Simple Circular Curves for kenya-road-design-manual-for highway and expressway design

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

1. 1

Curves
Curves are regular bends provided in the lines of communication like roads, railways etc. and
also in canals to bring about the gradual change of direction.
Curve ranging refers to the setting out, on the ground, of pegs marking a curved line in
accordance with a plan.
Types of Curves
Curves can be broadly classified as follows.

A. Horizontal Curves
1. Circular Curves
i). Simple Circular Curve
A curve, connecting two intersecting straights having a constant radius all through is known
as a simple circular curve. It is tangential to the two straights at the joining ends.


Fig. 11.1: Simple circular curve
In Fig. 11.1, T1TT2 is a simple circular curve of radius R, joining the two straights T1I and T2I
intersecting at a point I.

2. 2

ii). Compound Curve
They’re curves formed when two or more simple circular curves, of different radii, turning in
the same direction join two intersecting straights, the resultant curve is known as a compound
curve.

Fig. 11.2: Compound curve
In Fig. 11.2, T1TT2 is a compound curve with two simple circular curves T1T and TT2 of radii
R1 and R2, respectively. ATB is the common tangent, and T is the common tangent point.
iii). Reverse/Serpentine Curve
They are curves formed when two simple circular curves, of equal or different radii, having
opposite direction of curvature join together.

Fig. 11.3: Reverse curve
In Fig. 11.3, T1TT2 is a reverse curve, formed from the curves T1TV and T2TU of radii R1 and
R2, respectively, joining the two straights T1U and T2V.
2. Transition Curve
It is a curve usually introduced between a simple circular curve and a straight, or between
two simple circular curves. It is also known as an easement curve.

3. 3

Its radius increases or decreases in a very gradual manner.
Usually employed in high-speed traffics whereby an abrupt change in direction could cause
the vehicle to overturn.

Fig. 11.4: Transition curve
In Fig. 11.4, T1TD is the transition curve introduced between the simple circular curve T4DD
T5 and the straight T1I.
B. Vertical Curve
These are curves, in a vertical plane, used to join two intersecting grade lines.
Vertical curves are classified as summit and sag vertical curve.
The reduced level of these curves changes from point to point in a gradual and systematic
manner.

Fig. 11.6: Vertical curves

4. 4

A vertical summit curve is provided when a rising grade (T1 I) joins a falling grade (T2 I), as
shown in Fig. 11.6(a), and a vertical sag curve is provided when a falling grade (T3 I) joins a
rising grade (T4 I), as shown in Fig. 11.6(b).
Horizontal Curves
Simple Circular Curves
Elements of a Simple Curve

5. 5


Point of Intersection (PI)
The point of intersection marks the point where the back and forward tangents intersect. The
surveyor indicates it as one of the stations on the preliminary traverse.
Intersecting Angle (I)
It’s the angle by which the forward tangent deflects from the back tangent.
The intersecting angle is the deflection angle at the PI. The surveyor either computes its value
from the preliminary traverse station angles or measures it in the field.
If the deflection angle is clock wise, the curve is a right hand curve.
Radius (R)
The radius is the radius of the circle of which the curve is an arc.
Point of Curvature (PC)
The point of curvature is the point where the circular curve begins. The back tangent is
tangent to the curve at this point.
Point of Tangency (PT)
The point of tangency is the end of the curve. The forward tangent is tangent to the curve at
this point.

6. 6

Length of Curve (L)
The length of curve is the distance from the PC to the PT measured along the curve.
Tangent Distance (Muskett)
The tangent distance is the distance along the tangents from the PI to the PC or PT. These
distances are equal on a simple curve.
Central Angle (Δ)
The central angle is the angle formed by two radii drawn from the center of the circle (O) to
the PC and PT. The central angle is equal in value to the I angle.
Long Chord (LC)
The long chord is the chord from the PC to the PT.
External Distance (E)
The external distance is the distance from the PI to the midpoint of the curve. The external
distance bisects the interior angle at the PI.
Middle Ordinate (M)
The middle ordinate is the distance from the midpoint of the curve to the midpoint of the long
chord. The extension of the middle ordinate bisects the central angle.
Calculation of Various Elements of a Simple Curve

