Gliding, Climbing, and Turning Flight Performance Robert Stengel, Aircraft Flight Dynamics, MAE 331, 2018

Learning Objectives • Conditions for gliding flight • Parameters for maximizing climb angle and rate • Review the V-n diagram • Energy height and specific excess power • Alternative expressions for steady turning flight • The Herbst maneuver

Reading: Flight Dynamics Aerodynamic Coefficients, 130-141

Review Questions

§ How does air density decrease with altitude? § What are the different definitions of airspeed? § What is a “lift-drag polar”? § Power and thrust: How do they vary with altitude? § What factors define the “flight envelope”? § What were some features of the first commercial transport aircraft? § What are the important parameters of the “Breguet Range Equation”? § What is a “step climb”, and why is it important?

1 Gliding Flight

Equilibrium Gliding Flight

1 C ρV 2S = −W sinγ D 2 1 C ρV 2S = W cosγ L 2 h! = V sinγ

r! = V cosγ 4

2 Gliding Flight • Thrust = 0 • Flight path angle < 0 in gliding flight • Altitude is decreasing • Airspeed ~ constant • Air density ~ constant

Gliding flight path angle

D C h dh # D & # L & tanγ = − = − D = = ; γ = −tan−1 % ( = −cot−1 % ( L C r dr $ L ' $ D ' L Corresponding airspeed 2W V = glide 2 2 ρS CD + CL 5

Maximum Steady Gliding Range

• Glide range is maximum when γ is least negative, i.e., most positive

• This occurs at (L/D)max 6

3 Maximum Steady Gliding Range

• Glide range is maximum when γ is least negative, i.e., most positive

• This occurs at (L/D)max

−1 # D & −1 # L & γ max = −tan % ( = −cot % ( $ L 'min $ D 'max ! h (h − ho ) tanγ = = negative constant = r! (r − ro ) Δh −Δh L Δr = = = maximum when = maximum tanγ − tanγ D 7

Sink Rate, m/s Lift and drag define γ and V in gliding equilibrium

1 2 1 2 D = C ρV S = −W sinγ L = CL ρV S = W cosγ D 2 2 D 2W cosγ sinγ = − V =

W CL ρS Sink rate = altitude rate, dh/dt (negative)

h = V sinγ 2W cosγ $ D ' 2W cosγ $ L '$ D ' = − & ) = − & )& ) CL ρS %W ( CL ρS %W (% L ( 2W cosγ $ 1 ' = − cosγ & ) C ρS L D 8 L % (

4 Conditions for Minimum Steady Sink Rate

• Minimum sink rate provides maximum endurance

• Minimize sink rate by setting ∂(dh/dt)/∂CL = 0 (cos γ ~1)

2W cosγ $C ' h = − cosγ & D ) CL ρS % CL ( 2W cos3 γ $ C ' 2 $W '$ C ' = − & D ) ≈ − & )& D ) ρS C 3/2 ρ % S ( C 3/2 % L ( % L ( 3C C Do and C 4C LME = DME = Do ε 9

L/D and VME for Minimum Sink Rate

1 3 3 L = = L ≈ 0.86 L ( D)ME 4 C 2 ( D)max ( D)max ε Do

2W 2(W S) ε V = ≈ ≈ 0.76V ME 2 2 L Dmax S C C ρ 3CD ρ DM+E LME o

5 L/D for Minimum Sink Rate

• For L/D < L/Dmax, there are L ≈ 0.86 L two solutions ( D)ME ( D)max • Which one produces smaller V ≈ 0.76V sink rate? ME L Dmax

Historical Factoids Lifting-Body Reentry Vehicles

M2-F1 HL- 10

M2-F2 X-24A

M2-F3 X-24B

6 Climbing Flight

Climbing Flight

• Flight path angle • Required lift

(T − D −W sinγ ) V! = 0 = (L −W cosγ ) m γ! = 0 = mV T − D T − D ( ) −1 ( ) L W cos sinγ = ; γ = sin = γ W W Rate of climb, dh/dt = Specific Excess Power

(T − D) Pthrust − Pdrag h! = V sinγ = V = ( ) W W

Excess Power Pthrust − Pdrag Specific Excess Power (SEP) = ≡ ( ) Unit Weight W 14

7 Steady Rate of Climb

Climb rate L = CLqS = W cosγ * 2 - ⎛ W ⎞ cosγ C +εC q CL = ⎜ ⎟ " T % ( Do L ) ⎝ S ⎠ q h = V sinγ = V ,$ '− / ⎛ W ⎞ cosγ ,#W & (W S) / V = 2 + . ⎝⎜ ⎠⎟ S CL ρ Note significance of thrust-to-weight ratio and wing loading

