MODULE 3 CURVE SURVEYING Curves Necessity Types, Simple curves, Elements… · 2020. 3. 28. ·...

MODULE – 3 CURVE SURVEYING

G.RAVINDRA KUMAR, Associate Prof, CED, Government Engineering College, Chamarajanagar Page 1 of 46

Curves – Necessity – Types, Simple curves, Elements, Designation of curves,

Setting out simple curves by linear methods (numerical problems on offsets

from long chord & chord produced method), Setting out curves by Rankine’s

deflection angle method (numerical problems). Compound curves, Elements,

Design of compound curves, Setting out of compound curves (numerical

problems). Reverse curve between two parallel straights (numerical problems on

Equal radius and unequal radius). Transition curves Characteristics, numerical

problems on Length of Transition curve, Vertical curves –Types – (theory).

L1, L3, L5

Curves:

To avoid abrupt change of direction curves are introduced between two

straights both in the horizontal and vertical plane. Curves are generally used on

highways and railways.



Curves can be broadly classified as follows

Basic definitions and

Notations:

Point of Intersection

(B): This is the point of

intersection of the two

straights which are smooth

connected by the circular

curve.

Deflection angle (Δ):

when two straights are

connected by a curve the

angle of deviation of the

direction of the second

straight from the direction

of the first straight is called the deflection angle of the curve. It is also called

Intersection angle.

Back tangent or first tangent (AT1): This is the tangent AT1 previous to the

curve is called the back tangent.

CURVES

Horizontal Curve Vertical Curve

Circular Transition Combined

Simple

Compound

Reverse

Cubic

Parabola

Clothoid

Cubic Spiral

Lemniscate

T2(PT)

R

D

C

A

Δ

Δ/2 Δ/2

Back tangent

Forward tangent

900 E

0

R

B’

B

T

T1 900 PC

PI



Forward tangent (T2C): The tangent T2C following the curve is called

the forward tangent or second tangent.

Point of curve (PC / T1): It is the beginning of the curve where the alignment

changes from a tangent to a curve.

Point of Tangency (PT / T2): It is the end of the curve where the alignment

changes from a curve to a tangent.

Tangent distance (T): It is the distance between PC to PI (T1 to B). (and also

distance from PI to PT) (B to T2).

Long chord (T1ET2): This is the chord of the curve joining the starting point and

the end point of the curve.

Length of the curve (T1DT2): This is the actual length of the from PC to PT.

External distance or Apex distance (BD): This is the distance between the

intersection point (PI) and the apex of the curve.

Mid Ordinate (DE): This is perpendicular distance from the middle point of the

curve to the long chord. The distance is also called the versed sine of the curve.

Normal Chord (C): A chord between two successive regular station on the curve.

Sub Chord (c): Sub chord is any chord shorter than the normal chord.

Elements of a simple curve:

1. Length of the curve ( l ): This is the actual length of the curve touching

end straights, between the point of tangency.

180

RΔ

2. Tangent Length T: This is the length of the end tangent measured from the

intersecting point to either end of the curve.

2

ΔRtanBTBT 21

3. Chainage of tangent points: The chainage of intersection point B is

generally known.

Chainage of T1 = Chainage at B – Tangent length (T )

Chainage of T2 = Chainage at T1 + Length of curve ( l )



= Chainage of T1 + 180

RΔ

4. Length of Long chord: This is the chord of the curve joining the starting

point and the end point of the curve.

2

Δ2Rsin=ETT 21

5. External distance or Apex distance: This is the distance between the

intersection point and the apex of the curve.

R2

ΔRsec = OD - OB = BD

6. Mid ordinate : This is perpendicular distance from the middle point of the curve to the long chord. The distance is also called the versed sine of the curve.

2

Δcos1R

2

ΔRcosR = OE OD DE

Degree of curve: Degree of curve may be defined either with respect to a fixed length of an arc

of the curve or with respect a fixed length of a normal chord of the curve.

Fixed length of an arc:

The degree of curve may by defined as the central angle of the curve that is

subtended by an arc of 30metres (or 100ft) length. This definition is generally

adopted for the railway curves.

Fixed length of a chord:

The degree of a curve may be defined as fixed central angle of the curve that is

subtended by a chord of 30 metres (or 100 ft) length. This definition is generally

adopted for the road curves.

Derivation of the formula:

Relationship between radius and degree of curve

i) Let D0 be the angle subtended by an unit chord of 30

m length of a circle whose radius is R

T1ET2 is the chord of length 30m of the curve T1T2 and

radius T1O = T1O = R and D the degree of the curve.

Degree of the curve ‘D’ is the angle subtended by the chord

at the centre of the curve.

In the triangle T1ET2, T1E = ½ x T1T2 = 30/2 =15 m,

T1OE = D/2, T1O = R

T2 T1

R D/2

D

E

0

R

15 m

30 m



radians in 2

D to tends

2

DinS small, too is

2

DinS

R

15

OT

ET

2

DinS

1

1

meters in R and degrees in D R

1719D

D

1719

D

36015

1802

D

15

2

D

15R

0

0

ii) When the unit chord length is 20 m:

radians in 2

D to tends

2

DinS small, too is

2

DinS

R

15

OT

ET

2

DinS

1

1

D

1146

D

36010

1802

D

10

2

D

10R

0

0

meters in R and degrees in D

R

1146D

T2 T1

R D/2

D

E

0

R

10 m

20 m



Selection of curve:

Any two given straights can be connected by an infinite number of circular

arcs. The curve to be provided in a particular case is determined by assuming any

one of the quantities, viz., the radius, the degrees of the curve, the tangent length,

the apex distance, or the long chord. All these quantities are independent. If one of

them is chosen, the others can be calculated.

Problem:

Two straights intersect at

a deflection angle of 800 and are

connected by a circular curve of

radius 10 chains.

Find: i) Length of each end

tangent, ii) Length of the curve,

iii) Length of the long chord, iv)

Apex distance, v) Mid – ordinate

of the curve, vi) Degree of the

curve.

Solution:

Given: Deflection angle = 800,

Radius (R) = 10 chains.

1. Tangent Length T:

chains39.82

80tan10

2

ΔRtanBTBT

0

21

2. Length of the curve ( l ):.

chains96.13180

8010

180

RΔ0

0

3. Length of Long chord:

chains86.122

80sin102

2

Δ2Rsin=ETT

0

21

4. External distance or Apex distance (BD):

chains 3.05 01

2

80cos

10 R

2

Δcos

R R

2

ΔRsec = OD - OB = BD

0

5. Mid ordinate (DE):

chains34.22

80cos110

2

Δcos1R

2

ΔRcosR = OE OD DE

0

T2(PT)

R

D

C

A

Δ=800

Δ/2 Δ/2

Back tangent

Forward

tangent

900 E

0

R

B’

B

T

T1 900 PC

PI



6. Degree of curve (D):

R

2/L

2

DinS

chain 1 length chord L R2

L

2

Dsin

"54.57 '51 2 102

1sin

R2

1sin

2

D 011

"5.085 '34 5D 0

Problem:®

Two straight alignments AI and IC of a road intersect at

I at a chainage (250+15), angle of deflection being

1200. Calculate the chainages of the point of

commencement and the point of tangency if the radius

of the right handed circular curve is 200 m. Assume the

length of the chain as 30 m.

Solution: Chainage at the point of intersection = (250+15) i.e., 250 chains and 15 links. Note: 20 metre chain contains 100 links, each link is of length = 0.2 m and 30 metre chain contains 150 links, each link is of length = 0.2 m. Length of one chain = 30 m Chainage at the point of intersection = 250 x 30 + 15 x 0.2 = 7503 m

Deflection angle () = 1200 Radius of curve = 200 m.

Tangent Length T= m 346.412

120tan200

2

ΔRtan

Length of the curve (l)= m 418.88180

120200

180

RΔ

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent

length = 7503 – 346.41 = 7156.59 m

Chainage at point of tangency (T2) = Chainage at point of curve (T1) + Length of curve = 7156.59 + 418.88 = 7575.47 m

Setting out of a curve:

A circular curve can be set out by

(i) Linear or chain and tape method when no angle measuring instruments is used:

(ii) Instrument methods in which a theodolite, tacheometer or a total station

instrument is used.

