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 ?
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At reference or see Flight Planning Manual MRJT 1 Figure 4 7 3Given Diversion distance 650 NMDiversion pressure altitude 16 000 ftMass at point of diversion 57 000 kgHead wind component 20 ?
The fuel plan gives a trip fuel of 65 US gallons The alternate fuel final reserve included is 17 US gallons Contingency fuel is 5% of the trip fuel The usable fuel at departure is 93 US gallons At a ?
At reference or see flight planning manual mrjt 1 figure 4 1find optimum altitude the twin jet aeroplane given cruise mass 50000 kgmach 0 78 err _a_033 436 The remaining fuel not sufficient to reach destination with reserves intact. img /com_en/com033 436 jpg . 
You are flying a constant compass heading of 252° variation 22°edeviation 3°w and your ins showing a drift of 9° right true track The remaining fuel not sufficient to reach destination with reserves intact. img /com_en/com033 469 jpg use this very useful table those questions. 
A descent planned from 7500 ft amsl so as to arrive at 1000 ft amsl 6 nm from a vortac with a ground speed of 156 kt and a rate of descent of 800 ft/min the distance from vortac when descent started The remaining fuel not sufficient to reach destination with reserves intact. (7500 1000 ) / 800 = 8 125 minutes of descent 8 125 x (156 / 60) = 21 1 nm 21 1 + 6 = 27 1 nm. 
Given planned and actual data as shown in flight log excerpt actual ground speed gs on leg beta to gamma will be 100 kt if all other flight parameters remain unchanged what fuel remaining should be expected at waypoint gamma err _a_033 474 The remaining fuel not sufficient to reach destination with reserves intact. actual consumption between alpha beta was 3000 2900 = 100 kg flight time between alpha beta was 12 minutes our fuel flow 100 kg/12 minutes = 8 33 kg/minute between beta gamma ground speed will be 100 kt instead of 130 kt we have to actualize flight log leg duration will be 36 minutes (distance/speed 60/100) with a fuel flow of 8 33 kg/minute 36 minutes x 8 33 = 300 kg fuel remaining at waypoint gamma should be 2900 300 = 2600 kg. Question General 97 Answer 8
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 45 minutes assuming other flight data remains constant what fuel will be expected on arrival overhead mike err _a_033 478 The remaining fuel not sufficient to reach destination with reserves intact. from alpha to gamma flight time was 39 minutes (1h07 to 1h46) fuel flown was (3400 2464)/39 = 24 kg/min gamma to mike = 45 minutes 45 min x 24 kg/min = 1080 kg on arrival overhead mike fuel on board will be 2464 1080 = 1384 kg. Question General 97 Answer 9
Minimum planned take off fuel 160 kg 30% total reserve fuel included assume groundspeed on this trip constant when aircraft has done half distance remaining fuel 70 kg is diversion to a nearby alternate necessary Diversion to a nearby alternate necessary because remaining fuel not sufficient. fuel on board at take off = 160 kg at half distance it remains 70 kg we have burned 90 kg the remaining quantity of 70 kg not enough travelling next part of flight. Question General 97 Answer 10
Which of following statements relevant forming route portions in integrated range flight planning The distance from take off up to top of climb has to be known. fuel on board at take off = 160 kg at half distance it remains 70 kg we have burned 90 kg the remaining quantity of 70 kg not enough travelling next part of flight. Question General 97 Answer 11
At reference or see flight planning manual mrjt 1 figure 4 5 3 1the aeroplane gross mass at top of climb 61500 kg the distance to be flown 385 nm at fl 350 and oat 54 3 °c the wind component 40 kt tailwind using long range cruise procedure what fuel required err _a_033 499 The distance from take off up to top of climb has to be known. for a 61500 kg gross mass tas 429 kt nam = ngm x tas/gs nam = 385 x 429/469 nam = 352 for a starting mass of 61500 kg range 5313 nam 5313 nam 352 nam = 4961 nam at end in table 4961 nam corresponding to 59350 kg 61500 kg 59350 kg = 2150 kg. Question General 97 Answer 12