⎡ C q 2 ⎤ ⎛ T ⎞ Do ε (W S)cos γ h! = V ⎢⎜ ⎟ − − ⎥ ⎣⎝ W ⎠ (W S) q ⎦ C ρ h V 3 2 ⎛ T (h)⎞ Do ( ) 2ε (W S)cos γ = V − − ⎝⎜ W ⎠⎟ 2(W S) ρ (h)V 15

Condition for Maximum Steady Rate of Climb

C ρV 3 2 W S cos2 ! T $ Do ε ( ) γ h = V # &− − "W % 2 W S ρV ( ) Necessary condition for a maximum with respect to airspeed  ( + 3C ρV 2 2 W S cos2 ∂h " T % "∂T /∂V % Do ε ( ) γ = 0 = *$ '+V $ '-− + 2 ∂V )#W & # W &, 2 W S ρV ( ) 16

8 Maximum Steady Rate of Climb: Propeller-Driven Aircraft

( + • At constant power ∂Pthrust " T % "∂T /∂V % = 0 = *$ '+V $ '- ∂V )#W & # W &, • With cos2γ ~ 1, optimality condition reduces to

 3C ρV 2 2 W S ∂h Do ε ( ) = 0 = − + ∂V 2 W S ρV 2 ( )

• Airspeed for maximum rate of climb at maximum power, Pmax

2 4 ! 4 $ε (W S) (W S) ε V = # & ; V = 2 = V " 3% C 2 3C ME Do ρ ρ Do 17

Maximum Steady Rate of Climb: Jet-Driven Aircraft

Condition for a maximum at constant thrust and cos2γ ~ 1

3C ρ Do 4 ⎛ T ⎞ 2 2ε (W S) − V + ⎜ ⎟ V + = 0 ∂h 2 W S ⎝ W ⎠ ρ = 0 ( ) ∂V 3C ρ 2 Do 2 ⎛ T ⎞ 2 2ε (W S) − (V ) + ⎜ ⎟ (V ) + = 0 2(W S) ⎝ W ⎠ ρ Quadratic in V2 Airspeed for maximum rate of climb at maximum thrust, Tmax

0 = ax2 + bx + c and V = + x 18

9 Optimal Climbing Flight

What is the Fastest Way to Climb from One Flight Condition to Another?

10 Energy Height

Total Energy • Specific Energy Specific Energy ≡ Unit Weight • = (Potential + Kinetic Energy) per Unit Weight mgh + mV 2 2 V 2 = = h + • = Energy Height mg 2g

≡ Energy Height, Eh , ft or m

Can trade altitude for airspeed with no change in energy height if thrust and drag are zero 21

Specific Excess Power Rate of change of Specific Energy

dE d ! V 2 $ dh !V $ dV h = #h + & = +# & dt dt " 2g % dt " g % dt

⎛ V ⎞ ⎛ T − D − mgsinγ ⎞ (T − D) = V sinγ + ⎜ ⎟ = V ⎝⎜ g ⎠⎟ ⎝ m ⎠ W

= Specific Excess Power (SEP) 1 2 (CT − CD ) ρ(h)V S 2 Excess Power (Pthrust − Pdrag ) = V = ≡ W Unit Weight W

11 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)] max[SEP(h)] = 0 wrt V dV

Subsonic Minimum-Time Energy Climb Objective: Minimize time to climb to desired altitude and airspeed • Minimum-Time Strategy:

• Zoom climb/dive to intercept SEPmax(h) contour • Climb at SEPmax(h) • Zoom climb/dive to intercept target SEPmax(h) contour

Bryson, Desai, Hoffman, 1969 24

12 Subsonic Minimum-Fuel Energy Climb Objective: Minimize fuel to climb to desired altitude and airspeed

• Minimum-Fuel Strategy:

• Zoom climb/dive to intercept [SEP (h)/(dm/dt)] max contour • Climb at [SEP (h)/(dm/dt)] max • Zoom climb/dive to intercept target[SEP (h)/(dm/dt)] max contour

Bryson, Desai, Hoffman, 1969 25

Supersonic Minimum-Time Energy Climb Objective: Minimize time to climb to desired altitude and airspeed • Minimum-Time Strategy:

• Intercept subsonic SEPmax(h) contour • Climb at SEPmax(h) to intercept matching zoom climb/dive contour • Zoom climb/dive to intercept supersonic SEPmax(h) contour • Climb at SEPmax(h) to intercept target SEPmax(h) contour • Zoom climb/dive to intercept target SEPmax(h) contour

26 Bryson, Desai, Hoffman, 1969

13 Checklist

q Energy height? q Contours? q Subsonic minimum-time climb? q Supersonic minimum-time climb? q Minimum-fuel climb?

