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Question 102-1 : A turbojet aeroplane in straight and level flight is cruising at fl290. for a given mass and cg position, how will the high speed buffet and low speed buffet as an ias vary when the flight is operated at fl330 ? [ Exam pilot ]
The low speed buffet increases and the high speed buffet decreases.
..before the days of fms, the buffet boundary was displayed on a chart. the buffet boundary has both low and high speed limits... the low speed buffet boundary describes the ias at which we find the onset of the pre stall buffet at a 10% margin above vs.. the high speed buffet boundary describes the ias at which we find the onset of mach related buffet, normally mmo, but sometimes slightly below that....at low altitudes, these speeds are far apart, leaving a wide range of operating speeds. although stalling ias remains constant at first as altitude increases, it starts to increase at higher altitude...also, the ias at which we get mach buffet will steadily decrease, because tas increases in the climb and local speed of sound decreases. the two buffet boundaries will move closer toogether and eventually meet at an altitude where no margins exists, the coffin corner or aerodynamic ceiling, i.e. you cannot fly faster or slower because you will either stall or hit mmo...the right hand limit on the graph shows the maximum indicated airspeed vmo and then at high altitude the maximum mach number, mmo. the left hand limit is based on the stall speed, 1.1 vs...therefore, with increasing altitude, the low speed buffet will increase and the high speed buffet will decrease.
Question 102-2 : A difference between a reduced flex and a derated thrust take off is that... ?
On a flex take off, the toga thrust remains available.
When an aircraft has excess performance during takeoff, using maximum thrust is unnecessary and choosing a lower thrust setting offers benefits in engine life and costs. two procedures exist flex takeoff and derated thrust takeoff....the flex take off is achieved by calculating an assumed or flex temperature that is higher than the ambient which will, when entered into the fms, give a thrust setting lower than full thrust. the flex temperature will always be higher than the ambient temperature. on a flex take off the reduced thrust setting for take off will not need to be increased if there is an engine failure but, in emergency, it can be. full thrust may be applied at any time...according to amc 25 13 'when operating with reduced flex take off thrust, the thrust setting parameter, which establishes thrust for take off, is not considered a take off operating limit'...because full thrust can be applied at any time on a flex take off, the speeds vmcg and vmca must be corrected for full thrust, not the reduced thrust setting. a flex take off is not considered to be a normal take off, is always at the discretion of the pilot and is not authorised on wet runways, unless suitable performance accountability is made for the increased stopping distance on a wet runway. as most operators account for this, we can assume that flex thrust take offs are allowed on a wet runway, but not on a contaminated runway.... throughout the flex take off, a further thrust selection of the take off/go around toga is availablle.....amc 25 13 defines derated thrust as a take off thrust less than the maximum take off thrust, for which exists in the afm a set of separate and independent, or clearly distinguishable, take off limitations and performance data that complies with all the take off requirements of cs 25. when operating with a derated take off thrust, the value of the thrust setting parameter, which establishes thrust for take off, is presented in the afm and is considered a normal take off operating limit...a derated thrust take off is considered to be a normal take off and therefore can be carried out both on dry runwaysand on wet and contaminated runways, subject to the afm data and other operating restrictions...because the take off speeds are calculated at the lower thrust, the derated thrust setting is considered to be a limit and therefore full thrust or toga thrust cannot be selected, even in an emergency, until the hazard area for vmcg is well passed. airbus, for instance, does not allow toga to be selected on a derated take off, until the aircraft has achieved the minimum flap retraction speed.
Question 102-3 : Maximum endurance for a turboprop aircraft with a 10 kt headwind is achieved at 1 and the maximum endurance speed 2 ?
1 vmp. 2 remains the same
The best or maximum endurance speed for a propeller driven aircraft is the minimum power speed, vmp. . this is the minimum power required to maintain level flight. vmp is always less than the best range speed, vmd, and therefore speed unstable.the wind does not affect endurance flying, but it can affect the range flying.
Question 102-4 : What speed will be limiting with a descent at constant mach ?
