Question 104-1 : A single engine aircraft has a gross glide gradient of 8% determine the net glide distance from 10 000 ft to 2000 ft ? [ Exam pilot ]

Question 104-2 : What does vref stand for it is the speed in a specified ?

Landing configuration at the point where the aeroplane descends through 50 ft and is used to determine the landing distance for manual landings.

Vref is the landing reference speed at a point 50 feet above the landing threshold it is not less than 13 times the stall speed in the normal landing configuration in simple terms your final approach speed

Question 104-3 : The following runways are available 07250 and 01190 the wind is coming from 350°25 degrees which runway should the pilot choose ?

10.

Pilots prefer to land and take off in headwind because it increases the lift in headwind a lower ground speed and a shorter run is needed for the plane to become airborne since best performance is achieved with a headwind we should choose the most into wind take off condition in this case it is runway 010 alternatively the runway designation marking shown at the thresholds of paved runways is the whole number nearest one tenth the magnetic azimuth of the centerline of the runway measured clockwise from the magnetic north that means that the available runway 0119 has an approximate magnetic direction of 010º190º and the runway 0725 a magnetic direction of 070º250ºtaking off and landing into the wind or with the least possible crosswind component is a priority because the aerodynamic and performance characteristics of the aircraft are enhnacedcomparing the angle differences between the available runways' magnetic directions and the wind direction we get that the wind 350° m 25 kt is only 20° off the available runway 01 thus using this runway the least possible crosswind and the maximum possible headwind will be experienced during take offso the runway 01 should be preferred by the pilot

Question 104-4 : A multi engined performance class b aeroplane has a wingspan of less than 60 metres what is the semi width of the obstacle accountability area at a distance of d from the end of the toda a semi width… ?

Of at least 12 x wingspan plus 60 m plus d x0125.

Concerning class b take off obstacle clearance regulations for multi engined aeroplanes catpola310 a refers that the take off flight path of aeroplanes with two or more engines shall be determined in such a way that the aeroplane clears all obstacles by a vertical distance of at least 50 ft or by a horizontal distance of at least 90 m plus 0125 × d where d is the horizontal distance travelled by the aeroplane from the end of the toda or the end of the take off distance if a turn is scheduled before the end of the toda except as provided in b and c for aeroplanes with a wingspan of less than 60 m a horizontal obstacle clearance of half the aeroplane wingspan plus 60 m plus 0125 × d may be used assuming that this is the distance laterally from the line of flight to the edge of the accountability area it is the semi width or half width of the total accountability area

Question 104-5 : Which of the following options correctly describes the effects on aircraft performance of landing on wet or contaminated runways ?

The lower effective braking force has a much greater effect on increasing the landing distance than the benefit of higher drag.

All contaminants make the braking action less effective and all except ice act to resist acceleration on the take off run or landing ops requires an additional 15% factor to be imposed on the landing distance required when the runway is wet or contamianted unless flight manual information allows a reduction below thislanding on wet and contaminated runways has three effects there is a risk of hydroplaning on water and slush surfaces leading to a marked reduction in the coefficient of friction which leads to longer landing distances on ice and packed snow the coefficient of friction is markedly reduced without hydroplaning which leads to longer landing distances where hydroplaning is not present there is a displacement drag caused by the tyres pushing contaminant out of the way and an impingement drag as the contaminent is thrown up to hit the airframe this tends to reduce landing distancebut the #1 and #2 effects of the wet and contaminated runways above which increase the landing distance required far outweigh the effects of drag in #3

Question 104-6 : Following a take off limited by the 50 ft screen height a light twin engine aircraft climbs on a gradient of 9% it will clear a 800 ft obstacle in relation to the runway at sea level horizontally situated at 2 nm from the 50 ft point with an obstacle clearance margin of ?

344 ft.

Climb gradient is the ratio between the ground distance travelled and the altitude gained expressed in percentage and it is given by the formula gradient % = change in height distance travelled x 100solving for change in height we get change in height = gradient x distance travelled = 009 x 2 nm x 6080 ft = 1094 ftwe assume that the required climb gradient of 9% will be achieved at the screen height of 50 ft and thereafter thus 2 nm after the screen height the gained altitude will be 1094 ft + 50 ft = 1144 fttherefore the aircraft will clear the obstacle by 1144 ft 800 ft = 344 ftnote use this equation for nm conversion into feet 1 nm = 6080 ft

Question 104-7 : For a piston engine at what altitude will it have the best endurance ?

About mean sea level.

Endurance is the time that the aircraft can remain airborne so the best endurance is the maximum time that it can remain airbornethe specific fuel consumption sfc for a piston engine aircraft is given by the formula sfc = fuel flow power and expresses the overall efficiency of the engine in terms of its specific fuel consumptionthus the lower the sfc the more the engine's fuel efficiency the greater best the endurancethe lowest sfc for a piston aircraft is when the intake pressure is high the engine rpm is low the throttle is wide openwhich means in theory that sfc is the lowest possible at the mean sea level

Question 104-8 : What will be the vref speed for a class b performance category single engined piston aeroplane with a maximum take off mass of 1633 kg given the following stall speeds vso 60 kt vs1 63 kt vs 68 kt ?

