SIGHT DISTANCES
4.1 Overview
4.2 Stopping Sight Distance
4.2.1 Perception-Reaction Distance
4.2.2 Braking Distance
4.2.3 Sample Problems
4.3 Decision Sight Distance
4.4 Passing Sight Distance
4.4.1 Elements of Passing Sight Distance
4.4.2 Sample Problems
4.5 Sight Distance on Vertical Curves
4.5.1 Sight Distance on Summit Curves
4.5.2 Sight Distance on Sag Curves
4.5.3 Sight Distance sag curve with obstructing overpass
4.6 Sight Distance on Horizontal Curves
4.1 Overview
This section provides descriptions and information on sight distance, one of several principal elements of design that are
common to all types of highways and streets. Of utmost importance in highway design is the arrangement of geometric
elements so that there is adequate sight distance for safe and efficient traffic operation assuming adequate light, clear
atmospheric conditions, and drivers' visual acuity. For design, the following five types of sight distance are considered:
• Stopping Sight Distance
• Decision Sight Distance
• Passing Sight Distance
• Sight Distance on Vertical Curves
• Sight Distance on Horizontal Curves
4.2 Stopping Sight Distance
Sight distance is the length of roadway ahead that is visible to the driver. The available sight distance on a roadway should
be sufficiently long to enable a vehicle traveling at or a near the design speed to stop before reaching a stationary object
in its path. Although greater lengths of visible roadway are desirable, the sight distance at every point along a roadway
should be at least that needed for a below-average driver or vehicle to stop. Stopping sight distance is the sum of two
distances: (1) the distance traversed by the vehicle from the instant the driver sights an object necessitating a stop to the
instant the brakes are applied; and (2) the distance needed to stop the vehicle from the instant brake application begins.
These are referred to as brake reaction distance (perception-reaction distance) and braking distance, respectively. In
computing and measuring stopping sight distances, the height of the driver's eye is estimated to be 3.5 ft (1080mm] and
the height of the object to be seen by the driver is 2.0 ft (600 mm], equivalent to the taillight height of the passenger car.
Stopping sight distance consists of the following time intervals:
1. The time for the driver to perceive the obstruction.
2. The time to react.
3. The time for the vehicle to stop after brakes are applied (Braking Distance, BD)
The design standards of the American Association of State Highway and Transportation Officials (AASHTO) allow 1.5
seconds for perception time and 1.0 second for reaction time. The values of stopping sight distance used in design
represent a near worst-case situation. For design, a conservative distance is needed to allow a vehicle traveling at design
speed to stop before reaching a stationary object in its path. A generous amount of time is given for the perceptionreaction process, and a fairly low rate of deceleration is used. The design sight distance allows a below-average driver to
stop in time to avoid a collision in most cases.
Based on the results of many studies, 2.5 seconds has been chosen for a perception-reaction time. This time will
accommodate approximately 90 percent of all drivers when confronted with simple to moderately complex highway
situations. Greater reaction time should be allowed in situations that are more complex.
4.2.2 Braking Distance, BD
it is primarily affected by the original speed of the vehicle and the coefficient of friction between the tires and the road
surface, and negligibly by the tires rolling resistance and vehicle's air drag. The type of brake system in use only affects
trucks and large mass vehicles, which cannot supply enough force to match the static frictional force.
A. By Energy Equation.
BD =
( ± )
B. By Newton’s Second Law and Equation of motion
BD =
( ± )
where BD= breaking distance in meters
v = velocity in m/s
g = gravitational acceleration in m/sec2
f = coefficient of friction
G = grade of the road (+G
and -G
a = deceleration rate in m/sec2
)
The braking distance and the brake reaction time are both essential parts of the stopping sight distance calculations. In
order to ensure that the stopping sight distance provided is adequate, we need a more in-depth understanding of the
frictional force. The value of the coefficient of friction is a difficult thing to determine. The frictional force between your
tires and the roadway is highly variable and depends on the tire pressure, tire composition, and tread type. The frictional
force also depends on the condition of the pavement surface. The presence of moisture, mud, snow, or ice can greatly
reduce the frictional force that is stopping you. In addition, the coefficient of friction is lower at higher speeds. Since the
coefficient of friction for wet pavement is lower than the coefficient of friction for dry pavement, the wet pavement
conditions are used in the stopping sight distance calculations. This provides a reasonable margin of safety, regardless of
the roadway surface conditions. The table below gives a few values for the frictional coefficient under wet roadway
surface conditions (AASHTO, 1984).
