ENME 488: Flight & Spaceflight Mechanics PDF

ENME 488: Flight & Spaceflight Mechanics John Cater, [email protected] Department of Mechanical Engineering University of Canterbury, New Zealand Aerodynamic Moments The moment vector is given by Q = [L, M, N], where:...

ENME 488: Flight & Spaceflight Mechanics John Cater, [email protected] Department of Mechanical Engineering University of Canterbury, New Zealand Aerodynamic Moments The moment vector is given by Q = [L, M, N], where: → − 1. Rolling Moment, L [Nm], in the lift-crosswind plane 2. Pitching Moment, M [Nm], in the lift-drag plane 3. Yaw Moment, N [Nm], in the drag-crosswind plane 1 Introduction to the Theory of Flight - Damon Honnery, Monash University Australia Rotational Motion Moment Equations The moment equations relate the responses (LHS) to the applied moments (RHS): D " → −# d dt = → − Q inertial where: D Inertial r V → − − × m→  − =→  which represents the angular momentum of the vehicle. The time derivativeQ→ − in general can be expressed as: Q =→ → − D + −Ω × → −̇ → D − 2 Angular Acceleration If the inertia tensor is constant, then the moment becomes: Q → − = I − → → −̇ → ·Ω+Ω× − → I − → − ·Ω This can be rearranged to make the rate of change of angular velocity the subject: → −̇ I → − Q I − → h→ − → − → − i Ω = −1 · −Ω× ·Ω This can be used to determine the angular acceleration along the flight path in the body axis sytem, and expanded to include the thrust and aerodynamic moments: → −˙ I→ − −1 h− Ω Kf = f · Q M Q →F −−→− →A −−→ → f + fa a − ΩK × f · ΩK f I − −−→i f 3 Inertia Tensor The inertia tensor is given by:   Ix −Ixy −Ixz I→ − = −Ixy Iy −Iyx    −Ixz −Ixz Iz 4 Inertia Tensor Components Diagonal components are the moments of inertia that quantify the resistance to rotation Zaround each axis: y 2 + z 2 dm
Ix = Z x 2 + z 2 dm
Iy = Z x 2 + y 2 dm
Iz = The non-diagonal elements are called the products of inertia: Z Ixy = (x · y ) dm Z Ixz = (x · z) dm Z Iyz = (y · z) dm 5 Products of Inertia The products of inertia indicate asymmetry of the mass distribution: For a symmetric vehicle, terms containing y will be zero. Yechout, T. R. (2003). Introduction to aircraft flight mechanics. AIAA. 6 Symmetric Vehicles? By NASA/ - http://www.nasa.gov/centers/dryden/about/Organizations/Technology/Facts/TF-2004-01-DFRC. htmlhttp://www.nasa.gov/centers/dryden/images/content/86987main_TF-2004-01_popup1.jpghttp: //www1.dfrc.nasa.gov/Gallery/Photo/AD-1/HTML/ECN-13305.html,PublicDomain,https: //commons.wikimedia.org/w/index.php?curid=495396 7 Simplified Inertia Tensor For left-right symmetrical aircraft, Ixy = Iyz = 0. In this case, the inertia tensor beccomes:   Ix 0 −Ixz I → − =  0 Iy  0   −Ixz 0 Iz In the special case that the body-fixed axes are defined in the direction of the main inertia axes (quite unlikely): Ixz = 0 8 Angular Moment For linear momentum, Newton’s Law gives: F → − = m→ − a The angular moment can be found using the moments of → − I → −̇ inertia , and the angular acceleration Ω. In a reference frame attached to the vehicle, this is: Q =→ → − I · −̇Ω − → 9 Angular Acceleration The angular acceleration can be found by inverting the inertia tensor: → − −→ → Ω̇ = −1 · I− Q For a symmetric vehicle, this can be written:    Iz 0 Ix IzIxz−I 2   ṗ 2 Ix Iz −Ixz xz L 1 q̇  =  0 0  M       Iy Ixz Ix r˙ Ix Iz −I 2 0 Ix Iz −I 2 N xz xz 10 Angular Moment Equations The rate of change of angular momentum in an inertial frame can be found in a similar way to the forces, which gives the following for the moment responses: L = ṗIx + qr (Iz − Iy ) − (r˙ + pq)Ixz M = q̇Iy + pr (Iz − Ix ) + (p 2 − r 2 )Ixz N = r˙Iz + pq(Iy − Ix ) + (qr − ṗ)Ixz The three terms on the RHS side of the angular moment equations represent the angular acceleration, gyro precession and coupling terms. 11 Moment Response - Angular Accleration Consider the rolling moment: ṗIx + qr (Iz − Iy ) − (r˙ + pq)Ixz = L Assuming gyro and coupling terms are small: ṗIx = L This implies that if a rolling moment is applied, angular acceleration in roll ṗ will result. The larger the moment of inertia about the roll axis Ix , the smaller the roll acceleration. 