1000 Questions
AMSWERS AND EXPLANATIONS
FOR .,
JAR ATPL I N D CPL
PRINCIPLES OF FLIGHT
1000 Questions
ANSWERS AND ENPLANATIONS
FOR
JAR ATPL AND CPL
PRINCIPLES OF FLIGHT
Keith Williams
O Keith Williams, 2005
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out or otherwise circulated without the pr/or permission of the author.
The information contained in this publication is for private study purposes only. Whilst every effort has been made
to ensure its accuracy and validity, no responsibility is accepted for errors or discrepancies.
Printed at
Thakur Enterprises
ISBN 81-7002-094-8
A Himalayan Books Presentation
First Published in 2005 by Himalayan Books, New Delhi 110013
Distributed by
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CONTENTS
Section 1
Formulae and Shortcuts.
Section 2
Questions.
Section 3
Summary of Answers.
Section 4
Answers and explanations.
INTRODUCTION
-The purpose of this book is to assist students preparing to undertake the JAR
A-TPLand CPLAircraft performance examinations. The majority of the questions
are based upon feedback provided by students who have undertaken 'the JAR
A'TPL examination. By drawing feedback from the~widestpossible range of
sources, this book provides a more comprehensive range of questions than can
be achieved by any single ground school consulting only its own students. It
should however be noted that these questions are not verbatim copies of those
used in examination. Students should therefore use the material in this book as
an aid to developing an understanding of the subject, rather than as a list of
questions and answers to be memorised.
Other questions are intended to assist students in developing the
required level of understanding of the more important aspects of each part of the
syllabus. The explanations provided in Section 4 indicate the degree of
understanding required of students undertakingthese examinations.
The JAR CPL Principle of Flight syllabus is similar to that for the ATPL,
with the exceptions of matters relating to high speed flight and control flutter.
d
ignore all
Students preparing for the CPL examinations s h o ~ ~ ltherefore
questions relating to these subjects. Readers wishing to make comment on the
contents of this book or to provide feedback for future updates should contact the
author at e-mail address, keith.williams13@virgin.net
SECTION 1
FORMULAE AND SHORTCUTS
Subject
Whole Aircraft Lift:Drag Polar Diagram.
Airspeeds.
Lift.
Total Drag.
Profile Drag.
Lift Induced Drag.
Calculation of stalling speed following a weight change.
Shortcut calculation for small weight changes.
at any given airspeed.
Calculation of CL as a O h of CLMAX
Load Factor (n).
Load factor in balanced turns.
Calculation of gust load factor.
Changes in stalling speed caused by variations in load factor.
Changes in stalling speed caused by turning flight.
Calculation of new stalling speed following changes in bank angle.
Design manoeuvre speed VA.
Calculation of new VA following changes in weight.
Radius of turn calculations.
Rate of turn calculations.
Calculation of new fuel flow following a change in weight.
WHOLE AIRCRAFT L1FT:DRAG POLAR
DIAGRAM.
A
Stall speed
Best Prop endurance speed
Best glide endurance speed ( V M D ~ )
limb speed VyProp
Best prop range speed
Best jet endurance speed
Best jet angle of climb speed V X J ~ ~
Best glide range speed (VR~GA)
Best Lift:Drag ratio
Best jet range speed VMCR
Best jet rate of climb speed VY.J,~
IDENTIFICATION O F KEY POINTS
Vs occurs at the point of greatest CL at the top of the curve.
VMDoccurs where a tangent drawn from the origin touches the curve.
Airspeed increases in an anti-clockwise direction around the curve.
Vx Prop is slightly less than VMP.
Vy Prop is slightly greater than VMP.
VMPis greater than Vx Prop but less than Vy Prop.
Vy Prop is greater than VMpbut less than VMD.
VY Jet is VMD.
VY Jet is equal to ChlRCand is approximately 1.32 VMD.
AIRSPEED
The airspeed indicator produces an Indicated Airspeed output (IAS) that is
.
p is air density and V is TAS.
proportional to ?4 p ~ 2 Where
2 always produce the same IAS, regardless of
Any given value of % p ~ will
altitude. Climbing at constant IAS therefore means climbing at c o ~ ~ s t a?4
l ~P t ~ 2 .
But p decreases with increasing altitude, so The TAS equating to any given IAS
must increase, such that the rate of decrease in p is equal to the rate of increase
in (TAS)~.At 40000 feet in the standard atmosphere, p is approximately ?A of its
sea level value, so TAS is approximately twice IAS.
As altitude increases up to the tropopause at 36000 feet, air temperature and the
local speed of sound decrease. Above 36000 feet air temperature and the local
speed of sound remain constant. Climbing at constant mach number therefore
means IAS and TAS decrease up to 36000 feet. Above 36000 feet IAS continues
to decrease but TAS remains constant. The relation ship between EAS, CAS
TAS and Mach number in various atmospheric conditions can be determined
using the diagrams below:
v
EAS CAS TAS MACH
ALT
For altitudes below 36000 feet the speed lines
Move further apart as altitude increases.
Draw the chart such that the constant
parameter is vertical. The effect on the other
two is then indicated by the lines. This chart
Indicates a constant TAS climb with EAS and CAS
decreasing, while Mach increases.
For descents follow the lines down the chart. For
altitudes below se? level the lines may be extended
below their crossover point.
Correct for ISA to 36000 ft.
EAS :AS
ALT
\
TAS and MACH
In isothermal layers and above 36000 feet the
temperature is constant and-so the TAS: MACH ratio
is constant. So both are represented by a single line.
The EAS, CAS and the TAS+MACH lines move apart
with increasing altitude indicating that tlie EAS and
CAS equating to any given TAS or Mach number
decreases with increasing altitude.
This chart indicates that in a constant CAS climb
above 36000 feet, TAS and Mach number increase,
while EAS decreases.
Correct for Isothermal Layers and above the Tropopause.
EAS CAS MACH
TAS
In an illversion the normal temperature lapse process
is reversed, s u c l ~that temperature increases as altitude
increases. But the TAS at any given MACH number is
determined by temperature. So as altitude increases in
an in an inversion, the MACH a t any given TAS
decreases. The overall effect of this is a reversal of the
order of the TAS and MACH lines in the graph.
This effect is shown in the graph a t the left, which
indicates that in a constant CAS climb in an inversion,
the EAS decreases, while the MACH and TAS both
increase.
ALT
Correct for Inversions.
LIFT
Lift = C L ?4 pv2s
Where
CLis the coefficient of lift.
p is the air density.
V is the TAS.
S is the wing area.
Because p decreases with increasing altitude but % pvZis constant for any given
IAS, climbing a t constant IAS requires constant C L and hence constant angle of
attack to maintain constant lift.
Climbing at constant TAS requires increasing angle of attack to maintain
constant lift.
TOTAL DRAG
The equation for total drag is: D = CD%pv2s.
And CD= CDl+ CDp
Where
CDis the coeficient of total drag.
CDlis the coefficient of induced drag.
CDPis the coefficient of profile drag.
PROFILE DRAG
The coefficient of profile drag is approximately constant at normal angles of
attack, so profile drag a t any given speed does not vary with weight.
The equation for profile drag is: Dp = Cup % p ~ 2 ~ .
Dp is proportional to v2 where V = EAS.
INDUCED DRAG
The coefficient of induced drag is proportional to CL' SO induced drag a t any
given airspeed increases with increasing weight.
The equation for induced drag is: DI = Cl)l% P ~ 2 ~ .
DI is proportional to 1 / v 2 Where V .is the EAS.
CD1is proportional to 1/v4
Where V is the EAS.
CDIis also proportional to cL2,
CD, =
G~
Where k is the induced drag factor.
(k is 1 for elliptical plan forms)
A is the aspect ratio.
k nA
CALCULATION O F STALLING SPEED FOLLOWING WEIGHT CHANGES
Vs (at new weight) = Vs (at previous weight) x d(new weight / old weight)
SHORTCUT CALCULATION O F STALLING SPEED FOLLOWING SMALL
WEIGHT CHANGES
For small changes in weight the % increase in stalling speed is approximately
half the % increase in weight.
Also as VMDand other significant speeds are fixed fractions of Vs, small changes
in weight affect these speeds in the same manner.
For example a 10% increase in weight gives 5% increase in stalling speed and a
5% increase in VMD.
CALCULATION O F CI,AS % O F CLMAX
AT ANY GIVEN SPEED
To calculate CL as a % of CLMAX
at any given speed (V) use the equation:
CL at any V
= CLMAX
( ~ ~ 1/: v 2 ) x 100%
Where
For example:
CL at 1.2Vs =
V = the given speed.
Vsls is the basic l g stalling speed
(12/ 1.2~)x 100%
which is 69% of C L M ~ ~ .
LOAD FACTOR
Load factor = L/W
GUST LOAD FACTOR
Vertical gust cause random increases and decreases in CL. These cause random
changes in load factor. Load factor in a gust can be calculated using the
equation:
Gust load factor = CL in the gust 1 CL immediately before the gust
-
Gust Load intensity is a measure of the magnitude of the changes in load factor
caused by gusts.
Increases with decreasing aircraft weight.
Increases with increasing airspeed.
Is not affected by altitude.
Gust Load Intensity:
LOAD F.$ CTOR IN A TURN
Load factor in a banked turn = l/COS AOB
For example the Cosine of 60' is K, so load facto-r in a 60' bank is 1/(1/2) = 2.
EFFECT OF LOAD FACTOR ON STALLING SPEED
Where n is the load factor.
Stalling speed in a manoeuvre (VM)= Vslgdn
EFFECT OF TURNING ON STALLING SPEED
/COS
Stalling speed in a turn = V S ~ ~ ~ ( ~ AOB)
For example the Cosine of 60' is K so stalling speed in a 60' banked turn is
vslgd(2). That is 1.41 times the basic stalling speed.
CALCULATION OF STALLING SPEED AFTER CHANGE OF BANK
ANGLE
To calculate the stalling speed following a change in bank angle use:
VS at new AOB = Vs at old AOB ~ ( C O Sold AOB / COS new AOB)
THE DESIGN MANOEUFRE SPEED (VA
VAis the EAS at which the aircraft will stall at its limiting load factor.
For flaps up limiting load factor is 2.5g, for flaps down it is reduced to 2g.
CALCULATION OF VAFOLLOWING WEIGHT CHANGE
To calculate VAfollowing a weight change use:
VAat new weight = VAat old weight d(new weight / old weight)
'
RADIUS OF TURN
To calculate radius of turn use:
R = V* / TAN AOB
Where V is TAS.
Note that units of TAS, g and r must be compatible so to get r in meters, g must
be in mls2, and V must be in mls. To convert Kts into m/s multiply speed in Kts
by 0.515. For example 100 Kts = 51.5 m/s.
RATE OF TURN
To calculate rate of turn use:
ROT in radians per second = g TAN AOB / V Where V = TAS.
Notes.
1. Both g and V must be in the same units.
2. Unless otherwise stated g = 9.81 m/s2 and 1K t = 0.515 mls.
3. 1 radian = 57.3' so to calculate ROT in degreesls use:
ROT in degls = 57.3 g TAN AOB I V.
CALCULATING CHANGE IN FUEL FLOW FOLLOWING A CHANGE IN
WEIGHT
For jet aircraft the fuel tlow is proportional to the thrust. I n straight and level
flight thrust equals drag, so fuel flow is proportional to drag. But Induced drag
is proportional to the square of the coefficient of lift. When flying at Vmd, the
total drag is made up of equal parts of induced drag and profile drag. At this
speed any given OO/ change in weight will cause the same % change in drag and
hence the same O h change in fuel flow.
This can be summarised as Fuel Flow 1IWeight 1 = Fuel Flow 2 I Weight 2
Rearranging this equation gives:
Fuel Flow 2 = (Fuel Flow l/(Weight 11Weight 2)
Or:
Fuel Flow a t new weight = Fuel Flow at old weight x (New Weight 1 Old Weight)
The above equation is valid only for jet aircraft and only when flying a t VMD.
For any examination question stating that a jet is "in a hold" or "flying at
holding speed", students should assume that the aircraft is flying at VMD.
STABILITY
Lateral stability stronger that directional stability causes Dutch Roll.
Directional stability stronger than lateral stability causes Spiral Instability.
Downwash from the tailplane decreases the stabilising effect of the tailplane.
High tailplanes improve longitudinal stability but keeping the tailplane clear of
downwash from the wings.
High tailplanes increase the difficulty in recovering from deep stall.
Increasing o r decreasing camber does not affect longitudinal stability because
pitching moment about the aerodynamic centre is constant.
Ventral fins do not affect longitudinal stability, increase directional stability,
decrease lateral static stability, but increase lateral dynamic stability.
SECTION 2
Subject
Basics.
Aerofoil sections and wing planforms
Lift and drag.
High lift devices.
Stalling and stalling speed calculations.
Flying controls.
Climbing flight.
Turning flight.
Power available and power required.
Stability and manoeuvrability.
High speed flight.
Propellers.
Flight envelopes and aircraft performance.
Pages
to
20
to
25
to
43
to
62
to
76
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95
to
115
to
129
to
144
to
161
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1 1
to
194
to
213
BASIC 1.
3
Newton's second laws states that ..........
a.
b.
c.
d.
F = MA.
F = MIA.
A = MF.
M = AF.
BASIC 2.
The direction of static air pressure is .....?
a.
b.
c.
d.
Parallel to dynamic pressure.
Normal to dynamic pressure.
In all directions.
In no directions.
BASIC 3.
Bernoulli's theorem states that.. ....?
a.
b.
c.
d.
Air is incompressible.
Total pressure is constant.
Air has viscosity.
Air does not expand.
BASIC 4.
The SI units of mass and weight are
a.
b.
c.
d.
Kilograms,
Newtons,
Kilograms,
Newtons,
.... and .......
Newtons.
Kilograms.
Kilograms.
Newtons.
BASIC 5.
The SI units of density and velocity are
a.
b.
c.
d.
b.
c.
d.
....and .....?
Kilograms per cubic centimetre,
Kilograms per cubic metre,
Newtons per cubic gram,
Kilograms per cubic kilometre,
BASIC 6.
Bernoulli's theorem assumes that
a.
3
Air is compressible.
Temperature changes.
Density changes.
Air is inviscid.
Centimetres per second.
Metres per second.
Kilometres per minute.
Cubits per second.
........
3
BASIC 7.
Dynamic pressure acts ..........9
a.
b.
c.
d.
In all directions.
Downstream.
Upstream.
Upwards.
BASIC 8.
A line midway between the-upper and lower surfaces of a wing is the....
a.
I>.
c.
d.
Chord line.
Mean chord line.
Meail camber line.
Mean aerodynamic centre.
BASIC 9.
An imbalance of moments about a point will cause
a.
b.
c.
d.
......?
Co~istantrate rotation.
Decreasing rate rotation.
Constant angular acceleration.
Spinning.
BASIC 10.
The angle of attack is between the ...... ?
a.
b.
c.
d.
Flight path and chord line.
Fliglit path and longitudinal axis.
Relative airflow and longitudinal axis.
Relative airflow and chord line.
BASIC 11.
The SI unit of density is ........and that of energy is ......?
a.
b.
c.
d.
Kg/min3,
Kg/m3,
Kg/m3,
~ / m ~ ,
BTU.
Therm.
Joule.
Joule.
BASIC 12.
The weight of a mass is calculated using the constant
a.
b.
c.
d.
1N = 9.81 Kg.
1N = 1 Kg.
l N = 32Kg.
9.81N = 1Kg.
.......?
.......
9
BASIC 13.
A Newton is ...............7
a.
b.
c.
d.
The force required to give a mass of 1 Kg an acceleration of 1 m/s2.
The force required to give a mass of 1 Kg an acceleration of 9.81 m/s2.
The mass requiring a force of 1 Kg for an acceleration of 9.81 m/s2.
The mass requiring a force of 1 Kg for an acceleration of 1 m/s2.
BASIC 14.
A force of 1N applied to a mass of 1Kg it will cause an acceleration of .........7.
BASIC 15.
When flying below ISA MSL TAS will be ........IAS?
a.
b.
c.
d.
The same as.
Higher than.
Twice.
Lower than.
BASIC 16.
When flying a t 40000 feet above ISA MSL TAS will be approximately ........IAS?
a.
b.
c.
d.
Half.
Twice.
The same as.
Depends on load factor.
BASIC 17.
A pressure of 100 kPa acting on a surface of 1 m2 will exert a force of ...?
a.
b.
c.
d.
100,000 N.
1 N.
1,000,000 N
10,000,000 Kg.
BASIC 18.
The most efficient angle of attack of an aerofoil is ........7.
a.
b.
c.
d.
About -4'.
About 4'.
About 15'.
That which provides the best L:D ratio.
BASIC 19.
The zero lift angle of attack of a symmetrical aerofoil is
.....?
BASIC 20.
The zero lift angle of attack of an asymmetrical aerofoil is approximately.....?
BASIC 21.
Air density is increased by ........?
a.
b.
c.
d.
Increasing humidity.
Decreasing temperature.
Acceleration.
Acceleration o r deceleration depending on velocity.
BASIC 22.
Air viscosity is
a.
b.
c.
d.
............
3
Its ability to flow.
Its ability to expand and contract.
Its ability to resist acceleration.
Its resistance to flow.
BASIC 23.
When subsonic airflow passes through a convergent duct its velocity ..... and its '
.?
static pressure.
....
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
Increases.
Decreases.
Remains constant.
Decreases.
BASIC 24.
When subsonic airflow passes through a convergent duct its temperature
.......and its density ......?
a.
b.
Increases,
Increases,
Increases.
Decreases.
c.
d.
Decreases,
Decreases,
Remains constant.
Decreases.
BASIC 25.
When supersonic airflow passes through a convergent duct its velocity
static pressure.. ....?
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
.....its
Increases.
Decreases.
Remains constant.
Increases.
BASIC 26.
Accelerating airflow first reaches sonic speed in a convergent duct a t the
9
after which its velocity ...........
a.
b.
c.
d.
Centre,
Inlet,
Throat,
Inlet o r throat depending on temperature,
.......
Increases.
Decreases.
Remains constant.
Decreases.
BASIC 27.
In order to accelerate supersonic airflow a ........duct is required?
a.
b.
c.
d.
Convergent.
Parallel.
Convergent-divergent.
Divergent.
BASIC 28.
Sonic airflow will not accelerate through a parallel duct because.
a.
b.
c.
d.
............
3
It requires a convergent duct to accelerate.
It requires a divergent duct to accelerate.
Acceleration makes it supersonic after which it requires a divergent duct
to accelerate.
It becomes supersonic then requires a convergent duct to accelerate.
BASIC 29.
Acceleration of subsonic airflow through a convergent duct causes pressure and
temperature to reduce because.. ..?
a.
b.
c.
d.
It is incompressible.
Its total pressure remains constant.
Its total energy remains constant so increasing kinetic energy is matched
by decreasing static pressure and thermal energy.
It is compressible.
BASIC 30.
When supersonic airflow passes through a convergent duct its temperature .....
and its density .....?
a.
b.
c.
(1.
Increases,
Increases,
Decreases,
Decreases,
Increases.
Decreases.
Remains constant.
Decreases.
FORMS 1.
Two aircraft have the same wing shape and area if one has straight wings and
the other swept wings how will the aspect ratios of the two aircraft compare?
a.
b.
c.
d.
No change.
Aspect ratio will increase as wings stveepback angle increases.
Aspect ratio will decrease as wing sweepback angle increases.
Depends on the C of G position.
FORMS 2.
Which wing plan forms a r e most susceptible to tip stall?
a.
b.
c.
d.
High aspect ratio, elliptical, Low taper ratio, straight.
High aspect ratio, pointed, high taper ratio, swept.
Low aspect ratio, elliptical, low taper ratio, straight.
Low aspect ratio, pointed, low taper ratio, swept.
FORMS 3.
Which of the following are true of low aspect ratio rectangular wings?
a.
b.
c.
d.
They produce low induced drag and high profile drag.
They produce low prflile drag and high induced drag.
They are difficult to construct and their roots stall before their tips.
'They a r e easy to construct and their tips stall before their roots.
FORMS 4.
In what direction is the airflow deflected as it passes over the tips of lift
generating pointed wings.
a.
b.
c.
d.
Downwards.
Forwards.
Upwards.
Aft.
FORMS 5.
Which of the following statements are true of elliptical wings?
1.
Almost the entire wing reaches its stalling angle simultaneously.
,
2.
3.
4.
It poses structural problems making design and construction difficult.
The aircraft exhibits little o r no pre-stall'buffet.
Aileron effectiveness is reduced close to the stall.
a.
1, 2,3.
1,2,4.
2,3,4.
All of the above.
b.
c.
d.
FORMS 6.
What is the principal advantage of an elliptical wing plan form?
a.
b.
c.
d.
Reduced induced drag.
Reduced wing flutter.
Maximum strength : weight rati.0.
Reduced profile drag.
FORMS 7.
Where on a lift generating pointed wing is the greatest downwash?
a.
b.
c.
d.
At the tips.
Close to the roots
At the % span point.
At the leading edge.
FORMS 8.
Where on a CI/CL: semi-span distance chart do the curves for all non-elliptical
wings coincide?
a.
b.
c.
d.
Approximately a t the 1.1, 0.55 point.
Appl-oximately at the 0 , 0 point.
Approximately a t the 0 . 5 5 , l . l point.
Approximately at the 1 . 1 point.
FORMS 9.
Which of the following statements best describes the effect of reducing aspect
ratio?
a.
b.
c.
d.
The slope of the CL:a curve decreases and stalling angle increases.
The slope of the CL:a curve decreases and stalling angle decreases.
The slope of the CL:a curve increase and stalling angle increases.
The slope of the CL:a curve increase and stalling angle decreases.
FORMS 10.
Which of the following is the most accurate definition of dihedral angle?
a.
The horizontal angle between the 25% chord line and the lateral axis.
b.
c.
d.
The vertical angle between the longitudinal axis and the wing spar.
The vertical angle between the wing plane and tlie lateral axis.
The angle between the chord line and the longitudinal axis.
FORMS 11.
What a r e the angle of incidence and incidence respectively?
a.
b.
c.
d.
They a r e the same thing.
The angle between the chord line and the longitudinal axis and the angle
of attack respectively.
The angle of attack and the angle between the chord line and the normal
axis.
The angle between the chord line and the longitudinal axis, and between
the chord line and the free stream airflow respectively.
FORMS 12.
What is the relationship between the mean camber line and the chord line of a
symmetrical aerofoil?
a.
b.
c.
d.
They are the same thing.
They a r e closest a t the mid chord point.
They never meet.
They have nothing in common.
FORMS 13.
What is the principal cause of deep stall?
a.
b.
c.
d.
Tip stalling of swept back wings.
Tip stalling of straight wings.
Tip stalling of tapered wings.
High tailplanes.
FORMS 14.
How is the aspect ratio of a wing of complex plan form calculated?
a.
b.
c.
d.
Span / chord.
Chord / span.
By computer.
Span squared / wing area.
FORMS 15.
What is the angle of attack?
a.
b.
c.
d.
T h e angle between the chord line and the longitudinal axis.
The angle between the chord line and the relative airflow.
The angle between the chord line and the free stream airflow.
The angle between the total reaction and the chord line.
FORMS 16.
How does a supercritical aerofoil compare with a convention design?
a.
b.
c.
d.
Thinner, more cambered and convex under rear surface.
Thicker, less cambered and concave under rear surface.
Thinner, less cambered and concave under rear surface.
Thicker, more cambered and convex under rear surface.
FORMS 17.
What effect will increasing aspect ratio have on drag a t low a n d high speeds?
a.
b.
c.
d.
Decrease profile a t low speeds and increase induced a t high speeds.
Decrease induced a t low speeds, no effect at high speeds.
Increases induced a t low speed and decrease induced at high speeds.
Decrease induced and increase profile a t all speeds.
FORMS 18.
What effect will increasing aspect ratio have on stalling angle?
'
a.
b.
c.
d.
Decrease.
Increase.
None.
Increase o r decrease depending on weight.
FORMS 19.
What effect will increasing aspect ratio have on stalling speed?
a.
b.
c.
d.
Increase.
Decrease.
Noce.
Increase or decrease depending on weight.
FORMS 20.
Why do aircraft with swept back wings pitch u p in the stall?
a.
b.
c.
d.
C of P moves aft.
C of P moves outward and forward.
C of P moves inward and forward.
They do not pitch up.
FORMS 21.
How does the CL:a curve of a swept wing compare with that of a straight wing?
a.
b.
c.
d.
Steeper and lower maximum.
Shallower and higher maximum.
Steeper and higher maximum.
Shallower and lower maximum.
FORMS 22.
How does the stalling angle of a swept wing compare with that of a straight wing
of equal section?
a.
b.
c.
d.
Lower.
Higher.
The same.
Higher o r lower depending on weight.
FORMS 23.
How would the stalling speed of a swept wing compare with that of a straight
wing of equal section?
a.
b.
c.
d.
Higher.
Lower.
The same.
Higher or lower depending on altitude.
FORMS 24.
What is the mean aerodynamic chord?
\
a.
b.
c.
d.
The shortest.
The longest.
The flattest.
The average.
FORMS 25.
9
Increasing sweep back angle .........
a.
b.
c.
d.
Decrease induced drag.
Increases profile drag.
Increases MCRIT.
Increases wave drag.
FORMS 26.
Increasing sweep back angle ..........9
a.
b.
c.
d.
Increases induced drag.
Increases profile drag.
Increases wave drag.
Increases flutter.
FORMS 27.
Increasing sweep back angle
a.
b.
,
'
.......?
Increases gust response.
Decrease gust response.
c.
d.
Increases flutter.
Increases divergence.
FORMS 28.
Aerodynamic washout is ........9
a.
b.
c.
d.
Changing wing section from root to tip.
Increasing T:C ratio from root to tip.
Down wash.
Oscillating pitching motion.
FORMS 29.
Divergence is caused by
a.
b.
c.
d.
........9
C of P ahead of torsional axis, thin wings and excessive airspeeds.
C of P aft of torsioiial axis, thin wings and excessive thrust.
C of G aft of C of P, thick wings and excessive weight.
C of P aft of C of G, thin wings and excessive airspeed.
FORMS 30.
Angle of incidence is ......?
a.
b.
c.
d.
Between camber line and longitudinal axis.
Between chord line and longitudinal axis.
Between chord line and lateral axis.
Between camber line and lateral axis.
LD 1.
Which of the following statements is most accurate?
a.
b.
c.
d.
Dl is proportional to 1 / (EAS)~
Dl is proportional to 1 I (IAS)~
Dl is proportional to 1 I (RAS)~
Dl is proportional to 1 I (TAS)~
LD 2.
Which of the following statements is most accurate?
a.
b.
c.
d.
Dl is proportional to L.
Dl is proportional to LIV.
DI is proportional to 1/L.
Dl is proportional to V/L.
LD 3.
If a 50000 Ibf aircraft requires 25000 lbf of thrust maintain unaccelerated
straight and level flight at 250 Kts what is its L:D ratio?
LD 4.
What effect does increasing load factor have on power required at any given
aircraft weight and TAS?
a.
b.
c.
d.
No effect.
Increases in direct proportion to load factor.
Decreases in direct proportion to load factor.
Power required is inversely proportional to load factor.
LD 5.
What effect will increasing weight have on DI, Dp and DTotalat any given load
factor and airspeed?
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
No significant effect,
Decrease,
Increase,
No significant effect,
Increase.
No significant effect.
No significant effect.
Increase.
LD 6.
What effect will increasing aircraft weight have on minimum drag speed (VMD)
and speed stability.
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
Increase.
Decrease.
Increase.
Decrease.
LD 7.
What effect will lowering the landing gear have on VMDand speed stability?
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
Increase.
Decrease.
Increase.
Decrease.
LD 8.
What is the effect on D1 qnd Dp when an aircraft climbs to altitude a t constant
low IAS .
a.
b.
Increase,
Increase,
Increase.
Decrease.
c.
d.
No change,
Decrease,
No change.
Decrease.
LD 9.
-
What would be the effect on Dl, Dp and speed stability if the trailing edge flaps
of an aircraft were lowered to the 10 degree setting while maintaining constant
speed and wings level?
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
Increase,
Increase,
Increase,
Decrease,
Increase.
Decrease.
Increase.
Decrease.
LD 10.
What would be the effect on Dl, Dp and speed stability if the trailing edge flaps
of an aircraft were lowered to the 40 degree setting while maintaining constant
speed and wings level?
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
Increase,
Increase,
Increase,
Decrease,
Increase.
Decrease.
Increase.
Decrease.
LD 11.
What will be the effect on drag if humidity is increased if air temperature,
pressure and TAS remain constant?
a.
b.
c.
d.
Increase.
Decrease.
Remain constant.
Decrease o r increase depending on Mach number.
LD 12.
When flying at best L:D ratio what is the ratio of Dp: Dl?
a.
b.
C.
d.
2:l.
1:l.
1:2.
Depends on aerofoil section.
LD 13.
Which of the following occur at VhfD?
a.
b.
c.
d.
Minimum power required and best L:D ratio.
Minimum drag and greatest L:D ratio.
Minimum angle of attack and best rate of climb.
Minimum drag and greatest jet propeller aircraft range.
LD 14.
In what direction does the drag force act?
a.
b.
c.
d.
Parallel to relative airflow.
Parallel but opposite to the direction of flight.
Parallel to lift.
Parallel to weight.
..
LD 15.
Which of the following statements are true of the total reaction generated by a
wing flying at zero lift angle of attack.
a.
b.
c.
d.
It equals Dp.
It equals DI.
It equals weight.
It is twice Dp.
LD 16.
In what direction does the total reaction act when a wing is flying a t its zero lift
angle of attack?
a.
b.
c.
d.
Parallel but opposite to the direction of flight.
Vertically aft through the C of G.
Vertically upward through the aerodynamic centre.
Vertically aft such that it produces a nose up pitching moment.
LD 17.
-
What is the relationship between Dl and Dp at speeds below VMD?
a.
b.
C.
d.
Dl is greater than Dp.
Dl is less than Dp.
Dl = Dp.
Dl = Dp = D~otal.
LD 18.
Complete the following statement. As airspeed changes from VMOto CLMax
increases and .............. decreases.
................
a.
b.
C.
d-
DP,
Dl,
D~otal,
D Total,
D 1.
DP.
- Dp.
Dl.
LD 19.
Which of the following definitions of Dp is the most accurate?
a.
It is made up of form, friction and induced drag.
b.
c.
d.
It is made up of friction, induced and shock drag.
It is made up of friction, form and interference drag.
It is made up of interference, shock and form drag.
LD 20.
Which of the following statements concerning the generation of lift is most
accurate?
a.
b.
c.
d.
Lift is generated by a cambered aerofoil.
Lift is generated by an aerofoil at a positive angle.
Lift is generated by high speed airflow.
Lift is generated by the downward acceleration of air.
LD 21.
Which of the following causes induced drag (Dl)?
a.
b.
c.
d.
Shock waves above and below the wing.
Friction due to the air passing over the wing.
Rotating airflow caused by wing tip vortices.
The upwash of air caused by wingtip vortices.
LD 22.
Which of the following causes induced drag (Dl)?
a.
b.
c.
d.
Shock waves above and below the wing.
Friction due to the air passing over the wing.
Downwash of airflow over the trailing edge caused by wing tip vortices.
Upwash of airflow over the trailing edge caused by wingtip vortices.
LD 23.
A model of an aerofoil is placed in a rectangular wind tunnel such that its tips
are embedded in the sides of the tunnel. Which of the following statements best
represents the flow of air over the aerofoil when it is generating a lift force.
a.
b.
c.
d.
Because the wing effectively has no wingtips there will be no tip vortices
and no downwash.
Because the wing is generating lift there will be downwash but this will be
reduced by the effective lack of wingtips.
Downwash will not be reduced by the lack of wingtips.
The wing is effectively tw"o dimensional so its CL:a curve will be vertical.
I t will therefore produce no induced drag.
LD 24.
If it were possible to build and fly a wing of infinite length how much downwash
would it create and what would be the shape of its CL:a curve?
a.
No downwash,
Vertical CL:a curve.
b.
c.
d.
Little downwash,
No downwash,
Little downwash,
Shallow CL:a curve.
Shallow CL:a curve.
Vertical CL:a curve.
LD 25.
Which of the following is responsible for the creation of induced drag?
a.
b.
c.
d.
Angle of attack, camber, wing area and airspeed.
Pitch angle, camber, wing area and airspeed.
Pitch angle, camber, wing area and angle of attack.
Airspeed, wing area and pitch angle.
LD 26.
What proportion of total drag is made up of induced drag when flying at VMD?
L D 27.
If TAS is increased from 300 Kts to 400 Kts with no change in altitude,
configuration o r weight, by what percentage will power required change?
a.
b.
c.
d.
Decrease by 135%
Decrease by 35%
Increase by 135%
Increase by 235%
LD 28.
All other factors being equal, minimum drag is ...........
9
a.
b.
c.
d.
Constant.
Proportional to weight.
A function of density altitude.
A function of pressure altitude.
LD 29.
If air density is reduced by a factor of 4, by what factor will drag alter?
a.
b.
c.
d.
Decrease by a factor of 2.
Decrease by a factor of 4.
Increase by a factor of 4.
Decrease by a factor of 16.
LD 30.
If indicated airspeed is maintained constant while air density decreases by half
what will be the effect on total drag?
a.
b.
c.
d.
Increase by a factor of 2.
Decrease by a factor of 2.
Increase by a factor of 4.
Remain unchanged.
LD 31.
In straight and level flight an aircraft has a CLof 0.55. If an upward gust of
wind causing a loincrease in angle of attack would increase CLto 0.7, what
increase in CLwould be produced by an upward gust of twice the intensity?
LD 32.
If IAS is increased from 100 Kts to 200 Kts, by what factor would DI and CD1be
multiplied?
LD 33.
How does the drag produced by a turbulent boundary layer compare with that
of a laminar one?
a.
b.
c.
d.
Identical.
Greater friction drag.
Less friction drag.
Less wave drag.
LD 34.
What is the principal benefit of a turbulent boundary layer?
a.
b.
c.
d..
Less friction drag.
Less form drag.
Later separation.
Earlier transition.
LD 35.
In what direction does the weight of an aircraft act?
a.
b.
c.
d.
At right angles to the flight path.
Opposite lift.
Straight down.
Depends on rate of turn.
LD 36.
In what direction does lift act?
a.
b.
c.
d.
Straight up.
At right angles to the flight path.
At right angles to the relative airflow.
At right angles to thrust.
LD 37.
The majority of lift is produced by?
a.
b.
c.
d.
High pressure below the wing.
Low pressure above the wing.
Increased velocity below the wing.
Increased density below the wing.
LD 38.
How does static pressure below a wing at low positive angles of attack compare
with the local ambient static pressure?
a.
b.
c.
d.
Higher.
The same.
Lower.
Higher o r lower depending on speed.
LD 39.
Which of the following would give minimum glide gradient?
abc.
d.
C? 1 CD Min.
c D 2 1 CLMax.
c L 4 1 CDMi".
CL/CDMax.
LD 40.
Which of the following is the correct formula for CD1?
LD 41.
Entering ground effect causes?
a.
b.
c.
d.
Low pressure below the wings reduces lift.
Downwash is increased.
Downwash is reduced.
The aircraft slows down.
LD 42.
How does effective aspect ratio change when leaving ground effect?
a.
b.
c.
d.
Returns to actual value.
Increases.
No effect.
Increases o r decrease depending on speed.
LD 43.
In what way does ground effect influence induced drag?
a.
b.
c.
d.
lncreases it.
No effect.
Decreases it.
Angles it aft.
LD 44.
A low speed aircraft climbs from sea level to 40000 feet pressure altitude at
constant TAS. By what factor will its profile drag be multiplied if all other
factors remain unchanged?
LD 45.
At what height does grou-nd effect become significant?
a.
b.
c.
d.
1 wing span.
2 wing spans.
% wing span.
!4 wing span.
LD 46.
What will be the effect of raising the undercarriage of an aircraft?
a.
b.
c.
d.
Increased induced drag due to more efficient lift production.
Decreased induced drag due to less efficient lift production.
Increased profile drag and greater speed stability.
Decreased profile drag and lower speed stability.
LD 47.
What causes wing tip vortices?
a.
b.
c.
d.
Pressure differences in front of and behind the wings.
High pressure air leaking from below the wings.
Spanwise flow from tip to root under the wings.
Spanwise flow from root to tip above the wings.
LD 48.
How does aspect ratio affect wingtip vortex strength?
a.
b.
c.
d.
Shorter tip chord length.
Longer tip chord length.
Shorter root chord length.
Higher energy airflow.
LD 49.
If airspeed changes from 200 Kts to 400 Kts by what factor will CDIand CDP
change?
a.
b.
c.
d.
2,
1/16,
No change,
114,
2.
No change.
1/16.
114.
LD 50.
How will power required vary when leaving ground effect?
a.
b.
c.
d.
No change.
Increase.
Decrease.
Decrease o r increase depending on airspeed.
LD 51.
In what ways are DI and Dp proportional to V?
a.
b.
C.
Inversely.
v3,
v2,
11~3.
11v2.
LD 52.
'What happens to total drag when increasing speed to V2?
a.
b.
c.
d.
Decreases approximately with the square of speed.
Increases approximately with the square of speed.
Decreases approximately with the inverse of the square of speed.
Remains constant.
LD 53.
If IAS decreases by a factor of 5 how would drag vary?
LD 54.
If IAS decreased by a factor of 5 how would drag vary?
If CL increased by a factor of 5 how would CD1vary?
LD 56.
If aspect ratio increased by a factor of 2 how would CDIvary?
LD 57.
How does total drag vary with air density?
a.
Constant.
b.
c.
d.
Directly.
Inversely.
Conversely.
LD 58.
Ilow does laminar boundary layer compare with turbulent boundary layer?
a.
b.
c.
d.
Thinner.
Thicker.
Faster.
The same.
LD 59.
How does a vortex generator energise the boundary layer?
a.
b.
c.
d.
Mixes it wit11 free stream air.
Mixes it with laminar boundary layer.
Mixes it with turbulent boundary layer.
Prevents shock waves.
LD 60.
Which of the following are true of VMD?
a.
b.
c.
d.
Lower than VMp.
Gives best L:D ratio.
Gives best endurance in a,propeller aircraft.
Gives best range in a jet aircraft.
LD 61.
How is drag affected if pressure decreases with TAS and temperature constant?
a.
b.
c.
d.
Increases.
Decreases.
Constant.
Increase o r decrease depending on altitude.
LD 62.
How does retraction of flaps affect induced drag if IAS remains constant?
a.
b.
c.
d.
Increases.
Decreases.
Constant.
Increase or decrease depending on speed.
LD 63.
How does deployment of fowler flaps affect aspect ratio and angle of'incidence?
a.
b.
c.
d.
Increases,
Decrease,
Constant,
Increase,
Decrease.
Increase.
Constant.
Increase.
LD 64.
If for an elliptical wing the C L = 2.5, A = 4, what is CDI?
LD 65.
In straight and level flight C L= 0.44. If a gust of wind causing a 1' increase in
angle of attack would give a 0.06 increase in CL, what load factor would a gust
causing a 5' increase produce?
LD 66.
What would be the CDIof a wing of aspect ratio 4 and CLMar2.5 when operating
at zero lift angle of attack?
a.
b.
c.
d.
0.297.
0.497.
1.297.
None of the above.
LD 67.
As the angle of attack of a wing is increased above zero what happens to the
stagnation point, transition point and separation point?
a.
b.
c.
d.
Moves forward,
Moves aft,
Moves aft,
Moves aft,
Moves aft,
Moves aft,
Moves forward,
Moves aft,
Moves aft.
Moves aft.
Moves forward.
Moves forward.
LD 68.
How is lift created?
a.
b.
c.
d.
A symmetrical aerofoil.
A cambered aerofoil.
Downward acceleration of air.
Dynamic pressure.
LD 69.
Why does the downwash caused by wingtip vortices not create extra lift?
a.
b.
c.
d.
Because pressure differences create lift.
Because it is too small.
Because it is negated by equal upwash outboard of the tips.
Because lift is not created by downwash.
LD 70.
