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Refer to figure.
Complete line 1 of the 'FLIGHT NAVIGATION LOG'; positions 'A' to 'B'.

What is the HDG°(M) and ETA?

  • A
    282° - 11:28 UTC
  • B
    268° - 11:28 UTC
  • C
    268° - 11:14 UTC
  • D
    282° - 11:14 UTC

CAS to TAS with your flight computer:

  1. Align the Pressure Altitude of 18 000 ft inside the "AIRSPEED" window with the OAT of -20°C on the outer edge of the "AIR SPEED" window by rotating the inner disc of the flight computer.
  2. Then, find the Calibrated Airspeed (CAS) of 210 kt on the inner scale. Finally, read the True Airspeed (TAS) on the outer scale opposite the CAS of 210 kt on the inner scale. In this case, the True Airspeed (TAS) is 275 kt.

Determining Heading and G/S with the flight computer:

  1. Move the slide until the center dot falls on the arc marked i.e. 275 (you can choose any arc you like as this will not affect the final result, we choose the TAS).
  2. Next, rotate the window disc to align the True Wind Direction (050°) with the "TRUE INDEX" on the fixed scale.
  3. Draw a downward line starting from the ring along the centerline, with a length equal to the Wind Velocity (40 kt).
  4. Then, rotate the window disc to align the True Track (270°) with the "TRUE INDEX" on the fixed scale. The end of the line should be approximately 5° to the Left of the centerline, which corresponds to a 5° Left (Port) Drift Angle. Therefore, WCA is 5ºR.
  5. To determine the Ground Speed, find the speed arc that passes under the end of the line drawn. In this case, the Ground Speed is 305 kt.

Wind Correction Angle is the angular value opposite to the drift angle, and it refers to the heading angle correction in order to keep your aircraft in the desired track.

In this case, the WCA is 5º R (drift is 5ºL), the true heading = 270º + 5º = 275º (T). Let's determine the Magnetic Heading:

True heading
(T)

Variation
(V)

Magnetic heading (M)

Deviation
(D)

Compass heading
(C)

275º

7ºE

268º

Variation West, Magnetic Best. Variation East, Magnetic Least.

Assuming a Ground Speed of 305 kt, which is equivalent to 305/60 NM/min, the time required to travel a distance of 300 NM can be calculated using the formula:
Time = Distance/Ground Speed

In this case, the time would be (300 x 60) / 305 min = 59 min

Given that the Time of Departure is 10:15 UTC and the estimated flight time is 59 min, the Estimated Time of Arrival (ETA) can be calculated as 10:15 UTC + 00:59 = 11:14 UTC.

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