AE 430 - Stability and Control of Aerospace Vehicles
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AE 430 - Stability and Control of Aerospace Vehicles Longitudinal Static Stability Tailplanes Typically a wing alone will have –ve CM0 (Cm at zero lift) (positive camber airfoil) It may have a if CG is aft of AC By positioning a tailplane behind the wing and inclining it to give –ve lift – It improves the stability of the wing-tail combination -ve lift reduces overall lift of aircraft, so wings must carry more lift Results in increase of induced drag, plus drag of tailplane – Trim Drag Usually symmetrical airfoil are used for the tail because must produce both upward and downward airloads 1 Wing-Horizontal Tail Geometry lt Lt αwb - ε Tail zero z lift line Wing/body t i zero lift line t αwb Mac,t V∞ V∞ Dt V’ αt ε – Wing flow field interferes with horizontal tail downwash deflects V∞ downward local relative wind is reduced in magnitude; tail “sees” lower dynamic pressure – Fuselage blanks out part of the tail Downwash ε The value of the downwash at the tail is affected by fuselage geometry, flap angle wing platform, and tail position. It is best determined by measurement in a wind tunnel, but lacking that, lifting surface computer programs do an acceptable job. For advanced design purposes it is often possible to approximate the downwash at the tail by the downwash far behind an elliptically-loaded wing: 2CL ∂ε 2CL ε ≈ w So, ≈ α w π ARw ∂α π ARw 2 Tail Contribution – Aft Tail αFRL − ε lt z cgt Wing-tip vortex Wing-tip vortices are formed when high-pressure air spills up over the wing tips into the low-pressure space above the wing. 3 Pressures must become equal at the wing tips since pressure is a continuous function (figure a). The free stream flow combines with tip flow, resulting in an inward flow of air on the upper wing surface and an outward flow of air on the lower wing surface (figure b). Complete-wing vortex system. Flow field around an airplane created by the wing 4 Tail Contribution – Aft Tail αt = αw − iw −ε + it Angle of attach at the tail L = Lw + Lt Total lift wing-tail configuration 1 ρV 2S S C = C + 2 t t C = C +η t C L Lw 1 2 Lt Lw S Lt 2 ρVw S 1 2 2 ρVt Qt η = = Tail efficiency – ratio of dynamic pressure 1 2 Q 2 ρVw w Tail Contribution – Aft Tail αFRL − ε lt z cgt Mt = −lt Lt cos(αFRL −ε ) + Dt sin (αFRL −ε )⁄ + z »D cos α −ε − L sin α −ε ÿ + M cgt t ( FRL ) t ( FRL )⁄ act M = −l L = −l 1 ρV 2S C = −l Q S C t t t t 2 t t Lt t t t Lt 5 Tail Contribution – Aft Tail M = −l L = −l 1 ρV 2S C = −l Q S C cgt t t t 2 t t Lt t t t Lt M 1 2 cgt 2 ρVt St St Cm = = −lt CL = −ltη CL = −VHηCL cg t 1 2 1 2 Sc t Sc t t 2 ρV Sc 2 ρV St V = l Horizontal tail volume ratio H t Sc C = C α = C α − i − ε + i Lt Lαt t Lαt ( w w t ) Downwash dε ε = ε + α 0 dα w Downwash at zero angle of attach 2C Lw dε 2CL ε = (rad) Finite wing theory = α w π ARw dα π ARw ε 2ε 6 Tail Contribution – Aft Tail Cm = −VHηCL cg t t Cm = −VHηCL αw − iw −ε + it cg t α t ( ) dε ’ Cm = VHηCL (ε0 + iw − it ) −VHηCL ∆1− ÷αw cg t αt αt dα S V = l t H t Sc Cm = Cm + Cm αw cg t 0t αt To have C > 0 we can adjust the tail incidence angle m0t it C < 0 we can select properly V H and C to stabilize the aircraft mα t Lαt (CLαt can be increased for example, by increasing the aspect ratio of the tail) Example # 3 Consider the wing-body model in Example # 2 (previous class). Assume that a horizontal tail with no elevator is added to this model. The wing area and chord are 1.5 m2 and 0.45 m, respectively. The distance from the airplane's center of gravity to the tail's aerodynamic center is 1.0 m. The area of the tail is 0.4 m2, and the tail-setting angle is - 2.0°. The lift slope of the tail is 0.12 per degree. From experimental measurement, ε0= 0 and dε/dα = 0.42. If the absolute angle of attack of the model is 5°and the lift at this angle of attack is 4134 N, calculate the moment about the center of gravity. q = 6125N / m2 ; C = −0.003; i = 0 ∞ m