What happens to the total reaction when a wing enters ground effect?
a.
b.
c.
d.
Decreases and angles in a more rearward direction.
Increases in magnitude and is angled in a less rearward direction.
Increases and angles in a more rearward direction.
Decreases and angles in a more forward direction.
LD 71.
At low angles of attack the major component of total drag is .............drag?
a.
b.
c.
d.
Induced.
Vortex.
Shock.
Profile.
LD 72.
At high angles of attack the major component of total drag is .............drag?
a.
b.
c.
d.
LD 73.
Induced.
Vortex.
Shock.
Profile.
-
.............causes induced drag?
a.
b.
c.
d.
Angle of attack.
Aspect ratio.
Boundary layer separation.
The generation of lift.
LD 74.
Induced drag is directly proportional to .............9
a.
b.
c.
d.
Angle of attack.
(Lift force12.
(True airspeed)2'
Aspect ratio.
LD 75.
What is CLas a % of CLM~~a t 1.5Vs?
LD 76.
9
Induced drag is caused by .........
a.
b.
c.
d.
Upwash.
Tip tanks.
High tailplanes.
Wing tip vortices.
LD 77.
Entering ground effect. .......p ower required?
a.
b.
c.
d.
Increases.
Decreases.
Does not affect.
Increases or decreases depending on height.
LD 78.
Lowering the undercarriage will
a.
b.
c.
d.
........?
Increase induced drag and nose down pitching moment.
Decrease induced drag and nose down pitching moment.
Increase profile drag and nose down pitching moment.
Decrease profile drag and nose down pitching moment.
LD 79.
Doubling IAS multiplies drag by.. ..?
LD 80.
Tripling IAS multiplies drag by ....?
LD 81.
For a n elliptical wing if CDI= 0.6, A = 4, what is CL?
LD 82.
CDIvaries ....?
a.
b.
c.
d.
Directly with cL2
and inversely with aspect ratio.
Directly with speed and aspect ratio.
Directly with wing area and CL.
Directly with angle of attack and speed.
LD 83.
Increasing camber will. ...?
a.
b.
c.
d.
Increase the gradient of the lift slope.
Decrease the gradient of the lift slope.
Increase CL Max.
Decrease induced drag.
LD 84.
Increasing camber will....?
a.
b.
c.
d.
Increase the gradient of the lift slope.
Decrease the gradient of the lift slope.
Decrease CL Max.
Reduced stalling speed.
LD 85.
Increasing a i r temperature will.. ..?
a.
b.
c.
d.
Increase the gradient of the lift slope.
Decrease C L max.
Decrease lift a t any given C L and TAS.
Increase profile drag.
LD 86.
Increasing humidity will.. .
....?
a.
b.
c.
d.
Increase CLmax.
Increase required CL.
Decrease stalling speed.
Increase profile drag.
LD 87.
Increasing humidity will. ......?
a.
b.
c.
d.
Increase CLmax.
Decrease required CL.
Increase stalling speed.
Increase profile drag.
LD 88.
To maintain constant angle of climb, when climbing a t constant TAS..
a.
b.
c.
d.
.
Angle of attack must increase.
Angle of attack must decrease.
Angle of attack must remain constant.
IAS must increase.
1,D 89.
Which of the following statements are true?
1.
2.
3.
4.
a.
b.
c.
d.
Increasing aspect ratio reduces induced drag.
Increasing sweep back angle reduces induced drag.
Increasing EAS increases induced drag.
Increasing CLincreases induced drag.
1 and 4.
1 , 2 and 3.
1 , 3 and 4.
2 , 3 and 4.
LD 90.
..............aircraft weight will .........ground effect?
a.
b.
c.
d.
Increasing,
Increasing,
Decreasing,
Decrease,
Increase.
Decrease.
Increase.
Decrease.
....?
LD 91.
which two af the following statements are most accurate?
1.
2.
3.
4.
a.
b.
c.
d.
CDPis approximately constant and Dp is proportional to v2.
CDIis proportional to 1m4and DI is proportional to v2.
The V in a and b above is TAS.
The V in a and b above is EAS.
1 and 4.
1 and 3.
1 and 2.
2and3.
LD 92.
~
If an upward gust of wind increases CL by 25% by what % will C Dvary?
LD 93.
What would be the CDIof an elliptical wing if A = 5, and C L = 2?
LD 95.
.......... Landing speed a t constant weight will ........ ground effect?
a.
b.
,p c.
4- d.
Increasing,
Increasing,
Decreasing,
Decreasing,
Increase.
Decrease.
Increase.
Decrease.
LD 96.
..........Altitude at constant TAS will ......... Dl?
a.
b.
c.
d.
Increasing,
Increasing,
Decreasing,
Decreasing,
Increase.
Decrease.
Increase.
Decrease.
LD 97.
Induced drag is increased by
ratio?
a.
b.
c.
d.
Increasing,
Increasing,
Increasing,
Decreasing,
......... weight, ......... airspeed and ........aspect
Increasing,
Decreasing,
Decreasing,
Increasing,
Increasing.
Decreasing.
Increasing.
Decreasing.
LD 98.
Increasing load factor will ..........Dp,
stability?
a.
b.
c.
d.
Increase,
Decrease,
Not affect,
Not affect,
Increase,
Decrease,
Increase,
Decrease,
.............VhlDand ...........speed
Increase.
Decrease.
Decrease.
Increase.
LD 99.
At 0.9 Vs, the CLwill be ............ CLMax and CDwill be
VS?
a.
b.
c.
d.
Greater than,
Greater than,
Less than,
Less than,
Greater than.
Less than.
Less than.
Greater than.
LD 100.
3
CLis proportional to ............
FLAPS 1.
What is the effect of deploying trailing edge flaps?
a.
Increases CLand Vs
........... than at 1.1
b.
c.
d.
Decreases CLand Vs
Increases CLand decreases Vs
Decreases CLand increases Vs
FLAPS 2.
What is the effect of deploying leading edge slats?
a.
b.
c.
d.
Energising boundary layer and increasing Vs
De-energising boundary layer and decreasing Vs
Energising boundary layer and decreasing Vs
De-energising boundary layer and increasing Vs
FLAPS 3.
What is the effect on wing pitching moment, of deploying trailing edge flaps
prior to landing?
a.
b.
c.
d.
Increased nose down pitching moment.
Decreased nose down pitching moment.
Nose up pitching moment replaced by a nose down pitching moments.
Increased nose up pitching moment.
FLAPS 4.
What effect does deployment of trailing edge flaps have on stalling angle of
attack?
a.
b.
c.
d.
No change.
Increased stalling angle.
Decreased stalling angle.
Increased or decreased stalling angle depending on wing sweep angle.
FLAPS 5.
What is the effect of deployment of leading edge flaps in conjunction with
trailing edge flaps?
a.
b.
c.
d.
Moves C of P aft increasing nose down pitching moment caused by
trailing edge flaps.
Moves C of P forward reducing nose down pitching moment caused by
trailing edge flaps.
Moves C of G aft increasing nose down pitching moment caused by
trailing edge flaps.
Moves C of G fonvard reducing nose down pitching moment caused by
trailing edge flaps.
FLAPS 6.
What configuration of krueger flaps and slats would produce the best post-stall
handling characteristics in a swept wing aircraft?
a.
b.
c.
d.
Slats inboard and Krueger flaps outboard.
Full span slats.
Full span Krueger flaps.
Kruger flaps inboard and slats outboard.
FLAPS 7.
What trailing edge flap angle will give the minimum stalling speed?
a.
b.
c.
d.
Maximum deflection.
Zero degrees.
20 degrees.
30 degrees.
FLAPS 8.
What trailing edge flap angle will give best L : D ratio?
a.
b.
c.
d.
Zero angle.
Maximum angle.
20 degrees.
30 degrees.
FLAPS 9.
What will be the effect of deploying triple slotted fowler flaps to maximum
deflection?
a.
b.
c.
d.
L:D ratio, wing area, camber, stalling angle, stalling speed and angle of
incidence will all increase.
L:D ratio, wing area, camber, stalling angle, stalling speed and angle.of
incidence will all decrease.
L:D ratio, wing area, camber and stalling angle will increase but stalling
speed and angle of incidence will decrease.
Wing area, camber and angle of incidence will increase but stalling angle,
stalling speed and L:D ratio all decrease.
FLAPS 10.
What effect does the deployment of trailing edge flaps have on airflow over the
tailplane?
a.
b.
c.
d.
The design of modern aircraft is such that deployment of trailing edge
flaps has no effect on airflow over the tailplane.
Downwash is increased decreasing the effectiveness of the tailplane.
Downwash is increased increasing the effectiveness of the tailplane.
Downwash is decreased decreasing the effectiveness of the tailplane.
FLAPS 11.
What effect does the deployment of trailing edge flaps have on the intensity of
wingtip vortices?
a.
b.
c.
d.
Decrease.
Increase.
Decrease or increase depending on airspeed.
Decrease or increase depending on aircraft weight.
FLAPS 12.
Why is it necessary to ensure that trailing edge flaps are deployed
symmetrically?
a.
b.
c.
d.
To maintain C of G within limits.
To maintain C of P within limits.
To maintain lateral and directional control.
To maintain longitudinal control.
FLAPS 13.
What is the purpose of the slots in slotted flaps and how do they achieve this
purpose?
a.
b.
c.
d.
Reduce drag by allowing aii- to flow through the flaps.
Reduce nose down pitching moment by moving C of P forward.
Increase stalling speed by energising boundary layer over the upper
surfaces of the flaps.
Reduce stalling speed and increase stalling angle by energising boundary
layer over the flaps.
FLAPS 14.
Complete the following statement.
As trailing edge flaps move from fully retracted to fully deployed, both lift and
drag increase. Most of the additional drag is produced during the
of
The
deployment whilst most of the additional lift is produced during the
additional drag produced by the first half of the deployment is mainly
Whilst that produced during the second half is mainly .........
..........
......
.......
1.
2.
3.
4.
First half
Induced
Profile
Second half
FLAPS 15.
Deployment of trailing edge flaps in straight and level flight will
..............induced drag?
a.
b.
c.
d.
Not affect.
Decrease.
Increase.
Increase o r decrease depending on flap angle selected.
FLAPS 16.
Deployment of inboard trailing edge flaps will.. ... wing tip vortices?.
...
a.
b.
c.
d.
Decrease.
Not affect.
Increase.
Increase o r decrease depending on flap angle.
FLAPS 17.
Which of the following will reduce L:D ratio most?
a.
b.
c.
d.
15' trailing edge flap.
30' trailing edge flap.
45' trailing edge flap.
15'slat.
FLAPS 18.
What is the purpose of drooping ailerons?
a.
b.
c.
d.
To increase lift.
To prevent adverse yaw.
To maintain stability.
To increase roll rate.
FLAPS 19.
A split flap is ........compared to a plain flap?
a.
b.
c.
d.
More efficient.
Less efficient.
As efficient.
More o r less efficient depending on weight.
FLAPS 20.
Deployment of flaps in turbulence will
a.
b.
c.
d.
....................
3
Increase stalling speed 2nd risk of exceeding limiting load factor.
Decrease s&lling speed and risk of exceeding limiting load factor.
Increase stalling speed and decrease risk of exceeding limiting load factor.
Decrease stalling speed and increase risk of exceeding limiting load factor.
FLAPS 21.
Deployment of leading edge slats ..........9
a.
b.
c.
d.
De-energises boundary layer and moves C of P aft.
De-energises boundary layer and moves C of P forward.
Energises boundary layer and moves C of P aft.
Energises boundary layer and moves C of P forward.
FLAPS 22.
Deployment of fowler flaps
a.
b.
c.
d.
Increases wing area only.
Increases angle of incidence only.
Increases angle of incidence and wing area only.
Increases camber.
FLAPS 23.
Deployment of flaps
a.
b.
c.
d.
.............9
.......CL?
Increases.
Decreases.
Increases then decreases.
Decreases then increases.
FLAPS 24.
The first few degrees of flap deployment will..
a.
b.
c.
d.
Decrease.
Increase.
Increase o r decrease depending on speed.
Not affect.
FLAPS 25.
Full span Krueger flaps will.
a.
b.
c.
d.
........L:D ratio?
.........lateral stability?
Improve.
Degrade.
Not affect.
Improve o r degrade depending on angle.
FLAPS 26.
Split flaps?
a.
b.
c.
d.
Lower the underside of the trailing edge.
Lower the trailing edge.
Lower the leading edge.
Increases wing area.
FLAPS 27.
Deployment of trailing edge flaps..
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
......stalling angle and ......CLMax?
Increase.
Decreases.
Decreases.
Increases.
FLAPS 28.
The purpose of leading edge slats is to ............9
.
a.
b.
c.
d.
Decreases stalling angle.
Increase stalling angle.
Increase stalling speed.
Create turbulent boundary layer.
FLAPS 29.
Limiting load factor for a J A R certificated passenger aircraft with flaps
deployed is ....?
FLAPS 30.
Rlaximum speed for extending flaps is
.....?
FLAPS 31.
Maximum speed with extended flaps is..
FLAPS 32.
Deploying trailing edge flaps..
a.
b.
c.
Decreases.
Does not affect.
Increases.
...?
.........tailplane down force?
d.
i ~ ~ c r e a soe rs decreases depending on weight. '
FLAPS 33.
Deployment of flaps in icing conditions might..
a.
b.
c.
d.
Increase stalling angle.
Cause stalling.
Prevent flutter.
Increase rate of climb.
FLAPS 34.
Raising slats too soon after take-off might
a.
b.
c.
d.
......?
........?
Increase stalling angle.
Cause stalling.
Prevent flutter.
Increase rate of climb.
FLAPS 35.
Trailing edge flaps ............landing attitude?
a.
b.
c.
d.
Increase.
Decrease.
Do not affect.
Increase o r decrease depending on flap type.
FLAPS 36.
Leading edge flaps ............landing attitude?
a.
b.
c.
d.
Increase.
Decrease.
Do not affect.
Increase o r decrease depending on flap type.
FLAPS 37.
Trailing edge flaps.. ..........stalling angle?
a.
b.
c.
d.
Increase.
Decrease.
Do not affect.
Increase o r decrease depending on flap type.
FLAPS 38.
Leading edge slats ............stalling angle?
a.
Increase.
b.
c.
d.
Decrease.
Do not affect.
Increase o r decrease depending on weight.
FLAPS 39.
Leading edge flaps
a.
b.
c.
d.
............stalling angle?
Decrease.
Increase.
Do not affect.
Increase or decrease depending on flap type.
FLAPS 40.
Krueger flaps are ...... efficient than leading edge droop?
a.
b.
c.
d.
Less.
More.
No more nor less.
More o r less depending on angle of attack.
FLAPS 41.
Blown tlaps.
a.
b.
c.
d.
.........boundary layer and. .......stalling speed?
Energise,
Energise,
De-energise,
De-energise,
Increase.
Decrease.
Increase.
Decrease.
FLAPS 42.
Slotted flaps ..........boundary layer and ........stalling speed?
a.
b.
c.
d.
Energise,
Energise,
De-energise,
De-energise,
Increase.
Decrease.
Increase.
Decrease.
FLAPS 43.
7
Flap blowing ...........
a.
b,
c.
d.
Energises boundary layer and increases angle of incidence.
Energises boundary layer and decreases angle of incidence.
Requires more engine power.
Improves SFC.
FLAPS 44.
7
Asymmetric flap deployment.. ........
a.
b.
c.
d.
Improves lift performance.
Is used to prevent asymmetric yaw.
Causes loss of control.
Is not possible.
FLAPS 45.
Flap deployment improves C L most
a.
b.
c.
d.
b.
c.
d.
........downwash over the tailplane?
Decreases.
Increases.
Does not affect.
Increases o r decreases depending on angle.
FLAPS 47.
Fowler flaps a r e
a.
3
During the first few degrees.
During the last few decrees.
At high subsonic speeds.
During the take-off roll.
FLAPS 46.
Flap deployment..
a.
b.
. c.
d.
...........
...........than split flaps?
Less complex.
Less effective.
Lighter.
Slower.
FLAPS 48.
Failure of trailing edge flaps to deploy on landing will?
a.
b.
c.
d.
Decrease landing speed.
Decrease landing roll.
Increase nose u p attitude.
Increase angle of attack.
FLAPS 49.
Trailing edge flaps
a.
b.
c. \
d.
...........the CL:a curve.
Extend.
Shorten.
Flatten.
Do not affect.
FLAPS 50.
Leading edge slats
a.
b.
c.
d.
~
...........the C L :curve.
Extend.
Shorten.
Flatten.
Do not affect.
FLAPS 51.
Fowler flaps ...........the CL:acurve.
a.
b.
c.
d.
Straighten.
Shorten.
Extends.
Do not affect.
FLAPS 52.
The diagram below includes?
a.
b.
c.
d.
Double slotted plain flaps.
Leading edge flaps.
Double slotted split flaps.
Double slotted fowler flaps.
FLAPS 53.
The diagram below includes?
a.
b.
c.
d.
Blown flap and slat.
Plain flap and leading edge flap.
Slat and slotted split flap.
Slat and fowler flap.
FLAPS 54.
The diagram below includes?
a.
b.
c.
d.
Slats.
Leading edge flap.
Kruger flap.
Double slotted split flaps.
FLAPS 55.
The diagram below includes?
a.
b.
c.
d.
Kruger flap.
Leading edge slat.
Leading edge droop.
Drooped slat and plain flap.
FLAPS 56.
The diagram below includes?
a.
b.
c.
d.
Double slotted plain flaps.
Leading edge flaps.
Split flap.
Double slotted fowler flaps.
FLAPS 57.
The diagram below includes?
a.
b.
c.
d.
Drooped leading edge and plain flap.
Leading edge flap.
Double slotted split flaps.
Slat and plain flap.
FLAPS 58.
Pitch up on flap deployment is
a.
b.
c.
d.
........?
Due to downwash over tailplane.
Not possible.
Due to aft movement of C of P.
Due to forward movement of C of P.
FLAPS 59.
Retracting trailing edge flaps whilst leaving slats deployed in a climb will
a.
b.
c.
d.
Reduce drag.
Reduce lift, drag and L:D ratio.
Increase lift and reduce drag.
Reduce lift and drag and increase L:D ratio.
FLAPS 60.
9
Retraction of slats prior to flaps might .........
a.
b.
c.
d.
lncrease L:D ratio.
Increase stalling angle.
Reduce stalling speed.
Cause stall.
FLAPS 61.
9
Flap deployment causes pitch up due to .........
a.
b.
c.
d.
Decreased upwash.
Increased upwash.
Decreased downwash.
Increased downwash.
....?
FLAPS 62.
Using- slats to oppose the nose down moment caused by flap deployment ........
stalling speed compared to using the tailplane for the same purpose?
-
a.
b.
c.
d.
-
Increases.
Decreases.
Increases or decreases depending on C of G position.
Does not affect.
FLAPS 63.
Flap asymmetry causes ......?
a.
b.
c.
d.
Roll.
Roll and yaw.
Roll, pitch and yaw.
None of the above.
FLAPS 64.
Trailing edge flap deployment .....?
a.
b.
c.
d.
Increases stalling speed and stalling angle of attack.
Decreases stalling speed and angle of incidence.
Decreases stalling speed and stalling angle of attack,
Improves stability.
FLAPS 65.
Blown flaps ..... ?
a.
b.
c.
d.
Increase lift and decrease thrust.
Increase lift and increase thrust.
Decrease lift and increase thrust.
Decrease lift and decrease thrust.
FLAPS 66.
Spoiler deployment
a.
b.
c.
d.
...... ?
Increases L:D ratio.
Decreases L:D ratio.
Decreases separation over flaps.
Is not possible with flaps deployed.
FLAPS 67.
Deployment of trailing edge flaps .... ?
a.
b.
c.
Never improves L:D ratio.
Always improves L:D ratio.
Sometimes improves L:D ratio.
d.
Improves or reduces L:D ratio depending on C of G position.
FLAPS 68.
Trailing edge flap deployment
a.
b.
c.
d.
Move C of P aft.
Moves C of G forward.
Moves C of G aft.
Does not affect position of C of G or C of P.
FLAPS 69.
Leading edge flap deployment
a.
b.
c.
d.
.... ?
... ?
Move C of P aft.
Moves C of G forward.
Moves C of P forward.
Does not affect position of C of G or C of P.
FLAPS 70.
Leading edge flaps .... ?
a.
b.
c.
d.
Increase camber and angle of incidence.
Decrease camber and angle of incidence.
Increase camber and decrease angle of incidence.
Decrease camber and increase angle of incidence.
FLAPS 71.
Flap deployment ..... ?
a.
b.
c.
d.
Increases upwash and downwash.
Decreases upwash and downwash.
lncreases upwash and decreases downwash.
Decreases upwash and decreases downwash.
FLAPS 72.
Flap deployment ...... Dp and
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
...... Dl?
Increases.
Decreases.
Decreases.
Increases.
FLAPS 73.
Wing area and camber are increased by deployment of .... Flaps?
a.
Plain.
b.
c.
d.
Split.
Fowler.
Slotted.
FLAPS 74.
Split flaps are
a.
b.
c.
d.
.....than plain flaps?
Heavier.
Lighter.
More efficient.
Less efficient.
FLAPS 75.
Deployed flaps in a JAR certificated passenger aircraft must be capable of
withstanding ... . without permanent deformation?
..
a.
b.
c.
d.
lg.
2g.
2.5g.
3.8g.
FLAPS 76.
With trailing edge flaps deployed the stick shaker will activate a t
attack compared with the clean configuration?
a.
b.
c.
d.
A higher.
A lower.
A higher or lower depending on airspeed.
The same.
FLAPS 77.
Landing configuration is usually
a.
b.
c.
d.
........ 9.
Slats and flaps fully deployed.
Slats only deployed.
Flaps only deployed.
Slats fully deployed and flaps at minimum deflection angle.
FLAPS 78.
Flap deployment .....the landing run?
a.
b.
c.
d.
.......angle of
Increases.
Decreases.
Does not affect.
Increases or decreases depending on landing speed.
FLAPS 79.
Flap deployment ..... the take-off run?
a.
b.
c.
d.
Increases.
Decreases.
Does not affect.
Increases or decreases depending on deployment angle.
FLAPS 80.
Trailing edge flap deployment .....pitch attitude required in take-off and
landing?
a.
b.
c.
d.
Increases.
Decreases.
Does not affect.
Increases o r decreases depending on weight.
FLAPS 81.
Trailing edge flap deployment might cause nose down pitching due to ...... ?
a.
b.
c.
d.
Downwash over the tailplane.
Upwash over the tailplane.
Forward movement of the C of P.
Aft movement of the C of P.
FLAPS 82.
Trailing edge flap deployment . .. ... power required?
.. .
a.
b.
c.
d.
Increases.
Decreases.
Does not affect.
Increases or decreases depending on C of G position.
FLAPS 83.
With stabiliser trim stuck in cruise position deployment of landing flap will
......... ?
a.
b.
c.
d.
Increase stick forces in the flare.
Decrease stick forces in the flare.
Decrease nose down control authority.
Reduce trim drag.
FLAPS 84.
Deployment of flaps ......... ?
a.
b.
Increases longitudinal stability.
Decreases longitudinal stability.
c.
d.
Increases lateral stability.
Decreases lateral stability.
FLAPS 85.
Use of flaperons in take-off will ..... ?
a.
b.
c.
d.
Reduce roll authority.
Increase roll authority.
Not affect roll authority.
Reduce trim drag.
FLAPS 86.
Use of small angles of flap deflection
a.
b.
c.
d.
....?
Increases Dl more than Dp.
Increases Dp more than Dl.
Reduces Dl and increases Dp.
Increases DI and reduces Dp.
FLAPS 87.
Trailing edge flap deployment .... ?
a.
b.
c.
d.
Increases stalling angle.
Decreases stalling angle.
Does not affect stalling angle.
Increases or decreases stalling angle depending on C of G position.
FLAPS 88.
Full span Krueger flaps
a.
b.
c.
d.
.... ?
Encourage root stall.
Encourage tip stall.
Decrease stalling angle.
Increase stalling attitude.
FLAPS 89.
Use of outboard Krueger flaps alone would
a.
b.
c.
d.
Increase tip stall tendency.
Decrease tip stall tendency.
Not affect tip stall tendency.
Decrease wing root bending.
...... ?
FLAPS 90.
Premature slat retraction in climb out might .... ?
a.
b.
c.
d.
Cause stall.
Cause high speed buffet.
Increase rate of climb.
Decrease trim drag.
FLAPS 91.
Split flaps .... ?
a.
b.
c.
d.
Move upwards and downwards to act as airbrakes.
Are divided into sections to provide individual control.
Reduce flutter.
Are less prone to separation than plain flaps.
FLAPS 92.
Maximum flap deployment speed is
.... ?
FLAPS 93.
Full flap deployment in take-off will
..... ?
a.
b.
c.
d.
Increase climb performance.
Increase take-off distance required.
Increase acceleration rate.
Increase climb gradient.
FLAPS 94.
Full flap deployment in landing will ..... ?
a.
b.
c.
d.
Require a shallower approach.
Permit a steeper approach.
Increase landing speed.
Increase landing run.
FLAPS 95.
......... flaps offer the greatest increase in CL MA^?
a.
b.
c.
d.
Plain.
Split.
Fowler.
Slotted fowler.
FLAPS 96.
Blown flaps . .. ?
.
a.
b.
c.
d.
Increase power available.
Decrease power required.
Increase rate of climb.
Decrease power available.
FLAPS 97.
Spoiler deployment
.......The CL : a slope and ...... a stall?
a.
b.
c.
d.
Increases.
Decreases.
Increases.
Decreases.
Incl-eases
Decreases
Decreases
Increases
FLAPS 98.
Slat deployment ......
...the gradient of the CL :a curve?
a.
b.
c.
d.
Does not affect.
Increases.
Decreases.
Increase or decreases depending on airspeed.
FLAPS 99.
Slat deployment
a.
b.
c.
d.
Decreases
Decreases
Increases
Increases
....... a stall and trailing edge flap deployment ...... it?
Decreases.
Increases.
Increases.
Decrease.
FLAPS 100.
Stall angle is typically ... degrees with plain flaps and
STALL 1.
What % of CL
... degrees with split flaps?
would a wing produce when flying a t 1.3 Vs?
STALL 2.
What is the relationship in straight and level flight between the CLat any given
speed and CLMax?
STALL 3.
If the C of G is moved forward what will be the effect on stalling speed?
a.
b.
c.
d.
No effect
Increase
Decrease
Increase or decrease depending on aircraft weight
STALL 4.
If the C of G of an aircraft is moved forward what will be the effect on
longitudinal, directional and lateral stability?
a.
b.
c.
d.
No effect on longitudinal, increase lateral and directional.
Increase longitudinal and lateral, no effect on directional.
Decrease lateral, increase directional and longitudinal.
Increase longitudinal and directional, no effect on lateral.
STALL 5.
If the C of G is moved forward what will be the effect on longitudinal, directional
and lateral manoeuvrability.
ab.
c.
d.
No effect on any.
Increase lateral, decrease longitudinal and directional.
Decrease lateral and directional, increase longitudinal.
Decrease longitudinal and directional, no effect on lateral.
STALL 6.
If the weight of an aircraft is doubled what will be the effect on its stalling speed?
a.
b.
c.
d.
It will be doubled.
I t will be halved.
I t will be increased by approximately 41%.
It will be increased by approximately 50%.
STALL 7.
A JAR certificated commercial passenger aircraft is flying in straight and level
flight when the stall warning system activates the stick shaker. What will be its
CLas a percentage of CLMax?
STALL 8.
Which of the following statements are true of an aircraft as it climbs to high
altitude at constant IAS?
a.
b.
c.
d.
Its TAS will increase as a proportion to its IAS until it reaches Merit.
Its indicated stalling speed will increase throughout most of the climb
then reduce as it reaches Merit,
Its indicated stalling speed will decrease throughout most of the climb
then increase as it reaches Merit
Its indicated stalling speed will remain constant throughout most of the
climb then increase as it reaches Merit.
STALL 9.
The sketch below represents a typical whole aircraft L:D polar diagram. Which
of the points marked on it correspond to the low speed stall condition?
STALL 10.
The sketch below represents a typical whole aircraft L:D polar diagram. Which
of the points marked on it correspond to the best L:D ratio and a t what speed
does this occur?
STALL 11.
Which of the following represents the relationsltip between the design
manoeuvre speed (VA)and the straight and level l g stalling speed (Vsls)?
STALL 12.
If the zero lift angle of attack and stalling angle of attack for a particular aerofoil
section are -4 degrees and 16 degrees respectively, what will be its CLas a
percentage of CLM ~ when
,
the angle of attack is 6 degrees?
STALL 13.
In straight and level flight the CLof an aerofoil is 0.44. If a 1 degree increase in
angle of attack gives an increase of 0.06 in CLwhat will be the load factor when
subjected to an angle of attack increase of 5 degrees?
STALL 14.
An JAR certificated passenger aircraft has a CLMax of 2 and a l g (straight and
level) stalling speed of 100 Kts. What will be its 2.5g stalling speed?
a.
b.
c.
d.
138 Kts.
148 Kts.
158 Kts.
168 Kts.
STALL 15.
Which of the following best defines VAfor a JAR certified passenger aircraft?
a.
b.
The Design Manoeuvre Speed which is the maximum EAS a t which full
nose up control deflection can be employed without exceeding a load
factor of 2.5.
The Design Manoeuvre Speed which is the maximum TAS at which full
nose up control deflection can be employed without exceeding a load
factor of 3.75.
c.
d.
The Design Manoeuvre Speed which is the maximum EAS at which a load
factor of 2.5 can be achieved without stalling the aircraft.
The Design Manoeuvre Speed which is the minimum EAS a t which full
nose up control deflection can be employed without exceeding a load
factor of 2.5.
STALL 16.
What will be the effect of increasing altitude on the TAS a t which an aircraft will
stall assuming weight and load factor remain unchanged?
a.
b.
c.
d.
Increases.
Remains unchanged provided speed does not approach Merit.
Remains unchanged regardless of actual speed.
Decreases.
STALL 17.
Which of the following will give the smoothest ride in gusty conditions?
a.
b.
c.
d.
Low aspect ratio straight wing with flaps down.
High aspect ratio straight wing with flaps down.
Low aspect ratio swept wing flaps up.
High aspect ratio swept wing with flaps up.
STALL 18.
Which of the following configurations is best suited to flight in gusty conditions?
a.
b.
c.
d.
Flaps up.
Flaps in landing configuration.
Flaps and landing gear down.
Flaps up and landing gear down.
STALL 19.
If an aircraft stalls at 200 kts in a 3g pull up what will its stalling speed be in
straight and level flight?
a.
b.
c.
d.
67 Kts.
77 Kts.
115 Kts.
125 Kts.
STALL 20.
If CL= 2.5, CD1 = 0.35, What is the aspect ratio?
STALL 21.
If CDl = 0.6, A = 4, What is CL?
STALL 22.
If aircraft weight is doubled by what % will stalling speed increase?
STALL 23.
If stalling speed in a 3g pull up is 100 Kts what is the maximum bank angle in
level flight at 100 Kts?
a.
b.
c.
d.
50.5 Degrees.
60.5 Degrees.
70.5 Degrees.
80.5 Degrees.
STALL 24.
What angle of bank will produce limiting load factor in a J A R certificated
passenger aircraft?
a.
b.
c.
d.
46 Degrees.
56 Degrees.
66 Degrees.
76 Degrees.
STALL 25.
If Vs at a weight of 20000 Kg is 100 Kts what will Vs be a t a weight of 60000 Kg?
a.
b.
c.
d.
143 Ms.
153 Kts.
163 Kts.
173 Kts.
STALL 26.
If stalling speed in straight and level flight is 100 Kts what will it be in a 30
degree banked level turn?
a.
97 Kts.
b.
c.
d.
107 Kts.
117 Kts.
127 Kts.
STALL 27.
Which of the following aircraft is most likely to suffer wing drop in the event of a
single engine failure?
a.
b.
c.
d.
A twin propeller aircraft in the climb out.
A four engine propeller aircraft in the cruise.
A twin jet in the climb out.
A four engine jet in the cruise.
STALL 28.
What effect does propeller slipstream have on the stalling speed of a n aircraft?
1.
2.
By increasing airspeed over the wing it decreases stalling speed.
By providing the same effect as a boundary layer control system it
increases the value of CLMau.
a.
b.
c.
d.
1 only is true.
1 and 2 a r e true.
2 only is true.
Neither 1 nor 2 a r e true.
STALL 29.
If stalling speed in a 15' bank is 100 Kts what will it be in a 60' bank?
a.
b.
c.
d.
119 Kts.
129 Kts.
I39 I<ts.
149 Kts.
STALL 30.
What causes a swept wing to pitch up at the stall?
a.
b.
c.
d.
Turbulent boundary layer.
Flaps and slats.
Tip stall.
Chordwise flow.
STALL 31.
Correct stall and spin recovery action is?
a.
Minimum power, use ailerons to arrest roll, and stick forward to bring
the nose down.
b.
c.
d.
Maximum power, use ailerons to arrest roll, and stick forward to bring
the nose down.
Maximum power, stick in laterally neutral and fully forward, maximum
opposite rudder to arrest yaw.
Maximum power, Stick longitudinally neutral and full lateral deflection to
arrest roll and yaw.
STALL 32.
What happens to the C of P of a swept back wing and a straight wing
respectively in a stall?
a.
b.
c.
d.
Moves aft and forward respectively.
Both move aft.
Both move forward.
Moves forward and aft respectively.
STALL 33.
Iiow does fuel burn affect the low speed buffet?
a.
b.
c.
d.
Increases speed.
Decreases speed.
No change in speed.
Increase o r decrease depending on angle of attack.
STALL 34.
What is the highest permitted speed for the commencement of low speed buffet
in a J A R certificated passenger aircraft?
a.
b.
C.
d.
VS.
1.05 Vs.
1.2 vs.
1.3Vs.
STALL 35.
What is the stalling speed in a 1.5g banked turn?
a.
b.
c.
d.
Vslg d(l.5).
VsI.sg~ ( C OAOB).
S
VS1.sg~ ( ~ I C O
AOB).
S
Vslg d(111.5).
STALL 36.
An aircraft weight 24000 N stalls at 150 Kts. At what speed will it stall a t a
weight of 48000 N?
a.
b.
212Kts.
222 Kts.
c.
d.
232 Kts.
242 Kts.
STALL 37.
How is Vq calculated from Vslg?
STALL 38.
Deep stall is most probable with?
a.
b.
c.
d.
Ventral fin.
Canards.
High tailplane.
Swept back wings.
STALL 39.
Deep stall is most probable with?
a.
b.
c.
d.
Ventral fin.
Canards.
High tailplane.
Low tailplane.
STALL 40.
In a constant IAS climb?
a.
b.
c.
d.
Stalling TAS is constant.
Stalling IAS is constant.
Stalling TAS is constant before decreasing at high altitude.
Stalling IAS is constant before increasing at high altitude.
STALL 41.
C of G on the aft limit will?
a.
b.
c.
d.
Increases stalling speed but leave stalling angle unchanged.
Decrease stalling speed but leave stalling angle unchanged.
Increase stalling angle but leave stalling speed unchanged.
Leave both stalling angle and stalling speed unchanged.
STALL 42.
Deployment of trailing edge flaps?
a.
b.
c.
d.
Increases stalling speed and angle.
Decreases stalling speed and angle.
Decreases stalling speed but leaves stalling angle unchanged.
Decreases stalling angle but leaves stalling speed unchanged.
STALL 43.
At the stall?
a.
b.
c.
d.
Tips of swept back wings stall first, moving C of P inwards and forwards.
Roots of swept back wings stall first, moving C of P outwards and
rearwards.
Straight wing tips stall first, moving C of P inwards and forwards.
Straight wing roots stall first, moving C of P aft.
STALL 44.
Vortex generators?
a.
b.
c.
d.
Prevent tip stall.
Energise spanwise flow.
Energise shock waves.
Minimise shock induced separation.
STALL 45.
Stalling speed is increased by?
a.
b.
c.
d.
Weak longitudinal stability.
Strong longitudinal stability.
Tailplane lift.
Aft C of G.
STALL 46.
Canards?
a.
b.
c.
d.
Increase stall speed.
Decrease stall speed.
Do not affect stall speed.
Increase or decrease stall speed depending on C of G position.
STALL 47.
With a low thrust line?
a.
b.
c.
d.
Power decreases decrease stalling speed.
Power increases increase stalling speed.
Thrust increases increase stalling speed.
Thrust increases decrease stalling speed.
STALL 48.
Turning stalling speed is proportional to?
a.
b.
c.
d.
Load factor.
The angle of bank.
The square root of the load factor.
The square root of the cosine of the angle of bank.
STALL 49.
Flaps down in turbulence will?
a.
b.
c.
d.
Increases risk of stall.
Decrease risk of stall.
Not affect risk of stall.
Decrease risk of exceeding limiting load factor.
STALL 50.
As flight progresses?
a.
b.
c.
d.
Stalling speed increases.
Stalling speed decreases.
Stalling speed is constant.
Stalling speed increases or decreases with changing C of G position.
STALL 51.
If Vs in straight and level flight is 75 Kts what will it be in a 45' bank?
a.
b.
c.
d.
96 Kts.
89 Kts.
116 Kts.
126 Kts.
STALL 52.
The stalling speed in the landing configuration is.
.....?
STALL 53.
If Vsl, is 175 Kts a t a weight of 75000 N what will it be a t 100000 N?
a.
b.
c.
d.
202 Kts.
222 Kts.
232 Kts.
242 Kts.
STALI, 54.
If Vsl, is 175 Kts what will Vs be in a 47 degree level banked turn?
a.
b.
c.
d.
201 Kts.
211 Kts.
221 Kts.
231 kts.
STALL 55.
If Vs in a 47 degree level banked turn is 200 Kts what will it be in a 65 degree
level banked turn?
a.
b.
c.
d.
224 Kts.
234 Kts.
254 Kts.
264 Kts.
STALL 56.
If Vst, is 180 Kts a t a weight of 80000 N what wilI it be at 40000 N?
a.
b.
c.
d.
97 KTS.
107 Kts.
117 Kts.
127 Kts.
STALL 57.
If VS is 450 Kts at 7g what will Vslg be?
a.
b.
c.
d.
170 Iits.
180 Kts.
190 Iits.
200 Kts.
STALL 58.
If Vsl, is 100 Kts the stick shaker will activate at .........9
a.
b.
c.
d.
100 Kts.
105 Kts.
110 Kts.
115 Kts.
STALL 59.
If an aircraft stalls in 3g flight a t 100 Kts it will stall a t
level flight?
a.
b.
48 Kts.
58 Kts.
........ in straight and
c.
d.
68 Kts.
78 I<ts.
STALL 60.
In autorotation the CDof the ....... Wing increases and its CL
a.
b.
b.
c.
Up-going
Down-going
Up-going
Down-going
........?
Decreases.
Decreases.
Increases.
Increases.
STALL 61.
In autorotation the yaw is caused by dissymmetry of drag as the CDof the upAnd that of the down-going wing .......?
going wing
.......
a.
b.
c.
d.
Increases
Increases
Decreases
Decreases
Decreases.
Increases more.
Increases.
Decreases. More.
STALL. 62.
In autorotation the roll is caused by dissymmetry of lift as the CLof the up-going
wing ....... And that of the down-going wing
?
.......
a.
b.
c.
d.
Increases
Increases
Decreases
Decreases
Decreases.
Increases more.
Increases.
Decreases more.
STALL 63.
In autorotation the ......... wing becomes more stalled whilst the ....... Becomes
less so?
a.
b.
c.
d.
Inner
Inner
Outer
Outer
STALL 64.
Inner.
Outer.
Outer.
Inner.
STALL 65.
Stalling speed in a maximum g manoeuvre at the maximum operating altitude is
STALL 66.
Vso is?
a.
b.
c.
d.
Stalling IAS in take-off configuration.
Stalling EAS in landing configuration.
Stalling TAS in take-off configuration.
Stalling CAS in landing configuration.
STALL 67.
The minimum CAS a t which an aircraft can develop a lift force equal to its
weight a t an angle of attack which is not greater than its stalling angle is
?
