航空器的稳定与控制（英文版）：lecture_2

16.333: Lecture #2 Static Stability Aircraft Static Stability (longitudinal) Wing/Tail contributions 0 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 su?cient 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 C M cg = 0 – If then perturb α up, need a restoring moment that pushes nose back down (negative) Fall 2004 16.333 2–2 ? Classic analysis: – Eq at point B – A/C 1 is statically stable ? Conditions for static stability ?C M C M = 0; < 0 ?α ≡ C M α note that this requires C M | α 0 > 0 ? Since C L = C L α (α ? α 0 ) with C L α > 0, then an equivalent condition for SS is that ?C M < 0 ?C L Fall 2004 16.333 2–3 Basic Aerodynamics i w X cg X ac L w Z cg M ac w D w 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 i w so that, if α w = α F RL + i w , then α F RL = α w ? i w – With x k measured from the leading edge, the moment is: M cg = (L w cos α F RL + D w sin α F RL )(x cg ? x ac ) +(L w sin α F RL ? D w cos α F RL )(z cg ) + M ac w – Assuming that α F RL ? 1, M cg ≈ (L w + D w α AF RL )(x cg ? x ac ) +(L w α F RL ? D w )(z cg ) + M ac w – But the second term contributes very little (drop) 1 The aerodynamic center (AC) is the point on the wing about which the coe?cient of pitching moment is constant. On all airfoils the ac very close to the 25% chord point (+/- 2%) in subsonic ?ow. Fall 2004 16.333 2–4 Non-dimensionalize: ? L M C L = C M = 1 ρV 2 S 1 ρV 2 Scˉ 2 2 ? Gives: C M cg = (C L w + C D w α F RL )( x cg ˉc ? x ac ˉc ) + C M ac ? De?ne: – x cg = hˉc (leading edge to c.g.) – x ac Then ? = h ˉn cˉ (leading edge to AC) C M cg = (C L w + C D w α F RL )(h ? h ) + C ˉ Mn ac ≈ (C L w )(h ? h = C L α w ) + C ˉ Mn ac (α w ? α w 0 )(h ? h ) + C ˉ Mn ac ? Result is interesting, but the key part is how this helps us analyze the static stability: ?C M cg ?C L w = (h ? h ) 0> ˉn since c.g. typically further back that AC – Why most planes have a second lifting surface (front or back) Fall 2004 16.333 2–5 Contribution of the Tail FRL FRL ? FRL ? W i w i t L w L t D w D t M ac w M t Z cg w V V l t V' Z cg t ? c.g. ac ? Some lift provided, but moment is the key part ? Key items: 1. Angle of attack α t = α F RL + i t ? ?, ? is wing downwash 2. Lift L = L w + L t with L w ? L t and S t C L = C L w + η C L t S η = (1/2ρV t 2 )/(1/2ρV w 2 ) ≈0.8–1.2 depending on location of tail 3. Downwash usually approximated as d? ? = ? 0 + α w dα where ? 0 is the downwash at α 0 . For a wing with an elliptic distribution 2C L w d? 2C L α w =? ≈ πAR ? dα πAR Fall 2004 16.333 2–6 ? Pitching moment contribution: L t and D t are ⊥, ? to V ? not V ˉ– So they are at angle α = α F RL ?? to FRL, so must rotate and then apply moment arms l t and z t . α + D t sin ˉM t = ?l t [L t cos ˉ α] α ?L t sin ˉz t [D t cos ˉ α] + M ac t ? ˉ? First term largest by far. Assume that α ? 1, so that M t ≈?l t L t 1 L t = ρV 2 S t C L t 2 t 1 ρV 2 S t C L t M t l t 2 t C M t = = 11 ρV 2 Scˉ ? cˉ ρV 2 S 2 2 l t S t = ηC L t ? Scˉ De?ne the horizontal tail volume ratio V H = l t S t , so that Scˉ ? C M t = ?V H ηC L t ? Note: angle of attack of the tail α t = α w ?i w + i t ??, so that C L t = C L α t α t = C L α t (α w ?i w + i t ??) where ? = ? 0 + ? α α w So that ? C M t = ?V H ηC L α t (α w ?i w + i t ?(? 0 + ? α α w )) = V H ηC L α t (? 0 + i w ?i t ?α w (1 ?? α )) Fall 2004 16.333 2–7 ? More compact form: C M t = C M 0 t + C M α t α w = V H ηC L α t (? 