Principles of Flight Flashcards

Question

Pressure diagram at high AoA

Answer 1

When pressure gradient causes air pressure movement against air flow direction. i.e. at stall, high pressure area under wing "leaking" around trailing edge to upper surface, disturbing air flow and thus removing the negative pressure there

Answer 2

Lift = C(L) x 1/2 x rho x V^2 x S

Answer 3

CoP moves forwards as AoA increases and strong sucking force is created towards leading edge. It is furthest forward at peak C(Lmax) and then moves backwards after the stall. [Note: CoP for symmetrical aerofoil is static]

Answer 4

Centre of pressure for symmetrical aerofoils does NOT move with AoA.

Answer 5

Dimensionless numbers attributed to an aircraft (in a given state, e.g. flaps, AoA). They are ratios between wing loading and dynamic pressure, and drag and dynamic pressure respectively.

Answer 6

The point where all changes in magnitude of lift force effectively take place. AND The point about which pitching moment remains constant at 'normal' angles of attack.

Answer 7

As AoA increases, the lift force increases, but the CoP also moves forward. At normal angles of attack these balance to maintain the moment around the AC.

Answer 8

25% along chord line, regardless of aerofoil characteristics (as long as M < 0.4)

Answer 9

At zero lift/AoA a symmetrical aerofoil obviously has equal forces on both sides in equal positions, therefore also zero moment around AC. As AoA increases, force increases on one side relative to the other, however moment around AC must be unchanged therefore zero, therefore pitching moment for symmetrical aerofoil at 'normal' angles of attack always zero.

Answer 10

Roughness around the leading edge has a huge effect on C(Lmax), further back than about 20% from leading edge, the impact is minimal

Answer 11

At low angles of attack the impact is negligible (i.e. normal C(L) vs AoA profile) as airflow more easily follows required profile at lower "deflection". However as AoA increases, the icing/frost/roughness has starts to impact ability of air to smoothly follow the aerofoil and C(Lmax) is reached sooner.

Answer 12

Flaps increase lift and drag, but we're only looking at lift, so C(L) is shifted higher along whole AoA curve. Higher C(Lmax) will be achievable, but it is at a slightly lower AoA (i.e. critical angle) than clean configuration.

Answer 13

Wingspan (b) / Average Chord (c) or Wingspan (b) ^ 2 / Wing Area (S)

Answer 14

Tip Chord (C(T)) / Root Chord (C(R)) 1 for rectangular, 0 for pointed tip

Answer 15

Angle between root chord and the line 25% along chords (from leading edge)

Answer 16

Elliptical

Answer 17

Simply average chord

Answer 18

The chord of a hypothetical rectangular wing with the same span which would have the same pitching moment characteristics. Used to model longitudinal stability of aircraft (especially useful for swept wing aircraft).

Answer 19

Start underside wing, flowing outwards towards wing tip, then upwards and back towards the fuselage, before going back down to complete the spiral. For those at the wing tip the return towards fuselage flows over the top of the wing surface.

Answer 20

Trailing vortices create a downwash in the airflow under and behind the wing, which causes effective airflow to be at a higher angle than relative airflow. This requires an increased angle of attack to achieve the same amount of lift, compared to if there were no vortices.

Answer 21

The effective angle of attack is the angle between the chord and the effective air flow. The difference between total AoA and effective AoA (i.e. chord to relative air flow) is the induced angle of attack, in other words the amount of additional AoA required to maintain lift as a result of induced downwash.

Answer 22

For theoretical 2D study we use AoA = chord to relative airflow. Complications in 3D (wing twist, effective airflow) mean we take AoA to be longitudinal axis to relative airflow.

Answer 23

From when nose wheel first lifts off until it first contacts the ground on landing.

Answer 24

Weight - higher is stronger Wingspan - proximity of two trailing vortices Airspeed - slower is stronger Configuration - clean configuration stronger Attitude - high AoA is stronger

Answer 25

Remain around 3/4 of wingspan apart. Drift downwards and level off around 500 to 1000ft below aircraft. Up to 9nm behind large aircraft.

Answer 26

Within 1000ft of the ground they will contact the ground and drift outwards at about 5kts, +/- windspeed.

Answer 27

Ground effect reduces downwash therefore minimising vortex generation. This is why induced drag is lower when in ground effect.

Answer 28

Within about half a wing span

Answer 29

Experienced when taking off, this is the loss of extra lift you were getting from ground effect.

Answer 30

An increase in downwash (for example leaving ground effect) will deflect effective airflow over tailplane downwards (unless tailplane elevated high) decreasing effective AoA, creating a pitch up moment.

Answer 31

When entering ground effect pressure at static port will increase, causing under-read of altitude and under-read of ASI. Opposite when taking off (existing ground effect).

Answer 32

Thin wings generate less lift at given AoA. They will however fly faster before shock waves generated.

Answer 33

The distribution of pressure acting on the wing, NOT the airflow above and below the wing.

Answer 34

Form (aka Pressure drag) Friction Interference

Answer 35

Includes form (aka pressure) drag and skin friction drag. Doesn't include interference drag.

Answer 36

Airflow other than relative airflow, i.e. airflow impacted by the aircraft. In laminar boundary layer this means the air is being given velocity due to viscosity of air and skin friction against the air layer touching the wing.

Answer 37

Point at which boundary layer becomes turbulent

Answer 38

Higher kinetic energy and higher skin friction

Answer 39

- Surface condition - roughness causes turbulence downstream of that point - Adverse pressure gradient - reverse flow prevents laminar flow Transition point will be at the point of maximum curvature.

Answer 40

Further back than the transition point, this is where the turbulent boundary layer separates. Beyond this we get zero lift and high drag.

Answer 41

When boundary layer has insufficient kinetic energy to overcome adverse pressure gradient. Caused by increase in angle of attack or shock wave.

Answer 42

Adverse pressure gradient at trailing edge creates low pressure area, whilst there is relatively high pressure at the leading edge, resulting in a force in the direction of lower pressure area (i.e. backwards).

Answer 43

Turbulent layer actually has more kinetic energy than laminar thus more resistance to separation (due to adverse pressure gradient) than laminar flow. This delays separation and allows higher AoA, but cost of increased friction drag is high.

Answer 44

This is the ratio of length to depth of a body as it relates to streamlining and reduction of form drag. Optimal ratio is 3:1 (with round shape).

