AE 430 - Stability and ontrol of Aerospace Vehicles Static/Dynamic Stability Longitudinal Static Stability Static Stability We begin ith the concept of Equilibrium (Trim). Equilibrium is a state of an object hen it is at rest or in steady uniform motion, (i.e., ith constant linear and angular momenta). The resultant of all forces and moment about the G must both be equal to zero. Stability is defined as the ability of an aircraft to return to a given equilibrium state after a disturbance (it is a property of the equilibrium state) STATIALLY STABLE hen if it is disturbed from its equilibrium state by a small displacement, then the set of forces and moments so caused initially tend to return the aircraft to its original state 1
Trimmed Flight (or steady unaccelerated flight) Trimmed flight hen all the forces and moments are balanced (trust = drag; lift = eight; pitching moment = 0; yaing moment = 0 rolling moment = 0) Forces = 0; Moments = 0 The steady flight condition may involve a steady acceleration e.g. a correctly banked turn, or a steady dive or climb. Pitch trim ould be accomplished by deflecting the horizontal stabilizer, the elevator, or the elevator trim tab. M G = Trimmed state IS NOT NEESSARILY A STABLE STATE i.e. all the forces and moments may be balanced, but as soon as the state is perturbed the aircraft departs from equilibrium. 0 for trim Types of Stability 2
Static Stability Static stability of a body is an initial tendency of that body to return to its equilibrium state after a disturbance. Energy is being dissipated Positive damping Static longitudinal instability In this case there is no tendency to return to equilibrium Any disturbance from equilibrium leads to a larger disturbance, the motion is said to be divergent Artificial damping is needed Stability Augmentation System SAS Neutral static stability is the boundary beteen stability and instability, there is still no tendency to return to equilibrium, the motion is therefore not stable But, the motion does not diverge Energy is added to the system Negative damping E E Static Stability 3
Dynamic Stability DYNAMI STABILITY characterizes the time history of motion after a disturbance from equilibrium An aircraft is said to be dynamically stable if, after a disturbance, it eventually returns to its equilibrium state and remains there ABSOLUTE dynamic stability is not concerned ith ho long this return takes RELATIVE dynamic stability examines ho long it takes and hat the behavior of that return motion is To be dynamically stable, a system must first be statically stable A system can be dynamically unstable and be statically stable -- but not vice versa Dynamic Stability 4
Dynamic Stability Pilot-Induced Oscillation PIOs occur hen a pilot over-controls an aircraft and a sustained oscillation results Pilot-induced oscillations occur hen the pilot of an aircraft inadvertently commands an often increasing series of corrections in opposite directions, each an attempt to correct for the previous overcorrection ith an overcorrection in the opposite direction. The physics of flight make such oscillations more probable for pilots than for automobile drivers. An attempt to cause the aircraft to climb, say by applying up elevator ill also result in a reduction in airspeed. Another factor is the response rate of flight instruments in comparison to the response rate of the aircraft itself. An increase in poer ill not result in an immediate increase in airspeed. An increase in climb rate ill not sho up immediately on the vertical speed indicator. A pilot aiming for a 500 foot per minute descent, for example, may find himself descending too rapidly. He begins to apply up elevator until the vertical speed indicator shos 500 feet per minute. Hoever, because the vertical speed indicator lags the actual vertical speed, he is actually descending at much less than 500 feet per minute. He then begins applying don elevator until the vertical speed indicator reads 500 feet per minute, starting the cycle over. It's harder than it might seem to stabilize the vertical speed because the airspeed also constantly changes. The most dangerous pilot-induced oscillations can occur during landing. A bit too much up elevator during the flare can result in the plane getting dangerously slo and threatening to stall. A natural reaction to this is to push the nose don harder than one pulled it up, but then the pilot finds himself staring at the ground. An even larger amount of up elevator starts the cycle over again. http://.dfrc.nasa.gov/gallery/movie/f-8dfbw/html/em-0044-01.html 5
