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Question 249-1 : For a given aeroplane which two main variables determine the value of vmcg ? [ Question security ]

Airport elevation and temperature

.the vmcg is calculated at full rudder deflection this is not a variable but a fact airport elevation and temperature the two variables which determine the value of vmcg the vmcg speed is calculated at the maximum thrust available this is a predetermined parameter exemple 349 Airport elevation and temperature.

Question 249-2 : Given .aeroplane mass 50 000 kg.lift/drag ratio 10.thrust per engine 60 000n.assumed g 10m/s².for a straight steady wings level climb of a twin engine aeroplane the all engines climb gradient is ?

14%

Aerazur .50000kg = 500000 n.500000n / 10 = 50000 with 10 the lift/drag ratio .airplane thrust = 2 x 60000n = 120000n. 120000 50000 /500000 x 100 = 14% exemple 353 14%.

Question 249-3 : An aeroplane enters a horizontal turn with a load factor n=2 from straight and level flight whilst maintaining constant indicated airspeed the ?

Lift doubles

.if load factor is 2 you need twice the lift to maintain straight and level flight without increasing speed .the change in lift load factor can be found by 1/cos of the bank angle exemple 357 Lift doubles.

Question 249-4 : 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 m/s ² / g x tan bank angle .tas in m/s ==> 200 x 0 515 = 103 m/s. tas in m/s ² = 10609 m.we assume g = 10 and tan45° = 1.radius = 10609 / 1 x 10 = 1061.the formula to calculate the circumference is .circumference = 2 x pi x radius.circumference = 2 x 3 1415 x 1061 = 6666 m.time to fly a complete circle .6666m / 103 m/s = 64 71 secondes exemple 361 65 s.

Question 249-5 : An aeroplane is in a steady horizontal turn at a tas of 194 4 kt the turn radius is 1000 m the bank angle is . assume g = 10 m/s2 ?

45°

.radius = true air speed in m/s ² / g x tan bank angle .tas in m/s = 194 x 0 515 = 100 m/s.g acceleration = 10.1000 m = 10000 / 10 x tan bank angle.tan bank angle = 10000 / 10000 = 1.arctan 1 = 45° exemple 365 45°.

Question 249-6 : Given .aeroplane mass 50 000kg.lift/drag ratio 12.thrust per engine 50 000n.assumed g 10m/s².for a straight steady wings level climb of a twin engine aeroplane the all engines climb gradient is ?

11 7%

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 100000n/500000n 1/12 .climb gradient % = 100 0 2 0 08333 .climb gradient % = 100 x 0 11667 = 11 667% exemple 369 11.7%.

Question 249-7 : 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 1/v² and directly proportional to lift² or cl² or weight² .if 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 exemple 373 Induced drag increases by a factor of 4.

Question 249-8 : 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

exemple 377 The same turn radius.

Question 249-9 : 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 ?

0 25

.lift = 1/2 rho v² s cl.where .rho = density.v = velocity tas in m/s .s = surface.cl= coefficient of lift.if lift remains unchanged cl = 1/v² .cl = 1/2² = 1/2 x 1/2 = 0 25 exemple 381 0.25.

Question 249-10 : 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 factor .increasing the mass from 1000 kg to 2000 kg is the same as multiply load factor by 2 .vs = 100 kt x square root 2.vs = 100 kt x 1 41.vs = 141 kt exemple 385 141 kt.

Question 249-11 : 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 / weight.in descent lift is less than weight then load factor is less than 1 exemple 389 Lift is less than weight, load factor is less than 1.

Question 249-12 : 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 . 2769.you now have two solutions .push on the right rudder pedal or decrease the bank angle exemple 393 (i) too large, (ii) displaced towards the low wing.

Question 249-13 : 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 m/s ² / g x tan bank angle .tas in m/s = 200 x 0 515 = 103 m/s.g acceleration = 10.tan 45° = 1.radius = 103² / 10 x 1 .radius = 1060 m exemple 397 1 km.

Question 249-14 : When an aeroplane is flying at an airspeed which is 1 3 times its basic stalling speed the coefficient of lift as a percentage of the maximum lift coefficient clmax would be ?