Δ=1200

O

I

C

L

R

T

A



i) Linear or chain and tape method of setting out curve:

a) By ordinates or offsets from long chord

b) By successive bisection of arcs

c) By offsets from the tangents

d) By offsets from chords from long produced

By ordinates or offsets long chord:

Usually the lines AB

and BC are already plotted

on the ground. The

deflection angle Δ may be

set out very accurately by

means of a theodolite.

Lengths BT1, BT2 and T1T2

are calculated. Points T1, T2

and the midpoint E of T1T2

are obtained on the field. If

L is the length of long

chord T1E = L/2.

And

22 2/LROE

And

22O 2/LRROordinate MidDE

To get ‘OX’ at any distance ‘x’ from E, from the triangle OPP1,

222

X

2221

221

2

xROOE

xROP

ROPx

OExRO

xROOE

22X

22X

2

222X

2

LRxRO

022

X ORxRO

Dividing the long chord into an even number of parts, points on the curve can

be obtained with corresponding values of x, which is measured from centre.

A

B’

B

T

Δ

Δ

T1 T2

Back tangent

Forward

tangent

D P

0

L/2 L/2 X

OX P1

R

C E



Problem:

Two straights intersect at chainage 2056.44 m and the angle of intersection is 1300. If the radius of the simple curve to be introduced is 50 m, set out the curve by offsets from long chord for 5m interval. find the following:

(i) Chainage of the point of commencement (ii) Chainage at point of tangency (iii) Length of the long chord

Solution: Chainage at the point of intersection = 2056.44 m Angle of intersection = 1300

Deflection angle () = 1800 - 1300 = 500 Radius of curve (R) = 50 m.


50tan50

2

ΔRtan

Length of the curve (l) = m 63.34180

5050

180

RΔ 0


length = 2056.44 – 23.32 = 2033.12 m


Length of Long chord (L) = T1T2 = m 26.242

50sin052

2

Δ2Rsin

0

Starting from centre of long chord, for offset distances at 5 m interval, offsets are calculated for half of the long chord, i.e., 42.26/2 = 21.13m

32.45x05

42.26/205x05L/2RxRy

22

22222222

m43.432.45505y

m68.432.45005y

225

220

m38.332.455105y

m67.332.450105y

2215

2210

m032.4513.2105y

m51.032.452005y

2221.13

2220

]

Δ

1300

O

PI

T2

L

R

T

T1



Problem: ®

Two straights AC and CB intersect at C at a

chainage of 86.22 chains at a deflection angle of 620.

They are to be smoothly connected by a simple curve

of radius 12 chains. Find the tangent length, length of

curve and the chainages of the starting and end points

of the curve. Find also the length of the long chord.

Solution: Chainage at the point of intersection (C) = 86.22 chains

Deflection angle () = 620 Radius of curve = 12 chains.

Tangent Length T= chains 7.212

62tan12

2

ΔRtan

Length of the curve (l)

= chains 12.985180

6212

180

RΔ


length = 86.22 – 7.21 = 79.01 chains Chainage at point of tangency (T2) = Chainage at point of curve (T1) + Length of

curve=79.01+12.985 = 91.995 chains

Length of Long chord (L) = T1T2 = chains 12.3612

62sin122

2

Δ2Rsin

Problems: Two roads having a deviation angle of 450 48’ are to be joined by a 180 m

radius curve. Chainage at point of intersection is 3123.8 m. Calculate the necessary data if the curve is to be set by chain and offsets. Solution: By offsets from the long chords: Tangent Length T=

76m76.03mT

2

8445180tan

2

ΔRtan

Chainage of T1 = 3123.8 – 76 = 3047.8 m Length of the curve (L) =

m 143.88

180

8445180

180

RΔ

Chainage of T2 = 3047.8 + 143.88 = 3191.68 m Length of Long chord (L) T1T2 =

A

B’

B

T

Δ

Δ

T1 T2

Back tangent

Forward

tangent

D P

0

L/2 L/2 X

OX

P1

R

C E

Δ =620

O

C

B

L

R

T

A



140m140.08

2

8445sin1802

2

Δ2Rsin

Central offset ED

14.187m2

8445cos1180

2

Δcos1R

Starting from E offset distances at 10 m interval,

83.165x180y

70180x180L/2RxRy

22

22222222

13.89m83.16510180y

14.17m83.1650180y

2210

220

11.65m165.83130180y

13.05m165.83120180y

2230

2220

7.09m165.83150180y

9.67m165.83140180y

2250

2240

0.00m165.83170180y

3.875m165.83160180y

2270

2260

Problem: The length of the long chord of a simple curve of radius 400m is 100m. Find the lengths of the perpendicular offsets from the long chord at 10 m intervals. Solution: Radius = 400m, L= 100m, Interval = 10m

396.863x400y

100/2400x400y

L/2RxRy

22

2222

2222

3.012m396.86310400y

3.137m396.8630400y

2210

220

2.010m396.86330400y

2.637m396.86320400y

2230

2220

0m396.86350400y

1.132m396.86340400y

2250

2240



Problem:

Determine the ordinates of the points on a circular curve having a long chord of

100m and a versed sine of 5m. The ordinates are to be measured from the long

chord at an interval of 10m.

Solution:

Length of long chord (L) = 100m

Versed sine (O0) = 5m

Interval = 10m

The versined or midordinate

22O 2/LRRO

22 2/100RR5

m5.252

10

2550R

R525R5R50R2

22222

Ordinates at intervals of 10m

022

X ORxRO

Offsets from the tangents

1. Perpendicular offsets. 2. Radial offsets.

Perpendicular offsets:

22x xRRO

Radial offsets:

RxRO 22x



Problem:

If the approximate perpendicular offset for the mid-point of the circular curve deflecting through 760 38’ is 96.1m. Calculate the radius of the curve. Solution: Given: Δ = 760 38’, Ox = 96.1m Perpendicular offset method:

22x xRRO

The distance ‘x’ from T1 for locating the apex point.

m R62.02

38'67sinR

2

Δsin Rx

0

R215.0R785.0RR62.0RR1.6922

m98.446215.0

1.69R

Problems:®

Two roads meet at an angle of 1270 30’. Calculate the necessary data for setting out a curve of 15 chains radius to connect the two straight portions of the road if it is intended to set out the curve by chain and offsets only. Explain carefully how you would set out the curve in the field. Assume the length of chain as 20 meters. Solution:

Angle of intersection = 1270 30’ Deflection angle () = 1800 - 1270 30’ = 520 30’

Length of chain = 20 meter. Radius of curve = 15 chains x 20 = 300 m.


30'52tan300

2

ΔRtan

0

Radial offset method:

RxRO 22x

Assuming interval as 20 m First half of the curve is set from point of curve (T1)

m 0.6730020300O 2220

m 2.6630040300O 2240

m 5.9430060300O 2260

m 10.4830080300O 2280

m 16.23300100300O 22100

m 23.11300120300O 22120

m 31.06300140300O 22140

m 34.49300147.94300O 22147.94

Second half of the curve may be set from point of tangency (T2)



Perpendicular offset method:

22x xRRO

First half of the curve is set from point of curve (T1)

m 0.6720300300O 2220

m 2.6840300300O 2240

m 6.0660300300O 2260

m 10.8680300300O 2280

m 17.16100300300O 22100

m .2120300300O 22120 055

m 34.67140300300O 22140

m 39.01147.94300300O 22147.94

The distance ‘x’ from T1 for locating the apex point.

m 132.692

30'52sin300

2

Δsin Rx

0

m 30.94132.69300300O 22132.69

Second half of the curve may be set from point of tangency (T2)

Offsets from Chords Produced method:

This method is commonly

employed when a theodolite is not

available and it is necessary to set out a

curve only with a chain or a tape. The

curve is divided into number of chords

normally 20 or 30 metre length. As

continuous chainage is required along the

curve, two sub-chords generally occur,

one at the beginning and the other at the

end of the curve. The length of the first

sub-chord is equal to difference of the

chainage of all the full number of chains

just after commencement and the

chainage of the point of commencement.

Similarly, the length of the last sub-chord is equal to the chainage of full number of

chains just before the point of tangency and the chainage of point of tangency.

