16.333: Lecture #1 Equilibrium States Aircraft performance Introduction to basic terms Fall 2004 16.333 1–1 Aircraft Performance ? Accelerated horizontal ?ight - balance of forces – Engine thrust T – Lift L (⊥ to V ) – Drag D (? to V ) – Weight W dV D T mg L T ? D = m = 0 for steady ?ight dt and L ? W = 0 De?ne L = ? 1 2 ρV 2 SC L where – ρ – air density (standard tables) – S – gross wing area = ˉc × b, – cˉ = mean chord – b = wing span – AR – wing aspect ratio = b/cˉ c t c 0 kc ? k edge b = 2.s c Sweepback angle Trailing – Q = 1 2 ρV 2 dynamic pressure – V = speed relative to the air ? ? Fall 2004 16.333 1–2 – C L lift coe?cient – for low Mach number, C L = C L α (α ? α 0 ) 3 α angle of incidence of wind to the wing 3 α 0 is the angle associated with zero lift ? Back to the performance: 1 L = ρV 2 SC L and L = mg 2 2mg which implies that V = ρSC L so that V ∝ C ?1/2 L and we can relate the e?ect of speed to wing lift ? A key number is stall speed, which is the lowest speed that an aircraft can ?y steadily 2mg V s = ρSC L max where typically get C L max at α max = 10 ? Fall 2004 16.333 1–3 Steady Gliding Flight ? Aircraft at a steady glide angle of γ ? Assume forces are in equilibrium L ? mg cos γ = 0 (1) D + mg sin γ = 0 (2) Gives that D C D tan γ = L ≡ C L ? Minimum gliding angle obtained when C D /C L is a minimum – High L/D gives a low gliding angle ? Note: typically C D = C D min + 2 L C πARe where is the zero lift (friction/parasitic) drag – C D min – C 2 lift induced drag gives the L – e is Oswald’s e?ciency factor ≈ 0.7 ? 0.85 Fall 2004 16.333 1–4 ? Total drag then given by 1 1 ? ? D = ρV 2 SC D = ρV 2 S L (3)C D min + kC 2 2 2 1 (mg) 2 = 2 ρV 2 SC D min + k 1 ρV 2 S (4) 2 D V E V S 1 V Emd Total drag No-lift drag Lift-dependent drag ? So that the speed for minimum drag is ? ? ? 1/4 2mg k V min drag = ρS C D min Fall 2004 16.333 1–5 Steady Climb ? V R/C W L D T ? T - D ? Equations: T ? D ? W sin γ = 0 (5) L ? W cos γ = 0 (6) ? which gives L cos γ = 0 T ? D ? sin γ so that T ? D tan γ = L ? Consistent with 1–3 if T = 0 since then γ as de?ned above is negative ? Note that for small γ, tan γ ≈ γ ≈ sin γ R/C = V sin γ ≈ V γ ≈ (T ? D)V L so that the rate of climb is approximately equal to the excess power available (above that needed to maintain level ?ight) Fall 2004 16.333 1–6 Steady Turn R W L cos ? force L L sin ? ? ? Radius of turn Centrifugal ? Equations: L sin φ = centrifugal force (7) mV 2 = R (8) L cos φ = W = mg (9) ? tan φ = V 2 Rg V =Rω = V ω g (10) ? Note: obtain R min at C L max R min ( 1 2 ρV 2 SC L max ) sin φ = W V 2 g ? R min = W/S 1/2ρgC L max sin φ max where W/S is the wing loading and φ max < 30 ? Fall 2004 16.333 1–7 ? De?ne load factor N = L/mg. i.e. ratio of lift in turn to weight N = sec φ = (1 + tan 2 φ) 1/2 ? (11) tan φ = N 2 ? 1 (12) so that V 2 V 2 R = g tan φ = g √ N 2 ? 1 ? For a given load factor (wing strength) R ∝ V 2 ? Compare straight level with turning ?ight – If same light coe?cient L mg 1 C L = 1 = ρV 2 S ≡ N mg ρV t 2 S ρV 2 S 1 2 2 2 so that V t = √ N V gives the speed increase (more lift) ? Note that C L constant C D constant ? D ∝ V 2 C D ? ? T t ∝ D t ∝ V t 2 C D ～ N D so that must increase throttle or will descend in the turn