05 In-Flight Navigation

Question 1

Refer to IFR Chart E(LO) 1 and the winds aloft data below: An aircraft passes overhead the DUB VOR/DME at FL360 and descends to FL120 along Airway B39 at a constant CAS of 210 kt. What is (a) the aircraft magnetic heading and (b) groundspeed in the descent?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL80 135°/20 kt –10°C
FL40 120°/15 kt +00°C
SFC 100°/10 kt +10°C

Answer 1

137°M 275 kt

Question 2

An aircraft is on final approach to land at an aerodrome, on a 3° glidepath, at a range of 8.5 nm from touchdown. What is its height above ground level?

Answer 2

2584 ft
……………………………………
GP° = (Height x 60) / Distance

Question 3

An aircraft is climbing from FL40 to FL220. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 3

FL160 270°/25 kt –10°C

Question 4

An island is observed to be 15° to the left. The aircraft heading is 120°(M), variation 17°(W). The bearing °(T) from the aircraft to the island is:

Answer 4

088
…………………………………..
Heading: 120ºM
Variation: 17ºW
Heading: 103ºT
Bearing: 15º Left
Bearing: 088ºT from the aircraft to the island.

Question 5

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 050. At what distance from the airport will this be achieved?

Answer 5

3.6 NM

Question 6

Given: ETA to cross a meridian is 2100 UTC GS is 441 kt TAS is 491 kt At 2010 UTC, ATC requests a speed reduction to cross the meridian at 2105 UTC. The reduction to TAS will be approximately:

Answer 6

40 kt
…………………………………….
(At 2110; 50 minutes from meridian at 441 kts = 367½ nm to run.
Now need to cover 367½ nm in 55 minutes = 401 kts; reduction 40 kts.)

Question 7

Refer to IFR Chart E(LO)1 and the winds aloft data below: An aircraft passes overhead the CON VOR/DME at FL40 and climbs to FL250 along Airway B1 via RANAR at a constant CAS of 176 kt. What are the aircraft magnetic heading and groundspeed in the climb?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 7

121°M 213 kt

Question 8

An island appears 30° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 355° with the magnetic variation (VAR) 15°E?

Answer 8

220°
……………………………
Heading 355ºM, variation 15ºE, heading = 010ºT.
Heading 010ºT, bearing 30º right, bearing 040ºT to the island.
Bearing from the island to the aircraft = 040º + 180º =220ºT.

Question 9

An aircraft is making a descending turn onto final approach to land at an aerodrome, on a 3° glidepath, at a range of 7.6 nm from touchdown. By what height above ground level must the turn be completed?

Answer 9

2310 ft
……………………………..
GP° = (Height x 60) / Distance

Question 10

Complete LINE 2 of the ‘FLIGHT NAVIGATION LOG’, positions ‘C’ to ‘D’. What is the HDG°(M) and ETA?

Answer 10

HDG 193° - ETA 1239 UTC

Question 11

An aircraft is making a descending turn onto final approach to land at an aerodrome, on an 8% recommended gradient, at a range of 6.6 NM from the threshold.
By what height above ground level must the turn be completed?

Answer 11

3290 ft
……………………..

GR% = (Height x 100) / Distance

Question 12

An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position the TH should be:

Answer 12

112º

Question 13

Refer to the winds aloft data below: An aircraft descends at a rate of 1400 fpm from FL 190 to FL50 along a true track of 150°T and at a constant CAS of 240 kt. What is the approximate distance covered during the descent?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 13

43 nm

Question 14

Use Europe Low Altitude En-route Chart E(LO) 1A.

Two consecutive waypoints of a flight plan are Stornoway VORDME (N58°12.4’, W006°11.0’) and Glasgow VORDME (N55°52.2’, W 004°26.7’).

During the flight the Actual Time Over Stornoway is 11:15 UTC and the Estimated Time Over Glasgow is 11:38 UTC. At 11:21 UTC the fix of the aircraft is exactly over reporting point RONAR.

What is the Revised UTC over Glasgow, based on this last fix?

Answer 14

11:36
……………………………..
Current Speed @ RONAR = 44NM/6minutes = 440kts
Time to go from RONAR to Glasgow VOR: 107NM / 440kts = 14.59minutes = 15minutes
11:21 + 15mins = 11:36

Question 15

During approach the following data are obtained: DME 12.0 NM, altitude 3000 ft DME 9.8 NM, altitude 2400 ft TAS = 160 kt, GS = 125 kt The rate of descent is:

Answer 15

570 ft/min
………………………….
Rate of Descent = Change of Height (feet) / Time (minutes)

Question 16

An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL120. If the mean GS during the descent is 396 kt, the minimum rate of descent required is approximately:

Answer 16

1650 FT/MIN
……………………………………
100 nm at 396 kts = 15.15 minutes.
Aircraft descends through 25,000ft (FL370 – FL120)
25,000 feet in 15.15 minutes = 1,650 feet per minute.

Question 17

Refer to IFR Chart E(LO) 1 and the winds aloft data below: An aircraft passes overhead the STU VOR/DME at FL120 and climbs to FL300 along Airway G1 via SLANY at a constant CAS of 182 kt. What is (a) the aircraft magnetic heading and (b) groundspeed in the climb?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 17

286°M 270 kt

Question 18

Complete LINE 4 of the ‘FLIGHT NAVIGATION LOG’, positions ‘G’ to ‘H’. What is the HDG°(M) and ETA?