Design Speed (mph)
Coefficient of Friction (f)
20
0.40
30
0.35
40
0.32
60
0.29
The Stopping Sight Distance
SSD = PRD + BD
Substitution:
SSD = vt +
( ± )
and
SSD = vt +
( ± )
The calculated and design stopping sight distances are shown in Table 2-1.
The values given in Table 2-1 represent stopping sight distances on level terrain. As a general rule, the sight distance
available on downgrades is larger than on upgrades, more or less automatically providing the necessary corrections for
grade. Therefore, corrections for grade are usually unnecessary. An example where correction for grade might come into
play for stopping sight distance would be a divided roadway with independent design profiles extreme rolling of
mountainous terrain. A policy on Geometric Design for Highways and Streets, AASHTO, provides additional information
and suggested values for grade corrections in these rare circumstances.
4.2.3 Example Problems
Problem 1
A driver with a 2.5-second reaction time is travelling at 105 kilometers per hour up a 2% grade. Assuming AASHTO
standard values for all other information, what distance will be required for the driver to come to a complete stop if an
obstruction enters his or her field of vision? ANS. 191.19m
Problem 2
A car was traveling at a speed of 50 mph. The driver saw a roadblock 80 m ahead and stepped on the brake causing the
car to decelerate uniformly at 10 m/s2. Find the distance from the roadblock to the point where the car stopped. Assume
perception-reaction time is 2 seconds. ANS. 10.28m
Problem 3
The driver applied the brakes on a car that is running at a speed of 50 mph. In how many seconds will the car stop if the
average skid resistance is 0.70? ANS. 3.25 sec
Problem 4
A vehicle moving 45 kph was stopped by applying brakes and the length of the skid mark was 13.2 m. If the average skid
resistance of the level pavement is known to be 0.70, determine the brake efficiency of the test vehicle. ANS. 86.2%
4.3 Decision Sight Distance
Decision sight distance is the distance required for a driver to detect an unexpected or otherwise difficult-to-perceive
information source. recognize the source, select an appropriate speed and path, and initiate and complete the required
maneuver safely and efficiently. Because decision sight distance gives drivers additional margin for error and affords them
sufficient length to maneuver their vehicles at the same or reduced speed rather than to just stop. Its values are
substantially greater than stopping sight distance. Table 2-2 shows recommended decision sight distance values for
various avoidance maneuvers.
Examples of situations in which decision sight distance is preferred Include the following:
• Interchange and Intersection locations where unusual or unexpected maneuvers are required (such as exit
ramp gore areas and left-hand exits)
• Changes in cross section such as toll plazas and lane drops
• Areas Of concentrated demand where there is apt to be 'visual noise’ whenever sources of information
compete, as those from roadway elements, traffic, traffic control devices, and advertising signs
Decision Sight Distance Calculation
DSD = vt
where: DSD = decision sight distance, m
y = design speed, m/s
t = total time for the maneuver time (reaction +maneuver), s
4.4 Passing Sight Distance
Passing sight distance considerations are limited to 2-Iane, 2-way highways. On these facilities, vehicles may overtake
slower moving vehicles, and the passing maneuver must be accomplished on a lane used by opposing traffic.
The minimum passing Sight distance for 2-lane highways is determined from the sum of four distances as Illustrated in
the figure below, Element of Passing Sight Distance (2-Lane Highways). Passing Sight Distance on Two-Lane Highways and
the following provide the best assumptions used to develop passing sight distance values.
4.4.1 Elements of Passing Sight Distance:
1. Initial Maneuver Distance (𝒅𝟏 ). This is the distance traversed during the perception and reaction time and during the
initial acceleration to the point of encroachment on the left lane. For the initial maneuver, the overtaken vehicle is
assumed to be traveling at a uniform speed, and the passing vehicle is accelerating at the rate shown in the figure. The
average speed of the passing vehicle is assumed to be 15 km/h greater than the overtaken vehicle. Equation below is
used to determine 𝑑 :
𝑑 =
.