Vmo
..at sea level, the tas is comparitively low and the speed of sound is quite high. at high altitude for a given cas , the tas is much higher, because the air is thin and the local speed of sound lss is less, because it is colder...because the mach number is tas divided by lss, this means in turn that at altitude, your mach number is higher for a given cas, than it is at low level...the result of this is that, as you climb, the maximum speed is controlled at low level by cas and vmo, and at altitude you reach your limiting mach number mmo , before you reach vmo and therefore your maximum speed is controlled by mmo...in the descent the reverse is true. your maximum speed is controlled by mmo first, and then by vmo at lower altitudes...alternatively, lss increases during the descent, due to the rising temperature. therefore, tas must increase during descent to maintain constant mach no...ias is also increasing and will eventually reach its limit. vmo 'max operating' is the term for the maximum certified ias...note you can also advice the ectm lines to show that ias increases with constant mach.
Question 102-5 : What is the all engines climb gradient for a twin engine aeroplane with the following data given thrust per engine 33000 lbs. mass 89500 kg. lift to drag ratio 7.8 1. g = 10 m/s2 ?
20,7%
'the weight of an object is equal to its mass multiplied by the acceleration of gravity.'..so, weight = mass x g = 89500 kg x 10 m/s2 = 895000 n...also, the climb angles are very small, so, for the majority of such questions, we assume that weight = lift = 895000 n despite the fact that it's slightly more and that sin climb angle = tan climb angle...with a lift to drag ratio of 7.8 1, the drag is 895000/7.8 = 114743.6 n..since the thrust per engine is 33000 lbs or 33000 lbs / 2.2 = 15000 kg, then the total thrust produced by both engines will be 15000 kg x 10 m/s2 x 2 = 300000 n...we also know that sin climb angle = thrust drag / weight = 300000 n 114743.6 n / 895000 n = 0.207..thus, the climb gradient expressed in percentage will be climb gradient % = tan climb angle x 100 = sin climb angle x 100 = 0.207 x 100 = 20.7%.
Question 102-6 : Given the following information, what is the all engine climb gradient for a two engine aeroplane thrust per engine 118000 newtons. mass 76000 kg. lift to drag ratio 8 1. assume g = 10 m/s ?
18.5%
Climb gradient is the ratio of height gained to distance travelled => tangent to climb angle y.. sin y = t – d / w. sin y = t/w – d/w. sin y = t / m x g – d / l/cos y . sin y = t / m x g – d x cos y /l for small angles => cos is nearly 1. so sin y = t / m x g – d/l. climb gradient = t / m x g – d/l x 100. with regards to this question, climb gradient = 236000 / 76000 x 10 – 1/8 x 100 = 18.5%
Question 102-7 : When to use the formula 9 x square root of the tyre pressure in psi ?
For dynamic aqua planning speed
Either of the 'aquaplanning' or 'hydroplanning' terms can be used to refer to a condition where the tyre's tread is unable to clear all the surface water from under the wheel. water is then pushed ahead of the tyre and forms a wedge that lifts the tyre off the runway. this results in a loss of braking and directional control. the tyre can become scalded and blister, as friction causes the water to heat up and eventually boil, as it is pushed along the runway...there are three forms of hydroplanning that should be considered for contaminated runway operations... dynamic hydroplanning. viscous hydroplanning. reverted rubber hydroplanning...the point at which the aircraft starts to plane is called the dynamic hydroplanning speed. this can be approximated from one of the two formulae below... hydroplanning speed kt = 9 x square root p/ , for a rotating tyre take off. hydroplanning speed kt = 7.7 x square root p/ , for a non rotating tyre landing ,...where p is the tyre pressure in psi and is the specific gravity sg of the contaminant...if the contaminant is water, where sg=1, then the speed might be described as the dynamic aquaplanning speed and the formula given just as 9 x square root of the tyre pressure in psi or 7.7 x square root of the tyre pressure in psi...a thin film or fluid on a smooth runway surface prevents the tyre from making contact with the runway. viscous hydroplanning can occur at much lower speeds than dynamic hydroplanning and is most likely on a smooth, dirty surface. runway touchdown areas are likely to provide these conditions...reverted rubber hydroplanning occurs after a skid, often on the nosewheel which has neither braking nor anti skid protection. the high temperature created by friction, boil a thin layer of water and vaporised rubber to steam to support the tyre. this is likely if the nosewheel steering is used indiscriminately on a slippery runway. if the take off is rejected the contaminant which resisted acceleration now becomes as slippery as sin and braking action is drastically reduced.