78 kt.

Reference landing speed vref is the speed of the aeroplane in a specified landing configuration at the point where it descends through the landing screen height in the determination of the landing distance for manual landingsvso is the stall speed or the minimum steady flight speed in the landing configurationvs1 is the stall speed or the minimum steady flight speed with the aeroplane in a configuration appropriate to the case under considerationvs is the stall speed or the minimum steady flight speed at which the aeroplane is controllableaccording to cs 2373 amendment 4 a for normal utility and aerobatic category reciprocating engine powered aeroplanes of 2722 kg 6000 lb or less maximum weight the reference landing approach speed vref must not be less than the greater of vmc determined under cs 23149 b with the wing flaps in the most extended take off setting and 13 vsotherefore the vref for a class b performance category single engined piston aeroplane with a maximum take off mass weight of 1633 kg is 13 x vso = 13 x 60 kt = 78 kt

Question 104-9 : Calculate the net glide distance for a single engined aeroplane following an engine failureaeroplane altitude 8 000 ft terrain elevation 1 500 ft gross gradient 8% tas 200 kt headwind component 20 kt 1 nm = 6 080 ftstill air distance = height difference net gradients x 100 ground distance = ?

113 nm.

If an engine failes the pilot will put the aircraft into a glide at the best glide speed vmdaccording to the annex iv part cat gm1 catpola320 en route single engined aeroplanes engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 05% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landingthus the en route glide gradients will be steeper than the gross gradients by 05%the question asks for the net distance so we use the net gradient for our calculations the assumed en route gradient shall be the gross gradient of descent increased by a gradient of 05% worse than average gradient thus giving us a safety margin so we use a descent gradient of 85%using the formulae given we get 1 still air distance = height difference x 100 net gradient % = 8000 ft 1500 ft x 100 85 = 76 470 ft or 76 470 ft6080 = 1258 nm2 ground distance = still air distance x gstas = 1258 nm x 200 kt 20 kt 200 kt = 113 nmnote the gradient may also be given as a liftdrag ratio for example if you are given a ld ratio 12 1 then if gliding with vmd your gradient will be 112 = 0083 or 833%

Question 104-10 : A single engined piston aeroplane is flying at an altitude of 5 000 ft with a constant tailwind component of 15 kt given the following information if the engine fails at this altitude what should be the expected net glide distance over ground with an elevation of 300 ft aeroplane gross gradient 10% ?

83 nm.

If an engine fails the pilot will put the aircraft into a glide at the best glide speed vmdaccording to the annex iv part cat gm1 catpola320 en route single engined aeroplanes engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 05% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landingthus the en route glide gradients will be steeper than the gross gradients by 05%the question asks for the net distance so we use the net gradient for our calculations the assumed en route gradient shall be the gross gradient of descent increased by a gradient of 05% worse than average gradient thus giving us a safety margin so we use a descent gradient of 105%using the formulae given we get 1 still air distance = height difference x 100 net gradient % = 5 000 ft 300 ft x 100 105 = 44 762 ft or 44 762 ft6080 = 736 nm2 ground distance = still air distance x gstas = 736 nm x 120 kt + 15 kt 120 kt = 83 nmnote the gradient may also be given as a liftdrag ratio for example if you are given a ld ratio 12 1 then if gliding with vmd your gradient will be 112 = 0083 or 833%

Question 104-11 : A single engined piston aeroplane is flying at an altitude of 5 000 ft with a constant tailwind component of 15 kt given the following information if the engine fails at this altitude what should be the expected net glide distance over ground with an elevation of 250 ft ld 91 1 tas 120 kt assume 1 ?

77 nm.

If an engine fails the pilot will put the aircraft into a glide at the best glide speed vmdthe glide gradient may be given as a liftdrag ratio at vmda ld ratio of 91 1 implies a glide gradient of 191 = 011 or 11%according to the annex iv part cat gm1 catpola320 en route single engined aeroplanes engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 05% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landingthus the en route glide gradients will be steeper than the gross gradients by 05%the question asks for the net distance so we use the net gradient for our calculations the assumed en route gradient shall be the gross gradient of descent increased by a gradient of 05% worse than the average gradient thus giving us a safety margin so we use a descent gradient of 115%using the formulae given we get 1 still air distance = height difference x 100 net gradient % = 5 000 ft 250 ft x 100 115 = 41 304 ft or 41 304 ft6 080 = 68 nm2 ground distance = still air distance x gstas = 68 nm x 120 kt + 15 kt 120 kt = 765 nm closest answer is 77 nm

Question 104-12 : Which option describes for a class b performance category single engine piston aeroplane a restriction on determining the take off speed vr ?

Vr must be equal to or greater than vs1.

