the midpoint of the curve.
I 

M  R  1  cos  ; M  286 . 48 1  cos 24 o
2

M  24 . 76 m


SAMPLE PROBLEM 1
Simple Curve
c. Compute the area bounded by the tangents and the portion
outside the central curve acres.
T  R tan 24 o
T  286 . 48 tan 24 o
T  127 . 55
TR  2   R 2 I
Area 

2
360
127 . 55  286 . 48 ( 2 )  ( 286 . 48 ) 2 ( 48 )
Area 

2
360
Area  2162 . 8
SAMPLE PROBLEM 2
Simple Curve
1.
The offset distance of the simple curve from the PT to the tangent
line passing through the PC is equal to 120.20 m. The stationing of
PC is at 2+540.26. The simple curve has an angle of intersection of
50o.
a.
Compute the degree of curve.
120.20
Sin 50 
; T 156.91m
T
T  R tan 24 o ; T  R tan 25 o
o
156 . 91  R tan 25 o ; R  336 . 49 m
1145 . 916
D
; D  3 o 24 o
336 . 49
SAMPLE PROBLEM 2
Simple Curve
b. Compute the external distance.
I 

E  R  sec  1 ; E  336 .49 sec 25 o  1
2 

E  34 .79 m.

c. Compute the length of long chord.
L
 R Sin 25o ; L  2 (336.49) Sin 25o
2
L  284.41m.

SAMPLE PROBLEM 3
Simple Curve
1.
Two tangents making an angle of 62o from each other is connected
by a simple curve. A point “X” on the curve is located by a distance
along the tangent line from the PC equal to 240 m, and an offset
from the tangent equal to 60 m. The PC is at station 10+080.
a.
Compute the radius of the curve
60
tan  ;  14.04o ; 2  28.08o
40
R  60  R Cos 28.08o
0.1177R  60; R  509.70 m
SAMPLE PROBLEM 3
Simple Curve
b. Compute the tangent distance of the curve.
T  R tan 31 o ; T  509 . 70 tan 31 o
T  306 . 26 m
c. Compute the stationing of point “X” on the curve.
509.70 (28.08) 
S  R ; S  R 
180
S  249.80
Sta. of X  (10  080)  249.80
Sta. of X 10  329.80
CURVES
Compound Curves
A Compound Curve consist of 2 arcs of different radii bending
in the same direction and lying on the same side of their
common tangent. Then the center being on the same side of the
curve.
PCC = Point of Compound Curvature
RL = R1 = larger radius
RS = R2 = smaller radius
TL = T1 = larger tangent
TS = T2 = smaller tangent
CURVES
Compound Curves
A Compound Curve consist of 2 arcs of different radii bending
in the same direction and lying on the same side of their
common tangent. Then the center being on the same side of the
curve.
BD = T1 + T2 = common tangent
I1 = central angle of curve AE
I2 = central angle of curve EF
I = angle of intersection of tangents
AC and CF
I  I1  I 2
I
T1  R1 tan ( 1 )
2
I2
T2  R2 tan ( )
2
SAMPLE PROBLEM 1
Compound Curve
1.
The common tangent AB of a compound
curve is 76.42 m with an azimuth of
268o30’. The point of intersection PI or the
vertex is inaccessible. The azimuths of the
tangents T1 and T2 were measured to be
247o50’ and 282o50’ respectively. If the
stationing of A is 43+010.46 and the
degree of the first curve was fixed at 4o,
determine the stationing of PC, PCC, and
PT.
SAMPLE PROBLEM 1
Compound Curve
Stationing of the PC
I1  268 o 30 '  247 o 50 ' ; I1  20 o 40 '
I 2  282 o 50 '  268 o 30 ' ; I 2  14 o 20 '
D  4o
Sin
D1 10
10
 ; Sin 2 o  ; R1  286 .56
2 R1
R1
I1
T1  R1 tan ; T1  286 .56 tan 10 o 20 ' ; T1  52 .25 m
2
PC  43  010 .46   52 .25
PC  42  958 .21
SAMPLE PROBLEM 1
Compound Curve
Stationing of the PCC
T1  T2  76 .42
T2  76 .42  52 .25; T2  24 .17
I2
T2  R 2 tan ; 24 .17  R 2 tan 7 o10 ' ; R 2  192 .233 m
2
D 2 10
D2
10
D2
Sin

