16.333: Lecture #2 Static Stability Aircraft Static Stability(longitudinal) ing/Tail contributions
16.333: Lecture #2 Static Stability Aircraft Static Stability (longitudinal) Wing/Tail contributions 0
Fa2004 16.3332-1 Static Stabilit Static stability is all about the initial tendency of a body to return to its equilibrium state after being disturbed To have a statically stable equilibrium point, the vehicle must develop a restoring force/ moment to bring it back to the eq. condition Later on we will also deal with dynamic stability, which is concerned with the time history of the motion after the disturbance Can be ss but not ds, but to be ds. must be ss SS is a necessary, but not sufficient condition for ds To investigate the static stability of an aircraft, can analyze response to a disturbance in the angle of attack At eq. pt, expect moment about c g. to be zero CMa=0 If then perturb a up, need a restoring moment that pushes nose back down(negative)
Fall 2004 16.333 2–1 Static Stability • Static stability is all about the initial tendency of a body to return to its equilibrium state after being disturbed • To have a statically stable equilibrium point, the vehicle must develop a restoring force/moment to bring it back to the eq. condition • Later on we will also deal with dynamic stability, which is concerned with the time history of the motion after the disturbance – Can be SS but not DS, but to be DS, must be SS ⇒ SS is a necessary, but not sufficient condition for DS • To investigate the static stability of an aircraft, can analyze response to a disturbance in the angle of attack – At eq. pt., expect moment about c.g. to be zero CMcg = 0 – If then perturb α up, need a restoring moment that pushes nose back down (negative)
Fa2004 16.3332-2 Classic analysis Airplane 2 N ose up Nose down Equilibrium point Airplane 1 Eg at point b A/C 1 is statically stable Conditions for static stability aCM da note that this requires Cml >0 Since Cl= Clola-ao) with Clo >0, then an equivalent condition for Ss is that aC M
Fall 2004 16.333 2–2 • Classic analysis: – Eq at point B – A/C 1 is statically stable • Conditions for static stability ∂CM CM = 0; 0 • Since CL = CLα(α−α0) with CLα > 0, then an equivalent condition for SS is that ∂CM < 0 ∂CL
Fa2004 16.3332-3 Basic Aerodynamics OwI aFRL Fuselage Reference Line(FRL) Wing mean chord Take reference point for the wing to be the aerodynamic center (roughly the 1 /4 chord point) Consider wing contribution to the pitching moment about the c g Assume that wing incidence is iu so that, if aa= aFrl +iw, then aFRL aw-lu With ak measured from the leading edge, the moment Mcg=(Lu Cos aFrl du sin afrl)(aeg -aac) +(Lu sin aFrl -du cos aFrl(ecg)+ maca Assuming that aFrl 1 M≈(Lu+ Da, carl)(x-xa) +(LuaFrl-du) acg)+ Macu But the second term contributes very little(drop) I The aerodynamic er(AC)is the point on the wing about which the coefficient of pitching moment is constant. On all airfoils the ac very close to the 25% chord point(+/-2%)in subsonic flow
Fall 2004 16.333 2–3 Basic Aerodynamics iw Xcg Xac Lw Zcg Macw Dw c.g. � �FRL w Fuselage Reference Line (FRL) Wing mean chord • Take reference point for the wing to be the aerodynamic center (roughly the 1/4 chord point)1 • Consider wing contribution to the pitching moment about the c.g. • Assume that wing incidence is iw so that, if αw = αFRL + iw, then αFRL = αw − iw – With xk measured from the leading edge, the moment is: Mcg = (Lw cos αFRL + Dw sin αFRL)(xcg − xac) +(Lw sin αFRL − Dw cos αFRL)(zcg) + Macw – Assuming that αFRL � 1, Mcg ≈ (Lw + DwαAFRL)(xcg − xac) +(LwαFRL − Dw)(zcg) + Macw – But the second term contributes very little (drop) 1The aerodynamic center (AC) is the point on the wing about which the coefficient of pitching moment is constant. On all airfoils the ac very close to the 25% chord point (+/ 2%) in subsonic flow
Fa2004 16.3332-4 Non-dimensionalize M viSe Give CMcg=(Cla+ Cpu aFrl) )+C De tine cg=hc(leading edge to c g ac= hic(leading edge to AC) Ther +cD,aFRL h-hr)+Cr ≈(CLn)(h-hn)+CMo Cao(h-hn)+Cr Result is interesting but the key part is how this helps us analyze the static stabil C aC since c.g. typically further back that ac Why most planes have a second lifting surface(front or back)
Fall 2004 16.333 2–4 • Nondimensionalize: L M CL = 1 CM = ρV 2S 1ρV 2Sc¯ 2 2 • Gives: CMcg = (CLw + CDwαFRL)(xcg c¯ − xac c¯ ) + CMac • Define: – xcg = hc¯ (leading edge to c.g.) – xac • Then = hn¯c¯ (leading edge to AC) CMcg = (CLw + CDwαFRL)(h − hn¯) + CMac ≈ (CLw)(h − h = CLαw n¯) + CMac (αw − αw0)(h − hn¯) + CMac • Result is interesting, but the key part is how this helps us analyze the static stability: ∂CMcg ∂CLw = (h − hn¯) > 0 since c.g. typically further back that AC – Why most planes have a second lifting surface (front or back)
Fa2004 16.3332-5 Contribution of the tail Z aFRL Some lift provided but moment is the key part Key items 1. Angle of attack at=aFRL +it -E, E is wing downwash 2. Lift L= Lu+ lt with Lu > Lt and CL= CLm+nCCL 7=(1/2pV)/(1/ 2pV2)0.8-1.2 depending on location of tail Downwash usually approximated as de d where Eo is the downwash at ao. For a wing with an elliptic distribution C de 2C ar do
Fall 2004 16.333 2–5 Contribution of the Tail FRL FRL �W �FRL i w i t Lw Lt Dw Dt Macw M t Zcgw V V l t V' Zcgt � c.g. ac • Some lift provided, but moment is the key part • Key items: 1. Angle of attack αt = αFRL + it − �, � is wing downwash 2. Lift L = Lw + Lt with Lw � Lt and St CL = CLw + η CLt S η = (1/2ρVt 2)/(1/2ρVw 2) ≈0.8–1.2 depending on location of tail 3. Downwash usually approximated as d� � = �0 + αw dα where �0 is the downwash at α0. For a wing with an elliptic distribution 2CLw d� 2CLαw � ≈ = πAR ⇒ dα πAR
Fa2004 16.3332-6 o Pitching moment contribution: Lt and D are l, to v not v So they are at angle a= afrl -e to FRl, so must rotate and then apply moment arms lt and zt M lt [Lt cos a+D, sin a zt [Dt cos- Lt sin a+ Mac First term largest by far. Assume that a 1, so that Mte-llt v2S Define the horizontal tail volume ratio Vh- lt St so that Mt HcL Note: angle of attack of the tail at= aa-iwtit-E, so that where=60+∈a0m ● So that Mt HcL 2n,+ (1-∈a)
Fall 2004 16.333 2–6 • Pitching moment contribution: Lt and Dt are ⊥, � to V � not V – So they are at angle α¯ = αFRL − � to FRL, so must rotate and then apply moment arms lt and zt. Mt = −lt [Lt cos α¯ + Dt sin α¯] −zt [Dt cos α¯ − Lt sin α¯] + Mact • First term largest by far. Assume that α¯ � 1, so that Mt ≈ −ltLt 1 Lt = ρV 2 StCLt 2 t 1ρV 2 Mt StCLt lt 2 t CMt = = 1 1 ρV 2Sc¯ −c¯ ρV 2S 2 2 ltSt = − ηCLt Sc¯ Define the horizontal tail volume ratio VH = ltSt , so that Sc¯ • CMt = −VHηCLt • Note: angle of attack of the tail αt = αw − iw + it − �, so that CLt = CLαt αt = CLαt (αw − iw + it − �) where � = �0 + �ααw • So that CMt = −VHηCLαt (αw − iw + it − (�0 + �ααw)) = VHηCLαt (�0 + iw − it − αw(1 − �α))