2R

C Ochord sub initial thefor Offset

21

1



2R

CCC Ochord second thefor Offset 1

2

2R

CCC O ......O Ochord 1-n to3 thefor Offset 1-n 43

th rd

2R

CCC Ochord final thefor Offset nn

n

Problem:®

Tabulate the necessary data for getting out a circular curve with the following

data:

Angle of intersection = 1440

Chainage of point of intersection = 1390 m

Radius of the curve = 50 m

The curve is to be set out by offsets from chords produced with pegs at every 10 m

of through chainage.

Solution: Chainage at the point of intersection = 1390 m Angle of intersection = 1440 Deflection angle () = 1800 - 1440 = 360

Radius of curve = 50 m.


36tan50

2

ΔRtan

0


3650

180

RΔ


length = 1390.00 – 16.25 = 1373.75 m


Length of initial sub chord (C1) = 1380 -1373.75 = 6.25 m Length of final sub chord (Cn) = 1405.17 - 1400.00 = 5.17 m

Length of normal chord (C) = 10 m

210

1380 - 1400 chords Normal of No

Total no of chords = 1 + 2 +1 = 4

m 0.39502

25.6

2R


221

1

m 1.625

502

1025.610

2R


2



m 250

10

R

C Ochord3 thefor Offset

22

3 rd

m 0.78

502

5.17105.17

2R

CCC Ochord final thefor Offset 44

4

Problem: ®

Two tangent intersect at chainage 59+60, the deflection angle being 50030’.

Calculate the necessary data for setting out a curve of 2 chains radius to connect the

two tangents if it is intended to set out the curve by offset from chords produced.

Take peg interval equals to 50 links, length of the chain being to 20 metres (100

links).

Solution: Chainage at the point of intersection = (59 + 60) i.e., 59 chains and 600 links. Note: 20 metre chain contains 100 links, each link is of length = 0.2 m and 30 metre chain contains 150 links, each link is of length = 0.2 m. Length of one chain = 20 m Chainage at the point of intersection = 59 x 20 + 60 x 0.2 = 1192.00 m

Deflection angle () = 500 30’

Radius of curve = 2 chains = 2 x 20 = 40 m.


30'50tan40

2

ΔRtan

0


30'5040

180

RΔ 0


length = 1192.00 – 18.87 = 1173.13 m



Length of normal chord (C) = 50 links = 50 x 0.2 =10 m

210


Total no of chords = 1 + 2 + 1 = 4

m 0.59402

6.87

2R


221

1

m 2.11

402

106.8710

2R


2

m 2.5402

101010

2R

CCC Ochord3 thefor Offset 3

rd

m 1.93

402

8.38108.38

2R

CCC Ochord final thefor Offset 44

4



Rankine’s Deflection Angle method (Tangential Deflection

Angle method)

The angle between the back and the chord joining the point of commencement of

the curve and the other point on the curve is generally known as deflection angle.

degree R

C90 minutes

R

C1718.9 angle tangential angle deflectionFirst 11

1

degree R

C90

minutes R

C1718.9 angle tangential , chord 1)-(n to chord2 1-n32

nd

degree R

C90 minutes

R

C1718.9 angle tangential Final nn

n

C1, C, Cn are the initial, normal and final sub chord respectively.

11 angle tangential Δ angle deflectionFirst

212nd ΔΔ angle deflection 2

223nd ΔΔ angle deflection 3

234th ΔΔ angle deflection 4

-

-

22-n1-n

th ΔΔ angle deflection 1-n

n1-nn

th ΔΔ angle deflection n

The deflection angle for any chord is equal to the deflection angle for the previous

chord plus the tangential angle for that chord

Check: The total deflection angle for T2 = ½ x deflection angle of the curve



Point Chainage Chord

length Tangential

angle

" ' 0

Deflection angle

" ' 0

Actual Theodolite

reading

" ' 0

T1 P1 Initial sub

chord 1 11 Δ

P2 Normal chord 2 212 ΔΔ

P4 Normal chord 3 323 ΔΔ

T2 Final sub chord

4 434 ΔΔ

Problem:

Two tangents intersect at chainage 1190 m, the deflection angle being 360.

Calculate all the data necessary for setting out a curve with a radius of 60 m by

deflection angle method. The peg interval is 10 m.

Solution: Chainage at the point of intersection = 1190.00 m

Deflection angle () = 360 Radius of curve = 60 m. Peg interval = 10 m.


36tan60

2

ΔRtan

0


3660

180

RΔ 0

Chainage at point of curve (TC) = Chainage at the point of intersection – tangent

length = 1190.00 – 19.50 = 1170.50 m

Chainage at point of tangency (TT) = Chainage at point of curve (TC) + Length of curve = 1170.50 + 37.70 = 1208.20 m


Length of normal chord (C) = 10 m (peg interval)

210


Total no of chords = 1 + 2 + 1 = 4

9.3" 32' 460

9.590degree

R

C90 angle tangential angle deflectionFirst 01

1

" 28.73 46' 460

1090 angle Tangential 0

32



54.76" 54' 360

8.290degree

R

C90 angle tangential Final 04

4


length Tangential angle

" ' 0

Deflection angle

" ' 0

Actual Theodolite reading

" ' 0

TC

1170.50

0" 0' 00

0" 0' 00

0" 0' 00

P1

1180 9.5 9.3" 32' 4 0

1 9.3" 32' 4 Δ 011 0" 32' 40

P2

1190 10 " 28.73 46' 4 0

2 38.03" 18' 9 ΔΔ 0212 40" 18' 90

P3

1200 10 " 28.73 46' 4 0

3

6.76" 5' 14 ΔΔ 0323

0" 5' 140

TT

1208.20 8.2 54.76" 54' 3 0

4 1.52" 0' 18 ΔΔ 0

434

0" 0' 180

Check: 0" 0' 182

36

2

ΔΔ 0

4

Problem:

Two roads meet at an angle of 1470 30’ at chainage 75+35. Calculate the necessary data for setting out a curve of 3 chains radius to connect the two straight portions of the road if it is intended to set out the curve by tangential angle method. Explain carefully how you would set out the curve in the field. Assume the length of chain as 20 meters. Take peg interval as 5m Solution:

Deflection angle () = 180 - 1470 30’ = 320 30’ Length of one chain = 20 meter. One link = 0.2 m Peg interval = 5m, Radius of curve = 3 chains = 3 x 20 = 60 m. Chainage at the point of intersection = 75 + 35 i.e., 75 chains and 35 links. = 75 x 20 + 35 x 0.2 = 1507 m Angle of intersection = 1470 30’


30'32tan60

2

ΔRtan

0

Length of the curve (L) = m 34.04180

30'3260

180

RΔ 0

Chainage at point of curve (TC) = Chainage at the point of intersection – tangent

length = 1507.00 – 17.49 = 1489.51 m

Chainage at point of tangency (TT) = Chainage at point of curve (TC) + Length of curve = 1489.51+ 34.04 = 1523.55 m

Length of initial sub chord (C1) = 1490 - 1489.51= 0.49 m Length of final sub chord (Cn) = 1523.55 - 1520.00 = 3.55 m Length of normal chord (C) = 5m (peg interval)

65



2.25" 14' 060

0.4990degree

R


1



" 14.37 23' 260

590.... angle Tangential 0

732

42" 41' 160

3.5590degree

R


8


length

Tangential angle Deflection angle Actual

Theodolite reading

TC 1489.51

- 0" 0' 00

0" 0' 00

0" 0' 00

P1 1490.00

0.49

2.25" 14' 00

2.25" 14' 00

0" 14' 00

P2 1495.00 5 " 14.37 23' 20

16.62" 37' 20

20" 37' 20

P3 1500.00 5 " 14.37 23' 20

30.99" 0' 50

40" 0' 50

P4 1505.00 5 " 14.37 23' 20

45.36" 23' 70 0" 24' 70

P5 1510.00 5 " 14.37 23' 20 59.73" 46' 90

0" 47' 90

P6 1515.00 5 " 14.37 23' 20 14.1" 10' 21 0

20" 10' 21 0

P7 1520.00

5 " 14.37 23' 20 28.47" 33' 41 0

20" 33' 41 0

TT 1523.55

3.55 42" 41' 10

10.47" 15' 61 0

0" 15' 61 0

Check : '15162

'3032

2

00

8

Problem:

Two straights intersect at a chainage of 1764m and at a deflection angle of

320. They are to smoothly joined by a 50 curve. Taking the peg interval at 30m work

out the data required to set out the curve by the deflection angle method. Least

count of the theodolite is 20’. Take length of chain = 30m.

