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Question 111-1 : According to cs 25, the pilot may apply maximum braking after landing ? [ Training professional ]
Once the main gear has touched down provided you can lower the nose gear safely.
Amc 25.125 c landing..1 during measured landings, if the brakes can be consistently applied in a manner permitting the nose gear to touch down safely, the brakes may be applied with only the main wheels firmly on the ground. otherwise, the brakes should not be applied until all wheels are firmly on the ground...2 this is not intended to prevent operation in the normal way of automatic braking systems which, for instance, permit brakes to be selected on before touchdown.
Question 111-2 : According to part cat, what is correct regarding the declared runway distances ?
These do not account for runway line up, and are usually adjusted in case of a 90° entry or 180° turnaround.
..easa air ops..amc1 cat.pol.a.400 take off..loss of runway length due to alignment.. a the length of the runway that is declared for the calculation of toda, asda and tora does not account for line up of the aeroplane in the direction of take off on the runway in use. this alignment distance depends on the aeroplane geometry and access possibility to the runway in use. accountability is usually required for a 90° taxiway entry to the runway and 180° turnaround on the runway. there are two distances to be considered.. 1 the minimum distance of the main wheels from the start of the runway for determining toda and tora, ‘l’. and.. 2 the minimum distance of the most forward wheel s from the start of the runway for determining asda, ‘n’.
Question 111-3 : Select the correct statement when comparing a level runway to one with a 2% downslope ?
Vmbe will be lower than vmbe on a level runway.
..maximum brake energy speed vmbe..the maximum speed on the ground from which an airplane can safely stop within the energy capabilities of the brakes. v1 must not be greater than vmbe. at v1 the airplane must be able to stop or continue the take off, but above vmbe it is impossible to bring the airplane safely to a stop.... with an uphill slope it will be easier for the aircraft to stop, increasing the vmbe. increasing the vmbe limit, a higher v1 allowed permits a higher maximum take off mass.. on the other hand, a downslope will have an adverse effect on braking ability. vmbe and v1 will be reduced.
Question 111-4 : Select the combination of answers which reflect the reason for using reduced thrust..1. save engine life..2. reduced fuel consumption..3. v1 can be less than vmcg..4. noise abatement..5. allows contaminated runway operations..6. pmc inoperative ?
1 and 4
Advantages of using reduced take off thrust.using reduced thrust lowers the jet engine's internal operating pressure and temperature. the primary advantage of the use of reduced take off thrust is cost savings through increased engine life and reduced overhaul costs. the main benefits are reduced stress and wear on the engine. reduced cost on parts and mainenance. increased engine life. increased reliability, thus improving operating safety and efficency.... there are slipt view regarding statements 1 and 4. some sources state that reduced thrust secondary advantages include fuel savings, however assuming an uninterrupted climb, it will take longer to reach cruising altitude and consequently more fuel will be consumed.. concerning 4., as a result of the lower flight path, reduced take off thrust may increase the airplane to ground noise level but less thrust will also produces less noise....note we have received consistent feedback confirming the current correct option 1 and 4. nevertheless, the correct answer depends on the examiner's views and should therefore be appealed. please let us know if you come across this question in your official exam.
Question 111-5 : Complete the following sentence..for a take off on a runway contaminated by slush or loose snow the displacement and impingement drag will have a 1 effect on acceleration during the take off and a 2 effect on the deceleration during a rejected take off. ?
1 negative. 2 positive.
Loose contaminants result in additional drag due to the combination of displacement of the contaminant by the airplane tires and impingement of the contaminant spray on the airframe. this contaminant drag provides an additional force impeding acceleration during a takeoff, or assisting deceleration during a rejected takeoff....fluid contaminants contribute to stopping force by... resisting forward movement of the wheels i.e., causing displacement drag. and,. creating spray that strikes the landing gear and airframe i.e., causing impingement drag.
Question 111-6 : Consider a large commercial air transport turbojet aeroplane, when comparing the long range speed to the maximum range speed, the long range speed. assuming all other factors remain unchanged ?
Corresponds to a higher cost index.
Theoretically, the long range cruise should be the same speed and technique as maximum range. however, a higher power setting is usually recommended, which still gives a good specific range about 99% of the maximum range but increases the speed and decreases sector times. the long range cruise lrc speed is approximately 4% faster than maximum range speed with a fuel penalty of approximately 1%...cost index = cost of time ÷ cost of fuel... we can conclude that the lrc speed corresponds to a higher cost index than the maximum range speed.
