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At 10 15 the reading from a vor/dme station is 211°/ 90nm at 10 20 the reading ? [ Multiple protocol ]

Question 175-1 : 110°/70kt 100°/60kt 120°/50kt 110°/40kt

Admin .200° ch + 1°e 31°w = 170° th .211° mc 31°w = 180° tc .from 10 15 to 10 20 5 minutes has passed and 30 nm have been flown .5 min > 0 0833h 30 / 0 0833 = 360 kt gs .390 kt 360 kt = 30 kt hw component .now use your flight computer with 390 kt tas 180° tc 30 kt hw component and 10°crab angle to get 110°/70kt exemple 275 110°/70kt.110°/70kt.

An aircraft is departing from an airport which has an elevation of 2000 ft and ?

Question 175-2 : 10 3 nm 11 1 nm 13 3 nm 15 4 nm

Admin .2000 ft is at qnh 1003 hpa at 1013 hpa it is 2300 ft .to reach fl100 you must climb 7700 ft 10000 2300 .rate of climb is 1000 ft/min .7700/1000 = 7 7 min.at a ground speed of 80 kt it will take .7 7 x 80/60 = 10 26 nm exemple 279 10.3 nm.10.3 nm.

Two consecutive waypoints of a flight plan are stornoway vordme n58°12 4' ?

Question 175-3 : 11 36 11 34 11 38 11 33

Admin .distance stornoway to glasgow = 151 nm.distance stornoway to ronar = 44 nm.11 21 11 15 = 6 min.6 min = 44 nm so 60 min = 440 kt nm .151 44 = 107 nm.107/440 = 0 243h 0 243 x 60 = 14 6 min.11 21 + 14 6 min = 11 35 36 sec exemple 283 11:3611:36

An aircraft at fl360 is required to descent to fl120 .the aircraft should reach ?

Question 175-4 : 124 nm 88 nm 236 nm 166 nm

Admin .24000 ft to lose with 2000 ft/min this means descending 24000 ft in 12 min .the plane is flying 7 nm/min 12x7 84 nm .the plane needs 84 nm to reach fl120 .it also need to be leveled 40 nm before the next waypoint .that means we should start the descent 84 + 40 = 124 nm before next waypont exemple 287 124 nm.124 nm.

The distance between a and b is 90 nm at a distance of 15 nm from a the ?

Question 175-5 : 19° 16° 3° 21°

Admin .tke = distance off track x 60 / distance along track.tke = 4 nm x 60 / 15 nm.tke = 16°.to join back on our track .tke = distance off track x 60 / distance to go.tke = 4 nm x 60 /75 nm 90 nm 15 nm = 75 nm .tke = 3°.correction angle 16° + 3° = 19° to the left as we are right off the course exemple 291 19°.19°.

After 15 minutes of flying with the planned tas and true heading the aircraft ?

Question 175-6 : 292° 258° 287° 280°

Admin .draw the exercice . 1798.the dead reckoning position was at 15 min from a with a gs of 130 kt . 130/60 x 15 = 32 5 nm from a .the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° . it's the drift to applied in order to correct the wind .track error angle to join b from our current position = 3 nm x 60 / 15 nm = 12° .to reach destination b from this position the correction angle on the heading should be 5° + 12° = 17° .current heading is 275° new heading is 275° + 17° = 292° exemple 295 292°.292°.

An aircraft is flying from salco to berry head on magnetic track 007° tas 445 ?

Question 175-7 : 272° t 268° t 277° t 275° t

. /com en/com061 635 jpg.calculate the drift between our true track 002° and the true wind 050°/40 kt with your computer the drift is 4° left you have to apply a 4°right wind angle correction .true heading + relative bearing = true bearing of locator from the aircraft.006° + 266° = 272° exemple 299 272° (t).272° (t).

An aircraft is departing from an airport which has an elevation of 2000 ft and ?