Answer 45

C(Di) = C(L) ^ 2 / AR C(Di) = coefficient of induced drag C(L) = coefficient of lift AR = aspect ratio

Answer 46

Higher aspect ratio requires less AoA to generate the same C(L) due to decreased induced drag. The higher downwash of low aspect ratio decreases effective AoA, reducing C(L). However this also increases the critical AoA. Max C(L) is lower but max AoA is higher.

Answer 47

- Increased wing bending moment forces - Reduce rate of roll due to aerodynamic issues in roll - Reduce ground clearance in roll during take off

Answer 48

Aspect ratio Geometric washout Wing end plates (higher AR with less wing bending moment) Tip tanks Winglets - create negative drag (i.e. thrust force), block tip vortices and create new vortices that counter them Wing tip shape

Answer 49

Geometric washout is change in angle of incidence along the wing. Aerodynamic washout is a change in the aerofoil section along the wing.

Answer 50

Frontal area (i.e. configuration) Speed (increases with square of speed)

Answer 51

Parasite drag unaffected by weight, relates only to speed. Induced drag at a given speed is lower when weight is reduced, as the required lift is less. This reduces total drag and reduces the speed at which induced and parasite drag are equal (i.e. V(MD)).

Answer 52

No effect as V(MD) is based on IAS. TAS has to change to maintain IAS, but drag, V(MD) don't vary in terms of IAS.

Answer 53

Configuration doesn't impact induced drag significantly. However parasite drag is increased significantly at a given speed, pushing up total drag curve and reducing speed at which induced and parasite drag are equal (i.e. V(MD)).

Answer 54

V(MD) on the drag chart represents the break point for speed stability (speed unstable to left, stable to right). At approach we are close to V(MD) therefore speed instability is a concern. Extending flaps, gear etc increases drag, but reduces V(MD) therefore helps with speed stability.

Answer 55

Power = work / time = force * distance / time = force * speed Drag = force required Therefore power = drag * TAS

Answer 56

To get power vs speed curve, multiply each point on the drag vs IAS curve by TAS (assume IAS = TAS for simplicity) and the resulting low point will be at a lower speed. So V(MP) < V(MD). Flying at V(MP) results in more drag and less efficient flight, but lower power setting. Range will be reduced but endurance will increase.

Answer 57

In small craft use rudder, aileron use is dangerous as can cause spin. Larger craft have powerful rudder however and are designed to allow aileron usage up to "stall recognition", therefore aileron + small amount of rudder can be used.

Answer 58

1) Nose-down pitch that cannot be readily arrested. 2) Buffeting strong enough to deter further speed reduction. 3) Pitch control reaches aft stop with no further increase in pitch attitude possible.

Answer 59

V(SR) - reference stall speed

Answer 60

- zero thrust - propeller pitch controls in TO config - CoG position to maximise V(SR) - Trimmed for flight speed 1.13 * V(SR) to 1.3 * V(SR) - Decelerate by pulling back at max 1kt per second

Answer 61

At least greater of 5% or 5kts above V(SR) when decelerating at less than 1kt per second. If V(SR) is increased by 2% or 2kts above V(CLmax) due to stick pusher, only 3% or 3kts required (thus total 5%/5kts achieved vs V(CLmax)).

Answer 62

Warning margin must be sufficient to allow pilot to prevent stalling when recovery initiated within 1 second in slow turns at 1.5g load factor, 2kts per second deceleration and trim for 1.3 * V(SR). Must also operate in all normal configurations and also each abnormal configuration of high lift devices that might be in use due to system failures.

Answer 63

Flapper switch - activated when stagnation point moves past it Angle of attack vane - Attached to side of fuselage, vane sits in streamline of relative airflow, detecting AoA Angle of attack probe - Attached to side of fuselage, probe with slots sensitive to changes in relative airflow [Wing mounted vain positioned on lower surface of leading edge]

Answer 64

Also considers rate of change of AoA, thus allowing additional warning in the case that AoA is increasing rapidly, allowing time for recovery.

Answer 65

Roll occurring on recovery limited to: Wing level: 20 degrees 30 deg turn: 60 deg or 30 opposite direction, <1kt/sec deceleration 60 deg turn: 90 or 60 opposite, >1kt/sec deceleration

Answer 66

- Sharp leading edge ratio - Thin aerofoil relative to chord - Maximum camber and thickness to the aft

Answer 67

Large wing tips create strong, supported tip vortices, reducing effective AoA @ tips. Thus root stalls first, CoP moves rearward therefore: - Aileron's remain effective - Nose drops - Aerodynamic buffet (separated air over roots hits the tailplane) - No violent wing drop

Answer 68

Smaller wing tips mean relatively more vortices @ root and tips stall first. Smaller rearward movement of CoP. Problems therefore: - Ailerons ineffective - Limited pitch down - Limited tailplane buffet - Increased chance of wing drop

Answer 69

- Geometric washout - Variance of aerofoil section along wing (greater thickness & amber at root) - Leading edge slots towards tip - Stall strips near root - Vortex generators

Answer 70

Rows of small aerofoil shaped blades projecting around 2.5cm into airstream. They create small vortices which mix high energy free stream flow with boundary layer, increasing its kinetic energy to delay separation.

Answer 71

Flow of air over wing from root to tip collides with vortex direction at tip, creating slow moving air and stall at the TIP. Sweep back also means tips are aft of the root, so tip stall makes CoP move FORWARD. This creates a dangerous pitch UP effect.

Answer 72

Need to prevent spanwise (root to tip) flow of boundary layer over the wing, keep it straight in line with relative airflow. - Wing fences (boundary layer fences) sit on upper surface. - Vortilons do a similar job on underside of wing (engine nacelles help) - Saw tooth leading edge can create vortex over upper surface at high AoA that has the same effect.

Answer 73

Prevention of stall, thus "improving low speed handling characteristics". Prevention of stall is naturally connected to low speed phases of flight. At high speed these stall problems are less relevant.

Answer 74

On swept back wing aircraft with high mounted tailplane, the pitch up stall tendency then leads to the tailplane falling into the turbulent air from the wing. This reduces effectiveness of elevator control and thus prevents a recovery from the stall.

Answer 75

Necessary device for super stall characteristic aircraft, as it is impossible to recover from a stall. Provides 80lb forward push on elevator control in the stall before pilot can react. Can be "dumped" by the pilot.

