Flight Dynamics & Control Dynamics Near Equilibria

Similar documents
Satellite Orbits. Orbital Mechanics. Circular Satellite Orbits

+ r Position Velocity

Assistant Professor: Zhou Yufeng. N , ,

Radial geodesics in Schwarzschild spacetime

Electric Potential. and Equipotentials

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

1. The sphere P travels in a straight line with speed

defined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z)

On the Eötvös effect

Mark Scheme (Results) January 2008

Flight Dynamics & Control Equations of Motion of 6 dof Rigid Aircraft-Kinematics

Fluids & Bernoulli s Equation. Group Problems 9

2. The Laplace Transform

This immediately suggests an inverse-square law for a "piece" of current along the line.

13.5. Torsion of a curve Tangential and Normal Components of Acceleration

AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1

Optimization. x = 22 corresponds to local maximum by second derivative test

Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system

UNIT VII Central Force: Review Key

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information

10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =

9.4 The response of equilibrium to temperature (continued)

4-4 E-field Calculations using Coulomb s Law

Lectures # He-like systems. October 31 November 4,6

Energy Dissipation Gravitational Potential Energy Power

r a + r b a + ( r b + r c)

Chapter 28 Sources of Magnetic Field

Two dimensional polar coordinate system in airy stress functions

Chapter 2: Electric Field

AE 423 Space Technology I Chapter 2 Satellite Dynamics

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

Section 35 SHM and Circular Motion

APPENDIX 2 LAPLACE TRANSFORMS

π,π is the angle FROM a! TO b

B.A. (PROGRAMME) 1 YEAR MATHEMATICS

Solutions Practice Test PHYS 211 Exam 2

CHAPTER 2 ELECTRIC FIELD

Friedmannien equations

7.5-Determinants in Two Variables

Math 2142 Homework 2 Solutions. Problem 1. Prove the following formulas for Laplace transforms for s > 0. a s 2 + a 2 L{cos at} = e st.

DYNAMICS. Kinetics of Particles: Newton s Second Law VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

Important design issues and engineering applications of SDOF system Frequency response Functions

FI 2201 Electromagnetism

Course Updates. Reminders: 1) Assignment #8 available. 2) Chapter 28 this week.

Chapter 19 Webassign Help Problems

PHYS 601 HW 5 Solution. We wish to find a Fourier expansion of e sin ψ so that the solution can be written in the form

6. Gravitation. 6.1 Newton's law of Gravitation

Lecture 10. Solution of Nonlinear Equations - II

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD

STABILITY and Routh-Hurwitz Stability Criterion

LA0011_11GB. Formulas and Units. Rotation 2 W. W = work in Ws = J = Nm. = ang. velocity in rad./sec. f = frequency in rev./sec.

MAGNETIC EFFECT OF CURRENT & MAGNETISM

PHYSICS 211 MIDTERM I 22 October 2003

Design of an extended Luenberger observer for sensorless vector control of induction machines using virtual instrument

( ) ( ) Physics 111. Lecture 13 (Walker: Ch ) Connected Objects Circular Motion Centripetal Acceleration Centripetal Force Sept.

U>, and is negative. Electric Potential Energy

Chapter 6. NEWTON S 2nd LAW AND UNIFORM CIRCULAR MOTION. string

Chapter 6. NEWTON S 2nd LAW AND UNIFORM CIRCULAR MOTION

1. Viscosities: μ = ρν. 2. Newton s viscosity law: 3. Infinitesimal surface force df. 4. Moment about the point o, dm

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

The Wave Equation I. MA 436 Kurt Bryan

Electronic Supplementary Material

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

Topics for Review for Final Exam in Calculus 16A

1 Using Integration to Find Arc Lengths and Surface Areas

3.1 Magnetic Fields. Oersted and Ampere

Physics 1502: Lecture 2 Today s Agenda

PRACTICE EXAM 2 SOLUTIONS

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Forces and Accelerations. Seventh Edition CHAPTER

Boise State University Department of Electrical and Computer Engineering ECE470 Electric Machines

CONTRIBUTIONS TO THE THEORETICAL STUDY OF THE PRECISION SOWING MACHINES DYNAMICS

Effective stiffness of reinforced thin-webbed timber beams

ELECTRO - MAGNETIC INDUCTION

Solutions to Midterm Physics 201

Impulse and Momentum

Chapter 4 Kinematics in Two Dimensions

Physics 11b Lecture #11

DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

1.3 Using Formulas to Solve Problems

Spring-Pendulum Dynamic System

Multiple Experts with Binary Features

Homework: Study 6.2 #1, 3, 5, 7, 11, 15, 55, 57

( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )

1. The 0.1 kg particle has a speed v = 10 m/s as it passes the 30 position shown. The coefficient of kinetic friction between the particle and the

Data Structures. Element Uniqueness Problem. Hash Tables. Example. Hash Tables. Dana Shapira. 19 x 1. ) h(x 4. ) h(x 2. ) h(x 3. h(x 1. x 4. x 2.

