13 / 20

Using Mental Dead Reckoning (MDR), calculate the ground speed for the following situation:

Track: 337°(T)
TAS: 465 kt
Wind: 220°(T)/50 kt

  • A
    490 kt
  • B
    515 kt
  • C
    415 kt
  • D
    440 kt

Refer to figures.
This question can be solved either using an MDR technique or your Flight/Nav computer:

1. Mental Dead Reckoning (MDR) headwind/tailwind (HWC/TWC) component

The HWC or TWC can be estimated using the values from the following table:

90º - Wind angle 10º 20º 30º 40º 50º 60º
% of wind speed 0.2 0.3 0.5 0.6 0.8 0.9

Wind Angle is the difference between the wind direction and the track: 337º - 220º = 117º. When the wind angle is greater than 090º, then subtract it from 180º: 180º - 117º = 063º.

The wind comes 063º from the left of the track, but from behind. Therefore, the aircraft is experiencing a Tailwind Component (TCW).

  • Wing Angle (WA): 063º
  • 090º - 063 = 027º
  • Wind Speed: 50 kt

TWC = (0.5) x 50 = 25 kt

So, the Groundspeed is: GS = TAS + TWC = 465 kt + 25 kt = 490 kt.

2. Flight/Nav computer

  • Place the centre dot over TAS 465 kt.
  • Rotate the disk until wind direction 220o is under the True index.
  • Mark wind velocity 50 kt below the centre dot.
  • Rotate the disc until the track 337o is under the True index.
  • The mark shows a Groundspeed of 490 kt.

Your Notes (not visible to others)

This question has appeared on the real examination, you can find the related countries below.