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Question 183-1 : The equivalent of 70 m/sec is approximately ? [ Question security ]

136 kt

Admin .1 nm = 0 5 m/s.70 / 0 5 = 140 kt closest to 136 kt than 145 kt .if you want to find the exact answer .70 m/s x 3600 secondes = 252000 m/h.252000 m/h = 252 km/h.252 / 1 852 = 136 kt exemple 283 136 kt.

Question 183-2 : Given .runway direction 305° m surface w/v 260° m /30 kt .calculate the cross wind component ?

21 kt

.angle between the wind and the direction of the runway 305° 260° = 45° .crosswind = sine of the angle between the wind and the direction of the runway x windspeed.crosswind = sin45° x 30 kt = 21 2 kt exemple 287 21 kt.

Question 183-3 : 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 ?

6° right

Admin .use the one in sixty rule .for small angles for every 60 nm along 1 nm off track is equivalent to 1° deviation .4 nm deviation after having travelled 60 nm means there is 4° off track deviation caused by drift .in order to counteract drift you need to correct your heading by 4° .now to arrive at position b you have to recover those 4 nm deviation in the remaining 120 nm .2 nm per 60 nm > 2° alteration .4° + 2° = 6° .it's a right heading alteration since we are drifting left of the intended track exemple 291 6° right.

Question 183-4 : A flight is to be made from 'a' 49°s 180°e/w to 'b' 58°s 180°e/w the distance in kilometres from 'a' to 'b' is approximately ?

1000 km

Admin .you are travelling south along the greenwich anti meridian from 49°s to 58°s which is a 9° change of latitude .9° x 60 nm = 540 nm.540 nm x 1 852 = 1000 km exemple 295 1000 km.

Question 183-5 : 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' ?

8° right

Admin .use the one in sixty rule .track error angle from a = 3 nm x 60 / 30 nm = 6°. 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 / 90 nm = 2°.to reach destination b from this position the correction angle on the heading should be 6° + 2° = 8° exemple 299 8° right.

Question 183-6 : An aircraft was over 'a' at 1435 hours flying direct to 'b' given .distance 'a' to 'b' 2900 nm.true airspeed 470 kt.mean wind component 'out' +55 kt.mean wind component 'back' 75 kt .the eta for reaching the point of equal time pet between 'a' and 'b' is ?

1657

Ground speed out 470 + 55 = 525 kt.ground speed home 470 75 = 395 kt.pet = d x gsh / gso + gsh .pet = 2900 x 395 / 525 + 395 = 1245 nm .1245 nm / 525 kt = 2 37h.2 37 x 60 minutes = 142 minutes 2h42minutes .14h35 + 2h42 = 16h57 exemple 303 1657.

Question 183-7 : Given .distance 'a' to 'b' 2484 nm.groundspeed 'out' 420 kt.groundspeed 'back' 500 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?

193 minutes

Admin .ground speed out 420 kt.ground speed home 500 kt.pet = distance x gsh / gso + gsh .pet = 2484 x 500 / 420 + 500 = 1350 nm .1350 nm / 420 kt = 3 21 h.3 21 h x 60 minutes = 193 minutes exemple 307 193 minutes.

Question 183-8 : Given .distance 'a' to 'b' 2484 nm.ground speed out 420 kt.ground speed home 500 kt.safe endurance 08 h 30 minutes .the distance from 'a' to the point of safe return psr is ?

1940 nm

Admin .point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 420 kt.ground speed home = 500 kt.point of safe return psr = 8 5 x 500 / 420 + 500 .point of safe return psr = 4250 / 920.point of safe return psr = 4 62 h.distance of the psr from the departure point at a speed of 420 kt .4 62 h x 420 = 1940 nm exemple 311 1940 nm.

Question 183-9 : An aircraft was over 'q' at 1320 hours flying direct to 'r' given .distance 'q' to 'r' 3016 nm.true airspeed 480 kt.mean wind component 'out' 90 kt.mean wind component 'back' +75 kt .endurance 10 h.what is the eta at the point of equal time pet ?

1752

Admin .ground speed out 480 90 = 390 kt.ground speed home 480 + 75 = 555 kt.pet = d x gsh / gso + gsh .pet = 3016 x 555 / 390 + 555 = 1771 nm .1771 nm / 390 kt = 4 54 h.4 54 x 60 minutes = 272 minutes 4h32minutes .13h40 + 4h32 = 17h52 exemple 315 1752.

Question 183-10 : Given .distance 'a' to 'b' 1973 nm.groundspeed 'out' 430 kt.groundspeed 'back' 385 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?

