Question 250-1 : The stick force per g of a heavy transport aeroplane is 300 n/g .what 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 ? [ Explanation maintenance ]
450 n
The limit load factor of a large transport aeroplane is 2 5 g .the question states straight and steady flight so we are at 1 g ..we only require +1 5 g to reach a load factor of 2 5 g..1 5 x 300 n/g = 450 n
Question 250-2 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 28 000n .assumed g 10m/s² .for a straight steady wings level climb of a three engine aeroplane the all engines climb gradient is ?
8 5%
Climb gradient = thrust drag / weight x 100 ..lift/drag ratio 12 > when lift is 1 2 drag is 0 08333..total trust = 3 x 28000 n = 84000 n..weight = 50000 kg x 10m/s² = 500000 n..drag = 500000 x 0 08333 = 41665 n..climb gradient = 84000 41665 / 500000 x 100.climb gradient = 8 467%..or.. 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%
Question 250-3 : Given .aeroplane mass 50 000kg .lift/drag ratio 12 .thrust per engine 28 000n .assumed g 10m/s² .for a straight steady wings level climb of a three engine aeroplane the one engine inoperative climb gradient is ?
2 9%
Climb gradient = thrust drag / weight x 100 ..lift/drag ratio 12 > when lift is 1 2 drag is 0 08333..total trust = 2 x 28000 n = 56000 n 2 engines only one is inoperative ..weight = 50000 kg x 10m/s² = 500000 n..drag = 500000 x 0 08333 = 41665 n..climb gradient = 56000 41665 / 500000 x 100.climb gradient = 2 867%..or.. climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 56000n/500000n 1/12 .climb gradient % = 100 0 112 0 08333 .climb gradient % = 100 x 0 02867 = 2 867%
Question 250-4 : 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 ?
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%
Question 250-5 : 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
Question 250-6 : 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
Question 250-7 : 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
Question 250-8 : 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
Question 250-9 : 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
Question 250-10 : When is a turn co ordinated ?
When the longitudinal axis of the aeroplane at the cg is tangential to the flight path
Question 250-11 : 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
Question 250-12 : Regarding the lift formula if airspeed doubles lift will ?
Be 4 times greater
Lift formula .lift = 1/2 rho cl x v² x s..where .'1/2 rho' is dynamic pressure .'cl' is coefficient of lift .'v' is speed in m/s .'s' is wing area .if speed v in m/s is double lift will be 4 times greater as speed is squared
Question 250-13 : 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
Question 250-14 : 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
Question 250-15 : The angle of attack of an aerofoil is the angle between the ?
Chord line and the relative airflow
Question 250-16 : The forward speed of helicopters is limited by ?
Retreating blade stall and advancing blade tip speed
Question 250-17 : Reverse airflow over part of the rotor is associated with ?
Flight at high forward speed
Question 250-18 : 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 250-19 : In the case of a symmetrical aerofoil ?
Pitching moment variations due to centre of pressure movement are small
Question 250-20 : 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
Question 250-21 : A 'vortex ring state' ?
Encountered when descending with power causes an even higher rate of descent
Question 250-22 : During autorotative descent main rotor rpm are maintained by ?
The lift force created by the upflow of air
Question 250-23 : 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
Question 250-24 : 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%
Question 250-25 : The forces acting on a symmetrical aerofoil element when the upstream airflow is parallel to the chordline is ?
Only drag
Question 250-26 : Thickness/chord ratio of an aerofoil section is expressed in percentage of ?
Chord
Question 250-27 : The center of pressure of an aerofoil element ?
Is the point where the resultant aerodynamic force is applied
Question 250-28 : 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
Admin .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 250-29 : The formula take off mass/rotor disc area is used to calculate ?
Rotor disc loading
Question 250-30 : The angle of attack is the angle determined by ?
The chord line and the relative airflow
Question 250-31 : The coning angle of a fully articulated rotor is defined by ?
Lift centrifugal force and weight of the blade
Question 250-32 : 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
Question 250-33 : For an autogyro the power from engines provided to the rotor is ?
Nil
Question 250-34 : In an autogyro the lift force comes from ?
The forward movement
Question 250-35 : On an autogyro the engine power output is used ?
To ensure forward motion
Question 250-36 : An anti torque rotor is necessary on ?
A single rotor helicopter
Question 250-37 : The rotor thrust is always ?
Perpendicular to the tip path plane
Question 250-38 : On an articulated rotor variations in attitude of the tip path plane are due to ?
Blades flapping angles
Question 250-39 : The rotation axis of the main rotor is always ?
Perpendicular to the hub plane
Admin .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
Question 250-40 : 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
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