Answer 18

HDG 344° - ETA 1336 UTC

Question 19

An aircraft at FL350 is required to cross a VOR/DME facility at FL110 and to commence descent when 100 NM from the facility. If the mean GS for the descent is 335 kt, the minimum rate of descent required is:

Answer 19

1340 FT/MIN
……………………………
100 nm at 335 kts = 17.9 minutes.
Aircraft descends through 24,000ft (FL350 – FL110)
24,000 feet in 17.9 minutes = 1,340 feet per minute.

Question 20

The distance between two waypoints is 200 NM, To calculate compass heading, the pilot used 2°E magnetic variation instead of 2°W. Assuming that the forecast W/V applied, what will the off track distance be at the second waypoint?

Answer 20

14 NM
……………………….
On CRP5:
Total error (4º) on outer scale next to 60 on inner scale.
Go to distance along track (200 nm) 0n inner scale and
Read distance off track (13.3 nm) on outer scale.

Question 21

An aircraft is making a descending turn onto final approach to land at an aerodrome, on a 8% recommended gradient, at a range of 6.6 nm from touchdown. By what height above ground level must the turn be completed?

Answer 21

3210 ft
………………………
GR% = (Height x 100) / Distance

Question 22

When transferring range position lines it is correct to:

Answer 22

transfer the origin and plot the range position lines from the transferred origin.

Question 23

An aircraft is at position (53ºN, 006ºW) and has a landmark at position (52º47’N, 004º45’W), with a relative bearing of 060º.

Given:
Compass Heading = 051º
Variation = 16ºW
Deviation = 2ºE

What is the true bearing of the position line to be plotted from the landmark to the aircraft on a Lambert chart with standard parallels at 37ºN and 65ºN?

Answer 23

278º

Question 24

True Heading of an aircraft is 265° and TAS is 290 kt. If W/V is 210°/35kt, what is True Track and GS?

Answer 24

271° and 272 kt.
…………………………..
Nav Computer: Centre dot on TAS (290 kts), rotate to put wind direction (210°T) on HEADING index and mark wind on centre line 35kts below centre dot (over 255 kts). Rotate to put heading 265ºT next to HEADING index; read drift and groundspeed under wind mark: 6° Stbd (right) (track 271°T) and 272 kts.

Question 25

An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 020° with the magnetic variation (VAR) 25°W?

Answer 25

145°
.........................
Heading:      020ºM
Variation:      25ºW
Heading:      355ºT
Bearing:       30ºLeft
Bearing:       325ºT from aircraft to island
                    180º
Bearing:       145ºT from island to aircraft

Question 26

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?

Answer 26

40 NM

Question 27

An aircraft is on final approach to land at an aerodrome, on a 3° glidepath, at a range of 6.3 NM from touchdown. What is its height above ground level?

Answer 27

1915 ft
…………………….
GP° = (Height x 60) / Distance

Question 28

Which formula can be used to calculate the rate of climb/descent?

Answer 28

Rate of climb/descent (feet/min) = Gradient (%) x GS (kt)

Question 29

TAS is120 kt. ATA ‘X’ 1232 UTC, ETA ‘Y’ 1247 UTC, ATA ‘Y’ is 1250 UTC. What is ETA ‘Z’?

Answer 29

1302 UTC
…………………….
X to Y = 30nm in 18 minutes (1232 – 1250) = ground speed 100kts

Y to Z = 20nm at 100 kts = 12 minutes 1250 + 12 minutes = 1302.

Question 30

An aircraft is flying from A to B and is 2 NM left of track at a distance of 30 NM from A.

Distance A to B: 60 NM
Track A to B: 090° (M)
Heading maintained since overhead A: 084° (M)
What is the required heading (°M) to fly directly to B?

Answer 30

092°

Question 31

An aircraft is on final approach to land at an aerodrome, on a 5% recommended gradient, at a range of 5.8 nm from touchdown. What is the aircraft’s height above ground level?

Answer 31

1763 ft
……………………..
GR% = (Height x 100) / Distance

Question 32

At 0422 an aircraft at FL370, GS 320kt, is on the direct track to VOR ‘X’ 185 NM distant.

The aircraft is required to cross VOR ‘X’ at FL80. For a mean rate of descent of 1800 FT/MIN at a mean GS of 232 kt, the latest time at which to commence descent is:

Answer 32

0445
………………………………
Descent from FL370 to FL80 is through 29,000ft and at 1,800ft/min this will take 16.1 minutes.
In 16.1 minutes the aircraft will cover 62.3nm at its ground speed in the descent of 232kts.
At 0422 distance to run to top of descent is 184 – 62.3 = 122.7nm and at cruise ground speed of 320kts this will take 23 minutes.
Commence descent at 0422 + 23min = 0445.

Question 33

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?

Answer 33

9 NM

Question 34

An aircraft is climbing from 2,000’ amsl to FL260. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 34

FL180 290°/15 kt –15°C

Question 35

The flight log gives the following data: “True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compass heading” The right solution, in the same order, is:

Answer 35

Tr ºT Drift Hdg ºT Var Hdg ºM Dev. Hdg ºC

119º 3º L 122º 2ºE 120º +4º(E) 116º

Question 36

An aircraft at FL370 is required to commence descent at 120 NM from a VOR and to cross the facility at FL130. If the mean GS for the descent is 288 kt, the minimum rate of descent required is:

Answer 36

960 FT/MIN
…………………………………
120 nm at 288 kts = 25 minutes. Aircraft descends through 24,000ft (FL370 – FL130)
24,000 feet in 25 minutes = 960 feet per minute.