During take off the aircraft must be accelerated to a rotate speed called vr at which point the rotation is initiateda class b aircraft has no v1 if an engine fails before vr the take off is rejected and if it fails after vr land ahead a multi engine class b aircraft may be able to climb following an engine failure after vr but without guarantees also the class b take off uses a screen height of 50 ft rather than 35 ft and the speed at that height is not usually referred to as v2 being described in more general termsthe most obvious constraint on vr is that the aircraft must be operating above the stall speed in the take off configuration which is called vs1 for a single engine aeroplane vr must not be less than vs1 the speed at 50 ft after take off must not be less than the highest of a speed that is safe under all reasonably expected conditions including turbulence or a critical loss of thrust or 12 vs1

Question 104-13 : What will be the effect on the landing distance required for a class b aircraft if the runway has a downhill slope of 2% ?

The landing distance required will increase by 10%.

Part cat for class b aeroplanes requires to take into account of a downslope but not an upslope 'unless otherwise specified in the aircraft flight manual or other performance or operating manuals from the manufacturer the landing distances required should be increased by 5% for each 1% of downslope'also according to cap 698 section 2 sep1 p9 and section 3 mep1 p17 'the landing distance should be increased by 5% for each 1% downslope no allowance is permitted for upslope'therefore for a 2% downslope runway the landing distance required should be increased by 10%

Question 104-14 : A multi engine turboprop aircraft with a max take off weight of 5 300 kg is under performance class b given the following speeds what is its vr and v2vso 63 ktvs1 69 ktvs 77 ktvmc 70 kt ?

Vr 76 kt v2 83 kt.

This question requires knowledge of the following rules of class b take off speeds for a multi engine aircraft vr the rotation speed must not be less than 105 vmc 11 vs1v2 the take off safety speed must not be less than 11 vmc 12 vs1 a speed that is safe under all reasonably expected conditionsthese speeds should be committed to memory for the exam as they will not be found in cap 698 thankfully there is a nice progression to the speeds which does make them much easier to remembervmc is the minimum control speed air vs1 is the stall speed for the given configuration vs0 is landing config therefore in this question the rotation speed vr must be the highest of 105 x 70 kt = 735 kt 11 x 69 kt = 759 kt rounded up to 76 ktthe take off safety speed v2 must be the highest of 11 x 70 kt = 77 kt 12 x 69 kt = 828 kt rounded up to 83 kt

Question 104-15 : A pilot performs two landings onto the same runway both are from 3º approaches with the same conditions presentlanding 1 the aeroplane passes the runway threshold at a height of almost 100 ft and the correct landing airspeedlanding 2 the aeroplane passes the runway threshold at 50 ft but with an ?

Both landings result in a greater landing distance than expected where landing 2 is less predictable as it depends on how excess energy is dissipated during the flare and subsequent ground roll.

In both these landing scenarios the pilot has too much energy in scenario 1 they are too high but with the correct speed in scenario 2 they are at the correct height but too fastboth these scenarios are going to end up with a longer landing distance than expected as the aircraft has to get rid of the excess energy somehowin the 1st scenario the aircraft will land long but at the correct speed stopping further along the runway than usualin the 2nd scenario the pilot is too fast so will either land correctly and just take longer to slow down or could flare too much due to their extra lift and potentially porpoise a little down the runway this is much worse than just using extra braking and can sometimes result in a go around if it gets bad this is why the correct answer includes a caveat about the unpredictability of landing 2 it very much depends on pilot actions

Question 104-16 : A high tension cable is located 1 nm from the der departure end of runway calculate the obstacle clearance for a class b single engine piston aircraft with the following conditionsclimb gradient 5% tas 133 kt tailwind component 10 kt cable height 104 ft height difference = gd x climb gradient ?

229 ft.

Single engine aircraft in the performance class b category do not have specific climb requirements after take off as they are not allowed to be flown into imc or at night in accordance with part cat therefore terrain and obstacle avoidance is done on a 'see and avoid' system however it is good airmanship to calculate obstacle clearance like thisthe aircraft should have reached a screen height of 50 ft by the der departure end of runway and the climb gradient in still air is 5% the reason we can tell that this is the still air climb gradient is that the formula at the bottom of the question that they have given us includes a correction for headtailwind so the original glimb gradient must not be correctedthe formula at the bottom of the question does make this a lot more simple as we can just input our numbers into that formula for the majority of the workinggd = ground distance = 1 nm = 6080 ft climb gradient = 5% tas = 133 kt gs = 133 kt + 10 kt = 143 kt note as this is not a legal calculation we do not apply the 50% of headwind150% of tailwind factor height difference = 6080 x 5 100 x 133143 = 2827 ft = 283 ftthen add on the 50 ft screen height we started with to get 283 + 50 = 333 ft333 ft is the height of our aircraft 104 ft is the height of the obstacle our obstacle clearance is 333 104 = 229 ft

Question 104-17 : A light twin piston engined aeroplane is cruising at 2000 ft when one of its engines fails whilst the pilot is performing the necessary actions to secure the engine they observe that the aircraft is in a very slow descent even with their operative engine at maximum continuous power what is the ?