The following equations can be derived from Fig. 12.1.
1. Length of curve (l): Length of the circular curve T1DT2 is given by 2
360 180
R
lR



=  =


2. Tangent length T: Tangent length is given by

7. 7
12
tan
2
T B T B R

==

3. Chainage of tangent points: The chainage of intersection point B is generally known,
Subtracting the tangent length T,
Chainage of T1 = chainage of B – T
Chainage of T2 = chainage of T1 + length of curve (l)
= 1
chainage of
180
R
T

+

4. Length of long chord = T1 ET2 = 2 x R sin ∆/2 2 sin
2
R

=

5. External distance or apex distance = BD. sec
2
OB OD R R

= − = −

6. Mid ordinate = OD - OE cos 1 cos
22
R R R

= − = −



Designation of a curve
Degree of Curve (D)
The degree of a curve (D) is defined as the angle subtended at the centre by an arc or chord of
standard length.
The degree of curve defines the “sharpness” or “flatness” of the curve (figure 3). There are
two definitions commonly in use for degree of curve, the arc definition and the chord
definition.

8. 8

Arc definition. The arc definition states that the degree of curve (D) is the angle subtended at
the center by an arc 30m long. In this definition, the degree of curve and radius are inversely
proportional:
Using a 30 meter length of arc and substituting D = 1°, we obtain
Let - R be the radius
- s be standard length
- Da be degree of the curve
Referring to Fig. 2.4 (a)
∴ 180
a
s R D

=  
or 180
a
s
R
D

= … (2.1)
If s = 20 m
20 180 1145.92
aa
R
DD
=  = … (2.2a)
If s = 30 m
30 180 1718.87
aa
R
DD
=  = … (2.2b)
Chord definition. According to the chord definition the degree of a curve is equal to the
angle subtended at the centre by a chord of 30 m length. The radius is computed by the
following formula:
Let Dc be degree of curve as per chord definition and s be the standard length of chord. Then
referring to Fig. 2.4(b). sin
22
c
Ds
R =
… (2.3)
When Dc is small, sin
2
c
D may be taken approximately equal to 2
c
D radians,
Hence, for small degree curves (flat curves).
2 180 2
c
D s
R

=
or 180
c
s
R
D
= … (2.4)

9. 9

Comparing equations (2.1) and (2.4), we find for flat curves, arc definition and chord
definitions give same degree of curve. As in railways flat curves are used, chord definition is
preferred.
Example
A circular curve has a 200 m radius and 65° deflection angle. What is its degree (i) by arc
definition and (ii) by chord definition. Also calculate:
(a) length of curve, (b) tangent length, (c) length of long chord, (d) apex distance, and (e)
mid-ordinate.
Solution
(i) Arc definition
Assuming a 30 m chord length,
30
180
RD

  =

30 180 30 180
8.595
200
D
R

= = = 

(ii) Chord definition
Assuming a 30 m chord length,
15
sin( / 2)
R
D
= (but since D is small, sin 22
DD
 radians)
Hence, 15
2 180
R
D
=
 =15 2 180
8.595
200

=

(a) Length of the curve, 200 65 226.89
180 180
L R m

=  =   =

(b) Tangent length, 65
tan 20tan 127.41
22
T R m

= = =
(c) Length of long chord, 65
2 sin 2 200sin 214.92 m
22
LR

= =  =
(d) Apex distance = 65
sec 1 200 sec 1 37.13 m
22
R
   
− =  − =
   
   
(e) Mid-ordinate =65
1 cos 200 1 cos 31.32 m
22
R
   
− =  − =
   
   
Tools and Equipment for Setting Out Curves
a) Theodolite/total station

10. 10

b) Tapes/chains
c) Pegs
Methods of Setting out simple curves
The various methods used for setting curves may be broadly classified as:
(i) Linear methods
(ii) Angular methods.
A. Linear Methods
Only a chain or a tape is used and no angle measuring instrument is used. The following are
some of the linear methods used for setting out simple circular curves:
(i) Offsets from long chord
(ii) Successive bisection of chord
(iii) Offsets from the tangents–perpendicular or radial
(iv) Offsets from the chords produced.
1. Offsets from Long Chord
In this method, long chord is divided into an even number of equal parts. Taking centre of
long chord as origin, for various values of x, the perpendicular offsets are calculated to the
curve and the curve is set in the field by driving pegs at those offsets.