Solution: Chainage at the point of intersection = 1764 m

Deflection angle () = 320, Degree of curve = 50,

Length of one chain = 30 meter. Peg interval = 30 m, For, 30m chain length,

m 8.3435

1719

curve of Degree

1719R


32tan8.433

2

ΔRtan

0

Length of the curve (L) = m 01.921180

328.433

180

RΔ 0

Chainage at point of curve (TC) = Chainage at the point of intersection – tangent length

= 1764 – 98.58 = 1665.42 m

Chainage at point of tangency (TT) = Chainage at point of curve (TC) + Length of curve = 1665.42 + 192.01

= 1857.43 m Length of initial sub chord (C1) = 1680 – 1665.42= 14.58 m



Length of final sub chord (Cn) = 1857.43 – 1830.00 = 27.43 m Length of normal chord (C) = 30m (peg interval)

530



53.68" 12' 18.433

58.4190degree

R


1

" 59.34 29' 28.433

0390.... angle Tangential 0

632

8.39" 17' 28.433

43.7290degree

R


7


length


Theodolite reading

TC 1665.42

- 0" 0' 00

0" 0' 00

0" 0' 00

P1 1680.00

14.58

53.68" 12' 10

53.68" 12' 10

0" 13' 10

P2 1710 30 " 59.34 29' 20

53.02" 42' 30

0" 43' 30

P3 1740 30 " 59.34 29' 20

52.36" 12' 60

0" 13' 60

P4 1770 30 " 59.34 29' 20

51.7" 42' 80 0" 43' 80

P5 1800 30 " 59.34 29' 20 51.04" 12' 11 0

0" 13' 11 0

P6 1830 30 " 59.34 29' 20 50.38" 42' 31 0

0" 43' 31 0

TT 1857.43 27.43 8.39" 17' 20

58.77" 59' 51 0

0" 0' 61 0

Check : 0" 0' 162

32

2

ΔΔ 0

0

7

Problem:

Two straights of a National Highway intersect at a chainage of 1534.5m and

at a deflection angle of 360. They are to smoothly joined by a 80 curve. Taking the

peg interval at 20m, work out the data required to set out the curve by the

deflection angle method.

Solution: Chainage at the point of intersection = 1534.5 m

Deflection angle () = 320, Degree of curve = 80, Peg interval = 20 m,

For, 20m Peg interval length, meters in R and degrees in D R

1146D

m 25.1438

1146

curve of Degree

1146R


36tan25.143

2

ΔRtan

0

Length of the curve (L) = m 90180

3625.143

180

RΔ 0



Chainage at point of curve (TC) = Chainage at the point of intersection – tangent length

= 1534.5 – 46.55 = 1487.95 m

Chainage at point of tangency (TT) = Chainage at point of curve (TC) + Length of curve = 1487.95 + 90

= 1577.95 m Length of initial sub chord (C1) = 1500 – 1487.95 = 12.05 m Length of final sub chord (Cn) = 1577.95 – 1560.00 = 17.95 m Length of normal chord (C) = 20m (peg interval)

320



35.36" 24' 225.143

05.1290degree

R


1

" 58.94 59' 325.143

0290 angle Tangential 0

432

23.05" 35' 325.143

95.1790degree

R


5


length


Theodolite reading

TC 1487.95

- 0" 0' 00

0" 0' 00

0" 0' 00

P1 1500 12.05

35.36" 24' 20

35.36" 24' 20

40" 24' 20

P2 1520 20 58.94" 59' 30

34.3" 24' 60

40" 24' 60

P3 1540 20 58.94" 59' 30

33.24" 22' 100

40" 22' 01 0

P4 1560 20 58.94" 59' 30

32.18" 24' 41 0 40" 24' 41 0

TT 1577.95 17.95 23.05" 35'30

55.23" 59' 71 0

0" 0' 81 0

Check : 0" 0' 182

36

2

ΔΔ 0

0

8



Compound curve:

A compound curve has two or more circular arcs of different radii, curving in

the same direction. It can be pictured as connecting three straights, the middle one

being the common tangent to the two arcs. Compound curves may be required

when space restrictions prevent a single

circular curve, and when there are property

boundaries.

A compound curve is tangential to the

three straights AB, BC and KM at T1, T2 and

N respectively.

The two circular arcs T1N and NT2

having centres at O1 and O2 meet at the

point of compound curvature denoted by N.

The points N, O1 and O2 are collinear.

Tangents AB and BC intersect at B, AB

and KM at K and BC and KM at M.

Let,

RS = Smaller radius (O1T1)

RL = Larger radius (O2T2)

TS = The Smaller tangent length (BT1)

TL = The Greater tangent length (BT2)

Δ = The deflection angle between the end tangents AB and BC.

= The deflection angle between the rear and common tangents i.e., AB and

KM = BKM

= The deflection angle between the common and forward tangents i.e., KM

and BC = BMK.

ts = The length of the tangent to the arc (T1N) having a smaller radius. i.e.,

T1K = KN.

tL = The length of the tangent to the arc (NT2) having a gretaer radius. i.e.,

T2M = NM.

Elements of Compound curve:

Δ = +

2

tanR tKNKT SS1

2

tanR tMNMT LL2

2

tanR 2

tanRNMKNKM LS

A

B Δ

T1

T2

N

RS C

M K

O2

O1

RL

TS TL

α

α

tL tS



Applying sine rule to the triangle BKM,

sin

sintt

sin

sintt

180sin

sinKM BK

KBMsin

KM

sin

BK LsLs

sin

sintt

sin

sintt

180sin

sinKM BM

KBMsin

KM

sin

BM LsLs

sin

sinttt BKKTBT T LengthTangent Ls

s11S

sin

sinttt BM MTBT TLengthTangent Ls

L22L

Length of the first curve = 180

RL S

s

Length of the Second curve = 180

RL L

L

Total length of curve (L) = LS + LL

Setting out the compound curve:

The compound curve can be set by the method of deflection angle.

The first curve is set out by setting the theodolite at T1 (Point of curve).

The Second curve is set out by setting the theodolite at N (Point of compound

curve).

Out of RS, RL, TS, TL, Δ, and , four parts of the curve must be known. The

remaining three can be calculated from equations.

Procedure of setting out curve:

1) B, T1 and T2 are located by linear measurements.

2) Theodolite at T1 set out the first branch curve by Rankine’s method.

3) Shift the instrument and set up at N. with vernier set to /2 behind zero, i.e.,

3600 -/2 taking a back sight on T1 and Plunge the telescope which is thus directed along T1N produced.

4) Set the vernier to the first deflection angle Δ1 as calculated for the second

branch , thus directing the line of sight to the first point on the second arc.

5) Continue the process until the end of the second arc is reached.



Problem:

Two straights BA and AC are intersected by a

line EF. The angles BEF and EFC measure 1400

and 1450 respectively. The radius of the first

arc is 600m and that of the second arc 400m.

find the chainages of the tangent points and

the point of compound curvature, given that

the chainages of the intersection point A is

3415 m.

Solution:

Given: RL = 600 m, RS = 400m, BEF = 1400,

EFC, Chainage of the intersection point

A = 3415m.