.....
STALL'68.
Which of the following is the lowest flying speed?
STALL 69.
If Vslc is 175 Kts VA will be .....?
a.
b.
c.
d.
246.7 Kts.
256.7 Kts.
266.7 Kts.
276.7 Kts.
STALL 70.
At the aerodynamic ceiling maximum attainable load factor is ...?
STALL 71.
In a climbing turn the
wing is likely to stall
......and yaw ..... the turn?
......
a.
b.
c.
d.
Outer
Outer
Inner
Inner
roll
pitch
roll
pitch
first
last
first
last
.....causing the aircraft to
out of.
in to.
in to.
out of.
STALL 72.
wing is likely to stall
In a descending turn the
......and yaw ..... the turn?
......
a.
b.
c.
d.
first
last
first
last
Outer
Outer
Inner
Inner
STALL 73.
wing is prone to
A swept
layer air a t the
?
......
a.
b.
c.'
d.
......
Back
Forward
Back
Forward
roll
pitch
roll
pitch
Down
Down
up
Up
low energy
high energy
high energy
low energy
tip
root
shock
shock
tips.
roots.
tips.
roots.
.....in low speed stall and pitch ..... in high
UP.
down.
UP.
down.
STALL 75.
Straight wings are prone to pitch
q e e d stall?
Down
Down
UP
UP
out of.
in to.
in to.
out of.
..... stall due to separation of .......... boundary
STALL 74.
Swept back wings are prone to pitch
speed staIl?
a.
b.
c.
d.
.....causing the aircraft to
.....in low speed stall and pitch ..... in high
UP.
up then down.
UP.
down then up.
CON 1.
What control inputs are required to maintain airspeed a d altitude when
turning in a jet aircraft?
a.
b.
c.
d.
Increase thrust and angle of attack.
Increase turn radius and decrease thrust.
Increase pitching angle.
Increase bank angle.
CON 2.
Torsional aileron flutter involves?
a.
b.
c.
d.
c of G aft of hinge line and cyclic twisting of wings.
C of G aft of C of P and cyclic twisting of wings.
C of P fwd of torsional axis and cyclic twisting of wings.
None of the above.
CON 3.
Flexural aileron flutter involves?
a.
b.
c.
d.
C of G aft of the hinge line and cyclic bending of wings.
C of G aft of C of P and cyclic twisting of wings.
C of p fwd of torsional axis and cyclic twisting of wings.
None of the above.
CON 4.
Swept back wings are ........ to divergence?
a.
b.
c.
d.
More prone.
Less prone.
Never prone.
None of the above.
CON 5.
Divergence is caused by?
a.
b.
c.
d.
Weak wings, high speeds and C of P forward of torsional axis.
Strong wings, lorn speeds, and C of G aft of torsional axis.
Weak wings, high speed and C of G aft of torsional axis.
Swept wings, high speed and C of G aft of C of P.
CON 6.
Why might trim tabs be employed in power assisted flying controls?
a.
b.
c.
To enable control to be maintained following hydraulic failure.
To reduce control forces to zero.
To prevent overstressing of hydraulic actuators.
77
d.
To provide mach trim.
CON 7.
What is VMcL?
2.
The minimum speed at which control can be maintained following a
critical engine failure in the air, in the landing configuration.
Limited by the maximum roll rate of the aircraft.
a.
b.
c.
d.
Neither 1 nor 2 are correct.
Both 1 and 2 are correct.
1 only is correct.
2 only is correct.
1.
CON 8.
What is roll?
a.
b.
c.
d.
Rotation of the aircraft about its longitudinal axis.
Rotation of the aircraft about its lateral axis.
Rotation of the aircraft about its normal axis.
Caused by adverse yaw.
CON 9.
How is control mass balance achieved?
a.
b.
c.
d.
By fitting weight aft of the hinge.
By fitting weights onto the leading edge.
By fitting weights into the tip cap.
By fitting a horn balance.
CON 10.
What is a flaperon?
a.
b.
c.
d.
A combined elevon and flap.
A combined aileron and flap.
A combined trim tab and flap.
A variable incidence tailplane.
CON 11.
Which of the following minimise adverse yaw?
a.
.b.
c.
d.
Yaw damper.
Dorsal fin.
Roll spoilers.
Trim damper.
CON 12.
Which of the following reduces the effects of asymmetric d --rgwhen rolling?
a.
b.
c.
d.
Dorsal fin.
Vortilons.
Trim damper.
Aileron-rudder coupling.
CON 13.
What action taken in the cockpit will cause a pitch trim tab to move up in a fully
powered flying control system?
a.
b.
c.
d.
The trim switch must be moved in the nose up direction.
The trim switch must be moved in the nose down direction.
The trim wheel must be turned to the right.
Trim occurs automatically so no action is required.
CON 14.
What action should be taken if the turn and slip indicator shows needle to the
left and ball to the left?
a.
b.
c.
d.
Reduce right bank.
Increase Right bank.
Increase speed.
Decrease speed.
CON 15.
Aerodynamic balance methods include?
a.
b.
c.
d.
Split flaps and split rudders.
Weights added to the leading edge of the control surfaces.
Inset hinge and horn balance.
Rudder-aileron coupling.
CON 16.
What will roll spoilers do when a n aircraft enters a high sp -ed turn?
a.
b.
c.
d.
Move down on the up going wing.
Move u p on the down going wing.
Move up on the u p going wing.
Move down on the down going wing.
CON 17.
What will happen when an aircraft employing combined roll spoilers and speed
brakes is in ;dive with speed brakes deployed, enters a left turn?
a.
Right roll spoiler will go up o r left roll spoiler will retract.
.
b.
c.
d.
Left roll spoiler will go u p o r right roll spoiler will retract.
Ailerons will provide roll control.
Elevons will provide roll control.
CON 18.
Adverse yaw in a turn may be corrected by?
a.
b.
c.
d.
Balance tabs.
Anti-balance tabs.
Differential ailerons.
Mass balance.
CON 19.
Which of the following parameters affects control in the transonic speed range?
s
a.
b.
c.
d.
TAS.
RAS.
Mach number.
IAS.
CON 20.
Which of the following is true of power assisted flying controls?
a.
b.
c.
d.
None of the aerodynamic loads a r e felt by the pilot.
Some proportion of the aerodynamic loads are felt by the pilot.
The full aerodynamic loads are felt by the pilot.
Control is lost in the event of hydraulic failure.
CON 21.
At what angle must a variable incidence tailplane be set to trim the aircraft?
a.
b.
c.
d.
I t depends on the speed, thrust and C of G position of the aircraft.
Slightly nose u p if trailing edge flaps are deployed.
Slightly nose down if leading edge flaps are deployed.
Zero incidence, the trim tabs provide trim.
CON 22.
At what angle relative to the variable incidence stabilizer will the elevator be
when the aircraft is in trim in straight and level flight?
a.
b.
c.
d.
Up if C of G is forward.
Down if C of G is forward.
Neutral.
Depends on speed.
CON 23.
What elevator movement will be required to trim for a thrust increase in a low
wing aircraft with engines mounted below the wings?
a..
b.
c.
d.
Down.
Up.
Neutral.
Depends on C of G.
CON 24.
What advantage is provided by a variable incidence tailplane in comparison to a
conventional tailplane with elevator and trim tab?
a.
b.
c.
d.
It is lighter, stronger and of more simple construction.
It makes the aircraft more stable about the lateral axis.
It provides more effective trimming whilst maintaining pitch control
authority.
It requires less power to operate.
CON 25.
Which of the following statements is true of servo tabs?
a.
b.
c.
d.
The controls a r e less effective at low speeds.
The servo tabs also provide trim control.
They enable the use of smaller hydraulic actuators.
They enable the use of smaller control surfaces.
CON 26.
What is motion about the longitudinal axis called?
a.
b.
c.
d.
Pitching.
Rolling.
Yawing.
Phugoid.
CON 27.
Where are Mass balance weights fitted?
a.
b.
c.
d.
Behind the hinge line.
Ahead of the hinge line.
At the tips.
At the trailing edge.
CON 28.
What benefit is provided by fitting the engines a t the rear of the fuselage?
a.
The wings can be lighter and thinner.
81
b.
c.
d.
Engine failure results in less control problems.
The wings are less liable to flutter.
Better response to deep stall.
CON 29.
Which of the followir ::methods can be used in the event of hydraulic failure in a
power assisted flying control system?
a.
b.
c.
d.
Balance tabs.
Mass balance.
Variable incidence tailplane.
Horn balance.
CON 30.
Which of the following methods can be used in the event of hydraulic failure in a
power assisted flying control system?
a.
b.
c.
d.
Anti-balance tabs.
Aerodynamic balance.
Trim tabs.
Mass balance.
CON 31.
Which of the following w31 happen if trim tabs are used for emergency control
following hydraulic failure?
a.
b.
c.
d.
The tab will act as a balance tab.
The tab will act as an anti-balance tab.
The controls will be reversed.
Control effectiveness will be unaffected.
CON 32.
What is the advantage of a stabilator compared to a conventional system?
a.
b.
c.
d.
More powerful control.
Less power required.
No power required.
Better longitudinal stability.
CON 33.
What is the immediate effect of pushing the left rudder pedal forward?
a.
b.
c.
d.
Right yaw about the longitudinal axis and roll to the right.
Left yaw about the normal axis and roll to the right.
Left yaw about the normal axis and roll to the left.
None of the above.
CON 34.
What is the purpose of the mach trim system in a high speed aircraft?
a.
b.
c.
d.
To damp out phugoid motion in yaw.
To prevent pitch up in shock stall.
To prevent pitch down in shock stall.
To prevent Dutch Roll.
CON 35.
How is roll control achieved in high speed aircraft when cruising?
a.
b.
c.
d.
Ailerons.
Inboard ailerons and/or lift spoilers.
Inboard ailerons and/or roll spoilers.
Spoilers.
CON 36..
What benefit is provided by fitting the tailplane at the top of the fin?
a.
b.
c.
d.
Improved longitudinal stability.
Improved directional stability.
Improved roll stability.
Improved response in deep stall.
CON 37.
What is roll?
a.
b.
c.
d.
Rotation about the normal axis.
Rotation about the lateral axis.
Rotation about the pitch axis.
Rotation about the longitudinal axis.
CON 38.
What is the standard spin recovery technique?
a.
b.
c.
d.
Minimum power, use ailerons to arrest roll, stick forward, rudder to
oppose yaw.
Maximum power, centralise ailerons, opposite rudder to oppose yaw, stick
forward.
Minimum or maximum power depending on propulsion system type,
ailerons neutral, opposite rudder to oppose roll, stick forward.
Minimum or maximum power depending on propulsion system type,
ailerons neutral, opposite rudder to oppose yaw, stick forward.
CON 39.
Control flutter is reduced by?
a.
b.
c.
d.
Mass balance aft of the hinge line.
Aerodynamic balance forward of the hinge line.
Mass balance forward of the hinge line.
Power assisted flying controls.
CON 40.
What is the purpose of a down spring in a control system?
a.
b.
c.,
d.
To increase hinge moments.
To increase stick force gradient
To decrease stick force gradient.
To prevent control flutter.
CON 41.
How is control flutter minimised?
a.
b.
c.
d.
Aerodynamic balance.
Control lock out at high speed.
Mass balance.
Power assisted control systems.
CON 42.
How do ailerons and r r 1 spoilers respond if the control wheel is moved to the
left.
a.
b.
c.
d.
Left aileron down, left spoiler does not move, right aileron and spoiler up.
Right aileron and spoiler down, left aileron and spoiler up.
Left aileron and spoiler down, right aileron and spoiler up.
Left aileron and spoiler up, right aileron down, right spoiler remains
retracted.
CON 43.
What would be the effect if the C of G were at the forward limit for take-off?
a.
b.
c.
d.
Longitudinal stick force would be higher.
Longitudinal stick force would be lower.
VMcc would be lower.
Longitudinal stick force would be unchanged.
CON 44.
Methods of providing aerodynamic balance include?
a.
b.
c.
d.
Weights in front of the leading edge.
Q feel units.
Power assisted flying controls.
Horn balance.
CON 45.
VMCA
.........with altitude and VMcc ..........with altitude?
a.
d.
c.
d.
Increases,
Increases,
Decreases,
Decrease,
Increases.
Is not relevant.
Decreases.
Decreases.
CON 46.
\IMCGis limited by?
Critical engine failure on take off.
Yaw control authority.
Slippery runways.
Engine power.
a.
b.
c.
d.
CON 47.
A yaw damper?
a.
b.
c.
d.
Improves lateral stability.
Prevents spiral instability.
Is useful a t low speeds only.
Is not necessary in swept wing aircraft.
CON 48.
VMCLis minimum control speed in the
by.. ..?
...
a.
b.
c.
d.
Landing,
Landing,
Take-off,
Take-off,
..........Configuration and is determined
Pitch authority.
Yaw and roll authority.
Pitch authority.
Yaw and roll authority.
CON 49.
Aerodynamic balance is provided to?
a.
b.
c.
d.
Minimise pitching moment changes.
Prevent control flutter.
Prevent reverse control.
Reduce control hinge moments.
CON 50.
Adverse yaw is caused by?
a.
b.
c.
Angle of attack changes.
Speed changes.
Spinning.
d.
Ailerons.
CON 51.
Spring tabs?
a.
b.
c.
d.
Reduce hinge moments a t low speeds.
Reduce hinge moments a t high speeds.
Reduce hinge moments a t all speeds.
Stiffen the controls to prevent flutter.
CON 52.
What components ensure correct response to control inputs when an aircraft
enters or pulls out of a turn?
a.
b.
c.
d.
Elevons. '
Ventral fin.
Yaw damper.
Frise ailerons.
CON 53.
What components ensure correct response to control inputs when an aircraft
enters o r pulls out of a turn?
a.
b.
c.
d.
Dorsal fin.
Elevons.
Yaw damper.
Roll spoilers.
CON 54.
I f an elevator is moved to counter a forward C of @ position, in what direction
will the trim tabs move in relation to the elevator?
a.
Up.
b.
Down.
Up or down depending on type.
c.
They will not move relative to the elevator.
d.
CON 55.
If a trim tab is used to reduce elevator hinge moments to zero, what will be the
response of the elevators?
a.
b.
c.
d.
Move up.
Move down.
Not move.
Up o r down depending on airspeed.
CON 56.
What components ensure correct response to control inputs when an aircraft
enters o r pulls out of a turn?
a.
b.
c.
d.
Dorsal fin.
Differential ailerons.
Yaw damper.
Flaperons.
CON 57.
What device ensures correct response to control inputs when an aircraft enters
or pulls out of a turn?
a.
b.
c.
d.
Vortex generators.
Vortilons.
Dorsal fin.
Rudder-aileron coupling.
CON 58.
If the total moments about an axis are not zero, what will be the result?
a.
b.
c.
d.
Constant angular acceleration about the axis.
Equilibrium about the axis.
Stability about the axis.
Continuous rate rotation about the axis.
CON 59.
How is aerodynamic balance achieved?
a.
b.
c.
d.
Fitting weights a t the leading edge of the control surface.
Balance tabs.
Placing the hinge aft of the C of P of the control.
Rudder-aileron coupling.
CON 60.
How is aerodynamic balance achieved?
a.
b.
c.
d.
Fitting weights a t the leading edge of the control surface.
Horn balance.
Hinge line aft of C of G.
Aileron-rudder coupling.
CON 61.
What will be the immediate effect of pushing the right rudder pedal forward?
a.
b.
Right yaw and left roll.
Left yaw and right roll.
c.
d.
Right yaw and right roll.
Left yaw and left roll.
CON 62.
How is aerodynamic balance achieved?
a.
b.
c.
d.
Hinge line aft of C of P.
Balance tabs.
Rudder-elevator coupling.
Fitting weights a t the leading edge of the control surface.
CON 63.
How is aerodynamic balance achieved?
a.
b.
c.
d.
Fitting weights a t the leading edge of the control.
Placing hinge line closer to, but in front of C of P of control surface.
Placing hinge line aft of C of G of control surface.
Hydraulic power systems.
CON 64.
Which of the following statements best describes the purpose of the T-tail
configuration?
a.
b.
c.
d.
Reduce interference between wing downwash and tailplane?
Prevent deep stall.
Enable engines to be located a t sides of fuselage.
Increase fin effectiveness.
CON 65.
How is aerodynamic balance achieved?
a.
b.
c.
d.
Fitting weights at the leading edge of the control surface.
Internal balance using pressure differences above and below the wing.
Hinge line aft of C of P.
Pneumatic o r electrical power assistance.
CON 66.
Servo tabs?
a.
b.
c.
d.
Make controls less effective a t low airspeeds.
Are balance tabs.
Are activated by motion of parent control surface.
Permit use of smaller control surfaces.
CON 67.
If after suffering a right engine failure, the pilot of a twin engine aircraft
maintains altitude whilst keeping the wings level and the aircraft on track as
indicated below, what will the turn and slip indicator will show?
a.
b.
c.
d.
Right turn,
Left turn,
No turn,
No turn,
ball to the left.
ball to the right.
ball central.
ball right.
Sideslip Angla
CON 68.
When establishing V Min ~
a prototype
~
aircraft the nose wheel is considered
inoperative in order to?
a.
b.
c.
d.
Prevent weather-cocking in strong cross-wind take-offs.
Allow for slippery runways.
Allow for take-off abandonment after nose wheel lift-off.
Allow for aft C of G take-offs.
CON 69.
Servo tabs?
a.
b.
c.
d.
Are always used as trim tabs.
Are never used as trim tabs.
Are sometimes used as trim tabs.
Do not require trimming.
CON 70.
Servo tabs?
a.
b.
c.
d.
Cannot be used following hydraulic power failure.
Can always provide control following hydraulic power failure.
Can always provide control following hydraulic power failure.
Can never be used in power-assisted control systems.
!
CON 71.
If an elevator becomes jammed its servo-tab will?
a.
b.
c.
d.
Act as an anti-balance tab, increasing control hinge moment.
Jam.
Provide limited control but acting in the opposite sense to the elevator.
Cause all longitudinal control to be lost.
CON 72.
The greatest control authority is required when dealing with?
a.
b.
c.
Single engine failure in a twin propeller aircraft.
Single engine failure in a single propeller aircraft.
Single engine failure in a four engine jet aircraft.
Single engine failure in a four engine propeller aircraft.
d.
CON 73.
The principal advantage of mounting engines on rear fuselage is?
a.
b.
c.
d.
Reduced wing flutter.
keduce wing strength required.
Engines less affected by turbulence in a stall.
Yaw and pitch moments a r e less affected by thrust changes.
CON 74. :.
When cruising a t high mach numbers high speed jet aircraft usually employ
:. ... for roll control?
..
a.
b.
c.
d.
Outboard ailerons and lift spoilers.
Inboard ailerons and lift spoilers.
Inboard ailerons and roll spoilers.
Outboard o r inboard ailerons.
CON 75.
When handling critical engine failure in a heavily loaded twin engine aircraft?
a.
b.
c.
d.
Use rudder to arrest yaw and to get back on heading, then use aileron to
eliminates side-slip.
Use aileron to counter both yaw and side-slip.
Use rudder to counter both yaw and side-slip.
Use rudder to arrest and to turn beyond the original heading so that the
aircraft side-slips along the desired track. Use ailerons to keep wings
level.
CON 76.
VhlCCmay be limited by?
1.
2.
Maximum roll rate available.
Engine failure during take-off roll.
a.
b.
c.
d.
1 and 2 a r e correct.
1 only is correct.
2 only is correct.
Neither 1 nor 2 are correct.
CON 77.
Pushing the control column to the right produces ....... in a n aircraft exhibiting
adverse yaw?
a.
b.
c.
d.
Left roll and left yaw.
Right roll and right yaw.
Right roll and left yaw.
Left roll and right yaw.
CON 78.
Fully powered flying controls?
a.
b.
c.
d.
Are not required in large aircraft.
Generate high stick forces a t high speeds.
Generate high stick forces a t low speeds.
Generate low stick forces.
CON 79.
Power assistance produces?
a.
b,
c.
d.
'
High stick forces at low speeds.
High stick forces a t high speeds.
No stick forces.
Constant stick forces.
CON 80.
The rudder trim wheel and the rudder bar?
a.
b.
c.
d.
Operate in opposite directions.
Operate in the same direction.
Operate together.
Cannot be operated together.
CON 81.
To trim for a left engine failure turning the trim wheel . . ... moves the rim tab
9
...... and the rudder
. .. .
........
a.
b.
Left,
Left,
Left,
Right,
Left.
Left.
c.
d.
Right,
Right,
Left,
Right,
Right.
Left.
CON 82.
When making a large change in airspeed
a.
b.
c.
d.
Trim,
Adjust power,
Adjust speed,
Trim,
,
..............then .............
Adjust speed.
Trim.
Trim.
Adjust attitude.
CON 83.
The principal method of speed control is?
a.
b.
c.
d.
Power.
Propeller RPM.
Attitude.
Thrust.
CON 84.
The principal method of altitude control is?
a.
b.
c.
d.
Power.
Pt-opeller RPM.
Attitude.
Thrust.
CON 85.
Positive pitching is?
a.
b.
c.
d.
Nose up.
Nose down.
Nose horizontal.
Nose up o r down depending on load factor.
CON 86.
Positive rolling is?
a.
b.
c.
d.
Right wing down.
Right wing up.
Left wing down.
Wings level.
CON 87.
Positive yaw is?
3
a.
b.
c.
d.
Nose moving left.
Nose moving right.
Nose central.
Tail moving right.
CON 88.
Aileron reversal?
a.
b.
c.
d.
Is not possible with fully powered flying controls.
Is most likely with power assisted controls.
Is most likely with fully powered controls.
Is most likely with manual controls.
CON 89.
Control reversal?
Occurs only in the transonic speed range.
Cannot occur in the transonic speed range.
Is most likely a t high speeds.
Is most likely at low speeds.
a.
b.
c.
d.
CON 90.
Control flutter?
a.
b.
c.
d.
Is prevented by powered flying controls.
Is prevented by power assisted flying controls.
Is minimised by anti-balance tabs.
Is minimised by mass balancing.
CON 91.
Ailerons produce?
a.
b.
c.
d.
Roll only.
Roll and yaw.
Roll, yaw and pitch.
Yaw only.
-.
CON 92.
Stabilators provide?
a.
b.
c.
d.
'
.
-
Longitudinal stability and pitch control.
Longitudinal stability arld roll control.
Longitudinal stability, pitch control and roll control.
All of the above plus trimming.
CON 93.
The three axes of motion?
a.
b.
c.
d.
Never cross.
Cross at the C of P.
Cross at the C of G.
Cross at the neutral point.
CON 94.
Fly-by-wire control systems?
a.
b.
c.
d.
Are heavier than conventional systems.
Are lighter than conventional systems.
Do not require hydraulic power.
Use servo tabs.
CON 95.
Ruddervators?
a.
b.
c.
d.
Replace rudders and ailerons.
Replace rudders and elevators.
Replace elevators and ailerons.
Are used only for trimming.
CON 96.
Servo tabs ..... in a fully powered flying control system?
...
a.
b.
c.
d.
Cannot be used.
Act as trim tabs.
Provide emergency control following hydraulic failure.
Act as balance tabs.
CON 97.
For JAR certification the stick force gradient?
a.
b.
c.
d.
Is not permitted.
Must be positive.
Must be negative.
Must be constant.
CON 98.
A down spring?
a.
b.
c.
d.
Decreases stick force gradient.
Increases stick force gradient.
Increases stick forces in a pull up manoeuvre.
Is not used in JAR certificated aircraft.
CON 99.
A bob weight?
a.
b.
c.
d.
Decreases stick force gradient.
Increases stick forces at high speeds only.
Increases stick forces in a pull up manoeuvre.
Is not used in JAR certificated aircraft.
CON 100.
Primary flying controls include?
a.
b.
c.
d.
Ailerons, elevons and roll spoilers.
Ailerons, air brakes and lift spoilers.
Elevons, rudder and stabilisers.
Rudder, roll spoilers and nose wheel steering.
CLIMB 1.
What speed is required to achieve maximum endurance in a piston engine
powered and jet engine powered aircraft respectively?
CLIMB 2.
Select the appropriate words to complete the following statement.
..........
Fuel flow in a piston engine aircraft is proportional to
whilst that in a jet
Thrust output of a jet engine
powered aircraft is proportional to
with increasing airspeed whilst that of a piston engine ................
..................
1.
2.
3.
4.
5.
.........
Is approximately constant
Thrust
Reduces rapidly
Power
RPM
CLIMB 3.
Select the correct words to complete the following statement.
........... This
To achieve the maximum possible glide range it is necessary to fly at
is achieved by flying the aircraft in a ....................condition and at .........
a.
b.
c.
d.
BEST L:D RATIO
BEST L*:D RATIO
BEST L:D RATIO
BEST L*:D RATIO
20' FLAPS
FLAPS UP
FLAPS UP
FLAPS UP
VMP
VMD
VMD
VMP
CLIMB 4.
What speed is required to achieve maximum angle of climb in a jet aircraft and a
piston aircraft respectively and for what purpose might this be required?
a.
b.
c.
d.
VMD Minimum safe speed
VMP Minimum safe speed
Minimum safe speed VMD
VMD Minimum safe speed
Obstacle clearance after take-off.
Obstacle clearance after take-off.
Gain height rapidly.
Gain height rapidly.
CLIMB 5.
What is the effect of increasing altitude on best rate of climb IAS and best climb
TAS respectively?
a.
b.
c.
d.
Increases
Constant
Decreases
Decreases
Decreases
Increases
Constant
Increases
CLIMB 6.
What happens to the range between minimum and maximum flight speeds for a
subsonic aircraft as altitude increases?
a.
b.
c.
d.
I t increases.
I t decreases.
I t remains constant.
It decreases then increases.
CLIMB 7.
Select the appropriate words to complete the following statement.
Thrust horsepower output of a propeller aircraft
whilst that of a jet
...
1.
2.
3.
4.
5.
Is approximately constant.
Is unchanged.
Increases rapidly then decreases rapidly.
Reduces slowly.
increases approximately linearly.
......with increasing airspeed
CLIMB 8.
When flying a t VMDan aircraft has a C1, of 0.45 and a CDof 0.0225. If its engines
fail when flying a t 36000 feet what will be its maximum glide range?
a.
b.
c.
d.
100 nm.
200 nm.
120 nm.
175 nm.
CLIMB 9.
When flying a t VMDan aircraft weighing 400000 Ibf has a CL of 0.45 and a CD of
0.0225. If its engines fail when flying a t 36000 feet what will be its maximum
glide range if the pilot immediately dumps 100000 lbf of fuel? What effect will
the reduced weight have on VMD?
a.
b.
c.
d.
~ 25%.
Reduce V I \ I by
Increase VMDby 25%.
No change in VhlD.
Reduce VMD.
100 nm
200 nm.
120 nm.
120 nm.
CLIMB 10.
What is the available rate of climb a t service ceiling for a piston and jet aircraft
respectively?
a.
b.
c.
d.
500 fpnl
100 fpin
50 fpm
250 fpm
100 fpm.
500 fpm.
100 fpm.
500 fpm.
CLIMB 11.
What speed is required to achieve best rate of climb in a piston and jet aircraft
respectively?
a.
b.
c.
d.
VMD
V~IP
Between Vhlp and VMD
Greater than VMD
VMP.
VMD.
Between VMDand VNE.
Less than VNE.
CLIMB 12.
How is the absolute ceiling indicated on a power available / power required
graph for a piston and jet aircraft respectively?
a.
The power available curve will be tangential to the power required curve.
b.
c.
d.
The distance between power available and power required will be just
sufficient to give rates of climb of 100 fpm and 500 fpm respectively.
For a piston aircraft the power available curve will be tangential to the
power required curve. A thrust available / thrust required curve must be
used for a jet aircraft.
At VhlDand VRIPrespectively.
CLIMB 13.
What will be the effect of a headwind on glide range and glide angle
respectively?
a.
b.
c.
d.
Decrease
No effect
Decrease
Increase
Decrease.
Increase.
Increase.
No effect.
CLIMB 14.
What will be the effect of a tailwind on glide range and rate of descent?
a.
b.
c.
d.
Increase
No effect
Decrease
Increase
Decrease.
Increase.
Increase.
No effect
CLIMB 15.
Which of the following best describes the effect of flap deployment?
a.
b.
c.
d.
Increased lift at any given speed, reduced L;D ratio, increased take-off
speed and landing speeds, increased angle of climb.
Increased lift at any given speed, increased L;D ratio, reduced take-off
speed and landing speeds, reduced angle of climb.
Increased lift a t any given speed, reduced L;D ratio, reduced take-off
speed and landing speeds, reduced angle of climb.
Increased lift at any given speed, reduced L;D ratio, increased take-off
speed and landing speeds, no effect on angle of climb.
CLIMB 16.
Best endurance for a piston aircraft is achieved at VhlP,whilst that for a jet
aircraft is achieved a t (VhZD).Why is this so?
a.
b.
c.
d.
Fuel consumption in a piston aircraft is proportional to power output,
whereas that of a jet aircraft is proportional to thrust and hence drag.
Fuel consumption in a piston aircraft is proportional to rpm, whereas that
of a jet aircraft is proportional to power and hence drag.
Fuel consumption in a piston aircraft is proportional to thrust, whereas
that of a jet aircraft is proportional to power and hence drag.
Fuel consumption in a piston aircraft is proportional to power output,
whereas that of a jet aircraft is proportional to rpm.
CLIMB 17.
Which of the following best describes the effect of increasing altitude on
maximum rate of climb?
a.
b.
c.
d.
Both power available and power required increase with increasing
altitude but the gap between them narrows so rate of climb reduces.
Botli power available and power required decrease with altitude but the
gap between them narrows so rate of climb reduces.
Power available increases with altitude whilst power required reduces so
the gap between them and hence rate of climb increases.
Power available reduces with altitude whilst power required increases so
the gap between them and hence rate of climb decreases.
CLIMB 18.
Which of the following best describes the manner in which best climb speed
varies with altitude?
a.
b.
c.
d.
Best climb EAS increases whilst best climb TAS reduces with increasing
altitude for both piston and jet aircraft. The effect on TAS is of the same
magnitude in both cases.
Best climb EAS and best climb TAS both increase with increasing altitude
for both piston and jet aircraft. The effect on TAS is greater for a piston
than for a jet.
Best climb EAS and TAS both decrease with increasing altitude for both
piston and jet aircraft. The effect on TAS is greater for a jet than for a
piston.
Best climb EAS reduces whilst best climb TAS increases with increasing
altitude for both piston and jet aircraft. The effect on TAS is greater for
a piston than for a jet.
CLIMB 19.
How will a headwind and a tailwind respectively affect best range glide speed?
a.
b.
c.
d.
Decrease
Decrease
Increase
Increase
Increase
Decrease
Decrease
Increase
CLIMB 20.
If an aircraft enters a banked turn whilst climbing to cruising altitude what
effect will this have on rate of climb a t constant power setting?
a.
b.
c.
Climbing and turning occur in different planes and hence are not related,
so rate of climb will remain constant regardless of bank angle.
Banking will increase load factor and hence induced drag. This will
reduce the rate of climb.
Although banking will increase load factor and hence induced drag, rate
of climb depends only on excess thrust. Rate of climb will therefore
remain unchanged.
d.
Banking will reduce the climb gradient, so more power will be available
to increase rate of climb.
CLIMB 21.
Select the appropriate words to complete the following statement.
..........
whilst that in a jet
Fuel flow in a piston engine aircraft is proportional to
Thrust available from a jet
powered aircraft is directly proportional to
engine ..................with increasing airspeed whilst that of a piston engine
.........
................
1.
2.
3.
4.
5.
1s approximately constant
Thrust
Reduces rapidly
Power
Increases approximately linearly
CLIMB 22.
An aircraft weighing 200000 Ibf has a maximum excess power available of 1500
Thrust Horse Power. What will be its maximum rate of climb?
a.
b.
c.
d.
512 fpm.
753 fpm.
248 fpm.
358 fpm.
CLIMB 23.
An aircraft weighing 50000 lbf requires a thrust of 20000 Ibf when flying straight
and level a t 250 Kts IAS. What will be its maximum angle of climb a t this speed
assuming its maximum thrust at this speed is 40000 lbf and the total drag force
does not change during the climb?
a.
b.
c.
d.
45.6 Degrees.
20.6 Degrees.
23.6 Degrees.
12.6 Degrees.
CLIMB 24.
In tlie climb described in question 23 what will be the lift force generated by the
aircraft?
a.
b.
50000 lbf.
35825 Ibf.
-
c.
d.
45825 lbf.
25825 Ibf.
CLIMB 25.
An aircraft weighing 50000 Ibf requires a thrust of 20000 Ibf when flying straight
and level at 250 Kts IAS. How many Thrust Horse Power will it be developing
when flying straight and level at this speed?
(Assume 1 nm = 6000 ft)
CLIMB26.
Which of the following describes the method of calculating the thrust horsk
power being developed by an aircraft in unaccelerated straight and level flight?
1.
2.
3.
4.
a.
b.
e.
d.
Convert EAS into feet per minute, multiply by total drag in Ibf then
divide by 33000 ft lbflmin.
Convert TAS into feet per second, multiply by total drag in Ibf then divide
by 550 ft Ibflsec.
Convert EAS into feet per second, multiply by total drag in Ibf then divide
by 550 ft Ibflsec.
Convert TAS into feet per minute, multiply by total drag in Ibf then
divide by 33000 ft Ibf lmin.
1only.
3 only.
2 and 4.
1and 3.
CLIMB 27.
An aircraft weighing 50000 Ibf requires a thrust of 20000 Ibf when flying straight
and level at 250 Kts IAS. If it were to climb to 40000 ft what would be its power
available in comparison to that at ISA msl?
1-
CLIMB 28.
What would be the effect of sweeping back the wings of a variable geometry
aircraft in gliding flight?
a.
b.
Profile drag would reduce unless glide angle was reduced.
Induced drag would increase unless glide angle was increased.
c.
d.
Glide range and rate of descent would be increased.
Glide range and rate of descent would be decreased.
CLIMB 29.
How is maximum rate of emergency descent achieved in a high speed jet
aircraft?
a.
b.
c.
d.
Pitch down, retract flaps, spoilers and undercarriage and increase thrust,
Deploy spoilers, reduce thrust to idle, use pitch to maintain speed within
limits
Deploy high lift device, retract undercarriage and spoilers. Increase
thrust to maximum, pitch 45.degrees nose dcwn.
Thrust to idle, increase pitch to stalling angle, maintain nose up attitude
to hold aircraft in the stall.
CLIMB 30.
What do points A and B represent on the whole aircraft polar
diagram right?
a.
b.
c.
d.
Best L;D ratio
Minimum drag
Best L:D ratio
Stalling
Minimum drag.
Minimum power.
Stalling.
Best L:D ratio.
CLIMB 31.
An aircraft weighing 50000 Ibf is able to achieve a maximum rate of climb of
1500 ft / min when climbing a t 250 Kts TAS. How much Excess Power will be
required to achieve this rate of climb?
a.
b.
c.
d.
45000000 ft lbf / mjn.
55000000 ft lbf / min.
6500d000 ft Ibf 1 min.
75000000 ft lbf / min.
CLlMB 32.
An aircraft weighing 50000 Ibf is able to achieve a maximum rate of climb of
1500 ft / min when climbing a t 250 Kts TAS. What will be the maximum all up
weight a t which it can achieve a 5% climb gradient a t this speed assuming its
power available and total drag do not change during the climb?
a.
b.
c.
d.
40000 Ibf.
50000 lbf.
60000 Ibf.
70000 Ibf.
CLIMB 03.
Increased weight reduces the rate of climb and climb gradient but?
a.
b.
c.
d.
Vx increases and Vy decreases.
Vx and Vy increase.
Vx and Vy decrease.
Vx decreases and Vy increases.
CLIMB 34.
Best climb gradient is achieved by flying at approximately?
a.
b.
c.
d.
1.1 Vs.
1.2 Vs.
Best CL : CD'.
BestCL:CD.
CLIMB 35.
Absolute ceiling occurs when?
a.
b.
c.
d.
ROC is 300 ft/min.
ROC is 750 ft/min.
For a jet ROC is 100 ft/min and 500 ft/min for a propeller aircraft.
ROC is zero for both jets and propeller aircraft.
CLIMB 36.
What airspeed will produces the greatest glide endurance? ,
a.
b.
C.
d.
VMPwhich is higher than VMD.
VMPwhich is lower than VMDs
VMD.
Vnlp or VMDdepending on engine type.
CLIMB 37.
What factors determine maximum glide range?
a.
b.
c.
d.
Wind, C L : C ~ratio and weight.
Wind, CL:CDratio and height.
Wind, Speed and weight.
Wind, CL:CDand altitude.
CLIMB 38.
What are VMCAand VMDR?
a.
I
b.
c.
d.
Maximum and minimum speeds for gliding.
~ i e i m u mand maximum speeds for gliding.
Speed for shallowest glide path and speed for lowest rate of descent.
Speed for maximum glide range and speed for minimum glide range.
CLIMB 39.
How do VhlGAand V M ~compare?
R
a.
b.
c.
d.
VMCAis greater than VMDR.
VMGAis less than VMDR.
VMGAis the sameas VMDR.
Depends on aspect ratio.
CLIMB 40.
Which of the following is equal to lift in a steady climb or descent?
a.
b.
c.
d.
W Cos angle of climb or descent.
W Sin angle of climb or descent.
W Tan angle of climb or descent.
None of the above.
CLIMB 41.
What proportion of thrust is employed in supporting the weight of an aircraft in
a steady climb?
a.
b.
c.
d.
W Sin angle of climb.
W Cos.angle of climb.
W Tan angle of climb.
None of the above.
CLIMB 42.
What provides maximum glide range?
a.
b.
c.
d.
Strong headwind.
Light headwind.
Strong tailwind.
Light tailwind.
CLIMB 43.
What would give maximum glide range in a headwind?
a.
b.
c.
d.
Flying faster than VMD.
Flying slower than VMP.
Flying slower than VMD.
Flying faster than VMp.
CLIMB 44.
What would give maximum glide range in a tailwind?
a.
b.
c.
Flying faster than VMB
Flying slower than VMD
Flying faster than VMp.
d.
Flying slower than Vs.
CLIMB 45.
Which of the points on the CL:CDpolar
would give maximum glide range?
a.
b.
c.
d.
A.
B.
C.
D.
CL
CLIMB 46.
In a steady climb?
a.
b.
c.
d.
Lift is less than weight and thrust is less than drag
Lift is less than drag and weight is less than thrust.
Weight is less than lift and drag is more than thrust.
Weight is more than lift and drag is less than thrust.
CLIMB 47.
What is the speed for minimum sink rate?
CLIMB 48.
What flap position would give maximum glide range?
CLIMB 49.
What speed gives best angle of climb in a jet aircraft?
CLIMB 50.
Which of the following occur a t VMD?
1.
2.
3.
4.
5.
L:D Max.
Max jet endurance.
prop range.
Max jet climb angle.
Max glide range all types.
a.
b.
c.
d.
1,2,3.
2,3,4.
3,4,5.
ax
All of the above.
CLIMB 51.
What will be the effect of an increase in weight?
1.
2.
3.
4.
VMDwill increase.
Glide range will decrease.
Glide angle will increase.
Glide range and angle will be unaffected.
a.
b.
c.
d.
1,2,3.
2,3,4.
None of the above.
1,4.
CLIMB 52.
In what direction does lift act in a steady climb?
a.
b.
c,
d.
Upwards.
Vertically.
At right angles to the flight path.
None of the above.
CLIMB 53.
What is the absolute ceiling of an aircraft?
1.
2.
3.
4.
The altitude where the low speed and high speed stall lines cross.
The altitude a t which power required is equal to power available.
The altitude a t which thrust available is equal to drag.
The altitude at which rate of climb is zero.
a.
1,2,3.
2,3,4.
b.
c.
d.
None of the above.
All of the above.
CLlMB 54.
Increasing aircraft weight. .........glide speed and...
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
....rate of descent?
Decreases.
Increases.
Decreases.
Increases.
CLIMB 55.
To descend a t constant glide angle and IAS, the pitch attitude must.. .