; Sin

;
 2 o 59 '
2 R2
2 192 .23 2
D 2  5 o 58 '
I 1 ( 20 )
20 o 40 ' ( 20 )
20 .667 ( 20 )
LC 1 
; LC 1 
; LC 1 
D1
4
4
LC 1  103 .34
PCC  ( 42  958 .21)  (103 .34 )
PCC  43  061 .55
SAMPLE PROBLEM 1
Compound Curve
Stationing of the PT
I 2 ( 20)
14 o 40' ( 20)
14.33( 20)
LC 2 
; LC 2 
;
L

C2
D2
5o 58'
5.966
LC 2  48.10
PT  ( 43  061 .55)  ( 48.10)
PT  43  109 .65
SAMPLE PROBLEM 2
Compound Curve
1. Given a compound curve with
a long chord equal to 135 m
forming an angle of 12o and
18o respectively with the
tangents. The common
tangent is parallel to the long
chord. Determine the radii of
the compound curve.
SAMPLE PROBLEM 2
Compound Curve
AO
135

; AO  81 .60
o
o
Sin 9
Sin 165
BO
135

; BO  54 .52
o
o
Sin 6
Sin 165
40 .8
Sin 6 
; R1  390 .32 m
R1
o
Sin 9 o 
27 .26
; R1  174 .26 m
R2
SAMPLE PROBLEM 3
Compound Curve
1.
A turnaround pattern which fits with
the topography is provided in a
highway by connecting four
tangents with a compound curve
consisting of three simple curves.
The azimuths of AB = 220o15', BC
= 264o30', CD = 320o24' and DE =
32o58'. The radius of the last curve
is four times sharper than the first
curve. The distance BC = 303 m
and CD = 200 m.
SAMPLE PROBLEM 3
Compound Curve
a.
Compute the radius of the 3rd curve
I1  264 o 30 '  220 o15' ; I1  44 o15'
I 2  320 o 24 '  264 o 30 ' ; I 2  55 o 54 '
I 3  360 o 24 '  320 o 24 '  32 o 58' ; I 3  72 o 34 '
T1  T2  303
R1 tan 22 o 7.5' R2 tan 27 o 57 '  303
0.407 R1  0.530 R2  303
T2  T3  200
R2 tan 27 o 57 ' R3 tan 36 o17 '  200
0.530 R2  0.734 R3  300
R1  4 R3
0 .407 ( 4 R3 )  0 .530 R 2  303
1 .628 R3  0 .530 R 2  303
0 .734 R3  0 .530 R 2  200
0 .894 R3  103
R3  115 .21
SAMPLE PROBLEM 3
Compound Curve
b.
Compute the radius of the 2nd curve
R1  4 ( 115 .21) ; R1  460 .84 m
0 .407 R1  0 .530 R2  303
0 .407 ( 460 .84 )  0 .530 R2  303
R2  217 .81 m
SAMPLE PROBLEM 3
Compound Curve
c. If PC is at 12+152.60, what is the stationing of the PT
R1 I1
460 .84( 44 o15' )
LC1 
; LC1 
180
180 o
LC1  355 .91 m
R2 I 2
217 .81(55 o 54' )
LC 2 
; LC 2 
180
180 o
LC 2  212 .50 m
R3 I 3
115 .21(72 o 34' )
LC 3 
; LC 3 
180
180 o
LC 2  145 .92 m
Sta. of PT  (12  152 .60)  355 .91  212 .50  145 .92
Sta. of PT  12  654 .43
CURVES
Reverse Curves
A Reverse Curve is formed by two circular simple curves having a
common tangent but lies on opposite sides. The method of laying
out a reverse curve is just the same as the deflection angle
method of laying out simple curves. At the point where the curve
