Transcript Horizontal Alignment

Transportation Engineering
I
Lec-03
Horizontal Alignment &
Sight Distances
Dr. Attaullah Shah
Sight Distance
For operating a motor vehicle safely and
efficiently, it is of utmost importance that
drivers have the capability of seeing clearly
ahead. Therefore, sight distance of
sufficient length must be provided so that
the drivers can operate and control their
vehicles safely.
Sight distance is the length of the highway
visible ahead to the driver of the vehicle
Aspects of Sight Distance
• The distances required by motor vehicles to
stop.
• The distances needed for decisions at
complex locations
• The distances required for passing and
overtaking vehicles, applicable on two-lane
highways
• The criteria for measuring these distances for
use in design.
Stopping Sight Distance
At every point on the roadway, the
minimum sight distance provided should be
sufficient to enable a vehicle traveling at
the design speed to stop before reaching a
stationary object in its path. Stopping sight
distance is the aggregate of two distances:
•brake reaction distance and
•braking distance.
Brake reaction time
It is the interval between the instant that the driver
recognizes the existence of an object or hazard ahead
and the instant that the brakes are actually applied.
Extensive studies have been conducted to ascertain
brake reaction time. Minimum reaction times can be as
little as 1.64 seconds: 0.64 for alerted drivers plus 1
second for the unexpected signal.
Some drivers may take over 3.5 seconds to respond
under similar circumstances. For approximately 90% of
drivers, including older drivers, a reaction time of 2.5
sec is considered adequate. This value is therefore
used in Table 3.1
The braking distance of a vehicle on a roadway
may be determined by the formula
2
v
d 
2a
Equation (1)
d = braking distance (ft) or (m)
v = initial speed (ft/s) or (m/s)
a = deceleration rate, ft/s²
(m/s²)
Studies (Fambro et aI., 1997) document
that most drivers decelerate at a rate
greater than 4.5 m/s² (14.8 ft/s²) when
confronted with an urgent need to stop-for
example, when seeing an unexpected
object in the roadway. Approximately 90
percent of all drivers displayed deceleration
rates of at least 3.4 m/s² (11.2 ft/s²).
Such deceleration rates are within a driver's
capability while maintaining steering control
and staying in a lane when braking on wet
surfaces. Most vehicle braking systems and
tire-pavement friction levels are also capable
of providing this level. Therefore, a
deceleration rate of 3.4 m/s² (11.2 ft/s²) is
recommended as a threshold for determining
stopping sight distance (AASHTO, 2004).
Design Values
The sum of the distance traversed during
the brake reaction time and the distance to
brake the vehicle to a stop is the stopping
sight distance. The computed distances for
wet pavements and for various speeds at
the assumed conditions are shown in
Exhibit 3-1 and were developed from the
following equation:
2
v
S 
 vt
2a
Equation 2
r
where tr is the driver reaction time (sec).
If speed is given in miles per hour or
kilometer per hour, Equation (2) can be
rewritten as
Note that the units for S are in feet and V is in miles
per hour, assuming that 1 ft/sec = 0.682 mph (or
1.466 ft/sec = 1 mph).
In computing and measuring stopping sight
distances, the height of the driver's eye is
estimated to be 1,080 mm [3.5 ft] and the
height of the object to be seen by the driver
is 600 mm [2.0 ft], equivalent to the tail-light
height of a passenger car.
Effect of Grade on Stopping
When a highway is on a grade, the equation
for braking distance should be modified as
follows:
2
v
S 
 vt
( 2a  G )
r
Eq - 3
Where G is grade or longitudinal slope of
the highway divided by 100.
Stopping sight distances calculated as based
on Equation (2)
2
v
S 
 vt
2a
r
However, at the end of long down- grades where
truck speeds approach or exceed passenger car
speeds, it is desirable to provide distances
greater than those recommended in Exhibit - 1 or
even those calculated based on Equation (3).
2
v
S 
 vt
(2a  G )
r
It is easy to see that under these circumstances
higher eye position of the truck driver can be of
little advantage.
Discussion
The driver's reaction time, the condition of the
road pavement, vehicle braking system, and
the prevailing weather all play a significant
role in this problem.
Decision Sight Distance
Although stopping sight distances are
generally sufficient to allow competent and
alert drivers to stop their vehicles under
ordinary circumstances, these distances are
insufficient when information is difficult to
perceive. When a driver is required to detect
an unexpected or otherwise difficult-toperceive information source, a decision sight
distance should be provided.
Interchanges and intersections, changes in
cross-section such as toll plazas and lane
drops, and areas with "visual noise" are
