Method of Curve Range | Civil Engineering SSC JE (Technical) - Civil Engineering (CE) PDF Download

Surveying (Part 10)

METHOD OF CURVE RANGE :
1. Linear or chain and tape method and 2. Angular or instrumental methods. 

• Linear methods is those in which the curve is set out with a chain and tape only.  Instrumental methods are those in which a theodolite with or without a chain is employed to set out the curve.

PEG INTERVAL.
The length of unit chord (peg interval) is, therefore, 30 m for flat curve 20 m for sharp curves and 10m or less for very sharp curves.
 

LOCATION OF TANGENT POINTS:
To locate the tangent points T₁ and T₂, proceeds as follows:

(i) Having fixed the direction of the tangents, produce them so as to meet at the point B
(ii) Find the deflection angle by theodolite.
(iii) Calculate the tangent length BT₁ = BT₂ = R tan OB = R sec f/2

(iv) Locate the point T₁ and T₂ on back tangent and forward tangent.

� Chainage of point T₁ = Chainage of point of intersection B – BT₁ Chainage of point T₂ = chainage of point T₁ + length of the curve Chain and Tape (or linear) methods  of setting out curves

1. By offsets or ordinates from the long chord (Linear method)
2. Offsets from the tangents (Linear method)
3. Successive bisection of the chords (linear method)

Angular or Instrumental Methods
1. Offsets from Chord produced (Linear method)
2. Deflection angles from the point of curve and normal chord (Angular method)
3. Deflection angles from the point of cure and point of tangency, using two theodolite (Angular method)

1. Offsets from the Long Chord :

Let AB and BC = tangents to the curve T₁DT₂ T₁ and T₂ ar tangent points
T₁ T₂ = Long chord of length L.
ED = O₀ = offset at the mildpoint of
T₁T₂ (the versed sine) PQ = Qx = offset at a distance x from E, so that
EP = x OT₁ = OT₂ = OD = R radius of the curve

In DOT₁E, OT = R, T₁E = L/2 OT = OD -ED = R-O₀

OT_1² = T₁E²+ OE²

 In DOQQ₁ OQ² = QQ₁² + OQ₁²

� R² = x² + (OE + O_x)² 
� OE + O_x =

[since OE = OD – ED = R-O]

2. OFFSET FROM THE TANGENTS :
(i) Perpendiculr offsets
(ii) Radial offsets

(i) Perpendicular off sets

(ii) Radial offset

DE² = NE² + ON²
� R² = x² + (R-O_x)³

3. SUCCESSIVE BISECTION OF CHORDS

Let T₁T₂ be the long chord of a curve whose angle of deflection is f.
Divide T₁T₂ at C. Joint OE and produce it to intersect the curve at Now DE = OD-OE = R-R cos f/2 Oo = DE = R (1-cos f/2) Now consider T₁D and T₂D independent portions of the curve having T₁D and T₂D as long chord. it can be proved that offset E₁D₁ and E₂D₂ are each equal to R (1–cos f/4).
By further successive bisection of the chords T₁D₁, D₁D₁, DD₂ and D₂T₂ we may obtain the locations of other points on the curve.

 4. OFFSET FROM CHORD PRODUCED:
This method is commonly adopted when a theodolie is not available and it is necessary to set out a curve only with a chain or a tape. The curve is divided into a number of chords normally 20 or 30 m in length. As continuous chainage is required along the curve, two sub-chords generally occurs one at the beginning and the other at the end of the curve.
Offsets from chords produced may be computed with the help of the formula derived under : 

Let AB be the back tangent, T₁a = C₁ be the first sub-chord and � BT₁a =d OT₁ = Oa = R The chord T₁a being very newly equal to are T₁a

 � ...............(i)

Similarly the chord aa’ is very neurly equal to the arc aa’

or 

 � a a ' =dC₁ .............. (ii)
From equations (i) and (ii)

.............. (iii)

Now mT₁ and ma both being tangents � mT₁a = � maT₁ =d = � b' an (opposite angles)

Now  .............(iv)

..............(v)

Putting the value of d and d1 in equation (iii)

..............(vi)

Similary we get the value of third offset O₃ i.e, O₃

All the chords, excepting the sub-chords are generally equal .............(vii)

i.e. C₂ = C₃ = C₄ = ----------C_n-1

 where C is the length of the normal chord

The offsets for the last sub chord Cn

5. Ranking Method of Trngential Deflection Angles: