Flight Dynamics and Control Lecture 5: Longitudinal stability Derivatives G. Dimitriadis University of Liege 1 Previously on AERO0003-1 • We developed linearized equations of motion Longitudinal direction " % "m 0 0 0 0% " v % −Yv −(Yp + mWe ) −(Yr − mUe ) −mgcosθe −mgsinθe " v % " Yξ Yς % $ ' $ ' $ ' $ ' $ ' 0 I x −I xz 0 0 p $ −L −L −L 0 0 ' p Lξ Lς $ ' $ ' v p r $ ' $ ' "ξ% $ ' $ ' $ ' $ ' $ ' 0 −I xz Iz 0 0 r + −N −N −N 0 0 r = Nξ Nς $ ' $ ' $ ' $ v p r ' $ ' $ ' #ς & $ 0 0 0 1 0' $ϕ ' $ 0 −1 0 0 0 ' $ϕ ' $ 0 0 ' $ ' #$ 0 0 0 0 1&' #$ψ &' − #$ψ &' $ 0 0 ' # 0 Lateral0 direction1 0 0 & # & " % " − − − − θ % m −Xw 0 0 "u % Xu Xw (Xq mWe ) mgcos e "u % " X X % $ ' $ ' η τ $ ' $ ' $ ' $ 0 m − Z 0 0' w $ − − − + θ ' w Zη Zτ "η% ( w ) $ ' + Zu Zw (Zq mUe ) mgsin e $ ' = $ ' $ ' $ ' $ ' $ ' − $ q ' $ q ' Mη Mτ #τ & $ 0 M w I y 0' $−M u −M w −M q 0 ' $ ' $ ' $ ' $ 0 0 0 1' #θ & $ − ' #θ & # 0 0 & 2 # & # 0 0 1 0 & Longitudinal stability derivatives • It has already been stated that the best way to obtain the values of the stability derivatives is to measure them. • However, it is still useful to discuss simplified methods of estimating these coefficients. • Such estimates can be used, for example, in the preliminary design of aircraft. • This lecture will treat longitudinal stability derivatives. 3 Simple example • We keep the quasi-steady aerodynamic assumption. • Assume that the lift of an aircraft lies entirely in the z direction: 1 2 Z = ρU SCL 2 • where CL is the lift coefficient, assumed to be constant in this simple example. • Now consider that there is a perturbation in the airspeed such that U=Ue+u, Ue being the equilibrium airspeed 4 Simple example (2) • The total aerodynamic force in the z direction becomes: 1 2 1 2 2 Z = ρ (Ue + u) SCL = ρ (U e + 2Ueu + u )SCL 2 2 • We linearize such that u2=0: 1 2 Z = ρ (U e + 2Ueu)SCL 2 • Now we can calculate the derivative of Z with respect to u: ! ∂Z Zu = = ρUeSCL 5 ∂u Lift and Drag • This simple example considered only one aerodynamic force and one perturbation. Now we need to consider all aerodynamic loads and all perturbations. • Lift and drag are defined in directions perpendicular and parallel to the direction of motion: U Lift e αe θe V0 γe Xa W Drag e mg Za Pitching Moment, Ma 6 Longitudinal geometry and loads S l = T T L VT T Lw Sc ac cg MT M0 l T mg hc h denotes the cg c position h0 denotes the h c From Aircraft Design Lectures 0 aerodynamic centre 7 Za, Xa and Ma • Clearly, the aerodynamic forces Za and Xa and the aerodynamic moment Ma can be written as: Za = −Lift × cosα − Drag × sinα X a = Lift × sinα − Drag × cosα Ma = M0 + Lift(h − h0)c − LiftT lT • The last equation was obtained during the Aircraft Design lectures • We have assumed that the wing lift is 8 approximately equal to the total lift. In coefficient form • The lift and drag can be