In the ATC flight plan Item 13 in a flight plan submitted before departure the departure time entered is the ?
Experience > landing
Given Planned and actual data as shown in the Flight Log excerpt Arriving overhead GAMMA you are cleared for direct routeing to MIKE The flight time for direct flight GAMMA to MIKE will be 1 h 08 min ?
A turbojet aeroplane flies using the following data Flight level FL 330Flight regime 'Long Range' LR Mass 156 500 kgTailwind component at this level 40 kt With a remaining flight time of 1 h 10 min ?
Given true course tc 017°w/v 340°/30 kttrue air speed tas 420 ktfind wind correction angle wca and ground speed gs Estimated off block time. Wind correction angle (wca) another term drift except drift expressed as left or right wind correction angle uses '+' or ' ' under index set true track 017° centre dot on tas 420 kt with rotative scale set wind 340°/30 kt img /com_en/com033 38a jpg drift always measured from heading to track so turn to set true heading 015 5° (017° 2 5° right drift) under index img /com_en/com033 38b jpg you now read a ground speed of 396 kt a drift of 2º right (wca ?2º). 
The true course 042° the variation in area 6° w and wind calm the deviation card reproduced in annex in order to follow this course pilot must fly a compass heading of err _a_033 41 Estimated off block time. Use this very useful table those questions img /com_en/com033 41 jpg the nearest compass deviation reading taken from 045° indicating 4° less than actual magnetic heading add 4° onto 048° (magnetic heading) giving answer of 052° (compass heading). 
At reference or see flight planning manual mrjt 1 figure 4 3 5 given a trip time of about 9 hours within limits of data given a temperature decrease of 30°c will affect trip time approximately err _a_033 67 Estimated off block time. img /com_en/com033 67 jpg a 9h trip becomes 8 6 isa +20°c a 9h trip becomes 9 2 isa 10°c difference 30°c will affect trip time approximately (0 6/8 6) x 100 = 7%. 
At reference or see flight planning manual mrjt 1 figure 4 5 4 planning an ifr flight from paris to london the twin jet aeroplane given estimated landing mass 49700 kgfl 280w/v 280°/40 ktaverage true course 320°procedure descent 74 m/250 kiasdetermine distance from top of descent to london elevation 80 ft err _a_033 75 Estimated off block time. from table img /com_en/com033 75 jpg time 19 minutes distance nam 86 nm tas = distance/time = (86/19) x 60 = 272 kt on nav computer true course 320° w/v 280°/40 kt ground speed is 240 kt distance (nm) from top of descent to london = 240 x 19/60 = 76 nm. Question 94-8
At reference or see flight planning manual mrjt 1 figure 4 7 3 given diversion distance 720 nmtail wind component 25ktmass at point of diversion 55000kgtemperature isadiversion fuel available 4250 kgwhat the minimum pressure altitude at which above conditions may be met err _a_033 76 Estimated off block time. img /com_en/com033 76 jpg . Question 94-9
Flight planning chart an aircraft states that time to reach cruising level at a given gross mass 36 minutes and distance travelled 157 nm zero wind what will be distance travelled with an average tailwind component of 60 kt Estimated off block time. we know that a headwind or a tailwind component does not modify time to climb to a specific level thus we only have to calculate our speed 157 x (60 / 36) = 262 kt adding tail wind component 262 + 60 = 322 kt the distance travelled in 36 minutes will be 322 x (36 / 60) = 193 nm. Question 94-10
At reference or see flight planning manual mrjt 1 figure 4 3 6 given distance to alternate 400 nmlanding mass at alternate 50 000kgheadwind component 25 ktthe alternate fuel required err _a_033 79 Estimated off block time. Answer img /com_en/com033 1012 jpg. Question 94-11
At reference or see flight planning manual mrjt 1 figure 4 3 6 given twin jet aeroplanedry operating mass 35500 kgtraffic load 14500 kgfinal reserve fuel 1200 kgdistance to alternate 95 nmtailwind component 10 ktfind fuel required and trip time to alternate with simplified flight planning alternate planning err _a_033 82 Estimated off block time. Img /com_en/com033 1015 jpg. Question 94-12
