Transition Curves-kenya-road-design-manual-for highway and expressway design

kenya-road-design-manual-for highway and expressway design

1. 1

Transition Curves
A transition curve is provided for the following advantages:
a) It allows a gradual transition of curvature from the tangent to the circular curve or
from the circular curve to the tangent.
b) The radius of curvature increases or decreases gradually.
c) Enables the gradual change in superelevation in a convenient manner.
d) It eliminates the danger of overturning or side-slipping of vehicles.
Requirements of a Transition Curve
1) It should originate tangentially from the straight (i.e., at T1 and T2), refer to Fig. 11.39.

Fig. 11.39 Elements of a transition curve
2) It should meet tangentially with the circular curve (i.e., at C and C′).
3) Its radius should be infinite at the origin (i.e., at T1 and T2) on the straight.
4) Its radius at the junction with the circular curve (i.e., at C and C′) should be the same as
that of the circular curve.
5) Its length should be such that full cant or superelevation is attained at the points C and C′.
6) Rate of increase of curvature along the transition should be same as that of increase of
cant/superelevation.
Types of Transition Curves
There are mainly three types of transition curves, namely:

2. 2

• Cubical spiral (clothoid)
• Cubic parabola
• The Lemniscate curve.
1. Cubical Spiral (Clothoid)
In it, the radius of curvature is inversely proportional to its distance, from the beginning of
the curve.

Fig. 17.1 Cubic spiral
The standard equation of a cubical spiral (Fig. 17.1) is given by
3
6
l
x
RL
=
where L = Total length of the transition curve
R = radius of the circular curve.
l = distance measured along the curve
x = perpendicular offset from the tangent.
2. Cubical Parabola

Fig. 17 Cubic parabola
The standard equation of a cubic parabola (Fig. 17.2) is given by:
3
6
y
x
RL
=
where x = perpendicular offset from the tangent
y = distance measured along the tangent
R = radius of the circular curve.

3. 3

L = length of the transition curve
3. Lemniscate Curve
The standard equation of a lemniscate curve (Fig. 17.3) is given by: 3sin 2
r


=
… (17.3)

Fig. 17.3 Lemniscate curve
where r = radius of the curvature
ρ = polar ray of any point
α = polar deflection angle i.e., angle between the polar ray and the straight.
Superelevation
Superelevation on curves is defined as the raising of the outer end of a road or the outer rail
over the inner one. C is a point on the outer edge and D is a point on the inner one (Fig. 11.40
(a)). The difference of elevation between C and D is called superelevation or cant at the point
C.

Fig. 11.40 Superelevation
When a vehicle moves on a curved path, two forces act on it: weight W of the vehicle and the
centrifugal force P.

4. 4

From Fig. 11.40 (b):
h = superelevation (in m)
B = width of pavement (in m)
θ = angle of superelevation W = weight of the vehicle P = centrifugal force
v = speed of the vehicle (in m/s)
g = acceleration due to gravity = 9.81 m/s
2
R = radius of the curve (in m)
W = weight of the vehicle
P = centrifugal force acting on the vehicle centrifugal force
Centrifugal ratio
weight of the vehicle
=

or 2
Wv
P
gR
=
or 2
Pv
W gR
=
From Fig. 11.40(b),
tan
P
W
=
Also, tan
h
B
=
Hence, 2
h P v
B W gR
== 2
hv
B gR
=

or 2
Bv
h
gR
=
In the railways, the width of the track is taken as the distance between the two rails and is
represented by a gauge G (in m). However, if speed of the vehicle in kilometres per hour, and
G and R in metres, then the cant h in centimetres is given by
2
cm
1.27
Gv
h
R
=

5. 5

Length of a Transition Curve
The following are methods for calculating the length of transition curves:
1. By rate of superelevation
If h is the amount of superelevation in centimetres 1 in n is the rate of superelevation over the
transition curve L is the length of the transition curve in metres.
then metres
100
nh
L= ... (17.9)
Example
Calculate the length of a transition curve to be introduced between a straight and a curve such
that 15 cm superelevation may be introduced over the circular curve. Assume the rate of
superelevation as 1 in 500. metres
100
nh
L=

Here n = 500, h = 15 cm
∴ 500 15
= 75 m
100
L

=
2. By time rate
The length of the transition curve may be calculated by assuming a suitable time rate say ‘‘x’’
cm per second.
Let v be the average speed of vehicles in m / sec
h be the amount of superelevation in centimetres
x be the time rate, where x varies from 2.5 to 5.0 cm per second
L be the length of the transition curve in metres.
Time taken by the vehicle to travel a distance sec
L
L
v
=
Superelevation = time rate × time taken by the vehicle i.e.
L
hx
v
=
or hv
L
x
= ... (17.10)
Example
Calculate the length of a transition curve to be inserted between a straight and a circular curve
such that a superelevation of 15 cm over a circular curve may be attained. Assume the rate of
attaining superelevation as 2.5 cm per second and average speed of the vehicles as 60 km / h.

