Aircraft Flight Dynamics Robert F. Stengel Lecture7 Gliding, Climbing, and Turning Performance

Gliding, Climbing, and Turning Flight Performance
 Robert Stengel, Aircraft Flight Dynamics, MAE 331, 2012 ! Copyright 2012 by Robert Stengel. All rights reserved. For educational use only.! http://www.princeton.edu/~stengel/MAE331.html ! http://www.princeton.edu/~stengel/FlightDynamics.html ! •  Flight envelope" •  Minimum glide angle/rate" •  Maximum climb angle/rate" •  V-n diagram" •  Energy climb" •  Corner velocity turn" •  Herbst maneuver" The Flight Envelope Flight Envelope Determined by Available Thrust" •  Flight ceiling defined by available climb rate" –  Absolute: 0 ft/min" –  Service: 100 ft/min" –  Performance: 200 ft/min" •  Excess thrust provides the ability to accelerate or climb" •  Flight Envelope: Encompasses all altitudes and airspeeds at which an aircraft can fly " –  in steady, level flight " –  at fixed weight" Additional Factors Define the Flight Envelope" •  Maximum Mach number" •  Maximum allowable aerodynamic heating" •  Maximum thrust" •  Maximum dynamic pressure" •  Performance ceiling" •  Wing stall" •  Flow-separation buffet" –  Angle of attack" –  Local shock waves" Piper Dakota Stall Buffet" http://www.youtube.com/watch?v=mCCjGAtbZ4g ! Boeing 787 Flight Envelope (HW #5, 2008)" Best Cruise Region" Gliding Flight D = C D 1 2 ρ V 2 S = −W sin γ C L 1 2 ρ V 2 S = W cos γ  h = V sin γ  r = V cos γ Equilibrium Gliding Flight" Gliding Flight" •  Thrust = 0" •  Flight path angle < 0 in gliding flight" •  Altitude is decreasing" •  Airspeed ~ constant" •  Air density ~ constant " tan γ = − D L = − C D C L =  h  r = dh dr ; γ = −tan −1 D L # $ % & ' ( = −cot −1 L D # $ % & ' ( •  Gliding flight path angle " •  Corresponding airspeed " V glide = 2W ρ S C D 2 + C L 2 Maximum Steady Gliding Range" •  Glide range is maximum when γ is least negative, i.e., most positive" •  This occurs at (L/D) max " Maximum Steady Gliding Range" •  Glide range is maximum when γ is least negative, i.e., most positive" •  This occurs at (L/D) max " tan γ =  h  r = negative constant = h − h o ( ) r − r o ( ) Δr = Δh tan γ = −Δh −tan γ = maximum when L D = maximum γ max = −tan −1 D L # $ % & ' ( min = −cot −1 L D # $ % & ' ( max Sink Rate " •  Lift and drag define γ and V in gliding equilibrium" D = C D 1 2 ρ V 2 S = −W sin γ sin γ = − D W L = C L 1 2 ρ V 2 S = W cos γ V = 2W cos γ C L ρ S  h = V sin γ = − 2W cos γ C L ρ S D W $ % & ' ( ) = − 2W cos γ C L ρ S L W $ % & ' ( ) D L $ % & ' ( ) = − 2W cos γ C L ρ S cos γ 1 L D $ % & ' ( ) •  Sink rate = altitude rate, dh/dt (negative)" •  Minimum sink rate provides maximum endurance" •  Minimize sink rate by setting ∂(dh/dt)/dC L = 0 (cos γ ~1)" Conditions for Minimum Steady Sink Rate"  h = − 2W cos γ C L ρ S cos γ C D C L $ % & ' ( ) = − 2W cos 3 γ ρ S C D C L 3/2 $ % & ' ( ) ≈ − 2 ρ W S $ % & ' ( ) C D C L 3/2 $ % & ' ( ) C L ME = 3C D o ε and C D ME = 4C D o L/D and V ME for Minimum Sink Rate" V ME = 2W ρ S C D ME 2 + C L ME 2 ≈ 2 W S ( ) ρ ε 3C D o ≈ 0.76V L D max L D ( ) ME = 1 4 3 ε C D o = 3 2 L D ( ) max ≈ 0.86 L D ( ) max L/D for Minimum Sink Rate" •  For L/D < L/D max , there are two solutions" •  Which one produces minimum sink rate?" L D ( ) ME ≈ 0.86 L D ( ) max V ME ≈ 0.76V L D max Gliding Flight of the P-51 Mustang" Loaded