Question 175-8 : 3 6 nm 4 4 nm 4 0 nm 5 4 nm

Admin .2000 ft is at qnh 1003 hpa at 1013 hpa it is 2300 ft .to reach fl050 you must climb 2700 ft 5000 2300 .rate of climb is 1000 ft/min .2700/1000 = 2 7 min.at a ground speed of 80 kt it will take . 2 7 x 80/60 = 3 6 nm exemple 303 3.6 nm.3.6 nm.

During approach the following data are obtained .dme 12 0 nm altitude 3000 ?

Question 175-9 : 570 ft/min 600 ft/min 730 ft/min 700 ft/min

Admin .12 nm 9 8 nm = 2 2 nm.3000 ft 2400 ft = 600 ft.2 2 nm / 125 kt = 0 0176 h >1 056 min.600 ft / 1 056 = 568 ft/min exemple 307 570 ft/min.570 ft/min.

The distance between a and b is 90 nm at a distance of 75 nm from a the ?

Question 175-10 : 3°r 6°r 19°r 22°r

Admin .use the one in sixty rule .track error angle from a = distance off track x 60 / distance along track.track error angle from a = 4 nm x 60 / 75 nm.track error angle from a = 3°r exemple 311 3°r.3°r.

The true course according to the flight log is 270° the forecast wind is 045° ?

Question 175-11 : 5°l 6°r 2°l 3°r

Admin . 1798.with forecasted wind our ground speed is 130 kt .at 130 kt and 15 minutes of flight we will be at 32 5 nm from a .but the question states 2 5 nm ahead of the dead reckoning position so we are at 35 nm from a .use the one in sixty rule .track error angle from a = 3 nm x 60 / 35 nm = 5° exemple 315 5°l.5°l.

An aircraft flies from waypoint 7 63°00'n 073°00'w to waypoint 8 62°00'n ?

Question 175-12 : 4 7 nm right 8 8 nm right 8 8 nm left 4 7 nm left

Admin .1° longitude at equator = 60 nm.1° long at 60°lat = 30 nm.10' off track is 5 nm 10' = 1/6 from 1h so 1/6 from 30 nm is 5 nm 30 / 6 .as we are heading along meridian from 63°n to 62°n out true course is 180° and as we have ended up at 73°10' this is right of the track so 5 nm right .mathematically .distance nm = chlong in minutes * coslat.distance = 10 x cos62°.distance = 4 7 nm exemple 319 4,7 nm right.4,7 nm right.

An aircraft is departing from an airport which has an elevation of 2000 ft and ?

Question 175-13 : 7 2 nm 8 8 nm 10 8 nm 6 6 nm

Admin .1013 hpa 1003 hpa = 10 hpa > 300 ft 10 x 30 ft .2000 ft + 300 ft = 2300 ft.5000 ft 2300 ft = 2700 ft 2700 ft to climb .2700 ft / 500 ft/min = 5 4 min > 0 09h 5 4 / 60 .gs = 80 kt 100 kt tas 20 headwind component .80 kt x 0 09h = 7 2 nm exemple 323 7.2 nm.7.2 nm.

You are departing from an airport which has an elevation of 1500 ft the qnh is ?

Question 175-14 : 800 ft/min 870 ft/min 730 ft/min 530 ft/min

Admin .7500 1500 = 6000 ft.6000 / 7 5 = 800 ft/min exemple 327 800 ft/min.800 ft/min.

An aircraft is flying at fl200 .the qnh given by a meteorological station at an ?

Question 175-15 : 10 500 ft 9 200 ft 11 800 ft 20 200 ft

Admin .find the qnh altitude 1013 998 2 = 14 8 x 27 = 400 ft.altitude is 19600ft qnh .with aviat 617 computer .against altitude pressure = 20 put °c oat = 40 .then read in the inner circle the altitude 19600 the.on the outer circle 18400 true altitude .18400 8000ft = 10400ft = approximate clearance over the obstacle .for information 061 general navigation learning objectives states for questions involving height calculation 30 ft/hpa is to be used unless another figure is specified in the question exemple 331 10 500 ft.10 500 ft.

The qnh given by a station at 2500 ft is 980hpa .the elevation of the highest ?

