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Question 104-1 : Integrated range' curves or tables are presented in the aeroplane operations manuals their purpose is ? [ Exam pilot ]

To determine the fuel consumption for a certain still air distance considering the decreasing fuel flow with decreasing mass

exemple 204 To determine the fuel consumption for a certain still air distance considering the decreasing fuel flow with decreasing mass.

Question 104-2 : At reference or see flight planning manual sep 1 figure 2 4 .using the range profile diagramm for the single engine aeroplane determine the range with 45 minutes reserve in the following conditions .given .oat isa +16°c.pressure altitude 4000 ft.power full throttle / 25 0 in/hg/ 2100 rpm . err a ?

865 nm

. /com en/com033 491 jpg. exemple 208 865 nm.

Question 104-3 : At reference or see flight planning manual mrjt 1 figure 4 4. the final reserve fuel taken from the holding planning table for the twin jet aeroplane is based on the following parameters . err a 033 493 ?

Pressure altitude aeroplane mass and flaps up with minimum drag airspeed

exemple 212 Pressure altitude, aeroplane mass and flaps up with minimum drag airspeed

Question 104-4 : Finish the endurance/fuel calculation and determine atc endurance for a twin jet aeroplane with the help of the table provided contingency is 5% of the planned trip fuel and fuel flow for extra fuel is 2400 kg/h . err a 033 509 ?

Atc endurance 04 07

. /com en/com033 509 jpg.contingency is 5% of the planned trip fuel = 290 kg.final reserve for a jet aeroplane is 30 minutes .minimum take off fuel = 5800 + 290 + 1800 + 1325 = 9215 kg.fuel flow for extra fuel is 2400 kg/h so 585 kg gives 14 minutes 38 seconds exemple 216 Atc endurance: 04:07

Question 104-5 : Given .oat +5°c.during climb average head wind component 20 kt.take off from msl with the initial mass of 3 650 lbs.find time and fuel to climb to fl75 . 2464 ?

9 min 3 3 usg

Admin . 2465 exemple 220 9 min. 3,3 usg

Question 104-6 : Maximum allowable take off mass 64 400 kg.maximum landing mass 56 200 kg.maximum zero fuel mass 53 000 kg.dry operating mass 35 500 kg.estimated load 14 500 kg.estimated trip fuel 4 900kg.minimum take off fuel 7 400 kg.find the maximum allowable take off fuel ?

11 100 kg

Admin .in order to find the maximum allowable take off fuel we must consider the mtow maximum take off weight and the mlw maximum landing weight ..mtow = dry operating mass + estimated load + maximum take off fuel.64400 = 35500 + 14500 + maximum take off fuel.maximum take off fuel = 14400 kg.mlw = dry operating mass + estimated load + maximum landing fuel.56200 = 35500 + 14500 + maximum landing fuel.maximum landing fuel = 6200 kg.4900 kg trip fuel + 6200 kg = 11100 kg.11100 kg is the lower limitation exemple 224 11 100 kg.

Question 104-7 : At reference or use flight planning manual sep 1 table 2 2 3 .given .fl 70.oat 19°c.lean mixture 2300 rpm .find fuel flow in gallons per hour gph and tas . err a 033 536 ?

12 35 gph tas 159 kt

.no need for calculation . /com en/com033 536 jpg. exemple 228 12.35 gph, tas: 159 kt.

Question 104-8 : At reference or see flight planning manual mrjt 1 figure 4 5 3 1. given flight time from top of climb to the enroute point in fl280 is 48 min .cruise procedure is long range cruise lrc .temp isa 5° c.take off mass 56 000 kg.climb fuel 1 100 kg.find distance in nautical air miles nam for this leg ?

345 nam. 2000 kg

56000 1100 = 54900 kg on table tas is 437 kt .adjustments for operation at non standard temperatures states to decrease tas by 1 knot per degree celsius below isa .isa 5°c > tas = 432 kt..tas x 48/60 = 345 6 nam..on table 54900 > 3736 346 = 3390.we can read for 3390 by interpolation 52900 kg.54900 52900 = 2000 kg .see section 5 4 2 method on caa cap697 flight planning manual for that kind of questions exemple 232 345 namxsx 2000 kg

Question 104-9 : Planning a flight from paris charles de gaulle to london heathrow for a twin jet aeroplane.preplanning .maximum take off mass 62 800 kg.maximum zero fuel mass 51 250 kg.maximum landing mass 54 900 kg.maximum taxi mass 63 050 kg.assume the following preplanning results .trip fuel 1 800 kg.alternate ?