Answer 76

Stall speed related to square root of weight. Therefore if aircraft weight changes, multiply V(s) by sqrt(new weight / old weight).

Answer 77

Lift factor increases by (1/cos(theta)). So stall speed increases by sqrt of that. To derive, consider that lift in the bank needs to be sufficient to keep the same vertical component. This makes the lift the hypoteneuse and vertical lift adjacent.

Answer 78

Forward CoG creates a nose down pitch (CoP is further back), requiring download from the elevators which has to be offset by increased lift from main wings, thus higher stall speed.

Answer 79

Profile drag of landing gear creates pitch down moment, thus download from the elevators, requiring increased lift from main wings and increasing stall speed.

Answer 80

1) Propeller craft create slipstream over wings at high power (low speed) which delays the stall. 2) Jet engines have a vertical component of thrust due to low placed jet engines, supporting part of aircraft weight and delaying stall speed.

Answer 81

Compressibility of air above mach 0.4 means upwash of streamline pattern is less effective and C(Lmax) reduces between Mach 0.4 and 1.0 (increases thereafter). Thus over mach 0.4 stall speed increases.

Answer 82

V(s) is an IAS (not TAS) so doesn't change with altitude at low altitude. However at some point (around 30k ft) V(S) will rise above mach 0.4, where compressibility of air is an issue. Above that altitude V(S) will increase with altitude.

Answer 83

Increases roughness of wing surface, reducing lift by up to 30% (stall speed by 10-15%) and increasing drag by 40%. Skin friction decreases the kinetic energy of the boundary layer.

Answer 84

Affects the local contour of wing leading to severe adverse pressure gradients, as well as increasing skin friction. Stall speed up to 30% larger. Added weight also increases stall speed, but this is a secondary factor.

Answer 85

Similar effect to frost, although can also hide a layer of ice. Snow will not blow off in taxi or takeoff!

Answer 86

Stall AoA will reduce, so stall warner useless. Buffeting may be experienced, but roll control is likely the first clue - increasing roll oscillation or violent wing drop.

Answer 87

Likely to happen before main wing icing as it is thinner, so layer of ice has bigger impact. Provides a down force (negative AoA) to balance moment forces, so effect is likely to be a pitch down. Can be triggered by extending flaps which increases downwash and thus increases the negative AoA of the tailplane. Fix by putting flaps back up.

Answer 88

- Film of water increases weight up to 1-2% - Rougher surface reduces C(Lmax) - Drag increases 5% in light rain, up to 30% in heavy - Downwards impact can also have a big effect

Answer 89

Approaching the stall aircraft attitude is nose up, requiring downforce from tailplane so must be negative AoA. Negative tail stall is when this AoA reaches critical angle (in negative direction), thus losing downforce suddenly and leading to uncontrolled pitch down.

Answer 90

Aerodynamic surfaces at nose of aircraft to provide balancing moment, alternative to tailplane forces.

Answer 91

Canards must be designed to stall before the main wings and tailplane, otherwise they will prevent pitch down recovery from the stall.

Answer 92

A stall must occur before a spin can take place. A spin will happen if one wing is more stalled than the other - leading it to drop and the aircraft to yaw in that direction, eventually an autorotation maintains the spin.

Answer 93

The downgoing wing in a roll has a different airflow and thus a higher angle of attack. This can initiate a spin.

Answer 94

Incipient spin - From the point of stalling until the spin is fully developed Fully developed spin - Starts once angular rotation rate, airspeed and vertical descent speed are stable (same from one turn to the next) Spin recovery - Starts when anti-spin forces overcome pro-spin forces

Answer 95

Higher mass means slower initial spin rate, but as spin progresses it may increase more and will take more time/altitude to recover.

Answer 96

Forward CoG is more stable so less likely to spin, but stall speed is higher (nose down pitch). Aft CoG makes the spin flatter (nose level pitch). Being aft of the limit can lead to flat spin, but at a very high yaw rate, which increases rate of descent and can render the rudder and elevator ineffective.

Answer 97

Different aircraft may have very different spin characteristics so you must follow spin recovery procedure for the specific aircraft.

Answer 98

i) Throttle to idle ii) Neutralise ailerons iii) Full rudder against spin (turn coordinator, NOT balance ball) iv) Elevator to neutral When rotation stops: v) Neutralise rudder vi) Gradual back pressure on pitch control (too fast can stall, too slow can overspeed)

Answer 99

Turn coordinator needle, NOT balance ball!

Answer 100

LOW! As we are stalled. Vertical speed could be high.

Answer 101

If flying out of balance (too much or too little rudder, balance ball out of centre) one wing can stall suddenly and unexpectedly. If ailerons used to lift stalled wing, instead of stall recovery process, spin can develop.

Answer 102

Stall at load factor greater than 1g, so at higher speed than usual. More violent than 1g stall.

Answer 103

A second stall during recovery due to failure to decrease AoA sufficiently, or trying to regain altitude too quickly.

Answer 104

Compressibility of air. High speed jet at high altitude cruise will be marginally above critical mach with small shockwave on the wing. If it overspeeds this shockwave grows creating high static pressure and boundary layer aft of that point separating. The turbulent air will engulf the tail leading to severe airframe buffeting which can damage the aircraft.

Answer 105

Supersonic air SLOWING DOWN to become subsonic, creates a normal shockwave.

Answer 106

Flaps and leading edge devices for increasing lift at low speed

Answer 107

Flaps will increase C(L) at a given AoA. They will increase C(Lmax). They will reduce AoA at C(Lmax) however, mostly because the original wing chord is used to reference AoA, but the new aerofoil shape (leading edge to tip of flap) is much steeper.

Answer 108

Plain: High drag for low speed aircraft Split: Higher C(Lmax) than plain, but higher drag Slotted: Higher C(Lmax) and less drag than plain/split, but complex (only type that increases critical angle) Fowler: Highest C(Lmax) and least drag (due low thickness to chord), but big pitching moment change due to more wing further aft

Answer 109

CoP moves aft (especially with fowler type) increasing the natural nose down pitching moment forces.

Answer 110

Tailplane effective AoA is determined by downwash from wings. Lowering flaps will increase downwash, decreasing AoA of tailplane, reducing lift and leading to a nose up pitch effect.

Answer 111

To increase camber on high speed aerofoils

Answer 112

Often used on inboard section to promote root stall on swept wings, as they are less efficient than variable camber devices.