Question 175-16 : 10 400 ft 10 000 ft 11 200 ft 9 700 ft

Admin .we need to be at 10000 ft to avoid the obstacle by 2000 ft .temperature correction formula 4° x 10 x 10° = 400 ft.the altimeter over reading in cold air and if we flew exactly at 10000 ft indicated our true altitude would be 9600 ft .we need to cruise at 10000 + 400 = 10400 ft indicated in order to maintain a clearance of 2000 ft exemple 335 10 400 ft.10 400 ft.

An aircraft is departing from an airport which has an elevation of 2000 ft ?

Question 175-17 : 276 kt 289 kt 244 kt 331 kt

Admin . by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .temperature is 0°c at 2000 ft .approximately 4° at sea level we 'gain' 2°c per 1000 ft while descending .21000 ft x 2°/1000ft = 42°c.4°+ 42° = 38°c.temperature is around 38°c at fl210 .on computer in airspeed window set press alt '21' in front of coat °c ' 38°c' on the outer scale in front of cas 200 kt you can read tas 272 kt exemple 339 276 kt.276 kt.

During visual navigation in freezing conditions after heavy snowfall which of ?

Question 175-18 : A large river a country road a railway an electrical line

Admin .after heavy snowfall roads will not have been cleared by snow ploughs neither country road or railway a large river may freeze but you will always be able to distinguish its path exemple 343 A large river.A large river.

During a climb at a constant cas below the tropopause in standard conditions ?

Question 175-19 : Both tas and mach number will increase tas will decrease but mach number will increase both tas and mach number will decrease tas will increase and mach number will decrease

Admin .for those questions use the very simple 'ertm' diagram . 1037.the cas line is vertical because the question states climb at a constant calibrated airspeed cas . ertm for e as/ r as rectified air speed or cas / t as/ m ach exemple 347 Both tas and mach number will increase.Both tas and mach number will increase.

An aircraft is descending down a 12% slope whilst maintaining a gs of 540 kt ?

Question 175-20 : 6500 ft/min 650 ft/min 4500 ft/min 3900 ft/min

Admin .vertical speed = 12% gradient x 540 kt.vertical speed = 6480 ft/min exemple 351 6500 ft/min.6500 ft/min.

The departure airfield is at 2000 ft elevation temperature at the field is ?

Question 175-21 : Fl 200 with temperature 20°c fl 290 with temperature 40°c fl 100 with temperature 10°c fl 150 with temperature 0°c

Admin .by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .29000 2000 = 27000 ft.2/3 de 27000 = 18000 ft .18000 + 2000 = 20000 ft fl200 .température à 20000 ft = 20°c + 20 x 2°c = 20°c exemple 355 Fl 200 with temperature -20°c.Fl 200 with temperature -20°c.

The departure is from an airfield at 2000 ft elevation temperature at the field ?

Question 175-22 : 249 kt 230 kt 221 kt 180 kt

Admin .1 29000 2000= 27000ft.2 27000* 2/3 = 18000.3 18000+2000 = 20000 your average alt in climb patern .4 if oat at 2000 alt is +20 so at 20000 will be +20 reference figure 2*18 18 height explanation 2*18 because temp decrease 2 deg per 1000 ft . .5 align temp 16 with 22000 ft in th air speed window cr 3 or iwa 11092 and read oposit 180 kt your 249 tas at outer scale.this is only the way to solve tasks like this exemple 359 249 kt.249 kt.

Given .w/v at arrival aerodrome at 1000 ft amsl is 230°/15kt w/v at tod at fl ?

Question 175-23 : 163 kt 155 kt 180 kt 174 kt

Admin .at fl130 isa condition .ias=170kt => tas=206kt .we have w/v 280°/45kt => drift 12°l so mh=232° to get an average track of 220°.so gs= 178kt with computer.at 1000ft amsl isa condition .ias=170kt => tas=172kt .we have w/v 230°/15kt => drift 1°l so mh=221° to get an average track of 220°.so gs= 157kt with computer.as a result the average gs for the descent is 157+178 /2 = 167 5kt exemple 363 163 kt.163 kt.

Given .w/v at arrival aerodrome at msl is 200°/20kt w/v at tod at fl 100 is ?

