04 Dead Reckoning Navigation Flashcards by kazim unal

Given: GS = 236 kt. Distance from A to B = 354 NM. What is the time from A to B?

1 HR 30 MIN

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Given: An aircraft is flying at FL100, OAT = ISA - 15ºC. The QNH, given by a meteorological station with an elevation of 100 ft below MSL is 1032 hPa. 1 hPa = 27 ft Calculate the approximate True Altitude of this aircraft.

9900 ft

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Given: TAS = 250 kt, HDG (T) = 029°, W/V = 035/45kt. Calculate the drift and GS?

1L - 205 kt

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CAS 120 kt, FL 80, OAT +20°C. What is the TAS?

141 kt

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An aircraft takes on 320 US gallons of fuel weighing 870 kg.

What is the specific gravity of the fuel?

0.72

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Given: TAS = 485 kt, True HDG = 226°, W/V = 110°(T)/95kt. Calculate the drift angle and GS?

9°R - 533 kt

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At reference. 1215 UTC LAJES VORTAC (38°46’N 027°05’W) RMI reads 178°, range 135 NM. Calculate the aircraft position at 1215 UTC?

40°55’N 027°55’W

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Given:

True Track 095°
TAS 160 kts
True Heading 087°
GS 130 kts
Calculate W/V

057°/36 kts

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An aircraft takes off from the aerodrome of BRIOUDE (altitude 1 483 FT, QFE = 963 hPa, temperature = 32°C). Five minutes later, passing 5 000 FT on QFE, the second altimeter set on 1 013 hPa will indicate approximately :

6 500 FT

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Given:

Groundspeed 240 kts
Distance to go 500 NM
Calculate the time to go:

2 hours 05 min
…………………………………….
To travel 500 NM at a speed of 240 kt will take 2 hr 5 min
Background Information
Simply, divide distance by speed to find time

500 nm ÷ 240 kt = 2 hr 5 min

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Given: True HDG = 002°, TAS = 130 kt, Track (T) = 353°, GS = 132 kt. Calculate the W/V?

095/20kt

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Given: TAS = 90 kt, HDG (T) = 355°, W/V = 120/20kt. Calculate the Track (°T) and GS?

346 - 102 kt

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Given:

True HDG 074°
TAS 230 kts
Track (T) 066°
GS 242 kts
Calculate the W/V.

180/35 kts

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CAS is: 320 kt Flight level: 330 OAT: ISA +15°C TAS is approximately (compressibility factor 0.939):

530 kt

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Given: TAS = 200 kt, Track (T) = 110°, W/V = 015/40kt. Calculate the HDG (°T) and GS?

099 - 199 kt

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Given: True course from A to B = 090°, TAS = 460 kt, W/V = 360/100kt, Average variation = 10°E, Deviation = -2°. Calculate the compass heading and GS?

070° - 450 kt

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Given: GS = 95 kt. Distance from A to B = 480 NM. What is the time from A to B?

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Given: TAS = 375 kt, True HDG = 124°, W/V = 130°(T)/55kt. Calculate the true track and GS?

123 - 320 kt

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Maximum allowable crosswind component is 20 kt. Runway 06, RWY QDM 063°(M).

Wind direction 100°(M). Calculate the maximum allowable wind speed?

33 kt

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Given: TAS = 190 kt, True HDG = 085°, W/V = 110°(T)/50kt. Calculate the drift angle and GS?

8°L - 146 kt

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The following information is displayed on an Inertial Navigation System: GS 520 kt, True HDG 090°, Drift angle 5° right, TAS 480 kt. SAT (static air temperature) -51°C. The W/V being experienced is:

320° / 60 kt

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Given: TAS = 370 kt, True HDG = 181°, W/V = 095°(T)/35kt. Calculate the true track and GS?

186 - 370 kt

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Given: CAS 120 kt, FL 80, OAT +20°C. What is the TAS?

141 kt

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How many NM would an aircraft travel in 1 min 45 secs if the GS is 135 kt?