.........
a.
b.
c.
d.
Increase.
Decrease.
Increase then decrease.
Remain constant.
CLIMB 56.
For a constant mach number descent, gradient must ......?
a.
b.
c.
d.
Decrease.
Increase.
Increase then decrease.
Remain constant.
CLIMB 57.
For a constant IAS descent, gradient must
a.
b.
c.
d.
......?
Decrease.
Increase.
Increase then decrease.
Remain constant.
CLIMB 58.
A headwind in a constant mach number climb will
a.
b.
c.
d.
Decrease.
Increase.
Not affect.
Increase o r decrease depending on mach number:
CLIMB 59.
For maximum glide range, TAS must
a.
b.
c.
d.
............angle of climb?
Decrease in a headwind.
Decrease in a tailwind.
Increase in a tailwind.
Remain constant.
.....?
9
CLIMB 60.
To descend at constant IAS the
a.
b.
c.
d.
pitch down,
Pitch down,
Pitch up,
. pitch up,
Best cHtnb angle is achieved at
a.
b.
c.
d.
9
Increased. ..
Maintained constant.
Increased.
Decrease.
; CLI~B
61.
.
............must be...........
.............
9
VMP.
VMD.
Vx.
Vy.
CLIMB 62.
Maximum rate of climb occurs at
a.
b.
c.
d.
...........speed?
Maximum excess thrust.
Minimum drag.
Minimum thrust.
Maximum excess power.
CLIMB 63.
Maximum glide range is achieved at
a.
b.
c.
d.
.............
cL4
:CDMax.
C L :~CDMax.
CL: C DMax.
C L :~CDMax.
CLIMB 64.
Load factor in a steady climb will be....
a.
b.
c.
d.
...?
1.
More than 1.
Less than 1.
More or less than 1 depending on climb angle.
CLIMB 65.
Lift in a steady climb is equal to
a.
b.
c.
d.
9
Thrust.
Weight.
W Cos angle of climb.
W Sin angle of climb.
........?
CLIMB 66.
The correct procedure for an emergency descent is?
a.
b.
c.
d.
Maximise drag, minimise thrust, push the nose hard down.
Minimise drag, maximise thrust, push the nose hard down.
Maximise drag, minimise thrust, control speed with pitch attitude.
Maximise drag, minimise thrust, bank steeply.
CLIMB 67.
Which of the following is correct for a jet aircraft?
CLIMB 68.
Rate of climb is
a.
b.
c.
d.
..........by a headwind?
Increased.
Decreased.
Not affected.
Increased or decreased depending on load factor.
CLIMB 69.
Angle of climb is
a.
b.
c.
d.
..........by a headwind?
Increased.
Decreased.
Not affected.
Increased or decreased depending on load factor.
CLIMB 70.
Climb gradient is
a.
b.
c.
d.
Increased.
Decreased.
Not affected.
Increased or decreased depending on load factor.
CLIMB 71.
Power required is
a.
b.
c.
d.
........try a headwind?
..........as altitude increases in a steady climb?
Increased.
Decreased.
Not affected.
Increased or decreased depending on load factor.
CLIMB 72.
Load factor is
a.
b.
c.
d.
.......in a steady climb?
Increased.
Decreased.
Not affected.
Increased o r decreased depending on weight.
CLIMB 73.
If excess power is 25000 ft Ibflmin and aircraft weight is 10000 Ibf what will be
the maximum rate of climb?
a.
b.
c.
d.
2.5 ftlmin.
25 ftlmin.
250 ftlmin.
2500 ftlmin.
CLIMB 74.
If maximum thrust is 25000 Ibf, drag is 15000 Ibf and weight is 10000 Ibf what
will be the maximum angle of climb?
CLIMB 75.
If a t 250 Kts, excess power is 50,000,000 ft Ibflmin, and weight is 10000 Ibf, what
is maximum angle of climb?
CLIMB 76.
What might the points C and D represent on the whole aircraft polar
Diagram right?
a.
b.
c.
d.
Best L;D ratio
Jet aircraft Vy
Best L:D ratio
Prop aircraft Vy
Minimum drag.
Prop aircraft Vy.
Stalling.
Jet aircraft Vy.
CLIMB 77.
What might points C and D represent on the whole aircraft polar
Diagram right?
CL
a.
b.
c.
d.
Prop aircraft Vy
Prop aircraft Vx
VMD.
Prop aircraft Vy
Prop aircraft Vx.
VMP.
Prop aircraft Vx.
VMD.
CLIMB 78.
Increasing weight will ........glide speed and
a.
b.
c.
d.
Increase
Increase
Decrease
Decrease
........maximum glide endurance?
Increase.
Decrease.
Decrease.
Increase.
CLIMB 79.
Increasing weight will ......... VMD,and ..........rate of descent for best glide
range?
a.
b.
c.
d.
Increase
Increase
Decrease
Decrease
Increase.
Decrease.
Decrease.
Increase.
CLIMB 80.
Deploying landing flap will
a.
b.
c.
d.
Increase
Increase
Decrease
Decrease
........'Vx, and ..........maximum angle of climb?
Increil\e.
Decrease.
Decrease.
Increase.
CLIMB 81.
Use of reheat in a climb will
maximum rate of climb?
a.
b.
c.
d.
Increase
Increase
Decrease
Decrease
........maximum angle of climb, and .........
Increase.
Decrease.
Decrease.
Increase.
CLIMB 82.
If use of reheat doubles thrust available this will
climb?
a.
b.
c.
d.
..........the maximum angle of
Half.
Double:
More than double.
Less than double.
CLIMB 83.
If TAS is 200 Kts and rate of climb is 1000 fttmin, what is the climb gradient?
CLIMB 84.
If climb gradient is 15% and TAS is 250 Kts, what is the rate of climb?
CLIMB 85.
For maximum rate of descent use
maintain airspeed within limits?
a.
b.
c
d.
Maximum
Maximum
Minimum
Minimum
Minimum
Minimum
Maximum
Maximum
....thrust, ..........Drag, and ...........to
Spoilers.
Attitude.
Spoilers.
Attitude.
CLIIMB 86.
In a constant IAS climb
.......might be inadvertently exceeded?
CLIMB 87.
In a constant TAS climb
........might be inadvertently exceeded?
CLIMB 88.
In a constant mach climb
.......might be inadvertently exceeded?
CLIMB 89.
In a constant mach descent
.......might be inadvertently exceeded?
CLIMB 90.
When climbing at constant '/zpv2
IAS will
a.
b.
c.
d.
........
?
Increase.
Decrease.
Remain constant.
Increase up to 36000 ft then remain constant.
CLIMB 91.
If at 40000 feet altitude air density is % of its sea level value, how will best climb
TAS compare with best climb IAS?
a.
b.
c.
d.
The same.
Double.
Quadruple.
Less than.
CLIMB 92.
If maximum thrust is 10000 Ibf, drag is 500 Ibf and weight is 5000 lbf, what will
be the maximum rate of climb at the absolute ceiling?
b.
100 ft/min.
500 ft/min.
c.
d.
zero.
More than 500 ftlmin.
a.
CLIMB 93.
How does Vx compare with Vy a t the absolute ceiling?
a.
b.
c.
d.
The same.
More than.
Less than.
More o r less than depending on aircraft type.
CLIMB 94.
How does the angle of climb of a propeller aircraft vary with increasing
airspeed?
a.
b.
c.
d.
Decreases.
Increases.
Remains constant.
Increases up to VMDthen decreases.
CLIMB 95.
How does maximum angle of climb for a jet aircraft vary with increasing
airspeed?
a.
b.
c.
d.
Decreases.
Increases.
Remains constant.
Increases up to VMDthen decreases.
CLIMB 96.
How does maximum rate of climb for a jet aircraft vary with increasing
airspeed?
a.
b.
c.
d.
Decreases.
Increases u p to VMPthen decreases.
Remains constant.
Increases u p to 1.3brMD
then decreases.
CLIMB 97.
If weight is 15000 Ibf, maximum thrust is 25000 Ibf and drag is 5000 lbf, what
will be the maximum climb angle?
CLIMB 98.
If excess power is 350 THP when TAS is 250 Kts and weight is 10000 Ibf, what
will be the maximum rate of climb at that speed?
CLIMB 99.
If the weight of the aircraft in question 98 is doubled how will this affect
maximum rate of climb?
a.
b.
c.
d.
Double it.
Half it.
Increase it.
Decrease it by 41%.
CLIR~Bloo.
If airspeed is maintained constant as aircraft weight reduces due to fuel use the
3
aircraft will be in a
.............
a.
b.
c.
d.
Constant rate climb.
Cruise climb.
Zoom climb.
Constant altitude flight.
TURN 1.
What control inputs are required to maintain airspeed and altitude when
turning in a jet aircraft?
a.
b.
c.
d.
Increase thrust and angle of attack.
Increase turn radius and decrease thrust.
Increase pitching angle.
Increase bank angle.
TURN 2.
What turn radius will be achieved in a balanced turn at a true airspeed of 250
Kts if the aircraft is banked to 45 degrees? Use g = 10 m/s2, 1 Kt = 0.51 m/s
a.
b.
c.
d.
1625.25 ni.
1725.25 m.
6250 m.
2675 m.
TURN 3.
By what % will the load factor increase if an aircraft enters a balanced 60'
banked turn?
TURN 4.
Two aircraft, one of high mass and the other of low mass enter a 30' balanced
banked turn at a TAS of 300 Kts. Which of the following statements best
describes the results?
a.
b.
c.
d.
Low mass aircraft will achieve the tightest (lowest radius) turn.
High mass aircraft requires less power than low mass aircraft for same
turn radius.
The high mass aircraft will achieve the highest rate of turn.
Both aircraft will achieve the same turn radius and rate of turn.
TURN 5. '
If radius of turn in a constant altitude banked turn a t 250 Kts is 1000 m, what
will it be at 500 Kts if all other factors remain unchanged?
TURN 6.
By what percentage is load factor increased when going from straight and level
flight into a constant altitude 45' banked turn?
TURN 7.
If at 200 Kts in a 2915 meter radius constant altitude banked turn, the angle of
bank is 20°, what will it be in the same radius turn at 500 Kts?
TURN 8.
At a TAS of 400 Kts, what would be the radius of a constant altitude 45' banked
turn if g = 10 m/s2?
TURN 9.
Increasing weight at constant bank angle will
a.
b.
c.
d.
.............turn radius?
Decrease.
Increase.
Not affect;
Increase or decrease depending on altitude.
TURN 10.
What % increase in lift force will be required to maintain a 60°banked constant
altitude urn?
TURN11.
3
The turn radius of a 50000 Kg aircraft flying at 250 Kts and 45' bank will be
the same aircraft at 60000 Kg flying a t 200 Kts and 45"ank.
...........
a.
b.
c.
d.
Greater than.
Less than.
The same as.
Less than if at a lower altitude.
TURN 12.
If Vn, is 200 Kts what will be Vs in a 2g turn?
a.
b.
c.
d.
150 Kts.
281 Kts.
241 Kts.
250 Kts.
TURN 13.
What speed would be required to achieve a 90"anked
constant altitude turn?
a.
b.
c.
d.
200 Kts.
300 Kts.
400 Kts.
None of the above.
TURN 14.
If Vsrgis PO0 Kts what wili be jrs in a 45' banked level turn?
a.
b.
c.
d.
200 Kts.
119 Kts.
250 Kts.
130 Kts.
TURN 15.
What bank angle will cause a JAR certificated passenger aircraft to reach its
limiting load factor in a constant altitude turn with flaps down?
TURN 16.
What bank angle will cause a J A R certificated passenger aircraft to reach its
limiting load factor in a constant altitude turn with flaps up?
TURN 17.
To maintain altitude on entering a banked turn, thrust must be
3
of attack must be
........
a.
b.
c.
d.
Increased,
Increased,
Decreased,
Decreased,
Decreased.
Increased.
Increased.
Decreased.
TURN 18.
Rate 1 constant altitude turn bank angle is determined by?
a.
b.
c.
d.
Weight.
C of G position.
TAS.
Load factor.
........and angIe
TURN 19.
By what % will stalling speed be increased in a 75' banked constant altitude
turn?
8
TURN 20.
By what % is stalling speed reduced when straightening up after a 60' constailt
altitude banked turn?
TURN 21.
Stalling speed in a constant altitude banked turn is proportional to?
a.
b.
c.
d.
d ~ o a factor.
d
Cos AOB.
l/load factor.
TAS.
TURN 22.
The % increase in stalling speed in a constant altitude banked turn will be?
a.
b.
c.
d.
(Load factor -1) x 100%.
( d ~ o a dfactor -1) x 100%.
$Load factor +1) x 100%.
~ ( C OAOB
S -1) x 100%.
TURN 23.
If a 40000 N aircraft enters a 35' constant altitude banked level turn, the lift
generated will?
a.
b.
c.
d.
Remain constant.
Increase to 48830 N.
Increase to 45195 N.
Double.
TURN 24.
If an aircraft enters a 45' constant altitude banked turn without changing speed,
the thrust required will?
119
a.
b.
c.
d.
Double.
Quadruple
Triple.
Remain constant.
TURN 25.
In a constant altitude banked turn at constant speed, the load factor depends on?
a.
b.
c.
d.
AOB and TAS.
Radius of turn and weight.
AOB only.
Altitude.
TURN 26.
Doubling TAS a t constant AOB multiplies ROT by?
TURN 27.
Doubling TAS a t constant AOB increases turn radius by?
TURN 28.
If TAS = 200 Kts and turn radius = 1000 meters, what AOB will give a balanced
constant altitude turn?
a.
b.
c.
d.
27.25 degrees.
37.25 degrees.
47.25 degrees.
57.25 degrees.
TURN 29.
If TAS = 200 Kts, turn radius = 10C0 meters and AOB = 27 degrees, the aircraft
will?
a.
b.
c.
d.
Slip.
Skid.
Be in balanced flight.
Lose altitude.
TURN 30.
If ROT = 1 Radls, g = 9.81 m/s2 and V = 100 Kts, AOB for a balanced constant
altitude turn will be?
a.
b.
c.
d.
64 degrees.
74 degrees.
84 degrees.
94 degrees.
TURN 31.
If AOB = 47.25 degrees, radius of turn = 1000 meters and TAS = 200 Kts, the
'aircraft wiII?
a.
b.
c.
d.
Slip.
Skid.
Be in balanced flight.
Descend.
TURN 32.
If turn radius = 1000 meters, TAS = 200 Kts and AOB = 55 degrees the aircraft
wiIl?
a.
b.
c.
d.
Slip.
Skid.
Be in balanced flight.
Descend.
TURN 33.
If turn radius = 3000 meters, g = 9.81 m/s2 and TAS = 250 Kts, AOB for a
constant aItitude balanced turn is'?
a.
b,
c.
d.
20 degrees.
30 degrees.
40 degrees.
50 degrees.
TURN 34.
If TAS = 100 Kfs, turn radius = 1500 meters and g = 9.81 m/s2, AOB for a
constant altitude balanced turn is?
a.
b.
c.
d.
10 degrees.
20 degrees.
30 degrees.
40 degrees.
TURN 35.
If AOB = 45 degrees, TAS = 100 Kts and g = 9.81 d s 2 , ROT will be
balanced constant altitude turn?
a.
b.
c.
d.
........ in a
6 degrees.
11 degrees.
16 degrees.
21 degrees.
TURN 36.
If turn radius = 3000 meters, g = 9.81 m/s2, TAS= 250 Kts and AOB is reduced
from 30 degrees to 25 degrees the aircraft will?
a.
b.
c.
d.
Slip out to a radius of 5554 meters.
Slip in to a radius of 2554 meters.
Skid out to a radius of 3554 meters.
Skid in to a radius of 1554 meters.
TURN -37.
If AOB = 30 degrees, g = 9.81 m/s2, and ROT = 15 degls, what TAS will give a
balanced constant altitude turn?
a.
b.
c.
d.
22.5 Kts.
32.5 Kts.
42.5 Kts.
52.5 Kts.
TURN 38.
If radius of turn = 2000 meters, AOB = 40 degrees and g = 9.81 m/s2, what TAS
will give a constant altitude balanced turn?
a.
b.
c.
d.
152 kts.
252 Kts.
352 Kts.
452 Kts.
TURN 39.
If ROT = 15 degls, g = 9.81 m/s2 and TAS = 75 Kts ,what AOB will give a
balanced constant altitude turn?
a.
b.
c.
d.
25.6 degrees.
35.6 degrees.
45.6 degrees.
55.6 degrees.
TUKN 40.
If AOB = 40 degrees, g = 9.81 m/s2 and turn radius = 2500 meters, what TAS will
give a constant altitude balanced turn?
a.
b.
c.
d.
281 Icts.
381 Kts.
481 Kts.
581 Kts.
TURN 41:
If TAS = 150 Kts, AOB = 30 degrees and g = 9.81 m/s2, what radius will be
achieved in a constant altitude balanced turn?
a.
b.
c.
d.
1033 meters.
1 133 m r ' rs.
1233 meters.
1433 meters.
TURN 42.
A rate one 360 degree turn takes
a.
b.
c.
d.
....... to complete?
1 minute.
2 minutes.
3 minutes.
4 minutes.
TURN 43.
For a rate 1 level balanced turn at 40000 feet ISA, at an AOB of 30 degrees, the
IAS must be? Use g = 9.81 mls2 and IAS = TASi2 a t 40000 ft ISA.
a.
b.
c.
d.
54 Kts.
104 Kts.
154 Kts.
204 I<ts.
TURN 44.
Skidding in a turn is caused by excessive?
a.
b.
c.
d.
AOB.
Weight.
Load factor.
TAS.
TURN 45.
Slipping in a turn is caused by excessive?
a.
AOB.
b.
c.
d.
Weight.
Load factor.
TAS.
TURN 46.
If no corrective action is taken skidding in a turn will?
a.
b.
c.
d.
Reduce turn radius until a balanced turn is established.
Increase turn radius until a balanced turn is achieved.
Result in low speed stall.
Result in high speed stall.
TURN 47.
Minimum turn radius a t high altitude is limited by?
a.
b.
c.
d.
Stalling due to low density air.
Reduced power available.
Increased TAS:IAS ratio.
High g-loading on crew.
TURN 48.
VMCA
G
......... with increasing altitude?
a.
b.
c.
d.
Increases.
Decreases.
Remains constant.
Remains constant then increases.
TURN 49.
A balanced constant altitude turn requires?
a.
b.
c.
d.
Increased angle of attack and power.
Decreased angle of attack and power.
Increased angle of attack and decreased power.
Decreased angle of attack and increased power.
TURN 50.
Load factor in a 60 degree banked descending turn is ...
banked constant altitude turn?
.... in a 60 degree
a.
b.
c.
d.
Greater than.
Less than.
The same as.
Greater o r less than depending on altitude.
TURN 51.
In a balanced turn dihedral effect will?
a.
b.
c.
d.
Be maximum.
Decrease bank angle.
Increase bank angle.
Be nil.
TURN 52.
In a turn using excessive TAS dihedral effect will?
a.
b.
c.
d.
Be maximum.
Decrease bank angle.
Increase bank angle.
Be nil.
TURN 53.
In a turn using excessive bank angle dihedral effect will?
a.
b.
c.
d.
Be maximum.
Decrease bank angle.
Increase bank angle.
Be nil.
TURN 54.
On releasing ailerons in a co-ordinated turn, a spirally unstable aircraft will?
a.
b.
c.
d.
Increase bank angle and pitch nose down.
Increase bank angle and pitch nose up.
Decrease bank angle and pitch nose down.
Decrease bank angle and pitch nose up.
TURN 55.
In a turn using excessive bank angle, an aircraft will ........until forces are
..........
'7
a.
b.
c.
d.
Slip
Slip
Skid
Skid
Balanced.
Equal.
Balanced.
Equal.
TURN 56.
In a turn using excessive TAS, an aircraft will
a.
b.
c.
Slip
Slip
Skid
Balanced.
Equal.
Balanced.
........until forces are ...........
'7
d.
Skid
Equal.
TURN 57.
The ball in a serviceable slip indicator is
Indicatels the state of slip?
a.
b.
c.
d.
..........by ........... and ..........
Gravity
Acceleration
Gravity
Acceleration
Held central
Positioned
Held central
Positioned
Does not always.
Does not always.
Always.
Always.
TURN 58.
Stalling angle .........with bank angle in a turn?
a.
b.
c.
d.
Increases.
Decreases.
Does not vary.
Increases o r decreases depending on C of G position.
TURN 59.
Stalling speed ........with bank angle in a turn?
a.
b.
c.
d.
Increases.
Decreases.
Does not vary.
Increases ul decreases depending on C of G position.
TURN 60.
Turning capability
a.
b.
c.
d.
...... with increasing altitude?
Increases.
Decreases.
Remains constant.
Increases or decreases depending on C of G position.
TURN 61.
At the maximum operating alti ~ d aeJAK certificated passenger aircraft,
maximum bank angle in a constant altitude turn is?
a.
b.
c.
d.
29.7 degrees.
39.7 degrees.
49.7 degrees.
59.7 degrees.
TURN 62.
The maximum allowable bank angle in a constant altitude banked turn in a
passenger aircraft built to minimum J A R certification structural standards is?
a.
b.
c.
d.
46.4 degrees.
56.4 degrees.
66.4 degrees.
76.4degrees.
TURN 63. '
Increasing aircraft weight will ........ minimum turn radius if power is not a
limiting factor?
a.
b.
c.
d.
Increase.
Decrease.
Not affect.
Increase or decrease depending on TAS.
TURN 64.
Followi~igsingle engine failure, increasing weight ......... minimum turn radius?
a.
b.
c;
d.
Might increase.
Might decrease.
Will not affect.
Might increase or decrease depending on C of G position.
TURN 65.
Maximum rate of turn ...... with altitude?
a.
b.
c.
d.
Increases.
Decreases.
Does not vary.
Increases u p to 36000 feet then decreases.
TURN 66.
Minimum radius of turn .....with increasing altitude?
a.
b.
c.
d.
Increases.
Decreases.
Does not vary.
Increases up to 36000 feet then decreases.
TURN 67.
Rate of turn depends on?
a.
b.
IAS.
CAS.
c.
d.
EAS.
TAS.
TURN 68.
Maximum turn radius?
a,
b.
c.
d.
Increases with increasing altitude.
Decreases with increasing altitude.
Decreases with temperature.
Js unlimited a t all altitudes.
TURN 69.
Turning at a 90 degree AOB?
a.
b.
c.
d.
Always causes loss of altitude.
Never causes loss of altitude.
Causes loss of altitude only if TAS is too low.
Causes loss of altitude only if angle of attack is too low.
TURN 70.
Maximum allowable rate of turn in a JAR certificated passenger aircraft is?
a.
b.
c.
d.
Rate 1.
Rate 2.
Rate 2.5.
Rate 3.
TURN 71.
Stalling speed illcreases by
.....in a constant altitude 75 degree banked turn?
TURN 72.
Skidding in a turn occurs when
a.
b.
c.
d.
.... exceeds .... ?
Horizontal component of lift
Centrifugal force
Load factor
Power required
TURN 73.
Minimum turn radius is limited by?
Centrifugal force.
Horizontal component of lift.
Limiting load factor.
Power available.
a.
b.
c.
d.
Power available.
CL M A X Wing area.
Weight.
TURN 74.
Slipping in a turn occurs when .... exceeds .... ?
a.
b.
c.
d.
Horizontal component of lift
Centrifugal force
Load factor
Power required
Centrifugal force.
Horizontal component of lift.
Limiting load factor.
Power available.
TURN 75.
When skidding in a turn the slip indicator ball?
a.
b.
c.
d.
Moves towards the outside of the turn.
Moves towards the inside of the turn.
Remains central.
Moves towards the inside o r outside of the turn depending on the
relationship between TAS and AOB.
POW1.
If TAS is increased from 300 Kts to 400 Kts with no change in altitude,
configuration, o r weight, by what percentage will power required change?
a.
b.
c.
d.
Decrease by 135%.
Decrease by 35%.
Increase by 135%.
Increase by 235%
POW2.
What is the relationship between power required and TAS, as a n aircraft
accelerates above VhlD?
a.
b.
c.
d.
Power required increases in direct proportion to TAS.
Power required increases in inverse proportion to TAS.
Power required increases in proportion to (TAS)~.
Power required increases in proportion to (TAS)'.
POW3.
At what % of its stalling speed must a jet aircraft fly to achieve maximum
endurance for a given fuel load?
POW4.
For a pistoil aircraft a t constant weight, angle of attack, and configuration, what
mill be the effect of Increasing altitude?
a.
b.
c.
d.
lncreased power and TAS will be required.
Increased power will be required but at the same TAS.
Lower power will be required at the same TAS.
The same power will be required but a t an increased TAS.
POW5
For a piston aircraft a t a constant altitude, angle of attack, and configuration,
what wiil be the effect of increasing weight?
a.
b.
c.
d.
Rlore power will be required but a t the same TAS.
iMore power will be required but a t a higher TAS.
Tile same power will be required but at a higher TAS.
Rlore power will be required but a t a lower TAS.
POW6.
Compared with still air, when flying for maximum range into a headwind, speed
should be?
a.
R.
e.
d.
Faster.
The same,
Slower.
Depends on weight.
PO\V7.
Wlaat flight condition requires least power a t a given IAS?
a.
b.
c.
d.
Zero flap high altitude.
Zero flap low altitude.
30'flap low altitude.
30' flap high altitude.
POW8.
~Waximumpropeller aircraft range occurs at?
POW9.
Power required for a given IAS at 40000 feet altitude, is .......... times that
required at ISA MSL?
POW10.
Power required is proportional to?
a.
b.
C.
d.
TAS.
IAS.
TAS~.
TAS~.
POW1 1.
Power required equals?
a.
b.
c.
d.
Drag x IAS.
Drag x CAS.
Drag x EAS.
D r a g x TAS.
POW12.
As altitude increases, power available from a piston o r turbo-prop ......... whilst
3
that of a turbo-jet ...........
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
Increases.
Decreases.
Decreases.
Increases.
POW13.
As altitude increases, the power required curve moves ..... whilst the power
3
available curve mores .........
a.
b.
c.
d.
Up and right,
Up and left,
Down and left,
Down and right,
Down.
UPUPDown.
POW14.
As altitude increases jet aircraft excess power?
a.
Increases at a constant rate.
b.
c.
d.
Decreases at a constant rate.
Remains constant.
Decreases a t a rate that decreases with increasing altitude.
POW15.
Doubling IAS at a given altitude multiplies power required by?
POW16.
Minimum power required speed is?
a.
b.
c.
d.
Less t h a i ~Vs.
More than V M ~ .
Less than VMD.
Vs of propeller aircraft.
POW17.
Maximum range for a jet aircraft occurs at?
a.
b.
c.
d.
Less than Vs.
More than Vhan.
Less than VhlD.
VMD.
f OWIS.
Maximum excess power IAS
a.
b.
c.
d.
.......... with increasing altitude?
Increases.
Decreases.
Remains constant.
Increases then decreases.
POW19.
Maximum excess power TAS ......... ...with increasing altitude?
a.
b.
c.
d,
Increases.
Decreases.
Remains constant.
Increases then decreases.
POW20.
At the absolute ceiling, excess power?
a.
b.
c.
d.
Is zero.
Is maximum.
Varies with engine type.
Varies with TAS.
POW21.
At the absolute ceiling, VMPand VMD?
a.
b.
c.
d.
Are the same.
Both increase.
Both decrease.
Diverge.
POW22.
The power available and power required curves?
a.
b.
c.
d.
Never meet.
Never cross.
Are parallel at the absolute ceiling.
Cross a t the absolute ceiling.
POW23.
The power available and power required curves?
a.
b.
c.
d.
Cross at maximum and minimum speeds.
Never cross.
Are always parallel.
Are never parallel.
POW24.
In gliding flight?
a.
b.
c.
d.
The power required curves are irrelevant.
The power available curves are irrelevant.
No power is consumed.
Power consumption is always at a minimum.
POW25.
The best ratio of TAS to power required occurs at?
a.
b.
C.
d.
VMP.
VMD.
vx.
- Vy.
POW26.
Maximum jet endurance occurs at?
POW27.
Decreasing weight moves the power required curve?
a.
b.
c.
d.
'
Down and right.
Down and left.
Up and right.
Up and left.
POW28.
VhlDis point
a.
b..
c.
d.
....on the diagram right?
A.
B.
C.
D.
POW29.
Power required is?
a.
b.
c.
d.
Drag x IAS.
Drag x TAS.
Thrust required x TAS.
Thrust required x IAS.
POW30.
Excess power I Weight = ?
a.
b.
c.
d.
The sine of the maximum angle of climb.
The cosine of the maximum angle of climb.
The sine of the maximum angle of descent.
The maximum rate of climb.
POW31.
Moving C of G forwnrd will?
a.
Decrease range.
b.
c.
d.
Increases range.
Decrease range only if C of G is forward of C of P.
Decrease range only if C of G is aft of C of P.
POW32.
If the aft C of G limit is aft of the C of P and the forward limit if forward of the
C of P, then moving the C of G from its aft limit to its forward limit will?
a.
b.
c.
d.
Decrease range.
Increases range then decrease range.
Increase range.
Decrease range then increase range.
POW33.
Range will be maximum when?
a.
b.
c.
d.
C of G is forward of C of P.
CofGisaftofCofP.
C of G and C of P coincide.
When flying a t VMD.
POW34.
An aircraft of weight of 120000 Ibf can achieve a climb gradient of 2.5% using
maximum climb power, At what weight will it achieve a gradient of 3% if the
change in excess power is ignored?
a.
b.
c.
d.
100000 Ibf.
110000 Ibf.
130000 Ibf.
95000 Ibf.
POW35.
An aircraft of weight 120000 Ibf can achieve a climb gradient of 2.5% using
maximum climb power. What gradient will it achieve a t a weight of 150008 Ibf if
the change in excess power is ignored?
POW36.
If weight is increased by 25% when flying a t VklDpower required will increase
by?
POW37.
Thrust equals drag when?
a.
b.
c.
d.
Descending a t constant IAS.
Flying in level flight a t constant IAS.
Climbing a t constant TAS.
None of the above.
POW38.
Maintaining altitude after an increase in mass without changing angle of attack
requires?
a.
b.
c.
d.
Increased power and airspeed.
Decreased power and airspeed.
Increased power and decreased airspeed.
Decreased power and increased airspeed.
PO W39.
Two identical turbojet aircraft flying a t the same airspeed and altitude have the
sime SFC. Aircraft A weighs 200000 Kg and consumes fuel a t a rate of 5000
Kgthr. If aircraft B weighs 250000 Kg what will be its fuel consumption?
a.
b.
c.
d.
5250 Kglhr.
6250 Kglhr.
8250 Kglhr.
9812 Kglhr.
PO W40.
VIMPfor a turboprop is?
a.
b.
c.
d.
Lower than Vx.
Lowest fuel consumption speed.
Higher than Vy.
Lower than V2.
POW41.
VIMPfor a turboprop is?
a.
b.
c.
d.
Higher than Vx.
Vx.
Higher than Vy.
Lower than V2.
POW42.
If service ceiling is 12000 ft at a weight of 50000 Kg, at a weight of 75000 Kg it
will be?
a.
b.
c.
d.
Higher.
The same.
Lower.
Higher o r lower depending on engine type.
POW43.
Increasing altitude causes the power required curve to?
a.
b.
c.
d.
Move to the left and upwards.
Move to the left and downwards.
Move to the right and upwards.
Move to the right and downwards.
POW44.
Decelerating at the back of the drag curve?
a.
b.
c.
d.
Increases drag and power required.
Decreases d r a g and power required.
Increases drag and decreases power required.
Decreases drag and increases power required.
POW45
A propeller aircraft is inherently
...... with increasing airspeed?
a.
b.
c.
d.
More
More
Less
More
.......speed stable than a jet because its thrust
increases.
decreases.
increases.
decreases.
POW46.
This diagram represents?
a.
b.
c.
d.
Drag.
CL:CD ratio.
Power required.
Power available.
VMD
EAS
POW47.
This diagram represents?
a.
b.
c.
d.
Drag.
CL:CD ratio.
Power required.
Power available.
LID Max
POW48.
VMpis most probably at point
EAS
..on the
A
B
D
EA
C
S
POW49.
Which point on the diagram best represents Vx for
a propeller aircraft?
a.
b.
C.
d.
A.
B.
C.
D.
Drag
EX
A B C D
E
A B C D E
EAS
POWSO.
Which point on the diagram best represents Vu for
a propeller aircraft?
a.
b.
C.
d.
A.
B.
C.
D.
POW51.
As speed decreases below VMp?
a.
b.
c.
d.
Drag and power required decrease.
Drag and power required increase.
Drag decreases and power required increases.
Drag increases and power required decreases.
POWSZL
As speed increases above VMP?
a.
b.
c.
d.
Drag and power required decrease.
Power required increases, whilst drag decreases then increases.
Drag decreases and power required increases.
Drag increases and power required decreases.
POW53.
As speed changes from CLM,X to VMO?
a.
b.
c.
d.
Both drag and power required decrease then increases at the same rate.
Both drag and power required decrease then increase but at different
rates.
Drag increases and power required decreases.
Drag decreases and power required increases.
POW54.
If air density at 40000 feet is A
! of that at sea level, how will power required
change when climbing at constant IAS from MSL to 40000 feet?
a.
b.
c.
d.
Remain constant..
Increase by a factor of 2.
Increase by a factor of 4.
Increase by a factor of 8.
POW55.
If air density at 40000 feet is % of that at sea level, how will power required
change when climbing at constant TAS from MSL to 40000 feet?
a.
b.
c.
d.
Remain constant.
Decrease provided IAS remains above VwD.
Decrease provided IAS remains below VMO.
Decrease provided IAS remains above VMp
POW56.
If power lever setting is maintained at a constant value when climbing at Vx?
a.
Gradient and ROC will increase.
b.
c.
d.
Gradient and ROC will decrease.
Gradient will increase and ROC will decrease.
Gradient will decrease and ROC will increase.
POW57.
If power lever setting is maintained at a constant value when climbing a t Vy?
a.
b.
c.
d.
Gradient and ROC will increase.
Gradient and ROC will decrease.
Gradient will increase and ROC will decrease.
Gradient will decrease and ROC will increase.
POW58.
This diagram represents?
a.
b.
c.
d.
Variation of power available with
increasing weight.
Variation of power available with
increasing altitude.
Variation of power required with
Increasing weight.
Variation of power required with
increasing altitude.
EAS
POW59.
This diagram represents?
a.
b.
c.
d.
Variation of power available with
increasing weight.
Variation of power available with
increasing altitude.
Variation of power required with
decreasing weight.
Variation of power required with
increasing altitude.
EAS
POW60.
This diagram represents?
a.
b.
c.
d.
Variation of power available with
increasing weight.
Variation of power available with
decreasing altitude.
Variation of power required with
increasing weight.
Variation of power required with
decreasing altitude.
I
EAS
POW61.
Jet power available a t high altitude is
indicated by line on this diagram?
...
a.
b.
C.
d.
A.
B.
C.
D.
Power
EAS
POW62.
Propeller power available a t high altitude is
indicated by line ... on this diagram?
a.
b.
C.
d.
A.
B.
C.
D.
Power
EAS
POW63.
Jet power available a t low altitude is
indicated by line ...on this diagram?
POW64.
Propeller power availabie a t low altitude is
Aindicated by line on this diagram?
...
a.
b.
C.
a.
A.
B.
C.
D.
Power
EAS
POW65.
Propeller thrust available at high altitude is
indicated by the line on this diagram?
..
a.
b.
c.
d.
A.
B.
C.
D.
Thrust
EAS
POW66.
Propeller thrust available at low altitude is
indicated by the line on this diagram?
..
a.
b.
c..
d.
A.
B.
C.
D.
Thrust
EAS
POW67.
Jet thrust available at high altitude is
indicated by the line on this diagram?
..
Thrust
EAS
POW68.
Jet thrust available at low altitude is
indicated by the line on this diagram?
I
..
a.
b.
c.
d.
A.
B.
C.
Thrust
D.
EAS
POW69.
If weight is increased by 50% at VRIDpower required a t the same speed will
increase by?
POW70.
If weight is increased by 50% at a speed considerably higher than VMDpower
required will be increased by?
a.
b.
c.
d.
42.5%.
More than 62.5%.
Less than 62.5%.
62.5%.
POW71.
If weight is increased by 50% at a speed considerably lower than VMDpower
required will?
a.
b.
c.
d.
Increase by more than 62.5%.
Increase by less than 62.5%.
Increase by more than 62.5% only if speed is lower than VMP.
Decrease by more than 62.5% only if speed is lower than VMP.
POW72.
A multi engine aircraft is flying at its absolute ceiling when it suffers a single
engine failure. In order to continue flying at that altitude it must?
a.
b.
c.
d.
Reduce speed.
Increase speed.
Reduce weight.
lncrease power setting.
POW73.
A multi engine aircraft is flying at its absolute ceiling when it suffers a single
engine failure. In order to continue flying at the same speed it must?
a.
b.
c.
d.
Climb.
Descend.
Accelerate.
Decelerate.
POW74.
An,aircraft weighing SOW0 Ibf is flying at its absolute ceiling when the pilot
activates the reheat system. If reheat produces an additional 5000 Ibf of thrust it
will enable theqaircraft to?
a.
b.
c.
Climb at a gradient of 10%.
Climb a t an angle of approximately 5.7 degrees.
Climb at a rate of 1000 feet per minute.
d.
Climb at a gradient of 10% and at an angle of approximately 5.7 degrees.
YOW75.
As altitude increases the power required to fly at any given IAS?
a.
b.
c.
d.
Increases because the drag force increases with altitude.
Decreases because the drag force decreases wit altitude.
Increases because the drag force remains constant while TAS increases
with altitude.
Decreases because the drag force remains constant while TAS decreases
with altitude.
STAB 1.
What is short period mode?
a.
b.
c.
d.
Pilot induced oscillation.
Oscillation about the lateral axis.
Oscillation about the longitudinal axis.
Oscillation about the normal axis.
STAB 2.
What is phugoid motion?
a.
b.
c.
d.
Short period longitudinal oscillations.
Long period longitudinal oscillations.
Long period lateral oscillations.
Long period yawing oscillations.
STAB 3.
Where is the aft C of G limit of an aircraft?
a.
b.
c.
d.
Behind the neutral point.
Ahead of the neutral point.
On the neutral point.
On the manoeuvre point.
STAB 4.
At what position on an aircraft must the C of G be to make it neutrally
longitudinally stable during pull up manoeuvres?
a.
b.
c.
d.
The neutral point.
The transition point.
The aerodynamic centre.
The manoeuvre point.
STAB 5.
Which of the following contribute to lateral stability?
a.
b.
c.
d.
Anhedral, sweepback, high wing, high fin.
~ i h e d r a lsweepback,
,
low wing, high fin.
Anhedral, aspect ratio, high wing, ventral fin.
Dihedral, sweepback, high wing, high fin.
STAB 6.
What purpose is served by a dorsal fin a t high sideslip angles?
a.
b.
c.
d.
Decreases effectiveness of the fin in order to reduce lateral instability.
Decreases effectiveness of the fin in order to reduce Dutch roll.
Increases effectiveness of the fin by increasing its area.
lncreases effectiveness of the fin by reducing its aspect ratio.
STAB 7.
What action should be taken in order to prevent Dutch roll if the rudder damper
system fails when flying a t high s p e d a t high altitude?
a.
b.
c.
d.
Reduce speed and altitude immediately.
Reduce altitude.
Reduce altitude then reduce speed.
Increase speed.
STAB 8.
What would be the effect of loading an aircraft such that its C of G was 6ll the
forward limit?
a.
b.
c.
d.
lncreased longitudinal, lateral and directional stability.
Decreased longitudinal, lateral and directional Uability.
lncreased longitudinal and directional stabiliry but no effect on lateral
stability.
Increased longitudinal and lateral stability but no effect on directional
stability.
STAB 9.
What is the relationship between stability and manoeuvrability?
a.
b.
c.
d.
In order to be manoeuvrable an aircraft must first be stable.
Manoeuvrability and stability are not connected in any way.
Depends on the design of the aircraft autopilot system.
