2. Provided Sight Distance
• Potential sight obstructions
– On horizontal curves: barriers,
bridge-approach fill slopes, trees,
back slopes of cut sections
– On vertical curves: road surface at
some point on a crest vertical
curve, range of head lights on a sag
curve
4. 4
Line of sight is the
chord AT
Horizontal distance
traveled is arc AT,
which is SD.
SD is measured along
the centre line of
inside lane around
the curve.
See the relationship
between radius of
curve, the degree of
curve, SSD and the
middle ordinate
S
R
M
O
TA
5. 5
Middle ordinate
• 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
6. 6
Middle ordinate
• Angle subtended at centre of circle by
arc AT is 2θ in degree then
• S / πR = 2θ / 180
• S = 2R θπ / 180
• θ = S 180 / 2R π = 28.65 (S) / R
• R-M/R = cos θ
• M = [1 – cos 28.65 (S) / R ]
θR
M
A T
B
T
O
θ
7. 7
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
12. 12
Definition
• The transverse slope provided by
raising outer edge w.r.t. inner edge
• To counteract the effects of C.
Force (overturning/skid laterally)
L
N
a
EB
M
13. 13
•S.E. expressed in ratio of height of outerS.E. expressed in ratio of height of outer
edge to the horizontal widthedge to the horizontal width
e = NL / ML = tan= NL / ML = tan θ
tantan θ = sin= sin θ,, θ is very smallis very small
e =NL / MN = E / Be =NL / MN = E / B
E = Total rise in outer edgeE = Total rise in outer edge
B total width of pavementB total width of pavement
M
L
N
θ
EB
14. 14
FgFW fp =+
α
α
C cos θ Cx
Mx =M sin
θ
My =M cos a
Ff
Ff
θ
C
M 1 ft
e
≈
Rv
1. C.F
2. Weight of Vehicle
3. Friction force
X-X
Y-Y
C
N
N
C sin θ
16. 16
Vehicle Stability on Curves
where:
gR
v
fe s
2
=+
(ft/s),speeddesign=v
(-),tcoefficienfrictionside=sf
).ft/s(32onacceleratigravity 2
=g
(-),tionsupereleva=e
(ft),radius=R
Assumed
Desig
n
speed
(mph)
Maximum
design
fs max
20 0.17
70 0.10
Must not be too short
(0.12)0.10-0.06max =e
17. 17
Selection of e and fs
• Practical limits on super elevation (e)
– Climate
– Constructability
– Adjacent land use
• Side friction factor (fs) variations
– Vehicle speed
– Pavement texture
– Tire condition
• The maximum side friction factor is the point at
which the tires begin to skid
• Design values of fs are chosen somewhat below
this maximum value so there is a margin of safety
20. 20
Maximum Superelevation
• Superelevation cannot be too large since an
excessive mass component may push slowly
moving vehicles down the cross slope.
• Limiting values emax
– 12 % for regions with no snow and ice conditions
(higher values not allowed),
– 10 % recommended value for regions without
snow and ice conditions,
– 8% for rural roads and high speed urban roads,
– 4, 6% for urban and suburban areas.
21. 21
Example
• A section of road is being designed as a high-speed highway.
The design speed is 70 mph. Using AASHTO standards,
what is the maximum super elevation rate for existing curve
radius of 2500 ft and 300 ft for safe vehicle operation?
• Assume the maximum super elevation rate for the given
region is 8%.
• max e = ?
• For 70 mph, f = 0.10
• 1. 2500 = V2/15(fs+e) = (70 )2/(0.10 + e) = 0.0306
• e = 3%
• 2. 300 = V2/(fs+e) = (70 x 1.47)2/32.2(0.10 + e) = 0.988
• e = 9.8%
22. • 300 = V2/g(fs+e) = (70)2/15(fs + 0.8)
• f = 1.008 > 0.10
• 300 = V2/15(fs+e) = (V )2/15(0.10+ 0.8)
• V = 28.46 mph
22
25. 25
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 3-37 p. 186)
a. Rotate pavement about centerline
b. Rotate about inner edge of pavement
c. Rotate about outside edge of pavement
27. 27
Common methods of developing the
transition to super elevation
• At (2)the out side edge is far below the centre
line as the inside edge
• At (3)the out side edge has reached the level of
the centre line
• At point (4) the out side edge is located as far
above as the inside edge is below the centerline.
• Finally , at point (5) the cross section is fully
super elevated and remain through out the
circular curve
• The reverse of these profiles is found at the
other end of circular curve.