3.94

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Route 'A' (44°N 026°E) to 'B' (46°N 024°E) forms an angle of 35° with longitude 026°E. Variation at A is 3°E. What is the initial magnetic track from A to B?

322° ................................ A quick diagram indicates that this track from A to B is going north (ish) (from 44ºN to 46ºN) and west (ish) (from 026ºE to 024ºE). This means that the track will be more than 270ºT but less than 360ºT. If the angle between the track and the local meridian is 35º then the track = 360º - 35º = 325ºT variation is 3ºE so the track is 322ºM.

An aircraft is flying at FL180 and the outside air temperature is -30°C. If the CAS is 150 kt, what is the TAS?

195 kt

An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30 kt. The GS and drift angle are:

192 kt, 7° right

TAS = 472 kt, True HDG = 005°, W/V = 110°(T)/50kt. Calculate the drift angle and GS?

6°L/490 kt

An aircraft flies at FL 250. OAT = - 45°C. The QNH, given by a meteorological station with an elevation of 2830 ft, is 1033 hPa. Calculate the clearance above a mountain ridge with an elevation of 20410 ft.

4 200 ft

Given: TAS = 140 kt, HDG (T) = 005°, W/V = 265/25kt. Calculate the drift and GS?

10R - 146 kt

Given: GS = 120 kt. Distance from A to B = 84 NM. What is the time from A to B?

00 HR 42 MIN

Given: TAS: 210 kts Fuel flow: 42 US Gal/hr Specific gravity: 0.72 What is the specific fuel consumption?

0.545 kg/NM air distance.

The ICAO definition of ETA is the:

estimated time of arrival at destination

Given: M 0.80, OAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right. Calculate the true W/V?

020°/95 kt

Given: TAS = 485 kt, OAT = ISA +10°C, FL 410. Calculate the Mach Number?

0.825

Given: | FL250, OAT -15 ºC, TAS 250 kt. Calculate the Mach No.?

0.40

How long will it take to travel 284 NM at a speed of 526 km/h?

``` 1 hour ............................. To travel 284 NM at a speed of 526 km/h will take 1 hour. Background Information First convert 284 nm into kilometers 284 nm x 1.852 = 526 km Next, divide distance by speed to find time 526 km ÷ 526 = 1 hr ```

``` Given: True HDG 133° TAS 225 kts Track (T) 144° GS 206 kts Calculate the W/V. ```

075/45 kts

An aircraft departs Waypoint ELSEY (N59° E120°) on a magnetic heading of 005°M and a TAS of 240 kt. W/V = 100°/40. Local magnetic variation = 5°E. What is the approximate position of the aircraft after 15 minutes of flight?

N60° E120°

Given: Magnetic track = 075°, HDG = 066°(M), VAR = 11°E, TAS = 275 kt. Aircraft flies 48 NM in 10 MIN. Calculate the true W/V °?

340°/45 kt

Given: M 0.80, OAT -50°C, FL 330, GS 490 kt, VAR 20°W, Magnetic heading 140°, Drift is 11° Right. Calculate the true W/V?

020°/95 kt

The reported surface wind from the Control Tower is 240°/35 kt. Runway 30 (300°). What is the cross-wind component?

30 kt

Given: True HDG = 233°, TAS = 480 kt, Track (T) = 240°, GS = 523 kt. Calculate the W/V?