Question 102-8 : With regard to turbojet aeroplanes, wind 1 endurance and 2 the maximum endurance speed. which option is the correct one ?
1 does not affect, 2 does not affect
Endurance is the time that the aircraft can remain airborne...maximum endurance speed is the airspeed at which the aircraft burns the least possible amount of fuel, the minimum drag is produced and the minimum thrust is required, thus the maximum airborne time maximum endurance is gained. for turbojet aeroplanes, this maximum endurance speed is the vmd...unlike range, endurance is not affected by the wind conditions, since it is only affected by the airspeed selection and availability of the fuel onboard...also, the maximum endurance speed, as an airspeed, is unaffected by the wind.
Question 102-9 : How does the maximum specific range over the ground srg for a propeller driven aeroplane vary with altitude ?
It increases until the increase in the ground speed is offset by the increase in power required.
Refer to figures...note we are looking for further feedback about the exact wording of this question and the options, as currently the feedback is stable, but to our knowledge, is not quite correct. please let us know if you see this in the exam and what you remember from it, thank you...maximum specific range over the ground srg means the highest ratio of groundspeed to fuel flow, as srg = gs/ff...therefore, we have to work out how the ratio of gs/ff changes with altitude. we haven't been given any wind information so we will assume calm winds, and therefore gs = tas...the airspeed for maximum range in a propeller driven aircraft is vmd, this is because fuel flow in a propeller driven aircraft is directly proportional to engine power, not thrust, so therefore minimum fuel flow is vmp and the speed for best range is where the tangent from the origin intersects the curve, which is the greatest tas/power required ratio, and this is mathematically the speed for minimum drag vmd...as for the curve's change with altitude, this is more complicated. the power required curve is based on the drag curve x tas, as power = thrust x velocity...the drag curve is drag thrust required vs eas eas is cas corrected for compressibility, but without density correction. so therefore at higher altitudes, flying the same eas for minimum drag is going to produce a higher tas, which is good for range due to the lower density of air , which means that the power required drag x tas is going to increase the same amount, which is equally bad for range. this means that the specific range has remained the same. this can be easily seen by the fact the power required vs tas graph curve moves upwards and to the right in equal proportions with increasing altitude...the only things left to account for are changes would be specific fuel consumption and propeller efficiency, then, and the specific fuel consumption of piston propeller engines is roughly constant with altitude, and the propeller efficiency is unknown at different altitudes...therefore it is our stance that the maximum specific range flown at vmd remains the same with altitude for a piston aircraft without knowing the change in propeller efficiency at different airspeeds until compressibility takes effect at higher altitudes/air speeds. this does not agree with the current correct answer, hence the need for up to date feedback on this question, thank you..possible differences that could change this are if specified that it is a turboprop aircraft higher is much more efficient , or perhaps flying at a constant power setting, where higher is more efficient initially, etc.
Question 102-10 : The specific fuel consumption sfc for a propeller engine is… ?
Fuel flow per unit of power.
The overall fuel efficiency of a jet engine is usually expressed in terms of its specific fuel consumption sfc and its propulsive and thermal efficiencies...the overall fuel efficiency of a piston engine is measured by its specific fuel consumption sfc only...sfc is the mass of fuel required to produce a given amount of power/thrust for a given period of time...the specific fuel consumption sfc formulae are... for a propeller aircraft sfc = fuel flow / power.. for a jet aircraft sfc = fuel flow / thrust....so, for a propeller aircraft, fuel flow = sfc x power, which means that for a given sfc the fuel flow is proportional to the power required...thus, the lower the sfc, the better the efficiency.
Question 102-11 : At what altitude does a propeller driven aeroplane reach its maximum endurance ?
A piston engine aeroplane reaches its maximum endurance at about sea level.
Maximum endurance is the maximum time that an aircraft can remain airborne...the lowest specific fuel consumption sfc for a piston aircraft is when the intake pressure is high, rpm low and the throttle wide open, which means that sfc is at its lowest at about sea level...also, sfc = fuel flow / power, which means that for a given sfc, the fuel flow is proportional to the power required, which is drag x tas. with decreasing altitude, tas also decreases, so power required decreases too...therefore, as sfc and power required decreases with decreasing altitude, then the fuel flow decreases too...the lowest sfc permits to remain airborne as much as possible, giving the maximum endurance.