Stability and manoeuvrability are inversely related.
STAB 10.
What is the effect of increasing wing sweep on the stability of an aircraft?
a.
b.
c.
d.
lncreased longitudinal, lateral and directional stability.
lncreased longitudinal and directional stability.
Increased lateral and directional stability.
Increased longitudinal and lateral stability.
STAB 11.
What is the pitching moment generated by a cambered wing about its
aerodynamic centre in flight?
a.
b.
c.
d.
Always nose up and of constant magnitude.
Always nose down and of varying magnitude with angle of attack.
Nose up or nose down depending on angle of attack.
Always nose down at its zero lift value.
STAB 12.
What effect does an increase in angle of attack have on the distance between the
aerodynamic centre and centre of pressure of a wing in flight?
a.
b.
c.
d.
The distance remains constant.
The distance becomes greater.
z- The distance becomes smaller. The distance becomes smaller up to fhe stall then becomes greater.
%-
I-
STAB 13.
Which of the following combinations of stability are acceptable in a JAR
certificated commercial aircraft?
a.
b.
c.
d.
Statically unstable, dynamically stable.
Statically stable, dynamically unstable.
Dynamically and-statically unstable.
Dynamically and sl6atically stable.
-.::,:&<"*
a33~~k1;
STAB 14.
Which of the following conditions would maximise stability?
a.
b.
c.
d.
C of G on aft limit.
C of G on forward limit.
C of G on the manoeuvre point.
C of G on the neutral point.
STAB 15.
What is the position of the aft C of G limit in relation to the manoeuvre point of
the aircraft?
a.
b.
c.
Always behind the manoeuvre point.
Always ahead of the manoeuvre point.
Ahead or behind the manoeuvre point, d w n d i n g on the fuel load.
d.
Always directly above the manoeuvre point.
STAB 14L
Which of the following combinations contribute to lateral stability?
a.
b.
c.
d.
Anhedral, low wings, ventral fins, wing sweep back.
Dihedral, high wings, dorsal fins, wing sweep back.
Anhedral, low wings, dorsal fins, straight wings.
Dihedral, high wings, ventral fins, low C of G.
STAB 17.
What is phugoid motion?
a.
b.
c.
d.
Weakly damped long period oscillations in pitch attitude, altitude and
airspeed with approximately constant angle of attack.
Highly damped short period oscillations in pitch attitude, altitude and
airspeed with approximately constant angle of attack.
Pilot induced long period oscillations in pitch attitude, altitude and
airspeed with approximately constant angle of attack.
Not possible in swept wing aircraft.
STAB 18.
How will a change of C of G position affect the period of oscillations?
a.
b.
c.
d.
Forward movement of C of G will increases period.
Aft movement of C of G will increase period.
None of the above.
It depends on weight.
STAB 19.
What will be the effect of excessive directional stability?
a.
b.
c.
d.
Phugoid motion.
It will not be possible to yaw the aircraft.
Dutch roll.
Spiral instability.
STAB 20.
Which of the following are possible in an aircraft?
a.
b.
c.
d.
Statically unstable and dynamically stable.
Statically stable and dynamically stable.
Statically unstable and neutrally dynamically stable.
Neutrally statically table and positively dynamically stable.
STAB 21.
Which of the following conditions imposes the greatest stick force during
landing?
a.
b.
c.
d.
C of G aft and flaps down.
C of G forward and flaps up.
C of G aft and flaps up.
C of G forward and flaps down.
STAB 22.
What would be the effect of forward movement of the C of G ?
a.
b.
c.
d.
Lower stick force gradient.
Positive stick force gradient.
Negative stick force gradient.
Neutral stick force gradient.
STAB 23.
How would forward movement of C of G affect manoeuvre stability?
a.
b.
c.
d.
Increase.
Decrease.
Increases or decreases pending on weight.
None of the above.
STAB 24.
Which of the following would make stalling most likely in low speed flight?
a.
b.
c.
d.
Aft C of G.
Forward C of G.
No C ofG.
Central C of G.
STAB 25.
Which of the following are true of an aircraft that is positively statically stable?
a.
b.
c.
d.
Jt is never dynamically stable.
It is always dynamically stable.
Jt might be dynamically stable.
It cannot exhibit phugoid motion.
STAB 26.
What would be the effect of increasing camber?
a.
b.
c.
CL would decrease.
Angle of attack would increase.
Longitudinal stability would be unaffected.
d.
Lateral stability would be increased.
STAB 27.
How does anhedral affect stability?
a.
b.
c.
d.
Increases directional stability.
Decreases lateral stability.
Decreases longitudinal stability.
None of the above.
STAB 28.
An aircraft that is directionally statically unstable?
a.
b.
c.
d.
Will exhibits spiral instability.
Will be directionally dynamically stable.
Will yaw into sideslip.
Will yaw away from sideslip.
STAB 29.
What effect does roll damping have on stability?
a.
b.
c.
d.
Increases lateral dynamic stability.
Increases lateral static stability.
Increases frequency of oscillations.
Decreases frequency of oscillations.
STAB 30.
What effect do wing-mounted engine pods tend to have on stability?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on weight.
None of the above.
STAB 31.
Which of the following is essential for longitudinal stability?
a.
b.
c.
d.
A serviceable autopilot system.
A tailplane.
A canard.
Longitudinal dihedral.
STAB 32.
What effect does longitudinal dihedral have on a canard configuration?
a.
Canard stalls after wing.
b.
c.
d.
Canard stalls before wing.
Canard cannot stalt.
Canard and wing stall together.
STAB 33.
What effect does longitudinal dihedral have on a tailplane configuration?
a.
b.
c.
d.
Tailplane stalls first.
Wing stalls first.
Tailplane cannot stall.
Wing and tailplane stall together.
STAB 34.
What is short period motion?
a.
b.
c.
d.
Low frequency oscillations.
High frequency oscillations about the lateral axis.
High magnitude oscillations about the lateral axis.
Rapid oscillations about the longitudinal axis.
STAB 35.
The aft C of G limit is?
a.
b.
c.
d.
Aft of the neutral point.
Forward of the neutral point.
Above the neutral point.
The neutral point.
STAB 36.
Wing sweep back?
a.
b.
c.
d.
Increases lateral and directional stability.
Increases longitudinal and directional stability.
Increases longitudinal and lateral stability.
None of the above.
STAB 37.
Wing sweep back?
a.
b.
c.
d.
Increases speed stability.
Decreases speed stability.
Decreases lateral stability.
None of the above.
STAB 38.
If an aircraft is longitudinally statically unstable?
a.
b.
c.
d.
It will be speed unstable.
It will be speed stable.
Longitudinal and speed stability are not related.
I t will exhibit phugoid motion.
STAB 39.
What effect does increasing altitude have on stability?
a.
b.
c.
d.
Stability increases with altitude.
Stability decreases with altitude, all modes being affected to the same
degree.
Stability decreases with altitude all modes not being affected to the same
degree.
Altitude has no effect on stability.
STAB 40.
If the yaw damper fails when flying a t cruising speed at maximum altitude?
a.
b.
c.
d.
Descend to prevent Dutch roll.
Slow down to prevent Dutch roll.
Slow down then descend to prevent Dutch roll.
Descend then slow down to prevent Dutch roll.
STAB 41.
Spiral instability is likely when?
a.
b.
c.
d.
Directional stability is weaker than longitudinal stability.
Lateral stability is weaker than directional stability.
Lateral stability is stronger than directional stability.
The aircraft is directionally unstable.
STAB 42.
In phugoid motion?
a.
b.
c.
d.
Airspeed is constant.
Angle of attack is constant.
Attitude is constant.
Altitude is constant.
STAB 43.
What effect does CDhave on phugoid characteristics?
a.
b.
c.
d.
High CDgives high tendency to phugoid.
Low CDgives high tendency to phugoid.
CDhas no effect on phugoid.
None of the above.
STAB 44.
How can a pilot stop short period motion?
a.
b.
c.
d.
Use the controls.
Switch off the autopilot.
Release the controls.
Dive the aircraft.
STAB 45.
Constant amplitude oscillations about any axis are?
a.
b.
c.
d.
Dynamic instability.
Static instability.
Static stability plus neutral dynamic stability.
Neutral static stability plus positive dynamic stability.
STAB 46.
A ventral fin will?
a.
b.
c.
d.
Increase dynamic lateral stability.
Increase static lateral stability.
Decrease directional stability.
None of the above.
STAB 47.
A ventral fin will?
a.
b.
c.
d.
Increase directional stability in sideslip.
Decrease directional stability in sideslip.
Increase longitudinal stability in sideslip.
Be unaffected by sideslip.
STAB 48.
A ventral fin will?
a.
b.
c.
d.
Not affect Dutch Roll.
Reduce tendency to Dutch Roll.
Increase tendency to Dutch Roll.
Reduce tendency to spiral instability.
STAB 49.
Longitudinal stability requires?
a.
b.
c.
d.
A tailplane.
An autopilot.
An aft C of G.
Longitudinal dihedral.
STAB 50.
What effect does dihedral have on the stick force necessary to maintain sideslip?
a.
b.
c.
d.
Decreases it.
Increases it.
Increase o r decrease depending on directional stability.
None of the above.
STAR 51.
Dutch Roll tendency is increased by?
a.
b.
c.
d.
Forward movement of C of G.
Aft movement of C of G.
Forward movement of C of P.
Aft movement of C of P.
STAB 52.
C of G on forward limit?
a.
b.
c.
d.
Increases VMCC.
Decreases VMCC.
Increases stick force.
Decreases stick force.
STAB 53.
What is short period mode?
a.
b.
c.
d.
Oscillation about the lateral axis.
Pilot induced.
Oscillation about the longitudinaj axis.
Oscillation about the normal axis.
STAB 54.
Dutch Roll occurs when ........ stability is greater than
a.
b.
c.
d.
Lateral dynamic,
Lateral static,
Lateral static,
Lateral dynamic,
Lateral static.
Lateral dynamic.
Directional.
Longitudinal.
STAB 55.
Strong lateral dynamic stability will.
a.
b.
c.
d.
.......... stability?
.......Dutch Roll?
Increase.
Decrease.
Not affect.
Increase o r decrease depending on directional stability.
STAB 56.
increases manoeuvrability?
.........
a.
b.
c.
d.
An aft C of P.
An aft C of G.
A High wing.
A High tailplane.
STAB 57.
Sweepback produces ..........9
a.
b.
c.
d.
Strong longitudinal stability.
Strong lateral stability.
High roll rates.
High pitching rates.
STAB 58.
Sweepback produces?
a.
b.
c.
d.
Strong longitudinal stability.
Strong directional stability.
High yaw rates.
Low pitching rates.
STAB 59.
................increases manoeuvrability?
a.
b.
c.
d.
Dihedral.
Anhedral.
Low C of G.
Forward C of G.
STAB 60.
Dutch Roll is?
a.
b.
c.
d.
Stable cyclical rdling from one wing down to the other.
Unstable cyclical rolling from one wing down to the other.
Long period pitching with constant angle of attack.
Short period yawing with constant bank angle.
STAB 61.
Which of the following statements is true?
a.
b.
c.
d.
Positive yaw is nose to the left.
Positive yaw is nose to the right.
Positive sideslip is to the left.
Positive pitching is nose down.
STAB 62.
Which of the following statements is true?
a.
b.
c.
d.
The positive longitudinal axis is aft.
The positive lateral axis is left.
The positive lateral axis is right.
Positive roll is to the left.
STAB 63.
Which of the following is true?
a.
b.
c.
d.
Left yaw causes positive sideslip.
Left roll causes positive sideslip.
Positive sideslip causes positive bank.
Positive sideslip is to the left.
STAB 64.
Positive longitudinal stability requires a
a.
b.
c.
d.
Positive.
Negative.
Curved.
Flat.
STAB 65.
Positive lateral stability requires a
a.
b.
c.
d.
............Cl:$ slope?
positive.
Negative.
Curved.
Flat.
STAB 66.
Positive directional stability requires a
a.
b.
c.
d.
..............C,: P slope?
Positive.
Negative.
Curved.
Flat.
STAB 67.
Stick free stability is
a.
b.
c.
d.
..............CM:a slope?
............stick fixed stability?
Greater than.
Weaker than.
The same as.
Less important than.
STAB 68.
An in limits aft C of G will
a.
b.
c.
d.
..........stick force gradient?
Increase.
Decrease.
Not affect.
Increase or decrease depending on neutral point location.
STAB 69.
7
A dorsal fin ................ the effectiveness of the fin by ..............
a.
b.
c.
d.
Increases,
Increases,
Decreases,
Decreases,
decreasing its aspect ratio.
increases its aspect ratio.
decreasing aspect ratio.
increasing aspect ratio.
STAB 70.
A fin with a dorsal fin will have a ........... CL:pslope and a
compared to one without ?
a.
b.
c.
d.
Steeper,
Steeper,
Shallower,
Shallower,
lower.
higher.
lower.
higher.
STAB 71.
Increasing fin height will
lateral stability?
a.
b.
c.
d.
Increase,
Increase,
Decrease,
Decrease,
.........stalling angle
...........static lateral stability and ..........dynamic
increase.
decrease.
decrease.
increase.
STAB 72.
Increasing fin height will
directional stability?
............static directional stability and .........dynamic
a.
Increase,
increase.
Increase,
decrease.
c.
Decrease,
decrease.
d.
Decrease,
increase.
STAB 73.
Adding a ventral fin will
stability?
b.
....... ...static lateral stability and .. .......directional
a.
b.
Increase,
Increase,
increase.
decrease.
P
c.
d.
Decrease,
Decrease,
decrease.
increase.
STAB 74.
Adding a dorsal fin will .... . ... The tendency to spiral'instability?
..
a.
b.
c.
d.
Increase.
Decrease.
Increase o r decrease depending on size of dorsal fin.
Not affect.
STAB 75.
Adding a ventral fin will
a.
b.
c.
d.
...... tendency to Dutch Roll?
Increase.
Decrease.
Increase o r decrease depending on size of dorsal fin.
Not affect.
STAB 76.
Longitudinal static stability requires a?
a.
b.
c.
d.
Negative Cb,:P gradient.
Positive CM:P gradient
Negative Chq:agradient.
Positive CM:a gradient.
STAB 77.
Lateral static stability requires a?
a.
b.
c.
b.
Negative CI:P gradient.
Positive Cl:p gradient
Negative C1:agradient.
Positive Cl:a gradient.
STAB 78.
Directional static stability requires a?
a.
b.
c.
d.
Negative Cn:P gradient.
Positive Cn:P gradient
Negative C,,:a gradient.
Positive Cn:agradient.
STAB 79.
Directional stability requires that the aircraft generates a yawing moment that is
................
3
a.
b.
c.
d.
Directly proportional to yaw angle.
Negatively proportional to yaw angle.
Directly proportional to sideslip angle.
Negatively proportional to sideslip angle.
STAB 80.
Positive static longitudinal stability requires?
a.
b.
c.
d.
An aft C of G.
A low C of G.
A forward C of P.
Longitudinal dihedral.
STAB 81.
Positive static longitudinal stability requires?
a.
b.
c.
d.
A forward C of G.
A low C of G.
A forward C of P.
Longitudinal anhedral.
STAB 82.
A highly stable aircraft requires
a.
b.
s.
d.
.............. to manoeuvre?
A low tailplane.
A high tailplane.
Fly-by wire controls.
High stick forces.
STAB 83.
Strong dynamic stability requires?
a.
b.
c.
d.
Weak aerodynamic damping.
Strong aerodynamic damping.
No aerodynamic damping.
Mass balancing of its control surfaces.
STAB 84.
Winglets?
a.
b.
c.
d.
Improve lateral and directional stability.
Reduce lateral and directional stability.
Do not significantly affect lateral and directional stability.
Improve roll rates.
STAB 85.
Deep stall is most likely in an aircraft with?
a.
b.
c.
d.
A canard.
A delta wing.
Straight wings.
Low aspect ratio wings.
STAB 86.
Single engine failure is more likely to cause rolling in a ...........aircraft than in
9
a
...............
a.
b.
c.
d.
Twin counter rotating propeller,
Single jet,
Contra-rotating propeller,
Twin jet,
twinjet.
single propeller.
twinjet.
twin propeller.
STAB 87.
Short period oscillation occurs around the
a.
b.
c.
d.
...........axis.
Longitudinal.
Lateral.
Normal.
Pitch.
STAB 88.
The yaw damper uses a .......... gyro to sense
a.
b.
c.
d.
.
Vertical,
Horizontal,
Vertical,
Horizontal,
roll rate.
roll range.
Yaw displacenient.
yaw rate.
STAB 89.
The yaw damper prevents
a.
b.
c.
d.
Dutch roll,
Dutch roll,
Spiral instability,
Spiral instability,
STAB 90.
Directional stability is
a.
b.
c.
Increased,
Increased,
Decreased,
..........9
...........by enhancing ............stability?
lateral.
directional.
lateral.
directional.
.............by ................
9
high wings.
wing sweep back.
high wings.
d.
Decreased,
wing sweep back.
STAB 91.
Lateral stability is ........... by ...............9
a.
b.
c.
(1.
Increased,
Increased,
Decreased,
Decreased,
increasing TAS.
increasing altitude.
decreasing TAS.
increasing altitude.
STAB 92.
Aerodynamic damping increases with increasing?
a.
b.
c.
d.
~ltitude.
TAS.
Aspect ratio.
Sweep back.
STAB 93.
Stick forces in a fly-by-wire system increase in proportion to?
a.
b.
c.
d.
Weight and C of G position.
Altitude and IAS.
Control deflection and IAS.
Control deflection and TAS.
STAB 94.
Lateral static stability is increased by?
a.
b.
c.
d.
High fin,
Low fin,
Anhedral,
Dihedral,
low C of G,
high C of G,
low C of G,
sweep back,
low wings.
high wings.
high wings.
high wings.
STAB 95.
A C of G aft of the after limit will?
a.
b.
c.
d.
Reduces manoeuvrability about the lateral axis.
Reduce stability and trim drag in cruise flight.
Reduce stability and increase trim drag in cruise flight.
Increase stability and stalling speed.
STAB 96.
A forward C of G will?
a.
Increase manoeuvrability about the lateral axis.
b.
c.
d.
Improve stability and increase trim drag in cruise flight.
Reduce stability and increase trim drag in cruise flight.
Reduce stability and stalling speed.
STAB 97.
Roll damping ............with roll rate?
a.
b.
c.
d.
Increases.
Decreases.
Increases o r decreases depending on TAS.
Does not vary.
STAB 98.
A statically unstable aircraft ............ dynamically stable?
a.
b.
c.
d.
Is always.
Is never.
Is sometimes.
Does not need to be.
STAB 99.
The f?lselage of an aircraft
a.
b.
c.
d.
.............its stability?
Tends to increase.
Tends to decrease.
Increases o r decreases depending on TAS.
Has no effect on.
STAB 100.
Longitudinal dihedral?
a.
b.
c.
d.
Is essential for stability.
Is desirable for stability.
Has no effect on.
Is not possible in a canard configuration.
IISF 1.
When an aircraft is flying a t supersonic speed, in what area will the pressure
disturbances caused by its motion be felt?
a.
b.
c.
d.
Within the mach cone.
I11 front of the normal shock wave.
I n front of the oblique shock wave.
In front of the mach cone.
HSF 2.
What happens to the location of the centre of pressure as an aerofoil accelerates
to M CRIT?
a.
b.
c.
d.
I t moves backwards as pitch is reduced to maintain constant lift then
moves forward as the first shock wave forms on the upper surface of the
wing.
It moves towards the mid chord position due to the effects of the upper
surface shock wave.
I t moves slowly backwards as pitch angle is reduced to maintain a
constant lift force with increasing airspeed, then forwards as the first
shock waves form.
It remains stationary due to the balancing effects of changing pitch angle
and shock wave formation.
HSF 3.
What happens to the location of the centre of pressure as an aerofoil accelerates
from just below MCRITto supersonic speed?
a.
b.
c.
d.
It moves forward as the second shock wave forms on the lower surface of
the wing.
It moves forward as the initial shock waves form, then moves aft towards
the mid chord position as the shock waves on the upper and lower
surfaces move towards the trailing edge.
It moves rapidly backwards as pitch angle is reduced to maintain a
constant lift force with increasing airspeed.
I t remains stationary due to the balancing effects of shock waves a t the
leading and trailing edges.
HSF 4.
What happens to the pressure distribution above the upper surface of the wing
as an aircraft accelerates through the transonic speed range?
a.
b.
c.
d.
It retains its smooth shape, but the pressure drop intensifies and its peak
moves forward as airflow over the front of the wing accelerates to
supersonic speed.
It retains its smooth shape, the pressure drop intensifying and moving
forward before collapsing as the first shock wave forms.
It retains its smooth shape, the pressure drop intensifying and moving
forward before becoming irregular as the first shock wave forms.
I t retains its smooth shape, the pressure drop intensifying and moving
forward before moving rapidly aft as the first shock wave forms.
HSF 5.
What causes the centre of pressure of a wing to move aft, as an aircraft
accelerates through the transonic speed range?
a.
b.
c.
d.
The shock wave on the upper surface causes a n abrupt deceleration of
airflow to subsonic speed. This produces a n instantaneous decrease in
pressure over the aft section of the wing.
The shock wave on the upper surface causes an abrupt deceleration of
airflow to subsonic speed, resulting in a n instantaneous increase in
pressure over the aft section of the wing.
As speed increases the upper shock wave moves towards the trailing edge.
The curvature of the wing upstream of the shock wave forms a series of
expansion corners, causing the airflow to accelerate to higher supersonic
speeds. This reduces the pressure over the aft section of the wing ahead
of the shock wave.
The shock wave on the lower surface causes a n abrupt pressure drop
under the aft section of the wing, causing the effective lifting force to
move aft.
HSF 6.
Why do shock waves form above and below wings as they accelerate through the
transonic speed range?
a.
b.
c.
d.
Airflow a t supersonic or sonic speeds always produces shock waves
The shock waves form because acceleration due to the curvature of the
wing surfaces, causes airspeed to exceed the local speed of sound.
The adverse pressure gradients above and below the wings move forward
with it in the form of pressure waves. These pressure waves pile u p a t the
point where airspeed equals the local speed of sound. This piling up of
the pressure gradient forms a n instantaneous pressure increase o r shock
wave.
The adverse pressure gradients above and below the wing move forward
with it in the form of pressure waves. These pressure waves pile up a t the
point where airspeed changes from sonic to supersonic, forming a n
instantaneous pressure increase o r shock wave.
HSF 7.
At what speed does the shock wave produced by an aircraft pass over the
ground?
a.
b.
c.
d.
Because pressure waves cannot move faster than the local speed of sound,
the shock waves move over the ground a t Mach 1.
Because the shock wave loses energy as it moves away from the aircraft, it
moves over the ground a t less than the local speed of sound.
Because the speed of sound is lower a t high altitude, the shock wave
moves over the ground a t less than the true airspeed of the aircraft.
Because the mach cone angle is constant for a given mach number, the
shock wave passes over the ground a t the aircraft's ground speed.
HSF 8.
If airflow is to be decelerated from high supersonic to subsonic level, what form
of shock wave o r shock waves will produce the lowest energy loss?
a.
b.
c.
d.
A normal shock wave reducing flow to subsonic in a single stage.
' A series of oblique shock waves, each causing flow to change from
supersonic to a lower supersonic level, followed by a normal shock wave,
taking flow from sonic to subsonic.
A series of normal shock waves, each resulting in a small proportion of
the required deceleration.
A single oblique shock wave, taking flow from supersonic to subsonic.
HSF 9.
Which of the following best describes the area rule principle?
a.
b.
c.
d.
By carefully contouring the fuselage of the aircraft, accelerations around
it are maximised, resulting in minimum drag.
By carefully contouring the fuselage of the aircraft to minimise rates of
change of cross sectional area, accelerations around it a r e reduced,
minimising drag.
By carefully contouring the fuselage of the aircraft, accelerations around
it are prevented, resulting in minimum drag.
By carefully contouring the wings of the aircraft, accelerations around it
are minimised resulting in minimum drag.
HSF 10.
High-speed tuck under occurs in the
speed range, causing the aircraft
movement of the
as the shock waves
to pitch down. I t is caused by
above and below the wings move
Changes in airflow over the
also
contribute to tlre problem. I t is counteracted by the
sysfem
....
.................
.......
a.
b.
c.
d.
Transonic, fwd, centre of pressure,
Subsonic, fwd, . centre of gravity,
Transonic, aft, centre of pressure,
Supersonic, aft,
centre of gravity,
............
............
.................
fwd,
aft,
aft,
aft,
tailplane,
fin,
tailplane,
elevons,
mach trim.
auto-stab.
mach trim.
autopilot.
HSF 11.
Static pressure is ......... proportional to volume and directly proportional to .......
so volume increases and density
with the square of velocity. As velocity
increases static pressure and temperature are converted into
energy, which
increases with the square of velocity. A given increase in velocity therefore has a
speed. I t is therefore possible to
greater effect on density a t ..... speed than a t
make calculations based on the assumption that air is .............a t low speeds but
not a t high speeds.
............
........
.....
a.
b.
c.
d.
Directly, density,
increases,
Inversely, density,
decreases,
Directly, temperature, decreases,
Inversely, density,
decreases,
kinetic, high, low,
kinetic, high, low,
kinetic, high, low,
kinetic, high, low,
incompressible
incompressible
incompressible
compressible
--
HSF 12.
What happens to the static pressure, velocity, density and temperature of
supersonic airflow as it passes through an expansion corner?
a.
b.
C.
d.
Decrease,
Increase,
Increase,
Decrease,
decrease, increase,
increase, decrease,
decrease, increase,
increase, decrease,
increase.
decrease.
increase.
decrease.
HSF 13.
What happens to the static pressure, velocity, density and temperature of
supersonic airflow as it passes through a compression corner?
a.
b.
C.
d.
Decrease,
Increase,
Increase,
Decrease,
decrease, increase,
increase, decrease,
decrease, increase,
increase, decrease,
increase.
decrease.
increase.
decrease.
HSF 14.
Which of the following statements is true?
a.
b.
c.
d.
A shock wave will form when supersonic airflow passes through an
expansion corner.
A shock wave will form when supersonic airflow passes through a
compression corner.
A shock wave might form when supersonic airflow passes through an
expansion corner.
A shock wave might form when supersonic airflow passes through a
compression corner.
HSF 15.
Select the correct words to complete the following statement.
As an aircraft accelerates from C
wings
Lthe centre
~
~ of pressure
~
of its
............................................... At M C ~ lthe
T centre of pressure ..........
1.
2.
3.
4.
Moves forward as flow over the leading edge area is increased.
Moves aft as pitch is reduced to maintain constant lift.
Moves forward as the shock waves increases pressure over the aft area.
Moves aft as flow accelerates into the slnock wave.
HSF 16.
As an aircraft continues to accelerate from MCRiTto mach 1, the shock waves
move to the trailing edge causing the upper surface of the wings to
........................................... At supersonic speeds the highest velocity and hence
lowest static pressure over the wing occur ....................................................
1.
2.
3.
4.
At the trailing edge and the centre of pressure is at about the mid chord
point.
At the trailing edge and the centre of pressure is immediately in front of
the shock wave.
Act as a series of compression corners.
Act as a series of expansion corners.
HSF 17.
--
At what speed do the shock waves produced by a supersonic aircraft appear to
move ever the ground?
a.
b.
c.
d.
At the local speed of sound.
At the speed of sound local to the altitude at which the aircraft is flying.
At the speed at which the aircraft moves over the ground.
At the TAS of the aircraft.
HSF 18.
What does the mach trim system do?
a.
b.
c.
d.
Adjusts the longitudinal trim of the aircraft.
Adjusts the elevator trim tabs o r servo tabs.
Moves C of G aft by pumping fuel into aft trim tanks.
Adjusts stabilator trim.
HSF 19.
What is the purpose of vortex generators in transonic flight?
a.
b.
c.
d.
To energies boundary layer, in order to prevent it spreading outward on
swept wing aircraft.
T o prevent the formation of shock waves on the upper surface of the wing.
To reduce the effects of shock induced boundary layer separation.
To slow down the boundary layer, to delay the formation of shock waves.
HSF 20.
What will be the effect on TAS, if an aircraft climbs at constant mach number to
50000 feet altitude in the international standard atmosphere?
a.
b.
c.
d.
TAS will increase.
TAS will decrease.
TAS will decrease then remain constant.
TAS will increase then remain constant.
HSF 21.
In what way will a mach trim system affect the stick force gradient for a powered
flying control system, as an aircraft accelerates through the transonic speed
range?
a.
b.
c.
d.
It will reduce .tick force gradient as speed increases, in order to enable
the pilot to overcome the high control hinge moments experienced at high
speeds.
It will ensure that the required negative stick force : speed gradient is
maintained.
Mach trim systems are not required in fully powered flying control
systems because the controls are irreversible.
It will adjust the stick force gradient to prevent high-speed pitch up.
HSF 22.
How do the velocity, static pressure, temperature and density of a supersonic air
stream vary as it passes through a convergent duct?
a.
b.
C.
d.
Decrease, increase, increase, decrease.
Decrease, increase, increase, increase.
Increase, decrease, decrease, decrease.
Increase, decrease, increase, remains constant.
HSF 23.
How does an aircraft with sweptback wings behave in shock stall?
a.
b.
c.
d.
It pitches nose up.
It becomes more longitudinally stable.
It pitches nose down.
Shock stall does not affect sweptback wings.
HSF 24.
How quickly does the shock wave produced by an aircraft move over the
ground?
a.
b.
c.
d.
Aircraft IAS.
Aircraft CAS.
Aircraft TAS
Aircraft ground speed.
HSF 25.
Where on a n aerofoil does shock stall occur?
a.
b.
c.
d.
At the trailing edge.
At the leading edge.
Beneath the leading edge.
Behind the shock wave.
HSF 26.
What effect does a normal shock wave have on airflow passing through it?
a.
b.
c.
d.
Accelerates to supersonic.
Accelerates to sonic.
Decelerates to subsonic.
Density and temperature decrease.
HSF 27.
If velocity is to be decreased using a shock wave or shock waves, what is the most
efficient method?
a.
b.
c.
d.
A single normal shock wave.
A single oblique shock wave.
A series of oblique shock waves.
Any of the above.
HSF 28.
If a series of shock waves is used to decelerate supersonic flow to subsonic speed,
how would the shock waves vary?
a.
b.
c.
d.
Increasing half angle from the first to the last.
Decreasing half angle from the first to last.
Constant half angle from first to last.
They would all be normal.
HSF 29.
How can energy loss due to shock waves be minimised?
a.
b.
c.
d.
Fly below MCRIT.
Fly a t mach 1.
Fly a t mach 2.
Fly at high supersonic speeds.
HSF 30.
How does the shock wave angle vary with mach number?
a.
b.
c.
d.
Increases.
Decreases.
Remains constant.
Increases then decreases.
HSF 31.
Why does lift decrease when shock waves form on a wing a t transonic speed?
a.
b.
c.
d.
The bow wave reduces free stream velocity.
The shock wave at the trailing edge reduces velocity over the upper
surface.
The boundary layer separates behind the shock wave.
The boundary layer separates ahead of the shock wave.
HSF 32.
Why does drag increase at high transonic speeds?
a.
b.
c.
d.
High-speed friction.
Increased induced drag a t the shockwave.
Increased parasite drag at the shockwave.
Separation behind the shockwave.
HSF 33.
What is the speed of the airflow immediately downstream of a normal
shockwave?
a.
b.
c.
d.
Supersonic.
Subsonic.
Lower supersonic.
Transonic.
HSF 34.
\+'hat airspeed would produce the lowest wave drag?
a.
b.
c.
d.
Mach 1.
Subsonic.
Supersonic.
Transonic.
HSF 35.
What will be the speed of the local airflow immediately downstream of an
oblique shockwave?
a.
b.
c.
d.
Subsonic
Transonic.
Lower supersonic than in front of the wave.
Higher supersonic than in front of the wave.
HSF 36.
What does the mach trim system do?
a.
Move fuel aft.
b.
c.
d.
Adjust the elevator trim tabs.
Adjust the variable incidence tailplane.
Adjust longitudinal trim.
HSF 37.
How can MCRITbe increased?
a.
b.
c.
d.
Sweep wings back.
Use high aspect ratio.
Use high thickness to chord ratio.
Use anhedral.
HSF 38.
What is M C ~ I - r ?
a.
b.
c.
d.
The highest free stream velocity at which a shock wave will form on the
wings.
The minimum free stream velocity a t which airflow first become sonic
anywhere on an aircraft.
The minimum supersonic speed.
The speed at which wave drag is at a minimum.
HSF 39.
What effect does aircraft attitude have on MCRIT?
a.
b.
c.
d.
None.
Nose level attitudes increase it.
I t decreases with nose u p attitude.
I t decreases wit11 roll.
HSF 49.
Which of the following is the best definition of mach number?
HSF 41.
When does mach tuck under occur?
a.
b.
c.
d.
At any mach number depending on LSS.
At high fractional mach numbers.
At low fractional mach numbers.
Above mach 1 only.
HSF 42.
Which of the following affect the local speed of sound?
a.
b.
c.
d.
Dynamic pressure.
Temperature.
Density.
Velocity.
HSF 43.
Which of the following occurs in swept wing aircraft as speed increases above
MCKIT?
a.
b.
c.
d.
Low speed buffet.
Pitch up and buffet.
Pitch down and buffet.
High-speed shimmy.
HSF 44.
Where on an aircraft would a normal shockwave form?
a.
b.
c.
d.
Wherever an adverse pressure gradient moving upstream meets sonic
airflow.
Wherever an adverse pressure gradient moving upstream meets
supersonic airflow.
At the trailing edge.
At the aerodynamic centre.
HSF 45.
Which of the following might be inadvertently exceeded when a n aircraft
descends a t constant mach number?
tISF 46.
How does airflow react when passing through a normal shockwave?
a.
b.
c.
d.
Accelerates to supersonic speed.
Decelerates to lower supersonic speed.
Pressure, temperature and density increase.
Pressure temperature increases, density decreases.
HSF 47.
How can shock induced separation be reduced?
a.
b.
c.
d.
Vortex generators.
Sweepback.
Anhedral.
- High aspect ratio.
HSF 48.
What will be the shape of the pressure distribution over a wing at supersonic
speeds?
a.
b.
r.
d.
Square.
Rectangular.
Triangular.
Irregular.
HSF 49.
What will be the shape of the pressure distribution over a wing in transonic
flight?
a.
b.
c.
d.
Square.
Rectangular.
Triangular.
Irregular.
HSF 50.
Which of the following is caused by high-speed buffet?
a.
b.
c.
d.
Shock stall.
Shock separation.
Wave drag.
Vibration.
HSF 51.
Why does lift decrease as M C ~islexceeded?
~
a.
b.
c.
d.
Free stream separation.
Boundary layer separation.
Shock waves.
Stagnation.
HSF 52.
In what condition would outboard ailerons, inboard ailerons and roll spoilers be
when cruising at M0.82?
a.
b.
c.
d.
Locked in neutral, active, active.
Locked in neutral, locked in neutral, active.
Active, active, locked in neutral.
Active, locked in neutral, active.
HSF 53.
What effect does mach trim have on stick forces?
a.
b.
c.
d.
Decreases stick forces to maintain manoeuvrability.
Decrease stick forces to prevent tuck under.
Increase stick forces to maintain negative gradient.
Increase stick forces to maintain positive gradient.
HSF 54.
What happens to air density as supersonic flow passes through an expansion
wave?
a.
b.
c.
d.
Increases.
Decreases.
Unchanged.
Depends on speed.
HSF 55.
How do VMoand MMOcompare when changing altitude?
a.
b.
c.
d.
The TAS at VMo and MMo are the same TAS at all altitudes.
The TAS at VMo is greater than that at Mnlo at all altitudes.
The TAS at VMo is less than that at MMOa t all altitudes.
The TAS at VMois greater than the TAS a t M Ma t~high altitudes and less
a t low altitudes.
HSF 56.
What happens to the angle of the shock waves as speed increases?
a.
b.
c.
d.
Increases.
Decreases.
Constant.
Collapses.
HSF 57.
At what angle does the shock wave produced by a supersonic aircraft reflect
from the ground?
a.
b.
c.
d.
Greater than before hitting it.
Less than before hitting it.
The same as before hitting it.
I t does is not reflected from the ground.
HSF 58.
What is the minimum number of shock waves that an object flying at supersonic
speed will produce?
a.
b.
C.
d.
None.
,
1.
2.
3.
HSF 59.
What will be the effect of airflow striking the front fuselage of a n aircraft when
sideslipping to the right in supersonic flight?
a.
b.
c.
d.
Destroy it.
Yaw to the left.
Yaw to the right.
Pitch down.
HSF 60.
How would the CL:a curves for an aircraft in subsonic and supersonic flight
compare?
a.
I>.
c.
d.
Identical.
Subsonic is steeper.
Supersonic is steeper.
They cannot be compared.
KSF 61.
How does the CLiLlarof an aerofoil at subsonic speed compare with that of the
same aerofoil at supersonic speed.
a.
b.
c.
d.
Grcatcr.
Smaller.
Identical because CL depends only on shape and angle of attack.
They cannot be compared because of the great speed difference.
HSF 62.
What happens to CLMax
as an aircraft approaches MCR~T?
a.
b.
c.
d.
Increases.
Decreases.
Constant.
Decreases then increases.
HSF 63.
How does CL at MCRITcompare with that at mach I?
a.
b.
c.
d.
Greater.
Smaller.
Identical.
Much greater.
HSF 64.
What happens to the shock wave angle as supersonic speed is increased?
a.
b.
c.
d.
Increase.
Decrease.
Unchanged.
Increase then decrease.
HSF 65.
How can energy loss due to shock waves be prevented?
a.
b.
c.
d.
Fly below MCRIT.
Fly a t MCRIT.
Fly a t mach 1.
Fly a t supersonic speeds.
HSF 66.
If air is to be compressed by passing it through a shock wave, what type of wave
will result in lowest energy loss?
a.
b.
c.
d.
Normal.
Swept forward.
Slightly oblique.
Highly oblique.
HSF 67.
If shock waves a r e unavoidable, what flight speed will result in lowest energy
loss?
ab.
c.
d.
MCRIT.
Mach 1.
High supersonic.
Low supersonic.
HSF 68.
How will an increase in weight affect &IcRlT?
a.
b.
c.
d.
Increase.
Decrease.
Increase o r decrease depending on speed.
None of the above.
HSF 69.
What affect will forward movement of C of G have on R/ICRIT?
a.
b.
Increase.
Decrease.
c.
d.
Increase o r decrease depending on weight.
None of the above.
HSF 70.
What is minimum speed for high-speed buffet in a JAR certificated passenger
aircraft?
HSF 71.
What is the aerodynamic ceiling?
a.
b.
c.
d.
Maximum attainable altitude.
Altitude where it is just possible to pull 1.3g.
Altitude a t which the margin between low and high speed buffet is zero.
Altitude where rate of climb is zero.
HSF 72.
At what speed does the over-speed warning operate in a JAR certificated
passenger aircraft?
a.
b.
C.
d.
10 Kts above VMO.
MCRIT.
110% M C u ~ ~ .
MI.
HSF 73.
At what mach number does the over-speed warning operate in a JAR certificated
passenger aircraft?
a.
b.
C.
d.
10 Kts above MhlO.
MCRIT110% MCRIT.
MI.
HSF 74.
Where on a wing does laminar flow break down at, or slightly above MCRJT?
a.
b.
c.
d.
On the fin.
Behind the normal shock wave above the wing.
Behind the normal shock wave below the wing.
Ahead of the oblique shock wave above the wing.
HSF 75.
Over what part of a wing does velocity increase in supersonic flight?
a.
b.
c.
d.
None of it.
All of it except the leading edge.
Most of it.
Forward area only.
HSF 76.
Where above the surface of a wing is the velocity greatest in supersonic flight?
a.
b.
c.
d.
Just behind the shock wave at the trailing edge.
Just in'front of the shock wave a t the trailing edge.
Just behind the shock wave at the 50% chord point.
Just ahead of the shock wave at the 50% chord point.
HSF 77.
Where above the wing is static pressure lowest in supersonic flight?
a.
b.
c.
d.
