s
ME307 1 /9
SCHOOL OF ENGINEERING
LEVEL 3
SEMESTER 2
2003 / 2004
AIRCRAFT PERFORMANCE AND CONTROL
Examiners: Mr. J. Hutchinson / Dr D F Pearce
Attempt FOUR questions only.
Open Book – Student may take their classroom
Time allowed: 2 hours.
Total number of questions: 6
notes into the examination room
All questions carry equal marks.
The figures in brackets indicate the relevant marks for each section.
The following aids are supplied and can be used:Table 1 I.C.A.O standard atmosphere table..Imperial units.
Table 2 I.C.A.O standard atmosphere table..SI units.
Table 3 Dynamic pressure ‘Q’ versus Equivalent airspeed in knots.
Table 4 Relative air density versus altitude and ambient sea level temperature.
Chart 1 Mach number versus calibrated airspeed and altitude.
Chart 2 Compressibility correction to calibrated airspeed versus calibrated
airspeed and altitude.
ME307 2 /9
1) An aircraft is on the approach in a crosswind and is side slipping at an angle of ten degrees.
The airspeed is 113 Kcas and the wind speed is 20 knots. The control calibrations are:-
Rudder deflection in degrees = the pedal deflection in centimetres
Aileron deflection in degrees = 0.333 times the wheel deflection in degrees
Maximum rudder deflection is 25 degrees and maximum aileron deflection is 20 degrees
The following lateral derivatives apply:Roll due to sideslip
CLL
= 0.002
Roll due to ailerons
CLLA = -0.01
Yaw due to sideslip
CN
= 0.002
Yaw due to rudder
CNR
= -0.004
Ignore the rolling moment due to the rudder and other derivatives.
If the aircraft is fully trimmed out find:a)
i) The rudder deflection.
(5)
ii) Pedal deflection.
(5)
iii) The aileron deflection.
(5)
iv) The wheel deflection.
(5)
b)
Explain the difference between sideslipping and crabbing.
c)
What technique would the pilot have to perform just before touch down
if the aircraft was crabbing on the approach?
(3)
(2)
ME307 3 /9
2)
a)
If the drag coefficient for an aircraft can be expressed in the form CD=CD0 + kCL2 show
that the maximum L/D ratio for an aircraft occurs when:-
CL 
Where:-
C D0
k
CL is the lift coefficient.
CD0 is the profile drag.
k is the induced drag factor.
b)
(8)
An aircraft has the following drag coefficient:C D  C D 0  kC 2L
where C D0 = 0.025 and k = 0.055
The wing area is 371.612 m2 and the weight is 2002 kN.
i) Calculate the speed in m/s for minimum drag.
(3)
ii) Calculate the drag in Newtons at this speed.
(3)
iii) The aircraft has four gas turbine engines in which the thrust varies with altitude
and over the speed range to be considered the thrust variation with airspeed is
negligible. The thrust variation with altitude is proportional to the ambient pressure
ratio ‘δ’. The static sea-level thrust is 61385.4 N per engine.
What is the climb gradient in degrees at the minimum drag speed at 3000 m
altitude?
c)
(7)
Explain how the maximum airspeed would be found at any altitude.
Amplify your answer with a diagram.
(4)
ME307 4 /9
3)
a) The distance for an aircraft to accelerate to a given speed can be calculated from
the equation below:-
W
V2
XG =
log e ( 1 +
)
2gK
N
metres
where:W= Aircraft weight (N)
V = Aircraft velocity (m/s)
g = gravitational constant = 9.81 ms-2
(T  μW)
N
K
1
K   ρSC Dg  μC Lg 
2
T = Engine thrust (N)
S = Wing area (m2)
ρ = Air density (kgm-3)
μ = Rolling coefficient of friction
The aircraft starts a take-off but at 51.5 m/s the aircraft suffers an engine failure and the
captain decides to abandon the take-off. After a two seconds engine failure recognition
time he applies full brakes, deploys the speed brakes and throttles back the engines to idle
where the thrust is essentially zero. Deploying the speed brakes reduces the lift coefficient
from 0.4 to 0.05 and increases the drag coefficient to 0.15.With full braking the braking
coefficient is 0.6
The following conditions are for the acceleration phase:CLg
= 0.4
CDg
= 0.05
T
= 444822 N (constant with forward speed).
W
= 2224110 N
S
= 321.0 m2
ρ
= 1.225 kg m-3 air density
μ
= 0.015 (rolling coefficient of friction).
The following conditions are for the deceleration phase:CLg
= 0.05
CDg
= 0.15
T
=0N
W
= 2224110 N
S
= 321.0 m2
ρ
= 1.225 kg m-3 air density
ME307 5 /9
μB
b)
i)
= 0.6 Braking coefficient of friction.
Question 8 continued on page 5/8
Calculate the acceleration distance to 51.5 m/s
(7)
ii)
Calculate the deceleration distance from 51.5 m/s
(7)
iii)
Calculate the distance travelled during the reaction time
(2)
iv)
What is the total accelerate/stop distance?
(2)
The aircraft acceleration can be shown to be:-


