Horizontal Alignment Circular Curves CTC 440 Objectives n

Horizontal Alignment – Circular Curves CTC 440

Objectives n n n Know the nomenclature of a horizontal curve Know how to solve curve problems Know how to solve reverse/compound curve problems

Simple Horizontal Curve n Circular arc tangent to two straight (linear) sections of a route

Circular Curves n n n 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 n n n 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 n n Highway agencies –arc definition Railroad agencies –chord definition

Arc Definition-Derivision n Dc/100’ of arc is proportional to 360 degrees/2*PI*r Dc=18, 000/PI*r

Circular Curves n E –External Distance n n n 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 n 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 n n n 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, 6 th Ed. 1996, Paul Wright, ISBN 0 -471 -00315 -8

Example Problem n n n Δ=30 deg E=100’ minimum to avoid a building Choose an even degree of curvature to meet the criteria

Example Problem n n 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 n n n 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 n n XX+XX. XX (english) – a station is 100’ XX+XXX. XXX (metric) – a station is one km

Compound Curves n n n 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

Compound Curves n n n 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 n n n 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

Reverse Compound Curves n n 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 n n n R=90 m L=35. 4 m What is width? L=2 RsinΔ and w=2 R(1 -cos Δ) Solve for Δ (first equation) and solve for w (2 nd equation) W-3. 515 m=11. 5 ft

In General n n Horizontal alignments should be as directional as possible, but consistent with topography Poor horizontal alignments look bad, decrease capacity, and cost money/time

Considerations n n n 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

Worksheet: Identifying Finding Tangents and PI’s

Deflection Angles-Practice Back Tangent Azimuth=25 deg-59 sec Forward (or Ahead) Tangent Azimuth=14 deg-10 sec Answer: 11 deg 00’ 49” Back Tangent Bearing=N 22 deg E Forward Tangent Bearing=S 44 deg E Answer: 114 deg Back Tangent Azimuth=345 deg Forward Tangent Azimuth=22 deg Answer: 370 deg

Next lecture n Spiral Curves