110/75kt

If it takes 132.4 minutes to travel 840 NM, what is the speed in km/hr?

``` 705 ............................... The speed in km/hr is 705 km/hr Background Information First convert 840 nm into kilometers 840 nm x 1.852 = 1,555.68 km Next, convert the time from minutes into hours. 132.4 mins ÷ 60 = 2.21 hr Next, divide distance by time to find speed 1,555.68 km ÷ 2.21 hr = 705 km/hr ```

Given: TAS = 470 kt, True HDG = 317°, W/V = 045°(T)/45kt. Calculate the drift angle and GS?

5°L - 470 kt

Given: GS = 345 kt. Distance from A to B = 3560 NM. What is the time from A to B?

10 HR 19 MIN

Consider the following factors that determine the accuracy of a DR position: 1. The flight time since the last position update. 2. The accuracy of the forecasted wind. 3. The accuracy of the TAS. 4. The accuracy of the steered heading. Using the list above which of the following contains the most complete answer?

1, 2, 3 and 4

What is the ISA temperature value at FL 330?

-51°C ......................... In an International Standard Atmosphere the temperature is +15ºC at sea level and has a lapse rate of 1.98ºC per 1,000 up to 36,090 feet. At FL330 (33,000 feet) the temperature is +15º - (33 x 1.98º) = +15º - 65.34º = - 50.34ºC

Given: TAS = 170 kt, HDG(T) = 100°, W/V = 350/30kt. Calculate the Track (°T) and GS?

109 - 182 kt

Given: GS = 105 kt. Distance from A to B = 103 NM. What is the time from A to B?

00 HR 59 MIN

Given: TAS = 290 kt, True HDG = 171°, W/V = 310°(T)/30kt. Calculate the drift angle and GS?

4°L - 314 kt

Given: GS = 122 kt. Distance from A to B = 985 NM. What is the time from A to B?

8 HR 04 MIN

Given: Maximum allowable tailwind component for landing 10 kt. Planned runway 05 (047° magnetic). The direction of the surface wind reported by ATIS 210°. Variation is 17°E. Calculate the maximum allowable wind speed that can be accepted without exceeding the tailwind limit?

10 kt

Given: GS = 435 kt. Distance from A to B = 1920 NM. What is the time from A to B?

4 HR 25 MIN

An aircraft is flying at FL100. The OAT = ISA - 15°C. The QNH given by a station at an elevation 3000 ft is 1035hPa. Calculate the approximate True Altitude.

10 200 ft

The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affected by:

the accuracy of the forecast wind.

Given: TAS = 132 kt, HDG (T) = 053°, W/V = 205/15kt. Calculate the Track (°T) and GS?

050 - 145 kt

Given: Runway direction 083°(M), Surface W/V 035/35kt. Calculate the effective headwind component?

24 kt

Given: TAS = 485 kt, HDG (T) = 168°, W/V = 130/75kt. Calculate the Track (°T) and GS?

174 - 428 kt

Given: True HDG = 054°, TAS = 450 kt, Track (T) = 059°, GS = 416 kt. Calculate the W/V?

010/50kt

Given: TAS: 154 mph Fuel flow: 28 Imp Gal/hr Specific gravity: 0.72 What is the specific range?

1.46 NM air distance/kg

Given: Required course 045°(M); Variation is 15°E; W/V is 190°(T)/30 kt; CAS is 120 kt at FL 55 in standard atmosphere. What are the heading (°M) and GS?

055° and 147 kt

The accuracy of the, manually calculated, DR-position of an aircraft is, among other things, affected by:

the flight time since the last position update.

The QNH, given by a station at 2500 ft, is 980hPa.The elevation of the highest obstacle along a route is 8 000 ft and the OAT = ISA -10°C. When an aircraft, on route has to descend the minimum indicated altitude (QNH on the subscale of the altimeter) to maintain a clearance of 2000 ft, will be:

