16.333 Lecture # 8 Aircraft Lateral Dynamics Spiral, Roll, and Dutch Roll Modes ? ? ? ? Fall 2004 16.333 7–1 Aircraft Lateral Dynamics ? Using a procedure similar to the longitudinal case, we can develop the equations of motion for the lateral dynamics ?? v ? ? ? ? δ a , u = δ r x˙ = Ax + Bu , x = ? ? ? ? p r φ and ψ ˙ = r sec θ 0 ?? A = ? ? ? ? ? ? ? ? ? ? Y v Y p Y r m ? U 0 g cos θ 0 m m ( I L ? v + I ? N v ) ( L p + I ? N p ) ( L r + I ? N r ) 0 zx zx I ? zx I ? xx xx xx (I ? L v + N v ) (I ? L p + N p ) (I ? L r + N r ) 0 zx I ? zx zx I ? I ? zz zz zz 0 1 tan θ 0 0 where I ? = (I xx I zz ? I 2 xx zx )/I zz I ? = (I xx I zz ? I 2 zz zx )/I xx I ? = I zx /(I xx I zz ? I 2 zx zx ) and ?? ? ? ? ? (m) ?1 0 0 0 (I ? xx ) ?1 I ? zx ? ? ? ? ?? Y δ a Y δ r ? ? ? ? ? ? ? ? ? ? ? ? B = L δ a L δ r zz ) ?1 · zx (I 0 I N δ a N δ r 0 0 0 ? ? ? ? Fall 2004 16.333 7–2 Lateral Stability Derivatives ? A key to understanding the lateral dynamics is roll-yaw coupling. ? L p rolling moment due to roll rate: – Roll rate p causes right to move wing down, left wing to move up → Vertical velocity distribution over the wing W = py – Leads to a spanwise change in the AOA: α r (y) = py/U 0 – Creates lift distribution (chordwise strips) 1 δL w (y) = ρU 0 2 C l α α r (y)c y dy 2 – Net result is higher lift on right, lower on left – Rolling moment: b/2 b/2 L = δL w (y)·(?y)dy = ? 2 1 ρU 0 2 ?b/2 C l α py 2 c y dy ? L p < 0 ?b/2 U 0 – Key point: positive roll rate ? negative roll moment. ? L r rolling moment due to yaw rate: – Positive r has left wing advancing, right wing retreating → Horizontal velocity distribution over wing U = U 0 ? ry – Creates lift distribution over wing (chordwise strips) 1 1 L w (y) ～ ρU 2 C l cdy ≈ ρ(U 0 2 ? 2U 0 ry)C l c y dy 2 2 – Net result is higher lift on the left, lower on the right. b/2 b/2 – Rolling Moment: L = L w (y)·(?y)dy ≈ ρU 0 r C l c y y 2 dy ?b/2 ?b/2 – For large aspect ratio rectangular wing (crude) 1 1 L r ≈ ( to )C L > 0 6 4 – Key point: positive yaw rate ? positive roll moment. Fall 2004 16.333 7–3 ? N p yawing moment due to roll rate: – Rolling wing induces a change in spanwise AOA, which changes the spanwise lift and drag. – Distributed drag change creates a yawing moment. Expect higher drag on right (lower on left) → positive yaw moment – There is both a change in the lift (larger on downward wing be- cause of the increase in α) and a rotation (leans forward on down- ward wing because of the larger α). → negative yaw moment – In general hard to know which e?ect is larger. Nelson suggests that for a rectangular wing, crude estimate is that 1 N p ≈ ρU 0 2 Sb(? C L ) < 0 2 8 ? N r yawing moment due to yaw rate: – Key in determining stability properties – mostly from ?n. – Positive r has ?n moving to the left which increases the apparent angle of attack by rl f Δα f = (U 0 ) f – Creates increase in lift at the tail ?n by 1 ΔL f = ρ(U 0 2 ) f S f C L α f Δα f 2 – Creates a change in the yaw moment of 1 N = ?l f ΔL f = ? 2 ρ(U 0 ) f S f C L α f rl 2 f 1 – So N r = 2 ρ(U 0 ) f S f C L α f l f 2 < 0? – Key point: positive yaw rate ? negative yaw moment. L N p < 0 ? r > 0 < 0 Fall 2004 16.333 7–4 Numerical Results ? The code gives the numerical values for all of the stability derivatives. Can solve for the eigenvalues of the matrix A to ?nd the modes of the system. ?0.0331 ± 0.9470i ?0.5633 ?0.0073 – Stable, but there is one very slow pole. ? There are 3 modes, but they are a lot more complicated than the longitudinal case. Slow mode -0.0073 ? Spiral Mode Fast real -0.5633 ? Roll Damping Oscillatory ?0.0331 ± 0.9470i Dutch Roll ? Can look at normalized eigenvectors: Spiral Roll Dutch Roll β = w/U 0 0.0067 -0.0197 0.3269 -28 ? ?p = p/(2U 0 /b) -0.0009 -0.0712 0.1198 92 ? ?r = r/(2U 0 /b) 0.0052 0.0040 0.0368 -112 ? φ 1.0000 1.0000 1.0000 0 ? Not as enlightening as the longitudinal case. Fall 2004 16.333 7–5 Lateral Modes Roll Damping - well damped. – As the plane rolls, the wing going down has an increased α (wind is e?ectively “coming up” more at the wing) – Opposite e?ect for other wing. – There is a di?erence in the lift generated by both wings → more on side going down – The di?erential lift creates a moment that tends to restore the equilibrium. Recall that L p < 0 – After a disturbance, the roll rate builds up exponentially until the restoring moment balances the disturbing moment, and a steady roll is established. py V 0 -py ? ?' ?' Roll Rate p Disturbing rolling moment Restoring