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
Discussion:
15 comments Page 1 of 2.
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.
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)
Mayu said:
5 years ago
Longitudinal friction --------- 0.35 - 0.4.
Lateral friction --------------- 0.15.
Lateral friction --------------- 0.15.
(2)
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).
So as (25=25) (30=30) (40=45) (50=60) (60=80) (80=120) & (100=180).
(1)
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)
Dinesh said:
6 years ago
0.28 * V.t + 0.01V^2/brake efficiency.
V = 80
t = 2.5
And We get exact ans 120.
V = 80
t = 2.5
And We get exact ans 120.
(1)
Lucky Kumar said:
6 years ago
SSD = 0.278Vt+V^2/254(f+S%).
= 0.278*80*2.5+80*80/254(.35).
= 127.6 mtr.
Cofficent of friction (f) = 0.35.
= 0.278*80*2.5+80*80/254(.35).
= 127.6 mtr.
Cofficent of friction (f) = 0.35.
(1)
Ajay Sharma said:
5 years ago
This value is given directly in form of table.
For 80kmph = 120m.
For 80kmph = 120m.
(1)
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.
= 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.
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