Fa2004 16.3332-7 More compact form HnCLoEo+iw -it where we can chose VH by selecting lt, St and it Write wing form as Cm= cmo CMa aw CMac-Clo awo(h- hi Mo CLo (h-hn And total is Mo -CLo Qao(h-hm)+VHnClo Eo +iw -it) Mo at
Fall 2004 16.333 2–7 • More compact form: CMt = CM0t + CMαt αw = VHηCLαt ⇒ CM (�0 + iw − it) 0t = −VHηCLαt ⇒ CM (1 − �α) αt where we can chose VH by selecting lt, St and it • Write wing form as CMw = CM0w + CMαw αw = CMac − CLαw αw0 ⇒ CM (h − hn¯) 0w = CLαw ⇒ CM (h − hn¯) αw • And total is: CMcg = CM0 + CMααw ⇒ CM0 = CMac − CLαw αw0(h − hn¯) + VHηCLαt (�0 + iw − it) ⇒ CMα = CLαw (h − hn¯) − VHηCLαt (1 − �α)
Fa2004 16.3332-8 For static stability need Cm=0 and Cm 0 ,use it to trim the aircraft For cma <0, consider setup for case that makes 0 Note that this is a discussion of the aircraft cg location (" find h") But tail location currently given relative to c.g.(t behind it) which is buried in VH= need to define it differently Define lt=c(ht-h); ht measured from the wing leading edge, then Ives h)n CLo( h(Clo +nd Clo (1-Ea))-Clo hi-hiClo (l-Ea a bit messy but note that if Lr=lu+Lt, then CLn tn Lo +nCLo (Eo -iw+it+w(1-EaD) CLo- +Cl aa with CLar= CLau +m s Cla, (1 -Ea)
Fall 2004 16.333 2–8 • For static stability need CM = 0 and CMα 0 ⇒ use it to trim the aircraft • For CMα < 0, consider setup for case that makes CMα = 0 – Note that this is a discussion of the aircraft cg location (“find h”) – But tail location currently given relative to c.g. (lt behind it), which is buried in VH ⇒ need to define it differently • Define lt = c¯(ht − h); ht measured from the wing leading edge, then VH = ltSt cS¯ = St S (ht − h) which gives St n) − (ht − h)ηCLαt CM (1 − �α) α CLαw = (h − h¯ S St St = h(CLαw + η CLαt (1 − �α)) − CLαw hn¯ − htη CLαt (1 − �α) S S • A bit messy, but note that if LT = Lw + Lt, then St CLT = CLw + η CLt S so that St CLT = CLαw (αw − αw0) + η CLαt (−�0 − iw + it + αw(1 − �α)) S = CL0T + CLαT αw with CLαT = CLαw + η St CLαt (1 − �α) S
Fa2004 16.3332_9 Now have that Mo=hClor-Clo hi- Clo (1l-Ea Solve for the case with CM,=0, which gives h=h C +nchia (1-Ed L Note that with y CIo≈1,then h hn+(7 1) NP which is called the stick fixed neutral point ● Can rewrite as (1-∈a) T CLo(h- hNP which gives the pitching moment about the c g. as a function of the location of the c g. with respect to the stick fixed neutral point For static stability, c g. must be in front of NP (hnp> h)
� � • � � � � Fall 2004 16.333 2–9 • Now have that n¯ − htη St S CLαt CM (1 − �α) α = hCLαT − CLαw h • Solve for the case with CMα = 0, which gives CLαw St CLαt h = hn¯ + η ht (1 − �α) CLαT S CLαT n¯ + (γ − 1)ht CLαT • Note that with γ = CLαw ≈ 1, then h h = ≡ hNP γ which is called the stick fixed neutral point Can rewrite as CLαw St CLαt CMα = CLαT h − hn¯ − η ht (1 − �α) CLαT S CLαT = CLαT (h − hNP ) which gives the pitching moment about the c.g. as a function of the location of the c.g. with respect to the stick fixed neutral point – For static stability, c.g. must be in front of NP (hNP > h)