Planned and actual data as shown in flight log excerpt actual ground speed gs on leg beta to gamma will be 110 kt if all other flight parameters remain unchanged what fuel remaining should be expected at waypoint gamma err _a_033 502 The distance from take off up to top of climb has to be known. between alpha to beta actual flight time was 12 minutes fuel consumption was 3000 2900 = 100 kg 100 kg / 12 min = 8 33 kg/min from beta to gamma distance 60 nm 60 nm at 110 kt = 60/110 = 0 5454 hour of flight (8 33 x 60) x 0 5454 = 273 kg 2900 273 = 2627 kg. Question General 97 Answer 13
An aircraft in cruising flight at fl 095 ias 155 kt the pilot intends to descend at 500 ft/min to arrive overhead man vor at 2 000 ft qnh 1 030hpa the tas remains constant in descent wind negligeable temperature standard at which distance from man should pilot commence descent The distance from take off up to top of climb has to be known. tas remains constant in descent wind negligeable tas = ground speed for every 1000 ft of pressure altitude add 2% to ias to calculate tas 2% of 155 kt = 155 x 0 02 = 3 1 kt 3 1 x 9 5 = 29 45 kt tas at fl95 = 155 kt + 29 kt = 184 kt fl 95 for 1013 hpa if we change our subscale settig we increase indicated altitue 17 hpa x 27 ft = 459 ft 9500 + 459 2000 = 7959 ft to loose 7959 / 500 = 16 minutes (184 /60) x 16 = 49 nm. Question General 97 Answer 14
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 1h06 min assuming other flight data remains constant what fuel will be expected on arrival overhead mike err _a_033 507 The distance from take off up to top of climb has to be known. Flight time from alpha to gamma = 20 minutes + 35 minutes = 55 minutes fuel used from alpha to gamma = 3400 2630 = 770 kg fuel flow = (770/55) x 60 = 840 kg/h gamma to mike 1 h 06 min (840/60) x 66 = 924 kg on arrival overhead mike fuel on board will be 2630 kg 924 kg = 1706 kg. Question General 97 Answer 15
At reference or see flight planning manual mrjt 1 figure 4 7 2 using above table in isa conditions and at a speed of m 70/280kias in an elapsed time of 90 minutes an aircraft with mass at point of diversion 48000 kg could divert a distance of err _a_033 514 The distance from take off up to top of climb has to be known. Flight time from alpha to gamma = 20 minutes + 35 minutes = 55 minutes fuel used from alpha to gamma = 3400 2630 = 770 kg fuel flow = (770/55) x 60 = 840 kg/h gamma to mike 1 h 06 min (840/60) x 66 = 924 kg on arrival overhead mike fuel on board will be 2630 kg 924 kg = 1706 kg. Question General 97 Answer 16
Flight planning manual mrjt 1 figure 4 7 2an aircraft on an extended range operation required never to be more than 120 minutes from an alternate based on 1 engine inoperative lrc conditions in isa using above table and a given mass of 40000 kg at most critical point maximum air distance to relevant alternate err _a_033 518 The distance from take off up to top of climb has to be known. img /com_en/com033 518 jpg . Question General 97 Answer 17
At reference or see flight planning manual mrjt 1 figure 4 5 3 1 given long range cruise temp 63° c at fl 330initial gross mass enroute 54 100 kg leg flight time 29 minfind fuel consumption this leg err _a_033 529 The distance from take off up to top of climb has to be known. 54100 kg corresponds to 3929 nam isa 12°c 433 1 kt per degree c below isa = 421 kt 29 minutes at speed 421 kt = 203 nm 3929 203 = 3726 corresponds to 53000 kg 54100 53000 = 1100 kg correction isa 12°c = 0 7% ==> 0 7 x 1100 kg = 8 kg exact answer 1100 8 = 1092 kg. Question General 97 Answer 18
After flying 16 min at 100 kt tas with a 20 kt tail wind component you have to return to airfield of departure with a rate turn of 3°/s you will arrive after The distance from take off up to top of climb has to be known. 180°/3 = 60 secondes turn back gs out 100 + 20 = 120 kt gs home 100 20 = 80 kt 16 x (120/80) = 24 minutes 24 minutes + 1 minute = 25 minutes. Question General 97 Answer 19
