Instruction
- Total number of questions : 20
- Time allotted : 30:00 Minutes:Seconds.
- Each question carry 1 mark, no negative marks.
- DO NOT refresh the page.
- All the best
Aircraft Stability and Control [ Question No: 1 To 20 ] :
What does the hinge moment coefficient depend on most strongly?
Stick force in a steady pull-up comes from
An internal sealed balance works by
A bobweight in the control run is fitted to
The main practical reason for expressing derivatives non-dimensionally is that
Which axis system is normally used when quoting stability derivatives for dynamic analysis?
A reduced frequency is defined as \( k = \omega\bar c/2V \). For \( \omega = 3 \ \mathrm{rad/s} \), \( \bar c = 3 \ \mathrm{m} \) and \( V = 150 \ \mathrm{m/s} \), its value is
A Dutch roll oscillation is timed at four complete cycles in 16 seconds. Its undamped natural frequency, for light damping, is about
Statement I: A divergent spiral mode is normally acceptable in a certified aeroplane.
Statement II: The spiral is slow enough that a pilot or an autopilot corrects it long before the bank becomes large.
In the light of the above statements, choose the correct answer.
For a tail arm of \( 15 \ \mathrm{m} \) at \( 150 \ \mathrm{m/s} \), the downwash lag time that gives rise to the \( M_{\dot\alpha} \) term is
The derivative of pitching moment with rate of change of angle of attack exists mainly because
A Dutch roll damps to half amplitude in 3 cycles while a well damped mode takes 1. The poorly damped case takes
An aeroplane is reported as easy to fly but showing a slow tendency to tighten into a turn if the wings are not held level.
What balance of stabilities does this suggest?
Arrange these contributions to the yawing moment derivative in the order they are summed:
1. The fuselage, generally destabilising
2. Any propeller ahead of the centre of gravity, destabilising
3. The wing, contributing modestly if swept
4. The fin, strongly stabilising
5. The total, which must be positive for stability
Select the correct sequence.
An aeroplane's minimum control speed in the air is \( 92 \ \mathrm{kt} \) and its stalling speed \( 88 \ \mathrm{kt} \). The margin between them is
A Mach trim system
Match each point in List I with what vanishes there in List II.
List I
A. Stick fixed neutral point
B. Stick free neutral point
C. Stick fixed manoeuvre point
D. Stick free manoeuvre point
List II
1. Elevator angle gradient with speed
2. Stick force gradient with speed
3. Elevator angle required per g
4. Stick force required per g
Select the correct match.
Stick force per g is \( 30 \ \mathrm{N} \) at a manoeuvre margin of 0.12 chord. At a margin of 0.08 chord it would be about
The manoeuvre point differs from the neutral point in that it
A statically stable aeroplane is always dynamically stable. True or false?
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20
Aircraft Stability and Control [ Question No: 1 To 20 ] :
What does the hinge moment coefficient depend on most strongly?
Your Answer : (Not Answered)
Correct Answer : A
The coefficient is a function of the flow over the surface, so deflection and local incidence dominate, with Mach number and any tab position as further arguments. Weight and altitude affect the flight condition and therefore the dynamic pressure, which multiplies the coefficient rather than setting it. Fuel state is a mass matter. The number of hinges is a structural detail with no aerodynamic significance at all.
Stick force in a steady pull-up comes from
Your Answer : (Not Answered)
Correct Answer : A
In a steady manoeuvre the elevator is displaced from its trim position and the aeroplane has a pitch rate, so both the deflection and the rate contribute to the hinge moment the pilot must hold, which is why stick force per g depends on the manoeuvre margin rather than the static margin. Weight enters through the load factor demanded. The trim tab fixes the zero-force point rather than the gradient. Thrust plays no part in the hinge balance.
An internal sealed balance works by
Your Answer : (Not Answered)
Correct Answer : B
The surface's nose extends into a chamber sealed by a flexible diaphragm, and the pressure difference across that chamber acts ahead of the hinge to oppose the hinge moment. It gives effective balance with no external area and little drag penalty, which is why it is common on transports. No engine air is involved. A spring in the actuator would change the feel rather than the aerodynamic moment. Reducing the aft chord would reduce effectiveness as well as hinge moment.
A bobweight in the control run is fitted to
Your Answer : (Not Answered)
Correct Answer : D
A bobweight is a mass that responds to normal acceleration, so it adds a force proportional to g and raises the stick force per g without affecting the force at one g in level flight. It is the classic fix for an aeroplane whose manoeuvre forces are too light. It does nothing about the speed dependence of hinge moments. Flutter is prevented by mass balancing at the surface, which is a different weight in a different place. Artificial feel units are separate devices.