Just behind the shock wave a t the trailing edge.
Just in front of the shock wave a t the trailing edge.
Just behind the shock wave at the 50% chord point.
Just ahead of the shock wave at the 50% chord point.
HSF 78.
Where above the surface of a wing is the dynamic pressure greatest in supersonic
flight?
a.
b.
c.
d.
Just behind the shock wave at the trailing edge.
Just in front of the shock wave at the trailing edge.
Just behind the shock wave a t the 50% chord point.
Just ahead of the shock wave at the 50% chord point.
HSF 79.
What form of duct accelerates supersonic airflow?
a.
b.
c.
d.
Convergent.
Divergent.
Parallel.
None.
HSF 80.
What form of duct accelerates sonic airflow?
a.
b.
c.
Convergent.
Divergent.
Parallel.
d.
None.
HSF 81.
What form of duct increases static pressure of supersonic airflow?
a.
b.
c.
d.
Convergent.
Divergent.
Parallel.
None.
HSF 82.
What form of duct increases density of supersonic airflow?
a.
b.
c.
d.
Convergent.
Divergent.
Parallel.
None.
HSF 83.
What happens to CL as an aircraft accelerates from just below MCR~T
to
supersonic speeds?
a.
b.
c.
d.
Increases, decreases, increases, then,decreases again.
Decreases, increases, decreases, then increases again.
Remains constant.
Decreases then increases.
HSF 84.
What happens to CDas an aircraft accelerates from just below M C ~toT
supersonic speeds?
a.
b.
c.
d.
Increases, decreases, then decreases again.
Decreases, increases, then increases again.
Remains constant.
Increases then decreases.
HSF 85.
How does supersonic CL compare with subsonic CL?
a.
b.
c.
d.
Higher.
Lower.
The same.
Higher or lower depending on absolute temperature.
HSF 86.
How does supersonic CDcompare with subsonic CD?
a.
b.
c.
d.
Higher.
Lower.
The same.
Higher or lower depending on absolute temperature.
HSF 87.
What does supersonic buffeting cause?
a.
b.
c.
d.
Pitch down.
Pitch up.
Acceleration.
Vibration.
HSF 88.
Why is lift dramatically reduced just above MCRIT?
a.
b.
c.
d.
Shock waves.
High speed Buffet.
Attached bow wave.
Boundary layer separation.
HSF 89.
What effect does increasing wing sweep back have on MCRIT?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on thickness to chord ratio.
No effect.
HSF 90.
What effect does increasing thickness to chord ratio have on M C w ~ ?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on thickness to chord ratio.
No effect.
HSF 91.
What effect does increasing wing sweep back have on supersonic drag?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on thickness to chord ratio.
No effect.
NSF 92.
What effect does increasing wing sweep back have on the speed at which
maximum drag occurs?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on thickness to chord ratio.
'
No effect.
HSF 93.
What effect doe. increasing wing camber have on MCRIT?
a.
b.
c.
d.
Increase.
Decrease.
Increase o r decrease depending on thickness to chord ratio.
No effect.
HSF 94.
What effect does a supercritical wing have on MCRIT?
a.
b.
c.
d.
Increase.
Decrease.
Increase o r decrease depending on thickness to chord ratio.
No effect.
HSF 95.
What effect does a supercritical wing have on shock wave intensity at illCRLT?
a.
b.
c.
d.
Increase.
Decrease.
Increase or decrease depending on thickness to chord ratio.
No effect.
HSF 96.
At what altitude does the limiting factor change from VMOto MMO?
a.
b.
c.
d.
Critical altitude.
Absolute ceiling.
StabiIising altitude.
Crossover altitude.
HSF 97.
As air passes through an-expansion corner?
a.
b.
c.
A normal shock wave will form.
An oblique shock wave will form.
A mach cone will form.
d.
No shock wave will form.
HSF 98.
As air passes through a compression corner?
a.
b.
c.
d.
A normal shock wave will form.
An oblique shock wave will form.
A mach cone will form.
No shock wave will form.
HSF 99.
The high-speed buffet boundary
international standard atmosphere?
....................with increasing altitude in the
a.
b.
c.
d.
Increases up to 36000 feet then remain constant.
Decreases up to 36000 feet then remains constant.
Increases up to 36000 feet then decreases.
Decreases up to 36000 feet then increases.
HSF 100.
At what altitude does MMoequal Vsls?
a.
b.
c.
d.
Aerodynamic ceiling.
Absolute ceiling.
Cruise ceiling.
Service ceiling.
PROPS 1.
What is the most common reason for increasing the number of blades on a
propeller?
a.
b.
c.
d.
Increase aerodynamic efficiency.
Reduce propeller noise.
Increase power absorption capability.
Increase the effectiveness of the constant speed unit.
PROPS 2.
Which of the following will increase the gyroscopic precession effect of a
propeller?
a.
b.
c.
d.
lncreased angle of attack.
Decreased TAS.
Pitch and roll.
lncreased RPM.
PKOPS 3.
A constant speed propeller aircraft is descending with the throttle closed and
KPM lever set a t 2000 RPM. What would be the effect of retarding the propeller
lever?
a.
b.
c.
d.
Rate of descent would remain constant and RPM would increase.
Kate of descent would remain constant and RPM would decrease.
Kate of descent and RPM would increase.
Rate of descent and RPM would decrease.
PROPS 4.
What effect does the torque reaction of a single right handed tractor propeller
have during the take-off roll?
a.
b.
c.
d.
Weight on the right wheel increases while that on the left decreases.
Weight on the right wheel decreases while that on the left increases.
Weight on the right and left wheels decreases.
Weight on the right and left wheels increases.
PROPS 5.
What would be the result of propeller slipstream effect when taking-off using a
single right handed tractor propeller?
a.
b.
c.
d.
Pitch up.
Pitch down
Yaw to the right.
Yaw to the left.
PROPS 6.
Which of the following statements about propellersis true?
a.
b.
c.
d.
Angle of attack is the angle between the chord line of the blade and the
plane of rotation.
Critical tip speed is the RPM at which there is a risk of some part of the
blade stalling.
Geometric pitch is the distance that a propeller would move forward in
one turn if there were no slip.
Blade angle is the angle between the chord line of the blade and the shaft
axis.
PROPS 7.
Which of the following conditions will occur when a propeller is feathered?
a.
b.
c.
d.
The highest windmilling speed.
The maximum D:L ratio.
Minimum aerodynamic drag on the propeller.
Lowest blade angle.
PROPS 8.
Which of the following statements about propellers is true?
a.
b.
c.
d.
Angle of attack is the angle between the chord line of the blade and the
relative airflow.
Critical tip speed is the RPM at which there is a risk of some part of the
blade stalling.
Effective pitch is the distance that a propeller would move forward in one
turn if there were no slip.
Blade angle is the angle between the chord line of the blade and the shaft
axis.
PROPS 9.
What is the critical tip speed of a propeller?
a.
b.
c.
d.
The RPM at which the angle of attack at the blade tip is equal to the
stalling angle of the blades.
Critical tip speed is the lowest combination of RPM and inflow speed at
which the airflow over the tips of the blades first reaches the local speed
of sound.
The speed at which the blade tips experience low speed stall.
The speed a t which the blade tip angle is the equal to the staUing angle cf
the blades.
PROPS 10.
What is propeller blade angle?
a.
b.
c.
d.
The angle between the chord line of the blades and the axis of the
propeller shaft.
The angle between the chord line of the blades and the plane of rotation of
the propeller.
The angle between the chord line of the blades and the total reaction.
The angle between the chord line of the blades and the free stream
airflow.
PROPS 11.
Why is it necessary to vary blade angle from root to tip?
a.
b.
c.
d.
To prevent shock waves from forming at the tips.
To prevent shock waves from forming at the roots.
To ensure that angle of attack is acceptable over the whole blade.
To ensure that most thrust is produced at the tips.
PROPS 12.
Which of the following is true of a constant speed propeller?
a.
b.
RPM decreases with decreasing TAS.
RPM increases with increasing TAS.
c.
d.
Blade angle increases with increasing TAS.
Blade angle decreases with increasing TAS.
PROPS 13.
Which of the following will increase the angle of attack of a fixed pitch
propeller?
a.
b.
c.
d.
Decrease TAS and RPM.
Increase TAS and RPM.
Increase RPM and decrease TAS.
Decrease RPM and increase TAS.
PROPS 14
Which of the following will increase the angle of attack of a fixed pitch
propeller?
a.
b.
c.
d.
Pull up into a climb without increasing power.
Push down into a dive without decreasing power.
Decrease RPM while descending.
Decrease RPM while climbing.
PROPS 15.
How do propellers convert engine horsepower into thrust horsepower?
a.
b.
c.
d.
By using aerodynamic processes to generate forward facing lift which is
thrust.
By using aerodynamic means to generate lift, the forward facing
component of which is thrust.
By producing a low pressure area in front of the propeller disk which
causes the aircraft to be sucked forward.
By using aerodynamic means to generate a forward facing total reaction
which is thrust.
PROPS 16.
Which of the following produces best propeller efficiency?
a.
b.
c.
d.
Ensuring that the propeller wash has maximum velocity.
Ensuring that the propeller was11 velocity is less than TAS.
Ensuring that the propeller affects the minimum practicable mass flow of
air.
Ensuring that the velocity of the propeller wash is just higher than TAS.
PROPS 17.
An aircraft with a fixed pitch propeller is in straight and level flight. What will
happen if air density suddenly increases, assuming that engine power output is
not immediately affected?
a.
b.
c.
d.
The propeller RPM will decrease.
The propeller RPM will increase.
The propeller blade angle will increase.
The propeller blade angle will decrease.
PROPS 18.
How does drag on a windmilling propeller compare with that on a stationary
one?
a.
b.
c.
d.
Lower.
Higher.
Lower or higher depending on TAS.
Lower or higher depending on blade angle.
PROPS 19.
If propeller pitch is decreased in a glide, rate of descent and L:D ratio will?
a.
b.
c.
d.
Decrease,
Increase,
Decrease,
Increase,
Decrease.
Increase.
Increase.
Decrease.
PROPS 20.
Propeller efficiency is?
a.
b.
c.
d.
SHPITHP.
THPISHP.
SHPIBHP.
BHPITHP.
PROPS 21.
Propeller efficiency is?
a.
b.
c.
d.
THPIBHP.
THPIFHP.
SHPIBHP.
BHPITHP.
PROPS 22.
Increasing propeller RPM setting in a glide will
ratio.
a.
b.
C.
d.
Increase,
Decrease,
Increase,
Decrease,
Increase.
Decrease.
Decrease.
Increase.
.......range and ............L:D
PROPS 23.
During take-off using a right handed tractor propeller, torque reaction will
cause?
a.
b.
c.
d.
Right roll.
Left roll.
Right roll and yaw.
Left roll and yaw.
PROPS 24.
During the take-off run using a right handed tractor propeller, the torque
reaction will cause?
a.
b.
c.
d.
Increased load on right wheel and decreased on left wheel.
Decreased load on right wheel and increased on left.
Right roll and left yaw.
Left roll and right yaw.
PROPS 25.
During take-off using a right handed tractor propeller, the slipstream effect will
cause?
a.
b.
c.
d.
Right roll and left yaw
Left roll and right yaw.
Right roll and yaw.
Left roll and yaw.
PROPS 26.
During take-off using a tail wheel and right handed tractor propeller, the
asymmetric blade effect will cause?
a.
b.
c.
d.
Right roll.
Left roll.
Right yaw.
Left yaw.
PROPS 27.
During take-off using a nose wheel and right handed tractor propeller the
asymmetric thrust effect will cause?
a.
b.
c.
d.
Right roll.
Left roll.
Right yaw.
Left yaw.
PROPS 28.
During take-off using a tail wheel and right handed tractor propeller, gyroscopic
precession will cause?
a.
b.
c.
d.
Right roll.
Left roll.
Right yaw.
Left yaw.
PROPS 29.
During take-off using a nose wheel and right handed tractor propeller, the
gyroscopic precession effect will cause?
a.
b.
c.
d.
Right roll.
Left roll.
Right yaw.
Left yaw.
PROPS 30.
In an aircraft with twin right handed tractor propellers, the critical engine will
be?
a.
b.
c.
d.
The one which fails.
Right.
Left.
Right or left depending on stage of flight.
PROPS 3 1.
Propeller gyroscopic precession is altered by?
a.
b.
c.
d.
RPM change.
Right yaw and left roll.
Left yaw and right roll.
Pitch and roll.
PROPS 32.
Propeller gyroscopic precession force is induced by?
a.
b.
c.
d.
RPh1 change.
.Right yaw and left roll.
Left yaw and right roll.
Pitch and yaw.
PROPS 33.
Geometric pitch is?
a.
Blade angle minus angle of attack.
b.
c.
d.
Blade angle plus angle of attack.
The distance moved forward through the air in one revolution, if angle of
attack was zero.
The distance moved forward in one revolution.
--
-
PROPS 34.
Helix angle is?
a.
b.
c.
d.
Blade angle minus angle of attack.
Blade angle plus angle of attack.
The distance moved forward through the air in one revolution.
The distance moved forward in one revolution.
PROPS 35.
Blade angle is?
a.
b.
c.
d.
Helix angle minus angle of attack.
Helix angle plus angle of attack.
The distance moved forward through the air in one revolution.
The distance moved forward in one revolution.
PROPS 36.
A feathered propeller produces?
a.
b.
c.
d.
Minimum drag.
Maximum drag.
Maximum RPM.
Minimum glide speed.
PROPS 37.
For a fixed pitch propeller?
a.
b.
c.
d.
Blade angle increases with increasing TAS.
Angle of attack increases with increasing TAS.
Blade angle decreases with increasing RPM.
Angle of attack decreases with increasing TAS.
PROPS 38.
For a fixed pitch propeller?
a.
b.
a.
d.
Blade angle increases with increasing RPM.
Angle of attack increases with increasing RPM.
Blade angle decreases with increasing RPM.
Angle of attack decreases with increasing RPM.
PROPS 39.
For a constant speed propeller?
a.
b.
c.
d.
Blade angle increases with increasing RPM.
Angle of attack increases with increasing RPM.
Blade angle decreases wim increasing RPM.
Angle of attack decreases with increasing FWM.
PROPS 40.
For a constant speed propeller?
a.
b.
c.
d.
Blade angle increases with increasing TAS.
Angle of attack increases with increasing TAS.
Blade angle decreases with increasing TAS.
Angle of attack decreases with increasing TAS.
PROPS 41.
A coarse blade angle is ..........compared to a fine pitch angle?
a.
b.
c.
d.
More efficient at low TAS.
More efficient at high TAS.
More efficient at low RPM
Less efficient at high FWM.
PROPS 42.
Increasing power in a single right handed tractor propeller aircraft in the stall
will?
a.
b.
c.
d.
Make left wing drop worse.
Make right wing drop worse.
Not affect wing drop.
Increase stalling speed.
PROPS 43.
Increasing the number of propeller blades will?
a.
b.
c.
d.
Increase power absorption.
Improve efficiency.
Reduce propeller torque.
Reduce noise.
PROPS 44.
Increasing propeller blade length will?
a.
b.
c.
Decrease power absorption.
Increases efficiency.
Reduce torque reaction.
d.
Reduce tip speeds.
PROPS 45.
If the propeller RPM lever is pushed forward in a glide it will?
a.
b.
c.
d.
Increase range.
Increase drag.
Decrease RPM.
Improve cooling.
PROPS 46.
If propeller pitch is increased in a glide?
a.
b.
c.
d.
RPM and range will increase.
RPlM and range will decrease.
RPM will decrease and range will increase.
RPM will increases and range will decrease.
PROPS 47.
A constant speed propeller.. .. ...compared to a fixed pitch propeller?
..
a.
b.
c.
d.
Is lighter.
Is more efficient.
Is stronger.
Turns faster.
PROPS 48.
Maximum propulsive efficiency of a coarse pitch propeller is
a fine pitch propeller.
a.
b.
c.
d.
Higher.
Lower.
Identical.
Higher or lower depending on RPM.
PROPS 49.
Propeller propulsive efficiency?
a.
b.
c.
d.
Increases with increasing blade pitch.
Increases with increasing RPM.
Decreases with number of blades.
Decreases with increasing blade pitch.
PROPS 50.
Propeller blade twist?
.......compared to
a.
b.
c.
d.
Reduces blade angle from root to tip.
Increases blade angle from root to tip.
Is constant from root to tip.
Varies with RPM selection.
PROPS 51.
Centrifugal twisting moment?
a.
b.
c.
d.
Assists in unfeathering.
Always opposes aerodynamic turning moment.
Assists aerodynamic turning moment when in reverse thrust.
Assists aerodynamic turning moment when windmilling.
PROPS 52.
I n reverse thrust?
a.
b.
c.
d.
Angle of attack is greatest at the blade tips.
Angle of attack is greatest at the blade roots.
Angle of attack is greatest at the 70% radius point.
Angle of attack is zero.
PROPS 53.
Propeller torque?
a.
b.
c.
d.
Always opposes engine torque.
Never opposes engine torque.
Never assists engine torque.
Assists engine torque in windmilling flight.
PROPS 54.
Angle of attack is?
a.
b.
c.
d.
Lowest a t the tips.
Lowest a t the roots.
Lowest a t the 70% radius.
Uniform a t all points along the blade.
PROPS 55.
I n a constant IAS climb, a constant speed propeller will?
a.
b.
c.
d.
Decrease blade angle.
Increase blade angle.
Maintain constant blade angle.
Maintain constant angle of attack.
PROPS 56.
In a constant TAS climb, a constant speed propeller will?
a.
b.
c.
d.
Decrease blade angle.
Increase blade angle.
Maintain constant blade angle.
Maintain constant angle of attack.
PROPS 57.
If TAS increases at constant altitude, a constant speed propeller will?
a.
b.
c.
d.
Decrease blade angle.
Increase blade speed.
Maintain constant blade angle.
Maintain constant angle of attack.
PROPS 58.
If density decreases at constant TAS, a constant speed propeller will?
a.
b.
c.
d.
Decrease blade angle.
Increase blade angle.
Maintain constant blade angle.
Maintain constant angle of attack.
PROPS 59.
Left yaw of a propeller aircraft at VR might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Gyroscopic precession of a single left handed tractor propeller.
Asymmetric blade effect of a single left handed tractor propeller.
Asymmetric blade effect of a contra rotating propeller system.
PROPS 60.
Right yaw of a propeller aircraft at VR might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Gyroscopic precession of a single left handed tractor propeller.
Asymmetric blade effect of a single right handed tractor propeller.
Asymmetric blade effect of a contra rotating propeller system.
PROPS 61.
Right roll of a propeller aircraft at lift-off might be caused by?
a.
b.
c.
d.
Weathercock effect of a single right handed tractor propeller.
Weathercock effect of a single left handed tractor propeller.
Asymmetric blade effect of a single right handed tractor propeller.
Asymmetric blade effect of a contra rotating propeller system.
PROPS 62.
Left yaw and roll of a propeller aircraft in climb out might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Gyroscopic precession of a single left handed tractor propeller.
Asymmetric blade effect of a single right handed tractor propeller.
Asynlmetric blade effect of a contra rotating propeller system.
PROPS 63.
Right roll and yaw of a propeller aircraft in climb out might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Right engine failure.
Asymmetric blade effect of a single right handed tractor propeller.
Left engine failure.
PROPS 64.
Right yaw and roll of a propeller aircraft in climb out might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Asymmetric blade effect of a single left handed tractor propeller.
Asymmetric blade effect of a single right handed tractor propeller.
Left engine failure.
PROPS 65.
Right yaw of a propeller aircraft a t speeds approaching VR might be caused by?
a.
b.
c.
d.
Gyroscopic precession of a single right handed tractor propeller.
Asymmetric blade effect of a single right handed tractor propeller.
Asymmetric blade effect of a single left handed tractor propeller.
Asymmetric blade effect of a twin engine contra rotating propeller
system.
PROPS 66.
Sudden right yaw and roll of a propeller aircraft a t speeds above VKmight be
caused by?
a.
b.
c.
d.
Left engine failure.
Right engine failure.
Asymmetric blade effect of a single left handed tractor propeller.
Torque reaction in a twin engine contra rotating propeller system.
PROPS 67.
Centrifugal twisting moment?
a.
b.
Tends to drive the blades to coarse pitch.
Tends to drive the blades to fine pitch.
c.
d.
Tends to tear off the blades.
Tends to sweep back the blades.
ENV 1.
If \IA
at 60000 Kg is 250 Kts what will it be at 70000 Kg?
a.
b.
C.
d.
250 Kts.
260 Kts.
270 Kts.
280 Kts.
ENV 2.
If flaps up VA at 60000 Kg is 250 Kts what will it be at 70000 Kg with flaps
down?
a.
b.
C.
d.
230 Kts.
240 Ms.
250 Kts.
260 Kts.
ENV 3.
If weight is increased by 25% by what % will VAincrease?
ENV 4.
The positive limit load factor of a JAR certificated passenger aircraft is
.........flaps up and .......flaps down?
ENV 5.
At V,?
a.
b.
c.
d.
The aircraft will suffer permanent damage.
The aircraft cannot be damaged by control inputs.
The aircraft cannot be stalled before exceeding the limiting load factor.
Manoeuvring is not possible.
ENV 6.
If load factor is increased beyond.. .... at maximum operating altitude it will
cause.. .. ......3.
.
a.
b.
c.
d.
lg,
1.3g,
1.5g,
2.5g,
mach tuck under.
buffeting.
clear air turbulence.
overstressing of the structure.
ENV 7.
VAis?
a.
b.
c.
d.
Minimum speed a t which maximum nose up col~trolinput is permitted.
Maximum speed at which full nose up control input is permitted.
Minimum control speed.
Maximum control speed.
ENV 8.
Which of the following would increase VAfor a given aircraft structure?
a.
b.
c.
d.
Increased aspect ratio.
Increased camber.
Increased wing area.
Increased sweep back.
ENV 9.
Which of the following would improve ride quality in turbulence?
a.
b.
c.
d.
Increased aspect ratio.
Increased camber.
Increased wing area.
Increased sweep back.
ENV 10.
VAis?
a.
b.
c.
d.
The minimum speed at which it possible to attain limiting positive load
factor.
The maximum speed at which it is possible to attain limiting positive load
factor.
The recommended rough air flying speed.
The maximum operating speed.
ENV 11.
Deploying flaps in turbulence will?
a.
Reduce stalling hazard but increase overload hazard.
b.
c.
d.
Reduce stalling hazard and overloading hazard.
Increase stalling hazard but decrease overloading hazard.
Increase stalling hazard and overloading hazard.
ENV 12.
Aircraft manoeuvrability
a.
b.
C.
d.
Increases,
Increases,
Decreases,
Decrease,
........at high altitude because ............
9
Stability decreases.
TAS increases.
Buffet margins converge.
Buffet margins diverge.
ENV 13.
.............. aircraft are least affected by turbulence?
a.
b.
c.
d.
Dihedral winged.
Anhedral winged.
Swept winged.
Straight winged.
ENV 14.
The limiting load factor for a light utility aircraft in clean configuration is?
ENV 15.
A JAR certificated passenger aircraft must withstand ........p
without catastrophic failure?
a.
b.
c.
d.
2.5 time the limit load factor.
1.5 times the limit load factor.
The limit load factor.
3.8 times the ultimate load factor.
ENV 16.
At low all-up mass the positive limit load factor?
a.
b.
c.
d.
Protects the aircraft structure from overload.
Protects the passengers from excessive g loads.
Ensure structural integrity is maintained.
Preserves positive stability.
osilive loading
ENV 17.
VAis?
a.
b.
c.
d.
vsl,dn Limit.
Vsl 1 dn Limit.
~ ~ gLimit.
l h
~ c d Limit.
n
ENV 18.
A sudden up gust will?
a.
b.
c.
d.
Increase VA.
Decrease VA.
Increase a.
Decrease a.
ENV 19.
A sudden up gust will?
a.
b.
c.
d.
lncrease VB.
Decrease Vc.
Increase n.
Decrease n.
ENV 20.
Increasing CL..
.......load factor?
a.
b.
c.
d.
Decreases.
Does not affect.
Increases.
Increases or decreases depending on weight.
ENV 21.
Descending at constant mach number might cause?
a.
b.
c.
d.
MMo to be exceeded.
VMoto be exceeded.
Shock stall.
Tuck under.
ENV 22.
Climbing at constant IAS might cause?
a.
b.
c.
d.
MMo to be exceeded.
VMoto be exceeded.
Low speed stall.
TAS to reduce.
ENV 23.
The left boundary of the V-n and gust envelopes is?
a.
b.
c.
d.
High speed stall.
Low speed stall.
Mach tuck under.
VAand Ve respectively.
ENV 24.
The right boundary of the V-n and gust envelopes is?
a.
b.
Vc.
VD.
C.
Merit-
d.
VF and VRlcCrespectively.
s
ENV 25.
Exceeding maximum operating altitude is likely to cause?
a.
b.
c.
d.
Structural damage due to excessive load factor.
High speed buffet.
Shock stall.
Tuck under.
ENV 26.
An excessive gust-induced load factor at Vc could be remedied by?
a.
b.
c.
d.
Increasing aspect ratio.
Increased camber.
Increased sweep back.
Deployment of flaps.
ENV 27.
The margin between Vc and VD must be?
a.
b.
c.
d.
1.2 Vs.
1.3 Vs.
Sufficient to avoid unintentional over-speeds.
More than 50 Kts.
ENV 28.
Minimum limiting load factor with flaps down is?
ENV 29.
Maximum limiting load factor required for JAR passenger aircraft certification
is?
ENV 30.
Buffetingmust not occur between
a.
b.
c.
d.
..........And ............
9
VMOand M M ~ .
Vsrgand VMo.
Vslg and MMO.
1.2Vslg and VMO/lwMO.
ENV 31.
Increasing flap setting from 10 degrees to 30 degrees will
a.
b.
c.
d.
...... Vtmin?
Increase.
Decrease unless limited by V M ~ ~ .
Decrease or increase depending on altitude.
Decrease or increase depending on C of G position.
ENV 32.
At constant altitude the buffet boundaries will converge with?
a.
b.
c.
d.
1 Forward movement of C of G.
Aft movement of C of G.
Increasing temperature.
Decreasing load factor.
ENV 33.
At constant altitude the buffet boundaries will converge with?
a.
b.
c.
d.
Rearward movement of C of P.
Rearward movement of C of G.
Increasing temperature.
Increasing bank angle.
ENV 34.
Increasing IAS in a climb will ...... .. the altitude at which the mach limit is
reached?
.
a.
b.
Decrease.
Increase.
c.
d.
Not affect.
Increase or decrease depending on temperature.
ENV 35.
When descending at constant IAS a headwind will?
a.
b.
c.
d.
Increase distance covered over ground.
Decrease descent time.
lncrease descent time.
Decrease distance covered over ground.
ENV 36.
When descending at constant TAS a headwind will?
a.
b.
c.
d.
Decrease gradient.
Decrease descent time.
Increase descent time.
lncrease gradient.
ENV 37.
If weight decreases when cruising at constant altitude and CL?
a.
b.
c.
d.
Thrust and IAS required will decrease.
Thrust and IAS required will increase.
Thrust required will increases and IAS decrease.
Thrust required will remain unchanged but IAS will decrease.
ENV 38.
V h l Cwill
~ not be affected by
a.
b.
c.
d.
..........
3
Weight.
Altitude.
Power setting.
Humidity.
ENV 39.
When descending a t constant mach number below 35000 feet angle of attack
must?
a.
b.
c.
d.
Increase.
Decrease.
Remain constant.
Increase or decrease depending on temperature.
When descending a t constant IAS below 35000 feet angle of attach must?
a. - - Increase.
b.'
Decrease.
c.
Remain constant.
d.
Increase o r decrease depending on temperature.
ENV 41.
When descending a t constant TAS below 35000 feet, angle of attack must?
a.
b.
b.
c.
Increase.
Decrease.
Remain constant.
Increase o r decrease depending on temperature.
ENV 42.
As altitude increases a t constant weight and load factor, high speed buffet
9
margin ..........and low speed buffet margin
........
a.
b.
C.
d.
Increases,
Increases,
Decreases,
Decreases,
Decreases.
Increases.
Dacreases.
Increases.
ENV 43.
When climbing a t constant Mach number above the tropopause IAS will
?
TAS will
.....
a.
b.
c.
d.
Decrease,
Decrease,
Increase,
Increase,
.....and
Remain constant.
Decrease.
Remain constant.
Increase.
ENV 44.
At the maximum operating altitude?
a.
b.
c.
d.
Load factors above 1.3 will cause high speed buffet.
Load factor above 1 will cause high speed buffet.
Load factors above 2.5 will cause high speed buffet.
Any manoeuvres will cause high speed buffet.
ENV 45.
At the aerodynamic ceiling?
a.
b.
c.
d.
Load factors above 1.3 will cause high speed buffet.
Load factor above 1 will cause high speed buffet.
Load factors above 2.5 will cause high speed buffet.
Any increase in load factor above 1 o r speed change will cause buffet.
ENV 46.
At the absolute ceiling?
a.
b.
c.
d.
Load factors above 1.3 will cause loss of height.
Load factor above 1 will cause loss of height.
Load factors above 2.5 will cause loss of height.
Any manoeuvres will cause loss of height.
ENV 47.
At V,?
a.
b.
c.
d.
Stalling will occur a t limiting load factor.
Stalling will occur at ultimate load factor.
Stalling will occur at a load factor of 2 with flaps down.
Stalling is not possible.
ENV 48.
Vc is
......... than VB and, VB is ....... than VD?
a.
b.
c.
d.
Less,
Less,
More,
More,
Less.
More.
More.
Less.
ENV 49.
Vx is
......... than Vy and ...........than VMcc?
a.
b.
c.
d.
Less,
Less,
More,
More,
Less.
More.
More.
Less.
ENV 50.
VMCAis
.........than VR which is ..............than VMCC?
a.
b.
C.
d.
Less,
Less,
More,
More,
Less.
More.
More.
Less.
ENV 51.
The correct equation is?
ENV 52.
I n a constant mach number climb in the troposphere true airspeed?
a.
b.
c.
d.
Increases then decreases.
Decreases.
Remains constant.
Increases.
ENV 53.
In a constant mach number climb true airspeed?
a.
b.
c.
d.
Increases then remains constant.
Decreases then remains constant.
Decreases then decreases less slowly.
Decreases then decreases more slowly.
ENV 54.
V2 is?
a.
b.
c.
d.
The take-off safety speed.
The speed attained in the fourth segment.
The lowest speed at which directional control can be maintained following
an engine failure after take-off.
The minimum lift off speed with one engine out.
ENV 55.
At the intersections of the power available and power required curves?
a.
b.
c.
d.
Constant speed level flight is possible.
Accelerating level flight is possible.
Steady speed climbing flight is possible.
Accelerating climbing flight is possible.
ENV 56.
If fuel consumption is defined in Kglnm?
a.
b.
c.
d.
Forward movement of C of G decreases fuel consumption if C of G is
ahead of C of P.
C of G position has no effect on fuel consumption.
Forward movement of C of G increases fuel consumption if C of G is
ahead of C of P.
Aft movement of C of G reduces fuel consumption if C of G is aft of C of
P.
ENV 57.
The minimum and maximum values of V1 are?
ENV 58.
In level flight?
a.
b.
c.
d.
Maximum speed occurs when power required equals maximum power
available.
Minimum speed occurs when power required equals minimum power
available.
Maximum speed occurs when power required equals power available.
Maximum speed occurs when maximum excess power is available.
ENV 59.
In a constant TAS climb in the troposphere?
a.
b.
c.
d;
IAS increases.
IAS decreases.
IAS increases then remains constant.
IAS decreases then remains constant.
ENV 60.
In a constant IAS climb in the troposphere?
a.
b.
c.
d.
TAS increases.
TAS decreases.
TAS increases then remains constant.
TAS decreases then remains constant.
ENV 61.
In a constant TAS climb?
a.
b.
c.
d.
IAS increases.
IAS decreases.
IAS increases then increases more slowly.
IAS decreases then decreases more quickly.
ENV 62.
In a constant IAS climb in th.e troposphere?
a.
b.
2.-
d.
TAS increases and mach number decreases.
TAS increases and mach number increases.
TAS decreases and mach number increases.
TAS decreases and math number decreases.
ENV 63.
At low altitude?
a.
b.
c.
d.
The limiting variable is Vhl0.
The limiting variable is MMO.
The limiting variable is VDF..
The limiting variable is M C D ~ .
ENV 64.
At high altitude?
a.
b.
c.
d.
The limiting variable is VMo.
The limiting variable is MMo.
The limiting variable is VDF.
The limiting variable is MCDR.
ENV 65.
In a constant mach number descent there is a danger that?
a.
b.
c.
d.
MMowill be exceeded.
MCRIT
will be exceeded.
VMo will be exceeded.
Vs will be exceeded.
ENV 66.
VRmust be not less than ...... for a Class a aircraft?
a.
b.
C.
d.
Vs plus 10%.
VMc PIUS 5%.
VMCAPIUS 10%.
VMCLplus 15%.
ENV 67.
Increasing aircraft mass by 10%
a.
b.
c.
d.
Increases VMDby 5% but does not change parasite drag.
Increases VMDby 5% but does not change induced drag.
Decreases VMDand induced drag by 5%.
~ induced drag by 5%.
Increases V Mand
ENV 68.
Decreasing aircraft mass by 26%
a.
b.
c.
d.
......?
Decreases
Decreases
Increases
Increases
10
20
10
20
........VMDby ......
OO
/
but
............9
does not affect profile drag.
does not affect induced drag.
does not affect profile drag.
doe$ not affect induced drag.
ENV 69.
Flying at maximum range speed in a jet aircraft the angle of attack will be?
a.
b.
c.
d.
That giving CLMAx.
Higher than that giving best L:D ratio.
That giving optimum excess power available.
Lower than that giving best L:D ratio.
ENV 70.
Changing speed from VMPto V M ~ ?
a.
b.
c.
d.
Decreases drag and increases power required.
Increases drag and power required.
Decreases drag and power required.
Increases drag and decreases power required.
ENV 71.
The correct sequence of velocities is?
ENV 72.
The relationship between Vx and Vy is?
a.
b.
c.
d.
Vx is always less than Vy.
Vx is always greater than Vy.
Vx is always less than or equal to VY.
Vx can never equal Vy.
ENV 73.
The correct sequence is?
ENV 74.
Climb gradient in still air is closest to?
a.
b.
c.
d.
Height gained divide by distance moved through the air.
Distance moved through the air divided by height gained.
TAS divided by ROC.
ROC divided by CAS.
ENV 75.
As altitude increases?
a.
b.
c.
d.
Vx and Vy both increase.
Vx and Vy both decrease.
Vx decreases and Vy remains constant.
Vy decreases and Vx remains constant.
ENV 76.
If mach number and weight remain constant, increasing altitude requires?
a.
b.
c.
d.
Lower angle of attack.
Higher angle of attack.
Lower pitch attitude.
Lower CL.
ENV 77.
Vnlcc will be lowest with?
a.
b.
c.
d.
High ambient temperature humidity and pressure altitude.
Low ambient temperature, humidity and pressure altitude.
High ambient temperature and low humidity and pressure altitude.
Low ambient temperature and high humidity and pressure altitude.
ENV 78.
What is indicated in the buffet boundary chart?
a.
b.
c.
d.
Relationship between aircraft altitude, mass and low and high speed stall
mach numbers.
Relationship between aircraft altitude, mass and M C ~ ~ .
Relationship between aircraft altitude, mass and high and low speed stall
CAS.
Relationship between aircraft altitude, mass and low and high speed stall
IAS.
ENV 79.
When flying close to the maximum operating altitude?
a.
b.
c.
d.
Manoeuvrability is improved.
Manoeuvrability is degraded.
Manoeuvring is not possible.
Only one flight speed is possible.
ENV 80.
When flying close to the maximum operating altitude stability is?
a.
Improved.
b.
c.
d.
Degraded.
Not possible.
~egative.
ENV 81.
Flying at maximum range speed in a propeller driven aircraft the angle of attack
will be?
a.
b.
c.
d.
That giving best L:D ratio.
Higher than that giving best L:D ratio.
That giving optimum excess power available.
Lower than that giving best L:D ratio.
ENV 82.
At what speed must a jet aircraft and a propeller aircraft fly to maintain altitude
in straight and level flight at the absolute ceiling?
EIVV 83.
What happens as a propeller aircraft climbs to its absolute ceiling?
a.
b.
C.
d.
Vs, Vhlax, Vhlill, and Vy converge on Vx.
Vhlin, VM,,, VMPand VMDconverge on Vx.
Vklis, VMan and Vy Converge On Vx.
The CAS values equating to Vx and Vy both decrease.
ENV 84.
What happeiis to the CAS values of Vx and VMDas a propeller aircraft climbs to
its absolute ceiling?
a.
b.
c.
d.
Increase.
Decrease.
Remain constant.
Increase and decrease respectively.
ENV 85.
Which of the following occur when a propeller aircraft is a t its absolute ceiling?
a.
b.
Maximum thrust is equal to drag and maximum power available is equal
to power required.
Power available is equal to power required but maximum thrust available
is more than drag.
c.
d.
Maximum thrust available is equal to drag but power available is more
than power required.
Thrust, drag, power available and power required are all equal.
ENV 86.
Climb gradient is closest to?
a.
b.
c.
d.
Height gained divide by distance moved through the air.
Distance moved through the air divided by height gained.
TAS divided by ROC.
ROC divided by ground speed.
ENV 87.
Flying a t maximum endurance speed in a propeller driven aircraft the angle of
attack will be?
a.
b.
c.
d.
That giving best L:D ratio.
Higher than that giving best L:D ratio.
That giving optimum excess power available.
Lower than that giving best L:D ratio.
ENV 88.
Flying a t maximum endurance speed in a jet aircraft the angle of attack will be?
a.
b.
c.
d.
That giving best L:D ratio.
Higher than that giving best L:D ratio.
T h a t giving optimum excess power available.
Lower than that giving best L:D ratio.
ENV 89.
Point D on the diagram a t the right might be?
a.
b.
VMD.
VMP.
C.
Vxprup-
m
d
VYJ,~.
CD
ENV 90.
Point E on the diagram at the right might be?
a.
b.
C.
d.
CD
VMD.
VMP.
V~~rop
VyJet.
ENV 91.
Point C on the diagram at the right might be?
a.
b.
C.
VR~D.
Vhl~.
VXP~"~.
d.
v ~ ~ 1 . u ~ .
CD
ENV 92.
Point A on the diagram at the right might be?
C.
VS.
VA.
VMP.
d-
V ~ ~ e t .
a.
b.
ENV 93.
Point B on the diagram a t the right might be
if C is VYPrup?
CD
....
ENV 94.
Point B on the diagram at the right might be ... if
C is Vhlp?
a.
bC.
d.
V~II).
V~prop.
V~~rup.
VyJer.
ENV 95.
Point D on the diagram at the right might be?
a.
b.
C.
d.
Prop best endurance speed.
Prop best range speed.
VRIP.
Vy~et-
ENY 96.
Point D on the diagram at the right might be?
a.
b.
c.
d.
Prop best endurance speed.
Prop best climb speed.
VMP.
Best glide range speed.
CL
ENV 97.
Point D on the diagram at the right might be?
a.
b.
c.
d.
Prop best endurance speed.
Prop best climb speed.
VMP.
Jet best climb angle speed.
CL
ENV 98.
Point D,on the diagram at the right might be?
a.
b.
C.
d.
Prop best endurance speed.
Prop best climb speed.
Vhl~.
VMD.
CL
ENV 99.
Point D on the diagram a t the right might be?
a.
b.
c.
d.
Prop best endurance speed.
Jet best endurance speed.
VMP.
Vs.
CL
ENV 100.
Point B on the diagram at the right might be?
a.
b.
C.
d.
Prop best climb angle speed.
Prop best climb speed.
VMP.
VMD.
CL
ENV 101.
What might have caused the change from
curve 1 to 2 in the diagram a t the right?
a.
b.
c.
d.
Increasing altitude.
Increasing weight.
Decreasing altitude.
Decreasing weight.
