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Geometric Design

Geometric Design. Horizontal Curve Fundamentals. PI. L. PT. PC. R. R. Vehicle Cornering. R v. ≈. F c. α. F cn. F cp. α. W. W n. F f. W p. F f. α. Superelevation. Divide both sides by Wcos( α ). Minimum radius that provides for safe vehicle operation

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Geometric Design

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  1. Geometric Design

  2. Horizontal Curve Fundamentals PI L PT PC R R

  3. Vehicle Cornering Rv ≈ Fc α Fcn Fcp α W Wn Ff Wp Ff α

  4. Superelevation Divide both sides by Wcos(α)

  5. Minimum radius that provides for safe vehicle operation Given vehicle speed, coefficient of side friction, gravity, and superelevation Rv because it is to the vehicle’s path (as opposed to edge of roadway) Superelevation

  6. R versus Rv • R is used when introducing the basic physical properties that relate radius of curve to curve length • With vehicle cornering we talked about a specific vehicle, so consider Rv, the radius to the center of the vehicle in question • When discussing superelevation we started with vehicle cornering principles, used Rv

  7. R versus Rv • A little more care is required when we move from thinking about a road as a line to a roadway with width. • Rv is the radius to the vehicle’s traveled path (usually measured to the center of the innermost lane of the road) • R is measured to the centerline of the road • SSD is measured from center of inside lane

  8. Stopping Sight Distance SSD (not L) Looking around a curve Measured along horizontal curve from the center of the traveled lane Need to clear back to Ms (the middle of a line that has same arc length as SSD) Ms Obstruction Rv Δs Assumes curve exceeds required SSD

  9. Example Problem • A horizontal curve on a 2-lane highway (12 ft lanes) has a PC station 123+50 and a PT at station 129+34. The central angle is 34 degrees, the superelevation is .08, and 20.3 feet is cleared from the edge of the innermost lane. • Determine a maximum safe speed to the nearest 5 mi/hr.

  10. Solution • PC and PT given so PT-PC=L=584 ft • Ms by definition is to the center of the innermost lane. • We are told 20.3 feet is cleared from the edge of the road, so Ms = 20.3+12/2=26.3 ft. • R=L*180/Pi*delta=984.139 ft. This is to the centerline of the roadway. • Rv= 984.139 – 6 = 978.139 ft.

  11. Solution • Know Rv, Ms. SSD a function of speed. • If 50mph design speed, SSD=425 ft. (table 3.1), Ms=22.99 ft. • So speed could be higher (Ms=26.3) • If 55mph design speed, SSD=495 ft (table 3.1), Ms=31.15 ft. • So design speed to nearest 5mph is 50 mph.

  12. Solution • Compare this to table 3.5. • Minimum Rv for 50 mph and e=.08 is 760 ft. • Our Rv=978 feet, exceeds this value. OK

  13. Combining Horizontal and Vertical Curves x xx

  14. Stationing – Linear Reference System Horizontal Alignment 2+00 0+00 1+00 3+00 Vertical Alignment 100 feet >100 feet

  15. Example Highest design speed to nearest 5mph, minimum D is 8.0 150 feet, +5% grade E 125 feet, -3% grade N

  16. Want stationing for PC, PT, PVC, and PVT PC 15+00 PT PVI2 PVI1

  17. solution • Using D, solve for R. • R=719.197 • From table 3.5, maximum design speed for this radius is 50 mph.

  18. solution • T=716.197 • L=1125 • Calculate elevations on the ramp using T and the grade • ElevEW=150+T*G2/100=185.81 • ElevNS=125+T*G1/100=146.486

  19. Want stationing and elevation for PC, PT, PVC, and PVT PC 15+00 PT PVI2 PVI1

  20. solution • From table 3.3 Ks=96 • G=(ElevEW-ElevNS)/L*100=3.495 • This is the grade of the flat parts of the ramp. • Using K=L/A calculate length of 2 sag curves: • L1=6.495*96=623.564 • L2=144.436

  21. solution • Lcon=L-(L1+L2)/2=741 feet. • PC=1500=15+00 • PT=PC+L=2625=26+25 • PVC1=PC-L1/2=11+88.2 • PVT1=PC+L1/2=18+11.8 • PVC2=PVT+Lcon=25+52.8 • PVT2=PT+L2/2=26+97.2

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