 
dV
T
 gV 2
 g 0  μ  
0.5 * ρ * S C D g  μC Lg  k
dt
W
 W
Show that, if parameters other than the lift and drag coefficients are constant, the
maximum ground acceleration is given by :-
C Lg 

2b
where C D g  C D0  bC 2Lg
(7)
ME307 6 /9
4)
a)
i) Derive the equation below for the range of an aircraft in nautical miles.
W 
L
a 0 θ M( )log e  1 
D
 W2 
Range 
sfc
Where:-
(8)
The initial and final weights are W1 and W2 respectively
M is the Mach number
L/D is the lift drag ratio
sfc is the specific fuel consumption in lb/lb.hr or kg/kgf.hr
a0 is the speed of sound at sea-level in knots
θ is the square root of the ambient temperature ratio
A military aircraft is starting on a long range cruise at 30000 ft (9144 m). The weight at
the start of cruise is 400000 lb (181437 kg) and it can carry 150000 lb (68039 kg) of fuel.
The specific fuel consumption is reasonably constant at 0.7 lb/lb.hr (0.7 kg/kgf.hr).
The aircraft has a wing area of 4000 ft2 (372 m2) and the drag equation is:CD = 0.02 + 0.06 CL2
NOTE: At 30000 ft (9144 m) the square root of the ambient density ratio
and the square root of the ambient temperature ratio
σ = 0.6116
θ = 0.891 The speed of
sound ‘a0’ = 661.5 knots at sea-level and the speed of sound ‘a’ = 589 knots
at 30000 ft (9144 m).
b)
ii) Find the cruise Mach number for maximum range.
(4)
iii) Find the the corresponding cruise speed in knots calibrated airspeed.
(4)
iv) Find the range in nautical miles if 100000 lb (45360 kg) of fuel is used.
(3)
ME307 7 /9
i) If the aircraft has to maintain a ground speed of 450 knots in a ten knot headwind
what is the Mach number at altitude?
(3)
ii) What is the corresponding calibrated airspeed Kcas?
(3)
5)
a)
i) Show on a diagram for a conventional aircraft configuration the tail-off and tail-on
pitching moment coefficient curves versus lift coefficient.
(3)
ii) Explain how putting the tailplane on an aircraft makes it stable.
(3)
iii) Indicate the trim point on the diagram.
(2)
b)
Explain what the neutral point is.
(2)
c)
Given that the pitching moment coefficient about the centre of gravity is:Cmcg 

1
x cg  0.25cC L  Cmo  V a 1 
c

 i t    a 2e

Derive the equation for the stick fixed Neutral Point.
d)
(8)
Find the stick fixed Neutral Point for the aircraft flight condition below.
The aerodynamic characteristics are:-
V is 0.6
The lift curve slope of the wing ‘a’ is 0.1 per degree.
The lift curve slope of the tailplane ‘a1’ is 0.06 per degree.
The downwash is given by:
e)
d 

1 
 = 0.6
 d 
Show on a diagram how the neutral point could be found graphically.
(5)
(2)
ME307 8 /9
6)
a)
i)
Calculate the ambient density ratio at an altitude of 13715 m in
the International Standard Atmosphere.
ii)
(2)
Calculate the ambient pressure ratio at an altitude of 13715 m in
the International Standard Atmosphere.
b) Explain the use of the International Standard Atmosphere.
(2)
(3)
c) On an ISA+30ºC day an aircraft is flying at 20000 ft (6096 m) pressure altitude. If the
aircraft is fitted with an altimeter that has no instrument errors what is the geometric
(true) altitude above mean sea level?
(5)
d) An aircraft is flying at Mach 0.8 at 35000 ft (10668 m) on a standard day. Calculate or use
the applicable charts to find the following parameters:i)
Calibrated airspeed in knots-Kcas.
(2)
ii)
Equivalent airspeed in knots-Keas.
(2)
iii)
True airspeed in knots-Ktas.
(2)
e)
Explain the difference between calibrated airspeed and equivalent airspeed.
f)
Explain how the pressure height can be found for a non-standard temperature day if
the geometric height is given.
(4)
(3)
ME307 9 /9