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Question 273-1 : Given.aeroplane mass 50 000kg..lift/drag ratio 10..thrust per engine 30 000n..assumed g 10m/s²..for a straight, steady, wings level climb of a three engine aeroplane, the all engines climb gradient is ? [ Analysis topography ]

8.0%.

Climb gradient = thrust drag / weight x 100...lift/drag ratio 10 > when lift is 1, drag is 0.1...total trust = 3 x 30000 n = 90000 n...weight = 50000 kg x 10m/s² = 500 000 n...lift = 500000 x 1 = 500000 n..drag = 500000 x 0.1 = 50000 n...climb gradient = 90000 50000 / 500000 x 100..climb gradient = 8%. exemple 373 8.0%.

Question 273-2 : What is the heading change after 10 seconds of an aeroplane performing a rate one turn ?

30 degrees.

.a standard rate turn is defined as a 3° per second turn, which completes a 360° turn in 2 minutes. this is known as a 2 minute turn , or rate one = 180°/minute...10 secondes x 3°/sec = 30 degrees. exemple 377 30 degrees.

Question 273-3 : An aeroplane has a stall speed of 78 kt at its mass of 6850 kg. what is the stall speed when the mass is 5000 kg ?

67 kt.

.change in mass ratio = new mass / old mass = 5000/6850 = 0.73.stall speed of an aircraft changes in proportion to the square root of the change in mass.square root of 0.73 = 0.855.original stall speed = 78 kt.stall speed for a weight of 5000 kg ==> 78 x 0.855 = 67 kt. exemple 381 67 kt.

Question 273-4 : In order to fly a rate one turn at a higher airspeed, the bank angle must be ?

Increased and the turn radius will increase.

.a standard rate turn is defined as a 3° per second turn, which completes a 360° turn in 2 minutes. this is known as a 2 minute turn , or rate one = 180°/minute...turn radius = tas² m/s / g.tan bank angle..rate of turn = tas kt / 20 x 3.14 x turn radius...in order to maintain a 3° per second turn at a higher speed, you must increase the bank angle, your turn radius will increase. exemple 385 Increased and the turn radius will increase.

Question 273-5 : Which of these statements about vmcg determination are correct or incorrect.i. during vmcg determination, nose wheel steering may be used..ii. during vmcg determination, the cg should be on the forward limit. ?

I is incorrect, ii is incorrect.

.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 389 I is incorrect, ii is incorrect.

Question 273-6 : An aeroplane maintains straight and level flight at a speed of 2 * vs. if a vertical gust causes a load factor of 2, the load factor n caused by the same gust at a speed of 1.3 vs would be ?

N = 1.65.

.1 x 1.3/2 = 0.65 g..0.65g + 1g = 1.65g.. in straight and level flight load factor is 1g. exemple 393 N = 1.65.

Question 273-7 : When is a turn co ordinated ?

When the longitudinal axis of the aeroplane at the cg is tangential to the flight path.

exemple 397 When the longitudinal axis of the aeroplane at the cg is tangential to the flight path.

Question 273-8 : An aeroplane flying at 100 kt in straight and level flight is subjected to a disturbance that suddenly increases the speed by 20 kt. assuming the angle of attack remains constant, the load factor will initially ?

Increase to 1.44.

.load factor is proportional to lift, and lift is proportional to v²...new load factor / old load factor = new speed / old speed ²..new load factor = old load factor x 120 / 100 ² = 1 x 1.2² = 1.44. exemple 401 Increase to 1.44.

Question 273-9 : Regarding the lift formula, if airspeed doubles, lift will ?

Be 4 times greater.

exemple 405 Be 4 times greater.

Question 273-10 : 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 30% ?

0.60.

.aircraft is maintaining level flight, only the speed will change by 30%, lift will not change..lift = cl 1/2rho v² s rho = density.v was 1, now v is 1.30..lift = cl 1/2rho 1.30² s.lift = cl 1/2rho 1.69 s..in order to maintain lift to its original value when v was 1 , we can only divide cl by 1.69.. surface and 1/2rho can't be changed.cl was 1, now cl = 1/1.69 = 0.5917 approximate value 0.60. exemple 409 0.60.

Question 273-11 : The centre of pressure is the point ?

On the chord line of an aerofoil section through which the resultant of the lift forces acts.

The centre of pressure moves with the angle of attack... 669 exemple 413 On the chord line of an aerofoil section through which the resultant of the lift forces acts.

Question 273-12 : The angle of attack of an aerofoil is the angle between the ?

Chord line and the relative airflow.

exemple 417 Chord line and the relative airflow.