written, as usual, in terms of the lift and drag coefficients, so that 1 2 Za = − ρQ∞S(CL cosα +CD sinα) 2 1 2 Xa = ρQ∞S(CL sinα −CD cosα) 2 = 1 ρ 2 + − − M a Q∞Sc (Cm0 CL (h h0 ) VTCLT ) 2 • Using the notation of Aircraft Design, C = C + aα L L0 a=lift curve slope 2 CL C = C + e=Oswald factor 9 D D0 eπAR Angle of attack • Consider an aircraft in equilibrium at angle of attack αe, perturbed by small airspeeds u, w and small pitch rate q. Ue Ue+u −1 W + w −1 W −1 w α = tan e ≈ tan e + tan αe α Ue + u Ue Ue or V0 We w α = αe + Ue We+w 10 Effective camber • A pitching wing can be seen to feature an additional camber distribution due to the pitching motion. Local flow velocity due to pitch rate Additional contribution to C.G. angle of attack q α = − = α eff (h1 h) h0 c /4 Ue h V0 h1 = 3c /4 11 Wing-tail flow geometry • For a static aircraft, the downwash effect of the wing deflects the free stream flow seen by the tailplane by an angle α. • Total angle of attack of tail: αT=α-ε+ηT Tailplane Wing ηT α αT α V0 η ε Elevator V0 βη Trim tab • The total lift on the tailplane is given by: = + α + η + β 12 CLT a0 a1 T a2 a3 η Unsteady wing downwash • Now, consider a dynamic aircraft. A change in downwash at the wing because of a change in pitch angle at time t will at the tail Δt seconds later. • The time difference Δt is the time it takes for air leaving the wing to reach the tail and can be approximated as Δt=lt/Ue. • Then: ∂ε ∂ε ε (t) = α − α!Δt ∂α ∂α • So that: ' ∂ε ) ∂ε l C = aη + a 1− α + a t q + a η LT 1 T 1( * 1 2 ∂α ∂α Ue 13 Lift, drag and moment • In terms of perturbations around a static equilibrium condition, the lift, drag and moment can be written as: = + a + − CL CLe (w (h1 h)q) Ue 2aC = + Le + − CD CDe (w (h1 h)q) Ueeπ AR " " ∂ε %% ∂ε = + − − a1 − a + − − lt − η Cm Cme $h h0 VT $1 '' (w (h1 h)q) VT a1 q VT a2 # a # ∂α &&Ue ∂α Ue Second order perturbation terms in the drag expression have been ignored 14 Added mass terms • The acceleration stability derivatives, e.g. M w , X w , Z w result from the added mass effect of the aircrafts motion. • This motion carries along a certain mass of air around the aircraft. • As a consequence, the aircraft has more inertia than it would have had if it was moving in a vacuum • This is known as the added mass or apparent mass effect 15 Added mass terms (2) • Added mass can be seen as the inertia of cylinder of diameter c and length b that is moving with the wing. • Some added mass terms: c Lift = ρπ S(w + U q) added 4 e c % c ( c % 3c ( π Moment = ρπ S' h − * w − ρπ S' − h* U q − ρU c 2Sq added 4 & 2) 4 & 4 ) e 16 e 16 Perturbation • Consider that, as the flight is perturbed, the total airspeed is: 2 2 2 2 Q∞ = (Ue + u) +(We + w) = V0 + 2Ueu + 2Wew • The cosine and sine terms of the angle of attack become: # w & w cosα = cos% αe + ( = cosα e − sinα e $ Ue ' Ue # w & w sinα = sin% α e + ( = sinαe + cosαe $ Ue ' Ue 17 Force Za • The total force Za becomes: # & 1 2 1 2 a 2aCL = − ρ α + α − ρ % + − α + e + − α( Za Q∞S(CLe cos CDe sin ) Q∞S (w (h1 h)q)cos (w (h1 h)q)sin 2 2 $Ue Ueeπ AR ' c − ρπ S(w! +Ueq)cosα 4 • After linearization: = − ρ + α + α − 1 ρ 2 − α + α Za Zae (Ueu Wew)S(CLe cos e CDe sin e ) V0 S( CLe sin e CDe cos e )w 2Ue " % 1 2 a 2aCLe c − ρV0 S$ (w +(h1 − h)q)cosαe + (w +(h1 − h)q)sinαe '− ρπ S(w! +Ueq)cosαe 2 #Ue Ueeπ AR & 4 18 Z stability derivatives • Differentiating Za with respect to w, u, q etc yields the Z stability derivatives as: ! = −ρ α + α Zu UeS(CLe cos e CDe sin e ) ! = −ρ α + α − 1 ρ 2 − α + α Zw WeS(CLe cos e CDe sin e ) V0 S( CLe sin e CDe cos e ) 2Ue " % 1 2 a 2aCLe − ρV0 S$ cosαe + sinαe ' 2 #Ue Ueeπ AR & " 2aC % ! 1 2 a Le c Zq = − ρV0 S$ (h1 − h)cosαe + (h1 − h)sinαe '− ρπ SUe cosαe 2 #Ue Ueeπ AR & 4 ! c Zw" = −ρπ S cosαe 4 19 Force Xa • The total force Xa becomes: # & 1 2 1 2 a 2aCL = ρ α − α + ρ % + − α − e + − α( Xa Q∞S(CLe sin CDe cos ) Q∞S (w (h1 h)q)sin (w (h1 h)q)cos 2 2 $Ue Ueeπ AR ' c + ρπ S(w! +Ueq)sinα 4 • After linearization: = + ρ + α − α + 1 ρ 2 α + α Xa Xae (Ueu Wew)S(CLe sin e CDe cos e ) V S(CLe cos e CDe sin e )w 2Ue " % 1 2 a 2aCLe c + ρV0 S$ (w +(h1 − h)q)sinαe − (w +(h1 − h)q)cosαe '+ρπ S(w! +Ueq)sinαe 2 #U U eπ AR & 4 e e 20 X stability derivatives • Differentiating Xa with respect to w, u, q etc yields the X stability derivatives as: ! = ρ α − α Xu UeS(CLe sin e CDe cos e ) ! = ρ α − α + 1 ρ 2 α + α Xw WeS(CLe sin e CDe cos e ) V S(CLe cos e CDe sin e ) 2Ue " % 1 2 a 2aCLe + ρV0 S$ sinαe − cosαe ' 2 #Ue Ueeπ AR & " 2aC % ! 1 2 a Le c Xq = ρV0 S$ (h1 − h)sinαe − (h1 − h)cosαe '+ ρπ SUe sinαe 2 #Ue Ueeπ AR & 4 ! c Xw" = ρπ Ssinαe 4 21 Moment Ma • The total moment Ma becomes: # # ∂ε && = 1 ρ 2 + 1 ρ 2 − − a1 − a + − M a Q∞ScCme Q∞Sc %h h0 VT %1 (( (w (h1 h)q) 2 2 $ a $ ∂α ''Ue 1 2 ∂ε lt 1 2 − ρQ∞ScVT a1 q − ρV ScVT a2η 2 ∂α Ue 2 c # c & c #3c & π 2 + ρπ S%h − (w! − ρπ S% − h(Ueq − ρUec Sq 4 $ 2 ' 4 $ 4 ' 16 • After linearization: " " ∂ε %% = + ρ + + 1 ρ 2 − − a1 − a + − M a M ae (Ueu Wew)ScCme V0 Sc $h h0 VT $1 '' (w (h1 h)q) 2 # a # ∂α &&Ue 1 2 ∂ε lt 1 2 − ρV0 ScVT a1 q − ρV0 ScVT a2η 2 ∂α Ue 2 c " c % c "3c % π 2 + ρπ S$h − 'w! − ρπ S$ − h'Ueq − ρUec Sq 22 4 # 2 & 4 # 4 & 16 M stability derivatives • Differentiating Ma with respect to w, u, q etc yields the M stability derivatives as: ! = ρ Mu UeScCme " " ∂ε %% ! = ρ + 1 ρ 2 − − a1 − a M w WeScCme V0 Sc $h h0 VT $1 '' 2 # a # ∂α &&Ue " " %% ! 1 2 a1 ∂ε a M q = ρV0 Sc $h − h0 −VT $1− '' (h1 − h) 2 # a # ∂α &&Ue " % 1 2 ∂ε lt c 3c π 2 − ρV0 ScVT a1 − ρπ S$ − h'Ue − ρUec S 2 ∂α Ue 4 # 4 & 16 ! c " c % M w" = ρπ S$h − ' 4 # 2 & 23 ! 1 2 Mη = − ρV0 ScVT a2 2 Discussion of longitudinal derivatives • Notice that this methodology for the stability derivatives is by no means exact.