Weight = 9,200 lb (3,465 kg) L / D ( ) max = 1 2 ε C D o =16.31 γ MR = −cot −1 L D $ % & ' ( ) max = −cot −1 (16.31) = −3.51° C D ( ) L/D max = 2C D o = 0.0326 C L ( ) L/D max = C D o ε = 0.531 V L/D max = 76.49 ρ m / s  h L/D max = V sin γ = − 4.68 ρ m / s R h o =10km = 16.31 ( ) 10 ( ) =163.1 km Maximum Range Glide" Loaded Weight = 9,200 lb (3,465 kg) S = 21.83 m 2 C D ME = 4C D o = 4 0.0163 ( ) = 0.0652 C L ME = 3C D o ε = 3 0.0163 ( ) 0.0576 = 0.921 L D ( ) ME =14.13  h ME = − 2 ρ W S $ % & ' ( ) C D ME C L ME 3/2 $ % & & ' ( ) ) = − 4.11 ρ m / s γ ME = −4.05° V ME = 58.12 ρ m / s Maximum Endurance Glide" Climbing Flight •  Rate of climb, dh/dt = Specific Excess Power " Climbing Flight"  V = 0 = T − D −W sin γ ( ) m sin γ = T − D ( ) W ; γ = sin −1 T − D ( ) W  γ = 0 = L −W cos γ ( ) mV L = W cos γ  h = V sin γ = V T − D ( ) W = P thrust − P drag ( ) W Specific Excess Power (SEP) = Excess Power Unit Weight ≡ P thrust − P drag ( ) W •  Flight path angle " •  Required lift" •  Note significance of thrust-to-weight ratio and wing loading" Steady Rate of Climb"  h = V sin γ = V T W " # $ % & ' − C D o + ε C L 2 ( ) q W S ( ) * + , , - . / / € L = C L q S = W cos γ C L = W S # $ % & ' ( cos γ q V = 2 W S # $ % & ' ( cos γ C L ρ  h = V T W ! " # $ % & − C D o q W S ( ) − ε W S ( ) cos 2 γ q * + , - . / = V T h ( ) W ! " # $ % & − C D o ρ h ( ) V 3 2 W S ( ) − 2 ε W S ( ) cos 2 γ ρ h ( ) V •  Climb rate " •  Necessary condition for a maximum with respect to airspeed" Condition for Maximum Steady Rate of Climb"  h = V T W ! " # $ % & − C D o ρ V 3 2 W S ( ) − 2 ε W S ( ) cos 2 γ ρ V ∂  h ∂ V = 0 = T W " # $ % & ' +V ∂ T / ∂ V W " # $ % & ' ( ) * + , - − 3C D o ρ V 2 2 W S ( ) + 2 ε W S ( ) cos 2 γ ρ V 2 Maximum Steady " Rate of Climb: " Propeller-Driven Aircraft" ∂ P thrust ∂ V = 0 = T W " # $ % & ' +V ∂ T / ∂ V W " # $ % & ' ( ) * + , - •  At constant power" ∂  h ∂ V = 0 = − 3C D o ρ V 2 2 W S ( ) + 2 ε W S ( ) ρ V 2 •  With cos 2 γ ~ 1, optimality condition reduces to" •  Airspeed for maximum rate of climb at maximum power, P max " V 4 = 4 3 ! " # $ % & ε W S ( ) 2 C D o ρ 2 ; V = 2 W S ( ) ρ ε 3C D o = V ME Maximum Steady Rate of Climb: " Jet-Driven Aircraft" •  Condition for a maximum at constant thrust and cos 2 γ ~ 1" •  Airspeed for maximum rate of climb at maximum thrust, T max " ∂  h ∂ V = 0 0 = ax 2 + bx + c and V = + x = − 3C D o ρ 2 W S ( ) V 4 + T W # $ % & ' ( V 2 + 2 ε W S ( ) ρ = − 3C D o ρ 2 W S ( ) V 2 ( ) 2 + T W # $ % & ' ( V 2 ( ) + 2 ε W S ( ) ρ Optimal Climbing Flight What is the Fastest Way to Climb from One Flight Condition to Another?" •  Specific Energy " •  = (Potential + Kinetic Energy) per Unit Weight" •  = Energy Height" Energy Height" •  Could trade altitude with airspeed with no change in energy height if thrust and drag were zero" Total Energy Unit Weight ≡ Specific Energy = mgh + mV 2 2 mg = h + V 2 2g ≡ Energy Height, E h , ft or m Specific Excess Power" dE h dt = d dt h + V 2 2g ! " # $ % & = dh dt + V g ! " # $ % & dV dt •  Rate of change of Specific Energy " = V sin γ + V g " # $ % & ' T − D −mgsin γ m " # $ % & ' = V T − D ( ) W = V C T − C D ( ) 1 2 ρ (h)V 2 S W = Specific Excess Power (SEP) = Excess Power Unit Weight ≡ P thrust − P drag ( ) W Contours of Constant Specific Excess Power" •  Specific Excess Power is a function of altitude