Given planning data as shown in flight log excerpt fuel planning section after a balked landing at destination airport you have to divert to alternate airport with gear extended the re calculated flight time to alternate due to reduced speed 1 h 10 min and fuel flow will be 900 kg/h final reserve fuel remains unchanged what will be estimated landing mass at alternate airport err _a_033 91 Estimated off block time. 900 x (70/60) = 1050 kg to reach alternate airport subtract 1050 kg from our estimated landing mass (at destination) 7037 1050 = 5987 kg. Question 94-13
Flight planning chart an aeroplane states that time to reach fl 190 at a given gross mass 22 minutes and distance travelled 66 nm no wind what will be distance travelled with an average head wind component of 35 kt Estimated off block time. the time to reach fl 190 22 minutes the distance travelled 66 nm (no wind) with a head wind component of 35 kt our ground distance travelled will be less but time to reach fl 190 remains unchanged we need first to know our ground speed without wind 66/22 x 60 = 180 kt with 35 kt head wind our ground speed becomes 180 35 = 145 kt in 22 minutes distance travelled will be 145/60 x 22 = 53 17 nm. Question 94-14
At reference or see flight planning manual mrjt 1 figure 4 5 3 1given twin jet aeroplanefl 330long range cruiseoutside air temperature 63°cgross mass 50500 kgtrue air speed tas err _a_033 126 Estimated off block time. img /com_en/com033 126 jpg isa temperature at fl330 is 15°c (2°c x 33) = 51°c adjustements operation at non standard temperatures decrease tas 1 knot per degree c below isa difference between 63°c 51°c 12°c 433 12 = 421 kt ninorr this make no difference to answer but on picture you have marked wrong mass (should be 50500 not 55000 as you marked). Question 94-15
At reference or see flight planning manual mrjt 1 figure 4 5 3 1 given long range cruiseoat 45°c at fl 350gross mass at beginning of leg 40000 kggross mass at end of leg 39000 kg find true air speed tas and cruise distance nam a twin jet aeroplane err _a_033 147 Estimated off block time. isa at fl350 is 35 x ( 2°) + 15° = 55°c gross mass at end of leg 39000 kg from table we can read a tas of 422 kt as we are at isa+9 we must increase tas 1 kt per degree c above isa (as indicated on bottom of table) 422 + 10 = 432 kt fuel consumption on leg was 1000 kg (40000 39000) table values are 40000 >1163 nam 39000 > 936 nam 1163 ? 936 = 227 nam. Question 94-16
During an ifr flight in a beech bonanza fuel indicators show that remaining amount of fuel 100 lbs after 38 minutes fuel at take off 160lbs for alternate fuel 30 lbs necessary the planned fuel taxi 13 lbs final reserve fuel estimated at 50 lbs if fuel flow remains same how many minutes can be flown to destination with remaining fuel Estimated off block time. Minimum take off fuel = trip fuel + contingency + alternate + final reserve after 38 min 60 lbs of fuel has been used fuel flow = 60/38 = 1 579 lb/min at landing at destination alternate fuel final reserve must be in your tank 100 lbs (30 + 50) = 20 lbs we can fly 20/1 579 = 12 66 min. Question 94-17
Given planning data as shown in flight log excerpt fuel planning section after a balked landing at destination airport you have to divert to alternate airport with gear extended the re calculated flight time to alternate due to reduced speed 2h 20 min and fuel flow will be 780 kg/h final reserve fuel remains unchanged what will be estimated landing mass at alternate airport err _a_033 163 Estimated off block time. dadoki 2 333(2h 20min) x 780 = 1820 1820 1100(alternate fuel) 990 (reserve) = 270 4370(dom) + 500(traffic load) + 90(contingency) + 210(final reserve) + 270(remaining of alternate extra fuel) = 5440 kg. Question 94-18
At reference or see flight planning manual mrjt 1 figure 4 3 1cfor a flight of 2400 ground nautical miles following apply temperature isa 10°ccruise altitude 29000 ftlanding mass 45000 kgtrip fuel available 16000 kgwhat the maximum headwind component which may be accepted err _a_033 164 Estimated off block time. img /com_en/com033 164 jpg . Question 94-19
