16.333: Lecture #1 Equilibrium States Aircraft performance Introduction to basic terms Fall 2004 16.333 1–1 Aircraft Performance • Accelerated horizontal ﬂight balance of forces – Engine thrust T L – Lift L (⊥ to V ) – Drag D (� to V ) D T – Weight W mg T −D =m dV = 0 for steady ﬂight dt and L−W =0 • Deﬁne L = 21 ρV 2SCL where ct – ρ – air density (standard tables) – S – gross wing area = c̄ × b, c Trailing edge – c̄ = mean chord c0 b = 2.s – b = wing span – AR – wing aspect ratio = b/c̄ Sweepback angle �k kc – Q = 12 ρV 2 dynamic pressure – V = speed relative to the air Fall 2004 16.333 1–2 – CL lift coeﬃcient – for low Mach number, CL = CLα (α − α0) 3 α angle of incidence of wind to the wing 3 α0 is the angle associated with zero lift • Back to the performance: 1 L = ρV 2SCL and L = mg 2 � 2mg which implies that V = ρSC so that L −1/2 V ∝ CL and we can relate the eﬀect of speed to wing lift • A key number is stall speed, which is the lowest speed that an aircraft can ﬂy steadily � 2mg Vs = ρSCLmax where typically get CLmax at αmax = 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 D + mg sin γ = 0 Gives that D CD ≡ L CL ⇒ Minimum gliding angle obtained when CD /CL is a minimum tan γ = – High L/D gives a low gliding angle • Note: typically CL2 CD = CDmin + πARe where – CDmin is the zero lift (friction/parasitic) drag – CL2 gives the lift induced drag – e is Oswald’s eﬃciency factor ≈ 0.7 − 0.85 (1) (2) Fall 2004 16.333 1–4 • Total drag then given by � 1 2 1 2 � 2 D = ρV SCD = ρV S CDmin + kCL 2 2 1 2 (mg)2 = ρV SCDmin + k 1 2 2 2 ρV S D Total drag No-lift drag Lift-dependent drag VS 1 VEmd VE • So that the speed for minimum drag is � � �1/4 k 2mg Vmin drag = ρS CDmin (3) (4) Fall 2004 16.333 1–5 Steady Climb T-D L V � T D � R/C W • Equations: T − D − W sin γ = 0 L − W cos γ = 0 � which gives T −D− (5) (6) L cos γ = 0 sin γ so that tan γ = T −D L • Consistent with 1–3 if T = 0 since then γ as deﬁned above is negative • Note that for small γ, tan γ ≈ γ ≈ sin 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 ﬂight) R/C = V sin γ ≈ V γ ≈ Fall 2004 16.333 1–6 Steady Turn L sin � L cos � L � Centrifugal force R Radius of turn � W • Equations: L sin φ = centrifugal force mV 2 = R L cos φ = W = mg V2 ⇒ tan φ = Rg V =Rω = (8) (9) Vω g • Note: obtain Rmin at CLmax WV 2 1 2 Rmin( ρV SCLmax ) sin φ = 2 g ⇒ Rmin = (7) W/S 1/2ρgCLmax sin φmax where W/S is the wing loading and φmax < 30◦ (10) Fall 2004 16.333 1–7 • Deﬁne load factor N = L/mg. i.e. ratio of lift in turn to weight N = sec φ = (1 + tan2 φ)1/2 � tan φ = N2 − 1 so that V2 V2 R= = √ g tan φ g N 2 − 1 • For a given load factor (wing strength) R ∝V2 • Compare straight level with turning ﬂight – If same light coeﬃcient L mg 1 CL = 1 2 = 1 2 ≡ N mg ρVt2S 2 2 ρV S 2 ρV S √ so that Vt = N V gives the speed increase (more lift) • Note that CL constant ⇒ CD constant ⇒ D ∝ V 2CD ⇒ Tt ∝ Dt ∝ Vt2CD ∼ N D so that must increase throttle or will descend in the turn (11) (12)

16.333: Lecture #1

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