Multiple > protocol : The distance between a and b is 90 nm at a distance of 15 nm from a the ?

Question 173-1 : 19° 16° 3° 21°

exemple 273 19°. — .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 19°.

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

Question 173-2 : 292° 258° 287° 280°

exemple 277 292°. — .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° 292°.

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

Question 173-3 : 272° t 268° t 277° t 275° t

exemple 281 272° (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° 272° (t).

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

Question 173-4 : 3 6 nm 4 4 nm 4 0 nm 5 4 nm

exemple 285 3.6 nm. — .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 3.6 nm.

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

Question 173-5 : 570 ft/min 600 ft/min 730 ft/min 700 ft/min

exemple 289 570 ft/min. — .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 570 ft/min.

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

Question 173-6 : 3°r 6°r 19°r 22°r

exemple 293 3°r. — .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 3°r.

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

Question 173-7 : 5°l 6°r 2°l 3°r

exemple 297 5°l. — . 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° 5°l.

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

Question 173-8 : 4 7 nm right 8 8 nm right 8 8 nm left 4 7 nm left

exemple 301 4,7 nm right. — .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 4,7 nm right.

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

Question 173-9 : 7 2 nm 8 8 nm 10 8 nm 6 6 nm

exemple 305 7.2 nm. — .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 7.2 nm.

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

Question 173-10 : 800 ft/min 870 ft/min 730 ft/min 530 ft/min

exemple 309 800 ft/min. — .7500 1500 = 6000 ft..6000 / 7 5 = 800 ft/min 800 ft/min.

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

Question 173-11 : 10 500 ft 9 200 ft 11 800 ft 20 200 ft

exemple 313 10 500 ft. — .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 10 500 ft.

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

Question 173-12 : 10 400 ft 10 000 ft 11 200 ft 9 700 ft

exemple 317 10 400 ft. — .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 10 400 ft.

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

Question 173-13 : 276 kt 289 kt 244 kt 331 kt

exemple 321 276 kt. — . 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 276 kt.

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

Question 173-14 : A large river a country road a railway an electrical line

exemple 325 a large river. — .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 a large river.

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

Question 173-15 : 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

exemple 329 both tas and mach number will increase. — .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 both tas and mach number will increase.

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

Question 173-16 : 6500 ft/min 650 ft/min 4500 ft/min 3900 ft/min

exemple 333 6500 ft/min. — .vertical speed = 12% gradient x 540 kt.vertical speed = 6480 ft/min 6500 ft/min.

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

Question 173-17 : Fl 200 with temperature 20°c fl 290 with temperature 40°c fl 100 with temperature 10°c fl 150 with temperature 0°c

exemple 337 fl 200 with temperature -20°c. — .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 fl 200 with temperature -20°c.

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

Question 173-18 : 249 kt 230 kt 221 kt 180 kt

exemple 341 249 kt. — .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 249 kt.

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

Question 173-19 : 163 kt 155 kt 180 kt 174 kt

exemple 345 163 kt. — .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 163 kt.

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

Question 173-20 : 135 kt 145 kt 120 kt 150 kt

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

Question 173-21 : 270°/40 kt 270°/20 kt 270°/35 kt 280°/50 kt

exemple 353 270°/40 kt. — .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 270°/40 kt.

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

Question 173-22 : 10 nm 15 nm 7 nm 20 nm

exemple 357 10 nm. — .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 10 nm.

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

Question 173-23 : 2 5° l 2 5° r 1° l 1° r

exemple 361 2.5° l. — .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 2.5° l.

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

Question 173-24 : More or less perpendicular to our track more or less parallel to our track curved across our track oblique to our track

exemple 365 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 173-25 : 3 70% 4 10% 3 50% 3 90%

exemple 369 3.70%. — .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% 3.70%.

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

Question 173-26 : 5 4% 6 4% 7 6% 4 5%

exemple 373 5.4% — .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% 5.4%

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

Question 173-27 : 283 kt 309 kt 261 kt 274 kt

exemple 377 283 kt. — . 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 283 kt.

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

Question 173-28 : 1916 ft 700 ft 1220 ft 1290 ft

exemple 381 1916 ft. — .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 1916 ft.

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

Question 173-29 : 262 kt 259 kt 254 kt 273 kt

exemple 385 262 kt. — .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 262 kt.

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

Question 173-30 : 280°/50 kt 285°/55 kt 290°/55 kt 270°/30 kt

exemple 389 280°/50 kt. — .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 280°/50 kt.

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

Question 173-31 : 273 kt 244 kt 254 kt 259 kt

exemple 393 273 kt. — .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 273 kt.

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

Question 173-32 : 034° 038° 030° 042°

exemple 397 034°. — 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° 034°.

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

Question 173-33 : 6% 6 3% 4 7% 5%

exemple 401 6%. — .we have to descend 9000 ft.25 nm in ft is 25 nm x 6000 ft/nm = 151900 ft. 9000 / 150000 x 100 = 6% 6%.

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

Question 173-34 : 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

exemple 405 it decreases from 900 ft/min to 750 ft/min. — .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 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 173-35 : 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

exemple 409 groundspeed (kt) x gradient (ft/nm) / 60 — .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 groundspeed (kt) x gradient (ft/nm) / 60

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

Question 173-36 : 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

exemple 413 (vertical distance x 100) / ground distance. — .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 (vertical distance x 100) / ground distance.

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

Question 173-37 : Vertical distance ft / 100 / ground distance nm vertical distance ft x 60 / ground distance nm rate of climb or descent x ground speed height difference / altitude difference x 100

exemple 417 (vertical distance (ft) / 100) / ground distance (nm)). — .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 (vertical distance (ft) / 100) / ground distance (nm)).

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

Question 173-38 : The wind at fl120 the wind at fl190 the average wind speed from ground to fl180 the average wind speed from ground to fl120

exemple 421 the wind at fl120. — the wind at fl120.

An aircraft is flying according the flight log at the annex after 15 minutes of ?

Question 173-39 : 078° 115 ° 107° 090°

exemple 425 078°. — .15 min / 60 = 0 25h..0 25 x 130 kt gs = 32 5 nm..32 5 nm + 2 5nm ahead of dr = 35 nm..50 nm dist 35 nm = 15 nm..tke = 3 x 60/35 and 3 x 60/15..tke = 5° and 12° ca = 5° + 12° = 17°..as we are south of 095° to get back we need to fly more to the north therefore 095 17° = 078° 078°.

The 'night effect' which causes loss of signal and fading resulting in bearing ?

Question 173-40 : Skywave distortion of the null position and is maximum at dawn and dusk interference from other transmissions and is maximum at dusk when east of the ndb static activity increasing at night particularly in the lower frequency band the effect of the aurora borealis

.navigation using an adf to track ndbs is subject to several common effects for 'night effect' radio waves reflected back by the ionosphere can cause signal strength fluctuations 30 to 60 nautical miles 54 to 108 km from the transmitter especially just before sunrise and just after sunset more common on frequencies above 350 khz

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