AEF = = 1800 - BEF =1800 - 1400 = 400

AFE = = 1800 - EFC =1800 - 1450 = 350

218.40m 2

40tan600

2tanR tENET LL1

126.12m 2

35tan400

2tanR tFNFT SS2

EF = EN + NF = 218.40 +126.12 = 344.52 m

Consider, Δle AEF,

AEF + AFE + EAF = 1800

EAF = 1800 – AEF – AFE = 1800 – 400 – 350 = 1050

Applying sine rule to the triangle AEF,

m 204.60 105sin

35sin52.344

EAFsin

sinEF AE

EAFsin

EF

sin

AE0

0

m 229.27 105sin

40sin52.344

EAFsin

sinEF AF

EAFsin

EF

sin

AF0

0

m 423 40.218 04.602 tAE )(T ATlengthTangent LL1

m 355.4 12.126 27.922 tAF )(T ATlengthTangent SS2

Length of the first curve = m 90.418180

40600

180

R 0L

Length of the Second curve = m 35.244180

35400

180

R 0S

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (TL)

= 3415 – 423 = 2992 m. Chainage at point PCC (N) = Chainage at the point of curve + Length of first curve = 2992 + 418.90 = 3410.9 m Chainage at point of tangency (T2) = Chainage at the point of PCC + Length of

second curve = 3410.9 + 244.35 = 3655.25 m

TS

B

3415 A

1400

T1

T2

N

RS =400m

C

F E

O2

O1

RL =600m

TL

α

α

1450

Δ

tL tS



June / July – 06CV44 – 10 marks

The following data refer to a right hand compound curve.

(i) Total deflection angle = 800.

(ii) Radius of the first arc = 200 m

(iii) Radius of the second arc = 250 m

(iv) Chainage of the point of

intersection = 1504.80 m

(v) Deflection angle of the first arc =

500.

Determine the chainages of the starting, the

point of compound curve and the point of

tangency.

Solution:

Given: RS = 200m, RL = 250 m, Total

Deflection angle (Δ) = 800.

Chainage of the intersection point =

1504.80m.

Let, BA and BC be the back and forward

tangent. BMN = = 500,

Deflection angle of the first arc =BMN = () = 500,

Δ = + = 800

Therefore, Deflection angle of the second arc BNM = = Δ - Deflection angle of

the first arc

Deflection angle of the second arc () = 800 - 500 = 300.

Let ‘P’ be the point on the line MN

m 93.26 2

50tan200

2tanR tMTMP

0

ss1

m 67 2

30tan250

2tanR tNTPN

0

LL2

MN = MP + PN = 93.26 + 67 = 160.26 m

Consider, Δle BMN,

BMN + BNM + MBN = 1800

MBN = 1800 – BMN – BNM = 1800 – 500 – 300 = 1000

Applying sine rule to the triangle BMN,

m 81.23 100sin

30sin26.160

MBNsin

sinMN BM

MBNsin

MN

sin

BM0

0

m 124.66 100sin

50sin26.160

MBNsin

sinMN BN

MBNsin

MN

sin

BN0

0

m 174.49 26.93 1.238 TMBM (T)BT lengthTangent 11

A

B

Δ = 800

T1

T2

P

RS =200m C

N M

O2

O1

RL = 250

m

TS TL

=500 = 300

tL tS



m 191.66 67 24.661 TNBN (T)BT lengthTangent 22


50200

180

R 0S


30250

180

R 0L

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (BT1)

= 1504.80 – 174.53 = 1330.27 m. Chainage at point of CC (P) = Chainage at the point of curve + Length of first

curve = 1330.27 + 174.53 = 1504.8 m Chainage at point of tangency (T2) = Chainage at the point of CC + Length of

second curve = 1504.8 + 130.90 = 1635.7 m

Problem:

Two straights AB and BC are intersected by a line D1D2. The angles BD1D2

and BD2D1 are 400 30’ and 360 24’

respectively. The radius of the first arc is 600

m and that of the second arc is 800 m if the

chainage of intersection point B is 8248.10m.

Determine the chainages of the starting, the

point of compound curve and the point of

tangency.

Solution:

Given: Radius of the first arc (RS) = 600m,

Radius of the second arc (RL) = 800 m,

Chainage of the intersection point =

8248.10m

Deflection angle of the first arc = BD1D2 =

= 400 30’,

Deflection angle of the second arc = BD2D1

= = 360 24’.

Total Deflection angle (Δ) = + = BD1D2

+ BD2D1 = 400 30’ + 360 24’ = 760 54’

Let ‘P’ be the point on the line D1D2

m 221.35 2

'3040tan600

2tanR tTDPD

0

ss111

m 263.03 2

'2436tan800

2tanR tTDPD

0

LL222

D1D2 = D1P + PD2 = 221.35 + 263.03 = 484.38 m

Consider, Δle BD1D2,

A

B

Δ = 760 54’

T1

T2

P

RS =600m C

D2 D1

O2

O1

RL = 800m

TS TL

=400 30’ = 36024’

tL tS P



BD1D2 + BD2D1+ D1BD2 = 1800

D1BD2 = 1800 – BD1D2 – BD2D1 = 1800 – 400 30’ – 360 24’ = 1030 6’

Applying sine rule to the triangle D1BD2,

m 295.12 '6103sin

'2436sin38.484

BDDsin

sinDD BD

BDDsin

DD

sin

BD0

0

21

211

21

211

m 323 '6103sin

'3040sin38.484

BDDsin

sinDD BD

BDDsin

DD

sin

BD0

0

21

212

21

212

m 516.47 35.221 95.122 TDDB (T)BT lengthTangent 1111

m 586.03 03.263 233 TDDB )(TBT lengthTangent 222L2


'3040600

180

R 0S


'2436800

180

R 0L

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (BT1)


curve = 7731.63 + 424.12 = 8155.75 m Chainage at point of tangency (T2) = Chainage at the point of CC + Length of

second curve = 8155.75 + 508.30 = 8664.05 m.

June / July 2018 – 15 CV46 – 10 marks:

A compound consisting of two simple

circular curves of radii 350 m and 500 m is

to be laid out between two straights T1I and

IT2. PQ is the common tangent and D is the

point of compound curvature. The angle

IPQ and IQP are 550 and 250 respectively.

Given the chainages of point of intersection

are 1800.00m, Calculate the chainages of

T1, T2, and D.

Solution:

Given: Radius of the first arc (RS) = 350m,


Let, IA and IC be the back and forward

tangent.

Deflection angle of the first arc = IPQ =

= 550.

A

I

Δ = 800

T1

T2

RS =350m C

Q P

O2

O1

RL = 500m

TS TL

=550 = 250

tL tS D



Deflection angle of the second arc = IQP = = 250.

Total Deflection angle (Δ) = + = IPQ + IQP = 550 + 250 = 800.

‘D’ be the point on the line PQ.

m 182.20 2

55tan350

2tanR tPDPT

0

ss1

m 110.85 2

25tan500

2tanR tQTDQ

0

LL2

PQ = PD + DQ = 182.20 + 110.85 = 293.05 m

Consider, Δle PIQ,

IPQ + IQP + PIQ = 1800

PIQ = 1800 – IPQ – IQP = 1800 – 550 – 250 = 1000

Applying sine rule to the triangle PIQ,

m 125.76 100sin

25sin05.293

PIQsin

sinPQ PI

PIQsin

PQ

sin

IP0

0

m 307.96 20.182 25.761 TPIP (T)IT lengthTangent 11


55350

180

R 0S


25500

180

R 0L

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (IT1)

= 1800 – 307.96 = 1492.04 m. Chainage at point of CC (D) = Chainage at the point of curve + Length of first curve = 1492.04 + 335.98 = 1828.02 m Chainage at point of tangency (T2) = Chainage at the point of CC + Length of

second curve = 1828.02 + 218.17 = 2046.19 m.

Problem: Two straights AC and BC meet at C at a

deflection angle of 63029’. They are to be

smoothly connected by a compound curve

consisting of two circular branches AD and DB

of radii 18 chains and 36 chains respectively. If

the length of the first tangent AC is 15.78

chains, find (i) The angles subtended by the

two branches at their respective centres (ii)

Length of the second tangent CB, (iii)

Chainages of the straight point, point of

compound curvature and the end point of the

curve if the chainage of the intersection point

A

C Δ = 630 29’

T1

T2

RS =18chains B

D2 D1

O2

O1

RL = 36chains

TS =15.78 chains TL

Δ1 Δ2

Δ1

Δ2

tL tS D



is 243.60 chains.

Solution: Given:

Deflection angle Δ = 630 29’, Radius of the first arc (RS) = 18 chains,

Radius of the second arc (RL) = 36 chains,

Length of the first tangent AC = 15.78 chains,

Chainage of the intersection point C = 243.60 chains.