Question 111-7 : The principles of pavement classification numbers pcns and aircraft classification numbers acns are described by icao annex 14. these numbers describe the strength of a surface and the impact an aircraft will have on a surface. select the correct option. ?
An aircraft with a given acn value may operate on a surface with a lower pcn value if certain operational limitations are met.
Icao annex 14.aircraft classification number acn. a number expressing the relative effect of an aircraft on a pavement for a specified standard subgrade category.pavement classification number pcn. a number expressing the bearing strength of a pavement for unrestricted operations.20. the acn pcn method of reporting pavement strength.a for flexible pavements, occasional movements by aircraft with acn not exceeding 10 per cent above the reported pcn should not adversely affect the pavement..b for rigid or composite pavements, in which a rigid pavement layer provides a primary element of the structure, occasional movements by aircraft with acn not exceeding 5 per cent above the reported pcn should not adversely affect the pavement.conclusion as a general rule, the acn must not exceed the pcn, except that, provided this represents less than 5% of the movements, on a flexible pavement an acn can exceed the pcn by up to 10%, and on a rigid pavement up to 5%.
Question 111-8 : When planning a take off from porto lppr at a take off mass of 89 tonnes, the efb performance software gives a flex temperature of 36° and a range of possible decision speeds with v1 max at 155 kts and v1 min at 142 kts. complete the following sentence using a v1 of 155 kts, the climb performance 1 ?
1 improved. 2 improved.
The take off decision speed v1 is the maximum speed at which a rejected take off can be initiated and a stop within the asda can be accomplished with the critical engine inoperative. v1 is also the minimum speed at which a pilot can continue takeoff following an engine failure...following an engine failure, the overall thrust is decreased and the drag is increased. consequently, the aircraft will accelerate more slowly.... with a higher v1, the acceleration to vlof to be achieved on one engine is smaller than with a lower v1. therefore, the aircraft will take off sooner closer to the line up position. it will then climb at the same shallow angle as with a lower v1. with a higher v1, the overall climb performance regarding obstacle clearance , is improved.. choosing a lower v1 will imply that an engine failure can be accepted at a lower speed. meaning that the aircraft will also initiate its deceleration at a lower speed improving the stopping performance.
Question 111-9 : On a transatlantic flight, the captain asks the first officer why a final step climb from fl340 to fl380 ten minutes before top of descent might not be an efficient use of the fuel. select a possible explanation. ?
A 4000 ft climb might consume more fuel than would be saved by being at the higher altitude for a short period.
Optimum altitudes on short stages..airbus philosophy assumes a minimum 5 minute cruise sector, because a climb followed immediately by the descent is not appreciated by pilots, passengers or atc...if the stage length is of sufficient length that the optimum flight level can be reached, but the cruise is of short duration, then the benefits at this flight level will be marginal. it may even be worthwhile to cruise at one flight level lower, as the increased climb consumption offsets any reduced cruise consumption.
Question 111-10 : During the initial climb out, the pilot is required to do a 15° turn to the left to avoid an obstacle. the total drag will ?
Increase and the climb gradient will decrease.
Any manoeuvre such as a turn calls for more lift. due to the bank angle of the aircraft, the lift vector is tilted, therefore only a component of lift counteracts weight. but since level flight must be maintained, the component of lift has to be equal to weight. consequently, lift must be increased. by increasing the bank angle, load factor increases. due to increased lift, induced drag also increases producing greater drag, this in turn increases total drag...for a given aircraft mass the maximum climb gradient will occur when the excess thrust over drag is greatest. as seen above, the drag increases in a turn, provided that thrust remains the same, the climb gradient will decrease.
Question 111-11 : The standard climb speeds for a medium range jet transport are 250 kts below 10 000 ft and subsequently 300 kts/0.80m. choose the speed profiles you may expect if you entered a lower than usual cost index into the fmc. ?