The main practical reason for expressing derivatives non-dimensionally is that
Your Answer : (Not Answered)
Correct Answer : A
Non-dimensional form collapses the whole dynamic-pressure dependence into one table entry per Mach number, so a single set covers many speeds and altitudes. The equations of motion actually need the dimensional forms, which is why the conversion is always the next step. Inertias are still needed for that conversion. And dimensional derivatives are perfectly measurable; they are simply less portable between conditions.
Which axis system is normally used when quoting stability derivatives for dynamic analysis?
Your Answer : (Not Answered)
Correct Answer : A
Stability axes are the usual choice: body axes rotated so that the x-axis lies along the relative wind in the trim condition, which makes the trim incidence zero and simplifies the equations. Earth axes are the reference for navigation rather than for aerodynamics. Axes that follow every gust would not give a fixed frame to linearise in. Principal axes remove the product-of-inertia terms and are sometimes used, but they are a separate choice and rarely coincide with stability axes.
A reduced frequency is defined as \( k = \omega\bar c/2V \). For \( \omega = 3 \ \mathrm{rad/s} \), \( \bar c = 3 \ \mathrm{m} \) and \( V = 150 \ \mathrm{m/s} \), its value is
Your Answer : (Not Answered)
Correct Answer : B
Substituting gives
\[ k = \frac{3 \times 3}{2 \times 150} = 0.03 \]
which is far below unity, so quasi-steady aerodynamics is justified for this short-period motion. That is the calculation behind the assumption rather than an assertion of it, and it is why flutter analysis, at much higher reduced frequency, cannot use the same derivative set. 0.06 omits the factor of two. 0.30 slips a decade. 0.003 divides by ten as well as by the factor.
A Dutch roll oscillation is timed at four complete cycles in 16 seconds. Its undamped natural frequency, for light damping, is about
Your Answer : (Not Answered)
Correct Answer : B
The period is \( 16/4 = 4.0 \ \mathrm{s} \), so
\[ \begin{aligned} \omega_n \approx \frac{2\pi}{T} &= \frac{6.283}{4.0} \\ &= 1.57 \ \mathrm{rad/s} \end{aligned} \]
giving about 1.57 rad/s. At light damping the damped and undamped frequencies differ by well under one per cent, so the approximation is safe here but would not be for a well-damped mode. 0.25 rad/s is the cycle rate in hertz rather than radians per second, a factor of \( 2\pi \) adrift. 4.0 rad/s is the period read back as a frequency, and 0.79 halves the correct answer.
Statement I: A divergent spiral mode is normally acceptable in a certified aeroplane.
Statement II: The spiral is slow enough that a pilot or an autopilot corrects it long before the bank becomes large.
In the light of the above statements, choose the correct answer.
Your Answer : (Not Answered)
Correct Answer : A
Both statements are correct. Specifications do not require spiral stability outright; they require that if the mode diverges, it does so slowly, typically quoting a minimum time to double amplitude. A pilot glancing at the attitude indicator every few seconds corrects a twenty-second divergence without conscious effort, and a wing leveller removes the workload altogether. What is not acceptable is a rapid spiral, which can lead to a steep descending turn if attention lapses.
For a tail arm of \( 15 \ \mathrm{m} \) at \( 150 \ \mathrm{m/s} \), the downwash lag time that gives rise to the \( M_{\dot\alpha} \) term is
Your Answer : (Not Answered)
Correct Answer : B
The lag is simply the transport delay from wing to tail:
\[ \tau = \frac{l_t}{V} = \frac{15}{150} = 0.10 \ \mathrm{s} \]
so a tenth of a second. Compared with a short period of a second or two that is small but not negligible, which is why the derivative is retained. 1.0 s is a decade out and would be comparable with the whole mode. 0.01 s slips the other way. 2.25 s is the product of the two numbers scaled, which has the wrong dimensions entirely. Note that the lag shrinks as speed rises, so the term matters most at low speed.
The derivative of pitching moment with rate of change of angle of attack exists mainly because
Your Answer : (Not Answered)
Correct Answer : A
The wing's downwash takes about \( l_t/V \) seconds to reach the tail, so when incidence is changing the tail is immersed in the downwash belonging to an earlier incidence. That lag makes the tail's effective incidence depend on the rate of change as well as the value, which is exactly the derivative in question. It normally supplies a tenth to a third of total pitch damping. Boundary layer response is far faster and is not what the term models. Elevator inertia matters only in stick-free dynamics. Engine lag affects the phugoid through thrust, not this derivative.
