Lecture 11 Overview of Flight Dynamics I. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Transcription

1 Lecture 11 Overview of Flight Dynamics I Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

2 Point Mass Dynamics Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

3 Basic Assumptions for Point Mass Model 1. All forces act at the CG of the airplane 2. Acceleration due to gravity (g) is constant 3. Atmosphere is at rest relative to earth 4. Atmospheric 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

4 Point Mass Model for Flat (and Non-rotating) Earth Kinematic Equations Resolving the velocity vector along the local horizontal x = cosγ Resolving the velocity vector along the local vertical h = sinγ 4

5 Point Mass Model for Flat (and Non-rotating) Earth Dynamic Equations Resolving forces along the velocity vector m = T cosα D W sinγ Resolving forces to the velocity vector m γ = T sinα+ L W cosγ 5

6 Point Mass Model for Flat (and Non-rotating) Earth x h = = cosγ sinγ Note : 1 = Tcos D mgsin m ( α γ ) 1 γ = Tsinα + L mgcosγ m x equation is not coupled with others. ( ) 6

7 Point Mass Model for Spherical, Non-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 r = sinγ r θ = cosγ 7

8 Point Mass Model for Spherical, Non-Rotating Earth Dynamic Equations Resolving force components along the velocity vector m = T cosα D mg sinγ Resolving the force components to the velocity vector m γ = T sinα+ L m mg cosγ + R 2 8

9 Point Mass Model for Spherical, Non-Rotating Earth r θ = = sinγ cosγ r 1 = T D mg m ( cosα sin γ) Note : θ equation is not coupled with others. 1 γ = Tsinα + L mgcosγ + m m 2 R 9

10 Point Mass Model for Spherical and Rotating Earth When thrust is absent, the kinematic and dynamic equations are cosγ sinψ cosγ cosψ r = sin γ, ϕ =, θ = r rcosϕ D 2 = gsinγ +Ω ercosϕ( sinγ cosϕ cosγsinϕsinψ) m Lcosσ g cosγ cosγ γ = + + 2Ωe cosϕcosψ m r 2 Ωer + cos ϕ(cosγ cosϕ+ sinγsinϕsin ψ) Lsinσ ψ = 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

11 Point Mass Model for Spherical and Rotating Earth Where r Radial Distance from Center of Earth Earth Relative elocity Ω e Earth Angular Speed γ Flight Path Angle σ elocity Roll / Bank Angle ψ elocity Yaw / Heading Angle m Mass of ehicle g Acceleration due to Gravity θ Longitude ϕ Geocentric Latitude σ 11

12 Six DOF Model Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

13 Basic Force Balance Weight Lift Drag Side force Thrust Side force Dr. Radhakant Padhi, AE Dept., IISc-Bangalore 13

14 Basic Moment Balance Rolling Y Pitching Yawing X Z Dr. Radhakant Padhi, AE Dept., IISc-Bangalore 14

15 Control Action: Aileron Roll Dr. Radhakant Padhi, AE Dept., IISc-Bangalore 15

16 Control Action: Elevator Pitch Dr. Radhakant Padhi, AE Dept., IISc-Bangalore 16

17 Control Action: Rudder Yaw Dr. Radhakant Padhi, AE Dept., IISc-Bangalore 17

18 Six-DOF Model Assumptions: 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

19 Six-DOF Model From Geometry: r = r + r P is center of mass/gravity. Hence: ( ) p r ρ A A d = r r ρ d = 0 p CG 0 r ρ d = r ρ d = r m A p A p 1 r p = r Ad m ρ dm ds d 19

20 Dynamic Equations Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

21 Force and Moment Equations Newton s Second Law of Motion: (valid in inertial reference frame) dm ds d d dt d dt dr ρ d = ρ g d + F ds A A dt S Linear momentum Gravity Force Aerodynamic/Thrust Force dr r ρ d = r ρa gd A dt Angular momentum Gravity Moment + r Fds S Aerodynamic/Thrust Moment 21

22 Force Equation (inertial frame) d dr d d d ( ρ = ρ r + r ) d dt dt dt dt A A p dm ds d d d = r p ρa d + r ( ρa d ) dt dt dm m = 0 d d dr ( ) d p dp = mr p = m = m dt dt dt dt dt p 22

23 Force Equation (inertial frame) Moreover S ρagd= g ρad= mg m FdS= X + X T Hence Aerodynamic force Thrust force dp m = mg+ X + X dt T Applied in inertial frame dm ds d 23

24 Force Equation (body frame) Thrust forces are typically applied in body frame Aerodynamic 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 NOT an inertial frame and the Newton s laws are not applicable directly. 24

25 Force Equation (body frame) A Standard Result : da dt Equivalent in inertial frame A = + ω A t As seen in rotating frame where ω : Angular velocity of rotating frame wrt. inertial frame. A : Any vector Hence the force equation in body frame becomes : p m ( + ω p) = mg+ X + XT t B Forced applied in rotating (body) frame 25

26 Force Equation (body frame) T ω = = B T X = X Y Z X = X Y Z T g = g g g [ P Q R], ( p) [ U W] [ ], [ ] [ ] ( ω ) X Y Z i j k p = P Q R U W T T T T = i QW R j PW UR + k P UQ Y X Z ( ) ( ) ( ) T T 26

27 Force Equation (body frame) ( + ) = X + ( + T) mu R WQ mg X X ( + ) = Y + ( + T) m UR WP mg Y Y ( + ) = Z + ( + T) m W UQ P mg Z Z where g g Y Z gx = gsin Θ = gcosθsin Φ = gcosθcosφ Note: The gravity components can also be Formally derived from Euler angle definitions. mg cos Θ 27