250/280/0.78m
The cost index ci is calculated as the ratio of time costs to fuel costs cost index = time costs/fuel coststhe lower the ci, the more 'importance' the machinery places on saving fuel. low cost indexes will result in lower climb speed, both indicated and mach , lower cruise speed, a generally higher cruise altitude, a later descent and a slower descent mach/speed. the higher ci's will result in the opposite.in this case the first speed 250 kt below 10 000 ft is an atc restriction, therefore it must remain unchanged.since we now have a lower cost index, we must accelerate to a speed lower than 300 kt. the same logic applied to the speed at th crossover altitude. the only correct option 250/280/0.78m
Question 111-12 : An aircraft descends at m 0.78 instead of m 0.75. what effect will this have on its rate of descent rod ?
It will increase.
Rate of descent rod is the vertical component of the aircraft's velocity, normally expressed in feet per minute. rod is unaffected by wind.in a descent, if speed is increased, the aircraft will cover a greater distance per minute. thus, rate of descent increases.
Question 111-13 : A pilot is asked by atc to descend to fl120 at a point 50 nm away. he/she realizes he/she won’t make it. what should the pilot do ?
Descend at mmo until changeover altitude, switch to ias as long as you don’t exceed vmo. use speed brakes if needed.
..to maximize the rod, the airspeed should be maximized within the operational limits of the aircraft. at cruising altitude an airliner will be flying mach number, so the operational limit is the mmo maximum operating mach number. however, if the aircraft keeps descending at a constant mach, the tas and ias will keep increasing, leading to a vmo overspeed situation. therefore, at the changeover altitude, the flight crew must stop flying at a mach number and fly at constant ias...thus, the aircraft would initially descend at mmo, at the crossover altitude the limiting speed will become vmo...with airbrakes in use, the gradient of descend increases, consequently the rod increases as well.
Question 111-14 : After a new software update, the engineer has changed the crossover data from 250 kt / m0.75 to 250 kt / m0.78. the pilot now must make sure he/she does not exceed ?
Mmo
..the crossover altitude is the altitude where a specific indicated airspeed ias becomes a specific mach number m. it is the altitude where a specified ias and mach number represent the same tas...below the crossover altitude, the climb/descent is performed maintaining a constant ias. whereas above the crossover altitude, the climb/descent is performed maintaining a constant mach number. at the crossover altitude, there is one value of tas which corresponds to the given mach number and the given ias at the same time...in this case, the crossover altitude of this aircraft has been incorrectly raised higher up. this is because the altitude where 250kias = m0.78 is higher than the altitude where 250 kias = 0.75. draw an ectm graph or use the finger trick to prove this to yourself if necessary...as this is a tricky question requiring a lot of thought, let us go through both the climb scenario and the descent scenario...in the climb scenario, before/below the ca crossover altitude , we are flying at a constant ias, with our mach no. increasing as we climb. as the ca is now higher than it should be, it means we spend more time increasing our mach no., and that means it may go over our mmo. we cannot breach our vmo here as we are flying at a constant ias before the ca, and then a reducing ias after the ca...in the descent scenario, we are initially descending with constant mach no., but increasing ias. as the ca is higher up than usual, we reach it earlier, and our ias is not able to increase up to its usual value, so our descent below the ca may be at a lower speed, however, there is no chance of overspeed in the descent scenario, either by breaching the vmo or the mmo. we start with a constant mach no., decreasing after the ca, and our ias initially increases, but to a lower value than usual, then remains constant...so to conclude, the only opportunity for any type of overspeed here is by breaching mmo, and that would be possible when in the climb.
Question 111-15 : In order to carry out the increased v2 procedure the length of runway required is 1 and this helps to 2 obstacle clearance by 3 the take off run. ?
1 long. 2 increase. 3 increasing.
Take off with increased v2 speed. this procedure is used when the performance limited mass is the climb limit mass. at v2 the climbing performance is poor and limits the maximum take off mass. it is important to understand that in the event of engine failure, the initial climb out speed is v2. however, v2 is not the best climb angle speed. v2 is considerably slower than the best angle of climb speed, which is vx. climbing with a speed closer to the best angle of climb greatly enhance the climb performance. if runway available is long enough, it would be possible to stay on the runway for longer during the take off to build up more speed to a higher v1, this will ensure that at the screen height, a faster v2 will be reached, closer to vx. as a result of the improved climb performance, the climb limit mass can be increased.
Question 111-16 : Which of the following would be considered to be a high cost index ?
