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where Vi is the touchdown velocity, approximately equal to VA. Then,
s3 d V2
ds:--,f,: __ (2.357)
S3 = 2V: (F~F ) V-:
where Fi - TR + D and Fo = TR +/r,W.
(2.358)
Example 2.11
A certain jet aircraft has the following data: W - 50,000 N, T - 14,500 N,
CD = 0.02+0.04C~., CL,max - 1.2, and S - 30 m2.Assuming an obstacle height
of 15 m and Lt, - 0.05, calculate the total takeoff distance and time at sea level.
CZ = 2Atk
0.05
= ~2) .00 )
- 0.625
VStHII -
- 47.6290 m/s
Vl : 1.2 Vsran
- 57.143 m/s
AIRCRAFT PERFORMANCE
Fo - T - /r,W
= 14,500 - 0.05(50,000)
- 12,000 N
Fi -T-D
= 14,500 _ 0.5(1.225)(57.1432)(30)[0.02 + 0.04(0.6252)]
- 12,362.489 N
For the ground run,
W a,Fo
s, = 2~ (F~F ) a.F,
50,000(57.132) a,. 12,000
= 2~9 8 )(12 000 -12,362 489~ a,'12,362.489
-. 684.1065 m
Let us calculate the time for ground run* We have
a - Fo
- 12,000
N
b = Fj_F
12,362,489 - 12,000
:-
5-7. 432
- O.lll
Because both a and b are positive,
t, = -,/:,t -, ~/-.,,]
\~ = 36.49
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