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AVDASI 1 > Aeronautics > Flashcards

Aeronautics Flashcards

(45 cards)

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1
Q

How are ailerons deflected?

to roll left, and which deflection direction is positive

A
  • to roll left, left aileron deflects up, right deflects down (differentially)
  • this increases lift on right side, and decreases lift on left, causing roll to the left
  • deflection down = positive
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2
Q

How are elevators deflected

A
  • deflection down = positive
  • both sides of the elevator deflect together
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3
Q

How is the rudder deflected

A
  • deflection to the left = positive
  • to the left means rear of plane swings right
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4
Q

Directions of the three control surface effects

elevator, rudder, aileron directions

A
  • elevator controls pitch M (+ve nose up)
  • rudder controls yaw N (+ve nose pointing right)
  • aileron controls roll L (+ve right wing down)
  • this means positive control deflections give negative moments

e.g. signs are opposite to the deflection directions

positive control input for elevator = elevator deflects down, which creates more lift on tail, which pushes tail up and nose down, hence a negative pitching moment

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5
Q

Secondary effects of the control surfaces

A
  • elevator: changes height AND speed, not just pitch angle
  • rudder: also generates roll due to sideslip (flying at an angle to oncomoing flow)
  • aileron: also generates yaw due to roll rate and differential aileron drag (drag on two ailerons is not the same, which induces a yaw moment)
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6
Q

Trapezoidal wing parameters

winger taper ratio, standard mean chord, aspect ratio, MAC

A
  • wing taper ratio, λ = tip chord / root chord
  • standard mean chord, SMC, c bar = S / b (total planform area / span)
  • aspect ratio, AR = b / c bar = b^2 / S (AR defines the “thinness” of the wing)
  • mean aerodynamic chord MAC, c bar bar = 1/S ∫c^2 * y dy from -b/2 to b/2 (driven by how the aerodynamic properties of 3D wings work)
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7
Q

Explanation of lift

A
  • fluid forces arise from the distribution of pressure and shear stresses
  • N3L: for wing to generate lift, it must force air down i.e. lift can be thought of as a reaction force
  • lift is proportional to the amount of air diverted per second, and change in vertical velocity of air
  • air diverted per second is porportional to speed of wing and air density, vertical velocity is proportional to speed of wing and angle of attack
  • hence lift is proprtional to the speed squared, density and angle of attack
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8
Q

Equation for coefficient of lift

2 ways

A

C_l=L/(1/2 ρv^2 S)

OR

C_l=a(α-α_0)
where alpha0 is the zero lift angle of attack and a is a constant (gradient of Cl-alpha curve)
a = 2pi / rad for a thin 2D airfoil in an ideal fluid

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9
Q

What must you be at to fly at minimum speed?

When straight and level

A

C_lmax:
V_min= √(W/(1/2 ρSC_lmax )) (at a specific AoA)

hence for takeoff and landing, we need greater C_lmax, by increasing camber using flaps

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10
Q

Different types of drag?

A

Two main: Skin friction drag and normal pressure drag

pressure drag consists of wave drag (supersonic), induced drag (due to wings being 3D and finite, not infinite 2D) and form drag

profile drag is the overal “2D” drag - combination of skin friction drag and form drag

generally low form drag means higher skin friction drag due to the shape of airfoil, or vice versa

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11
Q

What is induced drag?

A

A drag that acts on 3D wings, proportional to lift^2

driven by the pressure difference between upper and lower surface on the wing (i.e. lift), as it creates vortices when airlflow “rolls up” around the wing tip

the tip vorties induce a downwash, which is a downwards flow component of air. This as a result, with V_infinity being horizontal, rotatess the velocity vector downwards by a small angle epsilon, which effectively reduces AoA and so reduces lift

lift is perpendicular to velocity vector, so lift force now has a horizontal component, which is the induced drag

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12
Q

Equations for induced drag coefficient C_D_i

A

ε= C_L/(π∙AR)

C_(D_i )= (C_L^2)/(π∙e∙AR)
where e is the spanwise efficiency factor

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13
Q

What exactly is e?

A

There is a distribution of wing loading that gives the minimal amount of induced drag, called elliptical loading, which gives an efficiency factor of 1

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14
Q

Total drag equation

A

C_D=C_(D_0 )+KC_L^2
where C_(D_0) is profile drag (i.e. 2D drag or zero-lift drag) and KC_L^2 is induced drag

where K = 1//(π∙e∙AR)

also can be written as
C_D=C_(D_min )+K〖(C_L-C_(L_0 ))〗^2

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15
Q

What is the critical Mach number, and drag divergence Mach number

A

M_crit lowest freestream Mach number at which any part of the airfoil reacts to the speed of sound

M_dd is the freestream Mach number where C_D begins to rapidly increase due to the wave drag caused by shockwaves

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16
Q

How do swept wings affect M_dd?

