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:
- Circular or Simple Curves
- Spiral or Transition Curves
- 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:
- The radius of the curve
- The beginning of the station
- 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).
1. Subtangent T
tan (Δ/2) = T/R
T = Rtan (Δ/2)
2. Long Chord (C)
Sin (Δ/2) = (1/2C)/R
C= 2R sin(Δ/2)
3. Mid Ordinate (M)
cos (Δ/2) = OB/R
OB = Rcos (Δ/2)
But, OB = R - M
Rcos (Δ/2) = R - M
M = R { 1- cos ( Δ/2) }
4. External Distance (E)
Consider triangle O-PI-BC,
cos (Δ/2) = R/ (R+E)
E = R { sec (Δ/2)) -1 }
5. Length of Curve ( L)
L/2πR = Δ/360
L = 2πR (Δ/360)
6. Fractional Portion of Curve
From the figure, we can derive the following relation:
360/ Δ = πR2/As= 2πR/L
Δ/360 = As/ πR2= L/2πR
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 = Δ/360 = L/2πR = As/ πR2
7. Tangent Deflection Angle Δ/ Central Angle of the Curve
Δ = (L x 360 ) 2πR
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 = Δ/D
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