16.333 Lecture 4 Aircraft dynamics . Aircraft non linear eom Linearization -dynamics . Linearization- forces moments o Stability derivatives and coefficients
16.333 Lecture 4 Aircraft Dynamics • Aircraft nonlinear EOM • Linearization – dynamics • Linearization – forces & moments • Stability derivatives and coefficients
Fa2004 16.3334-1 Aircraft Dynamics Note can develop good approximation of key aircraft motion( Phugoid using simple balance between kinetic and potential energies Consider an aircraft in steady level flight with speed Uo and height ho. The motion is perturbed slightly so that U0→U=U70+ ho→h=ho+△h perturbation. It then follows that u N mi t before and after the Assume that e=muz+ mgh is constar e From Newton 's laws we know that. in the vertical direction mh=L-w where weight W= mg and lift L= PSclUZ(S is the wing area) We can then derive the equations of motion of the aircraft mh=l-w OPSCL(U2 PSCL((Uo+u)2-U0x psCi(2uUo)(4 PSCL/94h U ( osce)△h(5) Since h= Ah and for the original equilibrium flight condition L W=3(pSCLU=mg, we get that ScL=2 Combine these result to obtain △+Ω2△h=0,≈V These equations describe an oscillation(called the phugoid oscilla tion of the altitude of the aircraft about it nominal value Only approximate natural frequency (Lanchester ), but value close
Fall 2004 16.333 4–1 Aircraft Dynamics • Note can develop good approximation of key aircraft motion (Phugoid) using simple balance between kinetic and potential energies. • Consider an aircraft in steady, level flight with speed U0 and height h0. The motion is perturbed slightly so that U0 → U = U0 + u (1) h0 → h = h0 + Δh (2) • Assume that E = 1mU2 + mgh is constant before and after the 2 perturbation. It then follows that u ≈ −gΔh U0 • From Newton’s laws we know that, in the vertical direction mh¨ = L − W 1 where weight W = mg and lift L = 2ρSCLU2 (S is the wing area). We can then derive the equations of motion of the aircraft: ¨ 1 mh = L − W = ρSCL(U2 − U0 2 ) (3) 2 1 = ρSCL((U0 + u) 2 − U0 2 ) ≈ 1 ρSCL(2uU0)(4) 2 � � 2 gΔh ≈ −ρSCL U0 = −(ρSCLg)Δh (5) U0 ¨ ¨ Since h = Δh and for the original equilibrium flight condition L = W 1 = 2(ρSCL)U2 = mg, we get that 0 � �2 ρSCLg g = 2 m U0 Combine these result to obtain: Δh¨ + Ω2 Δh = 0 , Ω ≈ g √ 2 U0 • These equations describe an oscillation (called the phugoid oscillation) of the altitude of the aircraft about it nominal value. – Only approximate natural frequency (Lanchester), but value close
Fa2004 16.3334-2 The basic dynamics are F=mi and th 1F=元+Bb× u. Transport thm m B T=互+B×H e Basic assumptions are 1. Earth is an inertial reference frame 2. A/ C is a rigid body 3. Body frame B fixed to the aircraft(i, j Instantaneous mapping of ie and Blw into the body frame: BW=Pi+Qj+Rk 1c=Ui+Vj+Wk B Uc)B R W By symmetry, we can show that Iru= luz =0, but value of Irz depends on specific frame selected. Instantaneous mapping of the angular momentum H=Hri+hui+hzk into the Body Frame given by H Ix0Ix21「P H h 0I2 yy 0 H 1x201
� k) � � � Fall 2004 16.333 4–2 • The basic dynamics are: �˙ I � ˙ I F = mvc and T� = H 1 �˙ B F = vc + ω × �vc Transport Thm. BI ⇒ � m � ˙ B BI ⇒ T � � = H + ω × H� • Basic assumptions are: 1. Earth is an inertial reference frame 2. A/C is a rigid body 3. Body frame B fixed to the aircraft ( �i,�j, BI • Instantaneous mapping of �vc and ω� into the body frame: BIω� = P�i + Q�j + R� k �vc = U�i + V�j + W� k ⎡ ⎤ ⎡ ⎤ P U ⇒ BIω ⎦ B = ⎣ Q ⎦ ⇒ (vc)B = ⎣ V R W • By symmetry, we can show that Ixy = Iyz = 0, but value of Ixz depends on specific frame selected. Instantaneous mapping of the angular momentum H = Hx �i + Hy �j + Hz � k into the Body Frame given by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ Hx Ixx 0 Ixz P HB = ⎣ Hy ⎦ = ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ Hz Ixz 0 Izz R