280/0.76/279
The cost index ci is calculated as the ratio of time costs to fuel costs cost index = time costs/fuel coststhe lower the ci, the more 'importance' the machinery places on saving fuel. low cost indexes will result in lower climb speed, both indicated and mach , lower cruise speed, a generally higher cruise altitude, a later descent and a slower descent mach/speed. higher ci's will result in the opposite higher speeds.in this case the highest cost index will correspond to the highest speeds. correct option 280/0.76/279
Question 111-17 : A performance class a turbojet aeroplane is planned to be operated on contaminated runways and on wet runways with suitable performance accountability for the increased stopping distance. for planning purposes, the use of ?
Derated take off thrust is allowed on both types of runways.
Reduced take off thrust atm may be used for take off on a wet runway if approved take off performance data for a wet runway is used. however, reduced takeoff thrust atm is not permitted for takeoff on a runway contaminated with standing water, slush, snow, or ice...derated takeoff thrust fixed derate may be used for takeoff on a wet runway and on a runway contaminated with standing water, slush, snow, or ice.
Question 111-18 : An aeroplane is conducting a 6 hour sector flying from north africa to europe. the initial cruising altitude is fl340 with the recommended maximum being fl390..normally the optimum cruising altitude is 500 ft to 2500 ft below the recommended maximum and the recommended maximum altitude increases by ?
The wind component at higher altitudes is less favourable, offsetting the increased fuel burn at the lower altitude.
There are some situations in which the pilots may decide not to make use of the step climb method and yet obtain better overall performance. one of them would be to make use of advantageous meteorological conditions such as staying within the core of a tailwind jet stream to reduce the air distance to be travelled, flight time and associated fuel consumption...rather than blindly going +2 000 or 2 000 feet or whatever is required in rvsm from optimum, use your wind forecasts and tradeoff tables to figure out when you should climb. depending on the specific airplane, altitude range, and wind situation, staying low may actually use less fuel.
Question 111-19 : Which of the following statements are correct regarding the take off climb gradient requirements for a performance class a aeroplane the climb gradient requirement in....1. each take off climb segment is a flight path climb gradient,. 2. the second segment of a four engined aeroplane is higher than ?
2, 4, and 6.
Segments of the take off climb.for a class a aircraft, one engine inoperative, the t/o climb is divided into 4 segments the take off flight path starts once the take off is complete, at 35 ft with the airplane at v2 with one engine inoperative. on a wet runway, the screen height is reduced to 15 ft. operating engines are at take off thrust, the flaps/slats are in take off configuration and landing gear retraction is initiated once safely airborne with positive climb. the first segment ends when the landing gear is fully retracted.begins when the landing gear is fully retracted. engines are at take off thrust and the flaps/slats are in the take off configuration. this segment ends at the higher of 400 ft.begins at 400 ft or higher specified acceleration altitude. engines are at take off thrust and the aircraft is accelerated in level flight. slats/flaps are retracted on speed. the segment ends when aircraft is in clean configuration and the final take off speed has been achieved. once this has happened, thrust can be reduced from maximum take off thrust, toga, to maximum continuous thrust, mct.starts when the flaps are retracted, the final segment speed is achieved, and the thrust is set to maximum continuous thrust. from this point the airplane is climbed to above 1500 ft where the take off flight path ends. the climb gradient for this last stage must not be less than 1.2%. 1º segment2º segment3º segment4º segmentstarts35 ftgear up> 400 aglflaps up.vfto.mctactionselect gear upclimb to > 400 aglretract flaps.accelerate to vfto.set mctclimb to 1500 aglgradient for 2 engines.> 0%.> 2.4%.> 1.2%.> 1.2%gradient for 3 engines.> 0.3%.> 2.7%.> 1.5%.> 1.5%gradient for 4 engines.> 0.5%.> 3.0%.> 1.7%.> 1.7%.note while third segment is usually flown in level flight, the available gradient must be at least equal to that required in final segment 1.2%. during third segment the high lift devices are retracted.
Question 111-20 : When performing an increased v2 takeoff, our v1 speed will be increased. what related hazards shoud we pay most attention to ?
The stop distance may be compromised.
