Horizontal Alignment – Circular Curves CTC 440. Objectives Know the nomenclature of a horizontal...
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Transcript of Horizontal Alignment – Circular Curves CTC 440. Objectives Know the nomenclature of a horizontal...
![Page 1: Horizontal Alignment – Circular Curves CTC 440. Objectives Know the nomenclature of a horizontal curve Know how to solve curve problems Know how to solve.](https://reader030.fdocuments.in/reader030/viewer/2022032604/56649e6f5503460f94b6c336/html5/thumbnails/1.jpg)
Horizontal Alignment – Circular Curves
CTC 440
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Objectives
Know the nomenclature of a horizontal curve
Know how to solve curve problems Know how to solve
reverse/compound curve problems
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Simple Horizontal Curve
Circular arc tangent to two straight (linear) sections of a route
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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
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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)
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Degree of Curvature
Highway agencies –arc definition Railroad agencies –chord definition
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Arc Definition-Derivision
Dc/100’ of arc is proportional to 360 degrees/2*PI*r
Dc=18,000/PI*r
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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
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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
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From: Highway Engineering, 6th Ed. 1996, Paul Wright, ISBN 0-471-00315-8
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Example Problem
Δ=30 deg E=100’ minimum to avoid a
building
Choose an even degree of curvature to meet the criteria
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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)
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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
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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
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Compound Curves
Avoid if possible for most road alignments
Used for ramps (RS<=0.5*RL) Used for intersection radii (3-
centered compound curves)
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Use of Compound Curves
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Use of compound curves: intersections
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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
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Reverse Compound Curves
Avoid if possible for most road alignments
Used for design of auxiliary lanes (see AASHTO)
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Use of RCC: Auxiliary Lanes
Source: AASHTO, Figure IX-72, Page 784
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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
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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
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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
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Lab WorksheetFind Tangents and PI’s
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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
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Next lecture
Spiral Curves