TS sin Δ = RS versin Δ + (RL- RS) versin Δ2

15.78 x sin 630 29’ = 18 x (1 - cos630 29’) + (36 - 18) x (1 – cos Δ2)

14.12 = 9.9637 + 18 x (1 – cos Δ2)

(1 – cos Δ2) = 0.2309

012 73.392309.01cos

23.756 39.73 29' 63ΔΔΔ

ΔΔΔ

00021

21

TL sin Δ = RL versin Δ - (RL- RS) versin Δ1

TL x sin 630 29’ = 36 x (1 – cos 630 29’) - (36 - 18) x (1 – cos 23.7560)

chains567.20'2963sin

525.1928.19T

0L

Length of the first curve = chains463.7180

756.2318

180

R 01S

Length of the Second curve = chains963.24180

73.3936

180

R 02L

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (CT1)

= 243.6 – 15.78 = 227.82 chains. Chainage at point of Compound Curve (D) = Chainage at the point of curve +

Length of first curve = 227.82 + 7.463 = 235.283 chains Chainage at point of tangency (T2) = Chainage at the point of CC + Length of

second curve = 235.283 + 24.963 = 250.246 chains.



Problem: Two straights AC and BC meet at C at a deflection angle of 63029’. They are

to be smoothly connected by a compound curve consisting of two circular branches

AD and DB of radii 18 chains and 36 chains respectively. If the length of the first

tangent AC is 15.78 chains, find (i) The angles subtended by the two branches at

their respective centres (ii) Length of the

second tangent CB, (iii) Chainages of the

straight point, point of compound curvature

and the end point of the curve if the

chainage of the intersection point is 243.60

chains. Take length of one chain = 20 m

(100 links)

Solution: Given:

Deflection angle Δ = 630 29’, Radius of the

first arc (RS) = 18 chains = 18 x 20 = 360

m,

Radius of the second arc (RL) = 36 chains

= 36 x 20 = 720 m,

Length of the first tangent AC = 15.78

chains = 15.78 x 20 = 315.6 m,

Chainage of the intersection point C =

243.60 chains = 243.60 x 20 = 4872 m

TS sin Δ = RS versin Δ + (RL- RS) versin Δ2

315.6 x sin 630 29’ = 360 x (1 - cos630 29’) + (720 - 360) x (1 – cos Δ2)

282.40 = 199.27 + 360 x (1 – cos Δ2)

(1 – cos Δ2) = 0.2309

012 73.392309.01cos

23.756 39.73 29' 63ΔΔΔ

ΔΔΔ

00021

21

TL sin Δ = RL versin Δ - (RL- RS) versin Δ1

TL x sin 630 29’ = 720 x (1 – cos 630 29’) - (720 - 360) x (1 – cos 23.7560)

m33.411'2963sin

5.3056.398T

0L

Length of the first curve = m26.149180

756.23360

180

R 01S

Length of the Second curve = 26.499180

73.39720

180

R 02L


= 4872 – 315.6 = 4556.4 m. Chainage at point of Compound Curve (D) = Chainage at the point of curve +

Length of first curve = 4556.4 + 149.26 = 4705.66 m

A

C

Δ = 630 29’

T1

T2

RS =18chains B

D2 D1

O2

O1

RL = 36chains

TS =15.78 chains TL

Δ1 Δ2

Δ1

Δ2

tL tS D



Chainage at point of tangency (T2) = Chainage at the point of CC + Length of second curve

= 4705.66 + 499.26 = 5204.92 m.

Problem:

Two straights AC and CB have bearing of

590 30’ and 1270 30’ intersect at C at a

chainage of 1456.5 m. Two points M and N are

located on AC and BC so that the bearing of MN

is 1050 30’. The straights AC and CB are to be

connected by a compound curve consisting of

arcs of radii 220 m and 275 m respectively. If

the line MN is the common tangent to the two

branches, find the chainages of the starting

point, point of compound curvature and the

end point of the curve.

Solution:

Bearing of AC = 590 30’ and Bearing of CB =

1270 30’, Chainage of the intersection point =

1456.5 m. Bearing of MN = 1050 30’.

Radius of the first arc (RS) = 220 m,


Δ = Bearing of CB – Bearing of AC

Δ = 1270 30’ – 590 30’ = 680

CMN = = Bearing of MN – Bearing of

AC

= 1050 30’ – 590 30’ = 460.

Δ = + = 680.

= CNM = Δ – = 680 – 460 =

220.

Let ‘D’ be the point on the line MN

m 93.38 MD

2

46tan220

2tanR tMTMD

0

ss1

m 53.45 DN

2

22tan275

2tanR tNTDN

0

LL2

MN = MD + DN = 93.38 + 53.45 =146.83 m.

Consider, Δle CMN,

CMN + CNM + MCN = 1800

MCN = 1800 – CMN – CNM = 1800 – 460 – 220 = 1120

Applying, sine rule to the triangle CMN,

B

590 30’ D

A

Δ = ?

RS =220 m

M

O2

O1

RL = 275 m

1456.5m @C

N

Δ1

Δ2

1270 30’

1050 30’

B

O2

O1 RL = 275 m

=220

=460

=460 =220

D

tL tS

A

1456.5m @C

Δ = 680

T1 T2

N M

TS TL

RS = 220 m



m 59.32 112sin

22sin83.146

MCNsin

sinMN CM

MCNsin

MN

sin

CM0

0

m. 113.92 112sin

46sin83.146

MCNsin

sinMN CN

MCNsin

MN

sin

CN0

0

m 152.7 38.93 9.325 TMCM )(TCT lengthTangent 1S1

m 167.37 45.5392.311 TNCN )(TCT lengthTangent 2L2


46220

180

R 0S


22275

180

R 0L



curve = 1298.8 + 176.63 =1475.43 m. Chainage at point of tangency (T2) = Chainage at the point of CC + Length of

second curve = 1475.43 + 87.16 = 1562.59 m.

Problem:

Two straights AC and BC are intersected

by a third line MN such that CMN = 450 30’ and

CNM = 350 30’ and the distance MN = 320 m.

find the radius of the curve which will be

tangential to the three lines AC, MN and CB. If

the chainages of the intersection point C is

4875.5m. Calculate the chainages of the point of

curve A and the point of tangency B.

Solution: Given:

CMN = = 45030’, CNM = = 350 30’,

Chainage of the intersection point C = 4875.5m.

Deflection angle Δ = + = 45030’ + 35030’ =

810 0’. Let P be the point of MN.

m R 0.419 2

'3045tanR

2tanR tMP

0

L

m R 0.320 2

'3035tanR

2tanR tNP

0

S

MN = 320m = MP + PN = 0.419 R + 0.320 R = 0.739 R

m 433. 739.0

320 R

O

C

P N

TS TL

A

Δ

T1 T2

R

B

M

R

=450 30’

α

=350 30’

tL tS



m 181.43 2

'3045tan433

2tanR tMP

0

L

m 138.56 2

'3035tan433

2tanR tNP

0

S

Consider, Δle MCN,

CMN + CNM + MCN = 1800

MCN = 1800 – CMN – CNM = 1800 – 450 30’ – 350 30’ = 990

Applying sine rule to the triangle AEF,

m 188.14 99sin

'3035sin320

MCNsin

sinMN CM

MCNsin

MN

sin

CM0

0

m 231.10 99sin

'3045sin320

MCNsin

sinMN CN

MCNsin

MN

sin

CN0

0

m 369.6 43.181 88.141 tCM (T)CT lengthTangent L1

m 369.66 56.138 31.102 tCN (T)CT lengthTangent S2

Length of the curve =

m 14.612180

'3035'3045433

180

R 00

Chainage at point of curve (T1) = Chainage at the point of intersection – tangent length (T)

= 4875.5 – 369.6 = 4505.9 m. Chainage at point of tangency (T2) = Chainage at point of curve (T1) + Length of

curve = 4505.9 + 612.14 = 5118.04 m

Problem: Dec 2017 / Jan 2018 – 06 marks

Two straights with a total deflection angle of 720 are to be connected by a

compound curve of two branches of equal length. The radius of the first branch is

300 m and that of the second branch is 400m, chainage of intersection point is

1500m. Calculate the chainage of tangent points and that of point of compound

curvature (PIC).