A

swept wings can delay M_dd, because the velocity now hits the wing at an angle equal to the sweep angle, hence velocity perpendicular to the wing (which is the only velocity the airfoil ‘sees’) is lower

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17
Q

what is pitching moment and centre of pressure

A

Lift and drag can be combined into a single aerodynamic fore, R. The line of action of R crosses the chord line at the centre of pressure, x_CP, about which there is no pitching moment. but CP is not fixed, and it changes with C_L and hence AoA

Hence we need a point where C_M does not vary with C_L, called the aerodynamic centre

Aerodynamic centre is at the quarter-chord point for 2D thin airfoils

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18
Q

Different regions of the atmosphere

A

Troposphere (0-11km)
Stratosphere (11-48km)
Mesosphere (48-80km)
Thermosphere (80km above)

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19
Q

Assumptions of the international standard atmosphere and ISA sea level values

Study These Flashcards
A
  • air is a perfect gas
  • air is dry
  • g does not vary with altitude
  • hydrostatic equilibrium exists

pressure: 1.01325e5 Pa
temperature: 288.15K
density: 1.225 kg/m3
speed of sound: 340.29 m/s

20
Q

rate of temperature decrease in the troposphere

Study These Flashcards
A

rate = -6.5 degrees celsius / km
at the tropopause (11km), T = -56 degrees C
the lower part of the stratosphere (11-20km) is isothermal so also = -56

21
Q

Equivalent airspeed

Study These Flashcards
A

The speed if flying at standard sea level density

V_E=V√(ρ/ρ_0 )=V√σ

22
Q

Minimum drag coefficient at minimum drag, lift coefficient at minimum drag, and velocity at minimum drag

Study These Flashcards
A

C_(D_0) = K (C_L)^2 at minimum drag
therefore C_D_min = 2C(D_0) = 2K(C_L)^2

This means C_(L_MD) = √(C_(D_0) / K)

and subbing in to find V_MD:
V_MD=〖(2W/ρS)〗^(1/2)∙〖(K/C_(D_0 ) )〗^(1/4)

23
Q

Power for straight and level flight

Study These Flashcards
A

P = Thrust x Velocity = Drag x Velocity at S&L flight

therefore find total drag at straight and level (CD0 + KCL^2) and multiply by V

24
Q

Effect of flying at higher altitude on required power

Study These Flashcards
A

flying at higher altitude will require more power
but at higher velocities the power required actually reduces