The increased drag from the propeller of the failed engine and the required control inputs to maintain direction mean that the power from the remaining engine is insufficient to maintain altitude.

When a multi engined aircraft loses an engine not only does thrust or power reduce but the overall drag is increased so more thrust is required the drag increases due to the dead engine for both a jet or a propeller engine however propellers can be feathered to reduce that drag to a minimum drag also increases due to the control deflections that are required to keep the aircraft flying straight and by the sideslip induced when maintaining wings levelin this case it is a propeller aircraft so produces power this means that when the engine fails the power required will increase due to the increased drag mentioned above whilst the power available is decreased by 50% due to the loss of one engine's power output this has put the aircraft into a 'power deficit' where the power required is higher than the power available this means that there is not enough thrust to overcome the drag and the aircraft has begun to descend this is often called a drift down where an aircraft descends into the denser air until the remaining engine is powerful enough to maintain altitude on its ownto go through the available options 'the power required has reduced because drag is increased by the stationary propeller and the required control inputs to maintain direction and the remaining engine has sufficient excess power' incorrectthe remaining engine does not have sufficient excess power and the power required has increased'the power available has reduced because the feathered propeller only produces a small amount of thrust compared to normal and therefore the power required is less than the power available' incorrecta feathered propeller does not provide any thrust just less drag and the power required is more than the power available'the pilot must not have feathered the propeller of the failed engine because all twin engined aeroplanes that are certified in accordance with cs 23 must be able to climb within their environmental envelope' incorrectwhilst there are requirements for certain climb gradients after take off or go around for some operations no aircraft is expected to have a positive climb gradient with an engine failed throughout their whole environmental envelope which we are assuming to mean up to their maximum operating altitude'the increased drag from the propeller of the failed engine and the required control inputs to maintain direction mean that the power from the remaining engine is insufficient to maintain altitude' correctyes nothing wrong with that option drag increased so power required increased and power available was cut in half

Question 104-18 : Complete the statement with regards to a twin engine propeller aircraft certified under cs 23 in the event of an engine failure the thrust required will 1 and the excess thrust will 2 ?

1 increase 2 decrease by more than 50%.

when a twin engine propeller aircraft has an engine failure 3 main things happen to the aerodynamics firstly half of the aircraft's total thrust is lost secondly the slipstream from that propeller will no longer be flowing over the section of wing behind it so a significant amount of slipstream induced lift is also lost finally the drag of the aircraft increases as now instead of a spinning propeller creating thrust there are stopped propeller blades creating dragso therefore the total drag has increased therefore the thrust required has increased as thrust required equals drag for level unaccelerated flight also the thrust available has been cut in half these two items both have a detrimental effect on excess thrust which we need if we want to accelerate or climb at all see the graphs in the figure above and notice how they come together and remove the excess thrust to almost nothing as for this question the excess thrust is the amount of extra thrust available after overcoming drag so if the total available thrust is halved then the excess thrust is reduced by more than half more than 50% compound this with the increased drag and the effect is even greater

Question 104-19 : Given the following information calculate the net glide distance for a single engined aeroplane operating under performance class baeroplane altitude 10000 ft terrain elevation 500 ft gross glide gradient 10% tas 200 kt headwind component 20 kt 1 nm = 6080 ft still air distance = height ?

134 nm.

The question asks for the net distance which means that we can use the net gradient easa air ops states that for the purpose of flight planning the gross glide gradient to reach such places where the aeroplane can safely land must be increased by 05% gross to net gradient catpola320 net gradient = gross gradient + 05% = 105%now let’s calculate the net glide distance still air distance = height difference net gradient x 100 still air distance ft = 9500 105 x 100 = 90 476 ft 90 476 6080 = 1488 nm ground distance = still air distance x gs tas ground distance = 1488 x 180200 = 134 nm

Question 104-20 : A single engine class b aeroplane has the following wind limitations crosswind 17 kt headwind 30 kt tailwind 10 kt consider a take off from runway 20 with wind from 260° using trigonometry determine the wind speed which allows to stay below the crosswind limit of 17 kt and take off 2 decimal places ?

19 kt.

The wind components may be calculated using sin or cossin a = opposite side hypotenusethe angle between the wind and the runway direction is 60°we need to calculate the hypotenuse of a triangle whose opposite side crosswind equals 17 kt hypotenuse = 17 sin 60 = 1963 kt to stay below the limit 19 kt would be the correct answer

Question 104-21 : A pilot is planning to fly a light aircraft to an airport with a short runway but which is long enough to land safely the pilot is accustomed to landing at airports with long or very long runways and a mixture of small and large aeroplanes where atc often asks small aircraft to maintain a high ?

There is a large risk of runway excursion as a higher than optimum approach speed increases the risk of floating with significant additional kinetic energy dissipating.

