1. Geometric Design

2. Horizontal Curve FundamentalsPI L PC PT 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. Superelevation 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)

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

8. 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

9. 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

10. Example Problem
A horizontal curve on a 2-lane highway (12 ft lanes) has a PC station and a PT at station 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.

11. 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 = /2=26.3 ft.
R=L*180/Pi*delta= ft. This is to the centerline of the roadway.
Rv= – 6 = ft.

12. 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.

13. 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

14. Combining Horizontal and Vertical Curvesx xx

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

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

17. Want stationing for PC, PT, PVC, and PVT
PVI2
PVI1

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

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

20. Want stationing and elevation for PC, PT, PVC, and PVT
PVI2
PVI1

21. 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=
L2=

22. solution Lcon=L-(L1+L2)/2=741 feet. PC=1500=15+00 PT=PC+L=2625=26+25
PVC1=PC-L1/2=
PVT1=PC+L1/2=
PVC2=PVT+Lcon=
PVT2=PT+L2/2=