(𝑣 − 𝑚 +
)
Where: 𝑡 = time of initial maneuver, sec
a = average acceleration, km/hr/sec
v = average speed of passing vehicle, km/h
m = difference in speed of passed vehicle and passing vehicle, km/h
2. Distance of Passing Vehicle in Left Lane (𝒅𝟐 ). This ts the dlstance traveled by the passing vehicle while it occupies the
left lane. Assumed times for when the passing vehicle occupies the left lane are shown in the figure. Equation below is
used to determine 𝑑 :
𝑑 =
.
Where: 𝑡 = time passing vehicle occupies the left lane, sec
v = average speed of passing vehicle, km/h
3. Clearance Distance (𝒅𝟑 ). This is the distance between the passing vehicle at the end of its maneuver and the opposing
vehicle. Based on various studies, this clearance distance at the end of the passing maneuver is assumed to be between
30 m and 90 m.
4. Opposing Vehicle Distance (𝒅𝟒 ). This is the distance traversed by an opposing vehicle during 2/3 of the time passing
vehicle occupies the left lane. As shown in the figure, the opposing vehicle appears after approximately 1/3 of the passing
maneuver (d2) has been accomplished. The opposing vehicle is assumed to be traveling at the same speed as the passing
vehicle. Therefore, d4 = d2
Minimum Passing Sight Distance (PSD) = d1 + d2 + d3 + d4
The table below summarizes the results of field observations directed toward quantifying the various aspects of the
passing sight distance (AASHTO, 1994).
Now that we know how to calculate the required passing sight distance, how do we calculate the actual passing sight
distance that we have provided in our geometric design? To do this, we simply assume that the driver's eyes are at a
height of 1080 mm from the road surface and the opposing vehicle is 1300 mm tall. The actual passing sight distance is
the length of roadway ahead over which an object 1300 mm tall would be visible, if your eyes were at an elevation of
1080 mm.
4.4.2 Example Problem
1.0 Find the minimum passing sight distance at a section of highway assuming: The speed of the passing vehicle is 70 kph,
the speed of the slow vehicle is 55 kph, acceleration of the passing vehicle is 2.3 km/hr/sec, initial time of maneuver is 4
seconds and time of passing vehicle traveling in left lane to re-entry on the right lane is 10 seconds. Assume that the
clearance between the re-entry on the right lane of the passing vehicle and the opposing vehicle is 55 m. ANS. 445.30m
4.5 Sight Distance on Vertical Curves
4.5.1 Summit Curves
A. When S < L
L=
B.
When S > L
L = 2S -
(
)
where: A = algebraic difference in grade in percent = g1- g2
L = length of curve in meters
S = sight distance (SSD/PSD) in meters
STANDARD VALUES IN ROAD DESIGN:
For Stopping Sight Distance (SSD)
h1 = 3.75 feet (1.14 m)
h2 = 6 inches (0.15 m)
For Passing Sight Distance (PSD)
h1 = 3.75 feet (1.14 m)
h2 = 4.5 feet (1.37 m)
For Decision Sight Distance (DSD)
h1 = 3.75 feet (1.14 m)
h2 = 2 feet (0.60 m)
Length of Common Tangent, T
T=
Where: L = length of curve
Rate of vertical curvature, K
K Value represents the horizontal distance for a 1% of grade change
K=
4.6.1 Example Problems
1.0) A 5% grade intersects a -3.4% grade at station 1+990 of elevation 42.30 m. Design a vertical summit parabolic curve
connecting the two tangent grades to conform with the following safe stopping sight distance specifications.
Design velocity =70 kph
Height of driver's eye from the road pavement = 1.14 m
Height of an object over the pavement ahead = 150 mm
Perception-reaction time = 3/4 sec
Coefficient of friction between the road pavement and the tires = 0.15
a) Determine the stopping sight distance. ANS. 110.94 m
b) Determine the length of curve. ANS. 244.17 m
c) Determine the elevation of highest point on curve. ANS. 39.83m
2.0) A vertical summit curve has tangent grades of 3.5% and -2.5% intersecting at STA 12+460.12 at an elevation of 150 m
above sea level. If the length of curve is 200 m.