Question 37

Refer to annex. 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. To reach destination B from this position, the TH should be:

Answer 37

078º

Question 38

An aircraft flies from waypoint 7 (63°00’ N, 073°00’W) to waypoint 8 (62°00’ N, 073°00’ W). The aircraft position is (62°00’ N, 073°10’W). The cross track distance in relation to the planned track is:

Answer 38

4,7 NM R

Question 39

The distance between A and B is 90 NM. At a distance of 15 NM from A the aircraft is 4 NM right of course. To reach destination B, the correction angle on the heading should be:

Answer 39

19º

Question 40

An aircraft at FL120, IAS 200kt, OAT -5° and wind component +30kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to:

Answer 40

159 kt
……………………
To answer this question, firstly work out how long it will take to cover the 100nm at the current speed.
From the data given, the current TAS is 242kt, which makes the current GS 272kt. Thus, to cover 100nm at 272kt will take 22mins.
So, to arrive there 5 minutes later, you need to be taking 27mins to cover 100nm.
Working backwards from this makes the revised GS 222kt, so the revised TAS would need to be 222-30=192kt.
Going back to your nav computer & with the same data in the airspeed window, locate 192kt on the outer scale and read off the inner scale the revised airspeed required - i.e. 159kt.

Question 41

Given: ILS GP angle = 3.5 DEG, GS = 150 kt. What is the approximate rate of descent?

Answer 41

900 FT/MIN

…………………………………….

Question 42

Given: True Heading = 090° TAS = 180 kt GS = 180 kt *****Drift 5° right Calculate the W/V?

Answer 42

005° / 15 kt

Question 43

An aircraft at FL390 is required to descend to cross a DME facility at FL70. Maximum rate of descent is 2500 FT/MIN, mean GS during descent is 248 kt. What is the minimum range from the DME at which descent should commence?

Answer 43

53 NM
…………………………
From FL390 to FL70 = 32,000 ft; 32,000ft at 2,500ft/min = 12.8 minutes.
12.8 minutes at 248kts = 53nm.

Question 44

An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM from the next waypoint. The rate of descent is 2000 ft/min. The average GS is 420 kt.
The minimum distance from the next waypoint at which descent should start is:

Answer 44

124 NM
…………………………………………………..
FL360 to FL120 = 24,000’
Rate of Descent = 2,000’ fpm
Time to descend = 24,000’ / 2,000’ = 12 minutes (0.2 hours)
Ground Speed = 420 kts
Distance from Top of Descent to Bottom of Descent = 420 x 0.2 = 84 NM
Distance from Bottom of Descent to WP = 40 NM

Distance from Top of Descent to Waypoint = 84 NM + 40 NM = 124 NM

Question 45

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:

Answer 45

160°
……………………………………
Heading 355ºM Variation 15ºE = Heading 010ºT Heading 010ºT
Bearing 30º left = 340ºT from aircraft to feature
340ºT - 180º = 160ºT from feature to aircraft

Question 46

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?

Answer 46

M0.81

Question 47

Intended true course: 120°
TAS: 100 kts
W/V: 360/50
Position lines (LOP) are obtained at 1400hrs, 1403hrs and 1406 hrs.
When transferring the position lines (LOP):

Answer 47

1 LOP is transferred by 11.6 NM, 1 LOP by 5.8 NM, the third one is not transferred.

Question 48

An aircraft at FL330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?

Answer 48

1950 FT/MIN
……………………………………….
65 nm at 330 kts = 11.8 minutes.
Aircraft descends through 23,000ft (FL330 – FL100)
23,000 feet in 11.8 minutes = 1,950 feet per minute.

Question 49

The distance between positions A and B is 180 NM. An aircraft departs position A and after having travelled 60 NM, its position is pinpointed 4 NM left of the intended track. Assuming no change in wind velocity, what alteration of heading must be made in order to arrive at position B?

Answer 49

6° Right

Question 50

Given: A descending aircraft flies in a straight line to a DME. DME 55.0 NM, altitude 33000 ft DME 43.9 NM, altitude 30500 ft M = 0.72, GS = 525 kt, OAT = ISA The descent gradient is:

Answer 50

3.7%

Question 51

Given: Distance ‘A’ to ‘B’ is 90 NM, Fix obtained 60 NM along and 4 NM to the right of course. What heading alteration must be made to reach ‘B’?

Answer 51

12° Left

Question 52

Whilst flying a visual navigation exercise in controlled airspace, it is confirmed that the aircraft is exactly on track. ATC instructions are to turn left 30° to avoid conflicting traffic. After two minutes they advise ‘You are now two miles left of your original track, turn right to regain track in 30 miles.’

What is the heading change required to comply with the instructions from ATC?

Answer 52

034°

Question 53

Complete LINE 1 of the ‘FLIGHT NAVIGATION LOG’; positions ‘A’ to ‘B’. What is the HDG°(M) and ETA?

Answer 53

268° - 1114 UTC

Question 54

Refer to IFR Chart E(LO) 1 and the winds aloft data below: An aircraft passes overhead the SHA VOR/DME at FL60 and climbs to FL150 along Airway W13 via KORAK at a constant CAS of 210 kt. What are the aircraft magnetic heading and groundspeed in the climb?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 54

057°M 255 kt

………………………….

Question 55

An aircraft is making a descending turn onto final approach to land at an aerodrome, on an 4.2° glidepath, at a range of 8.2 NM from the threshold. By what height above ground level must the turn be completed?

Answer 55

3560 ft
…………………………….
GP° = (Height x 60) / Distance

Question 56

An aircraft at FL350 is required to commence descent when 85 NM from a VOR and to cross the VOR at FL80. The mean GS for the descent is 340 kt. What is the minimum rate of descent required?