Question 273-13 : The forward speed of helicopters is limited by ?

Retreating blade stall and advancing blade tip speed.

exemple 421 Retreating blade stall and advancing blade tip speed.

Question 273-14 : Reverse airflow over part of the rotor is associated with ?

Flight at high forward speed.

exemple 425 Flight at high forward speed.

Question 273-15 : In forward flight, if the rotor rpm increase above that specified ?

The large centrifugal forces impose severe and possibly excessive loads on the hub.


Question 273-16 : In the case of a symmetrical aerofoil ?

Pitching moment variations due to centre of pressure movement are small.

exemple 433 Pitching moment variations due to centre of pressure movement are small.

Question 273-17 : In still air hover ground cushion is normally of practical value up to a rotor height above ground ?

Equal to the diameter of the main rotor.

exemple 437 Equal to the diameter of the main rotor.

Question 273-18 : A 'vortex ring state' ?

Encountered when descending with power causes an even higher rate of descent.


Question 273-19 : During autorotative descent, main rotor rpm are maintained by ?

The lift force created by the upflow of air.


Question 273-20 : In hovering, for a single rotor helicopter whose main rotor turns clockwise from above, the thrust of the main rotor will be mainly vertical but with a slight orientation towards the ?

Right.

exemple 449 Right.

Question 273-21 : The center of the pressure of a symmetrical aerofoil section is behind the leading edge approximately at the following % of the section chord at ?

25%.

exemple 453 25%.

Question 273-22 : The forces acting on a symmetrical aerofoil element when the upstream airflow is parallel to the chordline is ?

Only drag.


Question 273-23 : Thickness/chord ratio of an aerofoil section is expressed in percentage of ?

Chord.

exemple 461 Chord.

Question 273-24 : The center of pressure of an aerofoil element ?

Is the point where the resultant aerodynamic force is applied.

exemple 465 Is the point where the resultant aerodynamic force is applied.

Question 273-25 : The blades of a rotor are in track. an upward bent outboard trim tab of one of one of the blades influences the rotor blade in rotation as follows ?

The blade rotates above the plane of rotation with neutral trim tab position.

.bending the tab up on the trailing edge of the blade would tend to increase pitch angle. the track of the blade would be higher.. 1628

Question 273-26 : The formula take off mass/rotor disc area is used to calculate ?

Rotor disc loading.

exemple 473 Rotor disc loading.

Question 273-27 : The angle of attack is the angle determined by ?

The chord line and the relative airflow

exemple 477 The chord line and the relative airflow

Question 273-28 : The coning angle of a fully articulated rotor is defined by ?

Lift, centrifugal force and weight of the blade.

exemple 481 Lift, centrifugal force and weight of the blade.

Question 273-29 : The resulting rolling moment on the main rotor hub due to the asymmetrical airflow on the blades in forward flight is reduced by ?

Blade flapping

exemple 485 Blade flapping

Question 273-30 : For an autogyro, the power from engines provided to the rotor is ?

Nil

exemple 489 Nil

Question 273-31 : In an autogyro the lift force comes from ?

The forward movement

exemple 493 The forward movement

Question 273-32 : On an autogyro, the engine power output is used ?

To ensure forward motion

exemple 497 To ensure forward motion

Question 273-33 : An anti torque rotor is necessary on ?

A single rotor helicopter

exemple 501 A single-rotor helicopter

Question 273-34 : The rotor thrust is always ?

Perpendicular to the tip path plane

exemple 505 Perpendicular to the tip path plane

Question 273-35 : On an articulated rotor, variations in attitude of the tip path plane are due to ?

Blades flapping angles.

exemple 509 Blades flapping angles.

Question 273-36 : The rotation axis of the main rotor is always ?

Perpendicular to the hub plane.

.the hub plane also called the reference plane is always 90° to the shaft axis. it does not move and is not affected by different types rotor head.. 1610 exemple 513 Perpendicular to the hub plane.

Question 273-37 : Helicopter rotor blades are designed with twist or taper to incorporate washout a design that ?

Ensures adequate lift distribution along the span of the blade

exemple 517 Ensures adequate lift distribution along the span of the blade

Question 273-38 : In autorotation the blades are free to rotate because ?

A freewheel unit disengages the rotor from the engine

exemple 521 A freewheel unit disengages the rotor from the engine

Question 273-39 : The effect of ground cushion on a hovering helicopter is greatest on ?

Level ground with no wind

exemple 525 Level ground with no wind

Question 273-40 : A free wheel is fitted to a helicopter to ?

Prevent the rotor from driving the engine

exemple 529 Prevent the rotor from driving the engine


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