25
Minimum power required equations
power is proportional to Cd/Cl^3/2 hence for drag, K(C_L)^2 = 3C(D_0) hence C_(D_MP) = 4C(D_0) C_(L_MP) = √(3C(D_0) / K) sub this in to velocity eqn to find V_MP
26
Forces in gliding, glide angle, when does minimum glide angle occur velocity at any glide angle θ
Thrust = 0, but drag != 0 L = Wcosθ D = Wsinθ glide angle θ = D/L = C_d / C_L best glid angle at minimum θ = maximum L/D maximum L/D ==> min glide angle at velocity of min drag V_MD velocity = √(Wcosθ/(1/2 ρSC_L )) = steady level flight velocity * sqrt(cosθ)
27
Velocity at min glide angle, for small angles
For small angles, cosθ = 1, sinθ = 0 (for θ < 10 degrees) min glide angle requires max L/D which requires min drag velocity therefore velocity at min glide = velocity at min drag
28
sink rate and velocity at min sink rate
sinking speed v = Vsinθ from force balance and small angle such that L approx = W: and sinθ = D/W which approx equals D/L = C_d / C_l, hence v = V * C_d/C_l therefore v prop to C_d / C_l^3/2 therefore V_min_sink = V_MP then v = V_min_sink * C_d / C_l glider performance v/V = sinθ = D/L
29
Force balance in climb + climb angle max climb angle and min rate of climb
Thrust T > Drag D so aircraft ascends at climb angle θ horizontal component of weight opposes thrust L = Wcosθ, T-D = Wsinθ rate of climb υ = V(T-D)/W = Vsinθ hence rate of increase of potential energy (power) Vυ = TV - DV Max climb angle θ_max requires maximum excess thrust (T-D), since sinθ = (T-D)/W Max rate of climb υ_max requires maximum excess power (TV-DV), since υ = V(T-D)/V
30
Jet aircraft vs properller aircraft graphs of thrust and power against true airspeed
JET THRUST: constant thrust for any airspeed JET POWER: linear increase in power from 0 PROPERLLER THRUST: 1/x curve PROPERLLER POWER: constant power for any airspeed
31
Max climb rate and max climb angle for a JET
Max climb angle θ_max_jet occurs at V_MD Max climb rate υ_max_jet occurs above both MD and MP - it just occurs at max TV-DV
32
Max climb rate and max angle for a PROPERLLER
Max climb angle θ_max_prop occurs below both MP and MD - just at max T-D Max climb rate υ_max_prop occurs at V_MP
33
Max altitude cieling for a jet vs propeller
Point on a power-velocity graph where the max P_available only touches the P_required curve - does not cross at 2 points. The altitude at which this happens is maximum cieling altitude Can also be seen on a drag-equivalent airspeed graph. The point where the (constant) thrust line only touches the drag curve gives you the max cieling This also implies that at the max cieling, aircraft must be flown at V_MD for propeller, look at a Psqrtsigma vs EAS graph. Max cieling must be flown at V_MP
34
Speed stability for aircraft at cruise
In cruise, T = D Flying at high velocity is stable: increase in speed will increase drag, decelerating aircraft flying at low velocity (before MD) is unstable: speed increase reduces drag, which accelerates flying at MD is neutrally stable since it is a turning point: change in speed does not change drag
35
Why can't planes fly at absolute cieling?
stability is unstabel at max ceiling, so a service cieling is a practical alternative. At a service cieling, the aircraft has a specified rate of climb
36
Key assumption and formula for Breguet range
Assumption: rate at which fuel is burnt = rate at which aircraft weight is reduced f = thrust specific fiel consumption (TSFC) = mass of fuel burnt per unit thrust per second dW/dt = -fgT
37
JETS Endurance equation, max endurance Range equation, max range (for cruise climb) Range equation, max range (for constant altitude) Implications of these assumptions
Endurance E = t_2- t_1= 1/fg C_L/C_D ln⁡(W_1/W_2 ) max endurance at max C_L / C_D = at V_MD cruise climb range R = S_2 - S_1 = V/fg C_L/C_D ln⁡(W_1/W_2 ), or can replace V with a*M constant altitude range R = √(8/ρS) 1/fg (C_L^0.5)/C_D (√(W_1 )- √(W_2 )) Max range occurs at max V * C_L / C_D = max C_L^1/2 / C_D for both C_D_0 = 3K(C_L)^2 i.e. at max range for both: C_D = 4K(C_L)^2, C_L = √(C_D0 / 3K) implications of cruise climb: true airspeed and CL/CD are constant so altitude increases. constant throttle setting must be applied implications of constant altitude: σ and C_L are constant, so V_TAS reduces along with throttle. This means range is less than for cruise climb as V is less. Also implies a variation in f, so a constant f must be used, or treat as a series of shorter steps
38
PROPELLERS Endurance equation, max endurance Range equation, max range
E_prop = η/fg (1/V * C_L/C_D )ln⁡(W_1/W_2 ) max endurance at V_MP (i.e. maximise CL^3/2 / CD) R_prop = η/fg (C_L/C_D )ln⁡(W_1/W_2 ) max range at V_MD
39
Payload-range diagrams different plotting stages
Stage 1: range at max payload - W_initial = MTOW - W_final = OEW + reserve + payload Stage 2: range at max fuel - W_initial = MTOW - payload = MTOW - max useable fuel (fuel capacity) - OEW - W_final = MTOW - max useable fuel ( = OEW + reserve + payload, but payload now is not max payload) Stage 3: range with no payload - W_initial = OEW + max fuel capacity - W_final = OEW (payload = 0kg)
40
Load factor
n = L/W, or n = (V/V_stall)^2 how many times greater the lift is compared to weight
41
Manouevre flight envelope (v-n) diagrams load factor at stall boundary load factor at vertical gust
look at week 21- manoeuvres notes at the stall boundary, L = nW hence n = (ρV^2 S C_Lmax)/2W with a vertical gust of velocity υ, wing incidence is increased with υ/V, and change in C_L = a_1 * υ/V hence load factor n = (ρ*a_1*υ*S*V)/2W + 1
42
Relationship between load factor and bank angle
cos ϕ = 1/n high load factors require large bank angles bank angle ϕ occurs at no sideslip
43
Turn radius, load factor and velocity at min turn radius
R = V^2/(g√(n^2-1)) min turn radius R_min requires us to fly at maximum load factor (without damaging aircraft), and minimum velocity that can sustain that load factor (i.e. top left intersection on V-n diagram) this corner velocity V* = √((2*n_max* W)/(ρSC_Lmax ))
44
Turn rate, load factor and velocity at max turn rate Turn time
omega = V/R (circular motion) so just sub in Radius from above omega = (g√(n^2-1))/V max turn rate requires us to fly at max load factor, and minimum velocity that can sustain that load factor (same as min turn radius) turn time = angle (rad) / omega
45
Limits on turn performance
Stall velocity limits: V_Stall at C_Lmax maximum load factor limits: n_max turns are limited by stall if velocity is between V_stall and V* (i.e. between start of curve and corner point on V-n diagram) turns are limited by load factor if velocity is above V* (i.e. straight line past corner point) At V* is the best turn performance

AVDASI 1 (4 decks)

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