At reference or see flight planning manual mrjt 1 figure 4 5 3 2 find fuel flow the twin jet aeroplane with regard to following data given mach 74 cruiseflight level 310gross mass 50000 kgisa conditions err _a_033 171 Estimated off block time. 50000 > 2994 nam (maximum cruise distance) tas 434 kt in one hour we travel 2294 434 = 2560 in table weight associated with a distance of 2560 nam 47700 kg 50000 kg 47700 kg = 2300 kg/h see section 5 4 2 method on caa cap697 flight planning manual that kind of questions. Question 94-20
At reference or see flight planning manual mrjt 1 figure 4 3 1c within limits of data given a mean temperature increase of 30°c will affect trip time approximately err _a_033 172 Estimated off block time. draw a line a ficticious flight duration of 5h example img /com_en/com033 172 jpg at isa 10°c trip time 5h07 min (5 12h) at isa +20°c trip time 4h50 min (4 83h) a 30°c mean temperature increase decrease trip time approximately 17 minutes (0 28h) (0 28/5 12) x 100 = 5 46%. Question 94-21
At reference or see flight planning manual mrjt 1 figure 4 3 6 in order to find alternate fuel and time to alternate aeroplane operating manual shall be entered with err _a_033 173 Distance in nautical miles (nm) wind component landing mass at alternate. draw a line a ficticious flight duration of 5h example img /com_en/com033 172 jpg at isa 10°c trip time 5h07 min (5 12h) at isa +20°c trip time 4h50 min (4 83h) a 30°c mean temperature increase decrease trip time approximately 17 minutes (0 28h) (0 28/5 12) x 100 = 5 46%. Question 94-22
During a vfr flight at a navigational checkpoint remaining usable fuel in tanks 60 us gallons the reserve fuel 12 us gallons according to flight plan remaining flight time 1h35min calculate highest acceptable rate of consumption possible the rest of trip Distance in nautical miles (nm) wind component landing mass at alternate. 60 12 = 48 us gallons available 1 h 35 min = 95 min (48 / 95) x 60 = 30 31 us gallons/hour this the highest acceptable rate of consumption possible the rest of trip. Question 94-23
At reference or see flight planning manual mrjt 1 figure 4 3 1cfor a flight of 2000 ground nautical miles cruising at 30000 ft within limits of data given a headwind component of 25 kt will affect trip time approximately err _a_033 190 Distance in nautical miles (nm) wind component landing mass at alternate. img /com_en/com033 190 jpg with a headwind component of 25 kt we find a trip time of 307 minutes (5h07) without headwind a trip time of 285 minutes (4h45) ((307 285)/ 285) x 100 = 7 72%. Question 94-24
At reference or see flight planning manual mrjt 1 figure 4 3 6given distance to alternate 450 nmlanding mass at alternate 45 000 kgtailwind component 50 ktthe alternate fuel required err _a_033 191 Distance in nautical miles (nm) wind component landing mass at alternate. img /com_en/com033 191 jpg start at 450 nm go to reference line enter condition 50 kt tailwind go to next condition landing mass 45 000 kg you get your answer 2500 kg. Question 94-25
An aeroplane on an ifr flight the flight to be changed from ifr to vfr is it possible Yes pilot in command must inform atc using phrase 'cancelling my ifr flight'. img /com_en/com033 191 jpg start at 450 nm go to reference line enter condition 50 kt tailwind go to next condition landing mass 45 000 kg you get your answer 2500 kg. Question 94-26
See flight planning manual mrjt 1 figure 4 5 2 and 4 5 3 1 given distance c d 680 nmlong range cruise at fl340temperature deviation from isa 0° cheadwind component 60 ktgross mass at c 44700 kgthe fuel required from c d err _a_033 218 Yes pilot in command must inform atc using phrase 'cancelling my ifr flight'. For 44700 kg we have a tas of 431 kt no temperature correction since it isa condition nam = ground distance x (tas/gs) nam = 680 x (431/371) = 790 nam at line 44700 cruise distance nautical air miles 2150 nautique air miles substract 790 you find 1360 nam what the mass 1360 nm img /com_en/com033 218 jpg we find 41000 kg (this our end mass) 44700 41000 = 3700 kg see section 5 4 2 method on caa cap697 flight planning manual that kind of questions. Question 94-27
