5. 5
Issues Relating to
Horizontal Curves
1. Need to coordinate with
vertical and topography
2. 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 1°15'
55 1°00'
60 1°00'
65 0°45'
70 0°45'
Source: Ohio DOT Design Manual, Figure 202-1E
6. 6
WHAT IS SUPERELEVATION(e) ?
1. The purpose of superelevation or banking of
curves is to counteract the centripetal
acceleration produced as a vehicle rounds a
curve.
2. Consider the force diagram in Figure 4.17. If the vehicle
is traveling around a curve with a radius R at a constant
speed v, there will be a radial acceleration toward the
center of the curv.
7. 7
WHAT IS SUPERELEVATION(e) ?
1. There are practical upper limits to the rate of
superelevation on a horizontal curve. These
limits relate to considerations of climate,
constructability, adjacent land use, and the
frequency of slow-moving vehicles.
2. Where snow and ice are a factor, the rate of
superelevation should not exceed the rate on
which vehicles standing or traveling slowly would
slide toward the center of the curve when the
pavement is icy.
8. 8
WHAT IS SUPERELEVATION(e) ?
1. AASHTO recommends that maximum
superelevation rates be limited to 12 percent for
rural roadways; 8 percent for rural roadways for
which snow or ice are likely to be present; and 6
percent or 4 percent for urban streets.
9. 9
SIDE FRICTION
1. With the wide variation in vehicle speeds on
curves, there usually is an unbalanced force
whether the curve is superelevated or not. This
force results in tire side thrust, which is
counterbalanced by friction between the tires
and the pavement surface. This frictional
counterforce is developed by distortion of the
contact area of the tire
14. 14
Attainment of Superelevation -
General
1. Tangent to superelevation
2. Must be done gradually over a distance without
appreciable reduction in speed or safety and
with comfort
3. Change in pavement slope should be consistent
over a distance
4. Methods (Exhibit 3-40 pgs. 194 & 195)
a. Rotate pavement about centerline
b. Rotate about inner edge of pavement
c. Rotate about outside edge of pavement
17. 17
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
18. 18
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
19. 19
Source: A Policy
on Geometric
Design of
Highways and
Streets (The
Green Book).
Washington, DC.
American
Association of
State Highway
and
Transportation
Officials, 2004
5th
Ed.
20. 20
Source: A Policy
on Geometric
Design of
Highways and
Streets (The
Green Book).
Washington, DC.
American
Association of
State Highway
and
Transportation
Officials, 2004
5th
Ed.
25. 25
Attainment Location -
WHERE
1. Superelevation must be attained over a
length that includes the tangent and the
curve
2. Typical: 66% on tangent and 33% on
curve of length of runoff if no spiral
3. Iowa uses 70% and 30% if no spiral
4. Super runoff is all attained in spiral if
used
26. 26
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)
27. 27
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)
28. 28
Length of Superelevation
Runoff
α = multilane adjustment factor
Adjusts for total width
Also note that e and G can be decimals
or percents, as long as consistent
r
29. 29
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
30. 30
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,
2001 4th
Ed.
32. 32
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
•
33. 33
Lr = 12eα = (12) (0.04) (1.5)
G 0.005
Lr = 144 feet
34. 34
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
35. 35
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,
2001 4th
Ed.
37. 37
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
38. 38
Spirals
1. Advantages
a. Provides natural, easy-to-follow path
for drivers (less encroachment,
promotes more uniform speeds), lateral
force increases and decreases
gradually
b. Provides location for superelevation
runoff (not part on tangent/curve)
c. Provides transition in width when
horizontal curve is widened
d. Aesthetic
39. 39
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)
40. 40
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)
41. 41
Maximum Length of Spiral
• Safety problems may occur when
spiral curves are too long – drivers
underestimate sharpness of
approaching curve (driver
expectancy)
42. 42
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)
43. 43
Length of Spiral
o AASHTO also provides desirable spiral
lengths based on driver behavior rather
than a specific equation
o Superelevation runoff length is set equal
to the spiral curve length when spirals are
used.
o Design Note: For construction purposes,
round your designs to a reasonable values;
e.g.
Ls = 147 feet, round it to
Ls = 150 feet.
48. 48
Attainment of superelevation
on spiral curves
See sketches that follow:
Normal Crown (DOT – pt A)
1. Tangent Runout (sometimes known as crown
runoff): removal of adverse crown (DOT – A to B)
B = TS
2. Point of reversal of crown (DOT – C) note A to B =
B to C
3. Length of Runoff: length from adverse crown
removed to full superelevated (DOT – B to D), D =
SC
4. Fully superelevate remainder of curve and then
reverse the process at the CS.
56. 56
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.
57. 57
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’ = 238+21.94 - 168.00
Station = 236+ 53.94
58. 58
Location Example – with spiral
• Speed, e and NC as before and
∀∆ = 55.417º
• PI @ Station 245+74.24
• R = 1,432.4’
• Lr was 144’, so set Ls = 150’
59. 59
Location Example – with spiral
See Iowa DOT design manual for more
equations:
http://www.iowadot.gov/design/dmanual/02c-01.pd
Spiral angle Θs = Ls * D /200 = 3 degrees
• P = 0.65 (calculated)
• Ts = (R + p ) tan (Δ /2) + k = 827.63 ft
60. 60
• TS station = PI – Ts
= 245+74.24 – 827.63
= 237+46.61
Runoff length = length of spiral
Tangent runout length = Lt = (eNC / ed ) x Lr
= 2% / 4% * 150’ = 75’
Therefore: Transition from normal crown begins
at (237+46.61) – (75.00) = 236+71.61
Location Example – with spiral
61. 61
With spirals, the central angle for the
circular curve is reduced by 2 * Θs
Lc = ((Δ – 2 * Θs) / D) * 100
Lc = (55.417-2*3)/4)*100 = 1235.42 ft
Total length of curves = Lc +2 * Ls = 1535.42
Verify that this is exactly 1 spiral length
longer than when spirals are not used
(extra credit for anyone who shows me
why; provide a one-page memo by Monday)
Location Example – with spiral
62. 62
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???)
Location Example – with spiral
63. 63
Notes – Iowa DOT
Source: Iowa DOT Standard Road Plans
Note: Draw a sketch and think about what the last para is saying