TAS
ENV 102.
What might have caused the change from
curve 1 to 2 in the diagram at the right?
a.
b.
c.
d.
Increasing altitude.
Increasing weight.
Decreasing altitude.
Decreasing weight.
TAS
ENV 103.
What might have caused the change from
curve 1 to 2 in the diagram at the right?
a.
b.
c.
d.
Increasing altitude.
Increasing weight.
Decreasing altitude.
Decreasing weight.
TAS
ENV 104.
What might have caused the change from
curve 1 to 2 in the diagram at the right?
a.
b.
c.
d.
Increasing altitude.
Increasing weight.
Flap deployment.
Gear deployment.
EAS
ENV 105.
What might have caused the change from
curve 1 to 2 in the diagram at the right?
a.
b.
c.
d.
Increasing altitude.
Increasing weight.
Flap deployment.
Gear deployment.
EAS
SECTION 3
SUMMARY OF ANSWERS
Subject
Basics.
Aerofoil sections and wing planforms
Lift and drag.
High lift devices.
Stalling and stalling speed calculations.
Flying controls.
Climbing flight.
Turning flight.
Power available a ~ power
d
required.
Stability and manoeuvrability.
High speed flight.
Propellers.
Flight envelopes and aircraft performance.
Pages
217
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to
217
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218
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218
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219
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219
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220
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221
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221
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222
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222
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223
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224
BASIC 1. a.
BASIC 2. c.
BASIC 3. b.
BASIC 4. a.
BASIC 5. b.
BASIC 6. d.
BASIC 7. b.
BASIC 8. c.
BASIC 9. c.
BASIC 10. d.
BASIC 11. c.
BASIC 12. d.
BASIC 13. a.
BASIC 14. b.
BASIC 15. d.
BASIC 16. b.
BASIC 17. a.
BASIC 18. d.
BASIC 19. a.
BASIC 20. b.
BASIC 21. b.
BASIC 22. d.
BASIC 23. b.
BASIC 24. c.
BASIC 25. d.
BASIC 26. c.
BASIC 27. d.
BASIC 28. c.
BASIC 29. c.
BASIC 30. a.
FORMS 1. c.
FORMS 2. b.
FORMS 3. b.
FORMS 4. c.
FORMS 5. d.
FORMS 6. a.
FORMS 7. b.
FORMS 8. a.
FORMS 9. a.
FORMS 10. c.
FORMS 11. b.
FORMS 12. a.
FORMS 13. a.
FORMS 14. d.
FORMS 15. b.
FORMS 16. b.
FORMS 17. d.
FORMS 18. a.
FORMS 19. b.
FORMS 20. c.
FORMS 21. d.
FORMS 22. b.
FORMS 23. a.
FORMS 24. d.
FORMS 25. c.
FORMS 26. a.
FORMS 27. b.
FORMS 28. a.
FORMS 29. a.
FORMS 30. b.
FLAPS 1. c.
FLAPS 2. c.
FLAPS 3. c.
FLAPS 4. c.
FLAPS 5. b.
FLAPS 6. d.
FLAPS 7. a.
FLAPS 8. a.
FLAPS 9. d.
FLAPS 10. c.
FLAPS 11. a.
FLAPS 12. c,
FLAPS 13. d.
FLAPS 14. b.
FLAPS 15. c.
FLAPS 16. a.
FLAPS 17. c.
FLAPS 18. c.
FLAPS 19. a.
FLAPS 20. d.
FLAPS 21. d.
FLAPS 22. d.
FLAPS 23. a.
FLAPS 24. b.
FLAPS 25. c.
FLAPS 26. a.
FLAPS 27. d.
FLAPS 28. b.
FLAPS 29. a.
FLAPS 30. b.
FLAPS 31. a.
FLAPS 32. c.
FLAPS 33. b.
FLAPS 34. b.
FLAPS 35. b.
FLAPS 36. a.
FLAPS 37. b.
FLAPS 38. a.
FLAPS 39. b.
FLAPS 40. a.
FLAPS 41. b.
FLAPS 42. b.
FLAPS 43. c.
FLAPS 44. c.
FLAPS 45. a.
FLAPS 46. b.
FLAPS 47. d.
FLAPS 48. c.
FLAPS 49. a.
FLAPS 50. a.
FLAPS 51. c.
FLAPS 52. d.
FLAPS 53. a.
FLAPS 54. a.
FLAPS 55. a.
FLAPS 56. c.
FLAPS 57. d.
FLAPS 58. a.
FLAPS 59. d.
FLAPS 60. d.
FLAPS 61. d.
FLAPS 62. b.
FLAPS 63. b.
FLAPS 64. c.
FLAPS 65. a.
FLAPS 66. b.
FLAPS 67. c.
FLAPS 68. a.
FLAPS 69. c.
FLAPS 70. c.
FLAPS 71. a.
FLAPS 72. a.
FLAPS 73. c.
FLAPS 74. c.
FLAPS 75. b.
FLAPS 76. b.
FLAPS 77. a.
FLAPS 78. b.
FLAPS 79. d.
FLAPS 80. b.
FLAPS 81. d.
FLAPS 82. a.
FLAPS 83. a.
FLAPS 84. d.
FLAPS 85. a.
FLAPS 86. a.
FLAPS 87. b.
FLAPS 88. d.
FLAPS 89. b.
FLAPS 90. a.
FLAPS 91. d.
FLAPS 92. c.
FLAPS 93. b.
FLAPS 94. b.
FLAPS 95. d.
FLAPS 96. d.
FLAPS 97. b.
FLAPS 98. a.
FLAPS 99. d.
FLAPS 100. c.
STALL 1. b.
STALL 2. c.
STALL 3. b.
STALL 4. d.
STALL 5. d.
STALL 6. c.
STALL 7. c.
STALL 8. d.
STALL 9. a.
STALL 10. b.
STALL 11. a.
STALL 12. c.
STALL 13. b.
STALL 14. c.
STALL 15. a.
STALL 16. a.
STALL 17. c.
STALL 18. d.
STALL 19. c.
STALL 20. c.
STALL 21. b.
STALL 22. a.
STALL 23. c.
STALL 24. c.
STALL 25. d.
STALL 26. b.
STALL 27. a.
STALL 28. b.
STALL 29. c.
STALL 30. c.
STALL 3 1. c.
STALL 32. d.
STALL 33. b.
STALL 34. c.
STALL 35. a.
STALL 36. a.
STALL 37. a.
STALL 38. d.
STALL 39. c.
STALL 40. d.
STALL 41. b.
STALL 42. b.
STALL 43. a.
STALL 44. d.
STALL 45. c.
STALL 46. b.
STALL 47. d.
STALL 48. c.
STALL 49. a.
STALL 50. b.
STALL 51. b.
STALL 52. b.
STALL 53. a.
STALL 54. b.
STALL 55. c.
STALL 56. d.
STALL 57. a.
STALL 58. b.
STALL 59. b.
STALL 60. b.
STALL 61. c.
STALL 62. d.
STALL 63.b.
STALL 64. c.
STALL 65. c.
STALL 66. d.
STALL 67. b.
STALL 68. a.
STALL 69. d.
STALL 70. a.
STALL 71. a.
STALL 72. c.
STALL 73. a.
STALL 74. d.
STALL 75. b.
CON 1. a.
CON 2. a.
CON 3. a.
CON 4. b.
CON 5. a.
CON 6. b.
CON 7. b.
CON 8. a.
CON 9. b.
CON 10. b.
CON 11. c.
CON 12. d.
CON 13. a.
CON 14. c.
CON 15. c.
CON 16. b.
CON 17. b.
CON 18. c.
CON 19. c.
CON 20. b.
CON 21. a.
CON 22. c.
CON 23. a.
CON 24. c.
CON 25. a.
CON 26. b.
CON 27. b.
CON 28. b.
CON 29. c.
CON 30. c.
CON 31. c.
CON 32. a.
CON 33. b.
CON 34. c.
CON 35. c
CON 36. b.
CON 37. d.
CON 38. d.
CON 39. c.
CON 40. b.
CON 41. c.
CON 42. d.
CON 43. a.
CON 44. d.
CON 45. d.
CON 46. b.
CON 47. a.
CON 48. b.
CON 49. d.
CON 50. a.
CON 51. b.
CON 52. d.
CON 53. d.
CON 54. d.
CON 55. c.
CON 56. b.
CON 57. d.
CON 58. a.
CON 59. b.
CON 60. b.
CON 61. a.
CON 62. b.
CON 63. b.
CON 64. d.
CON 65. b.
CON 66. a.
CON 67. c.
CON 68. b.
CON 69. c.
CON 70. a.
CON 71. c.
CON 72. a.
CON 73. d.
CON 74. c.
CON 75. d.
CON 76. b.
CON 77. c.
CON 78. d.
CON 79. b.
CON 80. a.
CON 81. c.
CON 82. c.
CON 83. c.
CON 84. a.
CON 85. a.
CON 86. a.
CON 87. b.
CON 88. c.
CON 89. c.
CON 90. d.
CON 91. b.
CON 92. a.
CON 93. c.
CON 94. b.
CON 95. b.
CON 96. a.
CON 97. c.
CON 98. b.
CON 99. c.
CON 100. a.
CLIMB 1. b.
CLIMB 2. c.
CLIMB 3. c.
CLIMB 4. a.
CLIMB 5. d.
CLIMB 6. b.
CLIMB 7. d.
CLIMB 8. c.
CLIMB 9. d.
CLIMB 10. b.
CLIMB 11. c.
CLIMB 12. a.
CLIMB 13. c.
CLIMB 14. d.
CLIMB 15. c.
CLIMB 16. a.
CLIMB 17. d.
CLIMB 18. d.
CLIMB 19. c.
CLIMB 20. b.
CLIMB 21. c.
CLIMB 22. c.
CLIMB 23. c.
CLIMB 24. c.
CLIMB 25. b.
CLIMB 26. c.
CLIMB 27. a.
CLIMB 28. b.
CLIMB 29. b.
CLIMB 30. d.
CLIMB 31. d.
CLIMB 32. c.
CLIMB 33. b.
CLIMB 34. d.
CLIMB 35. d.
CLIMB 36. b.
CLIMB 37. b.
CLIMB 38. c.
CLIMB 39. a.
CLIMB 40. a.
CLIMB 41. a.
CLIMB 42. c.
CLIMB 43. a.
CLIMB 44. b.
CLIMB 45. d.
CLIMB 46. d.
CLIMB 47. c.
CLIMB 48. b.
CLIMB 49. b.
CLIMB 50. d.
CLIMB 51. d.
CLIMB 52. c.
CLIMB 53. b.
CLIMB 54. b.
CLIMB 55. d.
CLIMB 56. b.
CLIMB 57. d.
CLIMB 58. b.
CLIMB 59 b.
CLIMB 60. b.
CLIMB 61. c.
CLIMB 62. d.
CLIMB 63. c.
CLIMB 64. c.
CLIMB 65. c.
CLIMB 66. c.
CLIMB 67 c.
CLIMB 68. c.
CLIMB 69. a.
CLIMB 70. a.
CLIMB 71 a.
CLIMB 72. b.
CLINIB 73. a.
CLIMB 74. d.
CLIMB 75. b.
CLIMB 76. d.
CLIMB 77. a.
CLIMB 78. b.
CLIMB 79. a
CLIMB 80. c.
CLIMB 81. a.
CLIMB 82. c.
CLIMB 83. a.
CLIMB 84. c.
CLIMB 85. d.
CLIMB 86. d.
CLIMB 87. c.
CLIMB 88. c.
CLIMB 89. a.
CLIMB 90. c.
CLIMB 91. b.
CLIMB 92. c.
CLIMB 93. a.
CLIMB 94. a.
CLIMB 95. d.
CLIMB 96. d.
CLIMB 97. d.
CLIMB 98. a.
CLIMB 99. b.
CLIMB 100. b.
TURN 1. a.
TURN 2. a.
TURN 3. c.
TURN 4. d.
TURN 5. c.
TURN 6. b.
TURN 7. d.
TURN 8. c.
TURN 9. c.
TURN 10. a.
TURN 11. a.
TURN 12. b.
TURN 13. d.
TURN 14. b.
TURN 15. b.
TURN 16. a.
T U R N 17. b.
TURN 18. c.
TURN 19. d.
T U R N 20. a.
TURN 21. a.
TURN 22. b.
TURN 23. b.
T U R N 24. a.
T U R N 25. c.
T U R N 26. a.
TURN 27. d.
TURN 28. c.
T U R N 29. b.
TURN 30. c.
TURN 31. c.
T U R N 32. a.
TURN 33. b.
TURN 34. a.
T U R N 35. b.
T U R N 36. c.
TURN 37. c.
TURN 38. b.
TURN 39. c.
T U R N 40. a.
T U R N 41. a.
T U R N 42. b.
TURN 43. a.
T U R N 44. d.
TURN 45. a.
TURN 46. b.
TURN 47. b.
TURN 48. b.
TURN 49. a.
TURN 50. b.
TURN 51. d.
TURN 52. c.
T U R N 53. b.
TURN 54. a.
TURN 55. a.
TURN 56. c.
TURN 57. b.
TURN 58. c.
TURN 59. a.
TURN 60. b.
TURN 61. b.
TURN 62. c.
TURN 63. c.
TVRN 64. a.
TURN 65. b.
TURN 66. a.
TURN 67. d.
TURN 68. d.
TURhr 69. a.
TURN 70. a.
TURN 71. c.
TURN 72. b.
TURN 73. a.
TURN 74. a.
TURN 75. a.
POW 1. c.
POW 2. c.
POW 3. b.
POW 4. a.
POW 5. b.
POW 6. a.
POW 7. b.
POW 8. a.
POW 9. a.
POW 10. d.
POW 11. d.
POW 12. c.
POW 13. a.
POW 14. d.
POW 15. c.
POW 16. c.
POW 17. b.
POW 18. b.
POW 19. a.
POW 20. a.
POW 21. a.
POW 22. c.
POW 23. a.
POW 24. b.
POW 25. b.
POW 26. a.
POW 27. b.
POW 28. c.
POW 29. c.
POW 30. d.
POW 31. c.
POW 32. b.
POW 33. c.
POW 34. a.
POW 35. a.
POW 36. b.
POW 37. b.
POW 38. a.
POW 39. c.
POW 40. b.
POW 41. a.
POW 42. c.
POW 43. c.
POW 44. a.
POW 45. b.
POW 46. c.
POW 47. c.
POW 48. a.
POW 49. a.
POW 50. c.
POW 51. b.
POW 52. b.
POW 53. b.
POW 54. d.
POW 55. d.
POW 56. b.
POW 57. b.
POW 58. b.
POW 59. d.
POW 60. c.
POW 61. c.
POW 62. d.
POW 63. a.
POW 64. b.
POW 65. d.
POW 66. c.
POW 67. b.
POW 68. a.
POW 69. d.
POW 70. c.
POW 71. a.
POW 72. c.
POW 73. b.
POW 74. d.
POW 75. c.
STAB 1. b.
STAB 2. b.
STAB 3. b.
STAB 4. d.
STAB 5. d.
STAB 6. d.
STAB 7. c.
STAB 8. c.
STAB 9. d.
STAB 10. c.
STAB 11. d.
STAB 12. d.
STAB 13. d.
STAB 14. b.
STAB 15. b.
STAB 16. b.
STAB 17. a.
STAB 18. b.
STAB 19. d.
STAB 20. b.
STAB 2 1. d.
STAB 22. c.
STAB 23. a.
STAB 24. b.
STAB 25. c.
STAB 26. c.
STAB 27. b.
STAB 28. d.
STAB 29. a.
STAB 30. b.
STAB 31. d.
STAB 32. b.
STAB 33. b.
STAB 34. b.
STAB 35. b.
STAB 36. a.
STAB 37. b.
STAB 38. a.
STAB 39. c.
STAB 40. d.
STAB 41. b.
STAB 42. b.
STAB 43.b.
STAB 44. c.
STAB 45. c.
STAB 46. a.
STAB 47. a.
STAB 48. b.
STAB 49. d.
STAB 50. b.
STAB 51. b.
STAB 52. c.
STAB 53. a.
STAB 54. c.
STAB 55. b.
STAB 56. b.
STAB 57. b.
STAB 58. b.
STAB 59. b.
STAB 60. b.
STAB 61. b.
STAB 62. c.
STAB 63. a.
STAB 64. b.
STAB 65. b.
STAB 66. a.
STAB 67. b.
STAB 68. b.
STAB 69. a.
STAB 70. d.
STAB 71. a.
STAB 72. a.
STAB 73. d.
STAB 74. a.
STAB 75.b.
STAB 76. c.
STAB 77. a.
STAB 78. b.
STAB 79. c.
STAB 80. d.
STAB 81. a.
STAB 82. d.
STAB 83. b.
STAB 84. c.
STAB 85. b.
STAB 86. a.
STAB 87. b.
STAB 88. d.
STAB 89. b.
STAB 90. b.
STAB 91. d.
STAB 92. c.
STAB 93. c.
STAB 94. d.
STAB 95. c.
STAB 96. b.
STAB 97. a.
STAB 98. b.
STAB 99. b.
STAB 100. a.
HSF 1. a.
HSF 2. c.
HSF 3. b.
HSF 4. c.
HSF 5. c.
HSF 6. c.
HSF 7. d.
HSF 8. b.
HSF 9. b.
HSF 10. c.
HSF 11. b.
HSF 12. d.
HSF 13. c.
HSF 14. b.
HSF 15. c.
HSF 16. b.
HSF 17. c.
HSF 18. a.
HSF 19. c.
HSF 20. c.
HSF 21. b.
HSF 22. b.
HSF 23. c.
HSF 24. d.
HSF 25. d.
HSF 26. c.
HSF 27. c.
HSF 28. a.
HSF 29. a.
HSF 30. b.
HSF 31. c.
HSF 32. d.
HSF 33. b.
HSF 34. b.
HSF 35. c.
HSF 36. d.
HSF 37. a.
HSF 38. b.
HSF 39. c.
HSF 40. c.
HSF 41. b.
HSF 42. b.
HSF 43. c.
HSF 44. a.
HSF 45. d.
HSF 46. c.
HSF 47. a.
HSF 48. c.
HSF 49. d.
HSF 50. d.
HSF 51. b.
HSF 52. a.
HSF 53. c.
HSF 54. b.
HSF 55. d.
HSF 56. b.
HSF 57. c.
HSF 58. c.
HSF 59. b.
HSF 60. b.
HSF 61. a.
HSF 62. a.
HSF 63. a.
HSF 64. b.
HSF 65. a.
HSF 66. d.
HSF 67. c.
HSF 68. b.
HSF 69. b.
HSF 70. b.
HSF 71. c.
HSF 72. a.
HSF 73. a.
HSF 74. b.
HSF 75. b.
HSF 76. b.
HSF 77. b.
HSF 78. b.
HSF 79. b.
HSF 80. c.
HSF 81. a.
HSF 82. a.
HSF 83. a.
HSF 84. d.
HSF 85. b.
HSF 86. a.
HSF 87. d.
HSF 88. d.
HSF 89. a.
HSF 90. b.
HSF 91. b.
HSF 92. a.
HSF 93. b.
HSF 94. a.
HSF 95. b.
HSF 96. d.
HSF 97. d.
HSF 98. b.
HSF 99. b.
HSF 100. a.
PROPS 1. c.
PROPS 2. d.
PROPS 3. d.
PROPS 4. b.
PROPS 5. d.
PROPS 6. c.
PROPS 7. c.
PROPS 8. a.
PROPS 9. b.
PROPS 10. b.
PROPS 11. c.
PROPS 12. c.
PROPS 13. c.
PROPS 14. a.
PROPS 15. b.
PROPS 16. d.
PROPS 17. a.
PROPS 18. b.
PROPS 19. d.
PROPS 20. b.
PROPS 21. a.
PROPS 22. b.
PROPS 23. d.
PROPS 24. b.
PROPS 25. a. then d.
PROPS 26. d.
PROPS 27. d.
PROPS 28. d. then c.
PROPS 29. c.
PROPS 30. c.
PROPS 31. a.
PROPS 32. d.
PROPS 33. c.
PROPS 34. a.
PROPS 35. b.
PROPS 36. a.
PROPS 37. d.
PROPS 38. b.
PROPS 39. a.
PROPS 40. a.
PROPS 41. b.
PROPS 42. a.
PROPS 43. a.
PROPS 44. b.
PROPS 45. b.
PROPS 46. c.
PROPS 47. b.
PROPS 48. b.
PROPS 49. c.
PROPS 50. a.
PROPS 51. d.
PROPS 52. a.
PROPS 53. d.
PROPS 54. a.
ENV 87. b.
ENV 88. a.
PROPS 55. b.
PROPS 56. b.
PROPS 57. d.
PROPS 58. b.
PROPS 59. b.
PROPS 60. a.
PROPS 61. a.
PROPS 62. c.
PROPS 63. b.
PROPS 64. b.
PROPS 65. c.
PROPS 66. b.
PROPS 67. b.
ENV 1. c.
ENV 2. b.
ENV 3. b.
ENV 4. b.
ENV 5. b.
ENV 6. b.
ENV 7. b.
ENV 8. d.
ENV 9. d.
ENV 10. a.
ENV 11. d.
ENV 12. c.
ENV 13. c.
ENV 14. c.
ENV 15. b.
ENV 16. b
ENV 17. a.
ENV 18. c.
ENV 19. c.
ENV 20. c.
ENV 21. b.
ENV 22. a.
ENV 23. b.
ENV 24. b.
ENV 25. b.
ENV 26. c.
ENV 27. c.
ENV 28. a.
ENV 29. d.
ENV 30. d.
ENV 31. b.
ENV 32. a.
ENV 33. d.
ENV 34. a.
ENV 35. d.
ENV 36. d.
ENV 37. a.
ENV 38. a.
ENV 39. b.
ENV 40. c.
ENV 41. b.
ENV 42. c.
ENV 43. a.
ENV 44. a.
ENV 45. d.
ENV 46. d.
ENV 47. a.
ENV 48. d.
ENV 49. b.
ENV 50. b
ENV 51. b.
ENV 52. b.
ENV 53. b.
ENV 54. a.
ENV 55. a.
ENV 56. c.
ENV 57. a.
ENV 58. c.
ENV 59. b.
ENV 60. a.
ENV 61. d.
ENV 62. b.
ENV 63. a.
ENV 64. b.
ENV 65. c.
ENV 66. b.
ENV 67. a.
ENV 68. a
ENV 69. d.
ENV 70. a.
ENV 71. a.
ENV 72. c.
ENV 73. b.
ENV 74. a.
ENV 75. d.
ENV 76, b.
ENV 77. a.
ENV 78. a.
ENV 79. b.
ENV 80. b.
ENV 81. a.
ENV 82. a.
ENV 83. c.
ENV 84. c.
ENV 85. a.
ENV 86. a.
ENV 89. a.
ENV 90. d.
ENV 91. d.
ENV 92. a.
ENV 93. b.
ENV 94. b.
ENV 95. b.
ENV 96. d.
ENV 97. d.
ENV 98. d.
ENV 99. b.
ENV 100. a.
ENV 101. a.
ENV 102. a.
ENV 103. b.
ENV 104. b.
ENV 105. d.
SECTION 4
EXPLANATIONS
Subject
Basics.
Aerofoil sections and wing planforms
Lift and drag.
High lift devices.
Stalling and stalling speed calculations.
Flying controls.
Climbing flight.
Turning flight.
Power available and power required.
Stability and manoeuvrability.
High speed flight.
Propellers.
Flight envelopes and aircraft performance.
BASIC 1. a.
Newton's second low of motion states that when a force acts upon a body the
change of momentum is proportional to the force and acts in the same direction.
This is usually simplified into the form, Force equals mass multiplied by
acceleration, or F = MA.
BASIC 2. c.
The static pressure a t any point in the atmosphere is caused by the weight of the
air above that point pressing down upon it. I t acts in all directions
simultaneously.
BASlC 3. b.
Bernouli's theorem states that the sum of static and dynamic pressure is constant
at all points along streamline. This sum of static and dynamic pressure is called
total pressure.
BASIC 4. a.
In the SI system the unit of mass is the Kilogram and the unit of force or weight
is the Newton. One Newton is the force required to give an acceleration of 1 m/s2
when acting on a mass of 1 Kg. I t should however be noted that in the JAR
system of weight and balance calculation, mass is often used in place of weight.
BASIC 5. h.
Density is a measure of how tightly the material of a body is packed together. In
the SI system its units are Kilograms per cubic metre, which is commonly
abbreviated to kg/m" Velocity is the rate of change of position of a body. In the
SI system it is measured in metres per second, which is commonly abbreviated to
mls.
BASIC 6. d.
In stating that the total pressure of air is constant at all points in a streamline
Rernouli assumed that air is inviscid and hence not subject to friction losses.
BASIC 7. b.
Dynamic pressure is caused when moving air is brought to rest on a surface.
This causes the air to give up its kinetic energy and this process exerts a force or
dynamic pressure on the surface. Dynamic pressure is equal to half the density
of the air multiplied by the square of its velocity. This is commonly stated as
.
pressure acts only in the downstream
dynamic pressure = % p ~ 2 Dynamic
direction.
BASIC 8. c.
A line from the trailing edge to the leading edge passing midway between the
upper and lower surfaces is called the mean camber line. A straight line from
the trailing edge to the leading edge is called the chord line. In a symmetrical
wing the mean camber line and chord line coincide at all points along their
lengths.
BASIC
9. c.
If the clockwise and anti-clock~risemoments about a point are in balance there
will be no resultant moment. If however the moments are not in balance then the
point will be subject to angular acceleration in the direction of the resultant
nioment. If no friction or velocity related damping forces act on the body the
acceleratio~~
rate will be constant.
BASIC 10. d.
The angle of attack of an aerofoil is the angle between the chord line and the
relative airflow. I t should be noted that when lift is being generated, the relative
airflow is not tlie same as the free stream airflow due to the lift-induced
clownwash of the airflow.
BASIC 11. c.
Density is a measure of how tightly the material of a body is packed together. In
the SI system its units are Kilograms per cubic metre, which is commonly
abbreviated to kg/m3. Energy is the ability to do work. The SI unit of energy is
the joule, which is the energy required to produce 1 Newton Metre of work.
BASIC 12. d.
The weight of a body is the force exerted by gravitational acceleration acting on
its mass. The standard gravitational acceleration on earth is 9.81 m/s2. The
equation for the weight of a body is based on Newton's second law, F = MA,
such that W (in Newtons) = 9.81 x Mass. Thus a mass of 1Kg weighs 9.81N.
The correct conversion factor is therefore 9.81N = 1Kg.
BASIC-13. a.
The S1 units of force, mass and acceleration are the Newton, The Kilogram and
the ~ e t r e l ~ e c o n d1 ~Newton
.
is the force required to give a mass of 1 Kg an
acceleration of 1 mls2.
BASlC 14. b.
The SI units of force, mass and acceleration are the Newton, The Kilogram and
the metrelsecond2. A force of IN acting on a mass of 1 Kg will produce an
acceleration of 1 m/s2.
BASlC 15. d.
An airspeed indicator produces an indicated airspeed (1.4s) that is proportional
to 1/2 p\12, where p is air density and V is TAS. For any given value of '/2 pv2 the
IAS will be constant. These instruments are calibrated such that at ISA MSL
JAS = TAS. But density decreases with increasing altitude such that for a given
IAS, the TAS must increase to balance the decreasing density so that pv2
remains constant. At altitudes below ISA MSL density is increased so the TAS
equating to any given IAS is decreased to maintain constant '/z pv2. When flying
below ISA MSL TAS will therefore be lower than IAS.
BASIC 16. b.
An airspeed indicator produces an indicated airspeed (IAS) that is proportional
to '/z pv2, where p is air density and V is TAS. For any given value of % p v 2 the
IAS will be constant. But density decreases with increasing altitude so that for a
given IAS, the TAS must increase such that the rate of increase in v2balances
the rate of decrease in p such that ?4p v 2 remains constant. At 40000 feet ISA,
density is '/4 of its MSL value so for any given IAS, v2must be 4 times its MSL
value. This means that V, which is TAS, must be twice its sea level value.
BASIC 17. a.
One Pascal o r Pa is a @essure of 1 ~ / so
m a~pressure of 1 kilopascal o r kPa
exerts a force of 1 0 0 0 ~ / m A
~ . 100 kPa pressure therefore exerts a force of 100,
000 N.
BASIC 18. d.
The angle of attack at which any given aerofoil achieves its greatest efficiency is
dependent upon aerofoil cross section. This means that although options a, b,
and c- might be correct for some specific aerofoil they cannot be correct for all
aerofoils. The purpose of a n aerofoil is to generate lift. The drag it produces in
generating this lift represents the losses caused by its inefficiency. The greatest
efficiency therefore occurs when the ratio of lift to drag is greatest. This occurs
at the angle of attack producing the best L:D ratio.
BASIC 19. a.
I n order to generate lift a symmetrical aerofoil requires a n airflow and a positive
angle of attack. At zero angle of attack such an aerofoil generates parasite drag
but no lift. The zero lift angle of attack of a symmetrical aerofoil is therefore
zero.
BASIC 20. b.
Because its upper surface is more convex than its lower surface, a cambered
aerofoil will generate lift a t all positive and slightly negative angles of attack.
The zero lift angle for such aerofoils depends on the degree of camber but is
typically in the order of -4'.
BASIC 21. b.
Density is a measure of how closely packed a material is. I t is specified in terms
of the mass of material in a unit volume. As temperature decreases the molecules
of a material move closer together causing it to contract. This contraction causes
any given mass of material to take up a smaller volume. This reduction in
volume for a given mass represents an increase in density.
BASIC 22. d.
The viscosity of a material is a measure of the shear forces set up within it when
it flows. These shear forces resist the flowing of the material so increasing
viscosity implies increased resistance to flow.
BASIC 23. b.
For steady flow of air through a duct the mass flow through all stages of the duct
must be constant. That is to say the mass flow into the duct must equal the mass
flow out of the duct. A convergent duct is one in which the cross sectional area of
the outlet is smaller than that of the inlet. The mass flow through any stage of a
duct can be calculated using the equation, Mass flow = pVA, where p is air
density, V is velocity and A is the cross sectional area. At low speed the changes
in air density caused by flow a r e very small so p can be assumed to be constant.
Under theses circumstances mass flow is proportional to VA. This means that as
air passes through a-convergent duct the reduction in A causes a n increase in V
such that mass flow remains constant. But increasing velocity increases dynamic
pressure and according to Bernouli's theorem dynamic plus static pressure is
constant a t all points in a stream line. This means that as increasing velocity
causes dynamic pressure to increase, the static pressure decreases. The overall
effect of subsonic flow through a convergent dust is therefore an increase in
velocity and a decrease in static pressure.
BASIC 24. c.
Theories relating to air flowing through ducts are based on the assumptions that
air is inviscid and hence suffers no friction losses and that no energy is put into
o r taken out of the system. This means that the total energy of the a i r remains
constant. The principal forms of energy possessed by air are heat, static
pressure and kinetic energy proportional to the square of its velocity. If the sum
of these energies remains constant then as one form increases in magnitude the
others must decrease. As the velocity of air flowing through a convergent duct
increases this causes its kinetic energy to increase. This increase in kinetic
energy causes its temperature (heat energy) and static pressure to decrease. At
low speeds the effects of these changes a r e insufficient to cause significant
changes in density. The overall effect is that temperature decreases but density
remains substantially constant.
BASIC 25. d.
Kinetic energy = % MV', where M is the mass and V is its velocity. This means
that for any given acceleration, the increase in kinetic energy will be greater at
high speeds than at law speeds. Because the energy of air passing through a duct
is assumed to be constant, then as acceleration causes kinetic energy to increase,
t l ~ etemperature and static pressure of the air decrease. This decrease in static
pressure causes the air to expand thereby reducing its density. This increase in
volume requires a larger area duct to maintain constant mass flow a t any given
velocity. The overall effect of these processes is that at high speeds a divergent
duct is required to produce an increase in flow velocity. At supersonic speeds the
rate of expansion with increasing velocity is such that a convergent duct causes
velocity to decrease and temperature and static pressure to increase. This is a
reversal of the effects a t low velocities.
BASIC 26. c.
The principal forms of energy possessed by air are heat, static pressure and
kinetic energy. Kinetic energy is proportional to the square of velocity. The sum
of these energies remains constant so as one form increases, the others must
decrease. As the velocity of air flowing through a convergent duct increases this
causes its kinetic energy to increase. This increase in kinetic energy causes its
temperature (heat energy) and static pressure to decrease. At low speeds the
effects of these changes are insufficient to cause significant changes in density.
The overall effect is that for subsonic flow through a convergent duct, velocity
increases whilst temperature and static pressure decrease.
At supersonic speeds the decrease in static pressure is large enough to cause the
air to expand thereby significantly reducing its density. This increase in volume
requires a larger area duct to maintain constant mass flow at any given velocity.
The means that a t supersonic speeds a divergeut $uct is required to produce an
increase in flow velocity. At these speeds the rate of expansion is such that a
convergent duct causes velocity to decrease and temperature and static pressure
to increase. This is a reversal of the effects at low velocities. The overall effects
of these processes is that subsonic flow passing through a convergent duct will
accelerate until reaching sonic speed a t the narrowest point (or throat) of the
duci. Further acceleration into the supersonic range mill then be impossible
unless the duct becomes divergent.
BASIC 27. d.
The principal forms of energy possessed by air are heaqstatic pressure and
kinetic energy. Kinetic energy is proportional to the square of velocity. The sum
of these energies remains constant so as one form increases in magnitude the
others must decrease. As the velocity of air flowing through a duct increases its
kinetic energy increases, causing its temperature (heat energy) and static
pressure to decrease. At supersonic speeds this decrease in static pressure is
large enough to cause the air to expand significantly reducing its density. This
increase in volume requires a larger area duct to maintain constant mass flow at
any given velocity. The means that at supersonic speeds a divergent duct is
required to produce an increase in flow velocity.
BASIC 28. c.
The behaviour of airflow is such that subsonic acceleration requires a convergent
duct and supersonic acceleration requires a divergent duct. At sonic speeds the
relationship between expansion rate and acceleration is balanced such that a
parallel duct will cause acceleration at constant mass flow. A sonic flow.passing
into a parallel duct would however immediately become supersonic requiring a
divergent duct for further acceleration. The overall effect of this process is that
acceleration of sonic flow through a parallel duct does not occur in reality.
BASIC 29. c.
The principal forms of energy possessed by air are heat, static pressure and
kinetic energy. Kinetic energy is proportional to the square of velocity. The sum
of these energies remains constant so as one form increases, the others must
decrease. As the velocity of air flowing through a convergent duct increases, its
kinetic energy to increases. This increase in kinetic energy causes its
temperature (heat energy) and static pressure to decrease. At low speeds the
effects of these changes are insufficient to cause significant changes in density.
The overall effect is that for subsonicflow through a convergent duct, velocity
increases whilst temperature and static pressure decrease.
BASIC 30. a.
The principal forms of energy possessed by air are heat, static pressure and
kinetic energy. Kinetic energy is proportional to the square of velocity. The sun]
of these energies remains constant so as one form decreases, the others must
increase. As supersonic airflow passes through a convergent duct it slows down.
This decreases its kinetic energy, causing its temperature (heat energy) and static
pressure to increase. At such speeds this increase in static pressure is sufficiently
large to cause the air to contract significantly increasing its density.
FORMS 1. c.
Wingspan is the distance from one wing tip to the other. And as indicated below,
span decreases with sweep angle. Aspect ratio is the ratio of (wingspan squared)
/ (wing area) so sweeping back the wings will result in a decrease in aspect ratio.
Span when swept back
i wings in swept back
+i position
4
Span when swept forward
Chord length is not
affected by sweep angle
FORMS 2. b.
High aspect ratio wings have weak wingtip vortices and hence little downwash at
the tips. This means that for any given aircraft attitude the tips of a high aspect
ratio wing experience a higher angle of attack than those of a low aspect ratio
wing of the same area. Pointed wings and those of high taper ratio exhibit
similar effects. Airflow over swept back wings tends to migrate outwards
causing thickening of the boundary layer a t the tips. This thick boundary layer
separates more easily than the thinner layer experienced by straight wings. The
overall effect is that stalling of swept back wings commences a t the tips and
moves inboard.
FORMS 3. b.
The rectangular section and short span makes low aspect ratio wings easy to
construct. Because of the short span and long chord they produce strong tip
vortices and hence high induced drag. The short span also reduces frontal area
thereby minimising profile drag. Because of the strong wingtip vortices the tips
of such wings experience low angles of attack and so stall first a t the root.
FORMS 4. c.
When pointed wings are subject to a positive angle of attack such that lift is
generated, their tips stall immediately. That is to say even the smallest lift
generating angle of attack will cause tip stall. In this condition the vortices form
further inboard resulting in upward deflection of airflow over the tips.
FORMS 5. d.
Because of their uniform lift distribution pattern all areas of an elliptical wing
experience the same levels of downwash and hence the same angle of attack..
This means that the entire wing stall simultaneously giving a very abrupt stall
with little o r no pre-stall buffet. Close to the stall even a very small deflection of
the ailerons results in localised stall thereby reducing aileron effectiveness. The
complex shape makes design and construction of elliptical wing more difficult
than for equivalent rectangular or tapered wings.
FORlMS 6. a.
Elliptical wing plan forms reduce wingtip vortices and hence induced drag.
Because an elliptical wing has a longer span than a rectangular wing of equal
area, profile drag is increased. Because of their complex shape elliptical wings
are difficult to design and construct and their strength :weight ratios and flutter
cl~aracteristicare inferior to many other plan forms.
FORMS 7. b.
The tips of pointed wings stall at any lift generating angle of attack causing the
wing tip vortices to form further inboard. This causes the tips to be subject to
upwash whilst the strongest downwash is close to the roots.
FORMS 8. a.
The CI/CL: semi-span distance chart illustrates the relationship between the
local coefficient of lift (CI) and the mean coefficient of lift (CL) for each point
along the span of wings of various plan forms. For an elliptical wing the CI:CL
ratio is one a t all points along the wing whilst for all other plan forms it varies
with semi-span distance. In most cases these charts indicate that the ratios for
all non-elliptical wings coincide at a value of approximately 1.1 a t about the 0.55
semi-span point.
FORMS 9. a.
Reducing aspect ratio increases the
intensity of wing tip vortices causing
an increase in downwash behind the
trailing edge of the wing. This
increased downwash reduces the
effective angle of attack thereby
reducing the coefficient of lift.
The intensity of the wingtip vortices
and resulting downwash increases
with angle of attack such that the
reduction in coefficient is greatest
a t high angles of attack. The overall
effect is a shallower CL::a curve and Stee
an increase in the stalling angle
as indicated in the diagram at the right.
Increased stalling angle
Shallower lift curve
FORMS 10. c.
The term dihedral refers to a design feature whereby the wings are inclined from
root to tip such that with the aircraft in a level attitude the wing tips a r e higher
than the wing roots. I n attempting to define the dihedral angle it must be noted
that any reference to horizontal angles mean angles parallel to the longitudinal
axis of the aircraft. Such terms cannot therefore be used to describe dihedral.
Of the options provided in this question the most accurate is that the dihedral
angle is the vertical angle between the plane of the wings and the lateral axis of
tlie aircraft.
FORMS 11. b.
Perliaps the most simple definition of the angle of incidence is that it is the angle
a t wliich the wing is attached to tlie fuselage. Such a definition is however
inadequate in that it does not provide a sufficiently clear distinction between the
angle of incidence and the sweep back angle. A more precise definition of the
angle of incidence is that it is the angle between the chord line of the wing and
the longitudinal axis of the aircraft. The term incidence refers to the angle of
attack, which is the angle between the chord line of the wing and the relative
airflow. The angle of incidence and the incidence are rarely the same.
4
FORMS 12. a.
The mean camber line is a n imaginary line wliich passes from the leading edge to
trailiug edge such that all points on it are midway between the upper and lower
surface of the aerofoil. In a cambered aerofoil the mean camber line will be
curved. The chord line is an imaginary straight line passing from the leading
edge to the trailing edge of an aerofoil. In the case of a symmetrical aerofoil the
curvature of the upper and lower surfaces are identical so the camber line is
straight. This means that the mean camber line and chord line coincide along
their entire lengths and hence are effectively the same thing.
