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► Given: Magnetic heading 311° Drift angle 10° left Relative bearing of NDB 270°. What is the magnetic bearing of the NDB measured from the aircraft?
A) 211°
B) 208°
C) 221°
D) 180°
Magnetic heading (311) + Relative bearing (270) = Magnetic bearing TO (221°)
► At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are:
A) 080° - 226 kt
B) 090° - 232 kt
C) 085° - 226 kt
D) 088° - 232 kt
Sketch the triangle. radial 310° FROM station means QDM 130°. 130° - 45° = 85°.
Distance (56,56) / Time (15 min) = 226 kts
► Given: An aircraft is flying a track of 255 degrees (M). 2254 UTC, it crosses radial 360° from a VOR station. 2300 UTC, it crosses radial 330° from the same station. At 2300 UTC, the distance between the aircraft and the station is:
A) The same as it was at 2254 UTC
B) greater than it was at 2254 UTC
C) randomly different than it was at 2254 UTC
D) less than it was at 2254 UTC
Sketch the triangle. It is an isosceles triangle meaning that both sides are the same.
► A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the feature is:
A) 160° T
B) 220°
C) 310°
D) 130°
The centreline of a airborne weather radar is our heading (355° M) so 30° to the left means 325° M. The reciprocal is 145° M and when we add variation (15°E) we get 160° T.
► A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°M. Variation 25°W, drift 10° Right and GS 360 kt. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was:
A) 30 NM and 240°
B) 40 NM and 110°
C) 40 NM and 290°
D) 30 NM and 060°
165° MH (subtract 25° VAR) = 140° TH
True heading (140) + Relative bearing (280) = True bearing TO station (60)
The reciprocal of 60° is 240°
Distance can easily be calculated using D = Speed x Time
► A ground feature was observed on a relative bearing of 315° and 3 MIN later on a relative bearing of 270°. The W/V is calm. Aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature?
A) 3 NM
B) 12 NM
C) 9 NM
D) 6 NM
Distance = Speed (180 kt) x Time (3min)
Distance = 9 NM
► A pilot received the following signals from a VOR DME station: radial 180° +/- 1° , distance = 200 NM. What is the approximate error?
A) +/- 3.5 NM
B) +/- 1 NM
C) +/- 2 NM
D) +/- 7 NM
Track error = distance off track / distance along track x 60
TE (1°) = DOT / DAT (200NM) x 60
DOT = 3.3 NM (Closest answer being +/- 3.5)
► An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing from the same position is 270°. Assuming no drift and a GS of 240 kt. What is the approximate range from the NDB at 0840?
A) 50 NM
B) 40 NM
C) 60 NM
D) 30 NM
Distance = Speed (240) x Time (10min)
Distance = 40 NM
► An aircraft at FL120, IAS 200kt, OAT -5°C and wind component +30kt, it is required to reduce speed in order to cross a reporting point 5min later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to:
A) 169 kt
B) 165 kt
C) 159 kt
D) 174 kt
On the navigation computer align 12.000ft with -5°C in the airspeed window. Find IAS 200 on the inner scale and read TAS 242
Now use the wind (+30kt) to convert TAS 242 in ground speed 272kt
Distance (100) / Speed (272) = Time (22min)
Now add 5 minutes (because we need to cross it 5 minutes later) = 27min and work all the way back to IAS:
Distance (100) / Time (27min) = 222kt GS
222 GS - 30 = 192 TAS
On navigation computer (still with 12000 and -5°C aligned) find TAS 192 on outer scale and read IAS 159kt on inner scale.
► Given: Distance A to B = 120 NM. After 30 NM aircraft is 3 NM to the left of course. What heading alteration should be made in order to arrive at point "B"?
A) 8° left
B) 6° right
C) 4° right
D) 8° right
Read these type Questions carefully! They are either asking for the error or the alteration that will allow you to arrive at point "B" , both will be between the possible answers!
TE = DOT / DAT x 60
TE = 3 (off track) / 30 (along track) x 60
TE = 6°
The answer 6° right is between the answers but is not correct! This is the error we've made after 30NM. If we would correct 6° to the right than we would fly parallel to our track but not arrive at B.
To arrive at B we need to double the correction, so do the formula again but this time with the distance along track which we need to go to arrive at B:
TE = 3 (off track) / 90 (120-30) x 60
TE = 2°
Now add both up: 6° right + 2° right = 8° right
(This last formula does not need to be calculated if the Question simply asks what the error was)
► An aircraft is planned to fly from position A to position B, distance 320 NM, at an average GS of 180kt. It departs A at 1200 UTC. After flying 70 NM along track from A, the aircraft is 3MIN ahead of planned time. Using the actual GS experienced, what is the revised ETA at B?