Statically Stable Response nose up Balanced positive pitch stiffness (restoring moment) nose don Equilibrium point Other necessary condition to trim at positive angle of attach, α m0 > 0 d dm dm dl mα = dα = For static stability m 0 dl dα d < L Longitudinal Static Stability nose up (+) nose don (-) Longitudinal static stability moments as a function of angle of attack. The curve is a composite of all the moment curves caused by the different components of the airplane, (the ing, fuselage, tail, thrust, etc). Longitudinal static stability components 6
Stable, neutral, and unstable static stability D-9. Note the contributions from the various components and the highly nonlinear post-stall characteristics 7
There are different degrees of stability Some aircraft tend to return to equilibrium faster An aircraft can be stable at loer angles of attack but may be unstable at higher angles of attack Wing ontribution 8
Wing ontribution Moments = M cg ( α ) sin ( α ) M cg = L cos i x cg xac+ D i x cg xac ( α ) cos( α ) + L sin i z D i z + M cg cg ac 1 2 Dividing for ρv Sc : 2 xac xac mcg = L cos( ) D sin ( ) α i + α i c c c c zcg zcg + L sin ( α i ) D cos( α i ) + mac c c Wing ontribution ( ) ( ) cos α i 1; sin α i α i ; L D xac zcg mcg = L ( ) + L α i + mac c c c zcg L ( α i ) negligible c x x ac cg xac mcg = m L ; m m ( L0 ) ac + cg = ac + + L α α c c c c = + α Lift oefficient α L L0 L Normal flight operation Well designed aircraft 9
Nonlinear contributions xac xac mcg = m L D ( ) ac + + α i c c c c zcg + L ( α i ) D c α α D = 0 + 2 α L L L 2 L = d + π ear Wind drag turn Wing ontribution xac m = 0 mac + L 0 c c mcg = m + 0 m α α xac m = α Lα c c α < < xac > 0 To have a ing alone statically stable m 0 To be able to trim the aircraft at positive angle of attach m0 10
Wing ontribution Positive camber give nose-don pitching moment Negatively cambered airfoil gives nose-up pitching moment and cancels nose-don moment caused by lift and eight vectors For straight-inged, tailless airplane, negative camber satisfies conditions for stable, balanced flight Not in general use Dynamic characteristics poor Drag and lmax poor Sept back ing ith tisted tips 11
onventional and forard tail arrangement Tailless Aircraft One example of a tailless aircraft that trims using a positive m0 airfoil section: the AeroVironment Pathfinder, solarpoered aircraft on a flight to over 50,000 ft (15.2 km). 12
Example # 1 For a given ing-body combination, the aerodynamic center lies 0.03 chord length ahead of the center of gravity. The moment coefficient about the center of gravity is 0.0050, and the lift coefficient is 0.50. alculate the moment coefficient about the aerodynamic center. xac, M = ( ) cg, M + ac, L c c xac, M = ( ) ac, M cg, L c c = 0.005 0.5(0.03) = 0.01 M ac, Example # 2 onsider a model of a ing-body shape mounted in a ind tunnel. The flo conditions in the test section are standard sea-level properties ith a velocity of 100 m/s. The ing area and chord are 1.5 m 2 and 0.45 m, respectively. Using the ind tunnel force and moment-measuring balance, the moment about the center of gravity hen the lift is zero is found to be -12.4 N m. When the model is pitched to another angle of attack, the lift and moment about the center of gravity are measured to be 3675 N and 20.67 N m, respectively. alculate the value of the moment coefficient about the aerodynamic center and the location of the aerodynamic center. 13
Example # 2 ont d 1 1 q V 0.225*100 6125 N / m 2 2 M cg, 12.4 m = = = 0.003 cg, q Sc 6125*1.5* 0.45 2 2 2 = ρ = = m = = 0.003 at zero lift m cg, ac, Example # 2 ont d L m L 3675 = = = 0.4 q S 6125*1.5 cg, Mcg, 20.67 = = = 0.005 q Sc 6125*1.5* 0.45 xac, m = m + ( ) cg, ac, L c c x ac, m m cg, ac, 0.005 ( 0.003) = = c c 0.4 xac, = 0.02 c c L 14