Question 272-1 : An aeroplane enters a horizontal turn with a load factor n=2 from straight and level flight whilst maintaining constant indicated airspeed the ? [ Analysis topography ]

Question 272-2 : Approximately how long does it take to fly a complete circle during a horizontal steady co ordinated turn with a bank angle of 45° and a tas of 200 kt ?

65 s.

You have to calculate the radius and with it calculate the circumference of the circle it will lead to the solution radius = true air speed in ms ² g x tan bank angle tas in ms ==> 200 x 0515 = 103 ms tas in ms ² = 10609 mwe assume g = 10 and tan45° = 1radius = 10609 1 x 10 = 1061the formula to calculate the circumference is circumference = 2 x pi x radiuscircumference = 2 x 31415 x 1061 = 6666 mtime to fly a complete circle 6666m 103 ms = 6471 secondes

Question 272-3 : An aeroplane is in a steady horizontal turn at a tas of 1944 kt the turn radius is 1000 m the bank angle is assume g = 10 ms2 ?

45°.

Radius = true air speed in ms ² g x tan bank angle tas in ms = 194 x 0515 = 100 msg acceleration = 101000 m = 10000 10 x tan bank angletan bank angle = 10000 10000 = 1arctan 1 = 45°

Question 272-4 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 50 000nassumed g 10ms²for a straight steady wings level climb of a twin engine aeroplane the all engines climb gradient is ?

117%.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 100000n500000n 112 climb gradient % = 100 02 008333 climb gradient % = 100 x 011667 = 11667%

Question 272-5 : An aeroplane transitions from steady straight and level flight into a horizontal co ordinated turn with a load factor of 2 the speed remains constant and the ?

Induced drag increases by a factor of 4.

Induced drag varies with lift speed and aspect ratio is inversely proportional to aspect ratio and v² so multiply by 1v² and directly proportional to lift² or cl² or weight²if speed is constant the induced drag varies as the square of the lift if load factor is 2 cl must be increased by 2 induced drag is 2² = 4

Question 272-6 : An aeroplane performs a steady co ordinated horizontal turn with 20 degrees of bank and at 150 kt tas the same aeroplane with the same bank angle and speed but at a lower mass will turn with ?

The same turn radius.

Question 272-7 : Whilst maintaining straight and level flight with a lift coefficient cl=1 what will be the new value of cl after the speed has doubled ?

025.

Lift = 12 rhov²sclwhere rho = densityv = velocity tas in ms s = surfacecl= coefficient of liftif lift remains unchanged cl = 1v² cl = 12² = 12 x 12 = 025

Question 272-8 : An aeroplane has a stall speed of 100 kt at a mass of 1000 kg if the mass is increased to 2000 kg the new value of the stall speed will be ?

141 kt.

Stall speed increases with the square root of the load factorincreasing the mass from 1000 kg to 2000 kg is the same as multiply load factor by 2vs = 100 kt x square root 2vs = 100 kt x 141vs = 141 kt

Question 272-9 : In a straight steady descent which of the following statements is correct ?

Lift is less than weight load factor is less than 1.

Load factor = lift weightin descent lift is less than weight then load factor is less than 1

Question 272-10 : In a slipping turn nose pointing outwards compared with a co ordinated turn the bank angle i and the ball or slip indicator ii are respectively ?

I too large ii displaced towards the low wing.

In a slipping turn the ball will be displaced in the direction of turn in the opposite direction for a skidding turn 2769you now have two solutions push on the right rudder pedal or decrease the bank angle

Question 272-11 : What is the approximate radius of a steady horizontal co ordinated turn at a bank angle of 45° and a tas of 200 kt ?

1 km.

Radius = true air speed in ms ² g x tan bank angle tas in ms = 200 x 0515 = 103 msg acceleration = 10tan 45° = 1radius = 103² 10 x 1 radius = 1060 m

Question 272-12 : When an aeroplane is flying at an airspeed which is 13 times its basic stalling speed the coefficient of lift as a percentage of the maximum lift coefficient clmax would be ?

59%.

