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Horizontal Alignment – Circular Curves. CTC 440. Objectives. Know the nomenclature of a horizontal curve Know how to solve curve problems Know how to solve reverse/compound curve problems. Simple Horizontal Curve. Circular arc tangent to two straight (linear) sections of a route. - PowerPoint PPT Presentation
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Horizontal Alignment – Circular Curves
CTC 440
Objectives Know the nomenclature of a
horizontal curve Know how to solve curve problems Know how to solve
reverse/compound curve problems
Simple Horizontal Curve Circular arc tangent to two straight
(linear) sections of a route
Circular Curves PI-pt of intersection PC-pt of curvature PT-pt of tangency R-radius of the circular arc Back tangent Forward (ahead) tangent
Circular Curves
T-distance from the PC or PT to the PI Δ-Deflection Angle. Also the central
angle of the curve (LT or RT) Dc -Degree of Curvature. The angle
subtended at the center of the circle by a 100’ arc on the circle (English units)
Degree of Curvature Highway agencies –arc definition Railroad agencies –chord definition
Arc Definition-Derivision Dc/100’ of arc is proportional to
360 degrees/2*PI*r
Dc=18,000/PI*r
Circular Curves E –External Distance
Distance from the PI to the midpoint of the circular arc measured along the bisector of the central angle
L-Length of Curve M-Middle Ordinate
Distance from the midpoint of the long chord (between PC & PT) and the midpoint of the circular arc measured along the bisector of the central angle
Basic Equations T=R*tan(1/2*Δ) E=R((1/cos(Δ/2))-1) M=R(1-cos(Δ/2)) R=18,000/(Π*Dc) L=(100*Δ)/Dc L=(Π*R*Δ)/180-------metric
From: Highway Engineering, 6th Ed. 1996, Paul Wright, ISBN 0-471-00315-8
Example Problem Δ=30 deg E=100’ minimum to avoid a
building
Choose an even degree of curvature to meet the criteria
Example Problem Solve for R knowing E and Deflection
Angle (R=2834.77’ minimum) Solve for degree of curvature (2.02
deg and round off to an even curvature (2 degrees)
Check R (R=2865 ft) Calc E (E=101.07 ft which is > 100’
ok)
Practical Steps in Laying Out a Horizontal Alignment POB - pt of beginning POE - pt of ending POB, PI’s and POE’s are laid out Circular curves (radii) are established Alignment is stationed
XX+XX.XX (english) – a station is 100’ XX+XXX.XXX (metric) – a station is one
km
Compound Curves Formed by two simple curves
having one common tangent and one common point of tangency
Both curves have their centers on the same side of the tangent
PCC-Point of Compound Curvature
Compound Curves Avoid if possible for most road
alignments Used for ramps (RS<=0.5*RL) Used for intersection radii (3-
centered compound curves)
Use of Compound Curves
Use of compound curves: intersections
Reverse Compound Curves Formed by two simple curves
having one common tangent and one common point of tangency
The curves have their centers on the opposite side of the tangent
PRC-Point of Reverse Curvature
Reverse Compound Curves Avoid if possible for most road
alignments Used for design of auxiliary lanes
(see AASHTO)
Use of RCC: Auxiliary Lanes
Source: AASHTO, Figure IX-72, Page 784
Example: Taper Design C-3 R=90m L=35.4m What is width? L=2RsinΔ and w=2R(1-cos Δ) Solve for Δ (first equation) and solve for
w (2nd equation) W-3.515m=11.5 ft
In General Horizontal alignments should be as
directional as possible, but consistent with topography
Poor horizontal alignments look bad, decrease capacity, and cost money/time
Considerations Keep the number of curves down to a
minimum Meet the design criteria Alignment should be consistent Avoid curves on high fills Avoid compound & reverse curves Correlate horizontal/vertical
alignments
Lab WorksheetFind Tangents and PI’s
Deflection Angles-PracticeBack Tangent Azimuth=25 deg-59 secForward (or Ahead) Tangent Azimuth=14 deg-10 secAnswer: 11 deg 00’ 49”
Back Tangent Bearing=N 22 deg E Forward Tangent Bearing=S 44 deg EAnswer: 114 deg
Back Tangent Azimuth=345 deg Forward Tangent Azimuth=22 deg Answer: 370 deg
Next lecture Spiral Curves