# Curve Surveying - Circular Curves

Most of the alignment in highways and railways is straight or tangent. Whenever there is a requirement to change the direction, a gradual change is brought in the alignment, which results in curves.

Curves are used in surveying alignments to change the direction of motion to lessen the forces involved when a vehicle changes direction is necessary.

Take a Quick Video: Components of Circular Curves

### Types of Curves in Surveying

The three main types of curves used in surveying are:

1. Circular or Simple Curves
2. Spiral or Transition Curves
3. Vertical or Parabolic Curves

### Circular or Simple Curves

A circular or simple curve is the simplest curve and it forms a segment of a circle. Circular curves are used for horizontal alignments as they can be laid out on the ground using basic surveying tools and techniques.

A circular curve is laid using a chain or EDM to measure the distance along the arc of the curve. A theodolite or transit is used to measure the horizontal angles from a reference line to the station to be set.

The parameter required for laying a circular curve are:
1. The radius of the curve
2. The beginning of the station
3. The distance along the arch between the instrument and the points to be set

### Geometry of  a Curve

The main components of a curve is given the figure below:

From the figure above, important curve parameters can be determined. The given details of the curve is the radius (R).

tan (Î”/2) = T/R

T = Rtan (Î”/2)

#### 2. Long Chord (C)

Sin (Î”/2) = (1/2C)/R

C= 2R sin(Î”/2)

#### 3. Mid Ordinate (M)

cos (Î”/2) = OB/R
OB = Rcos (Î”/2)
But, OB = R - M

Rcos (Î”/2) = R - M

M = R { 1- cos ( Î”/2) }

#### 4. External Distance (E)

Consider triangle O-PI-BC,
cos (Î”/2) = R/ (R+E)

E = R { sec (Î”/2)) -1 }

#### 5. Length of Curve ( L)

L/2Ï€R = Î”/360

L = 2Ï€R (Î”/360)

#### 6. Fractional Portion  of Curve

From the figure, we can derive the following relation:

360/ Î” = Ï€R2/As= 2Ï€R/L
Î”/360 = As/ Ï€R2= L/2Ï€R

Where, As is the area of the sector and L is the length of chord of sectior.
Hence the fractional part of the curve is:

Fractional Part =  Î”/360 = L/2Ï€R = As/ Ï€R2

#### 7. Tangent Deflection Angle Î”/ Central Angle of the Curve

Î” = (L x 360 ) 2Ï€R

#### 8. Degree of Curve (D)

As per Highway, D is the central angle subtended by a 100' arc and as per Railroad, it is the central angle subtended by a 100; chord.

From the figure above:
D/360 = 100/ 2Ï€R

D = 5729.58/R and L/100 = Î”/D

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