Airspeed = 13 vs and lift = 12 rhosv²cl => cl = lift 1v² => 1 13 ² = 059when we are at the stalling angle of attack we have clmax which means cl = 100% any increase in speed results in a decrease in cl reduction in angle of attack the cl is inversely proportional to the v² this can be written as cl = 1v²if cl was 100% then v would be vs which would be written as 1 lowest speed so as v is 13 x vs we can write it as cl = 113²cl = 13 x 13 = 1691 169 = 059 which is 059 of the maximummultiply by 100 to turn it in to a percentage

Question 272-13 : During a straight steady climb 1 lift is less than weight 2 lift is greater than weight 3 load factor is less than 1 4 load factor is greater than 1 5 lift is equal to weight 6 load factor is equal to 1 which of the following lists the correct statements ?

1 and 3.

During a straight steady climb the lift vector is acting at 90 degrees to the flight path and the weight vector will always act straight down perpendicular to the horizon only a part of this weight vector is acting in the direction opposite to the lift the other part of the weight vector acts in the same direction as the drag in order to keep the forces balanced lift is equal to the part of the weight vector acting perpendicular to the flight pathl=w cos gamma where gamma is the flight path angle to the horizontalso you can see that lift is now a fraction of the weight so lift is less than weight and mathematically this is shown because cos gamma can only have a value between 0 and 1 with it equalling 1 when gamma is zero ie horizontal flight and 0 when gamma is 90 degrees ie full on vertical climb the load factor bit comes from the definition of load factor which is liftweight now we've already determined that lift is less than weight in steady climb so this fraction has a numerical value less than oneagain being mathematical you can say from lift = weight x cos gamma liftweight = cos gamma henceload factor = liftweight = cos gamma and once again you can see that load factor can range from zero to one with it equalling one only in straight and level flight when the aircraft is climbing the thrust from the engine will overcome the rearward component of weight plus drag therefore the wings will not need to produce so much lift

Question 272-14 : Two identical aeroplanes a and b with the same mass are flying steady level co ordinated 20 degree bank turns if the tas of a is 130 kt and the tas of b is 200 kt ?

The load factor of a and b are the same.

Load factor = 1cos bank anglea and b are at 20 degree bank turns

Question 272-15 : Two identical aeroplanes a and b with the same mass are flying steady level co ordinated 20 degree bank turns if the tas of a is 130 kt and the tas of b is 200 kt ?

The turn radius of a is less than that of b.

Turn radius = tas² ms gtan bank angle aeroplane a radius = 65²10tan20° = 1160 maeroplane b radius = 100²10tan20° = 2747 m

Question 272-16 : Two identical aeroplanes a and b with the same mass are flying steady level co ordinated 20 degree bank turns if the tas of a is 130 kt and the tas of b is 200 kt ?

The lift coefficient of a is greater than that of b.

Lift between a and b is the same the more tas is high the less lift coefficient must be to remain in steady level flight

Question 272-17 : Which of these statements about vmcg determination are correct or incorrect i vmcg may be determined using both lateral and directional control ii during vmcg determination the lateral deviation from the runway centreline may be not more than 30 ft ?

I is incorrect ii is correct.

Vmcg the minimum control speed on the ground is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skillthe nosewheel steering is disconnected because the value of vmcg must also be applicable on wet andor slippery runwaysin the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 91 m 30 ft laterally from the centreline at any point

Question 272-18 : Which of these statements about vmcg determination are correct or incorrect i in order to simulate a wet runway nose wheel steering may not be used during vmcg determination ii during vmcg determination the cg should be on the aft limit ?

I is correct ii is correct.

Vmcg the minimum control speed on the ground is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skillthe nosewheel steering is disconnected because the value of vmcg must also be applicable on wet andor slippery runwaysin the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 91 m 30 ft laterally from the centreline at any pointvmc must be established with the most unfavourable centre of gravity at take off the cg should be on the aft limit

Question 272-19 : Which of these statements about the limiting value of 5 degrees bank angle during vmca determination are correct or incorrect i as the bank angle is decreased from 5 degrees to 0 degrees the value of vmca increasesii when the bank angle is increased beyond 5 degrees there is an increasing risk of ?