Solution:



Reverse (or Serpentine) curves:

A reverse curve is composed of two circular arcs curving in opposite directions

with a common tangent at their junction. The point at which the two arcs join is

called the point of reverse curvature.

Reverse curves are used when the straights are parallel or intersects at a very

small angle. They are frequently used in railway sidings, and sometimes on roads,

and railway tracks designed for low speeds. They should be avoided as far as

possible on highways and main railway lines where speeds are necessarily high for

the following reasons:

(1) They involve a sudden change of cant from one side to the other which is

practically impossible.

(2) The curves cannot be properly super elevated at the point of reverse

curvature.

(3) The sudden change in direction is objectionable.

(4) Steering is very dangerous in the case of highways. It is always preferable,

whenever practicable, to insert a short straight length or a reversed spiral

between the two branches of the reverse curve.

Reverse curve between parallel straights:

Let,

R1 = Smaller radius (O1A = O1D)

R2 = Greater radius (O2D = O2B)

1 = The angle subtended at the centre by the arc having a smaller radius R1.

2 = The angle subtended at the centre by the arc having a greater radius R2.

V = The perpendicular distance between straights AM and BN.

h = The distance between the perpendicular A and B

L = The length of the line joining the tangent points A and B.

Let, AM and BN be the parallel tangents.

A

Δ1

Δ2

B

D

E

F

Δ

Δ

V G H

O1

O2

M

N

h

R1

R2

Δ1/2

Δ2/2

J



Through the point of reverse curvature D, draw a line parallel to the

straights AO1 and BO2 in, cutting the perpendiculars at G and H.

1 =2 =

Perpendicular distance =(V) = AG + HB = (O1A – O1G) +(O2B – O2H)

O1A = R1, cosRcosDO GO DO

GO cos 111

1

1

versinR cos-1R cosR-R AG 1111

O2B = R2, cosRcosDO HO DO

HO cos 222

2

2

versinR cos-1R cosR-R HB 2222

versinRR cos-1R cos-1R V 2121

ΔversinRR V 21

21 RR

Vcos-1

ADB are collinear, AD + DB = L = Total length

AD = AP + PD

Consider triangle O1PA

2

sin R2

sinAO AP AO

AP

2sin 11

1

2

sinR2 DBy Similarlli , 2

sinR2AD 21

2

sinR2 2

sinR2 L length Total 21

2

ΔsinRR2 L length Total 21

Consider triangle ABJ

L

VRR2

2sinRR2 L

L

V

AB

AJ

2sin

2121

RR2V L 212

RR2VL 21

Distance between the end points of the reverse curve measured parallel to the

straights = h = GD + DH.

A

B

D

E

F

Δ/2

O1

O2

P

Q

R1

R2

Δ/2

Δ/2 Δ/2

Δ/2 Δ/2



Consider triangle GO1D

sinR sinDO GD

DO

GDsin

11

1

Consider triangle BO2D

sinR sinDO DH

DO

DHsin

22

2

sinR sinRh 21

sinΔRR h 21

Length of the first curve AD (l1) = 180

ΔR1

Length of the second curve DB (l2) = 180

ΔR 2

Note: If the radius of the two curves are equal

Δversin2R V

4V

LR curve of Radius

4VR 2R2VL2

2

Δsin4R L length Total

sinΔ2R h

Note: If the curve is short, it may be set out offsets from the long chords

AD and DB

Problem:

Two parallel straights 20 m apart are to be connected by a reverse curve consisting

of arcs of the same radius. The distance between the end points of the curve is 200

m. find the common radius. Find also the length of the whole curve.

Solution:

Given: V = 20 m, R1 = R2 = R, L =

200 m, R = ?, Length of the first curve (l) =

?

m 500 204

002

4V

L R curve the of Radius

22

A

Δ

Δ

B

C

R

O

O

2

20m 200 m



00

01-

11.48 5.742

5.740.1sin2

0.10 200

20

L

V

2sin

Length of the curve AB (l) = Length of the first curve AD (l1) + Length of the second

curve DB (l2) = m334.200180

48.115002

180

RΔ2

Problem:

A reverse curve AB is to be set out between two parallel railway tangents

32m apart. If the two arcs of the curve are to have same radius and the distance

between the tangent points A and B is 160 m, calculate the radius. The curve is to

be set out from AB at 10m intervals along that line. Calculate the lengths of the

offsets.

Solution:

Given: The Length (L) between tangent

points A and B, i.e., L = 160 m, V = 32 m

4RvL

m200324

601

4V

LR curve of Radius

22

The maximum ordinate at the mid-points of

AC and BC

2

20

2

LRRO

Here, L = Distance between AC = 160 / 2 = 80 m

m 4.042

80200200O

2

20

The various offsets from long chord may be calculated from Eqn.

n22

n ORxRO

X being measured from (F), the mid-point of AC.

m79.34.0420010200O 2210

m04.34.0420020200O 22

20

m78.14.0420030200O 2230

m04.0420040200O 2240

The ordinates of the other half of the arc AC are the same as above,

For each branch, offsets are:

0.00, 1.78, 3.04, 3.79, 4.04, 3.79, 3.04, 1.78, 0.00 m

A

Δ

Δ

B

C

R

O

O

2

32m 160 m

10m A C F

4.04

3.79 3.04 1.78 0 3.79

3.04 1.78 0

80m

20m



Problem:

A reverse curve is to be set out between two parallel tangents 18 m apart. The

distance between the tangent points C and D is 180 m and the two arcs of the curve

have the same radius. Calculate the radius, and the offsets at 7.5 m intervals, the

curve being set out by means of offsets from CD.

Solution:

Given: The length (L) between tangent

points C and D, i.e., L = 180 m,

V = 18m.

4RvL

m450814

801

4V

LR curve of Radius

22

The maximum ordinate at the mid-points

of CE and DE

2

20

2

LRRO

Here, L = Distance between CE = 180 / 2 = 90 m

m 26.22

90450054O

2

20

The various offsets from long chord may be calculated from Eqn.

n22

n ORxRO

X being measured from (F), the mid-point of CE.

m20.226.24505.7450O 227.5

m00.226.245051450O 22

15

m70.126.24505.22450O 2222.5

m26.126.245030450O 2230

m70.026.24505.37450O 22

37.5

m0.026.245045450O 2245

The ordinates of the other half of the arc CE are the same as above,

For each branch, offsets are:

0.00, 0.70, 1.26, 1.70, 2.00, 2.20, 2.26, 2.20, 2.00, 1.70, 1.26, 0.70, 0.00.

Problem:

Two parallel lines which are 469 m apart are joined by a reverse curve ABC

which deflects to the right by an angle of 300 from the first straight. If the radius of

the first arc is 1400 m and the chainage of A is 2500m, calculate the radius of the

second arc and the chainage of points B and C.

Solution:

C

Δ

Δ

D

E

R

O

O

2

18m 180 m

F 7.5m

90m

C E

1.70 2.26 2.20

1.26 0 0 0.70 1.70

2.20 1.26

0.70 2.0 2.0

15m



Let A and C be the points of tangencies and B point of reverse

curvature (PRC). The distance between the lines V = 469m

cos-1RR versinRR V 2121

30cos-1R4001 469 02

m 2100.66 1400

30cos-1

469R

02

m04.733 ABcurvefirst the of Length

180

301400

180

ΔR ABcurvefirst the of Length

01

m90.1099180

3066.2100

180

ΔR BC curve second the of Length

02

Chainage at B = Chainage at A + length of first curve

Chainage at B = 2500 + 733.04 = 3233.04 m

Chainage at C = Chainage at B + length of second curve

Chainage at C = 3233.04 + 1099.90 = 4332.94 m


A reverse curve ACB is to be set out between two parallel tangents 30 m

apart. The distance between the tangent points A and B is 120 m. Find

(i) The radius R if R1 = R2

(ii) The radius R2 if R1 = 100m

Also calculate the lengths of both the arcs of

reverse curve.

Solution:

Given: The length (L) between tangent points

C and D, i.e., L = 120 m, V = 30 m.

