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Given distance 'a' to 'b' is 90 nm fix obtained 60 nm along and 4 nm to the ? [ Multiple protocol ]

Question 173-1 : 12° left 16° left 4° left 8° left

Admin .to calculate the heading change at an off course fix to directly reach the next waypoint use the one in sixty rule . 4/60 x 60 = 4°. 4/30 x 60 = 8°.8° + 4° = 12° and left because we are to the right of the target course exemple 273 12° left.12° left.

Complete line 1 of the 'flight navigation log'.positions 'a' to 'b' what is the ?

Question 173-2 : Hdg 268° eta 1114 utc hdg 282° eta 1128 utc hdg 282° eta 1114 utc hdg 268° eta 1128 utc

Admin . 2496 exemple 277 Hdg 268° - eta 1114 utc.Hdg 268° - eta 1114 utc.

Complete line 2 of the 'flight navigation log' positions 'c' to 'd' what is the ?

Question 173-3 : Hdg 193° eta 1239 utc hdg 188° eta 1229 utc hdg 193° eta 1249 utc hdg 183° eta 1159 utc

Admin . 2496 exemple 281 Hdg 193° - eta 1239 utc.Hdg 193° - eta 1239 utc.

Complete line 3 of the 'flight navigation log' positions 'e' to 'f' what is the ?

Question 173-4 : Hdg 105° eta 1205 utc hdg 095° eta 1155 utc hdg 106° eta 1215 utc hdg 115° eta 1145 utc

Admin . 2496 exemple 285 Hdg 105° - eta 1205 utc.Hdg 105° - eta 1205 utc.

Complete line 4 of the 'flight navigation log' positions 'g' to 'h' what is the ?

Question 173-5 : Hdg 344° eta 1336 utc hdg 354° eta 1326 utc hdg 334° eta 1336 utc hdg 344° eta 1303 utc

Admin . 2496 exemple 289 Hdg 344° - eta 1336 utc.Hdg 344° - eta 1336 utc.

Complete line 5 of the 'flight navigation log' positions 'j' to 'k' what is the ?

Question 173-6 : Hdg 337° eta 1422 utc hdg 320° eta 1412 utc hdg 337° eta 1322 utc hdg 320° eta 1432 utc

Admin . 2496 exemple 293 Hdg 337° - eta 1422 utc.Hdg 337° - eta 1422 utc.

Complete line 6 of the 'flight navigation log' positions 'l' to 'm' what is the ?

Question 173-7 : Hdg 075° eta 1502 utc hdg 064° eta 1449 utc hdg 075° eta 1452 utc hdg 070° eta 1459 utc

Admin . 2496.you have to find tas .set 55°c in the airspeed window next to mkt .read on the outer main scale in front of 0 84 a tas of 485 kt .below center dot set tas 485 kt under index set true course 070° with the rotative scale set the wind 020°/60kt read below 60 kt a 6° right drift .true heading = 070° 6° = 064° .gs is 441 kt after having set the true heading on the computer .magnetic heading = 064° + 11° variation = 075° . 495/441 x60 = 67 minutes exemple 297 Hdg 075° - eta 1502 utc.Hdg 075° - eta 1502 utc.

Given .tas = 197 kt true course = 240° w/v = 180/30kt .descent is initiated at ?

Question 173-8 : 1400 ft/min 800 ft/min 950 ft/min 1500 ft/min

Admin .using flight computer you will get gs = 182 kt .distance = rate x time.39 nm = 182 x time.t = 39/182 = 0 214h > 12 8 min total time of descent .22000 ft 4000 ft = 18000 ft height to be flown in descent .18000 / 12 8 min = 1406 ft/min exemple 301 1400 ft/min.1400 ft/min.

Given .ils glide path angle = 3 5° ground speed = 150 kt .what is the ?

Question 173-9 : 900 ft/min 350 ft/min 700 ft/min 300 ft/min

Admin .1 in 60 rule is a rule of thumb . 3 5 x 150 x 100 /60 = 875 ft/min exemple 305 900 ft/min.900 ft/min.

Given aircraft height 2500 ft ils gp angle 3° at what approximate distance ?

Question 173-10 : 8 3 nm 7 0 nm 13 1 nm 14 5 nm

Admin .distance= height x 60 / angle° .distance= 2500x60/3° .distance= 50000 ft then divide by 6080 as 1nm=6080 ft exemple 309 8.3 nm.8.3 nm.

An island appears 60° to the left of the centre line on an airborne weather ?

Question 173-11 : 046° 086° 226° 026°

Admin .276° mh + 10°e = 286° th .286° 60° = 226° th to the island to get from the island simply reverse it .226° 180° = 046° exemple 313 046°.046°.

An island appears 45° to the right of the centre line on an airborne weather ?

Question 173-12 : 059° 101° 239° 329°

Admin .215° mh 21°w = 194° th .194° + 45° = 239° th to the island to get from the island simply reverse it .239° 180° = 059° exemple 317 059°.059°.