I is correct ii is correct.

When the bank angle is decreased from 5 degrees to 0 degrees the value of vmca increasesvmca is certified with a bank angle of not more than 5° towards the operating engine live engine low because a lower figure can be obtained compared with wings levelrudder deflection increases the camber of the vertical stabiliser which reduces its stalling angle of attack increasing bank angle could increase its angle of attack beyond that figure com encom080 807png

Question 272-20 : Which of these statements about the equilibrium of forces and moments at vmca are correct or incorrect i equilibrium of moments about the normal axis is provided by rudder deflectionii equilibrium of forces along the lateral axis requires either bank angle or side slip or a combination of both ?

I is correct ii is correct.

Vmca is the minimum ias at which directional control can be maintained following failure of the critical engine equilibrium of moments about the normal vertical axis is provided by rudder deflectionequilibrium of forces along the lateral axis requires either bank angle or side slip or a combination of both

Question 272-21 : Which of these statements about vmcg determination are correct or incorrect i vmcg must be determined using rudder control aloneii during vmcg determination the lateral deviation from the runway centreline may be not more than 30 ft ?

I is correct ii is correct.

Vmcg the minimum control speed on the ground is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skillthe nosewheel steering is disconnected because the value of vmcg must also be applicable on wet andor slippery runwaysin the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 91 m 30 ft laterally from the centreline at any point

Question 272-22 : Vmcl is the ?

Minimum control speed approach and landing.

Vmcl is the minimum control speed during landingapproach with all engines operating it is the calibrated airspeed at which when the critical engine is suddenly made inoperative it is possible to maintain control of the airplane with that engine still inoperative and maintain straight flight with an angle of bank of not more than 5 degrees

Question 272-23 : Given theta = pitch anglegamma = flight path anglealpha = angle of attackno wind bank or sideslipthe relationship between these three parameters is ?

Theta = gamma + alpha.

Jamacof we know that in normal level flight at cruise condition flight path angle is 0 because airplane is flying in horizontal plane at the same time we know that in positive cambered aerofoil during this condition we have positive angle of attack so we can say that gamma = theta alphaso theta = gamma + alpha

Question 272-24 : The four forces acting on an aeroplane in level flight are ?

Thrust lift drag and weight.

Question 272-25 : Whilst maintaining straight and level flight with a lift coefficient cl = 1 what will be the new approximate value of cl after the speed is increased by 41% ?

Question 272-26 : If an aeroplane performs a steady co ordinated horizontal turn at a tas of 200 kt and a turn radius of 2000 m the load factor n will be approximately ?

11.

Turn radius = tas² ms gtan bank angle tan bank angle = 100² 10 x 2000 = 0505 is the value of the bank angle tangent so bank angle = 265°=> n = 1 cos bank angle => n = 1 cos 265=> n = 111

Question 272-27 : If the stall speed of an aeroplane is 60 kt at what speed will the aeroplane stall if the load factor is 2 ?

85 kt.

Stall speed increases with the square root of the load factorvs = 60 kt x square root 2vs = 60 kt x 141vs = 8485 kt

Question 272-28 : For a straight steady wings level climb of a twin engine aeroplane the one engine inoperative climb gradient is aeroplane mass 50000 kgliftdrag ratio 12thrust per engine 60000 nassumed g 10ms² ?

37%.

50000 kg x 10ms² = 500000 newtonthrust only one engine operative = 60000 newtonclimb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 60000n500000n 112climb gradient % = 100 012 008333 climb gradient % = 100 x 003667 = 3667%

Question 272-29 : Consider the following statements about vmcg 1 vmcg is determined with the gear down2 vmcg is determined with the flaps in the landing position3 vmcg is determined by using rudder and nosewheel steering 4 during vmcg determination the aeroplane may not deviate from the straight line path by more ?

1 4.

