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
1
Quiz Answers
5. Increase clear zone
6. Improve horizontal and vertical
alignment
7. Assure adequate surface drainage
8. Increase skid resistance on downgrade
curves
2
Some of Your Answers







Decrease posted speed
Add rumble strips
Bigger or better signs
Guardrail
Better lane markers
Sight distance
Decrease radius
3
Superelevation and Spiral
Curves
CE 453 Lecture 18
4
Objectives
1.
Define superelevation runoff length and
methods of attainment (for simple and
spiral curves)
2. Calculate spiral curve length
5
Other Issues Relating to
Horizontal Curves
1.
2.
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
1°15'
55
1°00'
60
1°00'
65
0°45'
70
0°45'
Source: Ohio DOT Design Manual, Figure 202-1E
6
Attainment of Superelevation General
1.
2.
3.
4.
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)
a.
b.
c.
Rotate pavement about centerline
Rotate about inner edge of pavement
Rotate about outside edge of pavement
7
Superelevation
Transition Section
• Tangent Runout Section +
Superelevation Runoff Section
8
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
9
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 crosssection rotated about centerline
10
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.
11
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.
12
13
Source: CalTrans Design Manual online,
http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf
14
Same as point E of GB
15
Source: Iowa DOT Standard Road
Plans
1.
Attainment Location WHERE
Superelevation must be attained over a
length that includes the tangent and the
curve (why)
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 (see lab manual (Iowa Spiral length =
Runoff length)
16
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) 17
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)
18
Length of Superelevation
Runoff
r
α = multilane adjustment factor
Adjusts for total width
19
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
20
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.
21
Multilane Adjustment
• Runout and runoff must be adjusted for
multilane rotation.
• See Iowa DOT manual section 2A-2 and
Standard Road Plan RP-2
22
Length of Superelevation
Runoff Example
For a 4-lane divided highway with crosssection rotated about centerline, design
superelevation rate = 4%. Design speed
is 50 mph. What is the minimum length
of superelevation runoff (ft)
Lr = 12eα
G
•
23
Lr = 12eα = (12) (0.04) (1.5)
G
0.5
Lr = 144 feet
24
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
25
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
26
Association of State Highway
and Transportation Officials,
2001 4th Ed.
Spiral Curve
Transitions
27
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
28
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
29
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)
30
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)
31
Maximum Length of Spiral
• Safety problems may occur when
spiral curves are too long – drivers
underestimate sharpness of
approaching curve (driver
expectancy)
32
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)
33
Length of Spiral
o AASHTO also provides recommended spiral
lengths based on driver behavior rather
than a specific equation. See Table 16.12
of text and the associated tangent runout
lengths in Table 16.13.
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.
34
Source: Iowa35
DOT
Design Manual
36
Source: Iowa DOT
Design Manual
Source:
37Iowa
DOT Design
Manual
SPIRAL TERMINOLOGY
Source: Iowa DOT Design Manual
38
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.
39
With Spirals
Source: Iowa DOT Standard Road Plans RP-2
Same as point E of GB
40
With Spirals
Tangent runout (A to B)
41
With Spirals
Removal of crown
42
With Spirals
Transition of
superelevation
Full superelevation
43
44
Transition Example
Given:
• PI @ station 245+74.24
• D = 4º (R = 1,432.4 ft)
•  = 55.417º
• L = 1385.42 ft
45
With no spiral …
• T = 752.30 ft
• PC = PI – T = 238 +21.94
46
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.
47
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
48
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’
49
Location Example – with spiral
See Iowa DOT design manual for more
equations:
http://www.dot.state.ia.us/design/00_toc.ht
m#Chapter_2
• Spiral angle Θs = Ls * D /200 = 3 degrees
• P = 0.65 (calculated)
• Ts = (R + p ) tan (delta /2) + k = 827.63 ft
50
Location Example – with spiral
• TS station = PI – Ts
= 245+74.24 – 8 + 27.63
= 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) – (0+75.00) = 236+71.61
51
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 = (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 who can tell me why,
provide a one-page memo by Monday)
52
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???)
53
Notes – Iowa DOT
Source: Iowa DOT Standard Road Plans
54
Note: Draw a sketch and think about what the last para is saying