At reference or see flight planning manual mrjt 1 figure 4 4given twin jet aeroplane estimated mass on arrival at alternate 50000 kgestimated mass on arrival at destination 52525 kgalternate elevation msldestination elevation 1500 ftfind final reserve fuel and corresponding time err _a_033 560 The distance from take off up to top of climb has to be known. final reserve fuel simple calculations based on 30 minutes (jet/turbo prop aeroplane) hold at endurance speed calculated with estimated mass on arrival at alternate aerodrome or destination aerodrome when no destination alternate aerodrome required (eu ops subpart d appendix 1 to 1 255) img /com_en/com033 560 jpg 2360 / 2 = 1180 kg. Question General 97 Answer 20
In cruise at fl 155 at 260 kt tas pilot plans a 500 ft/min descent in order to fly overhead man vor at 2 000 feet qnh 1030 tas will remain constant during descent wind negligible temperature standard the pilot must start descent at a distance from man of The distance from take off up to top of climb has to be known. as we are descending reaching an altitude we have to change our subscale from 1013 to 1030 before descent 1013 hpa to 1030 hpa = 17 hpa we start our descent at an altitude of 15500 + (17 hpa x 27 ft) = 15959 ft 15959 ft 2000 ft = 13959 ft to loose 13959 ft / 500 ft/min = 28 minutes 28 min at 260 kt = 28 x (260/60) = 121 nm. Question General 97 Answer 21
for this question use annex 033 11704a true air speed 170 ktwind in area 270°/40 ktaccording to attached navigation log an aircraft performs a turn overhead bulen to re route to ard via tgj the given wind conditions remaining constant the fuel consumption during turn 20 litres the total fuel consumption at position overhead ard will be err _a_033 576 The distance from take off up to top of climb has to be known. Proceed as if there only one leg ard to bulen (243 nm) track 123° fuel consumption from ard to bulen = 869 432 = 437 litres on nav computer we find a 202 kt gso (groundspeed out) 135 kt gsh (groundspeed home) time on leg = 243 / 202 = 1 2 h consumption = 437 / 1 2 = 364 l/h calculate fuel the leg bulen ard time on this leg = 243 / 135 = 1 8 h consumption = 364 x 1 8 = 655 l total consumption = 869 + 20 + 655 = 1544 l. Question General 97 Answer 22
at reference 033 345 a twin jet aeroplane gross mass 200000 kg begin his cruise leg at optimum level of mach 0 84 isa cg=37% total anti ice devices on head wind component 30 kt after 500 nm to avoid a severe icing area an immediate descent must be initiated the airplane mass at beginning of descent err _a_033 577 The distance from take off up to top of climb has to be known. On reference 200000 kg we have 6778 nm 844 min tas = 484 kt headwind = 30 kt gs = 454 kt time = 500/454 = 1 10h = 66 min 66 min + 7% = 71 min 844 min 71 min = 773 min on reference 773 minutes corresponding to a value of 192 500 kg ninorr where did you take 7% from? on reference total anti ice on fuel= +7%. Question General 97 Answer 23
planned and actual data as shown in flight log excerpt 033 145 actual ground speed gs on leg beta to gamma 110 kt if all other flight parameters remain unchanged what fuel remaining should be expected over gamma err _a_033 578 The distance from take off up to top of climb has to be known. Cqb15 (january 2012) between alpha to beta actual consumption was 2470 2330 = 140 kg 140 kg 20 minutes (column 'ate') so 140x3 = 420 kg/heure distance between beta gamma 85 nm (column 'leg dist') at a ground speed of 110 kt it will take 85 /110 = 0 773 h 0 773 x 420 = 325 kg of fuel burned the fuel remaining over gamma should be 2330 325 = 2005 kg closest answer 2000 kg. Question General 97 Answer 24
