1. Horizontal Alignment
CE 453 Lecture 16

2. Objectives Identify curve types and curve components
See: (Chapter 5 from FHWA’s Flexibility in Highway Design)

3. Horizontal Alignment
Design based on appropriate relationship between design speed and curvature and their relationship with side friction and superelevation
Along circular path, vehicle undergoes centripetal acceleration toward center of curvature (lateral acceleration)
Balanced by superelevation and weight of vehicle (friction between tire and roadway)

4. Horizontal Alignment Tangents Curves Transitions
Curves require superelevation (next lecture)
Reason for super: banking of curve, retard sliding, allow more uniform speed, also allow use of smaller radii curves (less land)

5. Radius Calculation Rmin = ___V2______ 15(e + f) Where:
V = velocity (mph)
e = superelevation
f = friction (15 = gravity and unit conversion)

6. Radius Calculation Rmin related to max. f and max. e allowed
Rmin use max e and max f (defined by AASHTO, DOT, and graphed in Green Book) and design speed
f is a function of speed, roadway surface, weather condition, tire condition, and based on comfort – drivers brake, make sudden lane changes and changes within a lane when acceleration around a curve becomes “uncomfortable”
AASHTO: 20 mph with new tires and wet pavement to 60 mph
f decreases as speed increases (less tire/pavement contact)

7. Max e Controlled by 4 factors:
Climate conditions (amount of ice and snow)
Terrain (flat, rolling, mountainous)
Type of area (rural or urban)
Frequency of slow moving vehicles who might be influenced by high superelevation rates

8. Max e
Highest in common use = 10%, 12% with no ice and snow on low volume gravel-surfaced roads
8% is logical maximum to minimize slipping by stopped vehicles
For consistency use a single rate within a project or on a highway

9. Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, th Ed.

12. Radius Calculation (Example)
Design radius example: assume a maximum e of 8% and design speed of 60 mph, what is the minimum radius?
fmax = 0.12 (from Green Book)
Rmin = _____602________________
15( )
Rmin = 1200 feet

13. Radius Calculation (Example)
For emax = 4%? (urban situation)
Rmin = _____602________________
15( )
Rmin = 1,500 feet

14. Radius Calculation (Example)
For emax = 2%? (rotated crown)
Rmin = _____602________________
15( )
Rmin = 1,714 feet

15. Radius Calculation (Example)
For emax = -2%? (normal crown, adverse direction)
Rmin = _____602________________
15( )
Rmin = 2,400 feet

16. Curve Types Simple curves with spirals (why spirals)
Broken Back – two curves same direction (avoid)
Compound curves: multiple curves connected directly together (use with caution) go from large radii to smaller radii and have R(large) < 1.5 R(small)
Reverse curves – two curves, opposite direction (require separation typically for superelevation attainment)

17. Important Components of Simple Circular Curve
See: ftp:// /dotmain/design/dmanual/English/e02a-01.pdf
1.     See handout
2.     PC, PI, PT, E, M, and 
3.     L = 2()R()/360
4.     T = R tan (/2)
Direction of stationing
Source: Iowa DOT Design Manual

18. Sight Distance for Horizontal Curves
Location of object along chord length that blocks line of sight around the curve
m = R(1 – cos [28.65 S]) R
Where:
m = line of sight
S = stopping sight distance
R = radius

19. Sight Distance Example
A horizontal curve with R = 800 ft is part of a 2-lane highway with a posted speed limit of 35 mph. What is the minimum distance that a large billboard can be placed from the centerline of the inside lane of the curve without reducing required SSD? Assume p/r =2.5 and a = 11.2 ft/sec2
SSD = 1.47vt + _________v2____
30(__a___  G)
32.2

20. Sight Distance Example
SSD = 1.47(35 mph)(2.5 sec) +
_____(35 mph)2____ = 246 feet
30(__11.2___  0)
32.2

21. Sight Distance Example
m = R(1 – cos [28.65 S]) R
m = 800 (1 – cos [28.65 {246}]) = 9.43 feet
800
(in radians not degrees)

22. Horizontal Curve Example
Deflection angle of a 4º curve is 55º25’, PI at station Find length of curve,T, and station of PT.
D = 4º
 = 55º25’ = º
D = _ _ R = _ _ = 1,432.4 ft R

23. Horizontal Curve Example
D = 4º
 = º
R = 1,432.4 ft
L = 2R = 2(1,432.4 ft)(55.417º) = ft

24. Horizontal Curve Example
D = 4º
 = º
R = 1,432.4 ft
L = ft
T = R tan  = 1,432.4 ft tan (55.417) = ft

25. Stationing Example Stationing goes around horizontal curve.
For previous example, what is station of PT?
First calculate the station of the PC:
PI =
PC = PI – T
PC = – =

26. Stationing Example (cont)
PC =
L = ft
Station at PT = PC + L
PT = =

27. Suggested Steps in Horizontal Design
Select tangents, PIs, and general curves making sure you meet minimum radius criteria
Select specific curve radii/spiral and calculate important points (see lab) using formula or table (those needed for design, plans, and lab requirements)
Station alignment (as curves are encountered)
Determine super and runoff for curves and put in table (see next lecture for def.)
Add information to plans