During landing an aircraft has to go from having lots of kinetic energy and some height to being on the ground with no kinetic energy stopped this loss of energy is managed mostly by drag braking and thrust reversers if available the more energy the aircraft has faster speed over the threshold higher weight more height over the threshold the longer the runway needs to be to allow the aircraft to slow down sufficientlya pilot used to flying into long runways does not usually have to worry about having too much energy although an approach that is too fast is a bad approach and will usually lead to a poor landing or a very long flare if this pilot then flies to a much shorter runway and uses the same technique then they will land with a higher speed than they should which will invalidate the performance figures and mean they either flare for a large distance using up valuable runway and land at a normal speed or they might try and land at a speed which is too fast and then use up more runway braking from a higher speedeither of these could result in an overrun to land on a short runway the performance figures must be correct and the pilot must fly the short field approach and landing as per the procedure otherwise the performance figures are invalid this procedure often includes items such as a slower approach speed than usual a different flap setting a slightly different landing style etc

Question 104-22 : For a single engine aeroplane calculate the net glide distance following an engine failure given altitude 9 500 ftterrain elevation 500 ftgross gradient 11%tas 250 ktheadwind 42 ktstill air distance = height difference ft x 100 net gradient % ground distance = still air distance x gstas ?

107 nm.

The question asks for the net distance which means that we can use the net gradientair ops states that for the purpose of flight planning the gross glide gradient to reach such places where the aeroplane can safely land must be increased by 05%net gradient = gross gradient + 05% = 115%now let’s calculate the net glide distance still air distance = height difference ft x 100 net gradient % still air distance ft = 9 000 x 100 115 = 78 261 ft78 261 ft 6076 = 129 nm ground distance = still air distance x gstas ground distance = 129 x 208250 ground distance = 107 nm

Question 104-23 : What can be done to reduce any issues with asymmetric thrust in a twin engined propeller aircraft ?

Use counter rotating propellers.

For this question we can assume that the examiner means 'in the event of an engine failure' we don't know if that is included in the real question text or not due to feedback limitationsasymmetric blade effect also known as ‘p’ factor propeller blades are not flat they are in fact shaped as small wings as a consequence as the angle of attack of the airplane increases the air passing by hits the blades differently the down going blade viewed from the cockpit will have a larger angle of attack compared to the up going blade generating more thrust this then means that the down swinging blade exerts a greater force than the up going blade for this reason the thrust line will be displaced to the right of the engine centre line with clockwise rotating blades viewed from behind propellers rotating in the same direction clockwise if both engines rotate clockwise the right engine will have a longer thrust arm than the left engine this difference in thrust will give a yawing moment to the left with a clockwise rotating propeller in a nose up attitude the failure of the left hand engine will result in a larger yaw effect via the operating right hand engine rather than vice versa the left engine is therefore the critical engine the figure above shows this case counter rotating propellers they are called counter rotating propellers because the propeller on one wing turn in the opposite direction to the one on the other wing the principle advantage of counter rotation is to balance propeller torque effects and reduce p factor for both engines not just one thus eliminating a critical enginefor both counter rotating inwards or outwards the yawing moment will be the same regardless if the right or left engine fails however for a counter rotating inwards propeller the yawing moment will be smaller than that for a counter rotating outwards propeller the down going blade for the counter rotating inwards will be in the inner side and consequently the thrust vector will be close to the cg whereas for counter rotating outwards propellers the down going blade will be in the outer side resulting in a longer thrust arm and larger yawing momentthis is why counter rotating propellers are always assumed to be rotating inwards unless specified otherwise as it has no benefit to fit outwards counter rotating propsas for the other options placing the engines further from the aircraft's centre line longitudinal axis would make asymmetric thrust worse in the event of an engine failure the amount of blades per propeller does not really influence thrust asymmetry the movement of engines forwards or backwards does not affect thrust asymmetry only their distance from the longitudinal axis leftright

Question 104-24 : A single engine class b aeroplane has the following wind limitations crosswind 17 kt headwind 30 kt tailwind 10 ktconsider a take off from runway 20 with wind from 250° using trigonometry determine the wind speed which allows to stay below the crosswind limit of 17 kt and take off 2 decimal places ?

22 kt.

the wind components may be calculated using sin or cos sin a = opposite side hypotenusethe angle between the wind and the runway direction is 50ºwe need to calculate the hypotenuse of a triangle whose opposite side crosswind equals 17 kt hypotenuse = 17 sin 50º = 22 kt to stay below the limit 22 kt would be the correct answer

Question 104-25 : A single engine class b aeroplane has the following wind limitations crosswind 17 kt headwind 30 kt tailwind 10 ktconsider a take off from runway 20 with wind from 255°using trigonometry determine the maximum wind speed which allows to stay below the crosswind limit of 17 kt and take off 2 decimal ?

20 kt.