Take off with increased v2 speed..this procedure is used when the performance limited mass is the climb limit mass. at v2 the climbing performance is poor and limits the maximum take off mass. it is important to understand that in the event of engine failure, the initial climb out speed is v2. however, v2 is not the best climb angle speed. v2 is considerably slower than the best angle of climb speed, which is vx. climbing with a speed closer to the best angle of climb will greatly enhance the climb performance.... if extra runway length is available it would be possible to stay on the runway for longer during the take off to build up more speed to a higher v1, this will ensure that at the screen height, a faster v2 will be reached, closer to vx. as a result of the improved climb performance, the climb limited mass can be increased.....'the climb gradient is reduced, and we may not clear distant obstacles.' incorrect.... increased v2 procedure increases climb gradient, so obstacle clearance should be less of a problem...'the new v1 may be above controllability limits.' incorrect.... controllability limits are talking about vmcg, which is the minimum control speed on the ground, below which the rudder does not have enough airflow to fight an asymmetry if the engine fails. at higher speeds, this would not be a problem...'the braking efficiency during a rejected take off will be reduced.' incorrect.... the braking ability and efficiency will be the same. the increased v1 will mean that we are faster if we do a high speed rto rejected take off , so we will take further to stop, but our braking ability should be unaffected...'the stop distance may be compromised.' correct.... increasing our v1 reduces the safety margin of our asda accelerate stop distance available , and must be checked to make sure our asdr is within our asda.
Question 111-21 : A pilot is calculating the take off speeds for a contaminated runway and chooses a wet v1. should he/she expect this wet v1 to be higher or lower compared to v1 for dry runway conditions ?
Wet v1 would be lower for wet runway conditions.
A dry maximum v1 is the normal decision speed that following an engine failure allows the take off to be continued safely within the toda or to be stopped safely within the asda. a wet minimum v1 is the maximum speed for abandoning a take off on a contaminated runway. a wet v1 improves the stopping capabilities final stop point back to the dry conditions level but degrades the takeoff chances with a reduced screen height in the event of a take off being continued. a recommended wet v1 for contaminated conditions is the dry v1 10 knots. thus wet v1 is a lower speed than dry v1...note due to lower friction and consequently reduced braking action on a wet runway, longer braking distances are experienced, v1 must be decreased.
Question 111-22 : What is correct regarding the vertical obstacle clearance requirement for a performance class a aeroplane..if, in the engine out take off flight path, the bank angle is more than 150, all relevant obstacles must be cleared by at least 1 , in relation to the 2 take off flight path. ?
1 50 ft 2 net
Easa air ops. cat.pol.a.210 take off obstacle clearance b when showing compliance with a 3 any part of the net take off flight path in which the aeroplane is banked by more than 15° shall clear all obstacles within the horizontal distances specified in a , b 6 and b 7 by a vertical distance of at least 50 ft.
Question 111-23 : Aircraft type specific procedures for take off and landing on contaminated runways can be found in the operations manual , under normal procedures. ?
Part b
Easa air ops. amc3 oro.mlr.100 operations manual – general. b aircraft operating matters — type related2 normal procedures. the normal procedures and duties assigned to the crew, the appropriate checklists, the system for their use and a statement covering the necessary coordination procedures between flight and cabin/other crew members. the normal procedures and duties should include the following a pre flight,. b pre departure,. c altimeter setting and checking,. d taxi, take off and climb,. e noise abatement,. f cruise and descent,. g approach, landing preparation and briefing,. h vfr approach,. i ifr approach,. j visual approach and circling,. k missed approach,. l normal landing,. m post landing,. n for aeroplanes, operations on wet and contaminated runways.oro.mlr.101 operations manual – structure for commercial air transport. except for operations with single engined propeller driven aeroplanes with an mopsc of 5 or less or single engined non complex helicopters with an mopsc of 5 or less, taking off and landing at the same aerodrome or operating site, under vfr by day, the main structure of the om shall be as follows a part a general/basic, comprising all non type related operational policies, instructions and procedures.. b part b aircraft operating matters, comprising all type related instructions and procedures, taking into account differences between types/classes, variants or individual aircraft used by the operator.. c part c commercial air transport operations, comprising route/role/area and aerodrome/ operating site instructions and information.. d part d training, comprising all training instructions for personnel required for a safe operation.
Question 111-24 : What is correct regarding the effect of displacement drag and impingement drag on take off performance there is a 1 effect in case of a rejected take off and a 2 effect for continued take off. ?