6. 6
hv
L
x
=

Here h = 15 cm, x = 2.5 cm/sec.
60 1000 100
m/s
60 60 6
v

==

∴ 15 100
100 m
2.5 6
L

==

3. By rate of change of radial acceleration
Let v be the average speed of the vehicles in metres/sec
R be the radius of the circular curve in metres
2
v
R be radial acceleration.
L be the length of the transition curve in metres
‘a’ is the rate of development of radial acceleration.
Time taken to travel length L = L ⁄ v secs.
Time required to attain the maximum radial acceleration
2
secs
v
Ra
=
Comparing the values of time taken, we get
2

Lv
v Ra
=
or 3

v
L
Ra
= ... (17.11)
Example
The maximum allowable speed on a curve is 80 km/h and the rate of change of radial
acceleration is 30 cm / sec
2
. Calculate the length of the transition curve if the radius of the
circular curve is 200 metres. 3

v
L
Rc
=

Here R = 200 metres, c = 0.3 m/sec
2
80 1000
22.22 m/s
60 60
v

==


7. 7
( )
3
22.22
182.9 m
200 0.3
L==


Deflection Angles for Transition Curves
P and Q are two points on the transition curve and have coordinates (x, y) and (x + dx, y +
dy). The chord T1P makes an angle α and the tangent at P makes an angle ϕ, with the first
tangent at T1, as shown in Fig. 11.43(a). Thus,
3
/6
tan
y x LR
xx
==
or 2
tan
6
x
LR
= (11.26)

Fig. 11.43 Deflection and polar angles
and 3
tan
6
dy d x
dx dx LR


== 

or 2
tan
2
x
LR
= (11.27)
From Eqs. (11.26) and (11.27)
1
tan tan
3
=
Since α and ϕ are both small, we can write
1
3
=

8. 8

22
1
rad
3 2 6
LL
LR LR
=  =
Also, 22
180 573
60 min
6
Ll
LR RL



=   = (17.31a)
At point D when l = L;
2
radians
66
s
LL
LR R
==
Characteristics of a Transition Curve
The transition curves are introduced at both the ends of a circular curve by shifting the main
curve inwards.

Fig. 17.8
Let AK and KC be two straights.
Δ, the angle of deflection
t1, t2, the points of tangencies of the original curve
T1, T2 the points of tangencies of transition curve
E and F, the junction points of the transition curves with the circular curve.
R, the radius of the circular curve.
S, the shift of the circular curve
EN, the tangent at E meeting back tangent AK at N
ø the spiral angle
O, the centre of the main circular curve

9. 9

Construction. Drop EM perpendicular to O t1 and ED perpendicular to AK.
1. The spiral angle.
The angle between the back tangent and tangent at the junction of the transition curve with
the circular curve, is called spiral angle.
From ΔEMO, we know
∠MOE = 90° − ∠MEO
= ∠ MEN = END = ø = 180
radians
2
L
R


2. Shift.
The distance through which main circular curve is shifted inward to accommodate the
transition curves, is known as shift. Its value is; 2
24
L
R

Proof: Assume T1, the point of commencement as origin of the coordinates.
Back tangent as Y-axis and a perpendicular to back tangent as X-axis.
Let DE = X and T1, D = Y
Construction.
Prolong the shifted circular curve beyond E up to B.
Now, Angle DNE = angle NEM = angle EOM = ø.
The length of arc EB = R×ø
or =2
L
R
R

As EG is approximately equal to EB.
2
L
EG=
i.e. the shift t1B bisects the transition curve at G.
Shift S = t1B
= t1M − BM = DE − (OB − OM)
= X − (R − R cos ø) = X − R (1 − cos ø)
2
2sin
2
XR

= − •

10. 10

2
2 sin
2
XR

=−

 (ø/2 being small)
Substituting the values of 3
6
L
X
LR
= and 2
L
R
= , we get 2
3
2
64
LL
SR
LR R

= − 


22
68
LL
RR
=−

or 2
24
L
S
R
= ... (17.32)
3. The tangent length of the combined curve
Total tangent length = T1K = T1t1 + t1K
But, ( )
1
tan
2
t K R S

=+ and 11
2
L
Tt= … (17.33)
4. Length of the combined curve
The central angle for the circular curve = Δ − 2ø
The length of the circular curve ( )2
180
R−
=

∴ Total length of the combined curve = L + length of the circular curve + L ( )2
2
180
R
L
−
=+

… (17.34)
5. Chainages of main points of the curve
a) The chainage at the point of commencement (T1) of the combined curve = the
chainage at the point of intersection (K) − total tangent length.
b) The chainage at the first junction point (E) = Chainage at the point of commencement
(T1) + length of the transition curve (L).
c) The chainage at the second junction point (F) = chainage of the first junction point (E)
+ length of the circular curve.
d) The chainage at the point of tangency (T2) = chainage of the second junction point (F)
+ length of the transition curve (L).