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Continuous Charge Distributions

1.4 Using Newton s laws, show that r satisfies the differential equation 2 2

Lateral Dynamic Modeling of Hezarfen Unmanned Aerial Vehicle (UAV) and H Loop Shaping Robust Control System Design

Fourier-Bessel Expansions with Arbitrary Radial Boundaries

Chapter 4 Two-Dimensional Motion

Chapter #5 EEE Control Systems

Uniform Circular Motion

Michael Rotkowitz 1,2

SPH3UW/SPH4U Unit 3.2 Forces in Cetripetal Motion Page 1 of 6. Notes Physics Tool Box

The Formulas of Vector Calculus John Cullinan

Chapter 2. Review of Newton's Laws, Units and Dimensions, and Basic Physics

PhysicsAndMathsTutor.com

Transcription:

Flight Dynmic & Contol Dynmic Ne Equilibi Hy G. Kwtny Deptment of Mechnicl Engineeing & Mechnic Dexel Univeity

Outline Pogm Stight & Level Flight the pimy tedytte Petubtion Eqution fo Stight & Level Flight Longitudinl & Ltel Dynmic Coodinted Tun Stedy Sidelip

Pogm Detemine tim condition Define nd compute tedy-tte flight condition Stight nd level flight (cuie, climb, decend) Coodinted tun Stedy idelip Detemine petubtion model Smll petubtion fom tim condition Exmine dynmicl behvio

Reduction in Dimenion coodinte: x, y, z, φθψ,, velocitie: uvwpq,,,,, 12 tte Suppoe 1) the eth i flt nd 2) we ignoe vition of i denity Note: Thee popetie do not equie lineity ( x y z ) the dynmic e invint w..t. loction,, the dynmic e invint w..t. (inetil) heding Conequently, we could dop x, y, z, ψ nd tudy n 8 tte ytem Thee omitted vible cn lwy be included by dding the ppopite kinemtic eqution we often include z nd/o ψ

Reduced Eqution m u v qw inθ X + T m v = m pw u + mg coθinφ + Y m w qu pv coθcoφ Z ( I y I z ) q I xz pq 2 2 ( I x I y ) pq I xzq Ix Ixz p + L I y q = I z I x q + I xz p + M + Ixz Iz + N φ 1 inφtnθ coφtnθ p θ coφ inφ q = ψ inφecθ coφecθ T T

Equilibi The eqution e ognized in fit ode tndd fom, involving the tte vecto x, input vecto u, nd output vecto y. = = h( xu) E x x f xu, tte eqution y, output eqution A et of vlue ( x, u, y ) i clled n equilibium (o tim) point if it tifie: ( x u) (, ) = f, y = h x u We e inteeted in motion tht emin cloe to the equilibium point.

The Pimy Stedy Stte: Stight & Level flight The tight nd level flight condition equie equilibium flight long line pth with contnt flight pth ngle, contnt velocity V, zeo idelip nd wing level. Thu, we impoe the following condition fo tight nd level flight: Equilibium: ( α β ) u =, v =, w = V =, =, =, p =, q =, =, φ =, θ =, ψ = ω = Output: peed: V = V flight pth ngle: γ : = θ α = γ idelip: β = oll: φ = γ

Stight & Level Flight Popoition : An equilibium point tifying the tight nd level flight condition exit if nd only if thee exit α, δe, T which tify the equ (, α, δe ) i ( α γ ) T Z( V α δe ) M ( V, α, δe,) + TT = tion: X V mg n + + =,,, + mg co α + γ =, In thi ce the equilibium vlue of the tte nd contol e: longitudinl vible; tte: V = V, α = α, q =, θ = γ + α contol: δe = δe, T = T ltel vible; tte: β =, p =, =, φ =, contol: δ =, δ =

Lineiztion = + δ = + δ = + δ E( x ) δx = f x + δx( t), u + δu( t) Define: x t x x t, u t u u t, y t y y t The eqution become: Now, contuct Tylo eie fo f, h (, ) (, ) (, ) y + δy t = h x + δx t u + δu t (, ) (, ) f x u f x u f x + δx u + δu = f x u + δx + δu + hot x u h( x, u) h( x, u) h( x + δxu, + δu) = h( x, u) + δx + δu + hot x u Notice tht f x, u = nd h x, u = y, o f ( x, u ) f ( x, u ) δ = δ + δ x u h( x, u) h( x, u) δy = δx+ δu x u E x x x u Eδx = Aδx+ Bδu δy = Cδx+ Dδu