The wind components may be calculated using sin or cos sin a = opposite side hypotenusethe angle between the wind and the runway direction is 55°we need to calculate the hypotenuse of a triangle whose opposite side crosswind equals 17 kt hypotenuse = 17 sin 55° = 2075 kt to stay below the limit 20 kt would be the correct answer

Question 104-26 : A pilot is flying a twin engine piston aeroplane with all engines operating given the following information what will be the vertical clearance over the obstacle listed below mtom 4750 kg airport elevation 500 ft obstacle elevation 644 ft distance of the obstacle from the end of the tod 2000 m ?

168 ft.

Note this question has different answers in different caas unfortunately but according the latest feedback in many countries the answer is 168 ftmin gradient after take off for performance class b = 4%it is sometimes easier to put the distances all into the same units so 2000 m = 2000 x 328 = 6560 ft4% of 6560 = 262 ftthe aircraft ends the take off at screen height + runway elevation = 50 ft + 500 ft = 550 ftthe aircraft must legally be able to climb a further 262 ft new altitude = 550 + 262 = 812 ft do not make any adjustments for qnh we can work in true altitude here altimeter setting makes no difference therefore obstacle clearance given the 4% gradient = 812 644 = 168 ft

Question 104-27 : A class b light twin aeroplane is to take off on a commercial air transport flight from an airfield with data as given below what is the de factored take off distance that should be used to determine the maximum take off mass from the gross take off performance graph produced for a level paved and ?

1073 ft.

Slope correction 5 x 13 = 65% tora toda asda distances 1270 1655 1485 slope factor 1065 1065 1065 regulation factor 100 115 13 surface factor 100 100 100 de factored distance 1192 1351 1072 lowest value 1072 ftnote question asks for the de factored take off distance

Question 104-28 : The pilot of an aircraft has calculated a 4000 m service ceiling based on the forecast general conditions for the flight and a take off mass of 3250 kg if the take off mass is 3000 kg the service ceiling will be ?

Higher than 4000 m.

Considering question there is no limitation planned take off mass is lower than actual take of mass for gaining more altitude above service ceilingabsolute ceiling is the altitude at which the rate of climb theoretically is zeroservice ceiling the altitude where the rate of climb reduces to 100 ftminaerodynamic ceiling is the altitude at which the speeds for low speed buffet and for high speedbuffet are the same

Question 104-29 : Which engine is considered critical in the event of an engine failure during take off for an aeroplane propelled by two clockwise rotating piston engines ?

The left engine.

The critical engine of a multi engine fixed wing propeller driven aircraft is the one whose failure would result in the most adverse effects on the aircraft's handling and performancewhen one of the engines on a typical multi engine aircraft becomes inoperative a thrust imbalance exists between the operative and inoperative sides of the aircraftthis thrust imbalance causes several negative effects in addition to the loss of one engine's thrustfor reasons listed below the left engine of a conventional twin engine propeller driven aircraft is usually considered criticalfigure the operating right hand engine will produce a more severe yaw towards the dead engine thus making the failure of the left hand engine criticalasymmetrical yawwhen one engine becomes inoperative a torque develops which depends on the lateral distance from the center of gravity cg to the thrust vector of the operating engine multiplied by the thrust of the operating enginethe torque effect attempts to yaw the aircraft's nose towards the inoperative engine a yaw tendency which must be counteracted by the pilot's use of the flight controlsdue to the asymmetric blade effect p factor the right hand engine typically develops its resultant thrust vector at a greater lateral distance from the aircraft's cg than the left hand enginethe failure of the left hand engine will result in a larger yaw effect via the operating right hand engine rather than vice versa and it is termed the critical enginesince the operating right hand engine produces a stronger yaw moment the pilot will need to use larger control deflections in order to maintain aircraft controlthus the failure of the critical left hand engine is less desirable than failure of the right hand engineit is important to note however that this example depends upon both propellers turning clockwise as viewed from the rearon aircraft with counterclockwise turning engines such as the de havilland dove the right engine would be criticalaircraft which have counter rotating propellers rotating toward the cockpit on the top side such as the beechcraft duchess do not have a critical engine while both engines are critical on aircraft with counter rotating propellers turning away from the cockpit the lockheed p 38 was an example of the latter

Question 104-30 : The climb gradient of an aircraft after take off is 6% in standard atmosphere no wind at 0 ft pressure altitude using the following corrections ± 02% 1 000 ft field elevation± 01% °c from standard temperature 1% with wing anti ice 05% with engine anti icethe climb gradient after take off from an ?

39%.

Corrections to apply altitude 02% temperature 04 % isa + 4 wing and engine anti ice 15%therefore the new climb gradient = 6 02 04 15 => 39%

Question 104-31 : Given the following information what is the distance from the end of the todr to 1500 ft above reference zero for the purpose of obstacle clearance for a performance class b aeroplane cloud base 300 ft above reference zero tas 101 kt no wind all engines rate of climb at take off power 1830 ftmin ?

535 nm.