reversed in its direction is called the Point of Reversed Curvature.
CURVES
Elements of a Reversed Curve
PRC = Point of Reversed Curvature
R1 and R2 = radii of curvature
V1 and V2 = points of intersection of tangents
D1 and D2 = degree of curve
I2 - I1 = ϴ
PC = Point of Curvature
PT = Point of Tangency
PRC = Point of reversed curvature
LC = LC1 + LC2 = length of reversed curve
m = offset
P = distance between parallel tangents
CURVES
4 Types of Reversed Curve Problems
1. Reversed curve with equal radii and
parallel tangents.
2. Reversed curve with unequal radii and
parallel tangents.
3. Reversed curve with radii and converging
tangents.
4. Reversed curve with unequal radii and
converging tangents.
CURVES
Reversed Curve Sample Problem 1
1. In a railroad layout, the centerline of two parallel tracks are connected
with a reversed curve of unequal radii. The central angle of the first
curve is 16o and the distance between parallel tracks is 27.60 m
stationing of the PC is 15 + 420 and the radius of the second curve is
290 m.
a. Compute the length of the
long chord from PC to PT.
27 .60
Sin 8 
; L  198 .31
L
o
CURVES
Reversed Curve Sample Problem 1
b. Compute the radius of the first curve.
a  R1  R1Cos16o ; a  R1 (1  Cos16o )
b  R2  R2Cos16o ; b  R2 (1  Cos16o )
a  b  27.60
R1 (1  Cos16o )  R2 (1  Cos16o )  27.60
( R1  R2 )(1  Cos16o )  27.60
R1  R2  712.47; R1  712.47  290
R1  422.47 m
CURVES
Reversed Curve Sample Problem 1
c. Compute the stationing of the PT
422 .47 (16)
LC1 
; LC1  117 .98
180
290 (16)
LC 2 
; LC 2  80.98
180
Sta. of PRC  (15  420 )  117 .98
Sta. of PRC  15  537 .98
Sta. of PT  (15  537 .98)  80.98
Sta. of PT  15  618 .96
CURVES
Reversed Curve Sample Problem 2
2. Two tangents converge at an angle of 30O. The direction of the
second tangent is due east. The distance of the PC from the second
tangent is 116.50 m. The bearing of the common tangent is S 40O E.
a. Compute the central angle of
the first curve.
I 1  50 O  30 O
I 1  20 O
CURVES
Reversed Curve Sample Problem 2
b. If a reverse curve is to connect these two tangents, determine the
common radius of the curve.
a  R cos 30 O  R cos 50 O
a  R (cos 30 O  cos 50 O )
a  0 .223 R
b  R  R cos 50 O
b  R (1  cos 50 O )
b  0 .357 R
a  b  116 .50
0 .223 R  0 .357 R  116 .50
R  200 .86 m.
CURVES
Reversed Curve Sample Problem 2
c. Compute the stationing of the PT if station of the PC is 10 + 620.
R I 200 .86 ( 20)
L1 