Petubtion fo Stight & Level Flight ~ longitudinl eqution mcoα mv inα mv inα V minα mv coα mv coα d α = Iy dt q 1 θ 2 V QSCX ( αδ, e) QSCx (, e) mg co V α αδ θ 2 V QSCZ ( αδ, e) QSCz α ( αδ, e) V coα α 2 2 V QS CM ( αδ, e ) QS Cm (, e ) V QS Cmq (, e ) q α αδ αδ 1 θ 1 QSCx δ ( αδ, ) e e QSCz δ ( αδ, ) e e T + T QSCm δ ( αδ, ) δ e e e

Petubtion fo Stight & Level Flight ~ ltel eqution mv β Ix Ixy d p = Ixy Iz dt 1 φ b b 2V β 2V b b ( α) 2V ( α) 2V ( α) b b ( α) ( α) ( α) QSCyβ α QS Cy α QS Cy α mu mg coθ β QS Cl QS Clp QS Cl p β QS Cn QS 2V Cnp QS 2V Cn β 1 tnθ φ Cy δ ( α) C yδ α Clδ C lδ C ( α) C ( α) α α δ + QS δ nδ nδ

Longitudinl Dynmic Ne tight nd level flight the genel eqution of motion my be divided into two et which e lmot decoupled. Longitudinl Dynmic : o olling v =, φ =, ψ =, p =, =. Longitudinl Vible : Motion in body x-z plne, without ywing uwq,,, θ, x, y the x- z plne i plne of ymmety Uncoupled longitudinl motion exit povided oto gyocopic effect e bent the flt eth ppoximtion i vlid m u qw inθ X Foce eqution: m mg m w = qu + + coθ Z Moment eqution: Iq = M y x coθ inθ u Kinemtic: =, θ q z = inθ coθ v

Ltel Dynmic Ltel Dynmic : ( w, u u, θ = θ, q=, α α ) Ltel Vible : motion in body x-y plne, no pitching vp,,, ψφ,, y motion e mll-tjectoy emin cloe to tight & level eodynmic co-coupling tem e negligible Uncoupled ltel motion exit povided oto gyocopic effect e bent the flt eth ppoximtion i vlid Foce eqution: co in Moment eqution: Kinemtic: mv = mu + mg θ φ+ Y I x I xz p L = Ixz I z N φ 1 coφtnθ = ψ p coφecθ y = coθ inφu + inφinθ inψ + coφcoψ v

Bnked Coodinted Tun The bnked tun i defined by the following condition: 1) equilibium - V, αβ, uvw,,, pq,, e contnt v qw inθ X + T = m pw u + mg coθinφ + Y qu pv coθcoφ Z I y I z q + I xz pq L q+ I p + M + TT I N x I y pq + I xzq = Iz Ix ( 2 2 ) xz

Bnked Coodinted Tun~2 2) bnked tun condition: the (inetil) ngul velocity i veticl nd contnt p inθ ω = q = coθinφ ω ω co θcoφ 3) coodinted tun condition - um of gvity nd inetil foce lie in plne of ymmety ( x z plne) mpw mu + mg coθinφ = pv co βinα V co β coα + g coθinφ = 4) climb condition: V = V, γ = γ

Bnked Coodinted Tun ~ 3 Thee e 12 eqution in 12 unknown V, αβ,, pq,,, θφ,, T, δe, δ, δ The fct tht the velocity i contnt in body fme, with contnt ngul velocity bout z inuue tht gound tck i cicul. The coodinted tun condition inue tht pilot nd penge will not expeience ny ide foce. A pilot chieve coodintion by uing the udde in conjunction with n intument clled tun coodinto which meue the diffeence between the inetil nd gvity foce cting long the y xi. The udde lo induce moment which countect the dvee yw moment eulting fom inceed (deceed) dg on the outide (inide) wing poduced by ileon poition nd which cn be ignificnt duing the olling phe of the tun.

Bnked Coodinted Tun ~ 4 ( pqvα θ) 5 eqution explicitly yield 5 unknown,,,, o in tem of the othe, o thee cn be eliminted, leving: pv co βinα V co βcoα + g coθinφ = v qw inθ X + T m + mg + Y = qu pv coθcoφ Z ( I y I z ) q + I xz pq 2 2 ( I x I y ) pq + I xzq L I z I x q + I xz p + M TT + = N with pq,, known function of ω

