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 ∂α ≡ CMα note that this requires CM| α0 > 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)