Fa2004 16.3334-3 The overall equations of motion are then F Bl × 0-RQ1「U V+R 0-PIv W Q P 0w 「U+QW-Rv V+RU-Pn W+PV-QU T=互+B×H Imap+lzr 0-RQ1「Ix0Ix21「P →M +|R0-P 01m0Q 2+I2P QP0」LIx20 IP+I2R+QR(I2z-Iyy)+ pQlcz Q +PR(Iz -I2)+(R2- p2)Izz I22r+Ix P +PQ(Iwy-Ixa-QRlzz Clearly these equations are very nonlinear and complicated and we have not even said where f and t come from Need to linearize!! Assume that the aircraft is flying in an equilibrium condition and we will linearize the equations about this nominal flight condition
� 1 Fall 2004 16.333 4–3 • The overall equations of motion are then: 1 �˙ B F = vc + BIω� × �vc m ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ X U˙ 0 −R Q U ⇒ m ⎣ Y ⎦ = ⎣ V˙ ⎦ + ⎣ R 0 −P ⎦ ⎣ V ⎦ Z W˙ −Q P 0 W ⎡ ⎤ U˙ + QW − RV = ⎣ V˙ + RU − PW ⎦ W˙ + PV − QU � ˙ B BI T � � = H + ω × H� ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ L ˙ IxxP˙ + IxzR 0 −R Q Ixx 0 Ixz P ⇒ ⎣ M ⎦ = ⎣ ˙ IyyQ˙ ⎦ + ⎣ R 0 −P ⎦ ⎣ 0 Iyy 0 ⎦ ⎣ Q ⎦ N IzzR + IxzP˙ −Q P 0 Ixz 0 Izz R ⎡ ⎤ ˙ IxxP˙ + IxzR +QR(Izz − Iyy) + PQIxz = ⎣ I ⎦ yyQ˙ +PR(Ixx − Izz) + (R2 − P2)Ixz ˙ IzzR + IxzP˙ +PQ(Iyy − Ixx) − QRIxz • Clearly these equations are very nonlinear and complicated, and we have not even said where F� and T� come from. =⇒ Need to linearize!! – Assume that the aircraft is flying in an equilibrium condition and we will linearize the equations about this nominal flight condition
Fa2004 16.3334-4 Axes But first we need to be a little more specific about which Body Frame we are going use. Several standards 1. Body Axes-X aligned with fuselage nose. Z perpendicular to X in plane of symmetry(down). Y perpendicular to XZ plane, to le rig 2. Wind Axes-X aligned with vc. Z perpendicular to X(pointed down). Y perpendicular to XZ plane, off to the right 3. Stability Axes-X aligned with projection of vc into the fuselage plane of symmetry. Z perpendicular to X(pointed down). Same S(BODY) X-AXIS (STABILITY) BODY X-AXIS Z-AXIS (WIND Advantages to each, but typically use the stability axes In different flight equilibrium conditions, the axes will be oriented differently with respect to the a/c principal axes = need to trans form (rotate) the principal inertia components between the frames When vehicle undergoes motion with respect to the equilibrium Stability Axes remain fixed to airplane as if painted on
Fall 2004 16.333 4–4 Axes • But first we need to be a little more specific about which Body Frame we are going use. Several standards: 1. Body Axes X aligned with fuselage nose. Z perpendicular to X in plane of symmetry (down). Y perpendicular to XZ plane, to the right. 2. Wind Axes X aligned with �vc. Z perpendicular to X (pointed down). Y perpendicular to XZ plane, off to the right. 3. Stability Axes X aligned with projection of �vc into the fuselage plane of symmetry. Z perpendicular to X (pointed down). Y same. R E LATIV E WIND ( ) ( ) ( ) � � B ODY Z-AXIS B ODY Y -AXIS X-AXIS WIND X-AXIS S T AB ILIT Y X-AXIS B ODY • Advantages to each, but typically use the stability axes. – In different flight equilibrium conditions, the axes will be oriented differently with respect to the A/C principal axes ⇒ need to transform (rotate) the principal inertia components between the frames. – When vehicle undergoes motion with respect to the equilibrium, Stability Axes remain fixed to airplane as if painted on