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Q 11389

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  • Q 11389

    Given: Distance X to Y 2700 NM Mach Number 0.75 Temperature -45?C Mean wind component ''on'' 10 kt tailwind Mean wind component ''back'' 35 kt tailwind The distance from X to the point of equal time (PET) between X and Y is:

    TAS is 0.75*39*squareroot228 = 442 kt (whiz wheel gives round about 440 kt too)
    Therefore the PET = 2700*477/(477+435) = 1417 NM

    The answer given is 1386 NM

  • #2
    Re: Q 11389

    Don't worry, it's 10 kt TAILWIND and I used 10 kt HEADWIND

    Comment


    • #3
      Given: Distance X to Y 2700 NM Mach Number 0.75 Temperature -45?C Mean wind component ''on'' 10 kt tailwind Mean wind component ''back'' 35 kt tailwind The distance from X to the point of equal time (PET) between X and Y is:

      TAS = Mach no. * 38.95*squareroot(temperature in kelvin)

      TAS = 0.75 x 38.95 x √228 = 441kts

      GSO= (441+10) kts GSR= (441+35) kts

      PET= Distance x GSR/(GSO+GSR)===

      Therefore the PET = 2700*476/(476+431) = 1386 NM

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