10 400 ft

Given: Airport elevation is 1000 ft. QNH is 988 hPa. What is the approximate airport pressure altitude? (Assume 1 hPa = 27 FT)

1680 FT

Given: True altitude 9000 FT, OAT -32°C, CAS 200 kt. What is the TAS?

220 kt

What may cause a difference between a DR-position and a Fix?

The difference between the actual wind and the forecasted wind.

What is the ratio between the litre and the US-GAL ?

1 US-GAL equals 3.78 litres

Given: TAS = 155 kt, HDG (T) = 216°, W/V = 090/60kt. Calculate the Track (°T) and GS?

231 - 196 kt

Given: An aircraft is on final approach to runway 32R (322°); The wind velocity reported by the tower is 350°/20 kt.; TAS on approach is 95 kt. In order to maintain the centre line, the aircraft's heading (°M) should be:

328°V

Given: TAS = 198 kt, HDG (°T) = 180, W/V = 359/25. Calculate the Track(°T) and GS?

180 - 223 kt

Given: True HDG = 206°, TAS = 140 kt, Track (T) = 207°, GS = 135 kt. Calculate the W/V?

180/05kt

Given: Runway direction 305°(M),Surface W/V 260°(M)/30 kt. Calculate the cross-wind component?

21 kt

You are flying at FL80 and the air temperature is ISA +15°. What CAS is required to make TAS 240 kts?

208 kts

Given: ``` True track 352° Variation 11°W Deviation -5° Drift 10°L Calculate the compass heading? ```

018° .............................................................................. C D M V T Track 008º -5º(W) 003º 11ºW 352º Drift 10ºS Heading 358º -5º(W) 353º 11ºW 342º

Given: ``` True track 070° Variation 30°W Deviation +1° Drift 10°R Calculate the compass heading? ```

089° .......................... True Track = 070°T Drift = 10°R Drift is FROM Heading to TRACK, i.e. Right Drift = Heading LESS than Track Therefore, Heading = 060°T Now apply CDMVT (Cadbury's Dairy Milk Very Tasty) Read Bottom to Top Compass - 089°C Deviation - +1° (+ Positive Deviation = Easterly Deviation - Deviation East Compass Least) Magnetic - 090°M Variation - 30°W (Variation West Magnetic Best) True - 060°T

Given: TAS = 480 kt, HDG (°T) = 040°, W/V = 090/60kt. Calculate the Track (°T) and GS?

034 - 445 kt ................................. The question asks for track and groundspeed not heading and groundspeed as many of the questions do. It asks for Track and Groundspeed so once you set your HDG i.e. 040 read off your drift i.e. 6L so a HDG of 040 would give you a track of 034 and GS of 445.

Given: For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-wind limitation of 35 kt. The angle between the wind direction and the runway is 60°. Calculate the minimum and maximum allowable wind speeds?

20 kt and 40 kt

Given: Mach number: 0.8 Flight level: 330 OAT: ISA +15°C TAS is approximately (compressibility factor 0.94):

480 kt | .................................

An aircraft is flying at FL250, OAT = - 45°C. The QNH, given by a station at MSL, is 993.2 hPa. Calculate the approximate True Altitude.

23400 ft

At reference. 1300 UTC DR position 37°30'N 021°30'W alter heading PORT SANTO NDB (33°03'N 016°23'W) TAS 450 kt, Forecast W/V 360°/30kt. Calculate the ETA at PORT SANTO NDB?

1348

An aircraft is flying at FL 200. OAT = 0°C. When the actual air pressure on an airfield at MSL is placed in the subscale of the altimeter the indicated altitude is 19300ft. Calculate the aircraft's true altitude.

