CE 8 Circular Curves

CE 8: Highway Engineering
and Bridge Design

Geometric Design of
Highway Facilities
Horizontal Alignment
The horizontal alignment consists of straight sections of

the road (known as tangents) connected by curves. The
curves are usually segment of circles, which have radii that
will provide for a smooth flow of traffic. The design of the
horizontal alignment entails the determination of the radius,
the length of the curve, and the computation of the
horizontal offsets from the tangents to the curve to facilitate
locating the curve in the field.
There are four types of horizontal curves: simple,
compound, reversed, and spiral.
Simple Curve - is a layout of a simple horizontal curve. The curve is a
segment of a circle with radius R.
Compound Curves – consists of two or more simple curves in succession,

turning in the same direction, with any two successive
curves having a common tangent point.
Reversed Curves – usually consist of two simple curves with equal radii
turning in opposite directions with a common tangent.
They are generally used to change the alignment of a
highway.
• PC = Point of curvature. The beginning of curve • m = middle ordinate, the distance from midpoint of curve to
midpoint of chord.
• PT = Point of Tangency. The end of curve
• I = intersection angle (also called central angle. The angle of
• PI = Point of Intersection of the tangents. Also called vertex intersection of the tangents.
• T = length of tangent from PC to PI and from PI to PT • x = offset distance from tangent to the curve.
• R = radius of simple curve • θ = offset angle (deflection angle) subtended at PC between
PI and any point in the curve.
• L = length of chord from PC to PT.
• D = degree of curve. The central angle subtended by a length
• Lc = length of curve from PC to PT. of curve equal to one station (20m)
• E = external distance. The nearest distance from PI to the
curve
𝐼 𝐼
T = 𝑅𝑡𝑎𝑛 2 L = 2𝑅𝑠𝑖𝑛 2
𝐼 π𝑅𝐼
E = 𝑅𝑠𝑒𝑐 2 - 𝑅 Lc = 180°
𝐼 3600
m = 𝑅 - 𝑅𝑐𝑜𝑠 2 R= π𝐷
Example:
A circular curve with a radius of 280m, a central angle of 70°, and
its PC at station 10 + 300 must be staled by deflection angles and chords.
What is the deflection angle to station 10 + 450?
Ans. θ = 15.35°
Example:
The angle of intersection of a circular curve is 42°30’. If the
external distance is 25.42m, determine the length of the curve.
Ans. Lc = 258.47m
Example:
A 3-degree simple curve is used to connect two highways whose bearings are S 11° W and S 47° W .
• What is the length of the curve in meters?
• What is the external distance in meters?
• What is the tangent distance in meters?
• What is the middle ordinate in meters?
Ans. Lc = 240m
E = 19.66m
T = 124.11m
m = 18.69m
Example:
From point A on a simple curve, the perpendicular distance to the
tangent at point Q is “x”m. The tangent passes through PC. The distance
from Q to PC is 450m. Find x if the radius of the curve is 1600m.
Ans. x = 64.59m
Example:
The angle of intersection of a circular curve is 45°30’ and its
radius is 198.17m. PC is at Sta. 0 + 700. Compute the right angle offset
from Sta. 0 + 736.58 on the curve to the tangent through PC.
Ans. x = 3.37m
Example:
A 4-degree simple curve have tangents with bearings of N 20° E and N 60° E, respectively. The
stationing of PI is 3 + 980.
• Determine the middle ordinate in meters.
• Determine the stationing of PT.
• Determine the stationing of a point Q on the curve whose deflection angle from PC is 12°.
Ans. m = 17.28m
stationing of PT = 4 + 075.73
stationing of point Q = 3 + 995.73
Example:
The offset distances of two points A and B of a simple curve are 2m and 6m, respectively. The
chord length from PC to A is 30m.
• What is the radius of the curve?
• What is the chord length from PC to B in meters?
• What is the length of curve from PC to B in meters?
Ans. R = 225m
L = 51.96m
Lc = 52.08m
• PC = point of curvature • I2 = central angle of the second curve
• PT = point of tangency • I = angle of intersection = I1 + I2
• PI = point of intersection • L𝒄 = length of first curve
𝟏
• PCC = point of compound curve • L𝒄 = length of second curve

𝟐
• T1 = length of tangent of the first curve • L1 = length of first chord

• T2 = length of tangent of the second curve • L2 = length of second chord
• V1 = vertex of the first curve • L = length of long chord from PC to PT
• V2 = vertex of the second curve • T1 + T2 = length of common tangent measured from V1 to V2
• I1 = central angle of the first curve
Sta PT = Sta PC + L𝑐1 + L𝑐2
Sta PT = Sta PI – x – T1 + L𝑐1 + L𝑐2

Example:
The common tangent AB of a compound curve is 82.38m. The
angles the common tangent makes with the tangents through PC and PT
of the compound curve are 21°10’ and 15°20’, respectively. If the
degree of the first curve is 3°30’, what is the radius of the second curve?
Ans. R = 158m
Example:
The long chord of a compound curve makes an angle of 20° and 38°, respectively with the
tangents. The common tangent of the compound curve is parallel to the long chord that is 185m long.
• What is the radius of the smaller curve?
• What is the radius of the bigger curve?
• What is the length of the curve?
Ans. Rs = 101.76m
Rb = 357.72m
Lc = 192.36m
• PC = point of curvature • I1 = central angle of the first curve
• PT = point of tangency • I2 = central angle of the second curve
• PRC = point of reversed curvature • L𝒄 = length of first curve
𝟏
• T1 = length of tangent of the first curve • L𝒄 = length of second curve

𝟐
• T2 = length of tangent of the second curve • L1 = length of first chord

• V1 = vertex of the first curve • L2 = length of second chord
• V2 = vertex of the second curve • T1 + T2 = length of common tangent measured from V1 to V2
Sta PT = Sta PC + L𝑐1 + L𝑐2
Sta PT = Sta V1 – T1 + L𝑐1 + L𝑐2

Example:
A reversed curve of a railroad track has the following properties:
D1 = 3° I1 = 18°
D2 = 2° I2 = 24°
Stationing of PT = 32 + 121
• What is the stationing of PC?
Ans. Stationing of PC = 31 + 761
Example:
A reversed curve consists of two curves of equal radius. The curve is to connect two parallel
roads that are 60m apart. The length of the common tangent is 180m.
• What is the angle of intersection of the curve?
• What is the radius of the curve?
• What is the length of the curve from PC to PT?
Ans. I = 19.5°
R = 524.56m
Lc = 356.53m
Example:
In a railroad layout, the centerlines of two parallel tracks are
connected with a reversed curve of unequal radii. The central angle of
the first curve is 16° and the distance between parallel tracks is 27.6m.
Stationing of PC is 15 + 420 and the radius of the second curve is 290m.
Compute the stationing of PT.
Ans. Stationing of PT = 15 + 618.96
Thank you

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CE 8: Highway Engineering

and Bridge Design

The horizontal alignment consists of straight sections of

Compound Curves – consists of two or more simple curves in succession,

• PCC = point of compound curve • L𝒄 = length of second curve

• T1 = length of tangent of the first curve • L1 = length of first chord

Sta PT = Sta PI – x – T1 + L𝑐1 + L𝑐2

• T1 = length of tangent of the first curve • L𝒄 = length of second curve

• T2 = length of tangent of the second curve • L1 = length of first chord

Sta PT = Sta V1 – T1 + L𝑐1 + L𝑐2

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