An island appears 30° to the right of the centre line on an airborne weather ?

Question 173-13 : 220° 160° 130° 190°

Admin .355° mh + 15°e = 010° th .010° + 30° = 040° th to the island to get from the island simply reverse it .040° + 180° = 220° exemple 321 220°.220°.

An island appears 30° to the left of the centre line on an airborne weather ?

Question 173-14 : 145° 325° 205° 195°

Admin .020° mh 25°w = 355° th .island is on the left so 355° 30° = 325°.true bearing from the island so 325° 180° = 145° exemple 325 145°.145°.

Given an aircraft is flying a track of 255° m 2254 utc it crosses radial ?

Question 173-15 : The same as it was at 2254 utc greater than it was at 2254 utc randomly different than it was at 2254 utc less than it was at 2254 utc

Admin . 1724.this is an isosceles triangle .the first angle at 22 54 is 255° 180° = 75°.the second angle at the vor is 30° .the last one = 180° 75°+30° = 75° exemple 329 The same as it was at 2254 utcThe same as it was at 2254 utc

The distance between two waypoints is 200 nm to calculate compass heading the ?

Question 173-16 : 14 nm 7 nm 0 nm 21 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 200/60 3 33 nm off track for each degree of error .total error is 4° from 2°e to 2°w .4° x 3 33 nm = 13 33 nm .using goniometric functions .tan4° = / 200. = tan4° x 200. = 13 99 nm exemple 333 14 nm.14 nm.

Given .eta to cross a meridian is 2100 utc.gs is 441 kt.tas is 491 kt.at 2010 ?

Question 173-17 : 40 kt 90 kt 75 kt 60 kt

Admin .at 20h30 the airplane is at 50 minutes of the meridian with a speed of 441 kt thus at a distance of 367 5 nm 441/60 x 50 .now it has to travel 367 5 nm in 55 minutes . 367 5 / 55 x 60 = 401 kt .tas reduction is 441 401 = 40 kt exemple 337 40 kt.40 kt.

The flight log gives the following data 'true track drift true heading magnetic ?

Question 173-18 : 119° 3°l 122° 2°e 120° +4° 116° 115° 5°r 120° 3°w 123° +2° 121° 117° 4°l 121° 1°e 122° 3° 119° 125° 2°r 123° 2°w 121° 4° 117°

Admin . 2498.use this wonderful table for those questions exemple 341 119°, 3°l, 122°, 2°e, 120°, +4°, 116°119°, 3°l, 122°, 2°e, 120°, +4°, 116°

At 0020 utc an aircraft is crossing the 310° radial at 40 nm of a vor/dme ?

Question 173-19 : 085° 226 kt 090° 232 kt 080° 226 kt 088° 232 kt

Admin .draw the situation . 2499.it is a isosceles triangle with at least two equal sides .the angle at the vor is 90° and the other two angles are 45° .at 00 20 the bearing from the aircraft to the vor is 310° 180° = 130° .track is 130° 45° = 085° .it is not mandatory to calculate the groundspeed but you can use pythagoras .40² + 40² = distance between position at 00 20 and postion at 00 35 ².distance is = 56 567 nm.56 nm covered in 15 minutes . 56 567/15 x60 = 226 kt exemple 345 085° - 226 kt.085° - 226 kt.

Given .tas is 120 kt.ata 'x' 1232 utc.eta 'y' 1247 utc.ata 'y' is 1250 utc.what ?

Question 173-20 : 1302 utc 1257 utc 1300 utc 1303 utc

Admin .x to y = 30 nm in 18 minutes from 12h32 to 12h50 = 100 kt ground speed .y to z = 20 nm at 100 kt = 12 minutes .12h50 + 12 minutes = 13h02 exemple 349 1302 utc.1302 utc.

Given .fl120 oat is isa standard cas is 200 kt track is 222° m heading is ?

Question 173-21 : 050° t / 70 kt 040° t / 105 kt 055° t / 105 kt 065° t / 70 kt

Admin .oat at fl120 is 15° 2° x 12 = 9°c. 2501.magnetic heading is 215° variation is 15°w = true heading is 200° .our magnetic track is 222° minus 15°w = true track is 207° . 105 nm / 21 min x 60 = 300 kt ground speed .on the computer under centre dot set tas 239 kt under index set true heading 200° mark the point where drift 7°right crosses the ground speed 300 kt . 2502.wind is 050°/70 kt exemple 353 050°(t) / 70 kt.050°(t) / 70 kt.

A useful method of a pilot resolving during a visual flight any uncertainty in ?

Question 173-22 : Set heading towards a line feature such as a coastline motorway river or railway fly the reverse of the heading being flown prior to becoming uncertain until a pinpoint is obtained fly expanding circles until a pinpoint is obtained fly reverse headings and associated timings until the point of departure is regained

exemple 357 Set heading towards a line feature such as a coastline, motorway, river or railwaySet heading towards a line feature such as a coastline, motorway, river or railway

An aircraft is descending down a 6% slope whilst maintaining a ground speed of ?