51 629 kg

Take off mass = dry operating mass + traffic load + fuel trip + contingency + alternate + final reserve .contingency fuel is the higher of 5% of trip or 5 minutes holding at 1500ft .5% of trip = 90 kg.5 minutes holding at 1500 ft = 204 kg.the question states holding fuel final reserve 1 225 kg .holding fuel for a jet aircraft is 30 minutes at holding speed at 1500 ft above aerodrome elevation .contingency = 1225/30 x 5 = 204 kg .34000 + 13000 + 1800 + 204 + 1400 + 1225 = 51629 kg .. fuel policy .contingency fuel which shall be the higher of i or ii below . i either .5% of the planned trip fuel or in the event of in flight replanning 5% of the trip fuel for the remainder of the flight . ii an amount to fly for 5 minutes at holding speed at 1 500 ft 450 m above the destination aerodrome in standard conditions .note baggage mass is already included in traffic load .traffic load is the total mass of passengers baggage and cargo including any non revenue load exemple 236 51 629 kg.

Question 104-10 : You are to determine the maximum fuel load which can be carried in the following conditions . dry operating mass 2800 kg. trip fuel 300 kg. payload 400 kg. maximum take off mass 4200 kg. maximum landing mass 3700 kg ?

800 kg

. /com en/com033 548 jpg.we can take off with 1000 kg but we can land only after having burn 500 kg if not we will be over weight .the question states that trip fuel is 300 kg it means that we can take off with 800 kg 500 kg + 300 kg and then we will land at our maximum landing mass exemple 240 800 kg.

Question 104-11 : At reference or see flight planning manual mrjt 1 figure 4 5 1.given .brake release mass 62000 kg.temperature isa +15°c.the fuel required for a climb from sea level to fl330 is . err a 033 549 ?

1700 kg

. /com en/com033 549 jpg. exemple 244 1700 kg.

Question 104-12 : At reference or see flight planning manual sep 1 figure 2 1. given .take off mass 3500 lbs.departure aerodrome pressure altitude 2500 ft.oat +10°c.first cruising level fl 140.oat 5°c.find the time fuel and still air distance to climb . err a 033 557 ?

22 min 6 7 gal 45 nam

. /com en/com033 557 jpg. exemple 248 22 min, 6.7 gal, 45 nam

Question 104-13 : At reference or see flight planning manual sep1 figure 2 2 table 2 2 3.a flight has to be made with the single engine sample aeroplane.for the fuel calculation allow 10 lbs fuel for start up and taxi .3 minutes and 10 lbs of additional fuel to allow for the climb .10 minutes and no fuel correction ?

325 lbs

.total flight time .3h32 + 3 min + 10 min = 3h45 3 42h ..fuel comsumption at fl70 isa +20°c is . 68 5+69 /2 = 68 75 pph..3 42h x 68 75 = 235 pph..235 + 10 lbs for the climb + reserve fuel 30% of 235 + 10 for taxi = 325 lbs exemple 252 325 lbs.

Question 104-14 : A flight has to be made with a multi engine piston aeroplane mep 1 for the fuel calculations take 5 us gallons for the taxi and an additional 13 min at cruise condition to account for climb and descent calculated time from overhead .to overhead is 1h 47 min .powersetting is 45% 2600 rpm.calculated ?

47 us gallons

exemple 256 47 us gallons.

Question 104-15 : A flight has to be made with the sep for the fuel calculation .allow . 10 lbs fuel for start up and taxi. 3 min and 1 gallon of additional fuel to allow for the climb. 10 min and no fuel correction for the descent. planned flight time overhead to overhead is 3h 12 min. reserve fuel 30% of the trip ?

359 lbs

exemple 260 359 lbs.

Question 104-16 : A flight has to be made with the sep1 for the fuel calculation allow . 10 lbs fuel for start up and taxi . 3 min and 1 gallon of additional fuel to allow for the climb . 10 min and no fuel correction for the descent. planned flight time overhead to overhead is 3h 12. reserve fuel 30% of the trip ?