Fig. 11.9 Offsets from the long chord
Let it be required to lay a curve T1CT2 between the two intersecting straights T1I and T2 I
(Fig. 11.9). R is the radius of the curve, O0 the mid-ordinate, and Ox the offset at a point P at
a distance x from the mid-point M of the long chord.
From triangle OMT1

11. 11

( )
2
2 2 2
11
2
L
OM OT MT R


= − = − 



Now, CM = OC – OM
or O0 = R – OM =2
2
2
L
RR

−−


In triangle OP′G
22
OG R x=− and OM = R – O0
The required offset
PP′ = OG – OM
Hence, ( )
22
0
PP R x R O= − − −
or ( )
22
0x
O R x R O= − − − =1
2
2
02
1
x
R R O
R

− − +

24
024
1 ...
28
xx
R R O
RR

= − − − − +
 (binomial expansion usually given by ()
() ()()
1 2 2 3 3
1 1 2
...
2! 3!
n
n n n n n
n n n n n
a x a na x a x a x x
− − −
− − −
 =    + +
)
OR ()
()
2
2
1
1 1 ...
2!
n
n
nn
nny y y
x y x x n
x x x
−
 =  =    
  )
∴ 24
024
28
x
Rx Rx
O R R O
RR
= − − − +
∴ 2
0
2
x
x
OO
R
− (ignoring higher powers of x).
2. Offsets from the Tangents
The offsets from tangents may be calculated and set to get the required curve.
The offsets can be either radial or perpendicular to tangents.
(i) Radial offsets: If the centre of curve O is accessible from the points on tangent, this
method of curve setting is possible.

12. 12


Fig. 11.11 Radial offsets from tangent
Ox is the radial offset PP′ at any distance x along the tangent from T1
From the triangle OT1P 2 2 2
11
+OP OT T P=

(R + Ox)
2
= R
2
+ x
2 22
x
R O R x+ = +

22
x
O R x R= + −
1
2
2
2
1
x
RR
R

= + −

1
24
2
24
1 ...
28
xx
RR
RR

= + + + −

1
2
2
2
1
2
x
RR
R

= + −
 (neglecting higher powers)
2
2
x
x
O
R

(ii) Perpendicular offsets: If the centre of a circle is not visible, perpendicular offsets
from tangent can be set to locate the points on the curve.

13. 13


Fig. 11.10 Perpendicular offsets from tangent
Ox is the offset perpendicular to the tangent at a distance x from the point of curve T1.
In the triangle OEP′:
2 2 2
P O OE P E=+
or ( )
2
22
x
R R O x= − +
or ( )
2
22
x
R O R x− = −
or ( )
22
x
R O R x− = −
or ( )
1
2 2 2 2
2
x
O R R x R R x= − − = − −
1
2
2
2
1
x
RR
R

= − −

24
24
1 ...
28
xx
RR
RR

= − − + +

2
2
1
2
x
RR
R

= − −
 (neglecting higher powers)
22
2
22
x
Rx x
O R R
RR
= − + 

14. 14

3. Offsets from the chord produced
This method has the advantage that not all the land between T1 and T2 need be accessible.
However, to have reasonable accuracy the length of the chord chosen should not exceed
R/20.
In this method, a point on the curve is fixed by taking offset from the tangent taken at the rear
point of a chord.

Fig. 11.13 Offsets from chords produced
From figure 11.13:
T1 is the P.C. along the tangent T1I (Fig. 11.13). T1a is the first subchord, C1. From the P.C., a
length equal to first subchord C1 (T1a′ ) is taken. The perpendicular offset (O1) aa′ is set out,
thereby getting point a. T1a is joined and produced by distance C2
The second offset O2 (bb′) is set out to get point b. Points a and b are joined and produced
further by distance C3 (full chord length). The third offset O3 (cc′ ) is set out to get point c.
The procedure is repeated till the curve is completed.
 a′ T1a =??????1 = deflection angle of the first chord
So that 1 1 1 1 1
O a a Ta C= = = (11.7)
1 1 1
22
a
TO aTa  =  =
11
2Ta R=
or 11
1
22
T a C
RR
==
Putting this value of δ1 in Eq. (11.7) we get