examples where drivers need decision sight
distances. Exhibit 3-3, provides values used
by designers for appropriate sight
distances.
These values are applicable to most
situations and have been developed
from empirical data. Because of
additional maneuvering space needed
for safety, it is recommended that
decision sight distances be provided at
critical locations or critical decision
points may be moved to where adequate
distances are available.
If it is not practical to provide decision
sight distance because of horizontal or
vertical curvature or if relocation of
decision points is not practical, special
attention should be given to the use of
suitable traffic control devices for
providing advance warning of the
conditions that are likely to be
encountered.
Distances in Exhibit 3-3 for avoidance
maneuvers A and B are calculated as
based on Equation (2); however, with
modified driver reaction time as stated in
the notes for Exhibit 3-3. Decision sight
distances for maneuvers C, D, and E are
calculated from either 0.278Vt or 1.47 V t,
with t, modified as in the notes.
In computing and measuring stopping sight
distances, driver's eye height is estimated as
1080 mm (3.5 ft) and the height of the
dangerous object seen by the driver is 600
mm (2.0 ft), which represents the height of
tail-lights of a passenger car.
Stopping Sight Distance on plane Road
Braking Distance on sloping Road
Decision Distance
Passing Sight Distance for Two-Lane
Highways
On most two-lane, two-way highways,
vehicles frequently overtake slower-moving
vehicles by using the lane meant for the
opposing traffic. To complete the passing
maneuver safely, the driver should be able to
see a sufficient distance ahead. Passing
sight distance is determined on the basis
that a driver wishes to pass a single vehicle,
although multiple-vehicle passing is
permissible.
Based on observed traffic behavior, the
following assumptions are made:
1. The overtaken vehicle travels at a uniform
speed.
2. The passing vehicle has reduced speed
and trails the overtaken vehicle as it enters
a passing section.
3. The passing driver requires a short period
of time to perceive the clear passing
section, when reached, and to start
maneuvering.
4)The passing vehicle accelerates during the
maneuver, during the occupancy of the right
lane, at about 15 km/h (10 mph) higher than
the overtaken vehicle.
5)There is a suitable clearance length between
the passing vehicle and the oncoming
vehicle upon completion of the maneuver.
The minimum passing sight distance for twolane highways is determined as the sum of the
four distances shown in Figures on next slides.
Safe passing sight distances for various
speed ranges determined from distance
and time values observed in the field are
summarized in Exhibit 3-5
Initial maneuver distance d1
The distance d1 traveled during the initial
maneuver period is computed with the
following equation:
Distance while passing vehicle
occupies left lane (d2).
Passing vehicles were found in the study to
occupy the left lane from 9.3 to 10.4 s. The
distance d2 traveled in the left lane by the
passing vehicle is computed with the following
equation:
Clearance length (d3).
The clearance length between the
opposing and passing vehicles at the end
of the passing maneuvers was found in the
passing study to vary from 30 to 75 m [100
to 250 ft].
Distance traversed by an opposing vehicle
(d4)
The opposing vehicle is assumed to be
traveling at the same speed as the passing
vehicle, so
2d
d 
3
4
2
Initial maneuver distance d1
Distance while passing vehicle occupies left
lane (d2).
Clearance length (d3). The clearance length between the opposing and
passing vehicles at the end of the passing maneuvers was found in the passing
study to vary from 30 to 75 m [100 to 250 ft].
Distance traversed by an opposing vehicle (d4)
d4=2/3 d2
Estimation of Velocity of a Vehicle just Before
it is Involved in an accident
Some times it is necessary to determine the
velocity of a vehicle just before it is
involved in an accident. Following steps
are involved:
1) Estimate length of skid marks for all the
four tires of the vehicle and take the
average length. This is equal to breaking
distance.
2)Find out “f” by performing trial runs
under same environment /weather
conditions and using a vehicle having
similar conditions of tyres. Vehicle is
driven at a known speed and breaking
distance is measured.
3)The unknown speed is than determined
using the braking formula.
Horizontal Alignment
• Along circular path, vehicle undergoes
centripetal acceleration towards center of
curvature (lateral acceleration).
• Balanced by superelevation and weight of
vehicle (friction between tire and roadway).
• Design based on appropriate relationship
between design speed and curvature and
their relationship with side friction and
super elevation.
Vehicle Cornering
Fcp
Rv
Fcn
Fc