Flight planning manual mrjt 1 figure 4 5 4a descent planned at 74/250kias from 35000ft to 5000ft how much fuel will be consumed during this descent err _a_033 220 Yes pilot in command must inform atc using phrase 'cancelling my ifr flight'. img /com_en/com033 220 jpg from 35000 ft to 0 ft 290 kg from 5000 ft to 0 ft 140 kg from 35000 ft to 5000 ft 290 140 = 150 kg. Question 94-28
At reference or see flight planning manual mrjt 1 figure 4 7 3given diversion fuel available 8500kgdiversion cruise altitude 10000ftmass at point of diversion 62500kghead wind component 50kttemperature isa 5°cthe a maximum diversion distance and b elapsed time alternate are approximately err _a_033 253 (a) 86 nm (b) 3h 2 min. img /com_en/com033 253 jpg the reference quality not fantastic sorry that for information 'dash line over 'weigth at point of diversion' serve as adjustment variables we have to follow a 'continous line' slope because it more representative of our pressure altitude (continous lines are low pressure altitudes). Question 94-29
For a planned flight calculated fuel as follows flight time 2h42min the reserve fuel at any time should not be less than 30% of remaining trip fuel taxi fuel 9 kg block fuel 136 kg how much fuel should remain after 2 hours flight 25 kg trip fuel 8 kg reserve fuel. 127 kg at take off = 130/100 x trip fuel trip fuel = 127 x (100/130) trip fuel 98 kg after after 2 hours flight it remains 42 minutes (0 7 h) the fraction of trip fuel remaining is 0 7/2 7 x 98 = 25 kg the reserve fuel at any time should not be less than 30% of remaining trip fuel 30/100 x 25 = 8 kg. Question 94-30
An aircraft flying at mach 0 84 at fl 330 the static air temperature 48°c and headwind component 52 kt at 1338 utc controller requests pilot to cross meridian of 030w at 1500 utc given distance to go 570 nm reduced mach required 25 kg trip fuel 8 kg reserve fuel. 13h38 to 15h00 = 1h22 (1 37h) first step ground speed = distance/time = 570/1 37 = 416 kt second step true air speed = gs + headwind = 416 kt + 52 kt = 468 kt last step now on computer in airspeed window 48°c under mach index in front of 468 kt on outer scale you read reduced mach number 0 8 m. Question 94-31
You must fly ifr on an airway orientated 135° magnetic with a msa at 7800 ft knowing qnh 1025 hpa and temperature isa +10°c minimum flight level you must fly at 25 kg trip fuel 8 kg reserve fuel. 1025 1013 = 12 hpa 12 hpa x 27ft/hpa = 324 ft temperature correction 7 5 x 4 x 10 = +300 ft 7800 324 300 = 7176 ft it 'minu 324 ft because we decrease our pressure setting from 1025 to 1013 (indicated altitude will also decrease) it 'minu 300 ft because air mass warmer than isa on a 135° heading we need an odd level first available fl90. Question 94-32
An aircraft following a 215° true track must fly over a 10 600 ft obstacle with a minimum obstacle clearance of 1 500 ft knowing qnh received from an airport close which almost at sea level 1035 and temperature isa 15°c what the minimum fl aircraft should fly at allowing the temperature variation 25 kg trip fuel 8 kg reserve fuel. local qnh 1035 we gonna fly with a 1013 hpa setting 1035 1013 = 22 hpa 22 hpa x 27 ft/hpa = 594 ft (new learning objectives state 27ft per 1 hpa) we need 10600 + 1500 594 = 11506 ft we must correct temperature to determine true altitude/height following rule of thumb called '4% rule' shall be used the altitude/height changes 4% each 10°c temperature deviation from isa 4%/15 degrees = 6% 6% x 11506 = 690 ft air mass colder we need to flight higher 11506 + 690 = 12196 ft on a 215° true track we need an even flight level first one available fl 140. Question 94-33
Given dry operating mass 33000 kg traffic load 8110 kg final reserve fuel 983 kg alternate fuel 1100 kg contingency fuel not used 102 kg the estimated landing mass at alternate should be 25 kg trip fuel 8 kg reserve fuel. at alternate you must land with final reserve in your tanks contingency fuel if not used 33000 kg(dom) + 8110 kg(traffic load) + 983 kg(final reserve) + 102 kg(contingency fuel) = 42195 kg. Question 94-34
