PASSING SIGHT DISTANCE
Passing sight distance is the distance required to see an
oncoming vehicle of a certain minimum size. It is intended to
ensure that a passing maneuver can be completed safely
under certain assumptions as to vehicle speeds and
acceleration capabilities.
𝑷𝒂𝒔𝒔𝒊𝒏𝒈 𝑺𝒊𝒈𝒉𝒕 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝒅𝟏 + 𝒅𝟐 + 𝒅𝟑 + 𝒅𝟒
𝑑1 = distance traversed during perception and reaction time and during the
initial acceleration to the the point of encroachment on the left line.
𝑑2 = distance travelled while the passing vehicle occupies left lane
𝑑3 = distance between the passing vehicle and the end of its maneuver and the
opposing vehicle.
𝑑4 = distance traversed by opposing vehicle for two-thirds of the time the
2
passing vehicle occupies the left lane, or 𝑑1
3
Minimum passing sight distances for design of two-lane highways incorporate
certain assumptions about driver behavior. Actual driver behavior in passing
maneuvers varies widely. To accommodate these variations in driver behavior,
the design criteria for passing sight distance should accommodate the behavior
of a high percentage of drivers, rather than just the average driver.
The following assumptions are made concerning driver behavior in passing
maneuvers:
1. The overtaken vehicle travels at uniform speed.
2. The passing vehicle has reduced speed and trails the overtaken vehicle as it
enters a passing section.
3. When the passing section is reached, the passing driver needs a short
period of time to perceive the clear passing section and to react to start his
or her maneuver.
4. Passing is accomplished under what may be termed a delayed start and
a hurried return in the face of opposing traffic. The passing vehicle
accelerates during the maneuver, and its average speed during the
occupancy of the left lane is 15 km/h [10 mph] higher than that of the
overtaken vehicle.
5.When the passing vehicle returns to its lane, there is a suitable clearance
length between it and an oncoming vehicle in the other lane.
𝟏
𝒅𝟏 = 𝑽 − 𝒎 𝒕 𝟏 + 𝒂 𝒕𝟏
𝟐
Where: 𝑡1 = time for initial maneuver, sec
a = average acceleration rate, kph/sec
V = average speed of passing vehicle, kph
m = difference in speeds of passing and passed vehicles
𝒅𝟐 = 𝑽 𝒕𝟐
Where: 𝑡2 = time passing vehicle is traveling in left lane
and usually found to be 9.3 to 10.4 sec
V = average speed of passing vehicle, kph
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].
𝟐
𝒅𝟒 = 𝒅𝟐
𝟑
𝒅𝟐 = 𝑽 𝒕 𝟐
Where: 𝑡2 = time passing vehicle is traveling in left lane
and usually found to be 9.3 to 10.4 sec
V = average speed of passing vehicle, kph
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].
Problem
Calculate the passing sight distance for a two way traffic
highway for which the design speed is 60 kmph. The rate of
acceleration of the fast moving vehicle may be assumed as
3.6 kmph/second and the difference in speed between the
overtaken vehicle as 20 kmph. What will be the passing sight
distance?
Given:
Design speed, V= 60 kmph
Difference in speed, m = 20 kmph
Rate of acceleration, a = 3.6 kmph
Problem:
Calculate the passing sight distance for a one way traffic
highway for which the design speed is 65 kmph. The rate of
acceleration of the fast moving vehicle may be assumed
as 3.6 kmph/second and the difference in speed between
the overtaken vehicle as 20 kmph. What will be the passing
sight distance?
Given:
Design speed, V= 65 kmph
Difference in speed, m = 20 kmph
Rate of acceleration, a = 3.6 kmph
Problem:
Compute the passing sight distance that a vehicle A could
move while overtaking the slow moving
vehicle B before meeting the on-coming vehicle C.
Speed of car A = 97 kph
Speed of car B = 80 kph
Perception-reaction time = 2.5 sec
Average acceleration = 2.4 m/sec
Time the passing vehicle occupies the left lane = 10.4 sec
Distance between the passing vehicle at the end of its
maneuver and the opposing vehicle= 84 m