A Dutch roll damps to half amplitude in 3 cycles while a well damped mode takes 1. The poorly damped case takes
Your Answer : (Not Answered)
Correct Answer : B
Correct answer: Three times as many cycles
The comparison is a ratio of cycles to half amplitude:
\[ \frac{3}{1} = 3 \]
Cycles to half amplitude is the standard measure of oscillatory damping, and fewer is better. A Dutch roll taking three cycles to halve is poorly damped and tiring to fly, since the oscillation persists visibly after every disturbance. The usual causes are excessive dihedral effect relative to directional stability, and the usual remedy is a yaw damper rather than a change to the airframe.
An aeroplane is reported as easy to fly but showing a slow tendency to tighten into a turn if the wings are not held level.
What balance of stabilities does this suggest?
Your Answer : (Not Answered)
Correct Answer : A
The weathercock turns the nose into the developing turn faster than the dihedral effect can roll the wings back level. The result is the spiral mode, which diverges slowly and is easily controlled but must not be ignored.
Arrange these contributions to the yawing moment derivative in the order they are summed:
1. The fuselage, generally destabilising
2. Any propeller ahead of the centre of gravity, destabilising
3. The wing, contributing modestly if swept
4. The fin, strongly stabilising
5. The total, which must be positive for stability
Select the correct sequence.
Your Answer : (Not Answered)
Correct Answer : A
Correct answer: 1, 2, 3, 4, 5
The order runs through the airframe from nose to tail, ending with the sum. The fuselage comes first, its side area largely ahead of the centre of gravity and therefore destabilising. A tractor propeller is destabilising for the same geometric reason, its side force acting at the disc.
The wing contributes modestly, usefully so if swept, through the drag differential between forward and rear panels. The fin comes last and is by far the largest term, sized precisely to overcome everything before it. The total \( C_{n\beta} \) must be positive for stability, which is what the whole exercise is for.
An aeroplane's minimum control speed in the air is \( 92 \ \mathrm{kt} \) and its stalling speed \( 88 \ \mathrm{kt} \). The margin between them is
Your Answer : (Not Answered)
Correct Answer : C
Correct answer: 4 knots, with control speed the higher
The margin is the difference, and its sign is what the question tests:
\[ 92 - 88 = 4\ \mathrm{kt} \]
with \( V_{MCA} \) above \( V_S \). That ordering is required by certification, since an aeroplane that lost directional control before it stalled would be uncontrollable while still flying. It also explains why the take-off safety speed \( V_2 \) is set above both, and why on this aeroplane the controlling limit at low speed is the rudder rather than the wing - a small margin, so the engine-out case governs.
A Mach trim system
Your Answer : (Not Answered)
Correct Answer : B
It counters the nose down moment produced by the rearward shift of the aerodynamic centre. Without it the speed stability of the aeroplane would reverse, so that a small increase in speed would produce a tendency to accelerate further.
Match each point in List I with what vanishes there in List II.
List I
A. Stick fixed neutral point
B. Stick free neutral point
C. Stick fixed manoeuvre point
D. Stick free manoeuvre point
List II
1. Elevator angle gradient with speed
2. Stick force gradient with speed
3. Elevator angle required per g
4. Stick force required per g
Select the correct match.
Your Answer : (Not Answered)
Correct Answer : B
Correct answer: A-1, B-2, C-3, D-4
Each point is defined by the gradient that vanishes there, and two distinctions run through the set. Angle belongs to the stick fixed cases and force to the stick free ones; speed belongs to the neutral points and normal acceleration to the manoeuvre points.
So the stick fixed neutral point is where the elevator angle gradient with speed vanishes, the stick free neutral point where the stick force gradient does, and the two manoeuvre points where angle per g and force per g vanish respectively.
Stick force per g is \( 30 \ \mathrm{N} \) at a manoeuvre margin of 0.12 chord. At a margin of 0.08 chord it would be about
Your Answer : (Not Answered)
Correct Answer : A
Correct answer: 20 N
Stick force per g is proportional to the manoeuvre margin, so the two scale together:
\[ 30 \times \frac{0.08}{0.12} = 20\ \mathrm{N} \]
The trend matters as much as the number. Moving the centre of gravity aft shrinks the manoeuvre margin and lightens the stick, so the aeroplane becomes easier to overstress. Certification therefore sets a minimum stick force per g, and that minimum is one of the constraints fixing the aft centre of gravity limit.
The manoeuvre point differs from the neutral point in that it
Your Answer : (Not Answered)
Correct Answer : C
In a steady pull-up the aeroplane is also pitching, and the resulting damping adds a restoring moment. Stick force per g therefore survives a little beyond the centre of gravity at which trim stability has already vanished.
A statically stable aeroplane is always dynamically stable. True or false?
Your Answer : (Not Answered)
Correct Answer : C
Static stability guarantees only that the initial response opposes the disturbance. Whether the resulting oscillation grows, decays or persists depends on damping, which is an entirely separate property of the aeroplane.
Marks
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20