Question 102-12 : An aircraft is climbing at a maximum climb angle with constant speed and thrust. consider for theoretical purposes that the weight of the aircraft suddenly decreases but the flight otherwise continues as normal. for the pilot to maintain the same speed, all factors remaining the same, the angle of ?
Increased, to increase the backward component of weight in order to match the excess thrust.
.the balance of forces in a steady climb shows that thrust is acting upwards and an element of weight backward component of weight is adding to the total drag. as the thrust assists the lift, the lift required is less in level flight, thus less lift is required in a climb than in level flight.for a steady speed to be maintained, the thrust and the two retarding effects of the aerodynamic drag induced and parasite drag and the weight element must be equal t = d + wsin where , the climb angle.solving for sin , we get sin = t d /w, which shows that the climb angle depends on the excess thrust t d and the weight.thus, the speed for best angle of climb vx will be where the excess thrust is greatest.a decrease in mass, decreases the lift induced drag, increasing the excess thrust. therefore, so as to match the excess thrust and keep it constant, the total drag must be increased by increasing the backward component of weight, which is added to the drag. and since the sin is inversely proportional to the weight, as weight decreases the climb angle increases.
Question 102-13 : Which of the following options best describes the relationship between fuel flow and tas, for propeller driven aeroplanes ?
With increasing tas, fuel flow will increase, when flying above the minimum power speed.
. power required is the product of drag multiplied by the tas flown pwr required = drag x tas. thus, with increasing tas, parasite drag increases and as a consequence the power required increases too.the minimum power required is produced when the aircraft is flying with the minimum power required speed vmp , which exists at the lowest point of the power required curve.for a propeller driven aircraft, vmp is also the best endurance speed, which gives the minimum power required to maintain level flight and is the point where the product of speed and drag drag x tas are at a minimum.any increase or decrease of the tas above or below the vmp correspondingly, increases the power required and as a result the fuel flow and decreases the aircraft’s endurance.
Question 102-14 : What is vs1g ?
The 1 g stall speed at which the aeroplane can develop lift equal to its weight.
According to the easa cs definitions/amendment 1 'definitions and abbreviations used in certification specifications for products, parts and appliances'... vs1g means the 1 g stall speed at which the aeroplane can develop a lift force normal to the flight path equal to its weight....vs and vs1g are two stall speeds abbreviated for stalling conditions... vs corresponds to a conventional stall condition, when lift suddenly collapses and 'g' is less than one.. vs1g corresponds to the maximum lift coefficient clmax. clmax is the condition when the angle of attack is maximum and 'g' is equal to one...note vs is less than vs1g.
Question 102-15 : Which of the following statements about performance definitions and safety factors is correct ?
The net climb gradient is the gross climb gradient minus a safety margin.
When aircraft are being certified, the manufacturer must test the aircraft in many different ways to define certain levels of performance for many phases of flight, such as take off distance, maximum rate of climb, landing distance and many more.there are different types of performance categorisation.measured performance is the performance achieved by the manufacturer under test conditions for certification. it utilises new aeroplanes and the test pilots, and is therefore not representative of the actual performance that any single aircraft might be able to achieve in normal circumstances, flown by a normal pilot.gross performance is the expected fleet average of aeroplanes, given that they are sufficiently maintained and flown as per the techniques detailed in the flight manuals, by a normal pilot. this means that any aircraft in the fleet has a 50% chance of exceeding this performance, but also a 50% chance of not meeting this performance.net performance is the gross performance, reduced by the safety margin, to account for various possibilities that could commonly cause reduced performance, such as variations in pilot technique, below average aircraft performance, etc. the safety margin is set so that the chance of an aircraft not meeting the net performance level is extremely low. the safety margin is not fixed, it depends on the likelihood of the situation occurring, and is set to make sure that the total probability of an incident or accident occurring is at least lower than 1 in 1 million, when a passenger buys their ticket known as their remote probability. this might mean that the safety margin between gross and net performance is large for likely events such as a take off with all engines operating, but the safety margin is very low or even zero for unlikely events such as an engine failure at v1.