21 200 ft

Given: TAS = 227 kt, Track (T) = 316°, W/V = 205/15kt. Calculate the HDG (°T) and GS?

312 - 232 kt ....................................... For this one: put on the wind (205º/15kts - using 'wind down' method), centre dot on TAS (227kts) put track (316º) under HEADING index note drift is 4ºStbd and put track under 4ºStbd drift is still 4ºStbd so required heading is 312º and groundspeed is under wind mark: 232kts. Are you using something other than CRP5 ? All 'slide' type whizz wheels should give the correct answer. The circular Jeppessen whizz wheels only offer an approximate answer and are very 'fiddly' to use. There are a lot of questions in the nav exam which require quick and accurate use of the whizz wheel.

Given: Runway direction 210°(M), Surface W/V 230°(M)/30kt.Calculate the cross-wind component?

10 kt

Given: TAS = 190 kt, HDG (T) = 355°, W/V = 165/25kt. Calculate the drift and GS?

1L - 215 kt

Your on an airfield elevation 2000ft, QNH 1003. You want to climb to FL50, your rate of climb is 1000ft/min, your TAS is 100 and you have a headwind of 20. What is the distance it takes to get to FL50 ?

3.6 NM ............................... Difference between 1013hPa and QNH (1003hPa) = 10hPa x 27ft = 270ft. The airfield is at a pressure altitude of 2,270ft. Aircraft needs to climb 5,000 - 2,270 = 2,730ft to reach FL50. At 1,000ft per minute this will take 2.73minutes. Groundspeed = 100 - 20 = 80kts. In 2.73 minutes at a groundspeed of 80 kts the aircraft will cover a distance of 3.6nm.

Given: True Heading = 180°, TAS = 500 kt, W/V 225° / 100 kt, Calculate the GS?

435 kt

Given: TAS = 155 kt, Track (T) = 305°, W/V = 160/18kt. Calculate the HDG (°T) and GS?

301 - 169 kt

Given: Runway direction 230°(T), Surface W/V 280°(T)/40 kt. Calculate the effective cross-wind component?

31 kt

Given: TAS = 270 kt, True HDG = 145°, Actual wind = 205°(T)/30kt. Calculate the drift angle and GS?

6°L - 256 kt

Given: GS = 480 kt. Distance from A to B = 5360 NM. What is the time from A to B?

11 HR 10 MIN

21 200 ft

On a polar stereographic chart of the northern hemisphere whose grid is aligned with the zero meridian. Grid track 344° Longitude 115°00'W. Calculate the true course.

229°

Given: Magnetic track = 210°, Magnetic HDG = 215°, VAR = 15°E, TAS = 360 kt, Aircraft flies 64 NM in 12 MIN. Calculate the true W/V?

265°/50 kt

An aircraft travels 2.4 statute miles in 47 seconds. What is its groundspeed?

160 kt ....................................... 2.4 SM in 47 seconds is (2.4 ¸ 47) = SM per second X 60 = SM/minute X 60 = SM/hr = 183.8 mph X 5280 = ft/hr ¸ 6080 = 159.6 kts.

Given: course required = 085° (T), Forecast W/V 030/100kt, TAS = 470 kt, Distance = 265 NM. Calculate the true HDG and flight time?

075°, 39 MIN

The accuracy of the manually calculated DR-position of an aircraft is, among other things, affected by

the flight time since the last position update

If the TAS exceeds the CAS by 20% at FL100, what is the OAT?

+15º C

Given: TAS = 205 kt, HDG (T) = 180°, W/V = 240/25kt. Calculate the drift and GS?

6L - 194 kt

Given: GS = 154 kts Flight time = 20 mins Fuel used = 117 Kg What is the fuel used per NM?

2.28 ......................... Distance flown in one hour = 154 NM (154 kts) Distance flown in 20 mins = 154 x (20 ÷ 60) = 51.333 NM Fuel used = 176 ÷ 51.333 = 2.28 Kg/NM

