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An aircraft at FL370, M0.86, OAT -44°C, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned.

If the speed reduction were to be made 420 NM from the reporting point, what Mach Number is required?

  • A
    M0.73
  • B
    M0.75
  • C
    M0.81
  • D
    M0.79

First Step: We must calculate our groundspeed to find our flight time to cross reporting point. First a conversion from a mach no. to TAS must be carried.

M = TAS / LSS
LSS = 38.95 √K
LSS = 38.95 √273 - 44 = 589.42
TAS = M x LSS = 0.86 x 589.42 = 507 kts
GS = TAS – Headwind = 507 – 110 kt = 397 kt

Second Step: Calculate flight time
420 NM / 397 kt = 1 h 04 min to cross the reporting point

=> The A/C is requested to cross the reporting point 5’ later than planned, which makes a FT of 1h 9 min.

Third Step: Calculate the TAS in order to cross 5 min later.

420 NM / 1h 9 min = 365 kt (GS)
TAS = GS + Headwind
TAS = 365 + 110 = 475 kt

Forth Step: Calculate Mach no for TAS 475

M = 475 / 589.42 = 0.81

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