e
W

Ff
Wn
Wp
1 ft
Ff
• Figure illustrates the forces acting on
a vehicle during cornering. In this
figure,  is the angle of inclination, W
is the weight of the vehicle in pounds
with Wn and Wp being the weight
normal and parallel to the roadway
surface respectively.
• Ff is the side frictional force, Fc is the
centrifugal force with Fcp being the
centrifugal force acting parallel to the
roadway surface, Fcn is the
centrifugal force acting normal to the
roadway surface, and Rv is the radius
defined to the vehicle’s traveled path
in ft.
• Some basic horizontal curve
relationships can be derived by
summing forces parallel to the
roadway surface.
Wp + Ff = Fcp
• From basic physics this equation can
be written as
WV 2
WV 2
W sin   fs[W cos 
sin  ] 
cos
gRv
gRv
• Where fs is the coefficient of side
friction, g is the gravitational constant
and V is the vehicle speed (in ft per
second).
• Dividing both the sides of the
equation by W cos ;
sin 
cos WV 2 sin 
WV 2 cos
W
 fs[W

]
cos
cos
gRv cos
gRv cos
t an  f s
V2

(1  f s t an )
gRv
• The term tan is referred to as the
super elevation of the curve and is
denoted by ‘e’.
• Super elevation is tilting the roadway
to help offset centripetal forces
developed as the vehicle goes
around a curve.
• The term ‘fs’ is conservatively set
equal to zero for practical
applications due to small values that
‘fs’ and ‘’ typically assume.
• With e = tan, equation can be rearranged as
V2
Rv 
g ( f s  e)
• Here Rv in meters
• V speed in Km/hr
• G=9.8m/sec2
• e = coefficient of friction
For Imperial units:
V in mph
And Rv in feet
2
V
R 
15( f  e)
v
s
Superelevation
Road Section View
CL
2%
2%
Road Plan View
Superelevation
Road Section View
CL
1.5%
2%
Road Plan View
Superelevation
Road Section View
CL
1%
2%
Road Plan View
Superelevation
Road Section View
0.5%
CL
2%
Road
Plan View
Superelevation
Road Section
View
2
CL
%
-0.0%
Road Plan View
Superelevation
Road Section View Road Plan View
-0.5%
CL
2%
Superelevation
Road
Section
C
View
-1%
L
Road
Plan View
2
%
Superelevation
Road Section View
-.5% CL
2
%
Road Plan View
Superelevation
Road Section
View
C
-2%
L
Road Plan View
2
%
Superelevation
Road Section View
-3%
C
L
3
%
Road Plan View
Super elevation
Road Section View
-4%
CL
4%
Road Plan View
Superelevation
Road
Section
C
View
-3%
L
Road
Plan View
3
%
Superelevation
Road
Section
C
View
-2%
L
Road
Plan View
2
%
Superelevation
1.5
%
Road
Section
ViewC
L
Road
Plan View
2
%
Superelevation
Road
Section
C
View
-1%
L
Road
Plan View
2
%
Superelevation
0.5
%
Road
Section
ViewC
L
Road
Plan View
2
%
Superelevation
0.0
%
Road
Section
ViewC
L
Road
Plan View
2
%
Superelevation
Road
Section
C
View
0.5
L
%
Road
Plan View
2
%
Superelevation
Road
Section
C
View
1%
L
Road
Plan View
2
%
Superelevation
Road
Section
C
View
1.5
L
%
Road
Plan View
2
%
Superelevation
Road
Section
C
View
2%
L
Road
Plan View
2
%
• In actual design of a horizontal curve, the
engineer must select appropriate values of
e and fs.
• Super-elevation value ‘e’ is critical since
– high rates of super-elevation can cause vehicle
steering problems at exits on horizontal curves
– and in cold climates, ice on road ways can
reduce fs and vehicles are forced inwardly off
the curve by gravitational forces.
• Values of ‘e’ and ‘fs’ can be obtained from
AASHTO standards.
Horizontal Curve
Fundamentals
• For connecting straight tangent
sections of roadway with a curve,
several options are available.
• The most obvious is the simple curve,
which is just a standard curve with a
single, constant radius.
• Other options include;
– compound curve, which consists of two
or more simple curves in succession ,
– and spiral curves which are
continuously changing radius curves.
Basic Geometry
Tangent
Horizontal
Curve
Tangent
Tangent Vs. Horizontal Curve
• Predicting speeds for tangent and
horizontal segments is different
• May actually be easier to predict
speeds on curves than tangents
– Speeds on curves are restricted to a few
well defined variables (e.g. radius,
superelevation)
– Speeds on tangents are not as
restricted by design variables (e.g.
driver attitude)
Elements of Horizontal
Curves
PI