1 positive. 2 negative
Fluid contaminants contribute to stopping force by... resisting forward movement of the wheels i.e., causing displacement drag. and,. creating spray that strikes the landing gear and airframe i.e., causing impingement drag....a slower acceleration due to an increase in drag as the aircraft tyres plow through the contaminant displacement drag and a spray of contamination is created which strikes the landing gear and airframe impingement drag can be expected => it has a negative effect on take off distance and positive effect on rejected take off distance...note we assume that easa refers to the rejected take off distance only, rather than the asdr the runway required to accelerate to vr and bring the aircraft to a complete stop.
Question 111-25 : Cs 25.121 d 1 paragraphs i to iv states the pre requisite conditions for an aeroplane, during certification, demonstrating its one engine inoperative gradient of climb in the approach phase...which of the following options states the correct description of one of these prerequisites ?
The landing gear is retracted.
Cs 25.121 climb one engine inoperative d approach. in a configuration corresponding to the normal all engines operating procedure in which vsr for this configuration does not exceed 110% of the vsr for the related all engines operating landing configuration 1 the steady gradient of climb may not be less than 2·1% for two engined aeroplanes, 2·4% for three engined aeroplanes and 2·7% for four engined aeroplanes, with – i the critical engine inoperative, the remaining engines at the go around power or thrust setting.. ii the maximum landing weight.. iii a climb speed established in connection with normal landing procedures, but not more than 1·4 vsr. and. iv landing gear retracted.
Question 111-26 : This formula for calculating the dynamic hydroplaning speed distinguishes between rotating and non rotating tyres. which of the following statements best describes the most limiting speed for an aircraft operating on a wet runway ?
The hydroplaning speed of non rotating tyres is lower, because there is a higher risk of hydroplaning immediately following wheel spin up during touchdown.
. dynamic hydroplaning is an event which occurs when water forms a wedge between a tyre and the ground, effectively creating a barrier that stops the tyre gaining any traction, and therefore any braking ability, from the runway.it happens above the hydroplaning speed, which is calculated depending on the tyre pressure and whether the tyre is rotating or non rotating at the time. non rotating is referring to landing, before the wheels spin up. a higher hydroplaning speed is better.if rotating, hydroplaning speed = 9 x tyre pressure psi. if not rotating, hydroplaning speed = 7.7 x tyre pressure psi the non rotating tyres hydroplane easier e.g. at a slower speed as they require extra traction to spin up, which is more easily stopped by the ‘water barrier’ than if they were already spinning.note the aircraft mass does not directly affect hydroplaning speed.
Question 111-27 : A pilot is operating a series of short haul flights. the landing at one destination was performed at the maximum landing mass and the aircraft is only scheduled to spend a short time on the ground prior to the next departure. brake temperature… ?
May limit the turnaround time, to ensure that the wheel fuse plugs will not melt during the subsequent taxi out or take off.
Important note the current feedback from anac portugal provides option c fusible wheel plugs melting as the correct answer in the exam. we believe that option b absorbing brake energy is far more correct, and a much more important factor, whilst option c is still a possible scenario, but very unlikely and much less dangerous than a runway excursion during a rejected take off.nonetheless, if these two options are present in your exam, you should appeal this question on those grounds. we hope that a previous appeal will remove it from the portuguese exams at least, but we can never be sure.the quick turnaround weight limit is related to the heat of the aeroplane wheel brakes, which absorb the vast majority of the aeroplane's kinetic energy as heat during landing. these brakes are then hot and the temperature of them depends on the weight, approach speed, wind, use of reverse thrust, autobrake setting, etc..hot brakes introduce 2 major problems 1 if the aircraft attempts to take off soon after this, but has to reject the take off at or near v1, the brakes may still be hot, and not able to absorb enough energy to stop the aircraft. this occurs as 'brake fade', where the extremely hot brakes stop gaining much grip between the brakes and the pads other, more minor, issues can also occur..2 to prevent the intense heat from the brakes over pressuring the tyres, the tyres contain 'fusible plugs' that melt at a certain temperature, allowing a controlled escape of the gas, rather than an explosion. having hot brakes could cause these to melt, but this usually happens just after extreme braking, such as a rto rejected take off at a heavy weight, and they do not usually melt just after landing, and very rarely during taxi and take off unless the turnaround time has been very short indeed, and taxiing requires lots of braking these reasons mean that we would like to get our brakes down to a suitable temperature prior to the next flight, and why pilots may sometimes have to consult the 'minimum turnaround time' table/figure to find out when this will be. other methods to help cool the brakes quicker are to place brake fans pointing at them to provide fresh air for improved cooling.giving the above explanation, we strongly believe the brake fade absorbing the energy of an rto to be the most correct answer here, but see the note above and check the comments for the current feedback on what is considered correct in the exam, as this appears to be disputed by some authorities.