Given: Magnetic heading = 255° VAR = 40°W GS = 375 kt W/V = 235°(T) / 120 kt Calculate the drift angle?

6° left

Given: True HDG = 307°, TAS = 230 kt, Track (T) = 313°, GS = 210 kt. Calculate the W/V?

260/30kt

Given: GS = 135 kt. Distance from A to B = 433 NM. What is the time from A to B?

3 HR 12 MIN

Given: TAS = 220 kt; Magnetic course = 212 º, W/V 160 º(M)/ 50kt, Calculate the GS?

186 kt

Given: TAS = 135 kt, HDG (°T) = 278, W/V = 140/20kt. Calculate the Track (°T) and GS?

283 - 150 kt

Given: TAS = 95 kt, HDG (T) = 075°, W/V = 310/20kt. Calculate the drift and GS?

9R - 108 kt

An aircraft is flying at FL200. The QNH, given by a meteorological station at an elevation of 1300ft is 998.2 hPa. OAT = - 40ºC. The elevation of the highest obstacle along the route is 8 000 ft. Calculate the aircraft's approximate clearance above the highest obstacle on this route.

10 500 ft

Given: True Track 239° True Heading 229° TAS 555 kt G/S 577 kt Calculate the wind velocity.

130°/100kt

Given: Compass Heading 090°, Deviation 2°W, Variation 12°E, TAS 160 kt. Whilst maintaining a radial 070° from a VOR station, the aircraft flies a ground distance of 14 NM in 6 MIN. What is the W/V °(T)?

160°/50 kt ...................................... Heading: 090ºC, Dev 2ºW, 088ºM, Var 12ºE = 100ºT. TAS = 160 kts. Track = 070ºM, Var 12E = 082ºT. Groundspeed: 14 nm in 6 minutes = 140 kts. On the Nav Computer: Centre dot on TAS 160kts, put the heading, 100ºT, under the HEADING index. With heading 100ºT and track 082ºT the drift is 18º Port (left). Mark the wind vector where the drift (18º Port) crosses the groundspeed (140kts). Rotate to put this mark under the centre dot and read the wind: 160ºT/50kts.

Given: TAS = 200 kt, Track (T) = 073°, W/V = 210/20kt. Calculate the HDG (°T) and GS?

077 - 214 kt

For a landing on runway 23 (227° magnetic) surface, W/V reported by the ATIS is 180/30kt. VAR is 13°E. Calculate the cross wind component?

22 kt

Given: TAS = 130 kt, Track (T) = 003°, W/V = 190/40kt. Calculate the HDG (°T) and GS?

001 - 170 kt

The equivalent of 70 m/sec is approximately:

136 kt .......................... (70 x 3.28) = feet per second (70 x 3.28) ÷ 6,080 = nm per second ((70 x 3.28) ÷ 6,080) x 60 = nm per minute ((70 x 3.28) ÷ 6,080) x 60 x 60 = nm per hour (kts) = 135.95 kts

Given: TAS = 465 kt, HDG (T) = 124°, W/V = 170/80kt. Calculate the drift and GS?

8L - 415 kt

Given: True course A to B = 250°, Distance A to B = 315 NM, TAS = 450 kt. W/V = 200°/60kt. ETD A = 0650 UTC. What is the ETA at B?

0736 UTC

Given: Pressure Altitude 29000 FT, OAT -55°C. Calculate the Density Altitude?

27500 FT ................................... FL290 ISA =+15º - (2º x 29) Pressure altitude = 29,000 ft = -43ºC Diff = 120 ft x -12ºC = -1,440 ft OAT = -55ºC Density Altitude = 27,560 ft ISA deviation = -12ºC

Given: ``` Compass heading 242° True track 245° Variation 3°E Drift 5°R Calculate the deviation: ```

5°W

Given: TAS = 440 kt, HDG (T) = 349°, W/V = 040/40kt. Calculate the drift and GS?

4L - 415 kt

Given: Magnetic track = 315 º, HDG = 301 º(M), VAR = 5ºW, TAS = 225 kt, The aircraft flies 50 NM in 12 MIN. Calculate the W/V(°T)?

190 º/63 kt

Given the following: Magnetic heading 060° Magnetic variation 8°W Drift angle 4° right What is the true track?