a) Compute the length of the passing sight distance. ANS. 224.49 m
b) Compute the stationing of the highest point of the curve. ANS. 12+508.66
c) Compute the elevation of highest point on curve. ANS. 148.54 m
4.5.2 Sag Curves
Sag vertical curves are in the shape of a parabola. Typically, they are designed to allow the vehicular headlights to
illuminate the roadway surface (i.e., the height of object = 0.0 ft (0.0 mm)) for a given distance “S”. The light beam from
the headlights is assumed to have a 1° upward divergence from the longitudinal axis of the vehicle. These assumptions
yield the following basic equations for determining the minimum length of sag vertical curves:
STANDARD DESIGN VALUES:
H = 2 ft (600 mm)
β = 1°
A. When S < L
L=
(
)
At H= 0.60 m and β = 1°
L=
.
B. When S > L
L = 2S -
(
)
At H = 0.60 m and β = 1°
.
L = 2S -
Where: A = Algebraic difference in grade in percent = g1 – g2
L = Length of curve in meters
S = Sight distance in meters
Minimum Visibility, Vmin
Vmin =
Where: S = Sight Distance
L=
Where: v = velocity in kph
Maximum Velocity, Vmax
A = algebraic difference in grade in percent = g1 – g2
Minimum Radius of Sag Curve, Rmin
Rmin =
.
Where: v = velocity in kph
4.5.2 Example Problems
1.0) The length of sag parabolic curve is 130 m with a design speed of 80 kph. The back tangent has a slope of -3.5 %.
a) Find the slope of the forward tangent. ANS. 4.523%
b) Find the distance of the lowest point of curve from PC. ANS. 73.29 m
c) Find the length of the sight distance. ANS. 81.09 m
2.0) A vertical sag parabolic curve has a length of 120 m with tangent grades of -1.5% and +2.5% intersecting at station
12+640.22 and elevation of 240 m above sea level
a) Compute the length of the sight distance. ANS. 133.78 m
b) Compute the maximum speed that a car would travel to avoid collision. ANS. 108.86 kph
c) Compute the perception-reaction time e the driver to avoid collision.
Assume coefficient of friction between tires and pavement is 0.40. ANS. 0.57 sec
4.5.3 Sight Distance on vertical sag curve with obstructing overpass
A. When S < L
L=
B. When S > L
L =2S Where: A = algebraic difference in grade in decimal = g1 – g2
L = length of curve in meters
S = sight distance in meters
C = the vertical clearance between the sag curve and the obstruction
h1 = eye height (based on SSD/PSD)
h2 = object height (based on SSD/PSD)
H=C–z
z=
4.5.3 Example Problems
1) A -3% grade meets a +6% grade near an underpass. In order to maintain the minimum clearance allowed under the
bridge and at the same time introduce a vertical transition curve in the grade line, it is necessary to use a curve that lies
200 m on one side of the vertex of the straight grade and 100 m on the other. The station of the beginning of the curve
(200 m side) is 10+000 and its elevation is 228 m. If the uphill edge of the underside of the bridge is at station 10+220 and
at elevation 229.206 m, what is the stopping and passing sight distances of the curve? ANS. 303.11 m, 274.95 m
4.6 Sight Distance on Horizontal Curves
Use equation In Equation 4.6.1 to check stopping sight distance where sightline obstructions are on the inside of a curve. A
stopping sight distance sightline obstruction is any roadside object within the horizontal sightline offset (M) distance (such
as median barrier, guardrail, bridges, walls, cut slopes, buildings or wooded areas), 2.0 feet or greater above the roadway
surface at the centerline of the lane on the inside of the curve (h0). Equation 4.6.2 for use when the length of curve is
greater than the sight distance and the sight restriction is more than half the sight distance from the end of the curve.
Where the length of curve is less than the stopping sight distance or the sight restriction is near either end of the curve, the
desired sight distance may be available with a lesser M distance. When this occurs, the sight distance can be checked
graphically.
A. When S < L
M=
B. When S > L
M=
(
)
Where: M = clear distance from center of roadway to the obstruction
S = sight distance along the center of roadway
R = radius of centerline curve
L = length of curve
4.6 Example Problems
1) The clearance to an obstruction is 40 m and the desirable sight distance when rounding a horizontal curve is 600 m.