59%

.airspeed = 1 3 vs and lift = 1/2 rho s v² cl => cl = lift 1/v² => 1/ 1 3 ² = 0 59 .when 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 = 1/v².if cl was 100% then v would be vs which would be written as 1 lowest speed so as v is 1 3 x vs we can write it as .cl = 1/1 3².cl = 1 3 x 1 3 = 1 69.1 / 1 69 = 0 59 which is 0 59 of the maximum.multiply by 100 to turn it in to a percentage exemple 401 59%.

Question 249-15 : 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 path .l=w cos gamma where gamma is the flight path angle to the horizontal ..so 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 lift/weight now we've already determined that lift is less than weight in steady climb so this fraction has a numerical value less than one ..again being mathematical you can say from lift = weight x cos gamma ..lift/weight = cos gamma ..hence..load factor = lift/weight = 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 exemple 405 1 and 3

Question 249-16 : 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 load factor of a and b are the same

.load factor = 1/cos bank angle.a and b are at 20 degree bank turns exemple 409 The load factor of a and b are the same.

Question 249-17 : 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 m/s / g tan bank angle .aeroplane a radius = 65²/10 tan20° = 1160 m .aeroplane b radius = 100²/10 tan20° = 2747 m exemple 413 The turn radius of a is less than that of b.

Question 249-18 : 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 exemple 417 The lift coefficient of a is greater than that of b.

Question 249-19 : 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 skill .the nosewheel steering is disconnected because the value of vmcg must also be applicable on wet and/or slippery runways .in 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 9 1 m 30 ft laterally from the centreline at any point exemple 421 I is incorrect, ii is correct.

Question 249-20 : 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 skill .the nosewheel steering is disconnected because the value of vmcg must also be applicable on wet and/or slippery runways .in 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 9 1 m 30 ft laterally from the centreline at any point .vmc must be established with the most unfavourable centre of gravity at take off the cg should be on the aft limit exemple 425 I is correct, ii is correct.

Question 249-21 : 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 increases .ii when the bank angle is increased beyond 5 degrees there is an increasing risk ?

I is correct ii is correct

.when the bank angle is decreased from 5 degrees to 0 degrees the value of vmca increases .vmca 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 level .rudder 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 en/com080 807 png.

Question 249-22 : 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 deflection .ii 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 deflection .equilibrium of forces along the lateral axis requires either bank angle or side slip or a combination of both exemple 433 I is correct, ii is correct.

Question 249-23 : Which of these statements about vmcg determination are correct or incorrect .i vmcg must be determined using rudder control alone .ii 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 skill .the nosewheel steering is disconnected because the value of vmcg must also be applicable on wet and/or slippery runways .in 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 9 1 m 30 ft laterally from the centreline at any point exemple 437 I is correct, ii is correct.

Question 249-24 : Vmcl is the ?

Minimum control speed approach and landing

.vmcl is the minimum control speed during landing/approach 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 249-25 : Given .theta = pitch angle.gamma = flight path angle.alpha = angle of attack.no wind bank or sideslip .the 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 alpha..so ..theta = gamma + alpha

Question 249-26 : The four forces acting on an aeroplane in level flight are ?

Thrust lift drag and weight

exemple 449 Thrust, lift, drag and weight.

Question 249-27 : 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% ?

0 50

.aircraft is maintaining level flight only the speed will change by 41% lift will not change ..lift = cl 1/2rho v² s rho = density .v was 1 now v is 1 41..lift = cl 1/2rho 1 41² s.lift = cl 1/2rho 2 s..in order to maintain lift to its original value when v was 1 we can only divide cl by 2 . surface and 1/2rho can't be changed .cl was 1 now cl = 1/2 = 0 5 exemple 453 0.50.

Question 249-28 : 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 ?

1 1

.turn radius = tas² m/s / g tan bank angle .tan bank angle = 100² / 10 x 2000 = 0 5.0 5 is the value of the bank angle tangent so bank angle = 26 5° ..=> n = 1 / cos bank angle .=> n = 1 / cos 26 5.=> n = 1 11

Question 249-29 : 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 factor..vs = 60 kt x square root 2.vs = 60 kt x 1 41.vs = 84 85 kt exemple 461 85 kt.

