Aircraft Stability and Performance 2nd Year, Aerospace Engineering

Transcription

1 Aircraft Stability and Performance 2nd Year, Aerospace Engineering Dr. M. Turner March 6, 207

2 Aims of Lecture Part. To examine ways aircraft turn 2. To derive expressions for correctly banked turns 3. To compare turning and straight-level flight Part 2. To work through a steady turn example 2. To look briefly at height loss during turning

3 Turning Aircraft V R Centripetal acceleration: α = V2 R Centripetal force: m V2 R To execute a turn, aircraft can either Skid to turn - alter yaw of aircraft to create side-slip (basically using rudder) β = angle of side-slip β V (NB - slightly non-standard illustration)

4 Problems with turns using rudder. Passenger Airsickness Pilot typically sits close to centre of rotation. Passengers sitting behind pilot experience side-to-side motion Particularly uncomfortable for passengers sitting at the back of the aircraft 2. Yaw induced roll If rudder used to initiate turn, advancing wing experiences greater airspeed Greater airspeed implies greater lift (L = 2 ρc LSV 2 ) on advancing wing Greater lift tends to roll the aircraft - even with no aileron input! Main use of rudder Corrections of adverse yaw (during bank when induced drag on higher wing is greater, causing adverse yaw) Making small turns (especially at low speed where ailerons lose effectiveness)

5 Bank-to-turn Alternatively Bank-to-turn - roll the aircraft to alter the direction of the lift vector (using ailerons) φ = roll angle φ L t α Aircraft nose points into slide Bank-to-turn is typically more efficient (easier) in most aircraft Correctly banked turn: centripetal force provided solely by inward component of lift vector due to banking of aircraft mg NB: Rudder input necessary to prevent aircraft side-slipping

6 Turning aircraft - fundamental relationships Resolving horizontally and vertically: L t cosφ = mg L t sinφ = mα = m V2 R φ L t tanφ = V2 gr ( ) V 2 φ = arctan gr This implies mg α Aircraft nose points into slide The faster the aircraft is travelling, the more it must bank to adhere to a turn of a given radius For a fixed bank angle, the radius of the turn must increase as the square of the velocity

7 Load Factor An important parameter indicating the stress an aircraft (and its pilot/passengers) can be subjected to. Load Factor, N defined as N = = L t mg lift in turn weight = mg/(cosφ) mg = cosφ = secφ N is a ratio of forces (dimensionless) although often given in g. When bank angle is zero (φ = 0), N = Load factor a significant parameter in aircraft turning.

8 Load Factor - illustration φ=0 φ=30 φ=60 N= N=.2 N=2 Load factor indicates weight gain of passengers/pilots. Straight, level flight: normal sensation of weight Correctly banked turn at 60 o : sensation of twice normal weight Typical load factors: commercial aircraft 2; fighter aircraft 8 0.

9 Relationship between load factor and turning Bank angle sec 2 φ = +tan 2 φ tan 2 φ = sec 2 φ = N 2 tanφ = N 2 Bank angle increases with load factor (for small φ) Turn radius tanφ = N 2 = V2 gr R = V 2 g N 2 For a given load factor, radius of turn increases as square of speed Centripetal acceleration α = mv2 R = mv2 V 2 g N 2 = mg N 2 Centripetal acceleration increases with load factor.

10 Alternative expressions for turn radius Two expressions for lift: N = L t /(mg) L t = Nmg L t = 2 ρsc L,tV 2 Using this value in the expression for turn radius gives V2 = = L t 2 ρsc L,t Nmg 2 ρsc L,t R = V 2 g N 2 = = Nmg 2 ρsc L,tg N 2 Nm 2 ρsc L,t N2 Minimum radius turn, R min, attained at C L,max. R min = Nm 2 ρsc L,max N2

11 Rate of turn Rate of turn ω given by Using ω = V R = in the expression for ω gives V V 2 g N 2 V = = g V N2 Nmg 2 ρsc L,t ω = = 2 ρsc L,t Nmg g N 2 2 ρsc L,tg m N2 N Implication: turn rate increases as a function of C L,t

12 Turning flight vs Straight level flight When entering a turn, lift vector direction is altered. Thus to keep same C L during turn, velocity (throttle) must be adjusted Straight Level How does this compare to straight level flight? V 2 s = mg 2 ρsc L,s C L,s = mg 2 ρsv2 s Turn Thus for same lift coefficient, C L,s = C L,t : V 2 t = Nmg 2 ρsc L,t C L,t = Nmg 2 ρsv2 t V t = NV s i.e speed needs to be increased by N to keep C L the same in turn as in straight level flight

13 Turning flight vs Straight level flight Assume now that change in lift coefficient (incidence angle) is acceptable during turning flight. How can straight-level airspeed be maintained during turning flight? Straight Level V 2 s = mg Hence if V s = V t : 2 ρsc L,s Turn V 2 t = Nmg 2 ρsc L,t which implies C L,s = C L,t /N or C L,t = NC L,s For same speed in turn, lift coefficient must increase by a factor of N If aircraft is flying close to stalling speed, attempts to turn the aircraft while keeping velocity constant could stall the aircraft In practice, for gentle turns (φ 25 o ) effects of turning are small (sinφ φ). For φ 30 o, pilots tend not to adjust throttle and accept a slight loss in speed.

14 Turning vs straight level flight: remarks Constant C L Velocity must be increased: V t = NV s Increase in thrust required (or height loss) Turn radius increased: R = NV 2 s g N 2 Constant V Lift coefficient must be increased C L,t = NC L,s Increased C L increased α Turn radius decreased (danger of stall!)

15 Height loss during turning Assumptions: Thrust (throttle) is constant and independent of speed Turn takes place at same lift coefficient for initial level flight Velocity (V) increases to maintain this (hence some height loss must occur) Resolving normal to flight path L T L T = mg cosγ Lcosφ = mg cosγ Assuming γ small, cosγ and hence γ T T D T γ flight path angle Lcosφ = mg (N = secφ) mg

16 Height loss during turning L T Resolving along flight path T T +mg sinγ = D T γ T T D T γ flight path angle Since thrust = drag in straight level flight mg Also, Hence T = D = T T D T = ND (because V t = NV s ) D +mg sinγ = ND D(N ) sinγ = mg ( = ) (N ) C L /C D Implication: minimum angle occurs at minimum drag condition