AE 430 - Stability and Control of Aerospace Vehicles

AE 430 - Stability and Control of Aerospace Vehicles Atmospheric Flight Mechanics 1 Atmospheric Flight Mechanics Performance Performance characteristics (range, endurance, rate of climb, takeoff and landing distances, flight path optimization) Flight Dynamics Motion of the aircraft due to disturbances Stability and Control Aeroelasticity Static and Dynamic Aeroelastic phenomena (control reversal, wing divergence, flutter, aeroelastic response) 2 The aerodynamic forces and moment as well as the trust and weight have to be accurately determined 1

The aerodynamic forces and moment acting on the aircraft depend on the property of the atmosphere through which is flying Geometric shape Attitude to the flow Airspeed Property of the air mass (pressure, temperature, density, viscosity, speed of sound, etc.) 3 Overview of Units Mass and weight are often confused Here are some common units used for mass and weight Kilograms Newtons Pounds Slugs Which ones are mass and which ones are weight? What is the difference between mass and weight? 4 2

Overview of Units 5 Kilograms are a unit of mass (metric) Newtons are a unit of weight or force (metric) Slugs are a unit of mass (imperial) Pounds are a unit of weight or force (imperial) Weight = Mass * Gravity The weight of an object on the Earth and on the Moon is different The mass of an object on the Earth and on the Moon is the same Perfect Gas 6 A perfect gas is one in which inter-molecular forces are negligible thermodynamic state equation P = ρrt where P = pressure, ρ = density; T = temperature; R constant for a specific gas for normal air J ft lb f R = 287 = 1718 o o kg K slug R ( )( ) ( )( ) 2 2 m R = 287 = 1718 2 o 2 o ( sec) ( K) ( sec ) ( R) ft 3

Pressure 7 Pressure is the normal force per unit area exerted on a surface due to the time rate of change of momentum for gas molecules impacting that surface da is the incremental area around a point on the surface df is the force on one side of da due to pressure, so the pressure at the point on the surface is: df P lim F units da 0 da P A N/m 2 -Psf-Psi-Atm-dynes/cm 2 1 N/m 2 = 1.4504 x 10-4 lb f /in 2 = 2.0886 x 10-2 lb f /ft 2 1 lb f /in 2 = 6.8947 x 10 3 N/m 2 Ratio of the pressure P at altitude to sea-level standard pressure P δ P 0 Density/Specific Volume 8 density is the mass of a substance per unit volume dv incremental volume about point P dm the mass of the material (gas) inside dv the density, ρ, at a point P is: dm ρ = lim dv 0 dv units of density kilograms/cubic meter, kg/m 3 grams/cubic centimeter, gm/cm 3 pounds mass/cubic feet, lb m /ft 3 slugs/cubic feet, lb f sec 2 /ft 4 specific volume is volume per unit mass specific volume, v, is the reciprocal of density units for specific volume cubic meters/kilogram, m 3 /kg cubic feet/slug, ft 3 /slug cubic feet/slug, ft 4 /lb f sec 2 Ratio of the density ρ at altitude to sea-level standard density dv P ρ σ = ρ 0 4

Temperature Temperature is a measure of the average kinetic energy of particles making up the gas the temperature, T, of a gas is directly proportional to the average kinetic energy of the particles making up the gas Boltzmann s constant, K, is the constant of proportionality 3 Kinetic Energy = 2 KT 23 o K = 1.38 10 joules / Kelvin; 1 joule = 0.738 ft lb 9 units of temperature: Degrees Kelvin (absolute) Degrees Rankine (absolute) Degrees Celsius (not absolute) Degrees Farenheit (not absolute) Temperature affects the properties of the air such as density and viscosity o K o R C F o o Ratio of the temperature T at altitude to sea-level T θ = standard temperature T 0 Mach Number and Speed of Sound V airplane speed a speed of sound 0< M < 0.5 0.5 < M < 0.8 0.8 < M < 1.2 1.2 < M < 5 5 < M V M = a a = Incompressible subsonic flowfield Compressible subsonic flowfield Transonic flowfield Supersonic flowfield Hypersonic flowfield ( γ RT) 12 γ ratio of specific heats 10 5

Equations Summary 11 AV 2 2 V1 V2 1+ ρ = P2 + ρ P = AV 1 1 2 2 ρ AV 2 2 = ρ AV 1 1 1 2 2 2 γ γ 1 2 ρ2 T γ 2 P P = = ρ T 1 1 1 2 2 V1 V2 P 1+ = ct P 2 + ct P = ρ RT P 1 1 1 = ρ RT 2 2 2 2 2 CONTINUITY EQUATION (INCOMPRESSIBLE) BERNOULLI S EQUATION CONTINUITY EQUATION (COMPRESSIBLE) ISENTROPIC RELATIONS ENERGY EQUATION OF STATE Other isentropic relations 12 γ 0 ρ0 T0 p = = p1 ρ1 T1 T0 γ 1 2 = 1+ M1 T 2 1 then p p 0 1 1 γ 1 = 1+ M 2 ρ0 γ 1 = 1 M ρ + 2 γ γ 1 γ 2 γ 1 1 1 2 γ 1 1 6