Question 173-23 : 1800 ft/min 10800 ft/min 3600 ft/min 900 ft/min

Admin .1 nm = 6080 ft.6% ==> 0 06. 300 kt x 6080 ft / 60 min x 0 06 = 1824 ft/min exemple 361 1800 ft/min.1800 ft/min.

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

Question 173-24 : 258° 292° 270° 253°

Admin . 2517.15 min / 60 = 0 25h.0 25 x 130 kt gs = 32 5 nm.32 5 nm + 2 5 nm ahead of dr = 35 nm.50 nm distance 35 nm = 15 nm.tke = 3 x 60/35 and correction angle = 3 x 60/15.tke = 5° and 12° ca = 5° + 12° = 17° .as we are north of 275° to get back we need to fly more to the south therefore minus 17° so 275° 17° = 258° exemple 365 258°.258°.

An island is observed to be 30° to the right of the nose of the aircraft the ?

Question 173-25 : 330° 270° 250° 310°

Admin .magnetic heading 290°.variation east magnetic least .magnetic heading = true heading 10°e.true heading = magnetic heading + 10°e = 300° .the true bearing from the aircraft to the island is 300° + 30°right = 330° exemple 369 330°.330°.

An aircraft follows a radial to a vor/dme station at 10 00 the dme reads 120 nm ?

Question 173-26 : 10 24 10 27 10 18 10 21

Admin .15 nm in 3 minutes = 5 nm per minute.distance to station 105 nm.105 / 5 nm/min = 21 minutes .10 03 + 00 21 = 10 24 exemple 373 10:24.10:24.

You are departing from an airport which has an elevation of 2000 ft .the qnh is ?

Question 173-27 : 920 ft/min 1080 ft/min 590 ft/min 750 ft/min

Admin .total climb = 7500 ft 2000 ft = 5500 ft .10 nm at 100 kt = 10 nm/ 100/60 = 6 minutes .5500 ft / 6 min = 916 6 ft/min exemple 377 920 ft/min.920 ft/min.

At reference or see europe low altitude enroute chart e lo 1a.an aircraft is ?

Question 173-28 : 280 kt 385 kt 485 kt 180 kt

Admin .the distance between vor's is 59 nm.20 5 nm + 10 5 nm = 31 nm.59 nm 31 nm = 28 nm we need to make 28 nm in 6 minutes so .28 / 0 1 = 280 kt 6 minutes = 0 1h exemple 381 280 kt.280 kt.

The distance between point of departure and destination is 340 nm and wind ?

Question 173-29 : 1h and 49 min 1h and 30 min 1h and 37 min 1h and 21 min

Admin .psr = e x h / o + h e endurance h gs home o gs out .using flight computer you will get 20 kt headwind so gsout = 120 kt gshome = 160 kt .psr = 190 x 160 kt/120 + 160 3h 10min > 190 min .psr = 108 6 min > 1h 49 min exemple 385 1h and 49 min.1h and 49 min.

You are tracking the 200° radial inbound to a vor and your true heading is ?

Question 173-30 : 310°/65 320°/55 330°/50 300°/50

Admin .200° radial inbound = magnetic track 020° variation +5° = true track 025° .true heading = 010°.drift = 15° right.090° radial outbound = magnetic track 090° variation +5° = true track 095° .magnetic heading 080° variation +5° = true heading 085° .drift = 10° right .on the computer .centre dot on tas 240 kt put true heading 010° under the index and mark a line down the 15° right drift line .rotate to put true heading 085° under index and mark a line down the 10° right 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 310°/65 kt exemple 389 310°/65.310°/65.

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

Question 173-31 : 112° 080° 090° 107°

Admin .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 north of 095° to get back we need to fly more to the south therefore add 17° so 095 + 17° = 112° exemple 393 112°.112°.

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

Question 173-32 : 11 1 nm 10 3 nm 13 3 nm 16 6 nm

Admin .2000 ft 300 ft 10 hpa diff = 1700 ft.10000 ft 1700 ft = 8300 ft.8300 / 1000 ft/min = 8 3 min > 0 1383h .0 1383 x 80 kt gs = 11 1 nm exemple 397 11.1 nm.11.1 nm.

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

Question 173-33 : 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 401 110°/70kt.110°/70kt.

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

Question 173-34 : 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 405 10.3 nm.10.3 nm.

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

Question 173-35 : 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 409 11:3611:36

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

Question 173-36 : 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 413 124 nm.124 nm.

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

Question 173-37 : 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 417 19°.19°.

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

Question 173-38 : 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 421 292°.292°.

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

Question 173-39 : 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 425 272° (t).272° (t).

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

Question 173-40 : 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

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