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Question 104-1 : What does vref stand for it is the speed in a specified ? [ Exam pilot ]
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 1.3 times the stall speed in the normal landing configuration. in simple terms your final approach speed.
Question 104-2 : The following runways are available 07/250 and 01/190. 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 01/19 has an approximate magnetic direction of 010º/190º and the runway 07/25 a magnetic direction of 070º/250º.taking 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 enhnaced.comparing 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 off.so, the runway 01 should be preferred by the pilot.
Question 104-3 : 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 1/2 x wingspan plus 60 m, plus d x0.125.
Concerning class b take off obstacle clearance regulations for multi engined aeroplanes, cat.pol.a.310 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 0,125 × 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 0,125 × 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 104-4 : 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 this...landing 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 distance...but 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 104-5 : 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 100...solving for change in height we get change in height = gradient x distance travelled = 0.09 x 2 nm x 6080 ft = 1094 ft...we 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 ft...therefore, the aircraft will clear the obstacle by 1144 ft 800 ft = 344 ft...note use this equation for nm conversion into feet 1 nm = 6080 ft.
Question 104-6 : 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 airborne...the 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 consumption...thus, the lower the sfc, the more the engine's fuel efficiency, the greater best the endurance...the lowest sfc for a piston aircraft is when... the intake pressure is high. the engine rpm is low. the throttle is wide open,..which means in theory that sfc is the lowest possible at the mean sea level.
Question 104-7 : 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 landings.vso is the stall speed or the minimum steady flight speed in the landing configuration.vs1 is the stall speed or the minimum steady flight speed with the aeroplane in a configuration appropriate to the case under consideration.vs is the stall speed or the minimum steady flight speed at which the aeroplane is controllable.according to cs 23.73 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 23.149 b with the wing flaps in the most extended take off setting, and 1.3 vso.therefore, the vref for a class b performance category single engined piston aeroplane with a maximum take off mass weight of 1633 kg is 1.3 x vso = 1.3 x 60 kt = 78 kt.
Question 104-8 : Calculate the net glide distance for a single engined aeroplane following an engine failure.aeroplane 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 = ?
11.3 nm
If an engine failes, the pilot will put the aircraft into a glide at the best glide speed, vmd...according to the annex iv part cat gm1 cat.pol.a.320 en route single engined aeroplanes... engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 0.5% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landing....thus, the en route glide gradients will be steeper than the gross gradients, by 0.5%...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 0.5% worse than average gradient , thus giving us a safety margin. so, we use a descent gradient of 8.5%...using the formulae given, we get..1. still air distance = height difference x 100 / net gradient % = 8000 ft 1500 ft x 100 / 8.5 = 76 470 ft or 76 470 ft/6080 = 12.58 nm...2. ground distance = still air distance x gs/tas = 12.58 nm x 200 kt 20 kt / 200 kt = 11.3 nm...note the gradient may also be given as a lift/drag ratio. for example, if you are given a l/d ratio 12 1, then if gliding with vmd, your gradient will be 1/12 = 0.083 or 8.33%.
Question 104-9 : 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 ?
8.3 nm
If an engine fails, the pilot will put the aircraft into a glide at the best glide speed, vmd...according to the annex iv part cat gm1 cat.pol.a.320 en route single engined aeroplanes... engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 0.5% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landing....thus, the en route glide gradients will be steeper than the gross gradients, by 0.5%...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 0.5% worse than average gradient , thus giving us a safety margin. so, we use a descent gradient of 10.5%...using the formulae given, we get..1. still air distance = height difference x 100 / net gradient % = 5 000 ft 300 ft x 100 / 10.5 = 44 762 ft or 44 762 ft/6080 = 7.36 nm...2. ground distance = still air distance x gs/tas = 7.36 nm x 120 kt + 15 kt / 120 kt = 8.3 nm...note the gradient may also be given as a lift/drag ratio. for example, if you are given a l/d ratio 12 1, then if gliding with vmd, your gradient will be 1/12 = 0.083 or 8.33%.