Bnked Coodinted Tun ~ 5 Conide the poibility of olution in which ω i mll ( R lge). Popoition : Thee exit olution coeponding to ω = if nd only if thee exit α, δe, T tifying ( α δe ) mg ( α + γ ) + T = M ( V α δe ) X V,, in, Z,,, + T = T ( V α δe ) mg ( α γ ),, + co + =, Thi i the ce of tight nd level flight in which the equilibium vlue of the tte nd contol e: In thi ce the equilibium vlue of the tte nd contol e: longitudinl vible; tte: V = V, α = α, q =, θ = γ + α contol: δ = δ, T = T ltel vible; e e tte: β =, p =, =, φ =, contol: δ =, δ =

Bnked Coodinted Tun ~ 6 Popoition : eqution nd α δ T Suppoe, e, contitute egul olution of the longitudinl equilibium ( α) Cy δ ( α) ( α) ( α) ( α) ( α) ( α) ( α) C yβ det Clβ Clδ Clδ Cnβ Cnβ Cnβ Then thee exit olution to the coodinted bnk tun eqution, with βδ,, δ mll nd αδ α δ, e, T ne, e, T. To fit ode tem, the olution i ωv tnφ = coγ g Cyβ ( α) Cy δ ( α) β Cyp ( α) Cy ( α) Clβ ( α) Clδ ( α) C lδ α δ = Clp α Cl α Cnβ ( α) Cnβ ( α) Cnβ ( α) δ Cnp ( α) Cn ( α) ωb coφ 2V Cmα Cm δ e α Cmq ωc inφ C Clq 2 ( ec 1 l C l ) C α δ δ = + e e V φ lw

Cowind The tedy idelip i flight condition tht my be ued duing lnding ppoche in the peence of cowind. It i n equilibium condition in the ene tht ll cceletion nd ngul te e zeo. In ddition, the icft tck line gound pth coeponding to y =, i.e., ligned with the x xi. Flight pth ngle nd peed e lo pecified. cowind β = ψ = Advee wind condition wee involved in 33% of 76 lnding ccident between 1984 nd 1987. The key iue with the cbbed ppoch i the pottouchdown, on the gound dynmic. Some combintion i genelly ued.

Cowind ~ 2 Conide n inetil (flt eth fixed) fme with coodinte on the ufce nd z down. The intent i to lnd long the x -xi (o we equie y = ). Aume the cowind velocity i v. Thu, we hve the following condition: Regulted output: y = y = v + V co βcoαcoθinψ + Vin β inφinθinψ + coφcoψ V = V + V co βinα coφinθinψ inφcoψ = γ : = θ α = γ Equilibium condition: V =, α =, β =, q =, θ =, p =, =, φ =, ψ = x y

Cowind ~ 3 Popoition : (cbbing olution) A olution to the gound tcking poblem exit with β =, φ = if nd only if thee exit α, δe nd T which tify (, α, δe ) in ( α γ ), (, α, δe ) co ( α γ ) M ( V, α, δe,) + TT = X V mg + + T = Z V + mg + = nd V v coγ 1 :,

Cowind ~ 4 Popoition : (tedy idelip) A olution to the gound tcking poblem exit with ψ = if nd only if thee exit α, δe nd T which tify: (, α, δ ) in( ), (,, ) co( e α γ α δe α γ ) M ( V, α, δe,) + TT = X V mg + + T = Z V + mg + = nd φ, β, δ, δ tifying mg coθinφ + Cyβ ( α) β + Cy δ ( α) δ + Cy δ α δ = QS ( α) β ( α) δ C α β + C α δ + C α δ = lβ lδ lδ C + C + C α δ = nβ nδ nδ v V + coφin β =,

Cowind Exmple A hypotheticl ubonic tnpot (dpted fom Etkin) h the following dt C = -.168, C =.67, C =, yβ yδ yδ C = -.47, C =.3, C = -.4, lβ lδ lδ C =.3625, C = -.16, C = -.5 nβ nδ nδ Suppoe the cowind i v obtined (in degee) =.15 V. Then the following eult e β = -8.59437, φ=-.121212, δ = -19.749, δ = 8.61781 Slip to the left, oll to the left, hd ight udde, left ileon (into wind)

Comment fom Aibu Pilot I do not find cowind to be nymoe chllenging in thi iplne thn ny othe. You hve to undetnd tht you cnnot "lip" thi iplne becue you e commnding ROLL RATE with the Side Stick Contolle, not BANK ANGLE. Hee i uggetion: Allow the iplne to do n Auto Lnd in cowind when it i convenient, nd VFR. You will be hocked t the timing of when the iplne leve the CRAB nd pplie udde to lign the noe pllel with the unwy. You think it jut in't going to do it, nd t the vey lt econd, it lide it in pefectly. I would gue in the lt 2 feet o le. My technique i jut the me ny iplne I've eve flown. C-15 to B-767. Cb it down to the fle, pply enough udde to tighten the noe, dop the up-wind wing to pevent dift, lnd on the up-wind min fit. -- 5-Ye Aibu Cptin.