Optimal Control, Guidance and Estimation. Lecture 16. Overview of Flight Dynamics II. Prof. Radhakant Padhi. Prof. Radhakant Padhi

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1 Optimal Control, Guidance and Estimation Lecture 16 Overview of Flight Dynamics II Prof. Radhakant Padhi Dept. of erospace Engineering Indian Institute of Science - Bangalore Point Mass Dynamics Prof. Radhakant Padhi Dept. of erospace Engineering Indian Institute of Science - Bangalore

2 Basic ssumptions for Point Mass Model 1. ll forces act at the CG of the airplane 2. cceleration due to gravity (g) is constant 3. tmosphere is at rest relative to earth 4. tmospheric properties are functions of altitude only 5. Forces acting on airplane are thrust, aerodynamic forces and its weight 6. ehicle attitude is ignored only direction of velocity vector is considered. 3 Point Mass Model for Flat (and on-rotating) Earth Kinematic Equations Resolving the velocity vector along the local horizontal xɺ = cosγ Resolving the velocity vector along the local vertical hɺ = sinγ 4

3 Point Mass Model for Flat (and on-rotating) Earth Dynamic Equations Resolving forces along the velocity vector mɺ = cosα D W sinγ Resolving forces to the velocity vector mɺ γ = sinα + L W cosγ 5 Point Mass Model for Flat (and on-rotating) Earth xɺ = cosγ hɺ = sin γ ote : 1 ɺ = D mg m ( cosα sin γ ) 1 ɺ γ = α + γ m xɺ equation is not coupled with others. ( sin L mg cos ) 6

4 Point Mass Model for Spherical, on-rotating Earth Kinematic Equations Using the force balance equation m( cosγ ) = m r R R = r cosγ 2 2 cosγ Resolving velocity vector along local vertical and horizontal rɺ = sinγ rɺ θ = cosγ 7 Point Mass Model for Spherical, on-rotating Earth Dynamic Equations Resolving force components along the velocity vector mɺ = cosα D mg sinγ Resolving the force components to the velocity vector mɺ γ = sinα + L mg cosγ + m R 2 8

5 Point Mass Model for Spherical, on-rotating Earth rɺ = sin γ ɺ cosγ θ = r 1 ɺ = D mg m ( cosα sin γ ) 1 m 2 ɺ γ = sinα + L mg cosγ + m R ote : ɺ θ equation is not coupled with others. 9 Point Mass Model for Spherical and Rotating Earth When thrust is absent, the kinematic and dynamic equations are cosγ sinψ cosγ cosψ rɺ = sin γ, ɺ ϕ =, ɺ θ = r r cosϕ D 2 ɺ = g sinγ + Ω er cosϕ( sinγ cosϕ cosγ sinϕ sinψ ) m ɺ Lcosσ g cosγ cosγ γ = + + 2Ωe cosϕ cosψ m r 2 Ωer + cos ϕ(cosγ cosϕ + sinγ sinϕ sin ψ ) Lsinσ ψɺ = cosγ cosψ tanϕ + 2 Ωe(tanγ cosϕ sinψ sinϕ) m cosγ r 2 Ω sin cos cos cos e r ϕ ϕ ψ γ 10

6 Point Mass Model for Spherical and Rotating Earth Where r Radial Distance from Center of Earth Earth Relative elocity Ωe Earth ngular Speed γ Flight Path ngle σ elocity Roll / Bank ngle ψ elocity Yaw / Heading ngle m Mass of ehicle g cceleration due to Gravity θ Longitude ϕ Geocentric Latitude σ 11 Six DOF Model Prof. Radhakant Padhi Dept. of erospace Engineering Indian Institute of Science - Bangalore

7 Basic Force Balance Weight Lift Drag Side force hrust Side force 13 Basic Moment Balance Rolling Pitching Y Yawing X Z 14

8 Control ction: ileron Roll 15 Control ction: Elevator Pitch 16

9 Control ction: Rudder Yaw 17 Six-DOF Model Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ssumptions: Flat earth (spherical and rotational effects are negligible) Rigid body (no relative motion of particles, no spinning rotors) Constant mass and mass distribution (no fuel burning, no fuel slosh, no passenger movement etc.) Uniform mass density Constant gravity Body axis: rotating and moving Inertial axis: non-rotating and non-moving dm ds d 18