0 + i w ? i t )? C M 0 t = ?V H ηC L α t (1 ? ? α )? C M α t where we can chose V H by selecting l t , S t and i t ? Write wing form as C M w = C M 0 w + C M α w α w = C M ac ? C L α w α w 0 (h ? h ˉn )? C M 0 w = C L α w (h ? h ˉn )? C M α w And total is: ? = C M 0 + C M α α w C M cg ? C M 0 = C M ac ? C L α w α w 0 (h ? h ˉn ) + V H ηC L α t (? 0 + i w ? i t ) ? C M α = C L α w (h ? h ˉn ) ? V H ηC L α t (1 ? ? α ) Fall 2004 16.333 2–8 ? For static stability need C M = 0 and C M α < 0 – To ensure that C M = 0 for reasonable value of α, need C M 0 > 0 use i t to trim the aircraft ? ? For C M α < 0, consider setup for case that makes C M α = 0 – Note that this is a discussion of the aircraft cg location (“?nd h”) – But tail location currently given relative to c.g. (l t behind it), which is buried in V H ? need to de?ne it di?erently ? De?ne l t = ˉc(h t ? h); h t measured from the wing leading edge, then V H = l t S t ˉcS = S t S (h t ? h) which gives S t n ) ? (h t ? h)ηC L α t (1 ? ? α )C M α C L α w (h ? h ˉ = S S t S t = h(C L α w + η C L α t (1 ? ? α )) ? C L α w h ˉn ? h t η C L α t (1 ? ? α ) S S ? A bit messy, but note that if L T = L w + L t , then S t C L T = C L w + η C L t S so that S t C L T = C L α w (α w ? α w 0 ) + η C L α t (?? 0 ? i w + i t + α w (1 ? ? α )) S = C L 0 T + C L α T α w with C L α T = C L α w + η S t C L α t (1 ? ? α ) S ? ? ? ? ? ? ? Fall 2004 16.333 2–9 Now have that ? h η? ˉ tn S t S C L α t (1 ? ? α )C M α = hC L α T ? C L α w h ? Solve for the case with C M α = 0, which gives C L α w S t C L α t h = h + η h t (1 ? ? α ) ˉn C L α T S C L α T + (γ 1)h ˉ t ? n C L α T ? Note that with γ = C L α w ≈ 1, then h h = ≡ h N P γ which is called the stick ?xed neutral point Can rewrite as C L α w S t C L α t C M α = C L α T h ? h ? η h t (1 ? ? α ) ˉn C L α T S C L α T = C L α T (h ? h N P ) which gives the pitching moment about the c.g. as a function of the location of the c.g. with respect to the stick ?xed neutral point – For static stability, c.g. must be in front of NP (h N P > h) Fall 2004 16.333 2–10 ? Summary plot: (a = C L α T ) and (h n = h N P ) C m 0 0 C m C m m 0 (h - h n )? CG aft CG forward n n n ? = C + a h > h h = h h < h Observations: ? – If c.g. at h N P , then C M α = 0 – If c.g. aft of h N P , then C M α > 0 (statically unstable) – What is the problem with the c.g. being too far forward? Fall 2004 16.333 2–11 Control E?ects ? Can use elevators to provide incremental lift and moments – Use this to trim aircraft at di?erent ?ight settings, i.e. C M = 0 ? e Horizontal tail ? De?ecting elevator gives: ΔC L = dC L δ e = C L δ e δ e , with C L δ e > 0 dδ e ? C L = C L α (α ? α 0 ) + C L δ e δ e C m ?C m 0 ? e = 0 ? e > 0 ? Original trim ? Final trim ΔC m = dC m δ e = C m δ e δ e , with C m δ e < 0 dδ e ? C m = C m 0 + C m α α + C m δ e δ e C L 0 Final trim pt. ?C L ? e = 0 ? e > 0 ? Original trim pt. Fall 2004 16.333 2–12 ? For trim, need C m = 0 and C L TRIM ? ?? ? ? ? C m α C m δ e α T RIM ?C m 0 = C L α C L δ e (δ e ) TRIM C L TRIM + C L α α 0 So elevator angle needed to trim: α 0 )C L α C m 0 + C m α (C L TRIM + C L α =(δ e ) TRIM C m α C L δ e ? C L α C m δ e ? Note that typically elevator down is taken as being positive ? Also, for level ?ight, C L TRIM = W/(1/2ρV 2 S), so expect (δ e ) TRIM to change with speed.