Energy Maneuvering

22
G. Leng, ME Dept, NUS 4 : Climb and Turn Performance or how nimble is the aircraft ?

description

Energy Maneuvering and Management

Transcript of Energy Maneuvering

Page 1: Energy Maneuvering

G. Leng, ME Dept, NUS

4 : Climb and Turn Performance

or how nimble is the aircraft ?

Page 2: Energy Maneuvering

G. Leng, ME Dept, NUS

The rate of climb (ROC)

The rate of climb (ROC) is the rate at which an aircraft increases its altitude

Climb records

h (km) time (s) ROC (m/s)

Page 3: Energy Maneuvering

G. Leng, ME Dept, NUS

Climb model

γ : flight path (climb) angle

Page 4: Energy Maneuvering

G. Leng, ME Dept, NUS

Resolving forces parallel and perpendicular to flight path (velocity V)

T – D – W sinγ = m dV/dt

L – W cosγ = m V dγ/dt

The climb rate is then

dh/dt = V sinγ

=

or(T – D) V =

Page 5: Energy Maneuvering

G. Leng, ME Dept, NUS

The climb equation

(T – D) V = dh/dt + d [V2 / (2g) ] / dtW

Page 6: Energy Maneuvering

G. Leng, ME Dept, NUS

Specific energy

The quantity on the rhs of the climb equation can be written as :

dh/dt + d [V2 / (2g) ] / dt

=

=

Page 7: Energy Maneuvering

G. Leng, ME Dept, NUS

Energy perspective

Hence the climb equation is actually an ‘energy” equation

Ps = dhe/dt

The time to transit from one energy state to another is given by :

Page 8: Energy Maneuvering

G. Leng, ME Dept, NUS

Example : F104 Ps curves

How to fly from energy state A to state B as fast as possible ?

A

B

Source : NASA TN D6398

Page 9: Energy Maneuvering

G. Leng, ME Dept, NUS

Comparison F4H-1 vs F15A

F4H-1 2 x J79-GE-2A turbojets

thrust = 2 x 10,350 lb ( 2 x 4,693 kg)

MTOW = 54,600 lb (24,761 kg)

T/W =

F15A 2 x F100-PW-100 turbofan

thrust = 2 x 25,000 lb (2 x 11,250 kg)

MTOW = 56,000 lb (25,200 kg)

T/W =

Page 10: Energy Maneuvering

G. Leng, ME Dept, NUS

Energy-Maneuverability Theory

• Developed by USAF Col John Boyd at Georgia Tech in 1962

• Thesis :

• Proved that the F-4 could not out turn a MiG-21 except at low altitude and high speed

• E-M theory was applied to the design of the F15 and F16

USAF Col John Boyd 1927-1997

Page 11: Energy Maneuvering

G. Leng, ME Dept, NUS

Sustained level turn

Conventional aircraft change heading by banking

i.e. tilt the lift force so that the horizontal component provides the centripedal force for circular motion.

Page 12: Energy Maneuvering

G. Leng, ME Dept, NUS

Sustained level turn model

φ

L cosφ =

L sinφ =

Resolving forces in the vertical and horizontal directionsL

φ: : bank angle

dχ/dt : turn rateW = mg

Page 13: Energy Maneuvering

G. Leng, ME Dept, NUS

Define the load factor n = L/W

= 1 / cosφ ( n > 1 )

Dividing the 2 force equations

V/g dχ/dt = tan φ

=

=

Page 14: Energy Maneuvering

G. Leng, ME Dept, NUS

The turn equation is :

dχ/dt = (g/V) √ ( n2 - 1 )

The turn radius is :

R = V / dχ/dt

= V2/ ( g √ ( n2 - 1 ) )

Page 15: Energy Maneuvering

G. Leng, ME Dept, NUS

Example : Turn envelope

Page 16: Energy Maneuvering

G. Leng, ME Dept, NUS

Exercise : Performance

estimation

Assuming A has a corner velocity of 450 kts at a load factor of 8, what’s the turn radius and turn rate ?

Assuming B is flying at the same speed, what’s the load factor and turn rate for B ?

Page 17: Energy Maneuvering

G. Leng, ME Dept, NUS

Turn at maximum load factor

Question : What’s the bank angle ?

Page 18: Energy Maneuvering

G. Leng, ME Dept, NUS

load factor n = ⇒ bank angle =

airspeed V = 450 kts = 450*0.5151 = 232 m/s

turn rate =

=

=

=

turn radius = V / turn rate

=

Page 19: Energy Maneuvering

G. Leng, ME Dept, NUS

RA = V2 / (g √ (nA2 – 1 ) )

RB = V2 / (g √ (nB2 – 1 ) )

load factor nB =

=

turn rate =

=

Page 20: Energy Maneuvering

G. Leng, ME Dept, NUS

Comparison of turn rates

Page 21: Energy Maneuvering

G. Leng, ME Dept, NUS

Bottom Line

High turn rates require high load factors

Notes :

Page 22: Energy Maneuvering

G. Leng, ME Dept, NUS

V-n Diagram