Above reference zero reference zero is the point on the ground where the aircraft reaches 50 ft during the take off phase the gradient from 50 ft to the assumed engine failure height is the average all engine gradient x 077 if visual reference for obstacle avoidance is lost it is assumed that the critical power unit becomes inoperative at this point all obstacles encountered in the accountability area must be cleared by a vertical interval of 50 ft wind calm => tas = gs = 101 ktthe exercise can be solved in 2 steps 1 aeo gradient with fatorization of 077 from 50 ft to cloud base 300 ft net gradient = 1830 x 077 = 1410 ftminclimb = 300 cloud base 50 ft regulatory height = 250 ft250 ÷ 1410 ftmin = 1063 sec 1063 sec x 101 kt ÷ 3 600 = 03 nm 2 oei gradient from cloud base to 1500 ft 1500 ft 300 cloud base = 1200 ft1200 ft ÷ 400 ftmin = 3 min3 min x 101 kt 60 s = 505 nm total ground distance to climb to 1500 ft = 030 + 505 = 535 nm

Question 104-32 : Given the following information what is the minimum vertical clearance of the obstacle by the performance class b aeroplane cloud base 300 ft above reference zero no wind tas 101 kt obstacle at 15000 ft from the end of todr with a height of 600 ft above reference zero all engines rate of climb ?

215 ft.

Considerations performance class b from 50 ft to the cloud base the gradient = all engine gradient x 077 the critical power unit is assumed to become inoperative at the point where visual reference is lost for obstacle avoidance cloud base roc to climb to the cloud base all engine roc = 1 830 ftmin net all engine = 1 830 ftmin x 077 = 1 409 ftmin roc from cloud base to obstacle oei roc 400 ftminfirst step 1 calculate the time to climb to 300 ft beginning at 50 ft screen = 250 ft 1 409 fpm = 0177 min2 determine the distance covered in 0177 min = 0177 min x 101 kt 60 min = 0298 nm x 6 076 = 1 810 ftat this point distance to obstacle from the cloud base = 15 000 ft 1 810 ft = 13 190 ft1 nm = 6 076 ft 13 190 ft = 217 nmsecond step 1 calculate time to cover 217 nm = 217 nm 101 kt x 3 600 s = 773 s2 height climbed in 773 s = 773 s x 400 fpm 60 = 515 fttotal height climbed = 300 ft + 515 ft = 815 ftobstacle clearance = 815 ft 600 ft = 215 ft

Question 104-33 : At an aerodrome the minimum climb gradient for obstacle clearance is 63% what is the minimum rate of climb which the pilot has to maintain during the climb tas 135 kt tailwind 8 kt ?

910 ftmin.

Still air climb gradient % = roctas x 6 0006 080 ground climb gradient % = rocgs x 6 0006 080 roc = gs x gradient x 6 0806 000 roc = 135 kt + 8 kt x 63 x 6 0806 000 roc = 913 ftminnote generally speaking when talking about obstacle clearances groundspeed should be used instead of tas => when experiencing headwind it will just slow you down horizontally but not vertically therefore you will climb the same but have more time before you actually reach the obstacle leading you to be higher over the obstacle than in still air

Question 104-34 : Given a climb gradient of 33% and a groundspeed of 100 kts the rate of climb that will be achieved is ?

330 ftmin.

Rate of climb fpm = gradient % x tas kts rate of climb fpm = 33% x 100 kts = 330 ftminimportant formulas gradient in % = altitude difference ft × 100 ÷ ground difference in ft approximate gradient % = climbdescent angle ° × 100 ÷ 60 climbdescent angle in ° = arctg altitude difference ft ÷ ground distance covered in ft vertical speed ftmin = gradient % × tas kts vertical speed ftmin = groundspeed kts × gradient ftnm ÷ 60

Question 104-35 : Given the following information determine the defactored landing distance for a multi engined performance class b aeroplane operated for commercial air transportlanding distance available 4850 ft runway surface grass less than 20 cm long not very short factor 115 runway condition wet factor 115 ?

2564 ft.

For a multi engined performance class b aeroplane operated for commercial air transport the landing distance must not exceed 70% of the landing distance available lda which equates to a factor of 143for this question the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p17 143 for the regulations 115 for the grass runway up to 20 cm long 115 for the wet condition of the runway no allowance is permitted for upslopetherefore the defactored landing distance will be 4850 ÷ 115 ÷ 115 ÷ 143 = 2564 ftnote all the factors are cumulative and can be applied in any order

Question 104-36 : Given the following information determine the defactored landing distance for a multi engined performance class b aeroplane operated for commercial air transportlanding distance available 4300 ftrunway surface paved asphaltrunway condition wet factor 115 runway slope 05% downslopethe landing ?

2551 ft.