10 Six-DOF Model Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 From Geometry: r = r + r P is center of mass/gravity. Hence: ( p ) r ρ d = r r ρ d = 0 p CG 0 r ρ d = r ρ d = r m p p 1 r p = r ρ d m dm ds d 19 Dynamic Equations Prof. Radhakant Padhi Dept. of erospace Engineering Indian Institute of Science - Bangalore

11 Force and Moment Equations Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ewton s Second Law of Motion: (valid in inertial reference frame) dm ds d d dt d d r ρ d = ρ g d + F ds dt dt S Linear momentum Gravity Force erodynamic/hrust Force d r r ρ d = r ρ g d dt ngular momentum Gravity Moment + r F ds S erodynamic/hrust Moment 21 Force Equation (inertial frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 d d r d d ρ d = ρ ( r + r ) d dt dt dt dt p dm ds d d d = r p ρ d + r ( ρ d ) dt dt dm m = 0 d d d d r p dp = ( mr p ) = m = m dt dt dt dt dt p 22

12 Force Equation (inertial frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 Moreover S ρ g d = g ρ d = mg m F ds = X + X Hence erodynamic force hrust force dp m = mg + X + X dt pplied in inertial frame dm ds d 23 Force Equation (body frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 hrust forces are typically applied in body frame erodynamic forces act on wind frame, which is close to body frame (they are same when α=β=0) Body frame is a rotating frame. Hence it is O an inertial frame and the ewton s laws are not applicable directly. 24

13 Force Equation (body frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 Standard Result : d = + ω dt t Equivalent in inertial frame s seen in rotating frame where ω : ngular velocity of rotating frame wrt. inertial frame. : ny vector Hence the force equation in body frame becomes : p m + ( ω p) = mg + X + X t B Forced applied in rotating (body) frame 25 Force Equation (body frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ω = = B X = X Y Z X = X Y Z g = g g g [ P Q R], ( p) [ U W ] [ ], [ ] [ ] ( ω ) X Y Z i j k p = P Q R U W X ( ) ( ) ( ) = i QW R j PW UR + k P UQ Y Z 26

14 Force Equation (body frame) Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ( ɺ + ) = X + ( + ) ( ɺ + ) = Y + ( + ) ( ɺ + ) = Z + ( + ) m U R WQ mg X X m UR WP mg Y Y m W UQ P mg Z Z where g g Y Z g X = g sin Θ = g cosθsin Φ = g cosθcosφ ote: he gravity components can also be Formally derived from Euler angle definitions. mg cos Θ 27 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 d dt d r r ρ d = r F ds S dt Modified angular momentum (for rotating frame effect) Comment: pplied moment in the body frame = M + M erodynamic moment hrust moment his expression can be derived from the earlier expression. However, it is easier to visualize this equation directly in the body frame since, forces and moments act on the body frame and gravity force does not create any moment about CG. 28

15 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 d d r d d r r ρ d = r ρ d dt dt dt dt d r d r d d r = + r ρ d dt dt dt dt = 0 d = r rɺ + ω r ρ d dt = 0 29 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 d d r r ρ d = r ( ω r ) + ω ( ω r ) ρ d dt dt t = r ɺ ω r + ω rɺ + ω ( ω r ) ρ d = 0 = r ω r + ω ω r ρ d ( ɺ ( ) ( )) Hence, the moment equation is: r r + r d = M + M (( ɺ ω ) ω ( ω )) ρ 30

16 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 Standard Results : r ( ɺ ω r ) ρ d = ɺ ω ( r r ) r ( r ɺ ω) ρ d r ( ω ( ω r )) ρ d = r ω ( ω r ) r ( ω ω) ρ d However, r = x y z ω = [ ] [ P Q R] = r ω ω r ρ d ( ) (in body frame) 31 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ( ɺ ) ɺ ( ) ( ɺ ) r ω r ρ d = ω r r r r ω ρ d dm ( ɺ ω) = ɺ ω 2 r dm r r dm ( ) = Pɺ Qɺ Rɺ x + y + z dm [ ] ( ɺ ɺ ɺ ) x y z xp + yq + zr dm 32