15. 15

2
11
11
22
CC
OC
RR
=  =
For computing O2, a tangent PQ is drawn to the curve at a, and is produced both ways.
ab′ = ab = C2 (length of second chord)
2
O b b b Q Qb= = +
2
b Q O= and 2
Qb O= (offset between the tangent and the chord)
2
22
2 2 2 2
22
CC
O C C
RR
=  =  =
1
2 2 1 2 2
2
C
O C C
R
=  = 
So that 2
1 2 2
2 2 2
22
C C C
O O O
RR
 = + = +
( )
2
2
12
2
C
CC
R
=+
Similarly, the third offset ( )
3
3 2 3 2
2
C
O C C cc
R
= + =
Normally, 3 4 1
...
n
C C C C
−
= = = = (normal chord)
( )
1
2
n
n n n
C
O C C
R
−
=+
where Cn is the last subchord.
Field Procedure
1. Locate P.C. and P.T., i.e., T1 and T2.
2. Calculate the chainage of T1 by subtracting the tangent length from the P.I. and calculate
the length of the subchord so that the first peg is a full station.
3. Calculate the length of the curve and find the length of the last subchord.
4. Calculate all the offsets from O1 to On.
5. Put the zero mark of the chain (or tape) along the tangent T1I, take a length T1a′
(length of first subchord) and swing the chain such that the arc a′a = O1. The first
point a on the curve is thus fixed.
6. Pull the chain along T1a produced to the point b′, so that ab′ = C2 (normal chord). Put
the zero end of the chain at a and swing the chain with radius ab′ (C2). Cut off b′b equal
to second offset O2. The second point b is thus fixed.
7. Pull the chain along ab to the point c′ till, as discussed in step 6, the point of
tangency T2 is reached. The last point so fixed must coincide with T2 (fixed earlier).

16. 16

If the error is more (> 2 m), the curve is reset. If the error is less, it should be distributed
among all the points by shifting them parallel to the closing error by an amount
proportional to the square of the distance from the P.C., T1.

Example 1
Two straights AB and BC intersect at a chainage of 4242.0 m. The angle of intersection is
140°. It is required to set out a 5° simple circular curve to connect the straights. Calculate all
the data necessary to set out the curve by the method of offsets from the chord produced with
an interval of 30 m.
Solution
The chain used is of 30 m.
Radius of curve, 1718.9 1718.9
344 m
5
R
D
===
Deflection angle, 180 140 40 = −  = 
Tangent length, 1
tan 344tan 20 125.2 m
2
BT R

= =  =
Chainage of intersection point B = 4242.0 m
Chainage of ( )
1
4242.0 125.2 4116.8 mT= − =
Length of curve 180
R

= 

344 40 240.16 m
180

=   =

Chainage of T2 = chainage of T1 + length of the curve
= 4116.8 + 240.16 = 4356.96 m
Length of the chords
First subchord, C1 = 4140 – 4116.8 = 23.2 m
The last subchord, C9 = 4356.96 – 4350.0 = 6.96 m
There are seven unit chords of 30 m length.
Hence, there will be nine chords altogether.
The offsets are
2 2
1
1
23.2
0.78 m
2 2 344
C
O
R
= = =

( ) ( )
2 1 2
2
30 23.2 30
2.32 m
2 2 344
C C C
O
R
+  +
= = =


17. 17

22
3 4 5 8
30
... 2.62 m
344
C
O O O O
R
= = = = = = =
( ) ( )
9 8 9
9
6.96 6.96 30
0.37 m
2 2 344
C C C
O
R
+  +
= = =

Example 2
It is required to set out a curve of radius 100 m with pegs at approximately 10 m centres. The
deflection angle is 60°. Draw up the data necessary for pegging out the curve by each of the
following methods:
(i) Offsets from long chord (ii) Offsets from tangent.
Solution
(i) Offsets from long chord

Fig. 11.21
Since, mid-ordinate 0
60
100 1 cos 13.40 m
2
O

=  − =


Thus, 2
13.40
2
l
a
R
=−
Let peg interval, x = 10 m, so that 2
10
13.40 12.9 m
200
a= − =