Answer 56

1800 FT/MIN
………………………………..
85 nm at 340 kts = 15.0 minutes.
Aircraft descends through 27,000ft (FL350 – FL80)
27,000 feet in 15.0 minutes = 1,800 feet per minute.

Question 57

An aircraft is on final approach to land at an aerodrome, on a 3° glidepath, at a range of 8.5 nm from the threshold. What is its height above ground level?

Answer 57

2634 ft
…………………………….
You must be careful to establish whether the question states distance from touchdown or distance from the threshold. In the case of the latter, you must add the threshold crossing elevation onto the calculated solution.The formula to apply is:
GP° = (Height x 60) / Distance

Question 58

Given: Distance ‘A’ to ‘B’ is 100 NM, Fix obtained 40 NM along and 6 NM to the left of course. What heading alteration must be made to reach ‘B’?

Answer 58

15° Right

Question 59

An island appears 60° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 276° with the magnetic variation 10ºE.

Answer 59

046°
............................
Heading: 276ºM
Variation: 10ºE
Heading: 286ºT
Bearing: 60ºLeft
Bearing: 226ºT from aircraft to island 180º
Bearing: 046º from island to aircraft

Question 60

Given: Half way 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?

Answer 60

403 kt
……………………………..
C D M V T
237º -5º(W) 232º 19ºW 213º
On CRP-5: Place the centre dot on TAS (360 kts)
Rotate to put wind direction (330ºT) on HEADING index
Mark wind velocity 80 kts below centre dot on centre line (at 280 kts)
Rotate to put heading 213ºT under HEADING index
Read groundspeed under wind mark: 403 kts

Question 61

An aircraft at FL310, M0.83, temperature -30°C, is required to reduce speed in order to cross a reporting point five minutes later than planned. Assuming that a zero wind component remains unchanged, when 360 NM from the reporting point Mach Number should be reduced to:

Answer 61

M0.74

Question 62

Which formula can be used to calculate the rate of climb/descent?

Answer 62

Rate of climb/descent (ft/min) = (Groundspeed (kts) x Gradient (ft/NM)) ÷ 60

Question 63

On a True Heading of 090° the aircraft experiences drift of 5°S. On a True Heading of 180° the aircraft experiences no drift. On both headings the TAS is 200 kt and it is assumed that the wind is the same. What is the experienced wind speed and direction?

Answer 63

360° / 17 kt
………………………..
Use the CRP5 to find the W/V (wind down method). ……….

Question 64

Refer to the winds aloft data below: An aircraft climbs at a rate of 1000 fpm from an aerodrome at sea level (ISA conditions apply) to FL90 along a true track of 135°T and at a constant CAS of 140 kt. What is the approximate distance to the top of climb?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 64

19.5 nm

Question 65

Given: Distance ‘A’ to ‘B’ is 475 NM, Planned GS 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’?

Answer 65

340 kt
.........................
Groundspeed 315kts
Distance 475nm
ATD (Actual Time Departure) 10:00
Initial time to B = 91 mins
ETA B = 11:31
Fix at 10:40 190 nm along track - 190 nm in 40 mins = groundspeed 285 i.e. less than planned.
Distance to go 285
Time to original ETA of 11:31 = 51 mins
285 nm in 51 mins requires GS of 336kts.
Nearest answer 340kts

Question 66

The heading is 345° M, the variation is 20°E. A radar bearing of 30° left of the nose from an island is taken.

What bearing should be plotted?

Answer 66

155°T

Question 67

An aircraft descends from FL250 to FL100. The rate of descent is 1000 ft/min and the groundspeed is 360 kts.

Calculate the flight path angle:

Answer 67

1.6°
…………………..
The flight path angle is 1.6°

Question 68

An aircraft is flying from A to B a distance of 50 NM. The True course in the flight log is 090º, the forecast wind is 225º(T)/15kt and the TAS is 120 kt. 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.

To reach destination B from this position, the correction angle on the heading should be:

Answer 68

17º

Question 69

An island is observed by weather radar to be 15° to the left. The aircraft heading is 120°(M) and the magnetic variation 17°W. What is the true bearing of the aircraft from the island?

Answer 69

268°
.................................
MB bearing of island from aircraft: 15º left of 120º
= 120º – 15º = 105º
MB aircraft to island: 105º
Var:   W17
TB aircraft to island:  088º
Reciprocal:                180º
TB island to aircraft:  268º

Question 70

An aircraft is climbing from FL180 to FL300 at an average CAS of 240 kt. What is the average TAS in the climb?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 70

367 kt

Question 71

Given: FL120, OAT is ISA standard, CAS is 200 kt, Track is 222°(M), Heading is 215°(M), Variation is 15°W. Time to fly 105 NM is 21 MIN. What is the W/V?

Answer 71

050°(T) / 70 kt.

Question 72

An aircraft is descending from FL260 to FL100 at an average CAS of 220 kt. What is the average TAS in the descent?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 72

293 kt

Question 73

With reference to the Low Altitude En-route Chart E(LO)1A.

An aircraft flies from WICK VOR (58º27.6’N 003º05.9’W) to SOLA VOR (58º52.5’N 005º38.4’E). Groundspeed is 241 kts. After 35 minutes of flight the indicator shows a bearing from SOLA of 268º.

What heading correction is necessary to proceed directly to SOLA?

Answer 73

6º to the left.

