1. Quiz Answers
What can be done to improve the safety of a horizontal curve?
Make it less sharp
Widen lanes and shoulders on curve
Add spiral transitions
Increase superelevation

2. Quiz Answers Increase clear zone
Improve horizontal and vertical alignment
Assure adequate surface drainage
Increase skid resistance on downgrade curves

3. Some of Your Answers Decrease posted speed Add rumble strips
Bigger or better signs
Guardrail
Better lane markers
Sight distance
Decrease radius

4. Superelevation and Spiral Curves
CE 453 Lecture 18

5. Objectives
Define superelevation runoff length and methods of attainment (for simple and spiral curves)
Calculate spiral curve length

6. Other Issues Relating to Horizontal Curves
Need to coordinate with vertical and topography
Not always needed
MAXIMUM CENTERLINE DEFLECTION
NOT REQUIRING HORIZONTAL CURVE
Design Speed, mph
Maximum Deflection 25
5°30' 30
3°45' 35
2°45' 40
2°15' 45
1°15' 50 55
1°00' 60 65
0°45' 70
Source: Ohio DOT Design Manual, Figure 202-1E

7. Attainment of Superelevation - General
Tangent to superelevation
Must be done gradually over a distance without appreciable reduction in speed or safety and with comfort
Change in pavement slope should be consistent over a distance
Methods (Exhibit 3-37 p. 186)
Rotate pavement about centerline
Rotate about inner edge of pavement
Rotate about outside edge of pavement

8. Superelevation Transition Section
Tangent Runout Section + Superelevation Runoff Section

9. Tangent Runout Section
Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero
For rotation about centerline

10. Superelevation Runoff Section
Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa
For undivided highways with cross-section rotated about centerline

11. 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. 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.

14. Source: CalTrans Design Manual online, http://www. dot. ca

15. Same as point E of GB
Source: Iowa DOT Standard Road Plans

16. Attainment Location - WHERE
Superelevation must be attained over a length that includes the tangent and the curve (why)
Typical: 66% on tangent and 33% on curve of length of runoff if no spiral
Iowa uses 70% and 30% if no spiral
Super runoff is all attained in Spiral if used (see lab manual (Iowa Spiral length = Runoff length)

17. Minimum Length of Runoff for curve
Lr based on drainage and aesthetics
rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road)
current recommendation varies from 0.35% at 80 mph to 0.80% for 15mph (with further adjustments for number of lanes)

18. Minimum Length of Tangent Runout
Lt = eNC x Lr ed
where
eNC = normal cross slope rate (%)
ed = design superelevation rate
Lr = minimum length of superelevation runoff (ft)
(Result is the edge slope is same as for Runoff segment)

19. Length of Superelevation Runoff
α = multilane adjustment factor
Adjusts for total width

20. Relative Gradient (G) Maximum longitudinal slope
Depends on design speed, higher speed = gentler slope. For example:
For 15 mph, G = 0.78%
For 80 mph, G = 0.35%
See table, next page

21. Maximum Relative Gradient (G)
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.

22. Multilane Adjustment
Runout and runoff must be adjusted for multilane rotation.
See Iowa DOT manual section 2A-2 and Standard Road Plan RP-2

23. Length of Superelevation Runoff Example
For a 4-lane divided highway with cross-section rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft)
Lr = 12eα G

24. Lr = 12eα = (12) (0.04) (1.5) G
Lr = 144 feet

25. Tangent runout length Example continued
Lt = (eNC / ed ) x Lr
as defined previously, if NC = 2%
Tangent runout for the example is:
LT = 2% / 4% * 144’ = 72 feet

26. From previous example, speed = 50 mph, e = 4%
From chart runoff = 144 feet, same as from calculation
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.

