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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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