Sight Distance for

Horizontal Curves

1
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
 S
M

Sight
A R
T

Obstruction
on for O

Horizontal
Curves

3
 Line of sight is the
chord AT S
M

Horizontal distance A R
T
traveled is arc AT,
which is SD.

SD is measured along
the centre line of O
inside lane around
the curve.

See the relationship

between radius of
curve, the degree of
curve, SSD and the

middle ordinate

4
 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

5
 Middle ordinate

• Angle subtended at centre of circle by

arc AT is 2θ in degree then
• S / πR = 2θ / 180 A
M
T
B
• S = 2R θπ / 180 R θ
• θ = S 180 / 2R π = 28.65 (S) / R O

• R-M/R = cos θ
• M = [1 – cos 28.65 (S) / R ]
T

θ
6
 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
7
 Sight Distance Example
SSD = 1.47(35 mph)(2.5 sec) +
_____(35 mph)2____ = 246 feet
30(__11.2___ ± 0)
32.2

8
 Sight Distance Example
m = R(1 – cos [28.65 S])
R
m = 800 (1 – cos [28.65 {246}])= 9.43’
800

(in radians not degrees)

9
 N

B
E

a L
M

Superelevation
DR ABDUL SAMI QURESHI

10
 Motion on Circular
Curves
dv
at =
dt

2
v
an = C.F
Weight of Vehicle
R Friction force 11
 Definition
• The transverse slope provided by
raising outer edge w.r.t. inner edge
• To counteract the effects of C.
Force (overturning/skid laterally)

N
B E

a
M L 12
 N
B E

θ
L
M

•S.E. expressed in ratio of height of outer

edge to the horizontal width

e = NL / ML = tan θ
tan θ = sin θ, θ is very small
e =NL / MN = E / B
E = Total rise in outer edge
B total width of pavement
13
 Y-Y
1. C.F
2. Weight of Vehicle W p + F f = Fg
3. Friction force

Rv Cx

≈
θ
C co s
θ
C sin
C
α C

X-X α e
M cos a
My =
M 1 ft
Ff

θ
Ff = M si
n
Mx
θ N

14
Theo-
retical

Consi-
derati
on

15
Vehicle Stability on Curves
2
Desig Maximum
v n design
e + fs =
gR speed
(mph)
fs max
emax = 0.06 - 0.10 (0.12) 20 0.17
ewhere:
= superelevation (-),
70 0.10
f s = side friction coefficient (-),
v = design speed (ft/s), Assumed
R = radius (ft), Must not be too short
g = gravity acceleration (32 ft/s ).
2
16
 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

17
18
 New Graph

Side Friction Factor

19
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
 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. 20
 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%

21
• 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
Crown / super elevation
runoff length

23
Source: CalTrans Design Manual online,
http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

24
 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

25
Source: CalTrans Design Manual online,
http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

26
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.

27
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.

28
Attaining Superelevation (1)

•Location of inside edge, centre line, and out side edge are shown 29
relative to elevation of centerline
Attaining Superelevation (2)

30
Attaining Superelevation (3)

31
 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.

32
 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
 Superelevation Runoff and
Tangent Run out (Crown Runoff)
Fully superelevated cross section

Cross section with the adverse

crown removed
Normal cross section

34
Location of Runout and
Runoff

35
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 36
 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

37
Source: CalTrans Design Manual online,
http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

38
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)
39
Length of Superelevation
Runoff

α = multilane adjustment factor

Adjusts for total width

40
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)
41
 Design Requirements for
Runoffs
Maximum Relative Gradient

Relative gradient

Relative gradient is the rate of transition of edge line from

NC to full super elevation

42
 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

43
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, 44
2001 4th Ed.
Multilane Adjustment factor

• Runout and runoff must be adjusted for

multilane rotation.
45
 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
•
46
Lr = 12eα = (12) (0.04) (1.5)
G 0.5
Lr = 144 feet

47
 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

48
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
49
Association of State Highway
and Transportation Officials,
2001 4th Ed.
Transition curve

50
Spiral Curve
Transitions

51
 SPIRAL TERMINOLOGY

52
Source: Iowa DOT Design Manual
 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

53
 Tangent-to-Curve
Transition (Appearance)
• Improves appearance of the highways
and streets
• Provides natural path for drivers

54
 Tangent-to-Curve
Transition (Superelvation or curve
widening requirement)

• Accommodates distance needed to

attain super elevation
• Accommodates gradual roadway
widening

55
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

56
 Shape of transition
curves
• Spiral (Clotoid)= mostly used
• Lemniscates (rate of change of
radius not constant)
• Cubic parabola

57
 Transition Curves -
Spirals
The Euler spiral (clothoid) is used. The radius at any point of
the spiral varies inversely with the distance.

58
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)

59
• V= 50 mph
• C = 3 ft p c. sec
• R= 929
• Ls = 141 ft

60
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)
61
Maximum Length of Spiral
• Safety problems may occur when
spiral curves are too long – drivers
underestimate sharpness of
approaching curve (driver
expectancy)

62
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)

63
 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.
64
 Location of Runouts and
Runoffs
• Tangent runout proceeds a spiral
• Superelevation runoff = Spiral curve

65
66
Source: Iowa67
DOT
Design Manual
 68
Source: Iowa DOT
Design Manual
 69Iowa
Source:
DOT Design
Manual
 SPIRAL TERMINOLOGY

70
Source: Iowa DOT Design Manual
 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. 71
 Same as point E of GB
With Spirals

72
Source: Iowa DOT Standard Road Plans RP-2
 With Spirals

Tangent runout (A to B)

73
With Spirals

Removal of crown

74
With Spirals

Transition of
superelevation
Full superelevation

75
76
 Transition Example
Given:
• PI @ station 245+74.24
• D = 4º (R = 1,432.4 ft)
∀ ∆ = 55.417º
• L = 1385.42 ft

77
 With no spiral …
• T = 752.30 ft
• PC = PI – T = 238 +21.94

78
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.

79
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

80
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’

81
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

82
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

83
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)

84
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???)

85
Notes – Iowa DOT

86
 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

87
 Quiz Answers
5. Increase clear zone
6. Improve horizontal and vertical
alignment
7. Assure adequate surface drainage
8. Increase skid resistance on downgrade
curves

88
 Some of Your Answers
 Decrease posted speed
 Add rumble strips
 Bigger or better signs
 Guardrail
 Better lane markers
 Sight distance
 Decrease radius

89