130 minutes

Admin .ground speed out 430 kt.ground speed home 385 kt.pet = distance x gsh / gso + gsh .pet = 1973 x 385 / 430 + 385 = 932 nm .932 nm / 430 kt = 2 16 h.2 16 h x 60 minutes = 129 6 minutes exemple 319 130 minutes.

Question 183-11 : Given .distance 'a' to 'b' 2346 nm.ground speed out 365 kt.ground speed back 480 kt.safe endurance 8 h 30 minutes.the time from 'a' to the point of safe return is ?

290 minutes

Admin .point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 365 kt.ground speed home = 480 kt.point of safe return psr = 8 5 x 480 / 365 + 480 .point of safe return psr = 4080 / 845.point of safe return psr = 4 83 h.the time from 'a' to the point of safe return is 4 83 x 60 minutes = 290 minutes exemple 323 290 minutes.

Question 183-12 : Given .distance 'a' to 'b' 3623 nm.groundspeed 'out' 370 kt.groundspeed 'back' 300 kt.the time from 'a' to the point of equal time pet between 'a' and 'b' is ?

263 minutes

Admin .ground speed out 370 kt.ground speed home 300 kt.pet = distance x gsh / gso + gsh .pet = 3623 x 300 / 370 + 300 = 1622 nm .1622 nm / 370 kt = 4 38h.4 38 h x 60 minutes = 262 8 minutes exemple 327 263 minutes.

Question 183-13 : Given .magnetic track = 075°.magnetic heading = 066°.variation = 11°e.tas = 275 kt .aircraft flies 48 nm in 10 min .calculate the true wind ?

335°/45 kt

Admin .48 nm in 10 minutes > gs = 288 kt.on the computer .under index set true heading 077° and in center dot tas 275 kt .our true track is 086° so drift is 9° right .mark the point where the 9° right drift crosses the ground speed 288 kt .on the rotating scale you can read a wind of 335°/45kt . 1742 exemple 331 335°/45 kt.

Question 183-14 : Given .magnetic track = 210°.magnetic heading = 215°.variation = 15°e.tas = 360 kt.aircraft flies 64 nm in 12 minutes calculate the true w/v ?

265°/50 kt

Admin .64 nm in 12 minutes > gs = 320 kt.on the computer .under index set true heading 230° and over center dot tas 360 kt .our true track is 225° so drift is 5° left .mark the point where the 5° left drift crosses the ground speed 320 kt .on the rotating scale you can read a wind of 265°/50 kt . 2531 exemple 335 265°/50 kt.

Question 183-15 : Given .aircraft at fl150 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

Admin .at fl150 isa = 15°c 2°c x 15 = 15°c.oat is 5°c we are in air mass 10°c warmer than isa .changing subscale from 1013 hpa to 1003 hpa means that indicated altitude on the altimeter will decrease by 270 ft 10 hpa .15000 270 = 14730 ft .temperature correction .4 x 15 x 10 = 600 ft .14730 + 600 = 15330 ft 'plus' 600 ft because air mass is warmer than isa .maxscail .how do you find 4x15x10=600ft is it a formula . .yes this is the rule of thumb formula called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .this is an official formula given by easa for altitude/height calculations at the exam exemple 339 15 300 ft.

Question 183-16 : An aircraft takes off from the aerodrome of brioude altitude 1483 ft qfe = 963 hpa temperature = 32°c five minutes later passing 5000 ft on qfe the second altimeter set on 1013 hpa will indicate approximately ?

6 500 ft

Admin .difference between 963 hpa and 1013 hpa is 50 hpa 50 hpa x 30 ft/hpa = 1500 ft .5000 + 1500 = 6500 ft .your altimeter indicates your pressure altitude not your true altitude this is the reason why we do not correct the temperature we only want to know the reading of the altimeter .note 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 343 6 500 ft.

Question 183-17 : Given .distance a to b is 360 nm .wind component a b is 15 kt wind component b a is +15 kt tas is 180 kt .what is the distance from the equal time point to b ?

165 nm

Babar350 .e*o*h / o+h ..e endurance.o gs outnbound.h gs inbound..e = distance * o..eoh/ o+h = 2 18*165*195/ 360 is giving distance from etp from a 195 nm .the question is etp from b so 360 195 = 165 nm exemple 347 165 nm.

Question 183-18 : 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 ?

403 kt

Admin .first step find the true heading .237° 5° deviation on this heading = magnetic heading 232°.232° 19° variation west = true heading 213°.put 360 kt in center dot under index set true heading 213° on the rotating scale set wind 330°/80kt . 2532.you read a ground speed of 403 kt .the answer will be more accurate on a real computer such as the aviat 617 for example exemple 351 403 kt.