12 Moment Response - Gyroscopic Precession Gyroscopic precession is rotation at an angle to the axis. Precission occurs due to a combination of angular momentum and an applied moment/torque. ṗIx + qr (Iz −Iy ) − (r˙ + pq)Ixz = L 13 Moment Response - Gyroscopic Precession Example Consider a rolling pull-up manoeuvre with a resulting rolling moment, initially assuming that the angular acceleration and coupling terms are negligible, as is Iz : −Iy qr = L The angular momentum term for pull-up (with positive q) is Iy q, so the equation can be rewritten: −r (Iy q) = L If a positive rolling moment is applied (ailerons) there will be a negative yaw rate r , which contributes to adverse yaw: https://www.youtube.com/watch?v=D9cIof2O6Mc. 14 Moment Response - Coupling Terms The coupling terms describe the inertial coupling relationships for the vehicle; how disturbances in one axis create motion in others. These terms can be identified by the inertial products, e.g. Ixy , which describe asymmetry in intertial resistance. q̇Iy + pr (Iz − Ix ) + (p 2 − r 2 )Ixz = M 15 Moment Response - Coupling Terms Example Consider the pitching moment equation of motion with negligible gyro precession and applied pitching moment, and zero yaw rate r : p 2 Ixz = −Iy q̇ Many aircraft will have a negative Ixz , which means more off-axis mass distribution towards the back of the vehicle. This relationship predicts that a roll in either direction will cause a pitch up. This is called roll coupling. 16 Applied Moments The applied moments consist of roling, pitching and yawing moments due to aerodynamic forces, LA , MA , NA , and due to thrust LF , MF , NF. There is no contribution from the weight vector, as it passes through the centre of mass. If moments from rotating machinery (e.g. engines) are assumed to be balanced, the moment equation becomes: ṗIx + qr (Iz − Iy ) − (r˙ + pq)Ixz = LA + LF q̇Iy + pr (Iz − Ix ) + (p 2 − r 2 )Ixz = MA + MF r˙Iz + pq(Iy − Ix ) + (qr − ṗ)Ixz = NA + NF This gives six equations of motion in total (so far). 17 In-Plane Motion Longitudinal Equations of Motion Three of the equations apply to the longituinal (x-z) plane. These are the x force, z force and y moment equations: m (u̇ + qw − rv ) = −mg sin(Θ) + [−W cos(α) + A sin(α)] + F cos(iF ) q̇Iy + pr (Iz − Ix ) + (p 2 − r 2 )Ixz = MA + MF m (ẇ + pv − qu) = mg cos(Φ) cos(Θ) + [−W sin(α) − A sin(α)] − F sin(iF ) These are all that is required for wings-level flight. 18 Lateral Equations of Motion The remaining forces are the y force, x moment and z moment equations: ṗIx + qr (Iz − Iy ) − (r˙ + pq)Ixz = LA + LF m (v̇ + ru − pv ) = mg sin(Φ) cos(Θ) + RAY + RFY r˙Iz + pq(Iy − Ix ) + (qr − ṗ)Ixz = NA + NF Any lateral motion will cause the xz plane to move from an initial xz plane. 19 Kinematic Equations Kinematic Conditions Three additional equations are required to close the equations of motion, as there are more unknowns than equations, due to the Euler angles in the force equations. ˆ These are found from kinematic conditions, that relate the rotational moments to the Euler angles The first equation is from the equality: → − → −̇ →−̇ →−̇ Ω f = p iˆ + q jˆ + r k̂ = Ψ + Θ + Φ This says the magnitude of the three body rotation rates must equal the magnitude of the three Euler rates. Also: p p p 2 + q 2 + r 2 = Ψ̇2 + Θ̇2 + Φ̇2 20 Coordinate Transforms The coordinate transform matrices defined previously can be used to transform the angular rate vectors from the Earth axis system to the body axis system: → − → −̇ → −̇ → −̇ Ω f = p iˆ + q jˆ + r k̂ = Mx (Φ)My (Θ) Ψ + Mx (Φ) Θ + Φ These transformations are: → −̇ ˆ To transform Ψ requires a positive rotation through Θ, followed by a positive rotation through Φ → −̇ ˆ Θ requires a positive rotation only through Φ → −̇ ˆ Φ is already in the body axis 21 Kinematic Equations The required three kinematic equations are: p = − sin(Θ)Ψ̇ + Φ̇ q = sin(Φ) cos(Θ)Ψ̇ + cos(Φ)Θ̇ r = cos(Φ) cos(Θ)Ψ̇ − sin(Φ)Θ̇ 22 Kinematic Equations Example An aircraft has the following Euler angles and Euler rates: Ψ = 0 deg , Ψ̇ = 10 deg /s Θ = 0 deg , Θ̇ = 0 deg /s Φ = 90 deg , Φ̇ = 0 deg /s What are the body rates? Applying the kinematic equations: p = 0 deg /s q = 10 deg /s r = 0 deg /s In this case, a 10 deg/s Euler yaw rate is felt as a 10 deg/s pitch rate by the pilot. 23 Summary Key Concepts: ˆ Moments of inertia ˆ Moment equations ˆ Adverse yaw and rolling moment Key Skills: ˆ Explain the effects of disturbances on vehicle motion 24