``` 056° ....................... Heading: 060ºM Variation: 8ºW Heading: 052ºT Drift: 4º Right (Stbd) Track: 056ºT ```

Given: ``` True track 352° Variation 11°W Deviation -5° Drift 10°R Calculate the Compass heading: ```

358°

Given: TAS = 472 kt, True HDG = 005°, W/V = 110°(T)/50kt. Calculate the drift angle and GS.

6°L/490 kt

730 FT/MIN equals:

3.7 m/sec ............................ 730/60 = 12.1667 ft/second. 12.1667/3.28 = 3.7 m/sec.

Given: Groundspeed 540 kts Distance to go 72 NM What is the time to go?

8 min ........................ To travel 72 NM at a speed of 540 kt will take 8 min Background Information Simply, divide distance by speed to find time 72 nm ÷ 540 kt = 0.1333 hr Multiply by 60 to convert into minutes 0.1333 x 60 = 8 min

If 146 imperial gallons of fuel allows an endurance of 4 hours 26 minutes, what is the corresponding fuel flow?

39.5 US Gal/hr

Given: Course 040°T TAS 120 kts Wind speed 30 kts From which direction will the wind give the greatest amount of drift?

240°T

Given: TAS = 125 kt, True HDG = 355°, W/V = 320°(T)/30kt. Calculate the true track and GS?

005 - 102 kt

An aircraft is flying from SALCO to BERRY HEAD on Magnetic Track 007º, TAS 445 kt. The wind is 050º(T)/40 kt. Variation 5ºW, deviation +2º At 1000 UTC the RB of locator PY is 311º. At 1003 UTC the RB of locator PY is 266º. Calculate the distance of the aircraft from locator PY at 1003 UTC.

21 NM

Given: TAS 487kt, FL 330, Temperature ISA + 15.Calculate the MACH Number?

0.81 ............................... ISA FL330 = +15º - (2º x 33) = +15º - 66º = -51ºC deviation = +15ºC, OAT = -36ºC. In the AIRSPEED window put the temperature (-36ºC) next to the Mach No Index, go to the TAS (487kts) on the outer main scale and read the Mach No next to it; 0.81.

Given: GS = 510 kt. Distance A to B = 43 NM. What is the time (MIN) from A to B?

An aircraft travels 100 statute miles in 20 minutes, how long does it take to travel 215 NM?

50 mins

Given: Course 040°(T), TAS is 120 kt, Wind speed 30 kt. Maximum drift angle will be obtained for a wind direction of:

130° ......................... You will have maximum drift when there is a direct cross-wind. i.e. the wind is at 90º to your track.

Given: Magnetic heading = 255°, VAR = 40°W, GS = 375 kt, W/V = 235°(T) / 120 kt, Calculate the drift angle?

6° left .................................. Magnetic Heading = 255°M Magnetic Variation =40°W (Variation West Magnetic Best) True Heading = 255°M - 40°W = 215°T On CRP-5 Wind Side: Place Centre Dot over any convenient TAS, e.g. 500 kts Place Wind Direction = 235°(T) under True Heading Index Count down 120 kts and make a Wind Mark (over 380 kts) Place True Heading = 215°T under True Heading Index Note Wind Mark has moved to 6° Left indicating a Ground Speed of 390 kts Slide Wind Computer down to place Wind Mark over 375 kts Note Drift = 6° Left

Given: FL 290 CAS: 324 kts OAT: -46ºC What is the TAS?

487 kts

Given: FL 350, Mach 0.80, OAT -55°C. Calculate the values for TAS and local speed of sound (LSS)?

461 kt , LSS 576 kt

An aircraft is on final approach to land at an aerodrome (elevation 700 ft amsl) along a 6% descent profile. What is the aircraft's altitude, amsl, when it is 4 NM from the threshold?

2210 (ft amsl +/- 10) ............................................ Make sure all the units are the same, so: 4 NM x 6080 ft = 24 320 ft Align the red arrow (10) on the inner scale with 6 on the outer scale. Look up 24 320 ft on the inner scale and read 1460 ft on the outer scale. As the distance is measured from the threshold, add 50 ft for height ago at the threshold. As the aerodrome has an elevation of 700 ft, add 700 ft for height amsl. 1460 + 50 + 700 = 2210 ft amsl

An aircraft is flying at FL150, with an outside air temperature of -30°, above an airport where the elevation is 1660 ft and the QNH is 993 hPa. Calculate the true altitude. (Assume 30 ft = 1 hPa)

13 660 ft .................................... It is important to remember that calculation of a true altitude from a pressure altitude requires a two step approach:

How long will it take to fly 5NM at a groundspeed of 269Kts ?

1 min 07 secs

Given: True track 180°, Drift 8°R, Compass heading 195°, Deviation -2°, Calculate the variation?

21°W ........................................ C D M V T Course (track) 203º -2º(W) 201º 21ºW 180º Drift 8ºR (Stbd) Heading 195º -2º(W) 193º 21ºW 172º

Given: True HDG = 145°, TAS = 240 kt, Track (T) = 150°, GS = 210 kt. Calculate the W/V?