Question 74

At 10:15 the reading from a VOR/DME station is 211°/ 90NM, at 10:20 the reading from the same VOR/DME station is 211°/120NM. Compass Heading = 200º Variation in the area = 31ºW Deviation = +1º TAS = 390 kt The wind vector (T) is approximately:

Answer 74

110º/70kt

Question 75

An aircraft is descending from FL270 to FL100, following magnetic track 054° and maintaining CAS 250 kts.

Variation: 13°E
Temperature: ISA -10°C
W/V: 020/60
What is the GS?

Answer 75

277 kts

Question 76

You are tracking the 200° radial inbound to a VOR and your true heading is 010°. At the VOR you then track the 090° radial outbound and are showing a heading of 080°M The variation is +5° and the TAS is 240 kts. What is the wind (°T) has affected the aircraft ?

Answer 76

310°/65
………………………………….
200° radial inbound = Track 020°M, Variation +5° (E) = 025°T
Heading = 010°T
Drift = 15° Stbd
090° radial outbound = Track 090°M, Variation +5°(E) = 095°T
Heading 080°M, Variation +5°(E) = 085°T
Drift= 10° Stbd

Nav computer: centre dot on TAS (240 kts), put heading 010°T under the HEADING index and mark a line down the 15° Stbd drift line. Rotate to put heading 085°T under the HEADING index and mark a line down the 10° Stbd drift line. The point where these two lines intersect is the end of the wind vector; rotate to position it under the centre dot and read the wind.

Question 77

Given: Distance ‘A’ to ‘B’ is 325 NM, Planned GS 315 kt, ATD 1130 UTC, 1205 UTC - fix obtained 165 NM along track. What GS must be maintained from the fix in order to achieve planned ETA at ‘B’?

Answer 77

355 kt
…………………………….
325 nm at 315 kts would take 1hr 2mins not 1hr 18 mins (325/315 x 60) so planned ETA would be 11:30 + 1:02 = 12:32.

From Fix at 12:05 to B at 12:32 would be 160 nm (325 - 165) in 27 mins = 160/27 x 60 = 355 kts

Question 78

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’?

Answer 78

8° right
……………………..
3nm off track in 30nm along track = 6 in 60: initial track error 6º.
If the aircraft turns 6º to right it will fly parallel to track but 3nm to left of track.
It needs to regain the 3nm off track over the remaining 90nm to run: 3 in 90 = 1 in 30 = 2 in 60 so a further 2º alteration to the right is required.

Question 79

An aircraft is on final approach to land at an aerodrome, on a 3° glidepath, at a range of 7.6 NM from the threshold.

What is its height above ground level?

Answer 79

2360 ft
………………………
GP° = (Height x 60) / Distance

Question 80

Given: TAS = 197 kt, True course = 240°, W/V = 180/30kt. Descent is initiated at FL 220 and completed at FL 40. Distance to be covered during descent is 39 NM. What is the approximate rate of descent?

Answer 80

1400 FT/MIN

Question 81

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this be achieved?

Answer 81

10.3 NM
………………………………
Firstly they are both in different settings, get them both to flight levels.

2000’ QNH(1003) on 1013 would be 2300, so if your altimeter was always set to 1013(FL), then to fly to FL100 you would only have (10000-2300) left to fly.

7700/1000 = 7.7mins at 80kt GS

= 10.27 nm, 10.3 in the answer.

Question 82

An island appears 30° to the left of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading of 276° with the magnetic variation 12 °W?

Answer 82

054°
........................
Heading:          276ºM
Variation:        12ºW
Heading:         264ºT
Bearing:         30ºLeft
Bearing:         234ºT from aircraft to island
                      180º
Bearing:          054ºT from island to aircraft

Question 83

The track plot:

Answer 83

shows the path of the aircraft relative to the ground.

Question 84

An aircraft is climbing from FL180 to FL300. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 84

FL260 340°/40 kt –25°C

Question 85

The distance between two waypoints is 150 NM. When calculating the compass heading 2°E magnetic variation was used instead of 2°W.

Assuming the forecast W/V applied, what will the off-track distance be at the second waypoint?

Answer 85

10 NM

Question 86

During visual navigation in freezing conditions, after heavy snowfall, which of the following landmark will give the best reference for a visual checkpoint:

Answer 86

a large river

Question 87

An aircraft is climbing from FL40 to FL220 at an average CAS of 180 kt. What is the average TAS in the climb?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 87

233 kt

Question 88

An aircraft follows a radial to a VOR/DME station.
At 10:00 the DME reads 120 NM.
At 10:03 the DME reads 105 NM.
The estimated time overhead the VOR/DME station is:

Answer 88

10:24

Question 89

Which formula can be used to calculate the climb gradient?

Answer 89

% Gradient = altitude difference (ft) x 100 ÷ ground difference (ft)

Question 90

Given: Position NDB (55°10´N, 012°55´E) DR Position (54°53´N, 009°58´E) NDB on the RMI reads 090°. Magnetic variation = 10°W. The position line has to be plotted on a Lamberts conformal chart with standard parallels at 40°N and 48°N. Calculate the direction (T) of the bearing to be plotted from the NDB.

Answer 90

262°

Question 91

An island appears 45° to the right of the centre line on an airborne weather radar display. What is the true bearing of the aircraft from the island if at the time of observation the aircraft was on a magnetic heading (MH) of 215° with the magnetic variation 21ºW

Answer 91

059°
.........................
Heading 215ºM
Variation 21ºW
Heading 194ºT
Bearing 45º Right
Bearing 239ºT from aircraft to island 180º
Bearing 059ºT from island to aircraft

Question 92

An aircraft is flying from A to B a distance of 50 NM. The True Course in the flight log is 270º, the forecast wind is 045º(T)/15kt and the TAS is 120kt.