Question 102-16 : Cs 23 and cs 25 use the term net take off flight path to describe certain performance elements. the net take off flight path gradient is the… ?
Actual flight path of the aeroplane reduced by a safety margin.
The difference between the net and gross flight paths/performance is as follows.. the gross performance is the average performance that a fleet of aircraft should achieve if maintained satisfactorily and flown in accordance with the techniques established during flight certification and subsequently described in the aircraft performance manual. gross performance therefore defines a level of performance that any aircraft of the same type has a 50 percent chance of exceeding at any time.. the net performance is the gross performance diminished to allow for various contingencies that cannot be accounted for operationally, e.g., variations in piloting technique, temporarily below average performance, etc. it is improbable that the net performancel flight path will not be achieved in operation, provided the aircraft is flown in accordance with the recommended techniques, i.e., power, attitude, and speed...the net take off flight path ntofp is the gross take off flight path minus a safety margin, i.e. reduced by a fixed percentage depending on the number of engines the aircraft has or the aircraft category a or b and must clear all obstacles by at least 35 ft class a.
Question 102-17 : Vsr1 is the… ?
Reference stall speed in a specific configuration.
Vsr1 means reference stall speed in a specific configuration...vsr0 means reference stall speed in in the landing configuration...vs1g means 1 g stall speed at which the aeroplane can develop a lift force normal to the flight path equal to its weight...vs1 means stall speed or the minimum steady flight speed obtained in a specified configuration.
Question 102-18 : Which of the following characteristics is correct when describing compacted snow ?
Snow compressed into a solid mass such that aeroplane tyres will run on it without further compaction of the surface.
Eu ops allows contaminated runway operations, provided that the operator has approval...a contaminated runway is defined by eu ops as a runway of which a significant portion of its surface area within the length and width being used is covered by one or more of the substances listed under the runway surface condition descriptors which are, in turn... standing water. slush. wet snow. dry snow. compacted snow. ice..according to cs 25 amc 25.1591 'the derivation and methodology of performance information for use when taking off and landing with contaminated runway surface conditions'... 4.5 compacted snow 'snow which has been compressed into a solid mass such that the aeroplane wheels, at representative operating pressures and loadings, will run on the surface without significant further compaction or rutting of the surface'.
Question 102-19 : What will be the effect on the take off distance required, if both the aerodrome pressure altitude and the aerodrome temperature increase ?
Take off distance required will increase.
Among other factors, both temperature and pressure altitude affect the take off distance required.. the temperature has two effects. an increase in temperature reduces the engine thrust. an increase in temperature also means that the ias required for rotation is a higher tas and therefore a higher groundspeed. thus, as temperature increases, the take off distance required increases, too.. an increase in pressure altitude means also less thrust. it means too that a higher tas is required at vr, so the high pressure altitudes act like the high temperatures, they increase the take off distance required.
Question 102-20 : For a piston engine aeroplane, the speed for maximum range is ?
That which gives the maximum lift to drag ratio.
.if you want to stay in flight the longest time possible maximum endurance , you need to fly at the minimum power required speed vmp velocity for minimum power..if you want to travel the maximum distance possible maximum range , you need to fly at the speed that wich gives maximum lift to drag ratio vmd velocity for minimum drag.. 1066
Question 102-21 : Maximum endurance for a piston engine aeroplane is achieved at ?
The speed that approximately corresponds to the maximum rate of climb speed.
.maximum endurance in a piston engine aeroplane will occur at vmp minimum power speed , you want to use as little fuel as possible in order to stay airborne as longer as possible, you are not concerned with distance....the best rate of climb will occur where you have excess power available, that is vy best rate of climb speed and in a piston engine aeroplane, that is definitely slower than speed for minimum drag vmd. thus, maximum endurance for a piston engine aeroplane is achieved at a speed that approximates the best rate of climb speed vmp...a common mistake is done between the trhust required curve showing drags and power required curve showing required power. 1135.for the propeller driven aircraft curve, the lowest point of the curve is the tas at wich the least power is needed as opposed to producing the least drag and is therefore the best for endurance in level flight. it is also the maximum rate of climb speed because the gap between power required and power available is greatest more power is needed above and below the minimum power speed.