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 250 ft l/d 9.1 1. tas 120 kt. ?
7.7 nm
If an engine fails, the pilot will put the aircraft into a glide at the best glide speed, vmd...the glide gradient may be given as a lift/drag ratio at vmd...a l/d ratio of 9.1 1 implies a glide gradient of 1/9.1 = 0.11 or 11%...according to the annex iv part cat gm1 cat.pol.a.320 en route single engined aeroplanes... engine failure b the operator should first increase the scheduled engine inoperative gliding performance data by 0.5% gradient when verifying the en route clearance of obstacles and the ability to reach a suitable place for a forced landing....thus, the en route glide gradients will be steeper than the gross gradients, by 0.5%...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 0.5% worse than the average gradient , thus giving us a safety margin. so, we use a descent gradient of 11.5%...using the formulae given, we get..1. still air distance = height difference x 100 / net gradient % = 5 000 ft 250 ft x 100 / 11.5 = 41 304 ft or 41 304 ft/6 080 = 6.8 nm...2. ground distance = still air distance x gs/tas = 6.8 nm x 120 kt + 15 kt / 120 kt = 7.65 nm. closest answer is 7.7 nm
Question 104-11 : 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 initiated...a 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 terms...the 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. 1.2 vs1.
Question 104-12 : 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 p.9 and section 3 mep1 p.17... '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 104-13 : 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 v2...vso 63 kt..vs1 69 kt..vs 77 kt..vmc 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 1.05 vmc 1.1 vs1v2 the take off safety speed, must not be less than 1.1 vmc 1.2 vs1 a speed that is safe under all reasonably expected conditions.these 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 remember.vmc 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 1.05 x 70 kt = 73.5 kt 1.1 x 69 kt = 75.9 kt rounded up to 76 ktthe take off safety speed v2 must be the highest of 1.1 x 70 kt = 77 kt 1.2 x 69 kt = 82.8 kt rounded up to 83 kt
Question 104-14 : A pilot performs two landings onto the same runway. both are from 3º approaches, with the same conditions present...landing 1 the aeroplane passes the runway threshold at a height of almost 100 ft and the correct landing airspeed..landing 2 the aeroplane passes the runway threshold at 50 ft but with ?
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 fast...both 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 somehow...in the 1st scenario, the aircraft will land long, but at the correct speed, stopping further along the runway than usual...in 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 104-15 : 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 conditions.climb 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 this.the 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 head/tailwind, so the original glimb gradient must not be corrected.the 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 working.gd = 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 headwind/150% of tailwind factor height difference = 6080 x 5 /100 x 133/143 = 282.7 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 104-16 : 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 level.in 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 own.to 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.' incorrect.the 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.' incorrect.a 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.' incorrect.whilst 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.' correct.yes, nothing wrong with that option. drag increased so power required increased, and power available was cut in half.
Question 104-17 : 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 drag.so 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 104-18 : Given the following information, calculate the net glide distance for a single engined aeroplane operating under performance class b.aeroplane 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 ?
13.4 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 0.5%. gross to net gradient cat.pol.a.320 net gradient = gross gradient + 0.5% = 10.5%now let’s calculate the net glide distance still air distance = height difference / net gradient x 100 still air distance ft = 9500 / 10.5 x 100 = 90 476 ft 90 476 / 6080 = 14.88 nm ground distance = still air distance x gs / tas ground distance = 14.88 x 180/200 = 13.4 nm
Question 104-19 : 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 ?
19 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 60°.we need to calculate the hypotenuse of a triangle whose opposite side crosswind equals 17 kt hypotenuse = 17 / sin 60 = 19.63 kt to stay below the limit 19 kt would be the correct answer.
Question 104-20 : 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 sufficiently...a 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 speed...either 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 104-21 : For a single engine aeroplane, calculate the net glide distance following an engine failure. given..altitude 9 500 ft..terrain elevation 500 ft..gross gradient 11%..tas 250 kt..headwind 42 kt....still air distance = height difference ft x 100 / net gradient %..ground distance = still air distance x ?