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.

To reach destination B from this position, the correction angle on the heading should be:

Answer 92

17º

Question 93

What is the formula to work out gradient?

Answer 93

Gradient (%) = (Altitude Difference (ft) x 100)/Ground Distance

Question 94

An aircraft is planned to fly from position ‘A’ to position ‘B’, distance 320 NM, at an average GS of 180 kt. It departs ‘A’ at 1200 UTC. After flying 70 NM along track from ‘A’, the aircraft is 3 MIN ahead of planned time. Using the actual GS experienced, what is the revised ETA at ‘B’?

Answer 94

1333 UTC

Question 95

The distance between A and B is 90 NM. At a distance of 75 NM from A the aircraft is 4 NM right of course. The track angle error (TKE) is:

Answer 95

3ºR
…………………..
Calculate Track Error Angle =
TEA = Distance OFF Track x 60 / Distance ALONG Track

TEA = 4 x 60 / 75
TEA = 240 / 75
TEA = 3°

Question 96

An aircraft is on final approach to land at an aerodrome, on a 3° glidepath, at a range of 6.3 nm from touchdown. What is its height above ground level?

Answer 96

1915 ft
…………………….
GP° = (Height x 60) / Distance

Question 97

Where and when are the IRS positions updated?

Answer 97

Only on the ground during the alignment procedure.

Question 98

Transferring position lines (LOP) can be done using:

Answer 98

radials, DME, QDM/QDR.

Question 99

The True course in the flight log is 270º, the forecast wind is 045º(T)/15kt and the TAS is 120kt. 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 (TKE) is:

Answer 99

5ºL

Question 100

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1023 hPa. The TAS is 100 kt, the head wind component is 20 kt and the rate of climb is 1000 ft/min. Top of climb is FL 100. At what distance from the airport will this be achieved?

Answer 100

11.1 NM

Question 101

You are departing from an airport which has an elevation of 2000 ft. The QNH is 1013 hPa. 10 NM away there is a waypoint you are required to pass at an altitude of 7500 ft. Given a groundspeed of 100 kt, what is the minimum rate of climb?

Answer 101

920 ft/min
……………………..
Climbing from 2,000ft to 7,500ft = 5,500ft.

10nm at 100kts = 6 minutes.
5,500ft = 917ft/min
6mins

Question 102

Given:
Rate of climb: 740 ft/min
TAS: 200 kts
Headwind: 15 kts
Calculate the climb gradient:

Answer 102

4%

Question 103

An aircraft is making a descending turn onto final approach to land at an aerodrome, on a 4.2° glidepath, at a range of 8.2 nm from touchdown. By what height above ground level must the turn be completed?

Answer 103

3490 ft
……………………………
You must be careful to establish whether the question states distance from touchdown or distance from the threshold. In the case of the latter, you must add the threshold crossing elevation onto the calculated solution.

The formula to apply is:
GP° = (Height x 60) / Distance
Both measurements, height and distance, must be in the same units. It is suggested that you convert the ground distance into feet as this will produce the final height in feet.
8.2 nm x 6080 ft = 49 856 ft
4.2° = (Height x 60) / 49 856
Transpose the formula
(4.2 x 49 856) / 60 = Height

Height = 3490 ft above ground level

Question 104

An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:

Answer 104

2210 FT
………………..
An aircraft is maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:
The aircraft is 7 NM or 42,476 ft (7 × 6068) from the RWY.
The gradient is given as 5.2%; so it’s simply a case of multiplying the distance from the RWY by the gradient:
42476 × (5.2/100) = 2208.752 ft
Or do it on the whizz wheel; (much quicker) Put 10 on the inner scale next to the gradient (5.2%) on the outer scale, go to distance from touchdown (42,500ft) on the inner scale and read height above touchdown on the outer scale: 2,210ft.

Question 105

Refer to IFR Chart E(LO) 1 and the winds aloft data below: An aircraft passes overhead the TRN VOR/DME at FL300 and descends to FL60 along Airway B2 via GIRVA at a constant CAS of 310 kt. What is (a) the aircraft magnetic heading and (b) groundspeed in the descent?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 105

210°M 370 kt

Question 106

An aircraft is descending from FL260 to FL100. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 106

FL180 290°/15 kt –15°C

Question 107

Given: aircraft height 2500 FT, ILS GP angle 3°. At what approximate distance from THR can you expect to capture the GP?

Answer 107

8.3 NM

Question 108

Which formula can be used to calculate the climb gradient?

Answer 108

Gradient (°) = (Altitude difference (ft) x 60) ÷ Ground distance (ft)

Question 109

An aircraft at FL290 is required to commence descent when 50 NM from a VOR and to cross that VOR at FL80. Mean GS during descent is 271kt. What is the minimum rate of descent required?

Answer 109

1900 FT/MIN
…………………………
As both altitudes are flight levels there is no need to consider altimeter setting and pressure. Altitude to lose = 29 000 ft – 8000 ft = 21 000 ft Time to VOR = distance ÷ groundspeed = 50 nm ÷ 271 kts = 11 min Rate of descent = altitude to lose ÷ time to descent = 21 000 ft ÷ 11 min = 1909 ft/min

Question 110

An aircraft descends from FL250 to FL100. The rate of descent is 1000 ft/min and the groundspeed is 360 kts.

What distance will be covered during the descent?

Answer 110

90 NM

Question 111

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:

Answer 111

30 NM and 240°
……………………………………………….
A look at the four answers tells us that we only need to establish the bearing from the feature to the aircraft when the relative bearing was 280º.