Answer 113

Increase angle of attack due to shape of wing around the chord (opposite to trailing edge flaps). Thus C(L) is increased at all AoA, max AoA is increased and so C(Lmax) is significantly increased at the higher max AoA. Increase critical angle unlike most types of flap.

Answer 114

Allowing higher kinetic energy air over the wing delays separation of boundary layer, increasing critical AoA. C(L) profile at low AoAs below the normal critical angle is unchanged, but C(L) can keep going up by increasing AoA up to new critical angle. Other leading edge devices increase the C(L) line in general because of change in camber, but slots maintain same aerofoil with a bit more energy to delay separation.

Answer 115

Insignificant. Pressure profile over the wing changes (flattens, less peak suckage at front) but not in such a way as to change pitching moment.

Answer 116

On smaller aircraft, slats are allowed to be deployed by the suction pressure as the stagnation point moves up the leading edge, thus delaying the stall.

Answer 117

High drag Requirement for high AoA to achieve a high C(L), making landing visibility worse

Answer 118

Generally deploy the leading edge device first, to increase acceptable AoA. Deploying flaps first could increase upwash and effective AoA, forcing a stall. Likewise flaps should be retracted first.

Answer 119

Leading edge devices are less affected by asymmetry, not likely to get a sudden uncontrollable roll. Direct effect is instead a yawing effect. However at high angles of attack (e.g. take off or landing) the effect could be bigger as one wing might have lift and the other not.

Answer 120

Positive static stability: Will return to previous stable position if displaced Neutral: Will stay in the new position Negative: Will move further away from the stable position.

Answer 121

The CoP moves with angle of attack, whereas the Aerodynamic Centre (AC) is defined by staying in the same position (as AoA increases, CoP moves forward and lift increases, balancing the moment about the AC).

Answer 122

Obviously unstable. AC is forward of CoG, a gust will increase AoA, increasing lift, creating moment force at AC around CoG. Distance between AC and CoG determines size of moment force.

Answer 123

Tailplane generates positive stability. Gust will increase tailplane AoA and as tailplane AC is aft of CoG, a pitch down moment force is created. This is designed to be greater than the pitch up moment of main wing. Force will be lower, but distance from CoG greater, so moment can be greater.

Answer 124

If CoG is gradually moved backwards, there is a point where the moment force from main wing will equal moment force from tailplane in displacement. This CoG position is the neutral point, where the aircraft will have neutral stability.

Answer 125

The distance between the current CoG position and the neutral point. The minimum static stability margin will determine how far back the CoG is allowed to be.

Answer 126

Increased stability reduces control. If stability is negative, craft is impossible to control as every control movement then needs to be counteracted with an opposing one, like keeping a ball balanced on top of an upturned bowl! [NOTE: Forward CoG => more stability, less control => bigger elevator movements required]

Answer 127

For minimum control speeds, which relate to OEI condition and ability of rudder, the moment arm of the rudder is most important. So minimum control speeds determined at aft CoG limit. In general conditions however, aft CoG gives less aircraft stability and more controllability.

Answer 128

## Footnote Moment, Lateral, Normal

Answer 129

Positive coefficient of pitching moment is nose up, negative coefficient of pitching moment is nose down. Think about it as relating to the nose (+ve) pushed the nose up.

Answer 130

## Footnote Negative gradient means positive stability. Reduction in alpha moves us up the line to give a positive C(M), lifting the nose and reducing the alpha back down. In reality won't have a perfect straight line, could be a curve with negative stability at high alpha.

Answer 131

As AC is forward of CoG, the wing always contributes an unstable pitching moment. Moment distance (AC to CoG) doesn't change, so C(M) increases linearly with C(L). This positive slope leads to instability.

Answer 132

All symmetrical bodies at an angle of attack in an airflow create unstable moments without creating any lift. REAR FUSELAGE mounted nacelles have a POSITIVE effect.

Answer 133

The difference between angle of incidence of tailplane vs wing. Lower angle of incidence of tailplane means that a gust increasing AoA will have a different % impact on tailplane than wing, thus increasing effectiveness of tailplane stability vs wing instability. Also, we generally want in cruise a 4 deg AoA of wing and neutral tailplane.

Answer 134

Increase in AoA will increase downwash from the main wing which then reduces the effective AoA at the tailplane. Thus an increase in AoA is fully experienced by the main wing, but partially offset by downwash at the tailplane. This reduces stability effectiveness of tailplane.

Answer 135

CoG moving backwards primarily impacts moment force from the wing as the force is strong and moment arm is short there. The instability moment from wing increases as CoG moves rearwards, shifting the C(M) vs C(L) curve upwards, eventually reaching a flat "neutral point", then becoming positive gradient (i.e unstable).

Answer 136

Effect of prop slipstream over tailplane can be stabilising, other effects are destabilising: - Moment force of jet engines under wing - Moment force of prop engines - Slipstream over main wings (prop) - Downwash due to main wing slipstream - Jet stream near tailplane

Answer 137

Can be some stabilising effect on the main wing, but predominant effect is instability. This is due to increased downwash and reduced dynamic pressure at tail.

Answer 138

Elevator position doesn't affect stability (i.e. gradient of C(M) vs C(L)), but obviously impacts the point of trim, i.e. level of C(L) where C(M) = 0. Thus C(M) vs C(L) lines for different elevator deflections are parallel, but intersect C(M) = 0 at different points.

Answer 139

This characteristic means that the stick needs to be moved aft to increase AoA and trim at lower speed, and forward to decrease AoA and trim at higher speed. This is achieved if the CoG is forward of the neutral point.

Answer 140

When pitching the aircraft (e.g. up) the aircraft rotates around the CoG and the tailplane experiences a pitching velocity (downwards), meaning relative airflow is coming from a lower angle during the manoeuvre. This increases tailplane AoA and thus stability.

Answer 141

At low TAS manoeuvre stability effect will be greater, as the pitching velocity component of relative airflow velocity is relatively stronger.

Answer 142

As the CoG is moved aft, the manoeuvre stability effect will have a bigger impact on the wing and less on the tailplane. Thus there is a neutral point at which manoeuvre stability is neutral. This is usually aft of the neutral point however, so not of concern.

Answer 143

Naturally, high manoeuvring stability requires more stick force for a given load factor. CoG may be given a forward limit based on limiting the required stick force to increase load factor (i.e. pitch up).