27. Spiral Curve Transitions

28. Spiral Curve Transitions
Vehicles follow a transition path as they enter or leave a horizontal curve
Combination of high speed and sharp curvature can result in lateral shifts in position and encroachment on adjoining lanes

29. Spirals
Advantages
Provides natural, easy to follow, path for drivers (less encroachment, promotes more uniform speeds), lateral force increases and decreases gradually
Provides location for superelevation runoff (not part on tangent/curve)
Provides transition in width when horizontal curve is widened
Aesthetic

30. Minimum Length of Spiral
Possible Equations:
Larger of (1) L = 3.15 V3 RC
Where:
L = minimum length of spiral (ft)
V = speed (mph)
R = curve radius (ft)
C = rate of increase in centripetal acceleration (ft/s3) use 1-3 ft/s3 for highway)

31. Minimum Length of Spiral
Or (2) L = (24pminR)1/2
Where:
L = minimum length of spiral (ft)
R = curve radius (ft)
pmin = minimum lateral offset between the tangent and circular curve (0.66 feet)

32. Maximum Length of Spiral
Safety problems may occur when spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)

33. Maximum Length of Spiral
L = (24pmaxR)1/2
Where:
L = maximum length of spiral (ft)
R = curve radius (ft)
pmax = maximum lateral offset between the tangent and circular curve (3.3 feet)

34. Length of Spiral
AASHTO also provides recommended spiral lengths based on driver behavior rather than a specific equation. See Table of text and the associated tangent runout lengths in Table
Superelevation runoff length is set equal to the spiral curve length when spirals are used.
Design Note: For construction purposes, round your designs to a reasonable values; e.g.
Ls = 147 feet, round it to
Ls = 150 feet.

35. Source: Iowa DOT Design Manual

36. Source: Iowa DOT Design Manual

37. Source: Iowa DOT Design Manual

38. SPIRAL TERMINOLOGY
Source: Iowa DOT Design Manual

39. Attainment of superelevation on spiral curves
See sketches that follow:
Normal Crown (DOT – pt A)
Tangent Runout (sometimes known as crown runoff): removal of adverse crown (DOT – A to B) B = TS
Point of reversal of crown (DOT – C) note A to B = B to C
Length of Runoff: length from adverse crown removed to full superelevated (DOT – B to D), D = SC
Fully superelevate remainder of curve and then reverse the process at the CS.

40. With Spirals Same as point E of GB
Source: Iowa DOT Standard Road Plans RP-2

41. With Spirals
Tangent runout (A to B)

42. With Spirals
Removal of crown

43. With Spirals
Transition of superelevation
Full superelevation

45. Transition Example Given: PI @ station 245+74.24
D = 4º (R = 1,432.4 ft)
 = º
L = ft

46. With no spiral …
T = ft
PC = PI – T =

47. For:
Design Speed = 50 mph
superelevation = 0.04
normal crown = 0.02
Runoff length was found to be 144’
Tangent runout length =
0.02/ 0.04 * 144 = 72 ft.

48. Where to start transition for superelevation?
Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff:
Distance before PC = Lt + 2/3 Lr
=72 +2/3 (144) = 168
Start removing crown at:
PC station – 168’ = =
Station =

49. Location Example – with spiral
Speed, e and NC as before and
 = º
Station
R = 1,432.4’
Lr was 144’, so set Ls = 150’

50. Location Example – with spiral
See Iowa DOT design manual for more equations:
Spiral angle Θs = Ls * D /200 = 3 degrees
P = (calculated)
Ts = (R + p ) tan (delta /2) + k = ft

51. Location Example – with spiral
TS station = PI – Ts
= – =
Runoff length = length of spiral
Tangent runout length = Lt = (eNC / ed ) x Lr
= 2% / 4% * 150’ = 75’
Therefore: Transition from Normal crown begins at ( ) – ( ) =

52. Location Example – with spiral
With spirals, the central angle for the circular curve is reduced by 2 * Θs
Lc = ((delta – 2 * Θs) / D) * 100
Lc = ( *3)/4)*100 = ft
Total length of curves = Lc +2 * Ls =
Verify that this is exactly 1 spiral length longer than when spirals are not used (extra credit for who can tell me why, provide a one-page memo by Monday)

53. Location Example – with spiral
Also note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately ½ of the spiral length used. (good check – but why???)

54. Notes – Iowa DOT Source: Iowa DOT Standard Road Plans
Note: Draw a sketch and think about what the last para is saying