The distance could be calculated but it is not required to answer this question.
Heading 165ºM
Variation 25ºW
Heading 140ºT
Bearing 280ºRelative
Bearing 420ºT
Less 360º
Bearing 060ºT from the aircraft to the feature
Plus 180º
Bearing 240ºT from the feature to the aircraft

Question 112

You are departing from an airport which has an elevation of 1500 ft. The QNH is 1003 hPa. 15 NM away there is a waypoint you are required to pass at an altitude of 7500 ft. Given a groundspeed of 120 kt, what is the minimum rate of climb?

Answer 112

800 ft/min
…………………………..
Altitude to climb = altitude – elevation Altitude to climb = 7500 ft – 1500 ft = 6000 ft As the climb is to an altitude (altimeter setting remains on QNH) there is no need to consider any pressure error. Time to cover 15 nm at 120 kts: Time = distance ÷ speed = 15 nm ÷ 120 kts = 7.5 min Aircraft is required to climb 6000 ft in 7.5 min. Rate of climb = altitude to climb ÷ time to climb = 6000 ft ÷ 7.5 min = 750 ft/min

Question 113

Refer to the winds aloft data below: An aircraft descends at a rate of 2500 fpm from FL 260 to FL100 along a true track of 350°T and at a constant CAS of 230 kt. What is the approximate distance covered during the descent?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/25 kt –25°C
FL120 150°/20 kt –15°C
FL60 135°/15 kt –10°C
SFC 100°/10 kt +10°C

Answer 113

35 nm

Question 114

An aircraft is descending down a 6% slope whilst maintaining a G/S of 300 kt. The rate of descent of the aircraft is approximately:

Answer 114

1800 FT/MIN
………………………………..
Rates of descent are easily worked out on the whizz wheel (that wonderful device used by professionals; i.e. navigators). If the glide slope angle is given in degrees put the 60 on the inner scale next to glide slope angle on the outer scale. If it is given as a percentage put the 10 on the inner scale next to the glide slope percentage on the outer scale. Now go to the groundspeed in knots on the inner scale and next to it is rate of descent in feet per minute on the outer scale.

Question 115

An aircraft is descending from FL300 to FL220 at an average CAS of 320 kt. What is the average TAS in the descent?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 115

479 kt

Question 116

An aircraft is planned to fly from position ‘A’ to position ‘B’, distance 250 NM at an average GS of 115 kt. It departs ‘A’ at 0900 UTC. After flying 75 NM along track from ‘A’, the aircraft is 1.5 MIN behind planned time. Using the actual GS experienced, what is the revised ETA at ‘B’?

Answer 116

1115 UTC
……………………………………..
75nm at original ground speed (115kts) would take 39.1 minutes.
Aircraft has covered 75nm in 40.6 minutes = ground speed 111kts.
Remaining distance to run is 250 – 75 = 175nm and at revised ground speed of 111kts this will take 94.6 minutes.
Total elapsed time = 40.6 + 94.6 = 135.2 minutes (2h15m).
Departure time is 0900 add elapsed time 2h15m, ETA = 1115.

Question 117

Given:

Distance A to B: 415 NM

Planned GS: 295 kts

ATD: 1000 UTC

At 1020 UTC a fix is obtained 110 NM along the track. What GS must be maintained from the fix in order to achieve the planned ETA at B?

Answer 117

286 kts

Question 118

An aircraft is departing from an airport which has an elevation of 2000 ft and the QNH is 1003 hPa. The TAS is 100 kt, the head wind component is 20 kt and the rate of climb is 500 ft/min. Top of climb is FL 050. At what distance from the airport will this be achieved?

Answer 118

7.2 NM

Question 119

An aircraft tracks radial 200 inbound to a VOR station with a Magnetic Heading (MH) of 010º. After being overhead the VOR station the aircraft tracks radial 090 outbound with a MH of 080º. The TAS is 240 kt and the magnetic variation in the area is 5ºW. What is the wind vector (T)?

Answer 119

320º/50kt
………………………………
VOR Inbound - HDG 005ºT TRK 015ºT (Drift = 10º Right)

VOR Outbound - HDG 075ºT TRK 085ºT (Drift = 10º Right)

Place TAS = 240 knots under centre dot.

Place Inbound HDG under True Heading Index.

Draw a line down the length of the 10º Right Drift line.

Place Outbound HDG under True Heading Index.

Draw a line along the 10º Right Drift line where it intersects the Inbound Drift line.

Rotate the Wind Calculator until the intersection of the two lines is on the central wind speed line, where you read the wind speed and read the Wind Direction at the top of the calculator.

It should read 320 / 50.

Question 120

Refer to annex.

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.

To reach destination B from this position, the TH should be:

Answer 120

292º

Question 121

Complete LINE 3 of the ‘FLIGHT NAVIGATION LOG’, positions ‘E’ to ‘F’. What is the HDG°(M) and ETA?

Answer 121

HDG 105° - ETA 1205 UTC

Question 122

Refer to the winds aloft data below: An aircraft climbs at a rate of 1500 fpm from FL 120 to FL300 along a true track of 000°T and at a constant CAS of 180 kt. What is the approximate distance covered during the climb?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 122

61 nm

Question 123

An aircraft at FL350 is required to descend to cross a DME facility at FL80. Maximum rate of descent is 1800 FT/MIN and mean GS for descent is 276 kt. The minimum range from the DME at which descent should start is:

Answer 123

69 NM

Question 124

An aircraft is descending from FL180 to 2,000’ amsl. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C

FL80 235°/20 kt +05°C

FL40 220°/18 kt +12°C

2,000’ 220°/15 kt +15°C

SFC 200°/10 kt +20°C

Answer 124

FL100 250°/20 kt +00°C

Question 125

Complete LINE 6 of the ‘FLIGHT NAVIGATION LOG’, positions ‘L’ to ‘M’. What is the HDG°(M) and ETA?