Answer 144

At higher altitude, TAS increases to maintain steady flight (1g). Manoeuvring stability reduces with high TAS therefore stick force required for increase in load factor is reduced.

Answer 145

Is a force per "g", measuring in N/g. e.g. force required to create a 2.5g turn or climb. Often depicted in a linear graph. NOTE: Origin is 0 force, 1g, not 0g! NO stick force required at 1g, 150N/g implies 150N for a 2g turn!

Answer 146

This relates to mach tuck. Stick force stability means the stick doing what it should (back to slow down, forward to speed up). Mach tuck reverses this relationship so results in a "decrease in stick force stability".

Answer 147

- Stick centring spring - Down spring - Bobweight

Answer 148

Spring forces act to centre the stick, so force depends on amount of deflection. Contribution the greatest at low flight speeds where larger control deflections are required.

Answer 149

These work similarly, pulling the stick forward only (with a spring or a weight). Trimming will create a control force to offset the spring/weight and thus an increase in stick force regardless of position.

Answer 150

As the bobweight uses a weight to generate force, it will have a bigger effect during high g manoeuvres, thus creating more control resistance at those times and restricting the pilot.

Answer 151

Tailplane needs to be able to rotate the aircraft to take off attitude, with main gear on the floor. This requires it to offset 2 moment forces: - Resistance of main gear rolling on runway - CoG ahead of main gear On prop plane, slipstream over tailplane may help.

Answer 152

With flaps down, power off, CoG at most forward position, aircraft is at its most stable. Having sufficient elevator control in this configuration is a key limiting factor. Ground effect has an additional impact, reducing induced drag and downwash and so increasing AoA of tailplane, requiring elevator to be capable of a high angle to maintain pitch down.

Answer 153

Relates primarily to positive static stability where recovery from a disturbance is oscillatory. If oscillations are damped (gradually reduce), dynamic stability is positive. If they stay the same it is neutral, if they increase it is negative. In non-oscillatory behaviour, dynamic stability is same as static.

Answer 154

Subsidence (or dead beat return) describes static stability without oscillation, i.e. recover directly to the previous position without going past it. Divergence describes negative static stability, i.e. keep moving away from the previous position.

Answer 155

Long Period Oscillation Oscillations over a period of around 1-2 mins in pitch attitude, altitude and airspeed (with constant AoA). DO reduce to zero though so reflect positive static and positive dynamic stability. However because of longer periods do get significant variance in altitude (unlike short period oscillation).

Answer 156

This involves significant changes in AoA and thus load factor, whilst speed, height and pitch are constant. Oscillations are rapid, about 1-2 secs. This is dangerous as pilot input lag may lead to exacerbating the effect. Aircraft must demonstrate high level of damping of this effect, so take hands off controls and let is recover.

Answer 157

Dynamic stability is reduced at altitude due to reduced aerodynamic damping

Answer 158

Pitch: C(M) - positive when nose up Roll: C(L) - positive when rolling right Yaw: C(N) - positive when nose right

Answer 159

Sideslip angle designated as beta, it is angle between longitudinal axis and relative airflow in horizontal plane. Relative airflow coming from the right is described as a sideslip to the right and sideslip angle is +VE. [Note yaw angle is the opposite of sideslip angle]

Answer 160

The initial tendency of the aircraft to return to its equilibrium angle of sideslip (typically zero). Thus strong directional stability implies difficulty maintaining sideslip.

Answer 161

We require a positive relationship between beta and the C(N). e.g. Sideslip to the right => beta is +ve If C(N) is +ve a moment force yawing to the right will correct the sideslip

Answer 162

Fuselage is naturally unstable. AC is forward of its CoG so tends to get blown further around once turned.

Answer 163

Dorsol or ventral fins (back/belly) They are small and low aspect ratio (long and thin) so generate minimal "lift" at low angles, but create increasing stabilising effect at larger angles of yaw.

Answer 164

Dorsal fins generate positive lateral stability whilst ventral contribute negative lateral stability, so dorsal more common

Answer 165

Creates significant directional stability moment due to large moment arm from CoG. Large aspect ratio (relative to dorsal fin) increases impact at low angles of yaw (i.e. AoA) but the fin will stall before dorsal fin does, when dorsal fin will become more important.

Answer 166

Straight wing is negligible Nacelles usually destabilising, but depends. Swept back wing has a small positive stability effect due to higher lift on forward wing having more induced drag.

Answer 167

Generally destabilising. Higher for prop craft due to slipstream effect, negligible for jets. Biggest impact at high power, low speed.

Answer 168

Minimal (CoG limits for longitudinal stability are greater than any directional stability concerns).

Answer 169

High AoA blocks airflow over the fin so reduces directional stability. Ventral fins can combat this, but landing clearance can limit their size.

Answer 170

Positive rolling is taken to be when the right wing moves downwards

Answer 171

## Footnote +ve C(L) is roll to the right, +ve sideslip angle is to the right (airflow from the right). Chart shows that left sideslip gives correcting roll to the right.

Answer 172

If beta is +ve this means relative airflow is coming from the right. We want to roll to the left to recover therefore a -ve C(l) is required. Thus lateral stability requires a negative relationship between C(l) and beta.

Answer 173

Overly strong lateral stability can lead to control issues in crosswind landing and oscillatory coupling with directional stability, so we desire a mild level of lateral stability.

Answer 174

Creates high AoA of wing heading into sideslip and therefore a lifting moment on that side. Positive lateral stability.

Answer 175

Collective description for lateral stability contribution of wing position, flaps, power (etc.), based on the fact that dihedral is the overriding factor.

Answer 176

Sideslipping air is directed around the fuselage from the side, going below it at the bottom and above it at the top. This redirection creates lower AoA for sideslip side wing on a low wing aircraft, higher AoA on a high wing aircraft. So positive effect for high wing, negative stability effect for low wing.

Answer 177

Sweepback - more lift on wing facing airstream straight on (also more drag so directional stability too) Fin - small effect due to airflow resistance Ventral fin - negative effect Power - prop stream over inner area of wing reduces dihedral effect Flaps - Extra lift on inner area of wing reduces dihedral effect

Answer 178

Caused by high directional stability relative to dihedral effect. Gentle effect where sideslip causes a yaw into the turn, limited correction of the roll and gentle spiral starts. Eventually leads to a spiral dive. This is easy to identify and resolve.