Similarly, 2
20
13.40 11.40 m
200
b= − =

18. 18
2
30
13.40 8.9 m
200
c= − =
2
40
13.40 5.40 m
200
d= − =

(ii) Offsets from tangent

Fig. 11.23
Set out AC = CD = DE = EF = FG
where AG = PM = 100 × sin 30° = 50.0 m
Set pegs at 10 m interval.
Offset from tangent 22
R R x= − −
22
100 100 10 0.5 mCK= − − =
22
100 100 20 2.02 mDJ= − − =
22
100 100 30 4.60 mEH= − − =
22
100 100 40 8.35 mFL= − − =
22
100 100 50 13.40 mGM= − − =
Then, GM = IM cos 30°
( )
sec 1 cos30
2
100 sec30 1 cos30 13.40 m (checked)
R

= − 


=  −   =

19. 19

B. Angular Methods
Setting out with theodolite and tape
Rankine’s Method of Tangential or Deflection Angles
In this method, the curve is set out by the tangential angles (also known as deflection angles)
with a theodolite and a chain (or tape). The method is also called chain and theodolite
method.
A deflection angle to any point on the curve is the angle at P.C. between the tangent and the
chord from P.C. to that point. From the property of a circle, this deflection angle is equal to
half the angle subtended by the arc at the centre.

Fig. 11.14 Rankine’s method of deflection angle
In Fig. 11.14, T1 is the P.C.; a, b, c, etc., are the points on the curve; δ1, δ2, δ3, etc., are the
respective deflection angles between the chords and the respective tangents at T1, a, b, etc.,
∆1, ∆2, ∆3, etc., are the total deflection angles to the points a, b, c, etc.
From the property of a circle that the angle subtended by a chord at the centre is twice the
angle between the tangent and the chord. Then,
1 1 1
22
a
TO ITa  =  =
Now 1 1 1
2R Ta C = =
or 1
1
radians
2
C
R
=
or 11
1
180 180
degrees 60 minutes
22
CC
RR



=  =  

20. 20

i.e., 1
1
1718.9 minutes
C
R
=
Similarly, 32
23
1718.9 minutes, 1718.9 minutes
CC
RR
==
For the first chord T1a, the deflection angle ∆1 is its tangential angle δ1. For the second point
b on the curve, the deflection angle ∆2 = IT1b.
Let the tangential angle for chord ab = δ2, i.e., the angle between the tangent at a and chord
ab.
The angle subtended by the chord ab at T1 is  aT1b = δ2, so that
∆2 = IT1b = IT1a + aT1b.
1 2 1 2
  = + =  +
Similarly, 12
bTc= , so that
3 1 1 1
ITc ITb bTc =  =  +
1 2 3 2 3
   = + + =  +
and 12
...
nn
   = + + +
Last point of the curve is T2, so that
12
2
n
ITT

 =  =
In practice C1 is the first subchord and Cn the last subchord. C2 = C3 = ... = Cn-1 are full chain
lengths.
Check:
As a check the deflection angle ∆n, for the last point T2 is equal to ∆/2 where ∆ is the angle of
intersection.
Procedure of Setting out the Curve
1) Locate the tangent points (T1 and T2) and find out their chainages. From these
chainages, calculate the lengths of first and last sub-chords and the total deflection
angles for all points on the curve as described above.
2) Set up and level the theodolite at the first tangent point (T1).
3) Set the horizontal angle to zero and direct the telescope to the ranging rod at the
intersection point I and bisect it.
4) Loosen the vernier plate and set the vernier A (horizontal angle) to the first deflection
angle Δ1, the telescope is thus directed along T1a.

