9/27/2012

1

ADVANCE ENGINEERING SURVEYING (3+1)

Dr. Mohsin Siddique

Asst. Prof.

Dept. of Civil Engineering

FAST-NU

Lecture 5: Reverse Curve

25/09/2012

Reverse curve

• Reverse Curve: A reversecurve consists of two arcbending in opposite directionswith common tangent atjunction of arcs.

• Their centers lie on oppositesides of the curve.

• Their radii may be either equalor different and they have onecommon tangent.

• Reverse curves are generallyused to connect two parallelroads or railway lines or whentwo lines intersect at a verysmall angle.

9/27/2012

2

Reverse curve

• These curves are suitable for railways, city roads etc.

• But they should be avoided as far as possible for important tracks or highways for the following reasons:

▫ 1. Superevelations cannot be provided at the point of reverse curvature

▫ 2. A sudden change of direction would be dangerous for a vehicle

▫ 3. A sudden change of superevelation causes discomfort to passengers

▫ 4. Carelessness of the driver may cause the vehicle to overturn over a reverse curve.

• Setting out of reverse curve

• They are generally short, and hence they are set out by the chain and tape method.

Reverse curveNotations

• 1. AB and EF are the straight lines,

• 2.BE is the common tangent

• 3. C is the point of reverse curve

• 4. T1 and T2 are the tangent points

• Φ is the angle of intersection between the two straights

• Φ1 and Φ2 are the deflection angles of the common tangent

• R1 and R2 are the radii of the arcs

9/27/2012

3

Reverse Curves

Cases

• Case I – When the straights are non-parallel

• Suppose AB, BC and CD are lines of an open transverse along the alignment of a road as shown in figure.

• AB and CD when produced meet at a point E, where Φ is the angle of intersection.

• It is required to connect the lines AB and CD by a reverse curve with BC as the common tangent.

• Let

▫ Φ1= angle of deflection for the first arc

▫ Φ2= angle of deflection for the second arc

▫ Φ= angle of intersection between AB and CD

▫ T1 and T2=tangent points

▫ F=point of reverse curvature

▫ R=common radius for the arcs

Reverse Curves

Cases

• 1. Tangent length of first arc

• 2. Tangent length of second arc

• 3. Length of common tangent

• 4. Length of first curve

• 5. Length of second curve

• 6. Chainage of T1=chainage of B-T1B

• 7. Chainage of F=chainage of T1+1st curve length

• 8. Chainage of T2=chainage of F+2nd curve length

[ ]2/tan 11 φRBFBT ==

[ ]2/tan 22 φRCFCT ==

[ ] [ ]2/tan2/tan 21 φφ RR

CFBFBC

+=

+=

180/11

oRFT φπ=

180/22

oRFT φπ=

9/27/2012

4

Reverse Curves

Cases

• Case II – When the straight lines are parallel

• In figure, PQ and RS are two parallel lines a distance y apart. It is required to connect the lines PQ and RS by a reverse curve having equal radii. Line AB is drawn parallel to PQ or RS through point C.

• R= Common Radius

▫ C= point of reverse curvature

▫ T1 and T2=tangent points

▫ Φ= angle subtended at the centre by the curve

▫ T1T2=l=length of joining T1 and T2

▫ x= perpendicular distance between T1 and T2

▫ y=parallel distance between lines PQ and RS

Reverse Curves

Cases

• 1. Long chord for first curve

• 2. Long chord for second curve

• 3. Length T1T2=l

• From

[ ]2/sin21 φRCT =

[ ]2/sin22 φRCT =

KTT 21∆

[ ] [ ][ ]2/sin4

2/sin22/sin2

φ

φφ

Rl

RRl

=

+=

[ ] ly /2/sin =φ

4Ry

4Ry /42

=

=⇒×=

l

llyRl

k

9/27/2012

5

Reverse Curves

Cases

• 4.

• 5.

( )φ

φ

φ

cos12

cos

cos

21

2222

1111

−=+=

−=−=

−=−=

RBTATy

RRBOTOBT

RRAOTOAT

φ

φφ

sin2

sinsin

R

RRCBCAABx

=

+=+==

Numerical 7

• While surveying along the alignment of a road, the magnetic bearings of the line AB, BC and CD are measured as 80, 110 and 60 degrees respectively. The length of BC is 200m and the chainage of B is 950m. Calculate the necessary data for setting out a reverse curve connecting the lines AB and CD by taking BC as the common tangent. The radii of both curves may be assumed equal.