Question 175-24 : 135 kt 145 kt 120 kt 150 kt

exemple 367 135 kt.135 kt.

An aircraft is cruising in fl180 and thereafter descends to ground level the ?

Question 175-25 : 270°/40 kt 270°/20 kt 270°/35 kt 280°/50 kt

Admin .by convention average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude .average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude exemple 371 270°/40 kt.270°/40 kt.

The distance between two waypoints is 150 nm to calculate compass heading the ?

Question 175-26 : 10 nm 15 nm 7 nm 20 nm

Admin .for each degree of error that you have at every 60 nm of travel you will be 1 nm off track .you have 150/60 2 5 nm off track for each degree of error .total error is 4° from 2°e to 2°w .4° x 2 5 nm = 10 nm .using goniometric functions .tan4° = / 150. = tan4° x 150. = 10 nm exemple 375 10 nm.10 nm.

True track 085°.groundspeed 180 kt.wind 290°/30kt.variation 4°e .the ?

Question 175-27 : 2 5° l 2 5° r 1° l 1° r

Admin .tke = distance off track x 60 / distance along track.tke = 1 5 nm x 60 / 36 nm = 2 5°.the aircraft has drifted to the left therefore tke is 2 5° left exemple 379 2.5° l.2.5° l.

With only a visual straight line as visual cue a canal for example this line ?

Question 175-28 : More or less perpendicular to our track more or less parallel to our track curved across our track oblique to our track

exemple 383 More or less perpendicular to our track.More or less perpendicular to our track.

Given .a descending aircraft flies in a straight line to a dme .dme 55 nm ?

Question 175-29 : 3 70% 4 10% 3 50% 3 90%

Admin .33000 30500 = 2500 ft.55 nm 43 9nm = 11 1 nm.11 1 nm = 67488 ft.2500 = 67488 x x.x= 2500 / 67488 = 0 0370 3 7% exemple 387 3.70%.3.70%.

The descent gradient of an aircraft with the following data is . 60 nm norths ?

Question 175-30 : 5 4% 6 4% 7 6% 4 5%

Admin .total ground distance is 60+10 = 70 nm .altitude difference is 35000 12000 = 23000 ft .gradient in % = altitude difference in ft x 100 / ground distance in ft..ground distance in feet 70 nm x 6080 ft = 425600 ft.gradient in % = 23000 x 100 / 425600 = 5 4% exemple 391 5.4%5.4%

The average tas climbing from 1500 ft to fl180 with a given temperature of isa ?

Question 175-31 : 283 kt 309 kt 261 kt 274 kt

Admin . by convention at the exam easa specification average tas used for climb problems is calculated at the altitude 2/3 of the cruising altitude .1032 1013 = 19 hpa.19 hpa x 30 ft = 570 ft .18000 + 570 = 18570 ft.2/3 of 18570 = 12380 ft .to convert cas to tas 1% for each 600 ft and 0 2% for each degree of isa deviation.tas = cas x 12380/600 + 0 2% x 15°c = 230 x 20% + 3% = 282 9 kt exemple 395 283 kt.283 kt.

An aircraft is turning on a final approach to intercept a 3° glide slope which ?

Question 175-32 : 1916 ft 700 ft 1220 ft 1290 ft

Admin .1 in 60 rule is a rule of thumb . 3° x 4 nm /60 = 0 2 nm .0 2 nm x 6080 ft = 1216 ft .add 700 ft since we are looking for an altitude = 1216 + 700 = 1916 ft exemple 399 1916 ft.1916 ft.

Given .tas 220 kt.cruising level fl180.track during climb 080°.wind at msl ?

Question 175-33 : 262 kt 259 kt 254 kt 273 kt

Admin .we have to use the wind at the altitude 2/3 of the cruising altitude .fl180 x 2/3 = fl120.wind changes by 30° from ground to fl180 an speed increases from 30 kt .mean wind at fl120 is 260°+20° and 25kt+20kt = 280°/45kt .with your nav computer you will find 262 kt exemple 403 262 kt.262 kt.

An aircraft climbs from ground level to fl180 the following wind information is ?