21. 21

5) Hold the zero of the tape at T1, take a distance C1 (T1a) and swing the tape with an
arrow till it is bisected by the theodolite.
6) Loosen the upper clamp and set the vernier A (horizontal angle) to the second
deflection angle Δ2, the line of sight is now directed along T1b.
7) With the zero of the tape held at a and an arrow at the other end (chord distance = ab),
swing the tape about a, till the arrow is bisected by the theodolite at b. This
establishes the second point b on the curve.
8) Continue the process until the end of the curve is reached. The end point thus located
must coincide with the previously located point (T2). If not, the distance between them
is the closing error. If it is within the permissible limit, only the last few pegs may be
adjusted; otherwise the curve should be set out again.
Note:
In the case of a left-handed curve, each of the values Δ1, Δ 2 Δ 3 etc, should be subtracted
from 360
°
to obtain the required value to which the vernier is to be set i.e. the vernier should
be set to (360° – Δ1), (360° – Δ2), (360° – Δ 2) etc. to obtain the 1st, 2n, 3rd etc, points on the
curve.
Example 1
Two straights AI and BI meet at a chainage of 3450 m. A right-handed simple circular curve
of 250 m radius joins them. The deflection angle between the two straights is 50°. Tabulate
the necessary data to layout the curve by Rankine’s method of deflection angles. Take the
chord interval as 20 m.
Solution
Tangent length tan 250 tan 25 116.58 m
2
R

= =   =
Length of the curve 250 50
180 180
R

=  =  

= 218.166 ≈ 218.17 m
Chainage of starting point, T1 = 3450 – 116.58 = 3333.42 m
Chainage of the end point, T2 = 3333.42 + 218.17 = 3551.59 m
Length of the chords
First subchord, C1 = 3340 – 3333.42 = 6.58 m
Last subchord, C12 = Chainage of T2 – 3540 = 3551.9 – 3540 = 11.59 m
No. of normal chords ( )218.17 6.58 11.59
10
20
−+
==
There will be 12 chords altogether.

22. 22

1 1 1
1719 6.58
1719 45 15
250
C
R
  =  =  = = 
22
1719 20
1719 2 17 31
250
C
R
  =  =  = 
Hence, 2 2 1
2 17 31 45 15 3 2 46       = + =  + = 
3 3 2
2 17 31 3 2 46 5 20 17       = + =  +  = 
4 4 3
2 17 31 5 20 17 7 37 48       = + =  +  = 
5 5 4
2 17 31 7 37 48 9 55 19       = + =  +  = 
6 6 5
2 17 31 9 55 19 12 12 50       = + =  +  = 
7 7 6
2 17 31 12 12 50 14 30 21       = + =  +  = 
8 8 7
2 17 31 14 30 21 16 47 52       = + =  +  = 
9 9 8
2 17 31 16 47 52 19 5 23       = + =  +  = 
10 10 9
2 17 31 19 5 23 21 22 54       = + =  +  = 
11 11 10
2 17 31 21 22 54 23 40 25       = + =  +  = 
12
11.59
1719 1 19 42
250
  =  = 
12 12 11
1 19 42 23 40 25 25 07 25      = + =  +  =   

Point
Chainage
(m)
Chord
length
(m)
Deflection
angle (δ)
Total angle
(∆)
Setting out
angle
Remarks
T1 3333.42 ______ ______ ______ ______ Start point
1 3340 6.58 0°45′15″ 0°45′15″ 0°45′20″
The curve
is right-
handed.
Least
Count of
vernier =
20″
2 3360 20 2°17′31″ 3°02′46″ 3°02′40″
3 3380 20 2°17′31″ 5°20′17″ 5°20′20″
4 3400 20 2°17′31″ 7°37′48″ 7°37′40″
5 3420 20 2°17′31″ 9°55′19″ 9°55′20″
6 3440 20 2°17′31″ 12°12′50″ 12°12′40″
7 3460 20 2°17′31″ 14°30′21″ 14°30′20″
8 3480 20 2°17′31″ 16°47′52″ 16°48′00″
9 3500 20 2°17′31″ 19°05′23″ 19°05′20″
10 3520 20 2°17′31″ 21°22′54″ 21°23′00″
11 3540 20 2°17′31″ 23°40′25″ 23°40′20″
T2 3551.59 11.59 1°19′42″ 25°07′ 25° End point
Check: 12
50
25
22

 = = = 

23. 23

N/B The least count for a vernier theodolite is 20″ and so the deflection angles must be
adjusted to values that can be read off it (i.e., multiples of 20″). For optical/electronic
theodolites the least count is 1″ and thus the initial total deflection angles can be used for
setting out.