The document discusses different types of circular curves used in road design, including simple, compound, and broken-back curves. It notes that broken-back curves should be avoided if possible. The document also discusses using spirals to transition between tangents and circular curves, and defines elements of circular curves like radius, degree of curvature, tangent distance, and chord length. An example problem demonstrates computing values for a horizontal curve connecting two tangents. Finally, the document discusses laying out curves using deflection angles and computing subchords and subdeflections.

1. TARUNGEHLOTS
Horizontal Curves
Chapter 24
Types of Circular Curves
Simple Curve Compound Curves Broken-Back Curves
Broken-
Reverse Curves
Broken-Back Curves should be avoided if
Broken-
possible. It is better to replace the Curves
with a larger radius circular curve.
A tangent should be placed
between reverse Curves.
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2. Typical Configurations of Curves
Spirals are typically placed Spirals are typically
between tangents and used at intersections
circular curves to provide a to increase the room
transition from a normal for large trucks to
crown section to a make turning
superelevated one. movements.
Arc Definition
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3. Circular Curve Elements
Equations for Computing Properties of
Horizontal Curves
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4. Equations for Computing Properties of
Horizontal Curves
Example Problem
A tangent with a bearing of N 56° 48’ 20” E meets another tangent with a bearing
56° 48’ 20”
of N 40° 10’ 20” E at PI STA 6 + 26.57. A horizontal curve with radius = 1000
40° 10’ 20”
feet will be used to connect the two tangents. Compute the degree of curvature,
degree
tangent distance, length of curve, chord distance, middle ordinate, external
ordinate,
distance, PC and PT Stations.
Solution:
I = 56° 48’ 20” - 40° 10’ 20” = 16° 38’ 00”
56° 48’ 20” 40° 10’ 20” 16° 38’ 00”
D = 5729.578/R = 5729.578/1000 = 5° 43’ 46”
5° 43’ 46”
L = 100 (I/D) = 100 (16.63333/5.72944444) = 290.31’
290.31’
T = R tan (I/2) = 1000 tan (16.63333/2) = 146.18’
146.18’
LC = 2R sin (I/2) = 2(1000) sin (16.63333/2) = 289.29’
289.29’
E = R [ 1/cos (I/2) -1 ] = 1000 [ 1/cos 16.63333/2) – 1 ] = 10.63’
10.63’
M = R [ 1 – cos (I/2) ] = 1000 [ 1 – cos (16.63333/2) ] = 10.52’
10.52’
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5. Example Problem (continued)
A tangent with a bearing of N 56° 48’ 20” E meets another tangent with a bearing
56° 48’ 20”
of N 40° 10’ 20” E at PI STA 6 + 26.57. A horizontal curve with radius = 1000
40° 10’ 20”
feet will be used to connect the two tangents. Compute the degree of curvature,
degree
tangent distance, length of curve, chord distance, middle ordinate, external
ordinate,
distance, PC and PT Stations.
Solution:
PC STA = PI STA – T = 626.57 – 146.18 = PC STA 4 + 80.39
PT STA = PC STA + L = 480.39 + 290.31 = PT STA 7 + 70.70
Final thoughts:
Given the coordinates of the PI can you compute the coordinates of the PC?
How about the PT?
Can you compute the coordinates of the center of the circle?
Curve Layout by Deflection Angles
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6. Subchords and Subdeflections
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