This document discusses stopping sight distance (SSD) in highway engineering. It defines SSD as the minimum sight distance available to stop a vehicle safely at a certain design speed without collision. It lists factors that affect SSD like vehicle speed, braking efficiency, driver reaction time, road friction, and gradient. It presents the formulas used to calculate SSD for level roads, ascending gradients, and descending gradients. It provides examples of SSD calculations for different traffic and road conditions.

1. B.E. CIVIL ENGINEERING.
3rdYEAR
5th SEMESTER
HIGHWAY ENGINEERING(2150601)
SHAIKH MOHD.AMIR N. 150860106062
PRAJAPATI RIYA P. 150860106054
BHANDARI NEENAD H. 150860106006
RATHOD NEHA M. 150860106055

2. 1. Definition
2. Factors affecting stopping sight distance
3. Analysis of stopping sight distance
4. PIEV theory
5. Braking distance
6. Formulas for stopping sight distance
7. Examples

3.  The minimum sight distance available on a
highway to stop a vehicle at design speed,
safely without any collision with any other
obstruction is called stopping sight distance.
 For the purpose of measuring stopping sight
distance, IRC has suggested the height of eye
level of driver as 1.2 m and the height of the
object as 0.15m above the road surface.

4.  Factors affecting SSD are:-
1. Speed of vehicles
2. Efficiency of brakes
3. Total reaction time of drivers
4. Frictional resistance between the road and
the tyres
5. Gradient of the road

5.  The stopping sight distance of the vehicle is the
sum of :
1. Lag distance: The distance travelled by the
vehicle during the total reaction time is known
as lag distance.
lag distance = speed of the vehicle (m/sec)
* total reaction time (secs)
2. Reaction time:The time taken from the instant
the object is visible to the driver to the instant
the brakes are efficiently applied.

6.  According to PIEV theory the total
reaction of the driver is divided into
four parts:
1. Perception time:The time required
to transmits the sensations
received by the eyes and ears to
the brain with help of nervous
system.
2. Intellection time:The time
required for understanding the
situation
3. Emotion time:The time elapsed
during emotional sensations and
disturbance such as fear, anger,
etc. with reference to the situation.
4. Volition time:The time taken for
the final action.

7.  Braking distance = v2 / 2gf
where,
v = speed of the vehicle in m/sec
g = acceleration due to gravity = 9.8 m/ sec2
f = coefficient of friction (0.4 to 0.35)

8. 1. When road is without gradient
SSD = v * t + v2 / 2gf
2. When road is with ascending gradient
SSD = v * t + v2 / 2g(f+(n/100))
3.When road is with descending gradient
SSD = v * t + v2 / 2g(f-(n/100))
In case of single lane road with two way traffic,
the value of SSD is doubled.

9. Q: Calculate the safe SSD for design speed of 50kmph.
a.Two way traffic on a two lane road
b.Two way traffic on a single lane road
Ans: (a)Two way traffic on a two lane road:
v= 50kmph = 50/3.6 = 13.9 m/ sec2
t = 2.5 secs
f = 0.37
g = 9.8 m/ sec
:- SSD = v * t + v2 / 2gf
= 13.9 * 2.5 + 13.92 / 2*9.8*0.37
= 61.39 m
(b) Two way traffic on a single lane road:
:- SSD = 2* 61.39
= 122.78 m

10. Q: Two cars are approaching from the opposite directions at 90 & 60 kmph. Calculate the
minimum sight distance required to avoid head on collision.
Assume a reaction time of 2.5 secs, coefficient of friction of 0.7 and brake efficiency of
50%.
Ans: SSD = v * t + v2 / 2gf
v1 = 90 kmph = 90/3.6 = 25 m/sec
v2 = 60 kmph = 60/3.6 = 16.67 m/sec
Brake efficiency is 50%, therefore, f = 0.5 *0.7 = 0.35
:- for first car
SSD1 = v * t + v2 / 2gf
= 25 * 2.5 + 252 / 2 * 9.8 * 0.35
= 153.6 m
:- for second car
SSD2 = v * t + v2 / 2gf
= 16.67 * 2.5 + 16.672 / 2 * 9.8 * 0.35
= 82.20 m
SSD = SSD1 + SSD2
= 153.6 + 82.2
= 235.8 m

11. Q: Calculate the SSD on a road at a descending
gradient of 3% for a design speed of 80 kmph.
Take reaction time for driver as 2.0 secs and
f=0.30.
Ans: v = 80 kmph = 80/3.6 = 22.22 m/sec
t = 2.0 sec
f = 0.30
n = 3% (descending gradient)
:- SSD = v * t + v2 / 2g(f-(n/100))
= 22.22 * 2 + 22.22 / 2* 9.8 (0.3-(3/100))
= 44.44 + 93.29
= 137.73 m

12. THANKYOU…