Answer 179

Caused by large dihedral effect relative to directional stability. After yaw is introduced, aircraft will roll in same direction due to dihedral effect. Increased induced drag on the rising wing causes reverse yaw and thus oscillations between yawing and rolling.

Answer 180

Risk of pilot induced oscillation (PIO) if using the rudder to correct. Use ailerons instead. Aircraft will have yaw damping to reduce the effect.

Answer 181

Dutch roll more likely at high speed and high altitude (consider reducing speed/altitude to recover/prevent).

Answer 182

Oscillation of movement due to pilot attempting to correct short period motion, which can very quickly reach dangerous proportions. Solution is letting go of the controls.

Answer 183

High TAS reduces angle of attack changes and thus also aerodynamic damping. At ISA altitude of 40,000 ft, aerodynamic damping is half amount at sea level. Stability in general reduces with IAS being low relative to TAS, thus failure of dutch roll filter (for example) will lead to a lower operating height.

Answer 184

Or high speed tuck, or tuck under. Shock wave on swept back wing reduces lift forward of CoG and reduces downwash over tailplane. This creates a pitch down moment and aircraft becomes unstable in terms of speed.

Answer 185

A system to combat mach tuck, by deflecting elevators, moving CoG aft (fuel tanks) or adjusting tailplane incidence.

Answer 186

Hinge point is inside the control surface (rather than at the very end), reducing moment arm between aerodynamic force and hinge point.

Answer 187

If the hinge point were set far enough into the control surface that the aerodynamic force could move to the wrong side of it, reverse control movement would be required which would be very dangerous. Design needs to ensure this can't happen.

Answer 188

Sticks out on opposite side to main control surface to provide balancing aerodynamic force.

Answer 189

## Footnote Movement of control surface adjusts pressure to assist with control force

Answer 190

Small tab moving opposite direction to control surface at it's rear, acts to maintain the deflection of the control surface, reducing stick force required.

Answer 191

Moves in the same direction of control surface to magnify stick forces.

Answer 192

Similar to balance tab but pilot controls the tab directly, which creates the control surface deflection. WARNING: Need to be sure control locks are removed, can only check free movement of servo tab itself from the cockpit.

Answer 193

Modification of servo tab where a spring means maximum assistance is provided at high speed when stick forces are greatest.

Answer 194

A tab at edge of a control surface that doesn't move relative to the control surface when the control surface itself moves. Either set on the ground based on test flights, or adjustable via trim adjustments in the cockpit.

Answer 195

Trim tabs (and servo and spring tabs) can give some level of control, in the opposite sense to normal.

Answer 196

Will be slightly ahead of the nose. Compressibility causes kinetic heating of compressed air ahead of the nose. Higher temp means higher LSS. Further ahead of the nose, in the cooler air, LSS reduces and the shockwave forms.

Answer 197

Mach number at which the shockwave attaches (DET!) back to the aircraft as speed increases and offsets kinetic/compressibility effect. Will be around 1.3 x M(FS)

Answer 198

The mach number at which local flow becomes sonic anywhere on the aircraft (i.e. over accelerated surfaces such as cambered wing).

Answer 199

Angle of attack increases acceleration of air flow over surfaces, reducing speed at which air becomes sonic. Thus M(CRIT) decreases with AoA. Mass will require an increase in lift and AoA therefore higher mass reduces M(CRIT). A forward CG also increases lift required, so lower M(CRIT).

Answer 200

The transonic range is between M(CRIT) and M(DET) [NOT M(FS)]. This is the range in which we fly, where air over some parts of the aircraft are supersonic. Defined as the range where there will be subsonic, sonic and supersonic air on different parts of the aircraft.

Answer 201

All increase

Answer 202

Compressibility increases air density ahead of leading edge and thus speed over the wing. This increases C(L) initially. As shockwaves build up on the back of the wing however they will eventually cause separation and a "shock stall" causing C(L) to fall. NOT connected to AoA, shock stall point is the point of maximal C(L) in mach, for a given AoA.

Answer 203

Shockwaves start to occur on the bottom of the wing, but as speed increases both sets move backwards towards the trailing edge and lift stabilises. Trailing edge devices (ailerons) are less powerful, which can be a problem. At Mach 1.0 CP moves back to 50% and stabilises (has been moving a lot up to now), which can cause a big pitch down.

Answer 204

Just above mach 1.0

Answer 205

Shockwaves cause "wave drag", composed of 2 parts: i) Energy drag (energy lost to temperature increase through shockwaves) ii) Separation drag (separation of boundary layer behind shockwaves)

Answer 206

## Footnote Initially increases as compressibility accelerates airflow over wing. Drops once shockwave forms over large part of wing (shock stall). Recovers towards M1.0 as shockwaves move to back of wing.

Answer 207

## Footnote Drag divergence mach number is just above M(CRIT) and is when shockwaves cause significant drag increase.

Answer 208

Flatter upper surface delays shockwave, reducing drag and moving it more quickly to the trailing edge. Larger leading edge radius. Enables flight in transonic region. INCREASES M(CRIT)!

Answer 209

Thin wings have problems with strength and fuel capacity, but reduce the C(L) and drag profiles to stabilise flight in transonic region.

Answer 210

Path of the airflow over the wing is stretched out, so acceleration is reduced and C(L) and drag transonic profiles are smoothed. Drag in transonic range is reduced significantly. M(CRIT) increases. Downside to this is reduced lift so higher required take off speeds, higher drag at supersonic speed and other issues.

Answer 211

Sit on the upper wing surface just ahead of where significant shockwaves will start. They make the airflow behind turbulent, high energy, delaying the stall and pushing an upper shockwave to the back of the wing. This can increase critical AoA, but will also increase drag and gives a high attitude on approach (unlike flaps/slats).

Answer 212

Flight Path Angle + AoA = Pitch Angle

Answer 213

Sin (gamma) = (T - D) / W gamma = flight path angle [Assume D/W = D/L (inverse lift ratio) for questions]

Answer 214

Lift = w x cos (gamma)

Answer 215

Load Factor = 1 / cos (angle of bank) Note: Does not depend on weight, speed, etc.