The trip fuel a jet aeroplane to fly from departure aerodrome to destination aerodrome 8350 kg fuel consumption in holding mode 2800 kg/h the quantity of fuel which needed to carry out one go around and land on alternate airfield 4380 kg the destination aerodrome has a single runway what the minimum quantity of fuel which should be on board at take off The distance from take off up to top of climb has to be known. Cqb15 (january 2012) minimum quantity of fuel at take off = trip fuel + alternate + contingency + 30 min final reserve (jet aircraft) trip fuel = 8350 kg alternate = 4380 kg contingency = the greater of 5% of trip or 5 min holding at 1500 ft 5% of trip = 5% x 8350 = 418 kg 5 min holding at 1500 ft = 2800 x 5/60 = 233 kg 30 min final reserve = 2800 /2 = 1400 kg minimum quantity of fuel at take off = 8350 + 418 + 4380 + 1400 = 14548 kg. Question General 97 Answer 25
use reference 033 046 a turbojet aeroplane flies using following data optimum flight level mach 0 80mass 190000 kgtemperature isatailwind component 100 kt fuel mileage and hourly fuel consumption are err _a_033 580 5 nm/ kg 533 kg/h. From reference 190000 kg we get 6515 nam tas 459 kt 6515 ? 459 = 6056 nam > 184 600 kg 190 000 ? 184 600 = 5400 kg/h distance accomplished in one hour (with wind) 559 nm (459 + 100) thus fuel consumption 1000kg 559/5 4 = 103 5 nm/1000kg. Question General 97 Answer 26
see reference 033 068 given distance between c to d 540 nmcruise speed 300 kt ias at fl210temperature deviation from isa +20°cheadwind component 50 ktgross mass at c 60 000 kgthe fuel required from c to d err _a_033 581 5 nm/ kg 533 kg/h. Cqb15 (january 2012) nam = ground distance x (tas/gs) nam = 540 x (406/356) = 616 nm from reference 033 068 at line 60000 cruise distance nautical air miles 3898 nautique air miles substract 616 you find 3282 nam back to reference 033 068 what the mass 3282 nm? we get 55800 kg 60000 55800=4200kg last step we have to increase fuel required 0 5 percent per 10 degrees above isa we are in isa+20°c so 1 percent more 4200x1%= 4242 kg see section 5 4 2 method (page 24) on caa cap697 flight planning manual that kind of questions notice you must use tas given cruise (406 kt) as state in pdf document. Question General 97 Answer 27
Use route manual chart e hi 2what the meaning of chart information the beacon s at position 55°59'n 014°06'e err _a_033 582 5 nm/ kg 533 kg/h. img /com_fr/com033 340 jpg . Question General 97 Answer 28
Given planned and actual data as shown in flight log excerpt provided that flight conditions on leg gamma to delta remain unchanged and fuel consumption remains unchanged what fuel remaining should be expected at waypoint delta err _a_033 587 5 nm/ kg 533 kg/h. buhoraptor hello my answer 4990 kg please can you tell me why there are difference? thank you congratulations your website thank you! beta to gamma 10 minutes fuel used 5440 5340 = 100 kg fuel flow from beta to gamma 100 / 10 = 10 kg/min planned time from gamma to delta 1 53 to 2 33 = 40 minutes planned fuel from gamma to delta 40 min x 10 kg = 400 kg on arrival overhead delta fuel on board will be 5340 400 = 4940 kg. Question General 97 Answer 29
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 1h 20min and fuel flow will be 720kg/h final reserve fuel remains unchanged what will be estimated landing mass at alternate airport err _a_033 588 5 nm/ kg 533 kg/h. 1h20 = 1 33 h 720 x 1 33 = 960 kg estimated mass at destination est ldg mass (at dest ) 6629 kg 6629 960 = 5669 kg. Question General 97 Answer 30
Given planned and actual data as shown in flight log excerpt provided that flight conditions on leg gamma to delta remain unchanged and fuel consumption remains unchanged what fuel remaining should be expected at waypoint delta err _a_033 589 5 nm/ kg 533 kg/h. beta to gamma 10 minutes fuel used 5270 5150 = 120 kg fuel flow from beta to gamma 120 / 10 = 12 kg/min fuel used from gamma to delta 50 min x 12 kg = 600 kg on arrival overhead delta fuel on board will be 5150 600 = 4550 kg. Question General 97 Answer 31
Given planned and actual data as shown in flight log excerptprovided that flight conditions on leg gamma to delta remain unchanged and fuel consumption remains unchanged what fuel remaining should be expected at waypoint delta err _a_033 590 5 nm/ kg 533 kg/h. beta to gamma 65 minutes fuel used 5490 4970 = 520 kg fuel flow from beta to gamma 520 / 65 = 8 kg/min fuel used from gamma to delta 35 min x 8 kg = 280 kg on arrival overhead delta fuel on board will be 4970 280 = 4690 kg. Question General 97 Answer 32