10.7 nm
The question asks for the net distance, which means that we can use the net gradient.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 0.5%.net gradient = gross gradient + 0.5% = 11.5%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 / 11.5 = 78 261 ft78 261 ft / 6076 = 12.9 nm ground distance = still air distance x gs/tas. ground distance = 12.9 x 208/250. ground distance = 10.7 nm
Question 104-22 : 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 limitations....asymmetric 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 engine....for 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 moment...this 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 props...as 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 left/right
Question 104-23 : 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 ?
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 50º.we 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 104-24 : 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 255°.using trigonometry, determine the maximum wind speed which allows to stay below the crosswind limit of 17 kt and take off.. 2 ?
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° = 20.75 kt, to stay below the limit 20 kt would be the correct answer.
Question 104-25 : 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 ?
168 ft
Note this question has different answers in different caas unfortunately but according the latest feedback in many countries the answer is 168 ft......min 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 3.28 = 6560 ft...4% of 6560 = 262 ft..the aircraft ends the take off at screen height + runway elevation = 50 ft + 500 ft = 550 ft...the 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 104-26 : 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 1.3 = 6.5%... . . . tora. toda. asda. . . distances. 1270. 1655. 1485. . . slope factor. 1.065. 1.065. 1.065. . . regulation factor. 1.00. 1.15. 1.3. . . surface factor. 1.00. 1.00. 1.00. . . de factored distance. 1192. 1351. 1072. . ...lowest value 1072 ft..note question asks for the de factored take off distance.
Question 104-27 : 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 ceiling...absolute ceiling is the altitude at which the rate of climb theoretically is zero...service ceiling the altitude where the rate of climb reduces to 100 ft/min...aerodynamic ceiling is the altitude at which the speeds for low speed buffet and for high speed..buffet are the same.
Question 104-28 : 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 performance...when one of the engines on a typical multi engine aircraft becomes inoperative, a thrust imbalance exists between the operative and inoperative sides of the aircraft...this thrust imbalance causes several negative effects in addition to the loss of one engine's thrust...for reasons listed below, the left engine of a conventional twin engine propeller driven aircraft is usually considered critical...figure the operating right hand engine will produce a more severe yaw towards the dead engine, thus making the failure of the left hand engine critical...asymmetrical yaw..when one engine becomes inoperative, a torque develops which depends on the lateral distance from the center of gravity c.g. to the thrust vector of the operating engine, multiplied by the thrust of the operating engine...the 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 controls...due 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 c.g. than the left hand engine...the 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 engine...since the operating right hand engine produces a stronger yaw moment, the pilot will need to use larger control deflections in order to maintain aircraft control...thus, the failure of the critical left hand engine is less desirable than failure of the right hand engine...it is important to note, however, that this example depends upon both propellers turning clockwise as viewed from the rear...on aircraft with counterclockwise turning engines such as the de havilland dove , the right engine would be critical...aircraft 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 104-29 : 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..± 0.2% / 1 000 ft field elevation..± 0.1% / °c from standard temperature.. 1% with wing anti ice 0.5% with engine anti ice...the climb gradient after ?
3.9%
Corrections to apply.. altitude 0.2%. temperature 0.4 % isa + 4. wing and engine anti ice 1.5%..therefore the new climb gradient = 6 0.2 0.4 1.5 => 3.9%
Question 104-30 : 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 ?