297 lbs

exemple 264 297 lbs.

Question 104-17 : Given .dry operating mass 33500 kg .traffic load 7600 kg .maximum allowable takeoff mass 66200 kg .taxi fuel 200 kg .tank capacity 16100 kg .the maximum possible takeoff fuel is ?

15900 kg

exemple 268 15900 kg.

Question 104-18 : Given .brake release mass 58 000 kg.temperature isa 15.the fuel required to climb from an aerodrome at elevation 4000 ft to fl300 is . 1514 ?

1150 kg

exemple 272 1150 kg.

Question 104-19 : Given .distance from departure to destination 500 nm.true track 090°.wind 090°/20kt.tas 150 kt.what is the distance and time of the pet from the departure point ?

Distance 283 nm time 131 min

Ground speed out 130 kt.ground speed home 170 kt.pet = d x gsh / gso + gsh .pet = 500 x 170 / 130 + 170 = 283 nm exemple 276 Distance: 283 nm, time: 131 min.

Question 104-20 : Use route manual chart e hi 1.an aircraft has to fly from glasgow 55°52'n 004°27'w to benbecula 57°29'n 007°22'w cruising at 320 kt tas .assuming a headwind of 40 kt and cruise fuel consumption of 2300 kg/h what is the forecast fuel used for this flight . err a 033 6 ?

1117 kg

Ground speed = 320 40 = 280 kt .136 nm/ 280/60 = 29 14 minutes.2300/60 = 38 333 kg/min.29 14 x 38 333 = 1117 kg exemple 280 1117 kg.

Question 104-21 : Given .distance from departure to destination 500 nm.safe endurance 4 h.tas 140 kt.ground speed out 150 kt.ground speed home 130 kt.what is the distance and time of the psr from the departure point ?

Distance 279 nm time 111 min

Point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 150 kt.ground speed home = 130 kt.point of safe return psr = 4 x 130 / 150 + 130 .point of safe return psr = 65000 / 280.point of safe return psr = 1 86 h.1 86 x 60 = 111 6 min.distance of the psr from the departure point at a speed of 150 kt .111 6 min x 150/60 = 279 nm exemple 284 Distance: 279 nm, time: 111 min.

Question 104-22 : On a given path it is possible to chose between four flight levels fl each associated with a mandatory flight mach number m .the flight conditions static air temperature sat and headwind component hwc are given below .fl 370 m = 0 80 ts = 60°c hwc = 15 kt.fl 330 m = 0 78 ts = 60°c hwc= 5 ?

Fl 270

Set corresponding mark m kt against outside temperature at flight altitude read in front of mach number on the outer scale the true air speed .example . 60° and mach 0 8 => tas is 451 kt . 1137.fl 370 => tas = 451 kt => gs = 439 kt.fl 330 => tas = 440 kt => gs = 435 kt.fl 290 => tas = 456 kt => gs = 441 kt.fl 270 => tas = 445 kt => gs = 445 kt .the flight level allowing the highest ground speed is fl 270 exemple 288 Fl 270.

Question 104-23 : Given . distance from departure to destination 410 nm. safe endurance 3 6 h. true track 055°. w/v 180/35. tas 120 kt.what is the distance of the psr from the departure point ?

203 nm

First step find the gs out .given . true track. true air speed. wind direction and velocity.required wind correction angle and ground speed. computer solution .a set true track to true index .b turn the indicator to the wind direction in this case using the black azimuth graduation the angle being upwind counting anti clockwise .c shift the speed arc corresponding to the true air speed so as to coincide with the wind speed on the indicator .d read the wind correction angle at the same place read the ground speed under the center bore from the scale on the axis of the slide .setting 055° to true index.set the indicator to 180° on the black azimuth circle being upwind adjust the speed arc labelled 120 kt of the diagram slide to the wind speed 35 kt of the indicator scale .reading under the plotted point read the wind correction angle 14° under the center bore read the ground speed 136 kt . 1141.step two proceed on the same way for gs home you will find 97 kt .step three now apply the psr formula .psr = time x gs out x gs home / gs out + gs home .psr = 3 6 x 136 x 97 / 136 + 97 .psr = 203 8 nm .this is a 4 points question at the exam .mathematical calculation on this kind of exercise is valid only for one right angled triangle which is not the case here only the computer enables you to find the good answer exemple 292 203 nm.