Determine the minimum radius of horizontal curve if the length of curve is:
a) 650 m long. ANS. 1,125 m
b) 550 m long. ANS. 1,117.19 m
Exercises 5.0
1.0 Find the Stopping Sight Distance of the following roadway:
Design speed = 65 kph
Driver reaction time = 2 seconds
Grade = 0%
Deceleration rate = 3.4 m/s2
ANS. 84.05 m
2.0 Compute the minimum passing sight distance for the following data.
Speed of the passing car = 90 kph
Speed of the overtaken vehicle = 80 kph
Time of initial maneuver = 4 seconds
Average acceleration = 2.4 kph/sec
Time passing vehicle occupies the left lane = 9 seconds
Distance between the passing vehicle at the end of its maneuver and the opposing vehicle = 80 m
ANS. 549.22 m
3.0 A 4% grade intersects a -3.4% grade at station 1+990 of elevation 42.30 m. Design a vertical summit parabolic curve
connecting the two tangent grades to conform with the following safe stopping sight distance specifications.
Design velocity =80 kph
Height of driver’s eye from the road pavement = 1.14 m
Height of an object over the pavement ahead = 150 mm
Perception-reaction time = 3/4 sec
Coefficient of friction between the road pavement and the tires = 0.15
a) Determine the stopping sight distance. ANS. 149.14 m
b) Determine the length of curve. ANS. 388.74 m
c) Determine the elevation of highest point on curve. ANS. 38.72 m
4.0 The clearance to an obstruction is 9 m, the radius and length of the curve are 350 m and 500 m respectively. Determine
the desirable sight distance. ANS. 158.75 m
5.0 Find the length of the skid mark if the average skid resistance is 0.15 and the velocity of the car when the brakes were
applied was 50 kph. ANS. 65.55 m
6.0 A car runs 50 m from the brakes were suddenly applied until it stopped. The road grade is 4% up hill and the coefficient
of friction between the tires and the road surface is 0.35. What was the speed of the car in kph, just before the application
of the brakes? Compute the speed if the road grade is 4% downhill. ANS. 70.42 kph, 65.51 kph
7.0 The sag vertical curve has a back tangent grade of -3% and a forward tangent grade of +2% is to be designed on the
basis that the head lamp sight distance of a car traveling along the curve equals the minimum safe stopping distance with
the following specifications.
Design velocity = 100 kph
Height of an object over the pavement ahead = 150 mm
Perception-reaction time = 3/4 sec
Coefficient of friction between the road pavement and the tires = 0.15
a) Determine the heap lamp sight distance. ANS. 348.56 m
b) Determine the length of curve. ANS. 452.67 m
c) Determine the maximum velocity of the car that could pass through the sag curve. ANS. 126.58 kph
8.0 A vertical summit curve has grades of +3% and -2%, which intersects at station 20+040. If the stopping sight distance is
equal to 180 m
a) Compute the maximum speed that a vehicle could move along this curve to avoid collision if the perception
reaction time of the driver is 2.0 sec and the coefficient of friction is equal to 0.25. ANS. 87.62 kph
b) Determine the length of curve. ANS. 382.61 m
c) Determine the stationing of the highest point of curve. ANS. 20+078.26
9.0 The design speed of a vertical sag curve is equal to 100 kph. The tangent grades of the curve are -3% and +4%
respectively.
a) Compute the length of curve in meters. ANS. 177.22 m
b) Compute the sight distance in meters. ANS. 115.38 m
c) Compute the length of minimum visibility in meters. ANS. 57.69 m
10.0 A car traveling at 90 kph requires 60 m to stop after the brakes have been applied. Determine the slope of the road
surface if the average coefficient of friction is 0.50. ANS. 3.1% (upward)
11.0 While traveling at a constant speed, a driver saw a roadblock 120 m away. He applied the brakes and the car
decelerated uniformly at 3.4 m/s2. It stopped 9 m from the block. Find the initial constant speed of the vehicle in mph if the
driver's perception reaction time is 2.0 seconds. ANS. 48.09 mph
12.0 Two cars are approaching each other from the opposite directions at a speed of 100 kph and 80 kph respectively.
Assuming a reaction time 2.0 seconds and a coefficient of friction of 0.40 with a brake efficiency of 60%. Compute the
minimum sight distance required to avoid a head collision of the two cars. ANS. 368.74 m