Question 111-28 : You are the pilot of a jet transport aeroplane and are doing performance calculations for your departure on a contaminated runway with a strong crosswind component. which of the following statements best describes guidance for calculating take off performance ?
Special crosswind limitations may apply in case of contaminated runways. aircraft type specific limitations are to be found in the operations manual part b.
A contaminated runway is a runway where more than 25% of the surface area is covered in either.. more than 3 mm of water or its equivalent in wet snow or slush. compacted snow which will resist further compression and either hold together or break into lumps when picked up or. ice, including wet ice....taking off or landing on contaminated runways involves increased levels of risk related to deceleration and directional control...operating procedures may further restrict all such operations, or impose flight crew specific restrictions, requirements or limitations, such as by applying special crosswind limitations in case of contaminated runways...the operations manual, among other parts, contains part b which refers to... general information and units of measurement. limitations. abnormal and emergency procedures. performance. flight planning. mass and balance. loading. configuration deviation list. minimum equipment list. survival and emergency eqiopment including oxy emergency evacuation procedures. aeroplane systems
Question 111-29 : Which of the following statements describes typical characteristics of continuous descent arrivals cdas. by performing a cda the aircraft... ?
Remains higher for longer and operates at lower engine thrust.
..continuous descent arrival cda is an aircraft operating technique in which an arriving aircraft descends from an optimal position with minimum thrust and avoids level flight to the extent permitted by the safe operation of the aircraft and compliance with published procedures and atc instructions...the meaning of a cda is to avoid having to level off in stages, during the descent phase and to plan an approach/descent at idle thrust as far as is possible. the purpose of this is to reduce noise and save fuel, not time, by keeping the aircraft as high as possible for as long as possible...this implies that you start your descent at the correct point. if you start your descent too early, you have to either apply thrust in the descent to arrive at the bottom of descent at the right place, or level off and then apply thrust. if you start your descent too late, you have to apply the speedbrakes so as to increase the drag and arrive at the bottom of the descent point...despite the fact that the question does not take into consideration the 'bad' option of starting the descent too late, which requires speedbrakes, the wrong answers only consider a wasteful early descent...therefore, the statement 'remains higher for longer and operates at lower engine thrust', matches better.
Question 111-30 : Which flight profile corresponds to the highest cost index ?
Climb tas = 280 kts/mach 0.78 cruise tas = 350 kts/mach 0.8 descent tas = 280 kts/mach 0.75
The two major variable costs are the cost of fuel and the cost of flight time, excluding fuel. the longer the time airborne, the higher will be the scheduled maintenance, salary and leasing charges. flying faster will save on flight time costs, but increases the fuel burn and therefore fuel cost. the factor derived from the ratio of these two costs, the aircraft operating cost divided by the fuel cost, is called cost index ci and is given by the formula ci = costs incurred over time / fuel cost...'costs incurred over time' or 'time related costs' contain the sum of several components, such as ho y maintenance cost, flight crew and cabin crew cost per flight hour, the cost of ownership or aircraft rental, overtime, passenger dissatisfaction, hubbing or missed connections, etc....the ci is a number from 0 to a maximum which could, depending on type, be 100, 200, 500, 999 or even 9999. an index of 0 represents maximum range and the maximum value, whatever it is, represents minimum time. it is normal for a company route to have an attached ci. alternatively, the ci can be manually input bythe flight crew...airlines determine this ratio using a somewhat complex formula that is presented as a number on the flight plan which the crew then sets in the fms to control the speed. the higher the cost index the faster the speed, the greater the fuel burn and obviously the shorter the flight time. at a cost index of 000 the aircraft will be at maximum range speed vmr...the effect of a higher cost index is to increase the speeds, thus saving flight time, but increasing the fuel consumption. the effect of a lower cost index is the opposite, to decrease the speeds, thus saving fuel but increasing flight time...the only flight profile that corresponds to the highest cost index is climb tas = 280 kts/mach 0.78 cruise tas = 350 kts/mach 0.8 descent tas = 280 kts/mach 0.75, where all speeds at each phase of flight are the highest.