At reference or see flight planning manual mrjt 1 figure 4 5 4planning an ifr flight from paris to london the twin jet aeroplane given estimated landing mass 49700 kgfl 280wind 280°/40 ktaverage true course 320°procedure descent 74 m/250 kiasdetermine time from top of descent to london elevation 80 ft err _a_033 303 25 kg trip fuel 8 kg reserve fuel. img /com_en/com033 303 jpg no need calculations cmarzocchini whats reason you dont make any correction about wind coz descend at constant mach number ? if yes can you explain to me please? the wind will affect distance but not time. Question 94-35
At reference or see flight planning manual mrjt 1 figure 4 4given twin jet aeroplaneestimated mass on arrival at alternate 50000 kgelevation at destination aerodrome 3500 ftelevation at alternate aerodrome 30 ftfind final reserve fuel err _a_033 304 25 kg trip fuel 8 kg reserve fuel. img /com_en/com033 304 jpg 30 minutes at 1500 ft above alternate 2260/2 = 1180 kg ninorr why we do not take into account 3500ft of airfield elevation? we have to fly 1500ft above airfiled what makes 5000ft am i wrong with that calculations? the question states find final reserve fuel you do not have mass on arrival at destination aerodrome furthermore you have to land with this final reserve in your wing at alternate. Question 94-36
Given planned and actual data as shown in flight log excerpt arriving overhead gamma you are cleared direct routing to mike the flight time direct flight gamma to mike will be 1h10 min assuming other flight data remains constant what fuel will be expected on arrival overhead mike err _a_033 310 25 kg trip fuel 8 kg reserve fuel. from gamma you are cleared direct to mike gamma to mike 1h10 min from beta to gamma fuel consumption was 3200 2700 = 500 kg flight time from beta to gamma 1h42 1h17 = 25 min fuel flow between beta to gamma 500 kg / 25 min = 20 kg/min gamma to mike 1h10 x 20 kg/min = 1400 kg/ remaining fuel at gamma 2700 kg remaining fuel at mike = 2700 1400 kg = 1300 kg. Question 94-37
Given planned and actual data as shown in flight log excerpt arriving overhead gamma you are cleared direct routeing to mike the flight time direct flight gamma to mike will be 57 minutes assuming other flight data remains constant what fuel will be expected on arrival overhead mike err _a_033 327 25 kg trip fuel 8 kg reserve fuel. beta to gamma 25 minutes (1h37 to 2h02) fuel used 2950 2575 = 375 kg fuel flow from beta to gamma 375 / 25 = 15 kg/min fuel used from gamma to mike 57 min x 15 kg = 855 kg on arrival overhead mike fuel on board will be 2575 855 = 1720 kg. Question 94-38
Given fl 370mach 0 74oat 47°cthe tas 25 kg trip fuel 8 kg reserve fuel. img /com_en/com033 333 jpg by calculation mach number = tas/ lss lss= 39 square root t° kelvin lss= 39 square root (273 47)°k lss= 39 square root 226°k = 586 30 mach number = tas / 586 30 = 0 74 tas = 0 74 x 586 30 = 433 86 kt. Question 94-39
The still air distance in climb 189 nautical air miles nam and time 30 minutes what ground distance would be covered in a 30 kt head wind 25 kg trip fuel 8 kg reserve fuel. ground distance (nm) = air distance +/ (time x (effective wind/60)) ground distance (nm) = 189 nam (30 minutes x (30kt/60)) ground distance (nm) = 189 15 = 174 nm stanley is it correct counting 189/30 x 60 = 378 tas 378 30 kt = 348 348/2 = 174 nm yes it works. Question 94-40
At reference or see flight planning manual mrjt 1 figure 4 5 3 1given long range cruise outside air temperature oat 45°c in fl350mass at beginning of leg 40000 kgmass at end of leg 39000 kg find true airspeed tas at end of leg and distance nam err _a_033 345 25 kg trip fuel 8 kg reserve fuel. mass at end of leg 39000 kg we read 422 kt tas on table temperature isa+9 we must increase tas 1 kt per degree c above isa 422 + 9 = 431 kt we have burnt 1000 kg on leg (40000 39000) on table we read following values 40000 >1163 nam 39000 > 936 nam 1163 ? 936 = 227 nam.
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