Question 249-30 : For a straight steady wings level climb of a twin engine aeroplane the one engine inoperative climb gradient is .aeroplane mass 50000 kg .lift/drag ratio 12 .thrust per engine 60000 n .assumed g 10m/s² ?

3 7%

.50000 kg x 10m/s² = 500000 newton .thrust only one engine operative = 60000 newton .climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 60000n/500000n 1/12.climb gradient % = 100 0 12 0 08333 .climb gradient % = 100 x 0 03667 = 3 667% exemple 465 3.7%.

Question 249-31 : Consider the following statements about vmcg .1 vmcg is determined with the gear down .2 vmcg is determined with the flaps in the landing position .3 vmcg is determined by using rudder and nosewheel steering. 4 during vmcg determination the aeroplane may not deviate from the straight line path by ?

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 skill .the nosewheel steering is disconnected because the value of vmcg must also be applicable on wet and/or slippery runways .in 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 9 1 m 30 ft laterally from the centreline at any point

Question 249-32 : In general directional controllability with one engine inoperative on a multi engine aeroplane is favourably affected by .1 high temperature .2 low temperature .3 aft cg location .4 forward cg location . 5 high altitude . 6 low altitude . the combination that regroups all of the correct statements ?

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 condition .a forward cg position increases the moment arm and thus gives more directional controllability .air density is reduced at high altitude resulting in less thrust being produced from the remaining live engine s therefore a lesser turning moment exemple 473 1, 4, 5.

Question 249-33 : Given .aeroplane mass 50000 kg .lift/drag ratio 12 .thrust per engine 30000 n .assumed g 10m/s² .for a straight steady wings level climb of a three engine aeroplane the all engines climb gradient is ?

9 7%

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 90000n/500000n 1/12 .climb gradient % = 100 0 18 0 08333 .climb gradient % = 100 x 0 09667 = 9 667% exemple 477 9.7%.

Question 249-34 : 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 m/s ² / g x tan bank angle .tas in m/s = 500 x 0 515 = 257 5.g acceleration = 10.tan 30° = 0 577.radius = 257 5² / 10 x 0 577 .radius = 11492 m exemple 481 12 km.

Question 249-35 : 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 angle.load factor = 1 414.weight of the aircraft is 40000 newton .the lift is approximately 40000 x 1 414 = 56560 n exemple 485 56000 n.

Question 249-36 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 60 000n .assumed g 10m/s² .for a straight steady wings level climb of a twin engine aeroplane the all engines climb gradient is ?

15 7%

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 120000n/500000n 1/12 .climb gradient % = 100 0 24 0 08333 .climb gradient % = 100 x 0 15666 = 15 66% exemple 489 15.7%.

Question 249-37 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 20 000n .assumed g 10m/s² .for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

7 7 %

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 80000n/500000n 1/12 .climb gradient % = 100 0 16 0 08333 .climb gradient % = 100 x 0 07667 = 7 667% exemple 493 7.7 %.

Question 249-38 : Given .aeroplane mass 50 000kg .lift/drag ratio 10 .thrust per engine 20 000n .assumed g 10m/s² .for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

6 0 %

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 80000n/500000n 1/10 .climb gradient % = 100 0 16 0 1 .climb gradient % = 100 x 0 06 = 6% exemple 497 6.0 %.

Question 249-39 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 21 000n .assumed g 10m/s² .for a straight steady wings level climb of a four engine aeroplane the all engines climb gradient is ?

8 5%

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 84000n/500000n 1/12 .climb gradient % = 100 0 168 0 08333 .climb gradient % = 100 x 0 08467 = 8 467% exemple 501 8.5%.

Question 249-40 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 21 000n .assumed g 10m/s² .for a straight steady wings level climb of a four engine aeroplane the one engine inoperative climb gradient is ?

4 3 %

.climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 63000n/500000n 1/12 .climb gradient % = 100 0 126 0 08333 .climb gradient % = 100 x 0 04267 = 4 267% exemple 505 4.3 %.


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