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provides the study materials for most important topics in Civil engineering for
the candidates who are going to appear for the GATE-2016 examination. Here the
“TRANSPORTATION ENGINEERING-SIGHT DISTANCES” topics are explained. And also the
GATE-2016 practice sets, Example problems, previous year Question papers are
available in "jobghost.blogspot.com".
Sight
Distance:
It is the distance
visible to the driver at any instance of driving. And the sight distance is
most important in the road to avoid accidents due to insufficient visibility.
Three sight
distance situations are considered in the geometric design of road.
1. Stopping sight distance
2.
Over taking sight distance
3.
Safe sight distance for uncontrolled
intersection
1.
Stopping sight
distance:
It is the distance
required by the driver to stop the vehicle without colliding with other vehicle
visible at any instance of driving.
The Stopping sight
distance mainly depends on the horizontal alignment of road and vertical
profile of road, height of driver eye above ground level and height of object
visible to driver.
As per IRC the
height of driver’s eye and object above road surface is 1.2m and 0.15m
respectively.
t----> Perception time (2.5 sec)
v----> Speed of vehicle in m/sec
According
to PIEV theory “t” is the Total time required by the driver to take decision of
stopping the vehicle just after seeing the object. It is generally 2.5 seconds
for normal driver. It may vary depending up on the psychological factors of
driver.
2.
Overtaking
sight distance:
In all highway roads there is a fast moving and slow moving vehicle
exists. So the fast moving vehicle has to overtake the slow moving vehicle
without any collision. So the overtaking sight distance is the “distance
required by fast moving vehicle to overtake the slow moving vehicle without
colliding with upcoming vehicle or any other object safely”.
When there is only one way traffic no need to concentrate about
upcoming or opposing vehicle. But in two
way traffic system the design should also include the opposing vehicle
characteristics.
i. For One way traffic:
OSD = vb.t +vbT+2.S
ii. For two way traffic:
OSD = vb.t +vb.T+2S+v.T
a ----> Acceleration of vehicle in m/sec3
vb ----> Slow moving vehicle speed in m/sec
t
----> Reaction time (2 sec)
v ----> Fast moving vehicle
(If S is not given) S= (0.7vb +6)
(if vb is not given) vb=(v-4.5)
The actual
overtaking zone is provided for three times the required overtaking sight
distance.
Overtaking Zone = 3 x OSD
Example
Problems:
1. The design speed of vehicle is 80kmph.
Assume total reaction time=2.5 seconds, co-efficient of friction=0.35. The road
is in the falling gradient of 5%. Calculate the Safe Stopping sign distance?
Ans:
The speed of
vehicle is given in Km/h. To convert it to m/sec as given in formulas chapter.
Speed (v) = 80 x 5/18 = 22.5 m/sec
Time (t) = 2.5 seconds
friction (f) =0.35
Gradient (n)= 6% =0.06
So
N is a falling or
descending gradient so it should be denoted as negative.
-----------------------------------------------------------------------------------------------------
2. Calculate overtaking sight distance for the
following details of the highway road.
Speed of the vehicle= 90 km/hr
Acceleration
(a) = 1.9 km/hr
Assume necessary data
Ans:
i. For two lane one way traffic road:
For one way
traffic the opposing vehicle need not to be considered for the design.
OSD = vb.t +vbT+2.S
Convert
Km/hr units to m/sec
Vehicle Speed(v)=
90x5/18 = 25 m/sec
Acceleration(a)
= 1.9 x 5/18 = 0.5 m/sec
Assume t =2sec
So as
explained in formulas chapter find all
the required values
vb=
(v-4.5) =22.5 m/sec
S = (0.7 vb+6)=20.35
m
Hence
OSD = (22.5 x 2)+(22.5 x 12.4)+(2 x 20.35)
=335 m
ii. Single lane two way traffic road:
In single
lane two way traffic the opposing vehicle speed and distance need to be
calculated.
So the OSD
is
OSD = vb.t +vb.T+2S+v.T
=(22.5 x 2)+(22.5 x 12.4)+(2 x 20.35)+(25 x 12.4)
=648 m
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