28 d dt Moment Equation dr r ρ d = r Fds S A dt Modified angular momentum (for rotating frame effect) Comment: Applied moment in the body frame = M A + Aerodynamic moment M T Thrust moment This 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

29 Moment Equation d dr d dr r ρ d = r ρad dt dt dt dt A dr dr d dr = + r ρ A dt dt dt dt = 0 d = r r + ω r ρa d dt = 0 d 29

30 Moment Equation d dr r ρ d = r ( ω r) + ω ( ω r) ρad dt dt t A Hence, the moment equation is: = r ω r + ω r + ω ( ω r) ρa d = 0 = r ω r + ω ω r ρ d (( ) ( )) r r + r d = M + M (( ω ) ω ( ω )) ρa A T A 30

31 Moment Equation Standard Results : ( r ω r) ρ ( ) ( ) Ad = ω r r r r ω ρad r ( ω ( ω r) ) ρad = r ω( ω r) r( ω ω) ρad However, r = x y z ω = T [ ] [ P Q R] T = r ω ω r ρ d ( ) (in body frame) A 31

32 Moment Equation ( r ω r ) ρ ( ) ( ) Ad = ω r r r r ω ρ d A dm = ω ( ω) 2 rdm r r dm T ( = P Q R x + y + z ) dm x y z xp + yq + zr dm [ ] T ( ) 32

33 Moment Equation ( 2 2 P y + z ) dm Q xy dm R xz dm Ixx Ixy Ixz r ( ω r) d Q x z dm P yxdm R yzdm Iyy Iyx Iyz ( 2 2 R x + y ) dm P zx dm Q zy dm Izz Izx I zy ( 2 2 ρ ) A = + 33

34 Moment Equation I xxp IxyQ IxzR r ( r) ω ρad = IyyQ IyxP IyzR I zzr IzxP IzyQ Similarly r ω ω r ρ d = r ω ω r ρ d ( ( )) A ( ) A 2 2 xx + yz ( ) xz + ( zz yy ) 2 2 ( xx zz ) xz ( ) xy yz 2 2 ( ) + ( ) + I PR I R Q I PQ I I RQ = I I PR+ I P R I QR+ I PQ Iyy Ixx PQ Ixy Q P IxzQR IyzPR 34

35 Moment Equation Assumption: An airplane is symmetric about its XZ-plane I xy = I = yz 0 Note: Missiles and launch vehicles are typically symmetric about both XZ-plane as well as XY-plane I = I = I = xy yz zx 0 35

36 Moment Equation Aero Moment: [ ] T M A = L M N Thrust Moment: M = L M N [ ] T T T T T Q Hence, the moment equations are: I P I R I PQ+ I I RQ= L+ L ( ) xx xz xz zz yy T I Q + I I PR+ I P R = M + M ( ) ( 2 2) yy xx zz xz T I R I P + I I PQ+ I QR = N + N ( ) P R zz xz yy xx xz T 36

37 Force and Moment Equations mu R WQ mg X X ( + ) = sin Θ+ ( + ) T m UR WP mg Y Y ( + ) = cos Θsin Φ+ ( + ) T m W UQ P mg Z Z ( + ) = cos Θcos Φ+ ( + ) T I P I R I PQ+ I I RQ= L+ L ( ) xx xz xz zz yy T I Q + I I PR+ I P R = M + M ( ) ( 2 2) yy xx zz xz T I R I P + I I PQ+ I QR = N + N ( ) zz xz yy xx xz T 37

38 Force and Moment Equations 1 U = R WQ gsin Θ+ ( X + XT ) m 1 = WP UR+ gsin Φcos Θ+ Y + YT m 1 W = UQ P+ gcos Φcos Θ+ Z + Z m ( ) ( ) ( ) ( ) P = cqr+ c PQ+ c L+ L + c N + N T 4 Q = c PR c P R + c M + M ( 2 2) ( T ) ( ) ( ) R = c PQ c QR+ c L+ L + c N + N T 9 T U P T T Q W R 38

39 Force and Moment Equations where 2 c1 Izz( Iyy Izz) Ixz c 2 xz( zz xx yy I I + I I ) c3 1 Izz = ( 2 c ) 4 II xx zz Ixz Iyz 2 c8 Ixx( Ixx Iyy) + I xz c I 9 xx c = I I )/ I 5 c = I / I c 6 7 = 1/ I zz xx yy xz yy yy 39

40 Force and Moment Equations X Y N = T cos Φ cos Ψ T i Ti Ti i= 1 N = T cos Φ sin Ψ T i Ti Ti i= 1 N Z = T sin Φ T i Ti i= 1 N N ( cos sin ) ( sin ) L = T Φ Ψ z T Φ y T i Ti Ti Ti i Ti Ti i= 1 i= 1 N N ( cos cos ) ( sin ) M = T Φ Ψ z + T Φ x T i Ti Ti Ti i Ti Ti i= 1 i= 1 N N ( cos cos ) ( cos sin ) N = T Φ Ψ y + T Φ Ψ x T i Ti Ti Ti i Ti Ti Ti i= 1 i= 1 C 1 D C O D C D C X X s i ( ) ( )( ) h α Dδ T α T 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 δa δr δ A T( ) T( ) qsb N = α α β N = + s Cn Cn C β δ n δ R A δ R δ A Y = qs CY = qs CY β + CY C β δ Y A δr δ R [ 1 α ] M = qsc C = qsc C C C i + C T m m m m h m δ E o α i h δ E T ( α ) cosα sinα T = T σ i i T max sinα cosα i 40

41 Comment We have derived only the set of dynamic equations of the Six- DOF model, which describe the effect of forces and moments. A set of kinematic equations are also needed to complete the Six-DOF model, which will be discussed in the next class. 41

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