and airspeed" •  SEP is maximized at each altitude, h, when" d SEP(h) [ ] dV = 0 Subsonic Energy Climb" •  Objective: Minimize time or fuel to climb to desired altitude and airspeed" Supersonic Energy Climb" •  Objective: Minimize time or fuel to climb to desired altitude and airspeed" The Maneuvering Envelope •  Maneuvering envelope: limits on normal load factor and allowable equivalent airspeed" –  Structural factors" –  Maximum and minimum achievable lift coefficients" –  Maximum and minimum airspeeds" –  Protection against overstressing due to gusts" –  Corner Velocity: Intersection of maximum lift coefficient and maximum load factor" Typical Maneuvering Envelope: V-n Diagram" •  Typical positive load factor limits" –  Transport: > 2.5" –  Utility: > 4.4" –  Aerobatic: > 6.3" –  Fighter: > 9" •  Typical negative load factor limits" –  Transport: < –1" –  Others: < –1 to –3" C-130 exceeds maneuvering envelope" http://www.youtube.com/watch?v=4bDNCac2N1o&feature=related ! Maneuvering Envelopes (V-n Diagrams) for Three Fighters of the Korean War Era" Republic F-84" North American F-86" Lockheed F-94" Turning Flight •  Vertical force equilibrium" Level Turning Flight" L cos µ = W •  Load factor" n = L W = L mg = sec µ ,"g"s •  Thrust required to maintain level flight" T req = C D o + ε C L 2 ( ) 1 2 ρ V 2 S = D o + 2 ε ρ V 2 S W cos µ # $ % & ' ( 2 = D o + 2 ε ρ V 2 S nW ( ) 2 µ : Bank Angle •  Level flight = constant altitude" •  Sideslip angle = 0" •  Bank angle" Maximum Bank Angle in Level Flight" cos µ = W C L qS = 1 n = W 2 ε T req − D o ( ) ρ V 2 S µ = cos −1 W C L qS $ % & ' ( ) = cos −1 1 n $ % & ' ( ) = cos −1 W 2 ε T req − D o ( ) ρ V 2 S * + , , - . / / •  Bank angle is limited by " € µ : Bank Angle C L max or T max or n max •  Turning rate" Turning Rate and Radius in Level Flight"  ξ = C L qS sin µ mV = W tan µ mV = g tan µ V = L 2 − W 2 mV = W n 2 − 1 mV = T req − D o ( ) ρ V 2 S 2 ε − W 2 mV •  Turning rate is limited by " C L max or T max or n max •  Turning radius " R turn = V  ξ = V 2 g n 2 − 1 Maximum Turn Rates" “Wind-up turns”" •  Corner velocity" Corner Velocity Turn" •  Turning radius " R turn = V 2 cos 2 γ g n max 2 − cos 2 γ V corner = 2n max W C L mas ρ S •  For steady climbing or diving flight" sin γ = T max − D W Corner Velocity Turn" •  Time to complete a full circle " t 2 π = V cos γ g n max 2 − cos 2 γ •  Altitude gain/loss " Δh 2 π = t 2 π V sin γ •  Turning rate "  ξ = g n max 2 − cos 2 γ ( ) V cos γ Not a turning rate comparison" http://www.youtube.com/watch?v=z5aUGum2EiM ! Herbst Maneuver" •  Minimum-time reversal of direction" •  Kinetic-/potential-energy exchange" •  Yaw maneuver at low airspeed" •  X-31 performing the maneuver " Next Time: Aircraft Equations of Motion  Reading Flight Dynamics, 155-161 Virtual Textbook, Parts 8,9 . Gliding, Climbing, and Turning Flight Performance Robert Stengel, Aircraft Flight Dynamics, MAE 331, 2012 ! Copyright 2012 by Robert Stengel n max •  Turning rate" Turning Rate and Radius in Level Flight& quot;  ξ = C L qS sin µ mV = W tan µ mV = g tan µ V = L 2 − W 2 mV = W n 2 − 1 mV = T req − D o ( ) ρ V 2 S 2 ε − W 2 mV •  Turning. climb" •  Flight Envelope: Encompasses all altitudes and airspeeds at which an aircraft can fly " –  in steady, level flight " –  at fixed weight" Additional Factors Define the Flight