rolling moment V 0 Port wing Starboard wing Reduction in incidence Reduction in incidence Fall 2004 16.333 7–6 Spiral Mode - slow, often unstable. – From level ?ight, consider a disturbance that creates a small roll angle φ > 0 → This results in a small side-slip v (vehicle slides downhill) – Now the tail ?n hits on the oncoming air at an incidence angle β → extra tail lift → positive yawing moment – Moment creates positive yaw rate that creates positive roll mo- ment (L r > 0) that increases the roll angle and tends to increase the side-slip → makes things worse. – If unstable and left unchecked, the aircraft would ?y a slowly diverging path in roll, yaw, and altitude ? it would tend to spiral into the ground!! ? Can get a restoring torque from the wing dihedral ? Want a small tail to reduce the impact of the spiral mode. Fall 2004 16.333 7–7 Dutch Roll - damped oscillation in yaw, that couples into roll. ? Frequency similar to longitudinal short period mode, not as well damped (?n less e?ective than horizontal tail). ? Consider a disturbance from straight-level ?ight → Oscillation in yaw ψ (?n provides the aerodynamic sti?ness) → Wings moving back and forth due to yaw motion result in oscil- latory di?erential lift/drag (wing moving forward generates more lift) L r > 0 → Oscillation in roll φ that lags ψ by approximately 90 ? ? Forward going wing is low Oscillating roll ? sideslip in direction of low wing. Fall 2004 16.333 7–8 ? Do you know the origins on the name of the mode? ? Damp the Dutch roll mode with a large tail ?n. Fall 2004 16.333 7–9 Aircraft Actuator In?uence 10 ?2 10 ?1 10 0 10 ?2 10 ?1 10 0 10 1 10 2 |G b d a | Freq (rad/sec) 10 ?2 10 ?1 10 0 10 ?2 10 ?1 10 0 10 1 10 2 |G p d a | Freq (rad/sec) Transfer function from aileron to flight variables 10 ?2 10 ?1 10 0 10 ?2 10 ?1 10 0 10 1 10 2 |G r d a | Freq (rad/sec) 10 ?2 10 ?1 10 0 ?200?150?100 ?50 0 50 100150200 arg G b d a Freq (rad/sec) 10 ?2 10 ?1 10 0 ?350?300?250?200?150?100 ?50 0 arg G p d a Freq (rad/sec) 10 ?2 10 ?1 10 0 ?200?150?100 ?50 0 50 100150200 arg G r d a Freq (rad/sec) Figure 1: Aileron impulse to ?ight variables. Response primarily in φ. ? Transfer functions dominated by lightly damped Dutch-roll mode. ? Note the rudder is physically quite high, so it also in?uences the A/C roll. ? Ailerons in?uence the Yaw because of the di?erential drag Fall 2004 16.333 7–10 10 ?2 10 ?1 10 0 10 0 10 1 10 2 10 3 10 4 |G b d a | Freq (rad/sec) 10 ?2 10 ?1 10 0 10 ?2 10 ?1 10 0 10 1 10 2 |G p d a | Freq (rad/sec) Transfer function from rudder to flight variables 10 ?2 10 ?1 10 0 10 ?2 10 ?1 10 0 10 1 10 2 |G r d a | Freq (rad/sec) 10 ?2 10 ?1 10 0 ?200?150?100 ?50 0 50 100150200 arg G b d a Freq (rad/sec) 10 ?2 10 ?1 10 0 ?500?400?300?200?100 0 arg G p d a Freq (rad/sec) 10 ?2 10 ?1 10 0 ?300?200?100 0 100200 arg G r d a Freq (rad/sec) Figure 2: Aileron impulse to ?ight variables. Response primarily in φ. Fall 2004 16.333 7–11 0 5 10 15 20 25 30?4 ?2 0 2 x 10 ?4 b rad Aileron 1 deg Impulse ? 2sec on then off 0 5 10 15 20 25 30?4 ?2 0 2 x 10 ?3 p rad/sec da > 0 ==> right wing up 0 5 10 15 20 25 30?5 0 5 x 10 ?4 r rad/sec Initial adverse yaw ==> RY coupling 0 5 10 15 20 25 30?0.01 ?0.005 0 f rad time sec Figure 3: Aileron impulse to ?ight variables ? Aileron δ a =1deg impulse for 2 sec. – Since δ a > 0 then right aileron goes down, and right wing goes up → Reid’s notation, and it is not consistent with the picture on 6–4 (from Nelson). – In?uence of the roll mode seen in the response of p to application and release of the aileron input. – See e?ect of adverse yaw in the yaw rate response caused by the di?erential drag due to aileron de?ection. – Spiral mode harder to see. – Dutch mode response in other variables clear (1 rad/sec ～ 6 sec period). Fall 2004 16.333 7–12 0 5 10 15 20 25 30?0.01 0 0.01 0.02 b rad Rudder 1 deg step 0 5 10 15 20 25 30?0.1 0 0.1 p rad/sec 0 5 10 15 20 25 30?0.1 0 0.1 r rad/sec 0 5 10 15 20 25 30?2 ?1 0 1 f rad time sec Figure 4: Rudder step to ?ight variables ? Rudder step input 1deg step. – Dutch roll response very clear. Other 2 modes are much less pronounced. – β shows a very lightly damped decay. – p clearly excited as well. Doesn’t show it, but often see evidence of adverse roll in p response where initial p is opposite sign to steady state value. Reason is that the forces act on the ?n which is well above the cg → and the aircraft responds rapidly (initially) in roll. – φ ultimately oscillates around 2.5 ?