E

M
L

PC
PT
R
R




• Figure shows the basic elements of a
simple horizontal curve. In this figure
– R is the radius (measured to center line of the
road)
– PC is the beginning point of horizontal curve
– T is tangent length
– PI is tangent intersection
–  is the central angle of the curve
– PT is end point of curve
– M is the middle ordinate
– E is the external distance
– L is the length of the curve
Degree of Curve
• It is the angle subtended by a 100-ft arc
along the horizontal curve.
• Is a measure of the sharpness of curve
and is frequently used instead of the
radius in the actual construction of
horizontal curve.
• The degree of curve is directly related to
the radius of the horizontal curve by
5729 .6
D
R
• A geometric and
trigonometric analysis
of figure, reveals the
following relationships
PI

E

M
L

PC
PT
R
R





T  R t an
2


 1

E  R
 1
 cos(  ) 


2


 

M  R1  cos( ) 
2 

100
L
D
Stopping Sight Distance and
Horizontal Curve Design
SSD
Ms
Sight
Obstruction
Highway
Centerline
Rv
Critical
inside lane
s
• Adequate stopping sight distance
must also be provided in the design
of horizontal curves.
• Sight distance restrictions on
horizontal curves occur when
obstructions are present.
• Such obstructions are frequently
encountered in highway design due
to the cost of right of way acquisition
and/or cost of moving earthen
materials.
• When such an obstruction exists, the
stopping sight distance is measured
along the horizontal curve from the
center of the traveled lane.
• For a specified stopping sight
distance, some distance, Ms, must be
visually cleared, so that the line of
sight is such that sufficient stopping
sight distance is available.
• Equations for computing SSD
relationships for horizontal curves
can be derived by first determining
the central angle, s, for an arc equal
to the required stopping sight
distance.
• Assuming that the length of the horizontal
curve exceeds the required SSD, we have
100  s
SSD 
D
Combining the above equation with following
we get;
5729 .6
D
R
57.296SSD
s 
Rv
• Rv is the radius to the vehicle’s
traveled path, which is also assumed
to be the location of the driver’s eye
for sight distance, and is again taken
as the radius to the middle of the
innermost lane,
• and s is the angle subtended by an
arc equal to SSD in length.
• By substituting equation for s in equation
of middle ordinate, we get the following
equation for middle ordinate;