For a multi engined performance class b aeroplane operated for commercial air transport the landing distance must not exceed 70% of the landing distance available lda which equates to a factor of 1070 = 143for this question the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p17 143 for the regulations115 for the wet condition of the runway1025 for the downslopetherefore the defactored landing distance will be 4300 ÷ 143 ÷ 115 ÷ 1025 = 2551 ftnote all the factors are cumulative and can be applied in any order

Question 104-37 : From the screen height of 50 ft a single engined piston aeroplane performs a straight out climb in visual meteorological conditions with a climb gradient of 9% a tas of 60 kt experiencing a headwind component of 10 kt what will be the expected height clearance of a group of trees that are 8 m tall ?

34 m.

The gained height of the single engined piston aeroplane at a distance of 250 m is given by the formula height difference = gd x tas gs x gradient100 = 250 m x 60 kt 50 kt x 009 = 27 mfor a single engined aeroplane we assume that the required climb gradient will be achieved at the end of toda reference zero at the screen height of 50ft 15 m thus the height difference from the aerodrome elevation will be 27 m + 15 m = 42 msince the obstacle's height is 8 m then the aircraft will clear the obstacle by 42 m 8 m = 34 m alternatively the rate of climb is given by the formula roc = % climb gradient x tas = 9% x 60 kt = 540 ftminthe ground distance until the obstacle will be covered in climb time = gd gs = 250 m 1 852 m 60 kt 10 kt = 00027 hr or 016 min with a roc of 540 ftmin the aircraft will gain 540 ftmin x 016 min = 87 ft or 27 mfor a single engined aeroplane we assume that the required climb gradient will be achieved at the end of toda reference zero at the screen height of 50ft 15 m thus the height difference from the aerodrome elevation will be 27 m + 15 m = 42 msince the obstacle's height is 8 m then the aircraft will clear the obstacle by 42 m 8 m = 34 m

Question 104-38 : Calculate the de factored field length that should be used for input into the performance graphs for an sep aircraft from the data belowtora 4 500 ft surface and condition wet grass 1% upslope there is no stopway or clearway ?

2 637 ft.

an airfield is called 'balanced' when there is no stopway or clearway that is the case in this question so the field is balanced and the tora = toda = asda in this case the take off distance required must when multiplied by 125 not exceed the todain simple terms because there is only one number to use the factorisation of 125 is used for that figure instead of factorising multiple times for 1 115 and 13 tora toda adsa and choosing the most restrictivethe runway must be now factorised for surface condition and slope using the information from the annex above the 1% upslope shall reduce the maximum todr by 5% factor of 105 the surface and condition are one factor for take off and for wet grass it is a further 30% reduction in maximum todr factor of 13 so we start with the actual length of the tora toda and asda in this case those are all the same so we can just call it toda toratodaasda toda = 4500 ft regulation factor ÷125 slope 1% ÷105 surfacecondition wet grass ÷13 de factored maximum todr to use in graph 2637 ft for this question the de factored maximum todr that could be used to enter the take off performance graph for a sep aeroplane used for commercial air transport is 2637 ft

Question 104-39 : Given the following information what will be the distance covered from the end of runway to the point at which 1 500 ft will be reached above reference zero performance class b cloud base above reference zero 300 ft wind 10 kt headwind roc all engines 1 830 ftmin roc single engine 400 ftmin ?

482 nm.

considerations performance class b from 50 ft to the cloud base the gradient = all engine gradient x 077 the critical power unit is assumed to become inoperative at the point where visual reference is lost for obstacle avoidance cloud base roc to climb to the cloud base all engine roc = 1 830 ftmin net all engine = 1 830 ftmin x 077 = 1 409 ftmin roc from cloud base to obstacle oei roc 400 ftmin first step 1 calculate the time to climb to 300 ft beginning at 50 ft screen = 250 ft 1 409 fpm = 0177 min 1065 sec 2 determine the distance covered in 0177 min = 0177 min x 91 kt = 027 nmsecond step 1 calculate the time to climb 1 200 ft from 300 ft to 1 500 ft 1 200 ft ÷ 400 ftmin = 3 min 2 determine the distance covered in 3 min = 3 min x 91 kt = 455 nmtotal distance to reach 1 500 ft = 027 nm + 455 nm = 482 nm

Question 104-40 : Given the following informationdetermine the defactored landing distance for a piston aircraft operated for commercial air transportoat 10°c pressure altitude 3000 ft landing distance available lda 2200 ft runway downslope 05% runway surface paved wet ?

1307 ft.

For a performance class b aeroplane such as this aircraft which is piston powered operated for commercial air transport the landing distance must not exceed 70% of the landing distance available lda which equates to a factor of 143for this question the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p17 143 for the regulations 115 for the wet condition of the runway no correction of the paved runway 1025 for the downslope 5% factor per 1% downslope the downslope of 05% therefore equals 25% factor which is 1025 therefore the defactored landing distance will be 2200 ÷ 143 ÷ 115 ÷ 1025 = 13052 ft closest answer 1307 ftnote all the factors are cumulative and can be applied in any order no corrections need to be made for the pressure altitude or temperature de factoring the landing distance available allows us to use the graph to work out our maximum landing mass required wind etc