Question 104-24 : The surface weather system over england 53°n 002°w is . err a 033 22 ?

An occluded front moving east

. /com en/com022 33a png.over england the surface weather system is an an occluded front . /com en/com033 22 jpg.the arrow attached to the occluded front indicates direction and speed 20 km/h exemple 296 An occluded front moving east.

Question 104-25 : Which describes the intensity of icing if any at fl 150 in the vicinity of toulouse 44°n 01°e . err a 033 25 ?

Moderate or severe

.its due to the presence of cbs around toulouse always assume moderate or severe icing next to them exemple 300 Moderate or severe.

Question 104-26 : Given .distance a to b 2050 nm.mean groundspeed 'on' 440 kt.mean groundspeed 'back' 540 kt.the distance to the point of equal time pet between a and b is ?

1130 nm

.ground speed out gso = 440 kt.ground speed home gsh = 540 kt.distance to pet = distance x gsh / gso + gsh .distance to pet = 2050 x 540 / 440 + 540 .distance to pet = 1107000 / 980 = 1130 nm exemple 304 1130 nm.

Question 104-27 : Which is the heaviest type of precipitation if any forecast for bordeaux/merignac at 1000 utc . err a 033 27 ?

Light rain

. /com en/com033 27 jpg.at 1000 utc if any the heaviest type of precipitation forcast is light rain exemple 308 Light rain.

Question 104-28 : Given .distance from departure to destination 315 nm.true track 343°.w/v 015/15.tas 100 kt.what is the distance of the pet from the departure point ?

176 nm

..under index set true track 343° centre dot on tas 100 kt with the rotative scale set wind 015° / 15 kt .now drift is always measured from heading to track .turn to set true heading 345° 343° + 2° left drift under index you now read a ground speed out of 88 kt .proceed in the same way to find the ground speed home of 112 kt . right drift of 4° true heading of 159° .distance to pet = distance x gsh / gso + gsh .distance to pet = 315 x 112 / 88 + 112 .distance to pet = 35280 / 200 = 176 nm exemple 312 176 nm.

Question 104-29 : What lowest cloud conditions oktas/ft are forecast for johannesburg/jan smuts at 0300 utc . err a 033 31 ?

5 to 7 at 400 ft

.ft0900 prob30 0305 3000 bcfg bkn004 .30% chance that temporarily between 0300 utc and 0500 utc there will be patches fog with 3000 m visibility and 5 to 7 oktas of cloud at 400 ft exemple 316 5 to 7 at 400 ft.

Question 104-30 : Given .distance from departure to destination 210 nm.safe endurance 2 5 h.true track 035.w/v 250/20.tas 105 kt.what is the distance of the psr from the departure point ?

127 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .on the computer .under index set true track 035° centre dot on tas 105 kt with the rotative scale set wind 250°/20 kt we read a right drift of 5° .drift is always measured from heading to track so turn to set true heading 030° 035° 5° right drift under index you read a ground speed out of 122 kt .proceed on the same way for gs home .under index set true track 215° centre dot on tas 105 kt with the rotative scale set wind 250°/20 kt we read a left drift of 4° .drift is always measured from heading to track so turn to set true heading 219° 215° + 4° right drift under index you read a ground speed home of 88 kt ..point of safe return psr = 2 5 x 88 / 122 + 88 .point of safe return psr = 220 / 210.point of safe return psr = 1 04 h.distance of the psr from the departure point at a speed of 122 kt .1 04 x 122 = 127 nm exemple 320 127 nm.

Question 104-31 : Given the following .d = flight distance.x = distance to point of equal time.gso = groundspeed out.gsr = groundspeed return.the correct formula to find distance to point of equal time is ?

X = d x gsr / gso + gsr

exemple 324 X = d x gsr / (gso + gsr).

Question 104-32 : The flight crew of a turbojet aeroplane prepares a flight using the following data . flight leg air distance 2 700 nm. flight level fl 310 true airspeed 470 kt. tailwind component at this level 35 kt. initially planned take off mass without extra fuel on board 195 000 kg. fuel price 0 28 euro/l at ?