Vmcg the minimum control speed on the ground thus with the gear down is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skillthe nosewheel steering is disconnected because the value of vmcg must also be applicable on wet andor slippery runwaysin the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 91 m 30 ft laterally from the centreline at any point

Question 272-30 : In general directional controllability with one engine inoperative on a multi engine aeroplane is favourably affected by 1 high temperature2 low temperature3 aft cg location4 forward cg location 5 high altitude 6 low altitude the combination that regroups all of the correct statements is ?

1 4 5.

With one engine inoperative on a multi engine aeroplane the loss of directional control is due to assymetric thrust the turning moment at high thrust settings thrust will be less at high temperature reducing the turning moment regarding a cold air temperature conditiona forward cg position increases the moment arm and thus gives more directional controllabilityair density is reduced at high altitude resulting in less thrust being produced from the remaining live engine s therefore a lesser turning moment

Question 272-31 : Given aeroplane mass 50000 kgliftdrag ratio 12thrust per engine 30000 nassumed g 10ms²for a straight steady wings level climb of a three engine aeroplane the all engines climb gradient is ?

97%.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 90000n500000n 112 climb gradient % = 100 018 008333 climb gradient % = 100 x 009667 = 9667%

Question 272-32 : What is the approximate radius of a steady level co ordinated turn with a bank angle of 30 degrees and a tas of 500 kt ?

12 km.

Radius = true air speed in ms ² g x tan bank angle tas in ms = 500 x 0515 = 2575g acceleration = 10tan 30° = 0577radius = 2575² 10 x 0577 radius = 11492 m

Question 272-33 : An aeroplane with a mass of 4000 kg is performing a co ordinated level turn at a constant tas of 160 kt and a bank angle of 45°the lift is approximately ?

56000 n.

Load factor = 1 cos bank angleload factor = 1414weight of the aircraft is 40000 newtonthe lift is approximately 40000 x 1414 = 56560 n

Question 272-34 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 60 000nassumed g 10ms²for a straight steady wings level climb of a twin engine aeroplane the all engines climb gradient is ?

157%.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 120000n500000n 112 climb gradient % = 100 024 008333 climb gradient % = 100 x 015666 = 1566%

Question 272-35 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 20 000nassumed g 10ms²for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

77 %.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 80000n500000n 112 climb gradient % = 100 016 008333 climb gradient % = 100 x 007667 = 7667%

Question 272-36 : Given aeroplane mass 50 000kgliftdrag ratio 10thrust per engine 20 000nassumed g 10ms²for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

60 %.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 80000n500000n 110 climb gradient % = 100 016 01 climb gradient % = 100 x 006 = 6%

Question 272-37 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 21 000nassumed g 10ms²for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

85%.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 84000n500000n 112 climb gradient % = 100 0168 008333 climb gradient % = 100 x 008467 = 8467%

Question 272-38 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 21 000nassumed g 10ms²for a straight steady wings level climb of a four engine aeroplane the one engine inoperative climb gradient is ?

43 %.

Climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 63000n500000n 112 climb gradient % = 100 0126 008333 climb gradient % = 100 x 004267 = 4267%

Question 272-39 : The stick force per g of a heavy transport aeroplane is 300 ngwhat stick force is required if the aeroplane in the clean configuration is pulled to the limit manoeuvring load factor from a trimmed horizontal straight and steady flight ?

450 n.

The limit load factor of a large transport aeroplane is 25 gthe question states straight and steady flight so we are at 1 gwe only require +15 g to reach a load factor of 25 g15 x 300 ng = 450 n

Question 272-40 : Given aeroplane mass 50 000kgliftdrag ratio 12thrust per engine 28 000nassumed g 10ms²for a straight steady wings level climb of a three engine aeroplane the all engines climb gradient is ?

85%.

Climb gradient = thrust drag weight x 100 liftdrag ratio 12 > when lift is 12 drag is 008333total trust = 3 x 28000 n = 84000 nweight = 50000 kg x 10ms² = 500000 ndrag = 500000 x 008333 = 41665 nclimb gradient = 84000 41665 500000 x 100climb gradient = 8467%or climb gradient % = 100 total thrustweight 1 liftdrag ratio climb gradient % = 100 84000n500000n 112 climb gradient % = 100 0168 008333 climb gradient % = 100 x 008467 = 8467%