5.35 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 0.77. 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 kt...the exercise can be solved in 2 steps..1 aeo gradient with fatorization of 0.77 from 50 ft to cloud base 300 ft..net gradient = 1830 x 0.77 = 1410 ft/min..climb = 300 cloud base 50 ft regulatory height = 250 ft..250 ÷ 1410 ft/min = 10.63 sec.. 10.63 sec x 101 kt ÷ 3 600 = 0.3 nm....2 oei gradient from cloud base to 1500 ft..1500 ft 300 cloud base = 1200 ft..1200 ft ÷ 400 ft/min = 3 min..3 min x 101 kt / 60 s = 5.05 nm... total ground distance to climb to 1500 ft = 0.30 + 5.05 = 5.35 nm
Question 104-31 : 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 0.77.. 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 ft/min.. net all engine = 1 830 ft/min x 0.77 = 1 409 ft/min. roc from cloud base to obstacle.. oei roc 400 ft/min.....first step..1 calculate the time to climb to 300 ft beginning at 50 ft screen = 250 ft / 1 409 fpm = 0.177 min..2 determine the distance covered in 0.177 min = 0.177 min x 101 kt / 60 min = 0.298 nm x 6 076 = 1 810 ft..at this point, distance to obstacle from the cloud base = 15 000 ft 1 810 ft = 13 190 ft..1 nm = 6 076 ft. 13 190 ft = 2.17 nm..second step..1 calculate time to cover 2.17 nm = 2.17 nm / 101 kt x 3 600 s = 77.3 s..2 height climbed in 77.3 s = 77.3 s x 400 fpm / 60 = 515 ft.total height climbed = 300 ft + 515 ft = 815 ft..obstacle clearance = 815 ft 600 ft = 215 ft
Question 104-32 : At an aerodrome, the minimum climb gradient for obstacle clearance is 6.3%. what is the minimum rate of climb which the pilot has to maintain during the climb..tas 135 kt..tailwind 8 kt ?
910 ft/min
Still air climb gradient % = roc/tas x 6 000/6 080..ground climb gradient % = roc/gs x 6 000/6 080.. roc = gs x gradient x 6 080/6 000.. roc = 135 kt + 8 kt x 6.3 x 6 080/6 000.. roc = 913 ft/min..note 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 104-33 : Given a climb gradient of 3.3% and a groundspeed of 100 kts, the rate of climb that will be achieved is ?
330 ft/min
Rate of climb fpm = gradient % x tas kts..rate of climb fpm = 3.3% x 100 kts = 330 ft/min.important formulas... gradient in % = altitude difference ft × 100 ÷ ground difference in ft. approximate gradient % = climb/descent angle ° × 100 ÷ 60. climb/descent angle in ° = arctg altitude difference ft ÷ ground distance covered in ft. vertical speed ft/min = gradient % × tas kts. vertical speed ft/min = groundspeed kts × gradient ft/nm ÷ 60
Question 104-34 : Given the following information, determine the defactored landing distance for a multi engined performance class b aeroplane operated for commercial air transport.landing distance available 4850 ft. runway surface grass less than 20 cm long, not very short factor 1.15. runway condition wet factor ?
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 1.43...for this question, the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p.17... 1.43 for the regulations. 1.15 for the grass runway up to 20 cm long. 1.15 for the wet condition of the runway. no allowance is permitted for upslope...therefore, the defactored landing distance will be 4850 ÷ 1.15 ÷ 1.15 ÷ 1.43 = 2564 ft..note all the factors are cumulative and can be applied in any order.
Question 104-35 : Given the following information, determine the defactored landing distance for a multi engined performance class b aeroplane operated for commercial air transport.landing distance available 4300 ft.runway surface paved asphalt.runway condition wet factor 1.15.runway slope 0.5% downslope.the 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 1/0.70 = 1.43.for this question, the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p.17 1.43 for the regulations1.15 for the wet condition of the runway1.025 for the downslopetherefore, the defactored landing distance will be 4300 ÷ 1.43 ÷ 1.15 ÷ 1.025 = 2551 ftnote all the factors are cumulative and can be applied in any order.
Question 104-36 : 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 ?
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 gradient/100 = 250 m x 60 kt / 50 kt x 0.09 = 27 m...for 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 m...since 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 ft/min...the ground distance until the obstacle will be covered in climb time = gd / gs = 250 m / 1 852 m / 60 kt 10 kt = 0.0027 hr or 0.16 min. with a roc of 540 ft/min the aircraft will gain 540 ft/min x 0.16 min = 87 ft or 27 m...for 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 m...since the obstacle's height is 8 m, then the aircraft will clear the obstacle by 42 m 8 m = 34 m.
