SURVEYING
Simple Circular Curves
Presented By
Srikanth Samudrala
Department of CIVIL ENGINEERING
KAMALA INSTITUTE OF TECHNOLOGY AND SCINCE, SINGAPUR
Types of Curves:
Designation of a Curve:
 A) Arc definition:
L=DR
L=R*D*
𝜋
180
R=
𝐿∗180
𝐷∗𝜋
If L=20m
R=(1145.92/D)
If L=30m
R=(1718.87/D)
Designation of a Curve:
 B) Chard definition:
Sin
𝐷
2
=
𝐿/2
𝑅
Small angle Sin
𝐷
2
=
𝐷
2
𝐷
2
∗
𝜋
180
=
𝐿/2
𝑅
R=
𝐿∗180
𝐷∗𝜋
If L=20m
R=(1145.92/D)
If L=30m
R=(1718.87/D)
Circular Curve Geometry
Based on the instruments used in setting out the curves on the ground there are
two methods:
1) Linear method
2) Angular method
Methods of setting out simple
circular curve
Linear Method
 In these methods only tape or chain is used for setting out the curve . Angle
measuring instrument are not used.
Main linear methods are
 By offsets from the long chord.
 By successive bisection of arcs.
 By offsets from the tangents.
By offsets from the long chord
R = Radius of the curve
0o = Mid ordinate
0x = ordinate at distance x from the mid point
of the chord
T1 and T2 = Tangent point
0o = R -- 𝑅2 − (
𝐿
2
)2
0x = (𝑅2 −𝑥2 ) –(R – 0o)
Offset,
0x
(𝑅2 −𝑥2 ) –
(R – 0o)
0 3.515
2 3.347
4 2.828
6 1.907
8 0.459
8.485 0.000
Long chord length = 16.971
L/2 = 8.845m
Offset interval = 2 m
Scale
1m :1cm
1:100
W = 16.971 m
T1 T2
W/2 = 8.845m W/2 = 8.845m
2 m 2 m 2 m 2 m 0.459 m
O0m O2m O4m O6m O8m O8.845mO2mO4mO6mO8mO8.845m
3.515m
3.515m
2.828m
1.907m
2.828m
3.515m
2.828m
1.907m
2.828m
By successive bisection of arcs
 Join the tangent points T1,T2 and bisect the long chord at D.
 Erect perpendicular DC at D equal to the mid ordinate.
 Join T1C and T2C and bisect them at D1 and D2 respectively.
 D1 & D@ set out perpendicular offsets C1D1=C2D2=(1- cos
∆
4
) and obtain points
C1 and C2 on the curve.
By offsets from the tangents
 The offsets from the tangents can be of two types
1) Radial offsets
2) Perpendicular offsets
1) Radial offsets
 0X = 𝑅2 + 𝑥2 -- R
2) Perpendicular offsets
 Ox = R -- 𝑅2 − 𝑥2
Offset Ox R -- 𝑅2 − 𝑥2
0 0.000
2 0.168
4 0.686
6 1.608
8 3.056
10 5.367
12 12.000
Tangent length = 12m
Offset = 2 m
I
T1
Angular Method
 This methods are used when the length of curve is large.
The Angular methods are:
1) Rankine method of tangential angles
2) Two theodolite method
3) Tacheometric method
Rankine method of tangential angles
 this method also known as Tape and Theodolite method
 “A deflection angle to any point on the curve is the angle at p.c. between the
back tangent and the chord from p.c. to that point.”
Radius, R 24.7m
Deflection angle , q 600
Offset 5m
Chainage intersection point, I 20 m
Tangent length = 14.261m
Chainage T1 = 5.739m
Arc length = 25.866m
Chainage T2 = 31.605m
Stn. Chainage Chord length, C Deflection angle, Setting out angle, 
T1 5.739 0 00 0’ 0” 00 0’ 0”
1 10 4.261 40 56’ 32” 40 56’ 32”
2 15 5.000 50 47’ 57” 100 44’ 29”
3 20 5.000 50 47’ 57” 160 32’ 26”
4 25 5.000 50 47’ 57” 220 20’ 23”
5 30 5.000 50 47’ 57” 280 8’ 20”
T2 31.605 1.605 10 51’ 42” 300 0’ 2”
 = 25.866  = 300 00’ 2” θ / 2 = 600 / 2 = 300
I
T1
Scale
1m :1cm
1:100
T2
120000’ 00”
GIVEN DEFLECTION ANGLE , θ=60000’ 00”
I
T1
Scale
1m :1cm
1:100
T2
Two theodolite Method
Curves
Curves

Curves

  • 1.