The fuel transport operation is not recommended in this case

.the fuel is cheaper at destination the fuel transport operation is not recommended in this case exemple 328 The fuel transport operation is not recommended in this case.

Question 104-33 : From which of the following would you expect to find information regarding known short unserviceability of vor tacan and ndb ?

Notam

exemple 332 Notam.

Question 104-34 : The wind velocity over italy is . err a 033 54 ?

A maximum of 110 kt at fl380

. /com en/com033 54 jpg.a jet is crossing france and italy at fl 380 with a maximum of 110 kt exemple 336 A maximum of 110 kt at fl380.

Question 104-35 : Assuming the following data .ground distance to be covered 2 000 nm.cruise flight level fl 330.cruising speed mach 0 82 true airspeed 470 kt .head wind component 30 kt .planned destination landing mass 160 000 kg .temperature isa .cg 37.total anti ice on pack flow hi .time needed to carry out such ?

4 h 43 min

.nam = ngm x tas/gs = 2000 x 470/440 = 2136 nam . /com en/com033 56 jpg.between 2100 and 2200 we can find our answer 4 h 43 min . 210883 .i think that the answer is wrong because . 4 39 + 4 51 /2 = 4 45 h > 4h 27' . 4h 43' > 4 75h so the correct answer as i think is 4h 27' ..no times on the annex are shown as 4 39 for 4h39 and 4 51 for 4h51 .it's written on the top below fuel consumed kg time h min exemple 340 4 h 43 min.

Question 104-36 : When calculating the fuel required to carry out a given flight one must take into account .1 the wind.2 foreseeable airborne delays.3 other weather forecasts.4 any foreseeable conditions which may delay landing.the combination which provides the correct statement is ?

1 2 3 4

exemple 344 1, 2, 3, 4.

Question 104-37 : Which of the following flight levels if any is forecast to be clear of significant cloud icing and cat along the marked route from shannon 53°n 10°w to berlin 53°n 13°e . err a 033 64 ?

Fl250

. /com en/com033 64 jpg.there are isolated embedded cbs from below fl100 to fl220 so fl 210 is excluded .you have cat area n°2 from fl270 to fl400 so fl290 is excluded .only fl250 is forecast to be clear of significant cloud icing and cat along the marked route from shannon to berlin exemple 348 Fl250.

Question 104-38 : Given .distance from departure to destination 215 nm.safe endurance 3 3 h.true track 005°.w/v 290/15.tas 125 kt.what is the distance of the psr from the departure point ?

205 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .on the computer .under index set true track 005° centre dot on tas 125 kt with the rotative scale set wind 290°/15 kt we read a right drift of 7° .drift is always measured from heading to track so turn to set true heading 358° 005° 7° right drift under index you read a ground speed out of 120 kt .proceed on the same way for gs home .under index set true track 185° centre dot on tas 125 kt with the rotative scale set wind 290°/15 kt we read a left drift of 7° .drift is always measured from heading to track so turn to set true heading 192° 185° + 7° right drift under index you read a ground speed home of 128 kt ..point of safe return psr = 3 3 x 128 / 120 + 128 .point of safe return psr = 422 4 / 248.point of safe return psr = 1 70 h.distance of the psr from the departure point at a speed of 120 kt .1 7 x 120 = 204 nm exemple 352 205 nm.

Question 104-39 : Given .distance from departure to destination 200 nm.safe endurance 3 h.tas 130 kt.ground speed out 150 kt.ground speed home 110 kt.what is the distance psr from the departure point ?

190 nm

.point of safe return psr = endurance x homeward gs / outbound gs + homeward gs .ground speed out = 150 kt.ground speed home = 110 kt.point of safe return psr = 3 x 110 / 150 + 110 .point of safe return psr = 330 / 260.point of safe return psr = 1 2692 h.distance of the psr from the departure point at a speed of 150 kt .1 2692 h x 150 = 190 nm exemple 356 190 nm.

Question 104-40 : What minimum visibility m is forecast for 0600 utc at london lhr egll . err a 033 71 ?

1500 m

exemple 360 1500 m.


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