17 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, Pɺ ( y + z ) dm Qɺ xy dm Rɺ xz dm I xx I xy I xz r r d Q x z dm P yx dm R yz dm I yy I yx I yz 2 2 Rɺ ( x + y ) dm Pɺ zx dm Qɺ zy dm I zz Izx I zy 2 2 ( ɺ ω ) ρ = ɺ ( + ) ɺ ɺ 33 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 I xxp ɺ I xyq ɺ I xzr ɺ r ( ɺ ω r ) ρ d = I yyq ɺ I yxp ɺ I yzr ɺ I zzr ɺ I zxp ɺ I zyq ɺ Similarly r ω ω r ρ d = r ω ω r ρ d ( ( )) ( ) 2 2 xx + yz ( ) xz + ( zz yy ) 2 2 ( xx zz ) xz ( ) xy yz 2 2 ( yy xx ) + xy ( ) + xz yz I PR I R Q I PQ I I RQ = I I PR + I P R I QR + I PQ I I PQ I Q P I QR I PR 34

18 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ssumption: n airplane is symmetric about its XZ-plane I xy = I = 0 ote: Missiles and launch vehicles are typically symmetric about both XZ-plane as well as XY-plane yz I xy = I yz = I zx = 0 35 Moment Equation Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ero Moment: M = L M hrust Moment: M = L M [ ] [ ] Hence, the moment equations are: ( ) 2 2 ( ) ( ) ( ) I Pɺ I Rɺ I PQ + I I RQ = L + L xx xz xz zz yy I Qɺ + I I PR + I P R = M + M yy xx zz xz I Rɺ I Pɺ + I I PQ + I QR = + zz xz yy xx xz P Q R 36

19 Force and Moment Equations Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 ( ɺ + ) = sin Θ + ( + ) ( ɺ + ) = cos Θsin Φ + ( + ) ( ɺ + ) = cos Θcos Φ + ( + ) ɺ ɺ + ( ) = + ɺ ( ) + ( ) = + ɺ ɺ + ( ) + = + m U R WQ mg X X m UR WP mg Y Y m W UQ P mg Z Z I P I R I PQ I I RQ L L xx xz xz zz yy I Q I I PR I P R M M yy xx zz xz I R I P I I PQ I QR zz xz yy xx xz 37 Force and Moment Equations Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, Uɺ = R WQ g sin Θ + ( X + X ) m 1 ɺ = WP UR + g sin Φ cos Θ + Y + Y m 1 Wɺ = UQ P + g cos Φ cos Θ + Z + Z m ( ) ( ) ( ) ( ) Pɺ = c QR + c PQ + c L + L + c ( ) ( ) Qɺ = c PR c P R + c M + M ( ) ( ) Rɺ = c PQ c QR + c L + L + c Q U P W R 38

20 Force and Moment Equations Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 where c1 Izz( Iyy Izz) Ixz c2 Ixz Izz + Ixx Iyy ( ) c3 1 Izz = 2 c4 ( IxxIzz Ixz ) Iyz 2 c 8 Ixx Ixx Iyy I xz ( ) + c I 9 xx 2 c = I I ) / I 5 c = I / I c 6 7 = 1/ I zz xx yy xz yy yy 39 Force and Moment Equations Ref: Roskam J., irplane Flight Dynamics and utomatic Controls, 1995 X = cos Φ cos Ψ i i i i= 1 Y = cos Φ sin Ψ i i i i= 1 Z = sin Φ i i i= 1 ( cos sin ) ( sin ) L = Φ Ψ z Φ y i i i i i i i i i = 1 = 1 ( cos cos ) ( sin ) M = Φ Ψ z + Φ x i i i i i i i i i = 1 = 1 ( cos cos ) ( cos sin ) = Φ Ψ y + Φ Ψ x i i i i i i i i i i = 1 = 1 C 1 D C O D C D C X X s i ( ) ( )( ) h α Dδ α α qs α E E Z = δ Z = + s CL C O L C L C α i h Lδ i E h L Ls Cl Cl Cl β δ δ R δ ( ) ( ) qsb = α α β = + s Cn Cn C β δ n δ R δ R δ Y = qs CY = qs CY β + CY C β δ Y δr δ R [ 1 α ] M = qsc C = qsc C C C i + C m mo m mi h h m δ α δ E E ( α ) cosα sinα = σ i imax i sinα cosα 40

21 Comment We have derived only the set of dynamic equations of the Six- DOF model, which describe the effect of forces and moments. set of kinematic equations are also needed to complete the Six-DOF model, which will be discussed in the next class. 41 OPIML COROL, GUIDCE D ESIMIO Prof. Radhakant Padhi, E 42