A) 1401 UTC
B) 1333 UTC
C) 1347 UTC
D) 1340 UTC
We can make a quick sketch to visualise it better;
Actual time of departure at A = 1200 UTC
After 70NM at 180kt GS (23min) we should be at at 1223 UTC (ETA)
However, we are 3 minutes ahead so 1220 UTC (ATA)
With this time we can calculate the actual ground speed we experienced:
Distance (70nm) = V (?) x T (20min)
Actual GS = 210kt
We have (320NM - 70NM) 250NM still left to go to B at actual GS 210kt = 1h 11min
1220 + 0111 = 1331 (1333 UTC being the closest)
► Given: Distance A to B is 475 NM, Planned Ground speed 315 kt, ATD 1000 UTC. 1040 UTC - fix obtained 190 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at B?
A) 320 kt
B) 360 kt
C) 300 kt
D) 342 kt
475 NM / 315 kt = 1h30
Start at A was 10:00 so after 1h30 we should arrive at B at 11:30
After 190 NM we are at 10:40 . 50 minutes left to go until B , distance being 285 (475 - 190).
Speed = 285 / 50min = 342 kt
► Given: halfway between two reporting points the navigation log gives the following information: TAS 360 kt, W/V 330°/80kt, Compass heading 237°, deviation on this heading -5°, variation 19°W. What is the average ground speed for this leg?
A) 360 kt
B) 354 kt
C) 373 kt
D) 403 kt
First go from CH 237° to TH via deviation and variation:
True Heading = 213°
We now have the following values:
- True heading 213°
- TAS 360
- W/V 330/80
With these we can enter the wind side of our navigation computer for an accurate ground speed (inside the circle). However, the answers are far enough apart so we can also use the quick formula to save time:
Wind direction (330) - True heading (213) = 117
COS(117) = -0.45 (negative value so tailwind component)
0.45 x wind speed (80) = approx 36 TWC
TAS 360 + 36 TWC gives an approximate Ground speed of 396 kt (closest being 403kt)
► Given:
ETA to cross a meridian: 21:00 UTC
GS: 411 kts
TAS: 491 kts
At 20:10, ATC requests a speed reduction to cross the meridian at 21:05 UTC. The reduction to TAS will be approximately:
A) 60 kts
B) 90 kts
C) 75 kts
D) 40 kts
From 20:10 - 21:00 -> We have 367,5 NM to go (GS 441 x Time 50min)
Now we need to do it in 55min (20:10 - 21:05) so 367,5 / 55min = 401 kt
We can see a reduction in groundspeed of 40 kts (441 to 401). Since there is no information on the wind we can assume TAS being reduced by 40kt aswel.
► An island is observed to be 30 to the right of the nose of the aircraft. The aircraft heading is 290° (M). Variation 10°E. The bearing from the aircraft to the island is:
A) 330° (T)
B) 270° (T)
C) 250° (T)
D) 310° (T)
209°(M) + 30° to the right = 320° (magnetic bearing)
320° +10E = 330° True Bearing
► You are heading 345°(M), the variation is 20°E, and you take a radar bearing of 30° left of the nose from an island. What bearing do you plot?
A) 160°(T)
B) 155°(T)
C) 140°(T)
D) 180°(T)
We want to plot the true bearing FROM the island TO the aircraft.
345°(M) - 30° to the left = 315° magnetic bearing (ACFT TO ISLAND)
Reciprocal is 135° magnetic +20°E = 155° True (ISLAND TO ACFT)
► The true course in the flight log is 270°, the forecast wind is 045°(T)/15kts and the TAS is 120kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM south of the intended track and 2.5 NM ahead of the dead reckoning position. The track angle error (TK) is:
A) 5°L
B) 2°L
C) 3°R
D) 6°R
Find the Groundspeed via the navigation computer (or approximate value via the formula) = 130kt
Distance Along Track (DAT) = 130kt x 15min = 32.5 NM (+2.5 ahead) = 35NM
Distance Off Track (DOT) = 3NM
TE = 3 / 35 x 60
TE = 5°L (Left because we were heading towards the West and are South of that track)
A) 211°
B) 208°
C) 221°
D) 180°
Magnetic heading (311) + Relative bearing (270) = Magnetic bearing TO (221°)
► At 0020 UTC an aircraft is crossing the 310° radial at 40 NM of a VOR/DME station. At 0035 UTC the radial is 040° and DME distance is 40 NM. Magnetic variation is zero. The true track and ground speed are:
A) 080° - 226 kt
B) 090° - 232 kt
C) 085° - 226 kt
D) 088° - 232 kt
Sketch the triangle. radial 310° FROM station means QDM 130°. 130° - 45° = 85°.