Question 183-19 : 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°

Babar350 .maximum drift 60/95 x 20 = 12 63.actual drift max drift x sin 350 322 = 12 63 x sin 28 = 5 92°.mainting centre line is 322°+5 92 = 328° exemple 355 328°.

Question 183-20 : An aircraft takes off from an airport 2 hours before sunset the pilot flies a track of 090° t w/v 130°/ 20 kt tas 100 kt in order to return to the point of departure before sunset the furthest distance which may be travelled is ?

97 nm

Admin .resolve this question as a point of safe return question .point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .outbound gs on the computer .when you start with a track under index you must first apply drift before read the ground speed . 1944.drift=8° left .set heading 098° 090° + 8° under index read outbound gs 86 kt .proceed on the same way to finde homeward gs 116 kt ..point of safe return psr = 2h x 116 / 86 + 116 .point of safe return psr = 1 148 h .1 148 x 86 kt = 98 7 nm exemple 359 97 nm.

Question 183-21 : From the departure point the distance to the point of equal time is ?

Inversely proportional to the sum of ground speed out and ground speed back

.distance to the point of equal time = d x gsh/ gso + gsh .where .d = distance between departure and arrival .gso = ground speed out .gsh = ground speed home exemple 363 Inversely proportional to the sum of ground speed out and ground speed back.

Question 183-22 : 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

Admin .at flight level 55 temperature in standard atmosphere is .15° 2° x 5 5 = 4°c.on the computer in airspeed window put 4ºc next to fl55 go to cas 120 kt on inner scale and read tas on outer scale 131 kt .now centre dot on tas 131 kt under true index put 190° wind direction and mark wind speed 30 kt below at 101 kt .then rotate to put true track 060° 045° + variation east under true index drift is 9° left it means that on this heading 060° the aircraft would be on a true track of 051° .rotate to put true track 060° under drift and note that drift has changed to 10° left as we turn the drift changes and on a heading of 070° we will have a 10° left drift .magnetic heading = 070° minus variation east = 055° .ground speed is under wind mark 147 kt exemple 367 055° and 147 kt.

Question 183-23 : Given .airport elevation is 1000 ft .qnh is 988 hpa .what is the approximate airport pressure altitude assume 1 hpa = 30 ft ?

1750 ft

Admin .1013 988 = 25 hpa.25 hpa x 30 ft = 750 ft.1000 + 750 = 1750 ft exemple 371 1750 ft.

Question 183-24 : Given .true altitude 9000 ft.oat 32°c.cas 200 kt.the true air speed tas is ?

220 kt

Admin .you have first to convert true altitude in pressure altitude .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .deviation from isa = 15°c 2 x 9 = 3°c. 3°c to 32°c = 29°c we are in isa 29°c .4% x 9 x 29 = 1044 ft.pressure altitude = 9000 + 1044 = 10044 ft 10000 ft .now in the airspeed window set 29°c in front of 10000 ft pressure alitude . 2523.read cas 200 kt on inner scale and corresponding tas 220 kt on outer scale exemple 375 220 kt.

Question 183-25 : Given .course 040° t tas is 120 kt wind speed 30 kt .maximum drift angle will be obtained for a wind direction of ?

130°

Admin .maximum drift is obtained when the wind is at a right angle from our course .040° + 90° = 130° .or.040° 90° = 310° exemple 379 130°.

Question 183-26 : Given .cas 120 kt fl 80 oat +20°c .what is the tas ?

141 kt

Admin .in airspeed window set tempertaure +20° in front of pressure altitude fl80.on the outside scale in front of cas 120 kt you read tas=141 kt . 2533.cas = ias + correction for position and instrument error .instrument error is an eventual error of the airspeed indicator itself .position error is the error produced from the airflow around the static port wherever that is located and around the probe .at low speed we consider cas = ias exemple 383 141 kt.

Question 183-27 : 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°

.draw the situation . /com en/com061 399 jpg.initial true track from a to b will be 360° 35° = 325° .variation 3°e 325° 3° = 322° . variation west magnetic best variation east magnetic least exemple 387 322°.

Question 183-28 : 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 wind °t ?

160°/50 kt

.first step search for the ground speed .14 nm in 6 minutes = 14/6 x60 = 140 kt .next step calculate drift . /com en/com061 484a jpg.drift x is 18° left 100° 082° .put 160 kt tas on center dot under true index set 100° heading . /com en/com061 484 jpg.where the ground speed crosses the 18° left drift line you read on the red scale a wind coming from 160° for 50 kt exemple 391 160°/50 kt.