Question 111-31 : Complete the following phrase..any part of the 1 take off flight path in which the aeroplane is banked by more than 15° shall clear all obstacles by a vertical distance of at least 2 . ?
1 net. 2 50 ft.
Easa air ops..cat.pol.a.210 take off obstacle clearance.. 3 any part of the net take off flight path in which the aeroplane is banked by more than 15° shall clear all obstacles within the horizontal distances specified in a , b 6 and b 7 by a vertical distance of at least 50 ft.
Question 111-32 : Flying vfr from peiting 47°48.0'n 010°55.5'e to immenstadt 47°33.5'n 010°13.0'e determine the distance. 2440 ?
32 nm.
. 1136.report the track distance from peiting to immenstadt along the meridian, count the intervals one nautical mile per interval/minute.
Question 111-33 : Flying from edpj laichingen airport 48°30'n, 009°38'e to edtm mengen airport 48°03'n, 009°22'e.find magnetic course and the distance. 2215 ?
Magnetic course 202°, distance 28 nm.
. 1142.center your protractor over laichingen airport, you find a magnetic course of 202°..report the track distance along the meridian, count the intervals one nautical mile per interval/minute.
Question 111-34 : Aeronautical chart icao 1 500 000 stuttgart no 47/6 or route manual vfr gps chart ed 6..flying from erbach airport 48°21'n, 009°55'e to poltringen airport 48°33'n, 008°57'e..find magnetic course and the distance.. 1469 ?
Magnetic course 287°, distance 41 nm.
Img1148..report the magnetic north tick from sulz vor for example. it will allow you to measure the magnetic bearing of the track..distance calculation.track length is 1° of longitude..at latitude 48°n, 1° = 60 nm x cos48 = 40.14 nm.
Question 111-35 : Use route manual vfr gps chart ed 6..flying from edtm mengen airport 48°03'n, 009°22'e to edsz rottweil zepfenhan 48°12'n, 008°44'e..find magnetic course and the distance. 1470 ?
Magnetic course 288°, distance 27 nm.
Report the magnetic north tick from meg ndb, it will allow you to measure the magnetic bearing of the track 288°...distance calculation..track length is 009°22' to 008°44' = 0.38' of longitude.0.38' in degree. 100/60 x0.38 =0.63°of longitude...at latitude 48°n, 1° = 60 nm x cos48 = 40.14 nm...40.14 x 0.63 = 25.3 nm.
Question 111-36 : Aeronautical chart icao 1/500 000 stuttgart no 47 6 or route manual vfr+gps chart ed 6..flying vfr from peiting 47°48.0'n 010°55.5'e to immenstadt 47°33.5'n 010°13.0'e determine the magnetic course. 2441 ?
243°.
. 1136
Question 111-37 : Use route manual vfr gps chart ed 6..flying from edsz rottweil zepfenhan 48°12'n, 008°44'e to edtm mengen airport 48°03'n, 009°22'e..determine the highest obstacle within a corridor 5nm left and 5 nm right of the courseline. 1472 ?
3760 ft.
There is a tranmission tower located north/north west of rottweil zepfenhan.. 1158
Question 111-38 : For this question use trm, vfr directory for greece 7 3..what is the local time lt in greece in i winter and ii summer. 2442 ?
I lt = utc + 2 hours, ii lt = utc + 3 hours.
Img1171
Question 111-39 : Use route manual vfr gps chart ed 6..flying from edpj laichingen airport 48°30'n, 009°38'e to edsz rottweil zepfenhan 48°12'n, 008°44'e..find magnetic course and the distance.. 1471 ?
Magnetic course 243°, distance 41 nm.
. 1172.center your protractor over laichingen airport, you find a magnetic course of 243°...distance calculation.report the track distance along the mean latitude 48°20. the lenght is 62' 1°02'...at the equator, 62 minutes = 62 nm..at latitude 48.33° 48°20'.62' x cos48.33 = 41 nm.
Question 111-40 : For this question use trm vfr , aberdeen dyce area chart 19 5..what is the atis frequency that may not be used when in flight. 2443 ?
121.85.
.you can read next to 121.85 for ground use only.
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