Aerodynamic Forces A flow field is defined using a coordinate frame is specified using thermodynamic point properties like P, ρ, T, and V Pressure = P(x, y, z) Velocity = V(x, y, z) Pressure and shear distributions which exist on surfaces are the source of all aerodynamic forces 13 Pressure Shear Airfoils An airfoil is a section of a wing (or a fin, or a stabilizer, or a propeller, etc.) Cambered Symmetrical Laminar Flow Reflexed Supercritical 14 7

Review of Aircraft Metrics chord line straight line connecting the LE and TE mean camber line locus of points halfway between upper and lower surfaces camber maximum distance between mean camber line and chord line 15 Review of Aircraft Metrics Wing Area = S Wing Span = b Mean Chord = c = S b = ct + c Root Chord = c = c Tip Chord = 0 r ( ) 2 Taper Ratio = λ = cr ct 2 Aspect Ratio = AR = b c = b S c t r 16 8

More definitions 17 relative wind direction of v angle of attack angle between relative wind and chord line drag component of resultant aerodynamic force parallel to the relative wind lift component of aerodynamic force perpendicular to relative wind moment pressure distribution also produces rotational torque Airfoil with positive lift Geometric angle of attack 18 Airfoil at zero lift Absolute angle of attack 9

Center of Pressure c.p. 19 the resultant forces (lift and drag) acting at the center of pressure produce no moment since the pressure distribution over the airfoil changes with angle of attack, the location of the center of pressure varies with angle of attack the moment about the c.p. is not zero with L = 0 Aerodynamic Center a.c. 20 the a.c. is the point on the airfoil about which moments do not vary with angle of attack, assuming v is constant if L = 0, the moment is a pure couple equal to the moment about the aerodynamic center, m a.c. simple airfoil theory places the a.c. at the quarter chord point for low speed airfoils. at the half chord point for supersonic airfoils 10

21 a.c. and c.p. 22 http://142.26.194.131/aerodynamics1/stability/page7.html 11

Lift, Drag and Moment Coefficients Lift, Wing and Drag coefficients are written as C ab Where a describes the type of coefficient (lift, drag or moment) and b describes what the coefficient is a function of (reference coefficient, angle of attack, sideslip) b also describes the units of the coefficient For example if b is 0 (reference coefficient) then there are no units. If b is α (angle of attack) then the units are per degree or per radian 23 Lift Coefficients C l Lift Coefficient vs Angle of Attack Stall C l0 dc lα = dα C lα α 0L = zero lift angle of attack Slope of the linear region gives the infinite wing lift coefficient The reference lift coefficient is given by the point where the angle of attack is zero 24 α 0L Angle of Attack (Degrees) 12

Moment Coefficients C m Moment Coefficient vs Angle of Attack C m0 dc mα = dα C mα Slope of the linear region gives the infinite wing moment coefficient The reference moment coefficient is given by the point where the angle of attack is zero 25 Angle of Attack (Degrees) Drag Coefficients Drag Coefficient vs Angle of Attack C d C d0 The reference drag coefficient is given by the point where the angle of attack is zero The other drag coefficients can be determined using Excel, Matlab etc to perform a quadratic regression 26 Angle of Attack (Degrees) 13

Finite Wing Corrections All reference coefficients are not corrected Moment coefficients are NOT corrected Lift coefficient due to angle of attack is corrected AR is the aspect ratio of the wing e is the Oswald Efficiency Factor C C C C C Lα L0 D0 M 0 = C l0 = C = C = C d 0 m0 Mα mα Clα = Clα 1+ π ear 27 Note: do not forget 57.3 deg/rad conversion factor Finite Wing Corrections Drag coefficient due to angle of attack is corrected AR is the aspect ratio of the wing e is the Oswald Efficiency Factor C d,i is the induced drag coefficient C = C + C C Dα dα d, i Dα 2 Lα C = Cdα + π ear 28 14

Effect of flaps on lift Medium angle of attack High angle of attack Slat opened at high angle of attack With slats 24 degree angle of attack Without slats 15 degree angle of attack 29 Angle of attack Flaps 30 15

Nomenclature and Definitions 31 Positive direction of axes determined by right hand rule 32 16

Angle of attach, sideslip and flight speed 2 2 2 ( ) 1/2 V = u + v + w 33 Body-fixed Coordinates 34 The CG location is chosen as the origin of the xyz coordinate frame It is fixed to the airplane xz IS A PLANE OF SYMMETRY u, v, w are linear velocity components p, q, r are angular rates X, Y, Z, are the aerodynamic force components about the xyz axes L, M, N are the aerodynamic moments about the xyz axes xyz are the roll, pitch and yaw axes CG w N, r z b Z y b x b u X L, p v M, q Y 17