Question 175-34 : 280°/50 kt 285°/55 kt 290°/55 kt 270°/30 kt

Admin .by convention average wind velocity used for climb problems is wind velocity at the altitude 2/3 of the cruising altitude .average wind velocity used for descent problems is wind velocity at the altitude 1/2 of the descent altitude exemple 407 280°/50 kt.280°/50 kt.

An aircraft descends from fl240 to fl040 for the final approach .cas = 220 ?

Question 175-35 : 273 kt 244 kt 254 kt 259 kt

Admin .at the exam average tas used for descent problems is calculated at the altitude 1/2 of the descent altitude.at fl120 isa temperature = 15°c 2°c x 12 = 9°c .oat is isa +10°c thus oat is +1°c at fl120 .on the computer in airspeed window put +1ºc next to fl120 go to cas 220 kt on inner scale and read tas on outer scale 273 kt exemple 411 273 kt.273 kt.

When flying a visual navigation exercise in controlled airspace it is confirmed ?

Question 175-36 : 034° 038° 030° 042°

Question issue de ae this question asks about the on in sixty rule this rule states that after 60 nm a drift of 1° corresponds to an off track distance of 1 nm 2° = 2 nm 3° = 3 nm etc the question mentions an off track distance of 2 nm and if you cut the flight distance into half 30 nm instead of 60 nm angle and off track distance must double 30 nm => 4° => 4 nm the question states that the controller wants us to regain our original track after 30 nm so the correction angle should be 4° + 30° = 34° exemple 415 034°.034°.

An aircraft in cruise at fl120 is cleared to descend to 3000 ft .the distance ?

Question 175-37 : 6% 6 3% 4 7% 5%

Admin .we have to descend 9000 ft.25 nm in ft is 25 nm x 6000 ft/nm = 151900 ft . 9000 / 150000 x 100 = 6% exemple 419 6%.6%.

Given .descent from 15000 ft to 3000 ft msl.glide path angle during descent ?

Question 175-38 : It decreases from 900 ft/min to 750 ft/min 900 ft/min during the whole descent 825 ft/min during the whole descent 750 ft/min during the whole descent

Admin .rate of descent 3° = ground speed kt x 10/2.at 15000 ft with a ground speed of 180 kt rate of descent = 180 x 10/2 = 900 ft/min.approaching 3000 ft with a decelerating speed to reach 150 kt rate of descent = 150 x 10/2 = 750 ft/min exemple 423 It decreases from 900 ft/min to 750 ft/min.It decreases from 900 ft/min to 750 ft/min.

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

Question 175-39 : Groundspeed kt x gradient ft/nm / 60 altitude difference ft x 100 / ground difference ft climb/descent angle ° x 100 / 60 arctg altitude difference ft / ground distance covered ft

Admin .we must know the groundspeed to calculate a climb/descent gradient .calculate rate of descent rod on a given glide path angle or gradient using the following rule of thumb formulae .rod ft/min = gp degrees x gs nm/min x 100.or.rod ft/min = gp per cent x gs kt ..calculate climb/descent gradient ft/nm per cent and degrees gs or vertical speed according to the following formula .vertical speed ft/min = gs kt x gradient ft/nm / 60 exemple 427 Groundspeed (kt) x gradient (ft/nm) / 60Groundspeed (kt) x gradient (ft/nm) / 60

The correct formula for climb/descent gradient in % is .gradient in % = ?

Question 175-40 : Vertical distance x 100 / ground distance height difference / altitude difference x 100 rate of climb or descent x ground speed arctg altitude difference / ground distance

Admin .estimate average climb/descent gradient per cent or glide path degrees according to the following rule of thumb .gradient in % = vertical distance ft / 60 / ground distance nm .or.gradient in % = vertical distance x 100 / ground distance .gradient in degrees = arctan altitude difference ft / ground distance ft .or.gradient in degrees = vertical distance ft / 100 / ground distance nm .n b these rules of thumb approximate 1 nm to 6 000 ft and are based on the 1 60 rule exemple 431 (vertical distance x 100) / ground distance.(vertical distance x 100) / ground distance.

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