• Solution

( )o

o

BB

50240290

60180-BC of

3080110

2

2

1

=−=

+=

=−=

φ

φ

φ

curvature reverse ofpoint theFLet

950mB of Chainage

200

=

= mBC

9/27/2012

6

Numerical 7

• 1. Tangent length T1B

• 2. Chainage of T1=

• 3. First curve length

• 4. Chainage of F

• 5. Second curve length

[ ] mR 97.722/tan 1 =φ

m25.101962.14203.877 =+

( ) ( )

( ) ( )mR

RR

RRFCBFBCNow

4.272

2/50tan2/30tan200

2/tan2/tan, 21

=

+=

+=+= φφ

m03.87797.72950 =−

mRo

62.142180/1 =φπ

mRo

71.237180/2 =φπ

Numerical 7

• 6. Chainage of T2

• 7. Length of long chord for small curve

• 8. Length of long chord for large curve

• Setting the first curve

• The long chord T1F is divided into two equal parts, each of length 70.5m

• Offsets for left half are calculated, taking peg interval of 10 m

( ) mRFT 00.1412/sin2 11 == φ

m36.125771.23765.1019 =+

( ) mRFT 24.2302/sin2 22 == φ

9/27/2012

7

Numerical 7

( ) ( ) mOLRRO oo 28.92/5.7040.27240.2722/ordinate-Mid2222 =−−==−−==

( )

( ) mO

ORxRO O

09.928.940.2721040.272 22

10

22

10

=−−−=

−−−=

( ) mO 54.828.940.2722040.272 22

20 =−−−=

( ) mO 62.728.940.2723040.272 22

30 =−−−=

( ) mO 32.628.940.2724040.272 22

40 =−−−=

( ) mO 65.428.940.2725040.272 22

50 =−−−=

( ) mO 58.228.940.2721040.272 22

60 =−−−=

Numerical 7

• The offsets for right half are similar to those for the left half

• Setting the second curve

• The long chord T2F is divided into left and right halves.

• The offsets for left half are calculated exactly as described above. The offsets of the right half are similar o those of the left half.

• Then both the halves are set out in similar manner.

9/27/2012

8

Numerical 8

• A reverse is to be set out to connect two parallel railway line 30m apart. The distance between the tangent points is 150m. Both the arcs have the same radius. The curve is to be set out by the method of ordinates from long chord, taking peg interval of 10m. Calculate necessary data for setting the curve.

• Solution

• AB and CD represent the parallel lines. The line T1T2 is the distance between the tangent points, which is given as 150m.

Numerical 8

• We know that

• As

• Horizontal Distance between T1 and T2

[ ] [ ][ ]

( )

mR

R

RTT

RRFTFTTT

5.187

150/304150

2/sin4

2/sin22/sin2

21

2121

=

=

=

+=+=

φ

φφ

[ ] 150/302/sin =φ

[ ]o23

150/302/sin

=

=

φ

φ

[ ] [ ]

[ ] m

RR

FHGFGH

o52.14623sin5.1872

sinsin

=×=

+=

+=

φφ

9/27/2012

9

Numerical 8

• Setting out of curve

• The reverse curve is symmetrical, so the long chord for each curve =T1F=T2F=75m. The long chord of the first curve is divided into two halves (left and right) and ordinates are calculated for he left half. The ordinates for the right half will be similar to those for the left half. Taking peg interval of 10m, the ordinates for the left half are calculated as follows

( ) ( ) mOLRRO oo 78.32/5.375.1875.1872/ordinate-Mid2222 =−−==−−==

( )

( ) mO

ORxRO O

51.378.35.187105.18722

10

22

10

=−−−=

−−−=

( ) mO 71.278.35.187205.187 22

20 =−−−=

( ) mO 36.178.35.187305.187 22

30 =−−−=

Numerical 9

• While surveying along the alignment of a road, the magnetic bearings of the line AB, BC and CD are measured as 70, 120 and 50 degrees respectively. The length of BC is 250m and the chainage of B is 1050m. Calculate the necessary data for setting out a reverse curve connecting the lines AB and CD by taking BC as the common tangent. The radii of both curves may be assumed equal.

9/27/2012

10

THANK YOU

• Questions….

• Comments…

• Suggestions…

• Feel free to contact

mohsin.siddique@nu.edu.pk