Question 102-22 : The maximum indicated air speed of a piston engine aeroplane without turbo charger, in level flight, is reached ?
At the lowest possible altitude.
.maximum ias will always be achieved at sea level because this is where density is highest, the dynamic pressure is 1/2 rho v² which is sensed by the pitot system.
Question 102-23 : At reference or see performance manual sep1 1 figure 2.4..with regard to the graph for landing performance, what is the minimum headwind component required in order to land at helgoland airport.given.runway length 1300 ft.runway elevation msl.weather assume isa conditions.mass 3200 lbs.obstacle ?
10 kt.
.please find hereafter the full completed graph. 1072.be aware that we must assume a factored runway a piston engine airplane has to land in the 70% of the runway length..1300 ft x 0,7 = 910 ft... with this distance, none of the answers are correct.
Question 102-24 : How does the thrust of a propeller vary during take off run, assuming unstalled flow conditions at the propeller blades.the thrust ?
Decreases while the aeroplane speed builds up.
.the angle of attack as in the case of a aircraft wing is defined as the angle between the chord line and relative airflow. with a propeller however the relative airflow is the resultant of the airflow due to rotation and forward speed. change either of these values and the angle of attack will change.. 1079.the angle of attack decreases, similar to a decreases of angle of attack for an aircraft wing, less lift will be produce, for a propeller, it results in less thrust.
Question 102-25 : In a given configuration the endurance of a piston engine aeroplane only depends on ?
Altitude, speed, mass and fuel on board.
Oszklarska.the fuel on board weight is contained in airplane mass, is it not.therefore, the altitude, speed and mass answer is equivalent to altitude, speed, mass and fuel... .no, you need to know the quantity of fuel on board to be able to calculate endurance to stay airborne as longer as possible.
Question 102-26 : At a higher gross mass on a piston engined aeroplane, in order to maintain a given angle of attack, configuration and altitude ?
The airspeed must be increased and the drag will also increase.
.at a higher gross mass, we need to increase lift if we consider to maintain same altitude..lift = cl x 1/2 rho v² x s..cl = lift coefficient.rho = density.v = tas in m/s.s = surface..since density, surface and lift coefficient e.g. change in angle of attack do not change, only speed can changed..therefore the airspeed must be increased and the drag will also increase.
Question 102-27 : On a reciprocating engine aeroplane, to maintain a given angle of attack, configuration and altitude at higher gross mass ?
An increase in airspeed and power is required.
.at a higher gross mass, we need to increase lift if we consider to maintain same altitude..lift = cl x 1/2 rho v² x s..cl = lift coefficient.rho = density.v = tas in m/s.s = surface..since density, surface and lift coefficient e.g. change in angle of attack do not change, only speed can changed..therefore, power required increases, airspeed will increased and the drag will also increase..notice whatever the propulsion system, prop or jet , it doesn't play a role as seen by its absence in the lift formula.
Question 102-28 : An aeroplane with reciprocating engines is flying at a constant angle of attack, mass and configuration..with increasing altitude the drag ?
Remains unchanged but the tas increases.
.with increasing altitude, density rho decreases. angle of attack, mass and configuration remain constant, to maintain lift, tas has to increase...drag = cd x 1/2 x rho x v² velocity = tas...v increases while rho decreases, drag remains constant.
Question 102-29 : On a reciprocating engine aeroplane, with increasing altitude at constant gross mass, angle of attack and configuration the power required ?
Increases and the tas increases by the same percentage.
.this question compares an aircraft at a certain weight and angle of attack, for a first altitude, and then, in the same configuration, at a second altitude..you have to understand that the aircraft is in straight and level flight it is not climbing...our aircraft has to produce the same lift.lift l = 1/2 rho cl s v²..for a higher altitude, density reduces, cl and s remain unchanged, thus only tas v can and must increase..for that reason, the power required must increase.
Question 102-30 : At reference or see performance manual sep 1 figure 2.4..with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft..given.o.a.t 27 °c.pressure altitude 3000 ft.aeroplane mass 2900 lbs.tailwind component 5 kt.flaps landing position ?
Approximately 1850 feet.
