Civil Engineering - Highway Engineering - Discussion

Discussion Forum : Highway Engineering - Section 2 (Q.No. 23)
23.
The absolute minimum sight distance required for stopping a vehicle moving with a speed of 80 km ph, is
120 m
200 m
640 m
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Dinesh said:   1 decade ago
SSD = lag distance+braking distance.

= v.t + v^2/2gf.

= 22.22*2.5 + (22.2)^2/(2*9.81*0.35).

= 126.5 (near to 120) So answer is (a)120m.

Where v is in m/s and t = reaction time(2.5sec acc. to IRC).

f is longitudinal coefficient of friction.

Sarang mote said:   9 years ago
What is the lag distance?

Jitendra singh said:   9 years ago
Lag distance = vt.
V = m/ s.
T = sec.

Salman mohammed said:   9 years ago
SSD = 0.27vt + v.v/254(f +-g/100).

Where t = 2.5.
f = b/n 0.43 and 0.4 interpolate.

So to answer the question in meter we must use this equation.

Navi said:   9 years ago
Actually its from the direct values given in the books (for 20kmph = 20).
So as (25=25) (30=30) (40=45) (50=60) (60=80) (80=120) & (100=180).
(1)

Ankur said:   9 years ago
How to predict the value of f in such cases?

Asita said:   9 years ago
Yes, right @Navi.
(1)

ANKUR said:   8 years ago
We can predict because we have to calculate min SSD if take f= .15 we get more value but if we take f= .35 we get min value so we take f=.35.
(1)

Pakkirappa said:   8 years ago
V is given in kmph convert to m
I.e 1000/60*60=0.278
Ssd=vt+{v2/2gf}
Where t is time required for understanding the situation this is a constant value.
Ssd=0.27*80*2.5+{(0.27*80)^2/(2*9.81*0.35)}
Ssd=121.6m.
(4)

Shubham said:   7 years ago
Right @Navi.
(2)


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