Answer 216

AoB = 7 deg + TAS / 10 [Note: Bigger TAS => slower rate of turn @ given AoB]

Answer 217

R = TAS^2 / (g x tan(bank angle))

Answer 218

Opposite of climb (reverse thrust and drag, or angle, depending on how you look at it). Sin (gamma) = (D - T) / Weight

Answer 219

Roll in direction of yaw

Answer 220

Adverse yaw Later sideslip in direction of bank

Answer 221

Alternative to ailerons which don't work well for large jets. Spoiler causes separation on upper side of wing, causing flow separation and reducing lift. This causes a turn in the direction of that wing.

Answer 222

When rolling with aileron, have higher lift on upgoing wing, less on downgoing. Thus higher induced drag on outside of turn than inside, creating a yaw in opposite direction. Can resolve with: - Differential aileron (more deflection on upgoing aileron = downgoing wing) - Frise type ailerons

Answer 223

When roll is initiated, the upgoing wing will have lower AoA than the downgoing, creating a force that resists the turn. At a given displacement of the control column the turn will settle at a given rate of turn when the forces balance. This is a form of aerodynamic damping.

Answer 224

A stabiliser that is used as the pitch control surface. No separate elevator, but can have a trim tab. e.g. PA28

Answer 225

This is when the stabiliser is moved as a trimming mechanism. There is an elevator which is used for fine control and pilot input, whilst trimming adjusts the entire stabiliser and allows the elevator to move to a neutral position.

Answer 226

Turbulence pushes the wing up but because aileron CoG is behind its hinge, it lags as it is "dragged" up. This means it is in down position and creates an upwards force. When the wing rebounds back down the aileron will again lag and create a down force. Can be resolved with mass balances.

Answer 227

Similar to flexural, but instead of the wing moving directly up or down, the movement of the aileron causes it to twist around its torsional axis. Can still be resolved with mass balances.

Answer 228

Aileron movement from pilot creates a torsional movement in the wing (similar to torsional aileron flutter) which opposes the demanded roll. Can resolve by using inboard ailerons at high speed, where the wing is thicker and torsional force can be resisted.

Answer 229

This is movement solely in the wing without aileron movement, caused by CoG of the wing being different to torsional axis. Solution is mass balancing, ensuring that mass is appropriately positioned relative to the torsional axis.

Answer 230

Hard prevent control movements beyond limits. Soft might display a warning or an aural alert, but not physically prevent the control input.

Answer 231

Can have trim tabs on partially powered controls, but no point on fully powered (irreversible) as no pilot strength is required to move control surfaces.

Answer 232

Pitch or blade angle is the angle between the chorld line of the prop blade and the plane of rotation. Helix angle is the angle between the relative airflow over the blade and the plane of rotation. The difference is the angle of attack of the blade.

Answer 233

Pitch is the "distance" a blade would move through in one rotation, like a screw. Related to the propeller angles but a different way of viewing it. Geometric pitch is equivalent to blade angle, distance covered if the blade was screwed into something solid. Effective pitch is equivalent to helix angle is is the distance the blade moves forward through airflow. Slip is the difference (related to alpha).

Answer 234

Point 3/4 of the way along the blade where the pitch is measured as the representative pitch of the propeller.

Answer 235

Power is zero at zero TAS. Increases with TAS up to the peak point where alpha is 4 degrees, which is the most efficient point. Increasing speed beyond this point power reduces as the blade is inefficient and thrust falls.

Answer 236

Selecting a lower rpm causes the CSU to coarsen the blades (which increases drag to slow the propeller). The reduced rotational speed also affects the relative airflow however, reducing alpha, therefore further coarsening is required to maintain the rpm.

Answer 237

In a high speed dive, or in glide with zero power, forces on the propeller blade are a large component of drag and a torque force in the same direction as engine torque. This is maximised at fine pitch (high rpm) but coarsening will reduce the drag effect. So fine pitch is good for restarting engine, coarse for reducing drag (e.g. if engine is dead or planned high speed dive).

Answer 238

Loss of torque on failed engine leads to a fine pitch, therefore high drag which can create dangerous yaw on multi-engine. Feathering involves coarsening the blades to about 85 degrees, getting rid of the drag and windmill effect.

Answer 239

Especially on tailwheel. Get yaw to the left (for RH prop) when tail lifts up along runway (propeller circle tilted forwards precessed). Get yaw to the right on rotation [the only right yaw in all prop effects!].

Answer 240

Typically the prop circle faces upwards relative to direction of travel, so blade travels further when going down than when going up. Thus on right hand propeller, the right side of prop direction will produce more thrust than left side, creating asymmetrical thrust (P-factor).

Answer 241

Yaw to the left for right hand propeller due to slipstream effect throwing airflow from left to right over the fin.

Answer 242

Torque that attempts to turn aircraft opposite to direction of the prop rotation. Easily countered with aileron, creates a left yaw in RH propeller and increased force on left side wheels. Can be mitigated on multi prop with counter-rotating propellers. [NOT CONTRA-rotating, which are 2 opposite directed props infront of each other]

Answer 243

- Diameter (limited by tip speed approaching transonic range, plus ground clearance) - Disc solidity (more blades = more blade and less air in the disc, max around 5) - Max lift coefficient (camber increases this but increases drag so efficiency is lower) [2 bladed prop with long thin blades is very efficient, but low power absorption]

Answer 244

Most influenced by tip speed, so reduced rpm and reduced diameter help (but lower diameter affects power absorption). Decreasing number of blades does NOT reduce noise, as it requires longer blades for the same power, thus faster tip speeds.

Answer 245

When at high alpha. This is when air flow over the rudder and other surfaces is lowest so less torque force around to counter the yaw effect.

Answer 246

Thrust lost on failed engine and offset by P-factor on functioning engine (to the right for clockwise rotation) => Yaw to failed engine. Drag increases on failed engine (especially if not feathered) => Yaw to failed engine. Slipstream effect from prop creates a lot of lift, so lift reduces on failed side => Roll to failed engine.

Answer 247

The rudder creates a sideways force in order to create a moment to offset the moment from the single operational engine. There is no balance to this sideways force however so aircraft has a tendency to move sideways towards the failed engine. This means airflow approaches from that side and craft is in a sideslip.

Answer 248

i) Airflow over the fin creates an additional yawing effect towards the failed engine, requiring even greater rudder force to offset moment. ii) Fuselage blocks airflow over a portion of the wing on the operational engine side, reducing total lift.

Answer 249

"Five to the live" 5 degrees of bank towards the live engine will create a lateral component to lift (sideways, not yawing) which offsets part (or all) of the lateral force causing sideslip. This acts at or close to the CoG so no yaw effect.