28. 28
Common methods of developing the
transition to super elevation
• Location of inside edge, centre line, and out side edge are
shown relative to elevation of centerline
• The difference in elevation being equal to the normal crown
times the pavement width.
• At A the out side edge is far below the centre line as the
inside edge
• At B the out side edge has reached the level of the centre
line
• At point C the out side edge is located as far above as the
inside edge is below the centerline.
• Finally , at point E the cross section is fully super elevated
• The reverse of these profiles is found at the other end of
circular curve.
32. 32
Superelevation
Transition Section
• Tangent Runout (Crown Runoff)
Section + Superelevation Runoff
Section.
• Tangent runout = the length of highway
needed to change the normal cross section
to the cross section with the adverse crown
removed.
33. Super elevation runoff
• Super elevation runoff = the length of
highway needed to change the cross section
with the adverse crown removed to the
cross section fully super elevated.
33
34. 34
Superelevation Runoff and
Tangent Run out (Crown Runoff)
Normal cross section
Fully superelevated cross section
Cross section with the adverse
crown removed
36. 36
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
37. 37
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
39. 39
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)
41. 41
Minimum Length of Runoff
for curve
• Lr based on drainage and
aesthetics and design speed.
• Relative gradient is the 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)
43. 43
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
44. 44
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.
46. 46
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
•
47. 47
Lr = 12eα = (12) (0.04) (1.5)
G 0.5
Lr = 144 feet
48. 48
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
49. 49
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.
53. 53
Transition Curves –
Spirals (Safety)
• Provided between tangents and
circular curves or between two
circular curves
• It provides the path where radial
force gradually increased or
decreased while entering or leaving
the circular curves
56. 56
Ideal shape of transition curve
• When rate of introduction of C.F. is
consistent
• When rate of change C. Acceleration
is consistent
• When radius of transition curve
consistently change from infinity to
radius of circular curve
57. 57
Shape of transition
curves
• Spiral (Clotoid)= mostly used
• Lemniscates (rate of change of
radius not constant)
• Cubic parabola
58. 58
Transition Curves -
Spirals
The Euler spiral (clothoid) is used. The radius at any point of
the spiral varies inversely with the distance.
59. 59
Minimum Length of Spiral
Possible Equations: When consistent C.F is considered
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)
60. 60
• V= 50 mph
• C = 3 ft p c. sec
• R= 929
• Ls = 141 ft
61. 61
Minimum Length of Spiral
When appearance of the highways is considered
1.Minimum- 2.Maximum Length of Spiral
Or L = (24pminR)1/2
Where:
L = minimum length of spiral (ft) = 121.1 ft
R = curve radius (ft) = 930
pmin = minimum lateral offset between the
tangent and circular curve (0.66 feet)
62. 62
Maximum Length of Spiral
• Safety problems may occur when
spiral curves are too long – drivers
underestimate sharpness of
approaching curve (driver
expectancy)
63. 63
Maximum Length of Spiral
L = (24pmaxR)1/2
Where:
L = maximum length of spiral (ft) = 271
R = curve radius (ft)
pmax = maximum lateral offset between the
tangent and circular curve (3.3 feet)
64. 64
Length of Spiral
o AASHTO also provides recommended spiral
lengths based on driver behavior rather
than a specific equation.
o Super elevation 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 = 141 feet, round it to
Ls = 150 feet.
65. 65
Location of Runouts and
Runoffs
• Tangent runout proceeds a spiral
• Superelevation runoff = Spiral curve
71. 71
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.
72. 72
Source: Iowa DOT Standard Road Plans RP-2
With Spirals
Same as point E of GB
79. 79
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.
80. 80
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
81. 81
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’
82. 82
Location Example – with spiral
See Iowa DOT design manual for more
equations:
http://www.dot.state.ia.us/design/00_toc.htm#Ch
• Spiral angle Θs = Ls * D /200 = 3 degrees
• P = 0.65 (calculated)
• Ts = (R + p ) tan (delta /2) + k = 827.63 ft
83. 83
• 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
Location Example – with spiral
84. 84
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)
Location Example – with spiral
85. 85
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
87. 87
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
88. 88
Quiz Answers
5. Increase clear zone
6. Improve horizontal and vertical
alignment
7. Assure adequate surface drainage
8. Increase skid resistance on downgrade
curves
89. 89
Some of Your Answers
Decrease posted speed
Add rumble strips
Bigger or better signs
Guardrail
Better lane markers
Sight distance
Decrease radius