Distance (56,56) / Time (15 min) = 226 kts
► Given: An aircraft is flying a track of 255 degrees (M). 2254 UTC, it crosses radial 360° from a VOR station. 2300 UTC, it crosses radial 330° from the same station. At 2300 UTC, the distance between the aircraft and the station is:
A) The same as it was at 2254 UTC
B) greater than it was at 2254 UTC
C) randomly different than it was at 2254 UTC
D) less than it was at 2254 UTC
Sketch the triangle. It is an isosceles triangle meaning that both sides are the same.
► A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the feature is:
A) 160° T
B) 220°
C) 310°
D) 130°
The centreline of a airborne weather radar is our heading (355° M) so 30° to the left means 325° M. The reciprocal is 145° M and when we add variation (15°E) we get 160° T.
► A ground feature was observed on a relative bearing of 325° and five minutes later on a relative bearing of 280°. The aircraft heading was 165°M. Variation 25°W, drift 10° Right and GS 360 kt. When the relative bearing was 280°, the distance and true bearing of the aircraft from the feature was:
A) 30 NM and 240°
B) 40 NM and 110°
C) 40 NM and 290°
D) 30 NM and 060°
165° MH (subtract 25° VAR) = 140° TH
True heading (140) + Relative bearing (280) = True bearing TO station (60)
The reciprocal of 60° is 240°
Distance can easily be calculated using D = Speed x Time
► A ground feature was observed on a relative bearing of 315° and 3 MIN later on a relative bearing of 270°. The W/V is calm. Aircraft GS 180 kt. What is the minimum distance between the aircraft and the ground feature?
A) 3 NM
B) 12 NM
C) 9 NM
D) 6 NM
Distance = Speed (180 kt) x Time (3min)
Distance = 9 NM
► A pilot received the following signals from a VOR DME station: radial 180° +/- 1° , distance = 200 NM. What is the approximate error?
A) +/- 3.5 NM
B) +/- 1 NM
C) +/- 2 NM
D) +/- 7 NM
Track error = distance off track / distance along track x 60
TE (1°) = DOT / DAT (200NM) x 60
DOT = 3.3 NM (Closest answer being +/- 3.5)
► An aircraft obtains a relative bearing of 315° from an NDB at 0830. At 0840 the relative bearing from the same position is 270°. Assuming no drift and a GS of 240 kt. What is the approximate range from the NDB at 0840?
A) 50 NM
B) 40 NM
C) 60 NM
D) 30 NM
Distance = Speed (240) x Time (10min)
Distance = 40 NM
► An aircraft at FL120, IAS 200kt, OAT -5°C and wind component +30kt, it is required to reduce speed in order to cross a reporting point 5min later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to:
A) 169 kt
B) 165 kt
C) 159 kt
D) 174 kt
On the navigation computer align 12.000ft with -5°C in the airspeed window. Find IAS 200 on the inner scale and read TAS 242
Now use the wind (+30kt) to convert TAS 242 in ground speed 272kt
Distance (100) / Speed (272) = Time (22min)
Now add 5 minutes (because we need to cross it 5 minutes later) = 27min and work all the way back to IAS:
Distance (100) / Time (27min) = 222kt GS
222 GS - 30 = 192 TAS
On navigation computer (still with 12000 and -5°C aligned) find TAS 192 on outer scale and read IAS 159kt on inner scale.
► Given: Distance A to B = 120 NM. After 30 NM aircraft is 3 NM to the left of course. What heading alteration should be made in order to arrive at point "B"?
A) 8° left
B) 6° right
C) 4° right
D) 8° right
Read these type Questions carefully! They are either asking for the error or the alteration that will allow you to arrive at point "B" , both will be between the possible answers!
TE = DOT / DAT x 60
TE = 3 (off track) / 30 (along track) x 60
TE = 6°
The answer 6° right is between the answers but is not correct! This is the error we've made after 30NM. If we would correct 6° to the right than we would fly parallel to our track but not arrive at B.