Case 1: R = ?, if R1 = R2

4RvL

m120304

201

4V

LR curve of Radius

22

Case 2: R2 = ?, if R1 = 100m

RR2V L 212

m140100- 302

201 R-

2V

LR

2

1

2

2

cos-1RR versinRR V 2121

cos-1401001 30

A

Δ

Δ

B

C

R

O

O

2

30m 120 m

A

Δ=300

Δ=300

C

B

O1

O2

V=469m R1 =1400 m

R2 = ?

A

Δ=?

Δ=?

C

B

O1

O2

V=30 m R1 =100 m

R2 = ?



0.125

240

30cos-1

18" 57' 280.125-1 cos 01

m54.50180

18" 57' 28100

180

ΔR ACcurvefirst the of Length

01

m75.70180

18" 57' 28140

180

ΔR CB curvefirst the of Length

02

Problem:

Two parallel railway lines are to be connected by a reverse curve each having

the same radius. If the lines are 12 m apart and maximum distance between tangent

points measured parallel to the straight is 48 m (h). Find the maximum allowable

radius.

If however both the radii are to be different, calculate the radius of 2nd

branch if that of the 1st branch is 60 m. Also calculate the length of both the

braches.

Solution:

Let A and B be the points of tangencies and C point of reverse curvature

(PRC). The distance between the lines V = 12 m, h = 48 m.

25.048

12

h

V

2 tan

20.95" 4' 28225.0tan 0-1

Case 1:

sinR sinRh 21

R1 = R2 , sinR2h

m 51 20.95" 4' 28sin2

48

sin2

hR

0

Case 2:

sinR sinRh 21

20.95" 4' 28sinR 20.95" 4' 28sin0648 02

0

m 42 20.95" 4' 28sin

20.95" 4' 28sin0648R

0

0

2

m4.29180

20.95" 4' 2860

180

ΔR ACcurvefirst the of Length

01

m58.20180

20.95" 4' 2842

180

ΔR CB curvefirst the of Length

02

R1

Δ/2

Δ/2 J

A

Δ1

Δ2

B

C

E

F

Δ

Δ

V = 12 m

G H

O1

O2

H = 48m

R2



Transition curves:

On railways and highways it is the common practice to introduce a curve of

varying radius called a transition curve between the tangent and a circular curve.

The transition curve is also called the spiral or easement curve. It is also inserted

between the two branches of a compound or reverse curve.

The objects of introducing a transition curve at each end of the circular curve

are as follows:

(1) To accomplish gradually the transition from the tangent to the circular

curve, and from the circular curve to the tangent.

(2) To obtain a gradual increase of curvature from zero at the tangent point to

the specified quantity at the junction of the transition curve with the

circular curve.

(3) To provide a satisfactory means of obtaining a gradual increase of super

elevation from zero on the tangent to the specified amount on the main

circular curve so that the full super elevation is attained simultaneously

with the curvature of the circular curve at the junction of the transition

curve with the circular curve.

A transition curve should fulfil the following conditions when it is inserted between

the tangent and the circular curve.

(1) It should meet the original straight tangentially.

(2) It should meet the circular curve tangentially.

(3) Its radius at the junction with the circular curve should be the same as that

of the circular curve.

(4) The rate of increase of curvature along the transition curve should be the

same as that of increase of super elevation.

(5) Its length should be such that the full super elevation is attained at the

junction with the circular curve.

The types of the transition curve which are in common use are

(1) A cubic parabola – used in railways

(2) A clothoid or spiral – used in railways, and

(3) Bernoulli’s lemniscate – used in highways.

In order to admit a transition curve, the main circular curve requires to be

shifted inwards. When the transition curves are inserted at each end of the main

circular curve, the resulting curve is called the combined or composite curve.



A.M.I.E. 1987 Winter

Problem:

A road bend which deflects 800 is to be designed for maximum speed of 120 kmph,

If maximum centrifugal ratio is ¼ and the maximum rate of change of radial

acceleration is 30 cm/s2, the curve consisting of a circular arc combined with cubic

spirals. Calculate

i. The radius of circular curve arc

ii. The requisite length of the transition curve and

iii. The total length of composite curve.

June / July 2016 -10CV/CT44 – 08 marks

A road bend which deflects 760 is to be designed for maximum speed of 80 km per

hour if the maximum centrifugal ratio is ¼ and the maximum rate of change of

radial acceleration is 0.3m/s3. Calculate

(i) The radius of circular curve and

(ii) The length of the transition curve

(iii) Total length of transition curve.

June / July – 06CV44 8marks

A road bend which deflects 850 is to be designed for maximum speed of 80 km per

hour with a curve consisting of a circular are combined with cubic parabola. If

maximum centrifugal ratio is ¼ and the maximum rate of change of radial

acceleration is 0.3m/s3. Calculate

i. The radius of circular curve and

ii. The length of the transition curve.



Vertical curves:

A vertical curve is a curve lying in a vertical plane which connects two different

gradients.

Such a curve is introduced in highway and railway work in order to round off the

angle and to obtain a gradual change in grade. Abrupt changes in grade are thus

avoided at the apex of the curve. The vertical curve can be of any shape i.e., circular

or parabola but for simplicity of calculation work, the latter is preferred and used

invariably.

Grade:

The gradient or grade may be defined as a proportional rise or fall between two

points along a straight line. It is expressed either as a percentage or ratio.

1. As a percentage (%):- Vertical rise or fall per 20 m chain horizontals e.g:

1%, 2%, 5%. Etc.

2. As a ratio:– one vertical rise or fall in horizontals e.g: 1 in 200, 1 in 500

etc.

The grades are further classified into two categories.

a) Up-grades or positive grades.

b) Down grades or negative grades.

A grade is classified as upgrade if elevations along it increase whereas it is classified

as downgrade if the elevations decrease. It is important to note that these

classifications depend upon the direction of the movement of the vehicles. An up-

grade becomes a down grade if the direction of motion of the vehicle is reversed.

The gradient of a highway or railway is expressed in two ways.

(i) As a percentage, e.g: 3% and (ii) as 1 in n where n is the horizontal distance

in meters corresponding to 1 m rise or fall. E.g: 1 in 50. An ascending or up

grade rising to the right denoted by a plus (+) sign and a descending or down

grade falling to the right by a minus (-) sign.

The permissible rate of change in gradient for first class railways is

recommended as 0.06% per 20 m station at summits and 0.05% per 20 m station

for sags.

For second class railways permissible rate of change of gradient is 0.12% per

20 m station at summits and 0.1% per 20 m station for sags.

For small gradient angles there is no difference between a parabola and a

circular arc.

Suppose the gradient at the beginning of a summit curve is 1.25% and if the

rate of change of gradient is 0.05% per 20 m station, the gradients at the various

stations will be as follows:



Station Distance from the beginning of the vertical

curve (m)

Gradients

0 0 1.25%

1 20 1.20%

2 40 1.15%

3 60 1.10%

4 80 1.05%

5 100 1%

Change in gradient between two gradients:

When two gradients are like gradients

Change in gradient = Numerical difference between the gradients.

When two gradients are unlike gradients

Change in gradient = Numerical sum of the gradients.

Types of vertical curves:


With neat sketches explain the types of vertical

curves.

1) An upgradient (+g1%) followed by a down gradient

(-g1%). Such curve is called a convex curve or a

summit curve.

Change in gradient = Numerical sum of the gradients.

Change in gradient =(g1 + g2) %.

2) A downgradient (-g1%) followed by a

upgradient (+g2%). Such curve is also called a

sag curve or a concave curve.

Change in gradient = Numerical sum of the

gradients.

Change in gradient = (g1 + g2) %

-g1 % +g2 %



3) An upgradient (+g1%) followed by a steeper upgradient (+g2 %).

Such curve is also called a sag curve or a concave

curve.

Change in gradient = Numerical difference between

the gradients.

Change in gradient = (g2 – g1) %

4) A downgradient (-g1 %) followed by a another

steeper downgradient (-g2 %). Such curve is also called

a summit curve or a convex curve.

Change in gradient = Numerical difference between the

gradients.

Change in gradient = (g2 – g1) %

+g1 %

+g2 %

Date post:	07-Aug-2021
Category:	Documents
Upload:	others
View:	47 times
Download:	1 times

MODULE 3 CURVE SURVEYING Curves Necessity Types, Simple curves, Elements… · 2020. 3. 28. ·...

Documents