Vtoss the take off safety speed Category a helicopters. beta to gamma 65 minutes fuel used 5490 4970 = 520 kg fuel flow from beta to gamma 520 / 65 = 8 kg/min fuel used from gamma to delta 35 min x 8 kg = 280 kg on arrival overhead delta fuel on board will be 4970 280 = 4690 kg. Question General 97 Answer 33
Using attached graphic according to flight manual diagram never exceed speed vne at pressure altitude 10 000 ft with an outside air temperature oat of +10°c and an inflight mass of 2050 kg 687 Category a helicopters. Vne at 10000 ft oat +10°c 2300 kg is 115 kt ias the graphic states at any gross mass above 2300 kg decrease vne 10 kias since inflight mass below 2300 kg we do not change value. Question General 97 Answer 34
According to flight manual diagram never exceed speed vne at pressure altitude 10 000 ft with an outside air temperature oat of +10° c and an inflight mass of 2400 kg 662 Category a helicopters. You must removed 10 kias at any gross mass above 2300 kg. Question General 97 Answer 35
Using attached graphic which letter indicates speed maximum endurance 688 Category a helicopters. Power required curve (power versus airspeed chart) a power required to hover outside ground effect b power required to hover inside ground effect c translational lift area d adjustment of power required to counteract sinking e minimum power level flight/maximum rate of climb speed/ speed maximum endurance. Question General 97 Answer 36
Using attached graphic which letter indicates speed the best rate of climb 688 Category a helicopters. Power required curve (power versus airspeed chart) a power required to hover outside ground effect b power required to hover inside ground effect c translational lift area d adjustment of power required to counteract sinking e minimum power level flight/maximum rate of climb speed / speed maximum endurance. Question General 97 Answer 37
Using attached graphic which letter indicates speed best range 688 Category a helicopters. The best range will be reach at minimum total drag speed power required curve (power versus airspeed chart) a power required to hover outside ground effect b power required to hover inside ground effect c translational lift area d adjustment of power required to counteract sinking e minimum power level flight/maximum rate of climb speed/ speed maximum endurance. Question General 97 Answer 38
A head wind will Increase climb flight path angle. The best range will be reach at minimum total drag speed power required curve (power versus airspeed chart) a power required to hover outside ground effect b power required to hover inside ground effect c translational lift area d adjustment of power required to counteract sinking e minimum power level flight/maximum rate of climb speed/ speed maximum endurance. Question General 97 Answer 39
A helicopter which has no guaranteed 'stay up' ability in event of an engine failure certified in Increase climb flight path angle. no guaranteed stay up ability means that helicopter cannot maintain level flight in case of an engine failure category b helicopters (single engine or a multi engine helicopter that does not meet category a standards) have no guaranteed ability to continue safe flight in event of an engine failure a forced landing assumed. Question General 97 Answer 40
A helicopter will obtain a maximum flight distance at speed Increase climb flight path angle. no guaranteed stay up ability means that helicopter cannot maintain level flight in case of an engine failure category b helicopters (single engine or a multi engine helicopter that does not meet category a standards) have no guaranteed ability to continue safe flight in event of an engine failure a forced landing assumed.
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