115/35kt

Given: ``` True HDG 035° TAS 245 kt Track (T) 046° GS 220 kts Calculate the W/V. ```

340/50 kts

Given: True heading = 310°, TAS = 200 kt, GS = 176 kt, Drift angle 7° right. Calculate the W/V?

270° / 33 kt

Given: TAS = 465 kt, Track (T) = 007°, W/V = 300/80kt. Calculate the HDG (°T) and GS?

358 - 428 kt

Given: TAS = 270 kt, Track (T) = 260°, W/V = 275/30kt. Calculate the HDG (°T) and GS?

262 - 241 kt

An aircraft must fly 2000 ft above an obstacle of which the elevation is 13 600 ft. The QNH at the nearest airfield is 991 hPa, the elevation is 1500 ft and the temperature is - 20°C. Calculate the minimum altitude required.

17 400 ft

Given: ``` True Track 300° Drift 8°R Variation 10°W Deviation -4° Calculate the compass heading? ```

306° .......................... C D M V T Course (track) 314º -4º(W) 310º 10ºW 300º Drift 8ºR (Stbd) Heading 306º -4º(W) 302º 10ºW 292º

Given: True Heading = 090°, TAS = 200 kt, W/V = 220° / 30 kt. Calculate the GS?

220 kt

Given: TAS = 132 kt, True HDG = 257°, W/V = 095°(T)/35kt. Calculate the drift angle and GS?

4°R - 165 kt

Given: TAS = 270 kt, True HDG = 270°, Actual wind 205°(T)/30kt. Calculate the drift angle and GS?

6R - 259kt

Given: TAS = 140 kt, True HDG = 302°, W/V = 045°(T)/45kt. Calculate the drift angle and GS?

16°L - 156 kt

Given: TAS = 230 kt, HDG (T) = 250°, W/V = 205/10kt. Calculate the drift and GS?

2R - 223 kt

Given: TAS = 235 kt, HDG (T) = 076°, W/V = 040/40kt. Calculate the drift angle and GS?

7R - 204 kt

Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL. What is the endurance?

4 HR 32 MIN

An aircraft has to fly over a mountain ridge. The highest obstacle, indicated in the navigation chart, has an elevation of 9 800 ft. The QNH, given by a meteorological station at an elevation of 6200ft, is 1022hPa. The OAT = ISA+5ºC. Calculate the approximate indicated altitude to obtain a clearance of 2000 ft.

Between 11 500 ft and 11 700 ft

Given: Aircraft at FL 150 overhead an airport. Elevation of airport 720 FT. QNH is 1003 hPa. OAT at FL150 -5°C. What is the true altitude of the aircraft? (Assume 1 hPa = 27 FT)

15 300 FT

Given: TAS = 225 kt, HDG (°T) = 123°, W/V = 090/60kt. Calculate the Track (°T) and GS?

134 - 178 kt

Given: Hdg 265°, TAS 290 kt, W/V 210°/35 kt. Calculate Track and Groundspeed.

271° and 272 kt

265 US-GAL equals? (Specific gravity 0.80)

803 kg ................................. Put fuel volume (265USG) on inner main scale next to U.S.gal. index. Go to 0.8 on kgs. Sp.G scale and read weight on inner main scale (803kgs).

Calibrated Airspeed (CAS) is Indicated Airspeed (IAS) corrected for:

instrument error and position error.

An aircraft has to fly over a mountain ridge. The highest obstacle, indicated in the navigation chart, has an elevation of 9 800 ft. The QNH, given by a meteorological station at an elevation of 6 200 ft, is 1022 hPa. The OAT = ISA+5ºC. What is the approximate indicated altitude to obtain a terrain clearance of 2000 ft if your altimeter is set to QNH and, if your altimeter is set to 1013 hPa?

Between 11 500 ft and 11 700 ft

A useful method of a pilot resolving, during a visual flight, any uncertainty in the aircraft's position is to maintain visual contact with the ground and:

set heading towards a line feature such as a coastline, motorway, river or railway

04 Dead Reckoning Navigation Flashcards

ATPL GEN NAV