Answer 125

HDG 075° - ETA 1502 UTC

Question 126

An aircraft at FL140, IAS 210 kt, OAT -5°C and wind component minus 35 kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change, when 150 NM from the reporting point the aircraft should reduce IAS by:

Answer 126

20 kt
………………………
IAS (CAS) 210KTS FL140/-5ºC = TAS 262 KTS

Wind component = -35 KTS
Groundspeed = 227 KTS
150nm at 227 KTS GS = 39.6 mins
150nm in 44.6 mins = 202 KTS required GS
New TAS required = 202 + 35 = 237 KTS.
TAS 237 KTS FL140/-5ºC = New IAS 190 KTS
Old IAS = 210 KTS
New IAS = 190 KTS
Speed reduction = 210 - 190 = 20 KTS

Question 127

Complete LINE 5 of the ‘FLIGHT NAVIGATION LOG’, positions ‘J’ to ‘K’. What is the HDG°(M) and ETA?

Answer 127

HDG 337° - ETA 1422 UTC

Question 128

An aircraft is required to descend from FL230 to FL50 over a distance of 32 NM in 7 minutes.

What is the required ROD (when a wind component of -25 kts is expected during the descent)?

Answer 128

2570 ft/min
……………………………
These calculations can be performed very quickly on the CRP-5.

Time = 7 minutes
Total descent = 23,000 - 5,000 ft = 18,000 ft
On the CRP-5 Computer side

Align 7 minutes on the inner scale with 18,000 ft (18) on the outer scale - this represents 18,000 ft in 7 minutes

Look up 1 (10) at the red arrow on the inner scale - this represents 1 minute

Read 2,570 ft on the outer scale

The rate of descent is 2,570 feet per minute

Question 129

An aircraft is flying according the flight log at the Annex. After 15 minutes of flying with the planned TAS and TH the aircraft is 3 NM North of the intended track and 2.5 NM ahead of the dead reckoning position. To reach destination B from this position the TH should be:

Answer 129

258º

Question 130

During a low level flight 2 parallel roads that are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft:

Answer 130

groundspeed
…………………………………………
This would give distance travelled and time; speed = distance/time.

Question 131

An aircraft is descending from FL300 to FL220. What wind should be used for the calculation of average TAS?

FL300 350°/50 kt –35°C
FL260 340°/40 kt –25°C
FL220 320°/25 kt –20°C
FL180 290°/15 kt –15°C
FL160 270°/25 kt –10°C
FL100 250°/20 kt +00°C
FL80 235°/20 kt +05°C
FL40 220°/18 kt +12°C
2,000’ 220°/15 kt +15°C
SFC 200°/10 kt +20°C

Answer 131

FL260 340°/40 kt –25°C

Question 132

An aircraft is flying from A to B. The true course according to the flight log is 090º, the estimated wind is 225º(T)/15kt and the TAS is 120 kt.

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 (TKE) is:

Answer 132

5ºR

Question 133

Assuming zero wind, what distance will be covered by an aircraft descending 15000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN?

Answer 133

26.7 NM
…………………………
15,000ft at 3,000 ft/min = 5 minutes at 320kts = 26.7NM

Question 134

An aircraft is planned to fly from position ‘A’ to position ‘B’, distance 480 NM at an average GS of 240 kt. It departs ‘A’ at 1000 UTC. After flying 150 NM along track from ‘A’, the aircraft is 2 MIN behind planned time. Using the actual GS experienced, what is the revised ETA at ‘B’?

Answer 134

1206
……………………….
150nm at original ground speed (240kts) would take 37.5 minutes.
Aircraft has covered 150nm in 39.5 minutes = ground speed 228kts.

Remaining distance to run is 480 – 150 = 330nm and at revised ground speed of 228kts this will take 86.8 minutes.

Total elapsed time = 39.5 + 86.8 = 126.3 minutes (2h06m).

Departure time is 1000 add elapsed time 2h06m, ETA = 1206.

Question 135

Refer to the winds aloft data below: An aircraft descends at a rate of 2000 fpm from FL 360 to FL240 along a true track of 020°T and at a constant CAS of 280 kt. What is the approximate distance covered during the descent?

FL360 220°/80 kt –55°C
FL300 200°/65 kt –45°C
FL240 180°/45 kt –40°C
FL180 170°/35 kt –25°C
FL120 150°/25 kt –15°C
FL60 135°/20 kt –10°C
SFC 100°/10 kt +10°C

Answer 135

51 nm

Question 136

The descent gradient for an aircraft which is crossing a VOR (whilst descending on track) is:

60 NM north of the VOR at FL350

10 NM south of the VOR at FL120

Answer 136

5.4%
……………………………….
Gradient (%) = (Height / Distance) x 100

NOTE: Both Height and Distance must be in the same units

The total distance flown = 60 nm + 10 nm = 70 nm

The total descent = 35,000 - 12,000 ft = 23,000 ft

23,000 / 6,080 ft = 3.78 nm

Gradient (%) = (Height / Distance) x 100

Gradient (%) = (3.78 / 70) x 100

Gradient (%) = 0.054 x 100

Gradient (%) = 5.4%