Aerodynamic forces and moments Aerodynamic coefficients primarily functions of the Mach number (where is the speed of sound), the Reynolds number (where is the fluid density and is the viscosity), and the aerodynamic angles and. 35 reference length dynamic pressure reference area flight speed 36 Quantity Yaw N Forces and Moments Variable Dimensionless Coefficient Lift L C L = L/QS 'Up' normal to freestream Drag D C D = D/QS Downstream Sideforce Y C Y = Y/QS Right, looking forward Roll L C l = L/QSb Right wing down Pitch M C m = M/QSc Nose up C n = N/QSb Angles and Rates Positive Direction Nose right Quantity Symbol Positive Direction Angle of attack α Nose up w.r.t. freestream Angle of sideslip β Nose left Pitch angle Θ Nose up Yaw angle Ψ Nose right Euler angles Bank (roll) angle Φ Right wing down Roll rate p Right wing down Pitch rate q Nose up Yaw rate r Nose Right 18

Centroid / Center of Mass CM 37 The Significance of Weight and Balance http://aerosrv.atl.calpoly.edu/dbiezad/aero_ Courses/Aero_420/Articles/NASA_Dryden/ wt_and_balance.pdf Definition of Moment of Inertia 38 19

Diagonal Moments of Inertia 39 Products of Inertia Affect the lateral stability http://naca.larc.nasa.gov/reports/1947/nacatn-1193/naca-tn-1193.pdf 40 20

The Significance of Weight and Balance Higher Weight Higher Takeoff Speed Longer Takeoff and Landing Rolls Higher Fuel Consumption = Less Range Determined by Scales on the Landing Gear Weight Distribution along the Longitudinal Axis Represented by the Center of Gravity cg or cm expressed as Percent Chord Must be within Prescribed Limits for Safe Flight Changes as Fuel is Burned or Passengers Move Determined from the Weight Scale Data Download and Read wt_and_balance.pdf 41 Weight and Balance 42 21

Somebody's going to get fired 43 Weight and Balance Example 1. Lets assume that we want to know the weight and center of gravity of an empty passenger aircraft sitting on the ground at Kennedy International Airport in New York as depicted in Figure 2. Since the aircraft is at rest, we know that weight of the aircraft is supported by the forces exerted by the pavement beneath each landing gear. We also know that the sum of the clockwise moments produced by each gear about some axis of rotation, say, the tip of the nose of the aircraft, is exactly balanced by the counterclockwise moment produced by the weight of the aircraft acting through the center of gravity. If we pulled the aircraft onto a set of platform scales and measured the gear reaction forces as 100,000 lb for each of the two main gear and 25,000 lb for the nose gear and the distances of the main and nose gear aft of the nose of the aircraft were measured to be 50 ft and 10 ft, respectively, we would get the following results: 44 22

45 We now know that the aircraft, empty of fuel, passengers, and baggage weighs 225,000 lb and has a cg 45.56 ft aft of the nose. If we then fuel the aircraft with 40,000 gal of jet fuel weighing 6.25 lb/gal in fuel tanks that have a centroid (center of the volume or mass) location of 40 ft aft of the nose and load the aircraft with 200 passengers weighing an estimated total of 40,000 lb with a centroid of 55 ft aft of the nose and 10,000 lb of baggage in a baggage hold with a centroid of 50 ft aft of the nose, what would be the engine-start gross weight and cg? The results are: ft 46 So, the fully loaded aircraft has a weight of 525,000 lb and a cg 43.71 ft aft of the nose. Notice that the c.g. is at a location forward of the main gear. What would happen if the cg was aft of the main gear? 23

Weight and Balance 47 Example 2. Let's take the same fully loaded aircraft in the above example and look at forces acting on the aircraft after takeoff and after it has leveled off at cruise altitude with 4,000 lb of fuel having been burned to get there as shown in Figure 3. The aircraft is in stabilized, level flight, such that the weight of the aircraft is supported by the lift forces being generated by the wing and the tail. If the center of pressure on the wing (where the resultant wing lift force acts) is located at 45.0 ft aft of the nose of the aircraft and the center of pressure of the tail is 105 ft aft of the nose, what are the magnitude and direction of the forces acting on the wing and tail? Again, the sum of the lift forces on the wing and tail are equal and opposite to the weight of the airplane and the sum of the clockwise moments produced by the lift forces about the nose is exactly balanced by the counterclockwise moments produced by the weight of the aircraft acting through the center of gravity. We must first recalculate the weight and center of gravity of the aircraft after 4,000 lb of fuel have been consumed: 48 24

We must now express the equilibrium conditions of level flight in terms of the force and moment equations: Solving these equations simultaneously for the lift forces on the wing and tail, we get: 49 And by substitution, we can solve for the lift forces on the wing and tail: So, the lift force on the wing is 509,167 and the lift force on the tail is - 9,167 (a downward force). This is so, because the center of gravity of the aircraft was forward of the center of pressure of the wing, requiring a counterclockwise moment by the tail to balance the moment equation. 50 25