ac,w w xcg x − ac,w = 0.02; η = 1 (assumed) c c 7 L 4134 CLw = = = 0.45 q∞S 6125*1.5 dC 0.45 C = Lw = = 0.09 Lα ,w dα 5 l S 1.0*0.4 V = t t = = 0.593 H cS 0.45*1.5 xcg x CL d ’ … ac,wb α ,t ε Ÿ Cmcg = Cmac,wb + CLα ,wbαwb − −VH ∆1− ÷ c c CLα ,wb dα ⁄ +VH CLα ,t (ε0 − it ) » 0.12 ÿ Cm = −0.003+ 0.09*5…(0.02) − 0.593 (1− 0.42)Ÿ cg 0.09 ⁄ + 0.593*0.12(2) = −0.058 Mcg = q∞ ScCmcg = 6125*1.5*0.4*(−0.058) = −240N / m Example # 4 Consider the wing-body-tail model of Example # 3. Does this model have longitudinal static stability and balance? » ÿ ∂Cmcg xcg xac,wb CLα ,t ≈ ∂ε ’ = C … − −V ∆1− ÷Ÿ Negative Lα ,wb H « ◊ ∂αwb … c c CLα ,wb ∂α ⁄Ÿ slope; O K » 0.12 ÿ = 0.09 …0.02 − 0.593 (1− 0.42)Ÿ= −0.039 0.09 ⁄ C = C +V C ε − i m,0 mac,wb H Lα ,t ( 0 t ) = −0.003+ 0.593*0.12*(2.0) = 0.139 Positive C m ,0; O K Cm, = 0.139 − 0.039αe = 0 cg Equilibrium angle-of- 0 αe = 3.56 attack; reasonable; O K 8 Canard Alternatively put the tail ahead of the wing – known as foreplanes or canards Canard needs to generate +ve lift to create +ve pitching moment about CG This means that canard contributes to overall lift of aircraft Canard can create fast increase in lift, making aircraft very responsive Canard interferes with airflow over main wing – causes resultant force vector of wing to tilt backward – increasing drag Fuselage Contribution Streamlined fuselage has pressure distribution similar to body of revolution No net lift developed by pressure distribution Nose-up pitching moment developed by up-gust Pitching moment is destabilizing because not countered by net lift vector Fuselage is destabilizing component 9 Fuselage Contribution dM = fn Vol, Q = 1 ρV 2 dα ( 2 ) k − k l f C = 2 1 w2 α + i dx m0 — f 0w f f 36.5Sc 0 ( ) x=l k − k f C = 2 1 ƒ w2 α + i ∆x m0 f f ( 0w f ) 36.5Sc x=0 1 l f ∂ε C = w2 u dx deg-1 mα — f f 36.5Sc 0 ∂α ( ) x=l f 1 2 ∂εu -1 Cm = ƒ wf ∆x deg α f ( ) 36.5Sc x=0 ∂α See textbook (pp. 53-55) for details Fuselage Contribution 10 Fuselage Contribution Fuselage contribution Gilruth (NACA TR711) developed an empirically-based method for estimating the effect of the fuselage: ∂C 2 m fuse K f wf Lf = ∂CL SwcCL where: αw CLαw is the wing lift curve slope per radian Lf is the fuselage length wf is the maximum width of the fuselage Kf is an empirical factor discussed in NACA TR711 and developed from an extensive test of wing-fuselage combinations in NACA TR540. Kf is found to depend strongly on the position of the quarter chord of the wing root on the fuselage. In this form of the equation, the wing lift curve slope is expressed in -1 rad and Kf is given in the table. (Note that this is not the same as the method described in Perkins and Hage.) The data shown in table were taken from TR540 and Aerodynamics of the Airplane by Schlichting and Truckenbrodt: 11 Power Effects Propeller Jet Engine C Both will produce a contribution to mα Propulsive System Contribution The incremental pitching moment about the airplane center of gravity due to the propulsion system is: where T is the thrust and Np is the propeller or inlet normal force due to turning of the air. Another influence comes from the increase in flow velocity induced by the propeller or the jet slipstream upon the tail, wing and aft fuselage. 12 Propulsive System contribution In terms of moment coefficient, Since the thrust is directed along the propeller axis and rotates with the airplane, its contribution to the moment about the center of gravity is independent of αw. Then we have N = N qS C 1− ε α p prop p N pα ( α ) where the propeller normal force coefficient CNp/ α and the downwash (or upwash) εα are usually determined empirically Propulsive System contribution Nprop is the number of propellers and Sprop is the propeller disk area (πD2/4) and D is the diameter of the propeller. Note that a propeller mounted aft of the c.g. is stabilizing. This is one of the advantages of the pusher- propeller configuration. Note that n in is the propeller angular speed in rps .