Question 183-29 : Given .m 0 80 oat 50°c fl 330 gs 490 kt variation 20°w magnetic heading 140° drift is 11° right .calculate the true wind ?

020°/95 kt

Admin . 2064.tas is 462 kt .magnetic heading 140° variation 20°w = true heading 120° . 2061.read wind direction and force on the rotative scale exemple 395 020°/95 kt.

Question 183-30 : Given pressure altitude 29000 ft oat 55°c calculate the density altitude ?

27500 ft

. /com en/com061 486 jpg. exemple 399 27500 ft.

Question 183-31 : 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

Admin .in airspeed window put 30°c next to fl180 . 2056.in front of 150 kt on the inner scale read the tas 195 kt on the outer scale exemple 403 195 kt.

Question 183-32 : Calibrated airspeed cas is indicated airspeed ias corrected for ?

Instrument error and position error

exemple 407 Instrument error and position error.

Question 183-33 : An aircraft was over 'q' at 1320 hours flying direct to 'r' .given .distance 'q' to 'r' 3016 nm.true airspeed 480 kt.mean wind component out 90 kt.mean wind component back +75 kt.safe endurance 10 h 00 .the distance from 'q' to the point of safe return psr 'q' is ?

2290 nm

Point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 480 90 = 390 kt.ground speed home = 480 + 75 = 555 kt.point of safe return psr = 10 x 555 / 390 + 555 .point of safe return psr = 5555 / 945.point of safe return psr = 5 878h.distance of the psr from the departure point at a speed of 390 kt .5 878h x 390 = 2292 nm exemple 411 2290 nm.

Question 183-34 : 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

Admin .you have to turn your altimeter subscale setting knob counterclockwise from 1013 to 993 .indicated altitude will be decreased by 20 hpa x 30 ft = 600 ft .15000 600 = 14400 ft .now we must correct for temperature .outside temperature is 30°c at fl150 .isa at fl150 is 15°c 15 x 2°c = 15°c .we are in isa 15°c .you can use either the computer or with the following rule of thumb called the '4% rule' .the altitude/height changes by 4% for each 10°c temperature deviation from isa .an altimeter set to airport qnh will read correctly when on the ground at the airport irrespective of temperature .any temperature error therefore occurs due to non isa temperature in the layer of atmosphere between airport elevation and aircraft in flight .therefore .14400 1660 = 12740 ft.12740 x 0 04 x 1 5 = 764 ft .14400 764 = 13636 ft exemple 415 13 660 ft.

Question 183-35 : Given .true track 239°.true heading 229°.tas 555 kt.g/s 577 kt.calculate the wind velocity ?

130°/100 kt

Admin .229° to true index.555 kt to center bore.with a true headin of 229° and a true track of 239° we have 10° right drift .pencil mark the intersection of the 10° right drift with the ground speed arc 577 kt . 2053.read the wind speed and wind direction 130°/100 kt exemple 419 130°/100 kt.

Question 183-36 : Given .true track 245°.drift 5° right.variation 3°e.compass heading 242° .calculate the deviation ?

5°w

Admin . 2050.use this wonderful table for those questions exemple 423 5°w.

Question 183-37 : Given .true heading 090°.tas 180 kt.gs 180 kt.drift 5° right.the wind is ?

005° / 15 kt

Admin .set heading 090° undex index center dot on tas 180 kt . 2047.where right drift 5° crosses ground speed arc 180 kt read wind on the rotating scale 005°/15kt exemple 427 005° / 15 kt.

Question 183-38 : Given .magnetic heading = 255°.variation = 40°w.gs = 375 kt.w/v = 235° t / 120 kt.calculate the drift angle ?

6° left

Admin .the wind is coming in front of us our true air speed will be more than 375 kt .set wind index under true index and set 400 kt for example in center dot mark wind at 120 kt below the center dot on the rotating indicator. 2048.turn to put heading 215° 255° 40° under true index . 2046.shift the speed arc under the wind speed mark drift is 6° left and you also notice a tas of 490 kt . 2045 exemple 431 6° left.

Question 183-39 : Given .true track = 095° tas = 160 kt true heading = 087° gs = 130 kt .calculate the wind ?

057°/36 kt

exemple 435 057°/36 kt.

Question 183-40 : Given .true track 245°.drift 5° right.variation 3°e.compass heading 242° .calculate the magnetic heading ?

237°

Admin . 2043.use this wonderful table for those questions .note compass heading 242° is given for nothing exemple 439 237°.


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