dE dE dt 1 ⎡dh ⎛ V ⎞ dV ⎤ h = h = + dm dt dm m! ⎢ dt ⎝⎜ g ⎠⎟ dt ⎥ fuel fuel fuel ⎣ ⎦

SpaceShipOne Ansari X Prize, December 17, 2003

Brian Binnie, *78 Pilot, Astronaut

14 SpaceShipOne Altitude vs. Range MAE 331 Assignment #4, 2010

SpaceShipOne State Histories

15 SpaceShipOne Dynamic Pressure and Mach Number Histories

The Maneuvering Envelope

16 Typical Maneuvering Envelope: V-n Diagram

• 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 positive load factor limits – Transport: > 2.5 • Typical negative load factor limits – Utility: > 4.4 – Transport: < –1 – Aerobatic: > 6.3 – Others: < –1 to –3 – Fighter: > 9

Maneuvering Envelopes (V-n Diagrams) for Three Fighters of the Korean War Era

Republic F-84 Lockheed F-94

North American F-86

17 Turning Flight

Level Turning Flight

• Level flight = constant altitude • Sideslip angle = 0 • Vertical force equilibrium

L cos µ = W µ : Bank Angle • Load factor L L n = W = mg = sec µ,"g"s

• Thrust required to maintain level flight

2 1 2ε ⎛ W ⎞ T = C + εC 2 ρV 2S = D + req ( Do L ) 2 o ρV 2S ⎝⎜ cosµ ⎠⎟

2ε 2

= Do + 2 (nW ) ρV S 36

18 µ : Bank Angle

Maximum Bank Angle in € Steady Level Flight Bank angle

W ⎛ W ⎞ cosµ = µ = cos−1 ⎜ C qS ⎟ CLqS ⎝ L ⎠

1 −1 ⎛ 1⎞ = = cos ⎜ ⎟ n ⎝ n⎠ ⎡ ⎤ 2ε −1 2ε ⎢ ⎥ = W = cos W 2 2 ⎢ T − D ρV S ⎥ (Treq − Do )ρV S ⎣ ( req o ) ⎦ Bank angle is limited by C or T or n Lmax max max 37

Turning Rate and Radius in Level Flight Turning rate C qSsin µ ξ! = L mV W tan µ = mV g tan µ = V Turning rate is L2 −W 2 limited by = mV C or T or n Lmax max max W n2 −1 = Turning radius mV T − D ρV 2S 2ε −W 2 V V 2 ( req o ) R = = = turn  2 mV ξ g n − 1 38

19 Maximum Turn Rates

“Wind-up turns”

Corner Velocity Turn

2n W • Corner velocity V = max corner C S Lmas ρ T − D • For steady climbing sinγ = max or diving flight W V 2 cos2 γ • Turning radius Rturn = g n2 − cos2 γ max 40

20 Corner Velocity Turn

2 2 g(nma−x cos γ ) • Turning rate ξ = V cos γ V cosγ • Time to complete a t = 2π 2 2 full circle g nma−x cos γ

• Altitude gain/loss Δh2π = t2πV sinγ

Checklist

q V-n diagram? q Maneuvering envelope? q Level turning flight? q Limiting factors? q Wind-up turn? q Corner velocity?

21 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, Section 3.1, 3.2, pp. 155-161

Learning Objectives What use are the equations of motion? How is the angular orientation of the airplane described? What is a cross-product-equivalent matrix? What is angular momentum? How are the inertial properties of the airplane described? How is the rate of change of angular momentum calculated?

22 Supplemental Material

Gliding Flight of the P-51 Mustang

Maximum Range Glide Maximum Endurance Glide Loaded Weight = 9,200 lb (3,465 kg) 1 Loaded Weight = 9,200 lb (3,465 kg) (L / D) = = 16.31 max 2 C S 21.83 m2 ε Do =

−1 ⎛ L ⎞ −1 CD = 4CD = 4(0.0163) = 0.0652 γ = − cot = − cot (16.3) = −3.5° ME o MR ⎝⎜ D⎠⎟ max 3C Do 3(0.0163) CL = = = 0.921 (CD ) = 2CD = 0.0326 ME L/Dmax o ε 0.0576 C L D 14.13 Do ( )ME = (CL ) = = 0.531 L/Dmax ε ⎛ C ⎞ 2 ⎛ W ⎞ DME 4.11 76.49 h! = − = − m/s V = m/s ME ⎝⎜ ⎠⎟ ⎜ 3/2 ⎟ L/Dmax ρ S ⎝ CL ⎠ ρ ρ ME 4.68 γ ME = −4.05° h! = V sinγ = − m/s L/Dmax 58.12 ρ V = m/s ME ρ Rh =10km = (16.31)(10) = 163.1 km o 46