 28.65SSD  
 
Ms  Rv 1  cos

Rv



• Where Ms is the middle ordinate
necessary to provide adequate stopping
sight distance. Solving further we get;
Rv 
1  Rv  M s
SSD 
cos 
28.65 
 Rv



Max e
• Controlled by 4 factors:
– Climate conditions (amount of ice and snow)
– Terrain (flat, rolling, mountainous)
– Frequency of slow moving vehicles which
influenced by high superelevation rates
– Highest in common use = 10%, 12% with no
ice and snow on low volume gravel-surfaced
roads
– 8% is logical maximum to minimized slipping
by stopped vehicles
Example
A curving roadway has a design speed of 110
km/hr. At one horizontal curve, the
Super elevation has been set at 6.0% and the
coefficient of side friction is found to be 0.10.
Determine the minimum radius of the curve that
will provide safe
V2
Rv 
g ( f s  e)
R = 1102/9.8(0.10+0.06) = 595 meters
Radius Calculation (Example)
Design radius example: assume a
maximum e of 8% and design speed
of 60 mph, what is the minimum
radius?
fmax = 0.12 (from Green Book)
Rmin = _____602________________
15(0.08 + 0.12)
Rmin = 1200 feet
Radius Calculation (Example)
For emax = 4%?
Rmin = _____602_________
15(0.04 + 0.12)
Rmin = 1,500 feet
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 b/r
=2.5 sec and a = 11.2 ft/sec2
SSD is given as
SSD = 1.47vt + _________v2____
30(__a___  G)
32.2
SSD = 1.47(35 mph)(2.5 sec) + _____(35 mph)2____
30(__11.2___  0)
32.2
= 246 feet
Sight Distance Example
m = R(1 – cos [28.65 S])
R
m = 800 (1 – cos [28.65 {246}]) = 9.43
feet
800
(in radians not degrees)
Horizontal Curve Example
• Deflection angle of a 4º curve is 55º25’, PC
at station 238 + 44.75. Find length of
curve,T, and station of PC.
• D = 4º
•  = 55º25’ = 55.417º
• D = _5729.58_ R = _5729.58_ = 1,432.4 ft
R
4
Horizontal Curve Example
• D = 4º
•  = 55.417º
• R = 1,432.4 ft
• L = 2R = 2(1,432.4 ft)(55.417º) = 1385.42ft
360
360
Horizontal Curve Example
•
•
•
•
D = 4º
 = 55.417º
R = 1,432.4 ft
L = 1385.42 ft
• T = R tan  = 1,432.4 ft tan (55.417) = 752.29 ft
2
2
Stationing Example
Stationing goes around horizontal curve.
For previous example, what is station of
PT?
PC = 238 + 44.75
L = 1385.42 ft = 13 + 85.42
Station at PT = (238 + 44.75) + (13 +
85.42) = 252 + 30.17
Suggested Steps on
Horizontal Design
1. Select tangents, PIs, and general curves
make sure you meet minimum radii
2. Select specific curve radii/spiral and
calculate important points (see lab) using
formula or table (those needed for design,
plans, and lab requirements)
3. Station alignment (as curves are
encountered)
4. Determine super and runoff for curves and
put in table (see next lecture for def.)
5. Add information to plans
Geometric Design – Horizontal
Alignment (1)
• Horizontal curve
– Plan view, profile, staking,
stationing
– type of horizontal curves
– Characteristics of simple circular
curve
• Stopping sight distance on
horizontal curves
• Spiral curve
Lecture 8
Plan view and profile
+0
24
0
23+00
22+00
21+00
20+00
19+00
18+00
0
17+0
+0
16
0
+0
15
0
plan
700
700
600
500
profile
400
300
200
15+00
16+00
17+00
18+00
19+00
20+00
21+00
22+00
23+00 24+00
Surveying and Stationing
• Staking: route surveyors define the
geometry of a highway by “staking” out
the horizontal and vertical position of the
route and by marking of the crosssection at intervals of 100 ft.
• Station: Start from an origin by stationing
0, regular stations are established every
100 ft., and numbered 0+00, 12 + 00
(=1200 ft), 20 + 45 (2000 ft + 45) etc.
Horizontal Curve Types
Curve Types
1. Simple curves with spirals
2. Broken Back – two curves same direction
(avoid)
3. Compound curves: multiple curves
connected directly together (use with
caution) go from large radii to smaller radii
and have R(large) < 1.5 R(small)
4. Reverse curves – two curves, opposite
direction (require separation typically for
superelevation attainment)
1. Simple Curve
Circular
arc

