Horizontal Alignment – Circular Curves

1. Horizontal Alignment – Circular Curves CTC 440

2. Objectives • Know the nomenclature of a horizontal curve • Know how to solve curve problems • Know how to solve reverse/compound curve problems

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

5. 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

6. 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)

7. Degree of Curvature • Highway agencies –arc definition • Railroad agencies –chord definition

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

9. 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

10. 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

11. From: Highway Engineering, 6th Ed. 1996, Paul Wright, ISBN 0-471-00315-8

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

13. 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)

14. 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

15. 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

16. Compound Curves • Avoid if possible for most road alignments • Used for ramps (RS<=0.5*RL) • Used for intersection radii (3-centered compound curves)

17. Use of Compound Curves

18. Use of compound curves: intersections

19. 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

20. Reverse Compound Curves • Avoid if possible for most road alignments • Used for design of auxiliary lanes (see AASHTO)

21. Use of RCC: Auxiliary Lanes Source: AASHTO, Figure IX-72, Page 784

22. 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

23. 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

24. 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

25. Lab WorksheetFind Tangents and PI’s

26. 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

27. Next lecture • Spiral Curves