    SURVEYING Simple Circular Curves PresentedBy Srikanth Samudrala Department of CIVIL ENGINEERING KAMALA INSTITUTE OF TECHNOLOGY AND SCINCE, SINGAPUR
  • 2.
  • 5.
    Designation of aCurve:  A) Arc definition: L=DR L=R*D* 𝜋 180 R= 𝐿∗180 𝐷∗𝜋 If L=20m R=(1145.92/D) If L=30m R=(1718.87/D)
  • 6.
    Designation of aCurve:  B) Chard definition: Sin 𝐷 2 = 𝐿/2 𝑅 Small angle Sin 𝐷 2 = 𝐷 2 𝐷 2 ∗ 𝜋 180 = 𝐿/2 𝑅 R= 𝐿∗180 𝐷∗𝜋 If L=20m R=(1145.92/D) If L=30m R=(1718.87/D)
  • 7.
  • 11.
    Based on theinstruments used in setting out the curves on the ground there are two methods: 1) Linear method 2) Angular method Methods of setting out simple circular curve
  • 12.
    Linear Method  Inthese methods only tape or chain is used for setting out the curve . Angle measuring instrument are not used. Main linear methods are  By offsets from the long chord.  By successive bisection of arcs.  By offsets from the tangents.
  • 13.
    By offsets fromthe long chord
  • 14.
    R = Radiusof the curve 0o = Mid ordinate 0x = ordinate at distance x from the mid point of the chord T1 and T2 = Tangent point 0o = R -- 𝑅2 − ( 𝐿 2 )2 0x = (𝑅2 −𝑥2 ) –(R – 0o)
  • 15.
    Offset, 0x (𝑅2 −𝑥2 )– (R – 0o) 0 3.515 2 3.347 4 2.828 6 1.907 8 0.459 8.485 0.000 Long chord length = 16.971 L/2 = 8.845m Offset interval = 2 m Scale 1m :1cm 1:100 W = 16.971 m T1 T2 W/2 = 8.845m W/2 = 8.845m 2 m 2 m 2 m 2 m 0.459 m O0m O2m O4m O6m O8m O8.845mO2mO4mO6mO8mO8.845m 3.515m 3.515m 2.828m 1.907m 2.828m 3.515m 2.828m 1.907m 2.828m
  • 16.
  • 17.
     Join thetangent points T1,T2 and bisect the long chord at D.  Erect perpendicular DC at D equal to the mid ordinate.  Join T1C and T2C and bisect them at D1 and D2 respectively.  D1 & D@ set out perpendicular offsets C1D1=C2D2=(1- cos ∆ 4 ) and obtain points C1 and C2 on the curve.
  • 18.
    By offsets fromthe tangents  The offsets from the tangents can be of two types 1) Radial offsets 2) Perpendicular offsets
  • 19.
    1) Radial offsets 0X = 𝑅2 + 𝑥2 -- R
  • 20.
    2) Perpendicular offsets Ox = R -- 𝑅2 − 𝑥2
  • 21.
    Offset Ox R-- 𝑅2 − 𝑥2 0 0.000 2 0.168 4 0.686 6 1.608 8 3.056 10 5.367 12 12.000 Tangent length = 12m Offset = 2 m I T1
  • 22.
    Angular Method  Thismethods are used when the length of curve is large. The Angular methods are: 1) Rankine method of tangential angles 2) Two theodolite method 3) Tacheometric method
  • 23.
    Rankine method oftangential angles  this method also known as Tape and Theodolite method  “A deflection angle to any point on the curve is the angle at p.c. between the back tangent and the chord from p.c. to that point.”
  • 25.
    Radius, R 24.7m Deflectionangle , q 600 Offset 5m Chainage intersection point, I 20 m Tangent length = 14.261m Chainage T1 = 5.739m Arc length = 25.866m Chainage T2 = 31.605m Stn. Chainage Chord length, C Deflection angle, Setting out angle,  T1 5.739 0 00 0’ 0” 00 0’ 0” 1 10 4.261 40 56’ 32” 40 56’ 32” 2 15 5.000 50 47’ 57” 100 44’ 29” 3 20 5.000 50 47’ 57” 160 32’ 26” 4 25 5.000 50 47’ 57” 220 20’ 23” 5 30 5.000 50 47’ 57” 280 8’ 20” T2 31.605 1.605 10 51’ 42” 300 0’ 2”  = 25.866  = 300 00’ 2” θ / 2 = 600 / 2 = 300
  • 26.
    I T1 Scale 1m :1cm 1:100 T2 120000’ 00” GIVENDEFLECTION ANGLE , θ=60000’ 00”
  • 27.
  • 28.