Question 104-37 : Calculate the de factored field length that should be used for input into the performance graphs for an sep aircraft from the data below.tora 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 1.25, not exceed the toda.in simple terms, because there is only one number to use, the factorisation of 1.25 is used for that figure, instead of factorising multiple times for 1, 1.15 and 1.3 tora, toda, adsa and choosing the most restrictive.the 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 1.05. 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 1.3 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. tora/toda/asda toda = 4500 ft regulation factor ÷1.25 slope 1% ÷1.05 surface/condition wet grass ÷1.3 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 104-38 : 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 ft/min..roc single engine 400 ?
4.82 nm
..considerations performance class b... from 50 ft to the cloud base, the gradient = all engine gradient x 0.77.. 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 ft/min.. net all engine = 1 830 ft/min x 0.77 = 1 409 ft/min. roc from cloud base to obstacle.. oei roc 400 ft/min.....first step..1 calculate the time to climb to 300 ft beginning at 50 ft screen = 250 ft / 1 409 fpm = 0.177 min 10.65 sec..2 determine the distance covered in 0.177 min = 0.177 min x 91 kt = 0.27 nm...second step..1 calculate the time to climb 1 200 ft from 300 ft to 1 500 ft 1 200 ft ÷ 400 ft/min = 3 min..2 determine the distance covered in 3 min = 3 min x 91 kt = 4.55 nm...total distance to reach 1 500 ft = 0.27 nm + 4.55 nm = 4.82 nm
Question 104-39 : Given the following information,determine the defactored landing distance for a piston aircraft operated for commercial air transport...oat 10°c. pressure altitude 3000 ft. landing distance available lda 2200 ft. runway downslope 0.5%. 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 1.43...for this question, the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 p.17... 1.43 for the regulations. 1.15 for the wet condition of the runway. no correction of the paved runway. 1.025 for the downslope 5% factor per 1% downslope. the downslope of 0.5% therefore equals 2.5% factor, which is 1.025...therefore, the defactored landing distance will be 2200 ÷ 1.43 ÷ 1.15 ÷ 1.025 = 1305.2 ft, closest answer 1307 ft..note 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.
Question 104-40 : A commercial air transport flight is to be conducted in a single engined piston aircraft. at the most critical point during the flight, the distance to be travelled to a safe landing place is 10 nm. given the following information, what is the lowest altitude the aircraft should fly over this ?
5400 ft
Easa part cat.pol.a.320 en route – single engined aeroplanes.. a in the meteorological conditions expected for the flight, and in the event of engine failure, the aeroplane shall be capable of reaching a place at which a safe forced landing can be made, unless the operator is approved by the competent authority in accordance with annex v part spa , subpart l — single engined turbine aeroplane operations at night or in imc set imc and makes use of a risk period... b for the purposes of point a , it shall be assumed that, at the point of engine failure.. 1 the aeroplane is not flying at an altitude exceeding that at which the rate of climb equals 300 ft per minute, with the engine operating within the maximum continuous power conditions specified. and.. 2 the en route gradient is the gross gradient of descent increased by a gradient of 0,5%....net glide gradient = gross glide gradient + 0.5% = 10.5%..the glide gradient is a still air gradient. we must therefore work out the still air distance with the expected wind conditions...ground speed = 110 kt + 20 kt = 130 kt...ground distance = 10 nm..ground distance = still air distance x gs / tas..still air distance = ground distance / gs / tas = 10 / 130 / 110 = 8.46 nm.. from nm to feet 8.46 nm x 6080 = 51436.8 ft still air distance..still air distance = height difference / net gradient x 100.. still air distance / 100 x net gradient = height difference..height difference = 51436.8 / 100 x 10.5 = 5400.8 ft..the terrain is 0 ft, so the minimum altitude for the aircraft to glide clear at the critical point is 5400 ft.
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