. 1091
Question 102-31 : Performance manual sep 1 figure 2.4.with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft..given.o.a.t. isa +15°c.pressure altitude 0 ft.aeroplane mass 2940 lb.tailwind component 10 kt.flaps landing position down.runway tarred and dry. ?
Approximately 1930 feet.
. 1128
Question 102-32 : For this question use annex 032 005 or performance manual sep 1 figure 2.4..with regard to the landing chart for the single engine aeroplane determine the landing distance from a height of 50 ft...given.oat isa +15°c.pressure altitude 0 ft.aeroplane mass 2940 lb.headwind component 10 kt.flaps ?
Approximately 1800 feet.
Img /com en/com032 295.jpg.. oat isa +15°c , so oat is +30°c at 0 ft pressure altitude...runway wet grass..landing factor for short grass is 1.15 and landing factor for wet runway is 1.15..1275 ft x 1.15 = 1466 ft..1466 ft x 1.15 = 1686 ft..approximately 1800 feet is our answer.
Question 102-33 : With regard to the take off performance chart for the single engine aeroplane determine the maximum allowable take off mass.given.o.a.t isa.pressure altitude 4000 ft.headwind component 5 kt.flaps up.runway tarred and dry.factored runway length 2000 ft.obstacle height 50 ft. 2143 ?
3200 lbs.
.complete the graph with the data. 1093.you will find 3200 lbs.
Question 102-34 : At reference or see performance manual sep 1 figure 2.2..with regard to the take off performance chart for the single engine aeroplane determine the take off distance to a height of 50 ft..given.oat 7°c.pressure altitude 7000 ft.aeroplane mass 2950 lbs.headwind component 5 kt.flaps approach ?
Approximately 2050 ft.
Complete the graph with the data.. /com en/com032 297.jpg..you find 1900 ft, the closest answer is 2050 ft.
Question 102-35 : At reference or see performance manual sep 1 figure 2.1..with regard to the take off performance chart for the single engine aeroplane determine the take off speed for 1 rotation and 2 at a height of 50 ft..given.o.a.t. isa+10°c.pressure altitude 5000 ft.aeroplane mass 3400 lbs.headwind component 5 ?
71 and 82 kias.
Img /com en/com032 298.jpg..
Question 102-36 : At reference or see performance manual sep 1 figure 2.2..with regard to the take off performance chart for the single engine aeroplane determine the take off distance to a height of 50 ft..given.o.a.t. 38°c.pressure altitude 4000 ft.aeroplane mass 3400 lbs.tailwind component 5 kt.flaps approach ?
Approximately 3960 ft.
Img /com en/com032 299.jpg..do not forget to apply 'grass.correction factor'.3300 ft x 1.2 = 3960 ft.
Question 102-37 : With regard to the climb performance chart for the single engine aeroplane determine the climb speed ft/min..o.a.t. isa + 15°c.pressure altitude 0 ft.aeroplane mass 3400 lbs.flaps up.speed 100 kias. 2139 ?
1290 ft/min.
. 1129.. youssef92.for me i find that the answer is 1370ft/min... .if you start at 15°c instead of 30°c, you will find around 1350 1370 ft... but the question states o.a.t. isa + 15°c..it means an outside temperature of 30°c at sea level.
Question 102-38 : At reference or see performance manual sep 1 figure 2.2..with regard to the take off performance chart for the single engine aeroplane determine the take off distance over a 50 ft obstacle height..given.o.a.t. 30°c.pressure altitude 1000 ft.aeroplane mass 2950 lbs.tailwind component 5 kt.flaps ?
2275 ft.
Img /com en/com032 3587.jpg...1750 ft x surface factor 1.3 = 2275 ft
Question 102-39 : Using the landing diagram, for single engine aeroplane, determine the landing distance from a screen height of 50 ft , in the following conditions.given.pressure altitude 4000 ft.o.a.t. 5°c.aeroplane mass 3530 lbs.headwind component 15 kt.flaps down.runway tarred and dry.landing gear down. 2135 ?
1350 ft.
. 1097.1320 ft close to 1350 ft.
Question 102-40 : The pilot of a single engine aircraft has established the climb performance..the carriage of an additional passenger will cause the climb performance to be ?
Degraded.
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