11. 11

Check: The chainage at the point of tangency (T2) must be equal to the chainage at the point
of commencement (T1) + length of the combined curve i.e. ( )2
2
180
R
L
−
+


Example 1
It is required to join two straights having a total deflection angle 18° right by a circular curve
of 500 m radius, having cubic spiral transition curves at each end. The design velocity is 72
km per hour and the rate of change of radial acceleration along the transition curve is not to
exceed 25 cm / sec
3
. Chainage at the point of intersection is 840.0 m. Assume peg interval
along transition curve 10 m, and along circular curve 20 m, calculate the necessary data
required for setting out the curve.
Solution
Design velocity (Survey of Kenya) = 72 km/h= 20 cm/sec
Rate of change of radial acceleration a = 0.25 m/s
3

Radius of the circular curve = 500 m.
(i) Length of transition curve 2
v
L
Ra
=
•

or 3
20
64 m
500 0.25
L==

(ii) Shift of the curve
22
64
0.3413 m
24 24 500
L
R
= = =

(iii) Total tangent length
( )tan
22
L
RS

= + +
( )
64
500 0.3413 tan9
2
= + +
= 500.3413 × 0.158385 + 32 = 111.25 m
(iv) Chainage T1, T2, T3, and T4
Chainage at the point of commencement T1 = 840.0 − 111.25 = 728.75 m
Chainage at first junction point T3 = 728.75 + 64.0 = 792.75 m.
Spiral angle ø1 = 3δ1 = 3 × 73.344 = 219.987′ = 3° 667

12. 12

or 1
180 64 180
3.667
2 500 2
L
R



=  = =

Angle subtended by the circular arc = 18° − 2ø1 = 18° − 7.334 = 10°666
Length of circular arc T1 T2 = ( )
1
2
180
R
−
500
10.666 93.03 m
180

=  =
∴ Chainage at the end of circular curve T4 = 792.75 + 93.08 = 885.83 m
Chainage at the point of tangency T2 = 885.83 + 64 = 949.83 m
Chord length for circular arc is kept less than 20
R 500
25 m
20
==
The given chord length being less than 25 m, hence O.K.
(Survey of Kenya) Chord lengths and deflection angles
Cubic spiral transition curve with 10 m chord lengths. To locate a point on the transition
distance l metres from A, from Eqn. (17.31a), we get
2
573
min
l
RL
=
2573
min
500 64
l=

α = 0.01790625 l
2
min.
The various deflection angles are calculated in a tabular form below:
Chord
(m)
l (m) Chainage
(m)
Deflection Angle
0 0 748.75
1.25 1.25 730.00 0.01790625 = (125)² = 0.028′
10.00 11.25 740.00 0.01790265 (11.25)² = 2.267′
10.00 21.25 750.00 0.01790265 (21.25)² = 8.086′
10.00 31.25 760.00 0.01792265 (31.25)² = 17.487′
10.00 41.25 770.00 0.01790265 (41.25)² = 30.469′
10.00 51.25 780.00 0.01790265 (51.25)² = 47.032′
10.00 61.25 790.00 0.01790261 (61.25)² = 67.176′
2.75 64.00 792.75 0.01790265 (64.00)² = 73.34′

13. 13


Chord (m) Chainage
(m)
Deflection Angle = 1718.9C
R
Total Deflection Angle at
T1
0 792.75 0 0
7.25 800.00 7.25
1718.9 24.924
500
=

24.924 = 24′55′′
20.00 820.00 20
1718.9 68.756
500
=

93.680 = 1°33′39′′
20.00 840.00 20
1718.9 68.756
500
=

162.436 = 2°42′26′′
20.00 860.00 20
1718.9 68.756
500
=

231.192 = 3°51′12′′
20.00 880.00 20
1718.9 68.756
500
=

299.948 = 4°59′57′′
20.00 885.83 5.88
1718.9 68.756
500
=

319.990 = 5°20′0′′

(vi) The second Transition curve is to set out from point of tangency. For through chainage,
pegs may be placed as shown in the table below:
Chainage (m) Distance
from B (m)
Chord (m) Deflection Angle = 0.01790625 l²

Angle set on 20′′
theodolite
885.83 64.00 4.17 0.01790625 × 64² = 73.344
= 1° 13′ 21′′
358° 46′ 40′′
890.00 59.83 10.00 0.01790625 × (59.83)²
= 64.098 = 1° 04′06′′
358° 56′00′′
900.00 49.83 10.00 0.01790625 × (49.83)²
= 44.462 = 0°44′28′′
359° 15′40′′
919.00 39.83 10.00 0.01790265 × (39.83)²
= 28.407 = 0° 28′24′′
359° 31′40′′
920.00 29.83 10.00 0.01790625 × (29.83)²
= 15.933 = 0° 15′56′′
359° 44′00′′
930.00 19.83 10.00 0.01790625 × (19.83)²
= 7.041 = 0° 07′02′′
359° 53′00′′

14. 14

940.00 9.83 9.83 0.01790625 × (9.83)²
= 1.730 = 0° 01′44′′
359° 58′20′′
949.83 0.00 0 0 = 0° 00′00′′ 360° 00′00′′