Civil Engineering - Airport Engineering - Discussion
Discussion Forum : Airport Engineering - Section 1 (Q.No. 15)
15.
The reduced level of the proposed site of an air port is 2500 m above M.S.L. If the recommended length by I.C.A.O. for the runway at sea level is 2500 m, the required length of the runway is
Discussion:
10 comments Page 1 of 1.
Kishore kumble said:
1 year ago
7% ÷ 300 × 2500 = 58.3
2500 × 58.3 = 145.75.
2500 + 14575 = 3957,
Ans = 3957.
2500 × 58.3 = 145.75.
2500 + 14575 = 3957,
Ans = 3957.
Bishnu pokhrel said:
5 years ago
2500+2500*49÷100=3725.
(2500÷300)for 300 of MSL 7% is added (7*7=49).
(2500÷300)for 300 of MSL 7% is added (7*7=49).
(2)
Aditya said:
6 years ago
How 2500?
It should be 2100.
It should be 2100.
(1)
Amber said:
6 years ago
(7/300) x 2500 = 58.33 above m.s.l.
Basic runway length = 1.5833 x 2500.
= 3958.25.
= 3725 m (app).
Basic runway length = 1.5833 x 2500.
= 3958.25.
= 3725 m (app).
(6)
James said:
7 years ago
The Max length increase possible is 35%.
So, 2500 x 0.35 + 2500 = 3375.
Nearest safe length = 3725.
So, 2500 x 0.35 + 2500 = 3375.
Nearest safe length = 3725.
(3)
Chhaya said:
8 years ago
@Anfield.
300 m Elevation 7% Length is increased.
300 m Elevation 7% Length is increased.
(2)
Anfield said:
8 years ago
7%is correct but per 300m is it?
Please explain.
Please explain.
(1)
Suyagn vaghela said:
9 years ago
Basic length is to be increased at rate 7% per 300m, Above MSL;
So, correction is (7/100) X (2500/300) X 2500 = 1458 m.
Then, total length is 3958m.
So, correction is (7/100) X (2500/300) X 2500 = 1458 m.
Then, total length is 3958m.
Mandy said:
9 years ago
2500+2500(0.07/300) = 3958 (put the nearest one).
(2)
Aisha said:
10 years ago
How?
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