R

Straight road
sections
2. Compound Curve
Circular arcs
R1
Straight road
sections
R2
3. Broken Back Curve
Circular arc
Straight road
sections
4. Reverse Curve
Circular arcs
Straight road
sections
5. Spiral
R = Rn
R=
Straight road
section
Angle measurement
90
60
30
0
180
(a) degree
(b) Radian
1   / 180radians 0.0174532
1radian  (180/  )  57.2957
radians
As the subtended arc is proportional to the radius
of the circle, then the radian measure of the angle
Is the ratio of the length of the subtended arc to
the radius of the circle
radian measure R  arc length
• Define horizontal Curve:
• Circular Horizontal Curve Definitions
• Radius, usually measured to the centerline of the road, in
ft.
•  = Central angle of the curve in degrees
• PC = point of curve (the beginning point of the horizontal
curve)
• PI = point of tangent intersection
• PT = Point of tangent (the ending point of the horizontal
curve)
• T = tangent length in ft.
• M = middle ordinate from middle point of cord to middle
point of curve in ft.
• E = External distance in ft.
• L = length of curve
• D = Degree of curvature (the angle subtended by a 100-ft
arc* along the horizontal curve)
* Note: use chord in practice
• C = chord length from PC to PT
PI

T
L
PC

2
90 
C
E
M

2

2
90 
R
R
 
2 2

PT

2
Key measures of the curve
M  R[1  cos( / 2)]
100(180/  ) 18000 5729.57
D


R
R
R
or,
100
D
 57.2957
R

T  R tan
2
1
E  R[(
)  1]
cos( / 2)

L
R
180

C  2 R sin
2
180 / 
Note
converts from radians to degrees
Example:
A horizontal curve is designed with a 2000-ft
radius, the curve has a tangent length of 400 ft
and
The PI is at station 103 + 00, determine the
stationing of the PT.
Solution:
T  R tan

2
400  2000tan

2
  22.62
L

180
R  789.58 ft
PC  (103 00)  (4  00)  99  00
PT  PC  L  (99  00)  (7  89.58)  106 89.58
Sight Distance on Horizontal Curve:
Minimum sight distance (for safety) should
be equal to the safe stopping distance
Sight Distance
Highway Centerline
M
PC
PT
Line of sight
Sight Obstruction
Centerline of inside lane
R
R
To provide minimum sight
distance:
M  R[1  cos(
28 .65 d s
)]
R
Or, by the degree of
curvature, D
M
Where, ds = safe
stopping distance, ft.
and,
d D
5729 .57
[1  cos( s )]
D
200
Try yourself
v2
d s  1.47v * t 
30(
a
 0.01G )
32.2
v = design speed, mi/h
t = reaction time, secs
G = grade, %
ds = stopping distance, in ft.
a =deceleration rate, 11.2 ft/s2, recommended by Green Book
Example:
A 6 degree curve (measured at the centerline of
the inside lane) is being designed for a
highway
with a design speed of 70 mi/hr., the grade is
level, the driver reaction time is taken as 2.5 s
(ASSHTO’s standard value). What is the closest
place that a roadside object (trees etc) can be
Placed?
Solution:
2
70
d s  1.47* 2.5 * 70 
 726.6 ft
30(0.348 0.01* 0)
The closest place of a object is given by:
5729 .58
726 .6 * 6
M
[1  cos(
)]  68.3 ft
6
200
Spiral Curve:
Spiral curves are curves with a
continuously changing radii,
they are sometimes used on
high-speed roadways with
sharp horizontal curves and
are sometimes used to
gradually introduce the super
elevation of an upcoming
horizontal curve