Answer 250

With wings level and rudder used to get balanced flight, the balance ball will be central as forces are balanced. Using roll to eliminate sideslip will cause it to move towards live side (think gravity) as the aircraft isn't balanced.

Answer 251

Increase in mass requires increased speeds and thus asymmetric inbalanced forced are smaller. Biggest problems are with less mass and less speed.

Answer 252

The bank uses lift to create a lateral force. This reduces lift, requiring even higher alpha and higher drag. Given the already high drag, doing this too much is a risk. Risk of FIN STALL.

Answer 253

Higher temperature and higher altitude reduce maximum potential thrust, which allows lower minimum control speeds. This is because control speeds are based on the assumption that remaining engines operate at full thrust.

Answer 254

Minimum control speed on the ground. Requires control of aircraft (rudder only, not nosewheel steering) within 30ft (9.1m) laterally at most.

Answer 255

i) In each TO configuration (or most critical one) ii) Maximum thrust on other engines iii) Most unfavourable CoG iv) Most unfavourable mass v) Trim for TO [WIND IGNORED!]

Answer 256

CAS at which aircraft can be controlled in S&L flight with AoB no more than 5 degrees. Limited to 1.13 x V(SR) based on a set of assumptions.

Answer 257

1.13 x V(SR) assuming: i) Most critical TO configuration along flight path after becoming airborne, but with landing gear retracted ii) Maximum thrust on other engines iii) Most unfavourable CoG iv) Maximum sea level TOW v) Trim for TO vi) Ground effect negligible vii) Inactive prop windmilling, unless automatic feathering device

Answer 258

Only recovery is to reduce thrust on operating engine(s) and recover speed by pitching down. Obviously this is not a good position to be in, so great care must be taken to avoid speed reducing to this point.

Answer 259

V(MCG) is based on specific conditions (CoG position, altitude, temperature) based on a specific take off. V(MC) isn't recalculated for specific conditions, so in favourable conditions (forward CoG, high altitude, high mass) control may actually be possible below V(MC).

Answer 260

Minimum speed at which aircraft can be controlled in case of failure of critical engine during approach and landing. Need to be capable of 20 degree turn towards the live engine within 5 seconds.

Answer 261

i) Most critical configuration (or each configuration) ii) GA thrust on other engines iii) Most unfavourable CoG iv) Most unfavourable mass v) Trim for approach vi) Prop on inactive engine at position it achieves without pilot intervention

Answer 262

CS23: - V(NE): Structural never exceed speed - V(NO): Normal operating speed, gives a margin to ensure that V(NE) is not exceeded due to turbulence/windshear CS25: - M(MO): Maximum mach number - V(MO): Calculated IAS that varies with altitude to ensure M(MO) not exceeded

Answer 263

V(MO) <= V(C)

Answer 264

Stall speed at limit load factor (+2.5g or -1.0g for CS25). Calculate as V(A) = V(S) x sqrt(n) where n = 2.5g [Stall speed varies with sqrt of load factor] [Relates to maximum elevator - below V(A) max elevator stalls, above it exceeds 2.5g]

Answer 265

CS25: 2.5g Light: 3.8g Utility: 4.4g Aerobatic: 6.0g

Answer 266

V(D) - Design dive speed, absolute speed limit the aircraft is capable of sustaining V(C) - Design cruise speed, aircraft tested in level flight up to this speed, then put into manoeuvre to reach V(D). V(D) >= 1.25 V(C) These are limits for testing, they feed into the limits relevant to pilots.

Answer 267

## Footnote Maximum speed flaps extended. G limit reduced to 2.0.

Answer 268

Mach numbers can start to limit speeds and high load factors not possible, so the envelope starts to shrink. This includes increasing stall speed.

Answer 269

## Footnote Note: Gust lines originate from V=0, N=1. Shape symmetrical above V(B).

Answer 270

Design speed for maximum gust intensity

Answer 271

Maximum speed for turbulence penetration. Somewhere between V(A) and V(C). Must be 35kt below V(MO) if speed is NOT limited by M(MO). More complex calculation once speed is M(MO) limited as there are two buffet boundaries.

Answer 272

Change in load factor as a result of a gust (vertical) changing AoA of the wing. Higher alpha will reduce gust load factor as the change in AoA is proportionately less as a percentage.

Answer 273

Lower speed means higher AoA so lower GLF. Higher mass requires higher alpha. High wing loading requires higher alpha. Low wing area relates to high wing loading. High aspect ratio wings produce higher lift at lower AoA, so high aspect ratio gives low alpha and higher GLF. Swept wing is low aspect ratio High altitude increases TAS so reduces effective impact of the same speed gust (vector diagram).

Answer 274

Service ceiling: 100ft/min capable Absolute ceiling: 0ft/min capable

Answer 275

e.g. spoiler Increase drag and reduce lift

Answer 276

V(B): 66 ft/sec V(C): 50 ft/sec V(D): 25 ft/sec Reducing as high velocity reduces acceptable gust speed. Note: 66 ft/sec covers velocity from V(B) up to V(C) (and so on).

Answer 277

LSS = 38.95 sqrt(T) MN = TAS / LSS IAS = 1/2 rho TAS^2 So for constant IAS, temperature doesn't affect MN (and vice versa). However if TAS is constant, temperature will change MN (and vice versa).

Answer 278

Considered the stall limit on jet transport aircraft. Sufficient buffet that flight crew would be deterred from further AoA. There are lesser levels of initial buffet which a pilot might push through.

Answer 279

More effective on straight wing, simply as straight wings are more effective at lift than swept wings.

Answer 280

Drag parallel to RELATIVE AIRFLOW. Lift perpendicular to RELATIVE AIRFLOW.

Answer 281

Reduce interference drag. Relates to "area rule".

Answer 282

Half of the cone angle, supersonic cone you fly into. Sin (mach angle) = 1 / MN Closes in as you fly faster [Unsure if this is in exams]

Answer 283

Angle of relative airflow is flatter for the faster outside wing. So it has a higher AoA when climbing, lower when descending.

Answer 284

Contra: Weird ones on the same shaft Counter: Opposite direction normal props

Answer 285

Lift curve refers to lift vs AoA. If it is steep then a small change in AoA due to a gust will have a big impact on lift. Thus gust factor is BIGGER for a STEEPER lift curve.