o
180
180 o
L1  70.11 m
R I 200 .86 (50)

o
180
180 o
L1  175 .28 m
L1 
Sta. of PT  (10  620 )  70.11  175 .28
Sta. of PT  10  865 .39
CURVES
Reversed Curve Sample Problem 3
3. A perpendicular distance between
two parallel tangents of a reverse
curve is 8m and the chord distance
from the PC to the PT is equal to
30 m. Compute the central angle of

the reverse curve.
8
30
  15 . 47 o
Sin  
I  2  2 (15 . 47 o )  30 o 56 '
CURVES
Spiral Curves
A Spiral Curve is used to provide a gradual transition from a straight line or
tangent to the full curvature of a circular curve. It begins very flat with a radius of
infinity and increases in sharpness as the circular curve is approached along the
alignment.
When the circular curve is reached, the spiral curve will have a degree of
curvature equal to that of the circular curve. Although spirals are not used
universally for highway curves, they are used extensively by railroads.
CURVES
Spiral Curves
CURVES
Elements of Spiral Curves
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
L = Length of spiral from TS to any point
along the spiral
Ls = Length of spiral
PI = Point of intersection
I = Angle of intersection
Ic = Angle of intersection of the simple
curve
CURVES
Elements of Spiral Curves
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
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
CURVES
Formulas of Spiral Curves
L5
Distance along tangent to any point on the spiral: Y  L 
2
3
40 R 2 LS
LS
At L = Ls, Y = Yc, thus, YC  LS 
40R 2
L3
Offset distance from tangent to any point on the spiral: X 
6 RLS
2
LS
At L = Ls, X = Xc, thus, X C 
6R
CURVES
Formulas of Spiral Curves
2
LS
1
Length of throw: p  X C 
4
24 R
L2
Spiral angle from tangent to any point on the spiral (in radian):  
2 RLS
L
At L = Ls, θ = θs, thus,  S  S
2R
1
L2
Deflection angle from TS to any point on the spiral: i   
3
6 RLS
L
1
At L = Ls, i = is, thus, i   S  S
3
6R
CURVES
Formulas of Spiral Curves
i
L2
This angle is proportional to the square of its distance
 2
iS LS
LS
I
 ( R  P ) tan
Tangent distance: TS 
2
2
Angle of intersection of simple curve: I C  I  2 S
RP
R
I
cos
2
D
L
Degree of spiral curve:

DC LS
External distance: ES 
0.036 K
LS 
R
3
0.0079 K
e
R
2
K = maximum speed of the car that could pass through the spiral without skidding
e = superelevation of the road
CURVES
Spiral Curve Sample Problem 1
1. Two tangents having azimuths of 240o
and 282o are connected by an 80 m
spiral curve with a 6o circular curve.
The width of the roadway is 10 m. If
the design velocity is 60 kph.

a. Determine the super-elevation at
quarter points.
1145.916 1145.916
R

D
6
R  190.99 m
0.0079 K 2 0.0079 (60) 2
e

R
190.99
e  0.149 m / m width of roadway
e4  0.149(10) ; e4  1.49
e1  1 (1.49) ; e1  0.3725
4
e2  1 (1.49) ; e2  0.745
2
e3  3 (1.49) ; e3  1.118
4
CURVES
Spiral Curve Sample Problem 1
b. Determine the deflection angle at the end point.
LC 180
80
180
S
; 
2 RC 
2 (190 .99 ) 
S  12 o
S
12
i ; i
3
3
i4
CURVES
Spiral Curve Sample Problem 1
b. Determine the external distance.
XC
I
) sec  R C
4
2
I  282  240
E S  ( RC 
I  42 o
2
LC
(80 ) 2
XC 
;
6 R C 6 (190 .99 )
X C  5 .58
E S  (190 .99 
E S  15 .08 m
5 .58
) sec 21 o  190 .99
4
CURVES
Spiral Curve Sample Problem 2
2. A spiral curve was laid out in a certain
portion of the Manila-Cavite Coastal
Road. It has a length of spiral of 80 m
and an angle of intersection of the two
tangents of 42O degrees. If the degree

of curve is 6O, determine the following
elements of the spiral curve to be laid
out:
a. Determine the length of the long
and short tangent
1145.916 1145.916
R

D
6
R  190.99 m
S
LC 180
80
180
; 
2 RC 
2 (190 .99 ) 
S  12 o
CURVES
Spiral Curve Sample Problem 2
a. Length of long and short tangent
3
LC
(80) 3
YC  LC 
; YC  80 
2
40(190.99) 2
40 RC
YC  79.65 m.
2
LC
(80)2
XC 
; XC 
6RC
6(190.99)
X C  5.58 m
X
5.58
tan S  C ; h 
h
tan12O
h  26.25 m.
Long tan gent ( LT )  YC  h
LT  79.65  26.25
LT  53.4 m.
Short tan gent ( ST );
XC
5.58
Sin S 
; ST 
ST
Sin 12O
ST  26.84 m
CURVES
Spiral Curve Sample Problem 2
b. Determine the external distance.
XC
I
) sec  R C
4
2
5 .58
E S  (190 .99 
) sec 21 o  190 .99
4
E S  15 .08 m
E S  ( RC 
c. Determine the length of throw
X C 5 .58
P
;
4
4
P  1 .395 m
CURVES
Spiral Curve Sample Problem 2
d. Determine the maximum velocity
3
0.036 K
0.036 K
LC 
; 80 
R
190.99
K  75.15 kph