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Question 258-1 : The si unit of measurement for pressure is ? [ Explanation maintenance ]
N/m².
.icao annex 5 units of measurement attachment b. 2. mass, force and weight.2.1 the principal departure of si from the gravimetric system of metric engineering units is the use of explicitly distinct units from mass and force. in si, the name kilogram is restricted to the unit of mass, and the kilogram force from which the suffix force was in practice often erroneously dropped is not to be used. in its place the si unit of force, the newton, is used. likewise, the newton rather than the kilogram force is used to form derived units which include force, for example, pressure or stress n/m² = pa , energy nm = j , and power nm/s = w.
Question 258-2 : The polar curve of an aerofoil section is a graphic relationship between ?
Lift coefficient cl and drag coefficient cd.
Question 258-3 : True airspeed tas is ?
Lower than the indicated airspeed ias at altitudes below sea level, under isa conditions.
.true air speed tas is obtained from indicated air speed ias by correcting for the following errors instrument, position, compressibility and density..so, tas is greater than ias with increasing altitude as density reduces...for this question, they state at altitudes below sea level , so it is reverse tas will be lower than ias density is higher for altitude below mean sea level...keep in mind that.ias is proportional to 1/2 rho tas².. rho = density.
Question 258-4 : Total pressure is rho = density ?
Static pressure plus dynamic pressure.
Question 258-5 : The unit of measurement for density is ?
Kg/m³.
Question 258-6 : The transition point is the point where ?
The boundary layer changes from laminar to turbulent.
Question 258-7 : Considering a positively cambered aerofoil section, the pitching moment when the lift coefficient cl=0 is ?
Negative nose down.
.when a positively cambered airfoil is at zero lift it has to be at a small negative alpha to cancel out the camber, which gives some lift at zero alpha the angle of attack has a negative value when the lift coefficient equals zero..it is because the pressure distribution is producing negative lift at the front and positive lift at the back of the airfoil.. 1631.a symmetrical aerofoil need to have a positive pitch to produce lift. if the pitching moment is zero, the lift is zero..a positively cambered aerofoil produces lift with a zero pitching moment, thus, not to produce lift, it need a negative pitching moment.. 2416.in this example, lift coeffient = 0 when angle of attack = 4° the pitching moment is nose down.
Question 258-8 : An aerofoil is cambered when ?
The line, which connects the centres of all inscribed circles, is curved.
Question 258-9 : A flat plate, when positioned in the airflow at a small angle of attack, will produce ?
Both lift and drag.
Question 258-10 : Positive static stability of an aeroplane means that following a disturbance from the equilibrium condition ?
The initial tendency is to return towards its equilibrium condition.
Question 258-11 : Load factor is ?
Lift/weight.
Question 258-12 : Which statement is correct about an expansion wave in supersonic flow.1. the static temperature in front of an expansion wave is higher than the static temperature behind it..2. the speed in front of an expansion wave is higher than the speed behind it. ?
1 is correct and 2 is incorrect.
.the effect on pressure, temperature density and velocity is the opposite in expansion waves to oblique/normal shock waves... expansion wave.also known as a prandtl meyer expansion fan. when a supersonic airflow passes over a surface that suddenly drops away or rounds corner, an expansion fan of an infinite number of shock waves forms at the apex of the corner and diverts the airflow around the corner..airflowing through an expansion fan goes through the following changes..1.the airsteam accelerates, and the air behind the shock wave has higher supersonic velocity.2.the airflow direction follows the surface contour.3.the static pressure behind the expansion wave decreases.4.the density behind the expansion wave decreases.5.there is no loss of energy.. 2417. oblique/normal wave. 2418
Question 258-13 : Which rotor system is most effective in terms of control power ?
The rigidhead.
. 2405
Question 258-14 : What happens as a helicopter accelerates through 12 kt ?
It passes through translational lift.
Question 258-15 : With reference to translational lift, which of the following statements is true ?
Translational lift with associated reduction in induced flow and inflow angle becomes effective at about 12 knots
Question 258-16 : When a helicopter is in an ideal free air hover the trt is acting vertically and is equal and opposite to ?
Weight and parasite drag.
..trt total rotor thrust..the parasite drag comes from the airflow blowing downwards over the fuselage. it adds to the apparent weight, so you need a little extra thrust to overcome it...looking at forward flight, you have four main vectors. 2419.as you slow down, and the downwash slowly changes from being rear and down to just being down, the vectors move around a little lift and weight stay the same, drag moves to being a bit down, and thrust has to move a bit up..in the hover, weight and drag coincide, and lift and thrust coincide, so in the hover you get.lift + thrust = weight + drag.
Question 258-17 : As a helicopter accelerates the attitude becomes ?
Nose down.
Question 258-18 : Where on the surface of a typical aerofoil will flow separation normally start at high angles of attack ?
Upper side trailing edge.
Question 258-19 : Lift is the ?
Component of the total aerodynamic force, perpendicular to the undisturbed airflow.
.the component of the force perpendicular to the relative free stream i.e. undisturbed airflow is lift.
Question 258-20 : A helicopter is statically stable when it shows the ?
Tendency to return to its original attitude after having been displaced
Question 258-21 : Blade sailing can occur when ?
The rotor is starting up or slowing down in gusty wind conditions
Question 258-22 : Which statement is correct regarding the coning angle ?
With increased rotor rpm the coning angle decreases
Question 258-23 : Dissymmetry of lift due to forward flight is compensated for by the ?
Advancing blade flapping up and reducing its angle of attack and the retreating blade flapping down and increasing its angle of attack
Question 258-24 : The rotor disc plane is always ?
Perpendicular to the rotor thrust
Question 258-25 : The main induced air flow velocity is considered perpendicular to the ?
Tip path plane
Question 258-26 : For a rotor which turns in a clockwise direction seen from above, in a backward flight, with zero wind, the pilot will have the advancing blade ?
On his right
Question 258-27 : For a rotor which turns in an anti clockwise direction seen from above, in a backward flight, with zero wind, the pilot will have the retreating blade ?
On his right
Question 258-28 : For a rotor which turns in a clockwise direction seen from above, in a backward flight, with zero wind, the pilot will have the retreating blade ?
On his left
Question 258-29 : Separation of the airflow from a blade ?
Occurs with increasing mach number at decreasing angle of attack.
Question 258-30 : Drag dampers in a rotor system ?
May use the principle of damping by friction.
Question 258-31 : The main reason for the negative range of tail rotor pitch is ?
For yaw control in autorotation.
Question 258-32 : Why is sweep back often applied to the tip of modern rotor blades ?
To reduce the mach effects on the advancing blade.
Question 258-33 : One reason for increased rotor rpm in the flare is ?
Conservation of angular momentum coriolis effect
Question 258-34 : The advance angle is ?
Used to compensate for phase lag.
Question 258-35 : The ground effect is affected by.1 the hover height.2 the length of the helicopter.3 the ground texture.4 the diameter of the main rotor.the combination of correct statements is ?
1, 3 and 4.
Question 258-36 : As a result of a positive 'g' manoeuvre in a helicopter there is ?
An increase in coning angle.
Question 258-37 : Drag dampers in a rotor system ?
May use the principle of damping by means of oil.
Question 258-38 : The forces and moments caused by the dragging of the rotor blades on a hingeless rotor head are absorbed ?
Through bending at the blade root.
Question 258-39 : The recommended recovery action from vortex ring settling with power is to ?
Cyclic forward to increase airspeed, pause for airspeed then apply power
Question 258-40 : The centre of gravity of a single rotor helicopter will be ?
Close to the main rotor axis.
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