To arrive at B we need to double the correction, so do the formula again but this time with the distance along track which we need to go to arrive at B:
TE = 3 (off track) / 90 (120-30) x 60
TE = 2°
Now add both up: 6° right + 2° right = 8° right
(This last formula does not need to be calculated if the Question simply asks what the error was)
► An aircraft is planned to fly from position A to position B, distance 320 NM, at an average GS of 180kt. It departs A at 1200 UTC. After flying 70 NM along track from A, the aircraft is 3MIN ahead of planned time. Using the actual GS experienced, what is the revised ETA at B?
A) 1401 UTC
B) 1333 UTC
C) 1347 UTC
D) 1340 UTC
We can make a quick sketch to visualise it better;
Actual time of departure at A = 1200 UTC
After 70NM at 180kt GS (23min) we should be at at 1223 UTC (ETA)
However, we are 3 minutes ahead so 1220 UTC (ATA)
With this time we can calculate the actual ground speed we experienced:
Distance (70nm) = V (?) x T (20min)
Actual GS = 210kt
We have (320NM - 70NM) 250NM still left to go to B at actual GS 210kt = 1h 11min
1220 + 0111 = 1331 (1333 UTC being the closest)
► Given: Distance A to B is 475 NM, Planned Ground speed 315 kt, ATD 1000 UTC. 1040 UTC - fix obtained 190 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at B?
A) 320 kt
B) 360 kt
C) 300 kt
D) 342 kt
475 NM / 315 kt = 1h30
Start at A was 10:00 so after 1h30 we should arrive at B at 11:30
After 190 NM we are at 10:40 . 50 minutes left to go until B , distance being 285 (475 - 190).
Speed = 285 / 50min = 342 kt
► Given: halfway between two reporting points the navigation log gives the following information: TAS 360 kt, W/V 330°/80kt, Compass heading 237°, deviation on this heading -5°, variation 19°W. What is the average ground speed for this leg?
A) 360 kt
B) 354 kt
C) 373 kt
D) 403 kt
First go from CH 237° to TH via deviation and variation:
True Heading = 213°
We now have the following values:
- True heading 213°
- TAS 360
- W/V 330/80
With these we can enter the wind side of our navigation computer for an accurate ground speed (inside the circle). However, the answers are far enough apart so we can also use the quick formula to save time:
Wind direction (330) - True heading (213) = 117
COS(117) = -0.45 (negative value so tailwind component)
0.45 x wind speed (80) = approx 36 TWC
TAS 360 + 36 TWC gives an approximate Ground speed of 396 kt (closest being 403kt)
► Given:
ETA to cross a meridian: 21:00 UTC
GS: 411 kts
TAS: 491 kts
At 20:10, ATC requests a speed reduction to cross the meridian at 21:05 UTC. The reduction to TAS will be approximately:
A) 60 kts
B) 90 kts
C) 75 kts
D) 40 kts
From 20:10 - 21:00 -> We have 367,5 NM to go (GS 441 x Time 50min)
Now we need to do it in 55min (20:10 - 21:05) so 367,5 / 55min = 401 kt
We can see a reduction in groundspeed of 40 kts (441 to 401). Since there is no information on the wind we can assume TAS being reduced by 40kt aswel.
► An island is observed to be 30 to the right of the nose of the aircraft. The aircraft heading is 290° (M). Variation 10°E. The bearing from the aircraft to the island is:
A) 330° (T)
B) 270° (T)
C) 250° (T)
D) 310° (T)
209°(M) + 30° to the right = 320° (magnetic bearing)
320° +10E = 330° True Bearing
► You are heading 345°(M), the variation is 20°E, and you take a radar bearing of 30° left of the nose from an island. What bearing do you plot?
A) 160°(T)
B) 155°(T)
C) 140°(T)
D) 180°(T)
We want to plot the true bearing FROM the island TO the aircraft.
345°(M) - 30° to the left = 315° magnetic bearing (ACFT TO ISLAND)
Reciprocal is 135° magnetic +20°E = 155° True (ISLAND TO ACFT)
► The true course in the flight log is 270°, the forecast wind is 045°(T)/15kts and the TAS is 120kts. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM south of the intended track and 2.5 NM ahead of the dead reckoning position. The track angle error (TK) is:
A) 5°L
B) 2°L
C) 3°R
D) 6°R
Find the Groundspeed via the navigation computer (or approximate value via the formula) = 130kt
Distance Along Track (DAT) = 130kt x 15min = 32.5 NM (+2.5 ahead) = 35NM
Distance Off Track (DOT) = 3NM
TE = 3 / 35 x 60
TE = 5°L (Left because we were heading towards the West and are South of that track)