FORMS 13. a.
The term deep stall refers to a situation in which on reaching the stall, a wing
pitches leading edge upwards whilst simultaneously losing altitude due to loss of
lift. The ovel-all effect of this sequence is that the wing moves deeper into the
stall. The pitching upwards of the wing in the initial stages of the sequence is an
essential part of deep stall so any wing that pitches downward reduces its angle
of attack and hence pull out of the stall rather than moving deeper into it. This
pitching u p in the stall is peculiar to swept back wings and is caused by the
tendency of the wing tips to stall before the wing roots. Because the tips are
further aft than the roots this initial tip stall moves the C of P forward and it is
this that initiates the deep stall sequence. Tip stalling of straight and tapered
wings will not cause forward movement of the C of P and so cannot cause deep
stall. Whilst the position of the tailplane has a bearing on the ability of an
ail-craft to be pulled out of deep stall, it is not the fundamental cause of the
problem.
FORMS 14. d.
The simple equation for aspect ratio is the wing span divided by the chord length
(A = Span / Chord). In applying this equation to a wing of complex plan form a
difficulty arises in establishing the appropriate value for the chord length. In
such cases a more appropriate equation is the square of the wing span, divided
by the wing area. This equation is simply an adaptation of the first as shown
below.
Aspect ratio = Span / Chord
= (Span x Span) / (Chord x Span)
But Chord x Span = wing area so Aspect ratio = s p a n 2 / Wing area.
FORMS is. b.
The angle of attack of a n aerofoil is the angle between its chord line and the
airflow affecting the aerofoil in flight. When an aerofoil is generating lift, the
free stream airflow is deflected downwards and this alters the direction of the
airflow close to the aerofoil. The correct definition of angle of attack is therefore
the angle between the chord line and the relative airflow.
FORMS 16. b.
The purpose of employing a supercritical wing is to increase MCRIT,decrease the
~ to ease structural design problems by
drag rise at speeds above M C R land
enabling higher thickness to chord ratios to be employed. The increased MCRITis
achieved by reducing the camber of the upper surface such that acceleratian
over the wing is reduced. The reduction in lift caused by this lesser camber is
offset by the use of a concave o r reflex camber under the rear area of the wing.
The higher thickness to chord ratios are achieved by making the wings thicker.
A supercritical wing is therefore thicker and less cambered than a conventional
wing and is concave under the rear surface.
FORMS 17. d.
Aspect ratio is the ratio of wing span to chord
High aspec ratio
length, so increasing aspect ratio increases wing
span and decreases chord length. The increased
Low aspect ratio
wing span increases profile drag whilst the
decreased chord length reduces induced drag.
At low speeds the majority of drag is lift induced,
whereas a t high speed the majority is profile. At
low speeds the effect of increasing aspect ratio is
Drag
therefore to decrease the major component of
drag and increase the minor component,
EAS
decreasing total drag. At high speeds the major
(profile) component is increased and the minor
(lift induced) component decreased, causing an
overall increase in total rag. The overall effect
is therefore decreased drag a t low speeds and
increased drag a t high speeds as indicated in the diagram at then right.
L.l
FORMS 18. a.
Increasing aspect ratio decreases the
intensity of wing tip vortices, causing
a decrease in downwash behind the
trailing edge of the wing. This
decreased downwash increases the
effective angle of attack thereby
increasing-the coefficient of lift.
Tbe effect on wingtip vortices
and resulting downwash increases
with angle of attack such that the
increase in coefficient is greatest
a t high angles of attack. The overall
effect is a steeper CL:a curve and a
decrease in the stalling angle as
indicated in the diagram a t the right.
CL
Steeper lift
curve
Shallow lift curve
FORMS 19. b.
Increasing aspect ratio decreases the
intensity of wing tip vortices, causing Highe
a decrease in downwash behind the
CLMAX
trailing edge of the wing. This
decreased downwash increases the
effettive angle of attack thereby
increasing the coefficient of lift.
The effect on wingtip vortices
and resulting downwash increases
with angIe of attack such that the
increase in coefficient is greatest
a t high angles of attack. The overall
effect is a steeper CL:a curve and an
increased CLMAX.
In the stall condition CL = CLMAX
and lift = C L ~ A1/2Xp v 2 s where
p is the air density, V is the true stalling speed of the aircraft and S is the wing
area. By increasing CLMAX,
the higher aspect ratio enables the wing to produce
lift equal to the aircraft's weight at a lower airspeed. In this way the stalling
speed of the aircraft is reduced by increasing the aspect ratio.
FORMS 20. c.
The principal advantage of swept back wings is that they increase MCRIT
thereby
enabling aircraft to cruise at higher speeds without encountering the adverse
effects of transonic flight. The use of sweep back does however introduce a
number of disadvantages, including a tendency for airflow to migrate outwards
towards the wing tips. The outward migration causes the boundary layer at the
wing tips to thicken and lose energy. This thick, sluggish boundary layers tends
to separate easily as angle of attack is increased, causing the wing tips to stall
before the roots. Because the tips of swept back wings are outboard and to the
rear of the wing roots, tip stall causes the C of P to move inboard and forward.
Although the inward movement of the C of P causes no major problems, its
forward movement causes the wing to pitch nose up in the early stages of the
stall. This upward pitching together with the loss of altitude caused by the
reduction in total lift force cause is the primary cause of deep stall.
FORMS 21. d.
For any given wing area and plan form,
increasing sweep back angle decreases
aspect ratio. This has the same effeet as
reducing aspect ratio by any other means
in that wingtip vortices are intensified,
downwash increased and effective angle of
attack reduced. The overall effect is that
the gradient of the C L : a curve and
CLMAX
are reduced and stalling angle
increased as indicated in the diagram at
the right.
Steep gradient
..
.
..
.
FORMS 22. b.
Increasing sweep back angle decreases
Straight wing
aspect ratio, thereby-increasing the
intensity of wing tip vortices and causing
an increase in downwash behind the
trailing edge of the wing. This
increased downwash decreases the
effective angle of attack thereby
decreasing the coefficient of lift.
The effect on wingtip vortices
and resulting downwash increases
with angle of attack such that the
decrease in lift coefficient is greatest
Steep lift curve
a t high angles of attack. The overall
effect is a shallower CL: a curve, and Shallower lift curve
Increased T TALL
an increased stalling angle as
indicated in the diagram a t the right.
FORMS 23. a.
For any given wing area and plan form, increasing sweep back angle decreases
the aspect ratio. This has the same effect as reducing aspect ratio by any other
means in that wingtip vortices are intensified, downwash increased and effective
angle of attack reduced. The overall effect is that the gradient of the CL : a
curve and CLMAXare reduced. Lift = CL !h ~ P S
where
, p is the air density, V is
the true airspeed and S is the wing area. For any given combination of p and S,
the decreased CL caused by wing sweep means that a higher true airspeed is
required to produce the same amount of lift a t any given angle of attack. This is
equally true in the stall condition where the reduced value of C ~ , Mimposes
~X
a
higher stalling speed.
FORMS 24. d.
The chord line of an aerofoil is an imaginary straight line drawn from the
leading edge to the trailing edge. In the case of a rectangular wing the chord line
is of the same length a t all points along the wing. In the case of a tapered,
pointed o r complex shaped wing, the chord length is different a t each point along
the span. The mean aerodynamic chord is the average length of the chord lines
measured at all points along the entire wingspan.
FORMS 25. c.
The principal reasons for employing swept back wings are that they increase
MCRrTand decrease the magnitude of the wave drag affecting the wing in the
transonic speed range. Secondary benefits of swept wings include a reduction in
profile drag. One of the disadvantages of such wings is that they decrease aspect
ratio and in so doing, increase induced drag. Of the various options listed in this
question, only option c is correct.
FORMS 26. a.
The principal reasons for employing swept back wings are that they increase
MCRIT
and decrease the magnitude of the wave drag affecting the wing in the
transonic speed range. Secondary benefits of swept wings include a reduction if
profile d r a g . Also because the wing tips are behind the roots, upward bending
causes the tips to twist leading edge down, whereas downward bending twists the
tips edging edge up. This cyclic twisting tends to damp out wing flutter. One of
the disadvantages of such wings is that they decrease aspect ratio and in so doing
increase induced drag. Of the various options listed in this question, only option
a, is correct.
FORMS 27. b.
When flying in turbulence the wings are subjected to random upward and
downward gusts. Upward gust cause the angle of attack to increase, thereby
increasing both the CLand the magnitude of the lift force. Downward gusts have
the opposite effect, randomly decreasing CLand lift. The overall effect of these
random changes in lift is that the aircraft is accelerated upwards and downwards
producing an extremely rough ride for crew and passengers. For any given
vertical gust the magnitude of the change in CLis determined by the slope of the
CL : a curve such that shallow curves provide the lowest gust response and hence
smoothest ride.
As indicated in the diagram
at the right, increasing wing
sweep back angle decreases
the slope of the CL: a and
so decreases gust response.
Secondary effects of
increased sweep back include
self damping of flutter and
the prevention of divergence.
CL
1
Straight wing
Large increase in straight wing
/
Small increase in swept wing cL
wings
I
Increase in a is the same for both
FORMS 28. a.
If the magnitude of the lift force is uniform at all points along the span of a wing
a powerful upward bending moment will be exerted. In order to resist such
bending moments, the wing roots would need to be extremely thick. This
problem is sometimes overcome by tapering wings such that the outer areas
generate less lift than the inboard areas. This method incurs a secondary
problem however in that in generating less lift, the outboard areas also generate
less downwash and so operate ,t higher ekfective angles of attack. This
increasing effective angle of attack from root to tip'therefore makes the wings
more prone to tip stall. Aerodynamic washout is a process whereby the cross
section of the wings varies progressively from root to tip such that the outer
areas generate less lift but have a higher stalling angle. This overcomes the
problem of high bending moments whilst simultaneously reducing the
probability of premature tip stall.
FORMS 29. a.
When the lift force generated by a wing increases, the resulting bending
moments cause the outboard area of the wing to bend upwards. If the C of P of
the wing is forward of its torsional axis, the increasing lift force will also cause
the leading edge of the outboard section of the wing to twist upwards. This will
increase the angle of attack of the outboard section, further increasing lift and
hence bending and twisting the wing still further upwards. If the wing is
insufficiently strong to resist these bending and twisting moments it will diverge
and eventually break away from the aircraft. This process is termed divergence
and can occur only if the C of P is forward of the torsional axis, the aerodynamic
forces too high and the wing, sufficiently weak and flexible. The situation most
likely to cause divergence is therefore as described in option a.
FORMS 30. b.
The angle of incidence is the angle a t which the wing is attached to the fuselage.
More specifically it is the angle between the chord line of the wing and the
longitudinal axis of the aircraft. In most aircraft it varies only with changes in
flap and slat angle.
LD 1. a.
Induced drag = CDI'/t p v 2 s where:
CD,is the coefficient of induced drag.
p is the local air density.
V is the TAS of the relative airflow.
S is the wing area.
So Dl is proportional to all of the variables in this equation. IAS is derived from
% p v Zso as factors such as ambient temperature or altitude alter local air
density these changes will affect induced drag to the same degree. Induced drag
is therefore more closely proporHona1 to IAS than to TAS. High airspeeds tend
to increase air density above the local value, increasing both the induced drag
and IAS, so the more accurate EAS is employed to take account of these effects.
Increasii~gspeeds also enable the required lift to be generated at lower angles of
attack. This reduces wingtip vortices and hence induced drag as the square of
speed increases. Induced drag is therefore inversely proportional to EAS'.
LD 2. b.
Induced drag is produced as a direct result of the generation of lift. Increasing
the lift force using any given combination of wing, airspeed, and local
atmosphere, requires an increase in angle of attack. This increases the pressure
differences between the lower and upper surfaces of the wing and hence
increases wing tip vortices which are the cause of induced drag. Induced drag is
therefore directly proportional to lift. Increasing airspeed enables the required
lift to be generated at lower angles of attack, reducing both wingtip vortices and
induced drag. Induced drag is therefore inversely proportional to airspeed.
Combining the two proportionalities gives the result that induced drag is
proportional to LIV. Although options a, b and c are all correct, option b is the
most accurate.
LD 3. b.
L:D ratio is the ratio of lift to drag. In unaccelerated straight and level flight, lift
is equal to weight and drag is equal to thrust.
In this case L:D ratio = 50000:25000
Which is 2:l
LD 4. b.
Load factor is equal to lift divided by weight, so an increase in load factor at any
given weight results in an increase in lift. Increasing lift at any given TAS
increases induced drag and hence thrust required. Power required is equal to
thrust multiplied by TAS so an increase in load factor at any given TAS, will
result in a proportional increase in power required.
LD 5. a.
An increase in weight at any given load factor and airspeed requires an equal
increase in lift. This is achieved by increasing angle of attack, which results in
stronger wingtip vortices. These vortices are the principal cause of induced drag
(Dl), but have no significant effect on profile drag (Dp) within the normal angle
of attack range. Total drag DTotalis the sum of Dl and Dp SO increasing weight
also increase DToral.
LD 6. b.
Increasing weight requires a corresponding increase in lift. This results in an
increase in induced drag at any given airspeed. Induced drag is proportional to
~/(EAs') the magnitude of the increase diminishes with increasing airspeed. This
means that the increase in induced drag is most prominent at the low end of the
speed range. Increasing weight has no significant effect on profile drag. VMDis
the speed a t which Dl and Dp are equal so the increase in Dl in the low speed
range causes VMDto occur at a higher speed. Aircraft are speed stable a t all
speeds above VMDSO the increase in VMDreduces the speed range over which the
aircraft is speed stable. An increase in weight therefore increases VMDand
decreases speed stability.
LD 7. c.
Profile drag (Dp) is proportional to v2and induced drag (Dl) is proportional to
1/v2. VMDis the speed at which Dp and Dl are equal. Lowering of landing gear
increases Dr but has no significant effect on Dl. Because Dp is proportional to v2
the increase is greatest at the high end of the speed range. The increase in DP
therefore causes the crossover of the two curves at VhlDto occur at a lower
airspeed. Aircraft are speed stable at all speeds above V & ~soDany reduction in
V h qincreases
~
the speed range over which the aircraft is speed stable. The
@
lowering of landing gear therefore decreases VMDand increases speed stability.
LD 8. c.
Dl is proportional to ~ / ( E A s ~and
) Dp is proportional to EAS'. EAS is IAS
corrected for the effects of compressibility. These effects are small at low speeds
so because this question stipulates a constant low IAS it can be assumed that
EAS also remains constant. Both DI and Dp will therefore remain unchanged.
Had the question not stipulated a low IAS then the effects of compressibility
would need to be taken into account. Climbing at constant IAS results in an
increase in TAS so climbing through a large altitude range a t a constant high
IAS would result in the aircraft attaining speeds at which compressibility effects
became significant causing EAS to increase. Under these circumstance Dl would
reduce and Dp would increase with increasing altitude.
LD 9. b.
Although deployment of trailing edge flaps increases both Dl and Dp, the effect
on each is not constant a t all angles. Small angles of deflection produce a greater
increase in Dl than in Dp and a small increase in DTotal.VMDis the speed at which
Dl is equal to Dp so increasing Dl by a greater proportion than Dp will cause an
increase in VMD.Aircraft are speed stable at all speeds above VMDSO increasing
VPD will decrease the proportion of the speed range over which an aircraft is
speed stable. Deployment of trailing edge flaps to the 10 degree setting would
therefore increase both Dl and Dp and decrease speed stability.
LD 10.a.
Although deployment of trailing edge flaps increases both Dl and Dp, the effect
on each is not constant at all angles. Large angles of deflection produce a smaller
increase in Dl than in Dp and a large increase in DTotal. VMDis the speed at which
Dl is equal to Dp SO increasing Dl by a smaller proportion than Dp will cause a
decrease in VMD.Aircraft are speed stable at all speeds above VMD,so decreasing
VMDwill increase the proportion of the speed range over which an aircraft is
speed stable. Deployment of trailing edge flaps to the 40 degree setting would
therefore increase Dl, Dp and speed stability.
LD 11. b.
Drag = CDxpvZs Where
CDis the coefficient of total drag
p is the local air density
V is the TAS
S is the wings area
Water vapour is less dense than dry air so increasing humidity reduces the local
air density. If all other factors in the drag equation remain unchanged then total
drag will decrease.
LD 12. b.
Best L:D ratio is achieved when flying at VbID. At this speed Dp is equal to Dl so
the ratio of Dp:DLis 1 : l .
LD 13. b.
At any given
weight and ioad factor the lift requiredremains constant regardless of speed so
the greatest L:D ratio occurs when drag is minimum. VMDis therefore the speed
a t which drag is at a minimum and the L:D ratio is greatest. It is also the speed
a t which jet aircraft achieve best endurance, propeller aircraft maximum range
and all aircraft maximum glide range,
Vbrn
... is the speed a t which the total drag is at a minimum value.
LD 14. b.
Drag is the force which opposes the motion of an aircraft through the air so
except when the aircraft is climbing or descending vertically drag cannot be
parallel to lift or weight. Although it is commonly assumed that drag acts
parallel to the relative airflow it is more useful to consider it to act parallel but
opposite to the direction of flight. Strictly speaking drag only acts parallel to the
relative airflow when the aircraft is flying at its zero lift angle of attack. Under
these circumstances there is no downwash so the relative airflow is also parallel
but opposite to the direction of flight.
LD 15. a.
At zero lift angle of attack the wing generates no lift and hence no induced drag.
Under these circumstances the only force generated is profile drag so the total
,reaction is equal to Dp.
LD 16. a.
At zero lift angle of attack the wing produces no lift and hence no induced drag
or downwash. The total reaction is therefore the profile drag acting parallel but
opposite to the direction of flight.
LD 17. a.
Dl is proportional to 1 / v 2 and hence decreases with increasing airspeed. Dp is
proportional to v2and so increases with increased airspeed. V b l is
~ the speed a t
wl~ichDl and Dp are equal so a t all lower speeds Dl is greater than Dp. DTotPl
is
the sum of Dl and Dp at all speeds so Dl can never equal D~otaland Dp can do so
only a t the zero lift angle of attack where DI is zero.
LD 18. b.
Dl is proportional to 1 / v 2 so it increases as speed decreases. Dp is proportional to
v2SO it decreases as speed decreases. CLMaxoccurs a t the stall and Vblo is the
~ l l a s i m i ~speed
m
permitted under any circumstances. Changing speed from Vblo
to CI,,\~,,therefore represents a reduction in speed from maximum to minimum
values. This reduction in speed will cause Dl to increase and Dp to decrease.
LD 19. c.
An aircraft may be subject to five types of drag. These are friction, form,
interference, induced and shock. Induced drag is caused by the generation of lift
whilst sllock drag is the result of flight at transonic or supersonic speeds. Profile
drag is caused by the motion of the aircraft through the air and is not connected
with the generation of lift or any specific speed range. At low speeds profile drag
is the sum of friction, form and interference drags. At higher speeds shock drag
makes a further contrib-ltion to it.
LD 20. d.
Options a, b, and c are inadequate for the following reasons. A cambered
aerofoil and an aerofoil a t a positive angle will generate lift only when subject to
a moving airflow. A high speed airflow will generate lift only when passing over
an aerofoil possessing the correct combination of camber and angle of attack.
Option d represents the generation of lift as explained by Newton's second and
third laws of motion. That is to say that a force is generated when a mass of
matter is accelerated (F=MA) and the lift force generated is the equal and
opposite reaction to that which accelerates the mass. In this case the pressure
differences above and below the wing exert a force on the airflow causing it to
accelerate downwards. Lift is the equal an opposite reaction to this acceleration
and is equal to the mass of air multiplied by the downward acceleration given to
it. Option d is therefore the most accurate.
LD 21. c.
In order to create lift a wing must accelerate airflow downwards by generating
pressure differences above and below its surfaces. By leaking from the lower to
the upper surfaces the pressure differences produce rotating airflows or vortices
at the wing tips. The mass of air flowing upward in the part of the vortices
outboard of the tips is precisely equal to the mass of air flowing downwards
inboard of the tips. The vector sum of these flows is therefore zero so the
vortices do not generate any additional lift. The energy used to drive the vortex
flow is extracted from the aircraft and this represents additional drag. This
additional drag caused by the rotational airflow in the wingtip vortices is lift
induced drag or Dl.
LD 22. c.
In order to create lift a wing must accelerate airflow downwards by generating
pressure differences above and below its surfaces. By leaking from thc lower to
the upper surfaces the pressure differences produce rotating airflows or vortices
at the wing tips. The mass of air flowing upward in the parts of the vortices
outboard of the tips is precisely equal to the mass of air flowing downwards in
the parts inboard of the tips. The vector sum of these flows is therefore zero so
the vortices do not generate any additional lift. Inboard of the tips the
downward flow of the air in the vortices reduces the effective angle of attack of
the wing and so reduces the CLand hence lift. In order to maintain the required
lift force the wing must be pitched up to a higher angle in order to restore the
original angle of attack. Although this process restores the lift force, it also tilts
the total reaction further aft such that it includes a greater drag component.
This additional drag due to downwash of airflow over the trailing edge caused by
wingtip vortices is lift-induced drag or Dl.
LD 23. b.
Although the effective lack of wing tips will prevent the formation of tip vortices
the generation of lift will still involve the downward acceleration of the airflow.
There will therefore be just sufficient downwash to produce the lift force but not
the additional downwash that would be generated if tip vortices existed.
LD 24. a.
A wing of infinite span would have no wingtips and hence no wingtip vortices. It
would therefore have no wingtip vortex induced downwash. Also with infinite
span the wing would affect a n infinite mass flow of air at any speed. The lift
generated by a wing is equal to the mass airflow affected by it, multiplied by the
downward acceleration imparted to that air (F =MA). With a n infinitely large
mass flow of air a n infinitely small downward acceleration would be required to
produce lift. The wing would therefore produce no downwash. Any increase of
angle of attack would increase the lift force to a n infinitely large value so the
C L : curve
~
would be vertical. Such a wing would however also have infinite
mass thereby negating the benefits described above.
LD 25. a.
Induced drag is caused by the creation of lift. Lift is created by the downward
acceleration of air due to the pressure differences above and below the wing.
These pressure differences leak around the wingtips as vortices causing upwash
outboard of the wingtips and downwash at the training edges. The upwash is
equal to the downwash so no additional lift is produced by the vortices but the
downwash over the trailing edge reduces angle of attack and hence lift. T o
restore lift it is necessary to increase pitching angle and this tilts the total
reaction aft increasing the drag component, This additional component is
induced drag. T o create induced drag a wing must produce lift and a cambered
wing will produce lift a t zero angle of attack. Also a symmetrical (zero camber)
wing will also produce lift and induced drag when at a non-zero angle of attack.
Pitching angle alone does not create camber or angle of attack and hence cannot
create lift o r induced drag. Of the options provided only option a includes all of
the essential elements of induced drag.
LD 26. a.
V Mis~the speed a t which total drag is a t its minimum value for a given weight
and configuration. Its location on the Drag:EAS curve is indicated by the EAS
at which the induces drag and profile drag components a r e equal, and total drag
is the sum of the two. Induced drag is therefore 50% of total drag when flying a t
VMD.
LD 27. c.
Power required = Drag x TAS and Drag = CD1/2~ V ' S where V = TAS
SO -power required = CD1/2p
~ x v2
~
This means that power required is proportional to V' x V which is v3
So power required is proportional to V3
In accelerating from 300 Kts to 400 I<ts TAS is increased to 133% of its previous
value.
So power required is increased to (1.33)~times its previous value.
So power required is 2.35 time sit previous value which is an increase of 1350h
LD 28. b.
Increasing weight requires an increase in lift to maintain any given flight profile.
Increasing lift creates additional induced drag but has little effect on profile
drag. Total drag is the sum of induced and profile so as weight increases total
drag increases. Minimum drag is the total drag when flying at VMDSO increasing
weight also increases minimum drag.
LD 29. b.
Drag is equal to CD%p v 2 s where
CD is the coefficient of drag
p is the local air density
V is the true speed of the free stream
airflow
S is the surface area
CD, V and S are not affected by any change in air density. Qrag is therefore
directly proportional to air density so if density is reduced by a factor of 4, drag
will decrease by the same factor.
LD 30. d.
Drag is equal to CD%p v 2 s and indicated airspeed is roportional to % p v 2 If
air density is reduced by half with no change in CD,V or S, then both drag and
indicated airspeed will also reduce by half. If however indicated airspeed is
maintained constant while density reduces by half then v2must be doubled such
that '/2 p v 2 remains constant. Doubling of v2will affect drag in the same
nianner so if density is halved while indicated airspeed remains constant then v2
will double and drag will also remain constant.
P
.
LD~31.b.
The CL: angle of attack slope is a straight line up to the stall so doubling the
intensity of the upward gust would double the increase in CL. A gust causing a lo
increase in angle of attack increased CL by 0.15 from 0.55 to 0.7. A gust of twice
this intensity would therefore increase CL by twice as much which is a n increase
of 0.3.
LD 32. c.
Induced drag Dl, is proportional to 1/v2 and the coefficient of induced drag CDI,
is proportional to l / v 4 . I n this case V is multiplied by 2, so Dl is multiplied 112~
which is %, and CD1is multiplied by l/z4 which is 1/16.
LD 33. b.
In a laminar boundary layer the air flows over the surface in thin layers. The
layer closest to the surface does not move relative to it and each succeeding layer
moves with a slightly higher velocity. Because the average velocity of the layers
close to the surface is low, the friction drag forces are very small. I n turbulent
boundary layer all of the individual layers become mixed in a disorder random
flow pattern with the average velocity of any given air particle being higher than
in laminar layers. Because of this higher velocity the turbulent flow passing over
the surface produces much higher friction drag forces. Turbulent boundary
layer therefore produces more drag than laminar boundary layer.
LD 34. c.
Because of its higher mean velocity turbulent boundary layer produces greater
friction drag than laminar boundary layer. Form drag is pressure differences
acting on the forward facing and rear facing surfaces of the structure and hence
are not affected by the nature of the boundary layer. Separation of the airflow is
the principal cause of aerodynamic stall so early separation cannot be considered
a benefit. The higher energy of turbulent boundary layer enables it to penetrate
further into the adverse pressure gradient over the upper surfaces of wings a t
high angles of attack. This delays separation and hence enables the wing to
operate a t higher angles of attack without stalling.
LD 35. c.
Weight is the force generated by the earth's gravity acting upon the mass of an
aircraft. I t always acts directly downwards toward the centre of the earth.
Although lift acts in the opposite direction to weight in straight and level flight,
this is not the case when an aircraft climbs, descends, o r turns. Option b is
therefore correct only in straight and level flight. Similarly lift always acts at
right angles to the flight path but weight does so only in straight and level flight.
Rate of turn has no effect on the direction of the weight force.
LD 36. b.
It is convenient to consider two possible definitions of the direction of.the lift
force depending on what aspect of flight is being considered. I n straight and
level flight lift opposes weight and because weight acts vertically downwards lift
is said to act vertically upwards. Although this condition is true only in straight
and level flight it satisfies a more generally applicable definition that lift acts a t
right angles to the flight path. This definition is used when considering
manoeuvres such as climbing, descending, gliding, banking and turning. It is not
however generally applied when considering the aerodynamic processes affecting
the generation of lift. Under these circumstances it is commonly accepted that
lift acts at right angles to the relative airflow. Because the generation of lift
creates downwash the relative airflow direction is not exactly the same as the
reciprocal of the flight path. In effect this means that the lift force is angled
backwards slightly. The direction of the thrust line depends only on aircraft
structural considerations and has no direct relationship to the direction of the lift
force. Of the options offered in this question, a, b and c are all correct to some
extent. Option b however is the one specified in JAR 25.
LD 37. b.
Lift in generated by the downward acceleration of air. This downward
acceleration is produced by pressure differences created above and below the
wings, which are in turn generated by the acceleration of airflow around the
curved surfaces of the wings.. At all normal angles of attack the static air
pressure both above and below the wing is lower than ambient. The pressure
drop above the wing is however the greater of the two. Lift is the vector sum of
the upward and downward forces exerted by these pressure differences. I t is
only at very high angles of attack that static pressure above ambient levels is
exerted on the lower surface of tlie wing. The majority of lift is therefore
produced by low pressure above the wings.
LD 38. c.
At low positive angles of attack the acceleration of airflow over the lower surface
of a wing causes the air velocity to increase. This in turn causes its static
pressure and temperature to decrease. The static pressure below a wing a t low
angles of attack is therefore lower than ambient.
LD 39. d.
Minimum glide gradient gives maximum glide range. Glide range = L:D ratio
nlultiplied by height so for a given height, maximum range and minimum
gradient a r e achieved when flying a maximum L:D ratio. That is to say when CL
/ CDis a t a maximum.
L D 40. b.
CDI = k c L Z / n ~ Where
CDI is the coefficient of induced drag
CL is the coefficient of lift
k is a coefficient reflecting wing plan form (1
for elliptical)
A is the aspect ratio
LD 41. c.
At very low altitudes the downward flow of air is restricted by the proximity of
tlie ground. The overall effect is a slight increase in static pressure below the
wing and a reduction in downwash which increases the effective angle of attack.
The reduction in downwash tilts the total reaction into a less rearward direction
reducing the drag force. Unless thrust is reduced this reduction in drag will
cause the aircraft to accelerate.
LD 42. a.
At very low altitudes the down'ward flow of air is restricted by the proximity of
the ground. The overall effect is a slight increase in static pressure below the
wing and a reduction in downwash which increases the effective angle of attack.
The reduction in downwash caused by ground effect is similar to that caused by
increasing aspect ratio so ground effect increases the effective aspect ratio of the
wing. These effects are lost on climbing out of ground effect so effective aspect
ratio reduces, returning to its true value.
LD 43. c.
At very low altitudes the downward flow of air is restricted by the proximity of
the ground, The overall effect is a slight increase in static pressure below the
wing and a reduction in downwkh which increases the effective angle of attack.
The reduction in downwash tilts the total reaction into a less rearward direction
reducing the magnitude of the induced drag force.
LD 44. b.
As altitude increases air pressure decreases causing the air to expand. The
overall effect of this process is that in a standard atmosphere air density a t 40000
feet is approximately !4 of its sea level value. IAS is proportional to xpv2so
climbing at constant IAS causes TAS to increase. In a standard atmosphere the
rate of change of TAS is such that at 40000 feet IAS is a p p r o x i ~ a t e l yhalf of
TAS. Profile drag is proportional to EAS' and at low speeds EAS is
approximately equal to IAS. Profile drag is therefore approximately
proportional to IAS'. In this case IAS is multiplied by % so profile drag will be
multiplied by (%12. Profile drag will therefore be multiplied by %.
LD 45. c.
The magnitude of ground effect reduces rapidly with height above the ground,
such that it is significant only at or below % wingspan height.
LD 46. d.
Induced drag is caused by the generation of lift so raising o r lower the
undercarriage does not affect it in any way. Undercarriage position does
liowever affect the frontal area and streamlining of the aircraft so raising it will
reduce profile drag. Because profile drag is proportional to EAS', the drag
reduction is most significant a t high speeds. This means that VMD,where induced
drag and profile drag are equal, occurs at a higher speed. An aircraft is speed
stable a t all speeds above Vh1~so increasing VMDreduces the proportion of the
speed range over which it is speed stable. Raising the undercarriage therefore
reduces profile drag and makes an aircraft less speed stable.
LD 47. b.
In order to generate lift a wing alters static air pressures such that pressure
above the wing is lower than that below it. The vector sum of the forces
generated by these pressure differences is the total reaction and lift is a part of
this total reaction. These pressure differences leak around the wing tips such
that air flows from the comparatively high pressure area below the wing to the
lower pressure area above. This causes spanwise flow from root to tip below the
wing and from tip to root above it. Options c and d are incorrect because in that
they specify the wrong directions for these spanwise airflows. Although pressure
differences do exist in front of and behind the wing, these contribute to drag but
do not affect wingtip vortices.
LD 48. a.
Aspect ratio is the ratio of wingspan to mean chord length so a high aspect ratio
wing has a greater wingspan and shorter chord than one of lower aspect ratio
and similar wing area. At any given airspeed a high aspect ratio wing will
therefore affect a greater mass of air than its low aspect ratio equivalent. Lift is
the (Newton's third law) reaction to the downward acceleration of air and is
equal to the mass of air accelerated multiplied by the downward acceleration
given to it. To produce any given lift force the larger mass of air affected by a
high aspect ratio wing will require a smaller acceleration. The magnitude of the
downward acceleration is proportional to the pressure difference multiplied by
the time during which it acts on the air. Because of its longer span and shorter
chord length a high aspect ratio wing produces the same lift force by giving a
larger mass of air a smaller acceleration. Wingtip vortices are caused by the
pressure differences leaking around the wingtips. The strength of these vortices
is proportional to the pressure difference multiplied by the time during which
they act on the air. The shorter chord length of a high aspect ratio wing reduces
the time during which the pressure differences act thereby reducing vortex
strength.
LD 49. b.
CDlis proportional to l / v 4 and within normal angle of attack ranges CDPis
approximately constant. Doubling the airspeed from 200 to 400 Kts will
therefore multiply CD1by l/z4 which is 1/16. CDPwill be unchanged.
LD 50. b.
Power required is equal to thrust multiplied by TAS. Thrust required to
maintain any given flight profile is proportional to drag so an increase in drag a t
any speed will cause a n increase in power required. I n ground effect the
proximity of the ground restricts the down flow of air thereby causing a n
increase in the effective angle of attack. This tilts the total reaction to a less
rearward direction reducing the magnitude of the drag force. On leaving
ground effect these benefits are lost so both drag and power required increase.
LD 51. d.
Dl is caused by wing tip vortices due to the leakage of pressure differences from
the lower to the upper wing surfaces around the wing tips. Although the
magnitude of these pressure differences is constant for a given lift force,
increased airspeed reduces the time during which they act upon any given mass
of air. This reduces the strength of the wing tip vortices and hence the induced
drag. Dl therefore reduces with increasing airspeed such that it is proportional
to l/vZ: Because profile drag is caused by friction and pressure forces generated
by the flow of air over the surface of the aircraft, it increases with the square of
airspeed. Dyis therefore proportional to v'.
LD 52. c.
Below VnlDthe major component of total drag is induced drag which is
proportional to 11~'. Although profile drag increases with v', this is a very
small part of total drag at low speeds. V2 is lower than VMDSO total drag is
therefore approximately proportional to the inverse of the square of speed when
accelerating to V2.
LD 53. b.
Because this question does not state ~ v l ~ i cdrag
h is to be considered, both induced
and profile must be considered. Although this is a basic drag question it is
complicated by the fact that decreasing IAS by a factor of 5 means multiplying it
by 115. Dp is proportional to V' so decreasing IAS by a factor of five will
multiply Dp by (115') which is 1/25. Dl is proportional to l / v Zso this mill be
multiplied by (1/(1/5)'), which is 25.
LD 54. b.
Because this question does not state which drag is to be considered, both induced
and profile must be considered. Although this is a basic drag question it is
complicated by the fact that decreasiiig IAS by a factor of 5 means multiplying it
by 115. Dl is proportional to 1/v2so this will Re multiplied by (1/(1/5)'), which is
25. Dp is proportional to V' so decl-casing IAS by a factor of five will multiply
Dp by (115~)which is 1/25.
LD 55. d.
CDI= k ~ ~ , ' / n AIf
. k and A remain unchanged their magnitudes will not affect the
change in Cul so they can be ignored. So if only CL is changed the change in CD1
will be proportional to the square of the change in CL. Increasing CL by 5 will
therefore multiply CDIby 5', which is 25.
I,D 56. a.
CDI= ~CL'/XA.If CL and k remain unchanged their magnitudes will not affect
the change in C1,1 so they can be ignored. So if only A is increased the change in
C Dwill
~ be proportional to the inverse of the change in A. Increasing A by a
factor of 2 will therefore multipiy C m 5~ 11'2,
LD 57. b.
Drag = CD%~V'S where
CD is the coefficient of total drag
p is the air density
V is the free stream TAS
S is the surface area
If all other factors remain unchanged, the magnitude of the total drag force will
vary directly with air density.
LD 58. a.
In laminar flow the boundary layer is considered to move over the surface in thin
layers, each moving a t a higher velocity than that below it. The boundary layer
extends from the surface to a point where its velocity is 99% Of that of the free
stream airflow. Because of the low viscosity of air, the overall thickness of this
type of boundary layers is small. In turbulent boundary layer all of the
individual layers are mixed, producing a disordered flow, with air moving in all
directions. Because of its disordered nature this is typically much thicker than
laminar boundary layer.
LD 59. a.
A vortex generator protrudes through the boundary layer into the free stream
flow. By creating a vortex it takes some of the free stream flow closer to the
surface where it is mixed with the boundary layer. The overall effect is to
produce a thicker higher energy boundary layer.
LD 60. b.
P always lower than
Because power required is drag multiplied by TAS, V ~ I is
VMD.Also because fuel flow in a propeller aircraft is proportional to power, the
lowest fuel flow and hence maximum endurance for such aircraft occurs a t VMp.
Fuel flow in a jet aircraft is proportional to thrust and best range requires the
best ratio of low fuel flow :TAS. This occurs at a speed higher than VMD.The
VMDfor any given combination of weight, configuration and load factor, is the
speed a t which the total drag force is at a minimum. This mean that for any
given weight VMDproduces the lowest drag force and hence the greatest ratio of
lift to drag. That is to say VMDprovides the best L:D ratio.
LD 61. b.
Drag = C D ' / ~ ~ V ~ S
Where
C D is the coefficient of total drag
p is the air density
V is the free stream TAS
S is the surface area
Decreasing pressure causes air to expand thereby reducing its density. If all
other factors remain unchanged this reduction in air density will cause a
reduction in total drag.
LD 62. b.
Flap retraction a t constant IAS reduces the magnitude of the lift force produced.
Induced drag is directly proportional to the lift force so flap retraction also
decreases the magnitude of induced drag. Retraction also decreases wing
camber and angle of attack, producing a small decrease in profile drag. The
combii~edeffect of these processes is a reduction in total drag.
LD 63. b.
Deployinent of fowler flaps increases the chord length but does not change wing
span. Aspect ratio is wing span divided by mean chord length so deployment
decreases aspect ratio. As fowler flaps deploy they also move the trailing edge
downwards increasing the angle of incidence of the wing.
For an elliptical wing k = 1.
So for an elliptical wing CDI = CL' / nA.
Inserting the values of C L and A provided gives CuJ = figures given in this
question.
Gives CDI = 2.5' / 47c
LD 65. b.
The CL : a curve is a straight line up to the stalling angle, so within this range if a
loincrease in a produces a n increase of 0.06 in CL, then a 5' increase in a will
produce an increase of ( 5 x 0.06 = 0.3) in CL.
Load factor = lift 1 weight and in straight and level flight this equals 1. So
assuming the aircraft was in straight and level flight prior to the gust its load
factor would have been 1. Although the weight of the aircraft is not given, its
load factor of 1 can be expressed as a ratio of CL to some figure representing
weight, provided the ratio of the two is equal to 1. That is to say the load factor
before the gust equals 1 = 0.44 / 0.44.
On this basis the load factor in the gust is equal to the new C L / 0.44 which is =
(0.44 + 0.3) / 0.44.
The new load factor is therefore 0.74 / 0.44 which is 1.68.
LD 66. d.
SO when CL = zero any value of k and A ,
At zero lift CL = zero. CDI = kL2/7c~,
will give a CD1of zero. CDImust therefore be zero for any wing at its zero lift
angle of attack. Option d is therefore the most accurate.
LD 67. c.
The stagnation point is the point close to the leading edge, a t which the airflow is
brought to rest. As angle of attack increases this point moves downwards and aft.
The transition point is the point at which the laminar boundary layer flowing
over the wing becomes turbulent. Its location depends on a number of factors
including leading edge shape, surface condition and airflow velocity. As angle of
attack is increased the acceleration over the upper surface of the leading edge
increases causing the transition point to move forward towards the leading edge.
The separation point is the point at which the boundary layer separates from the
uppel- -srrrfuce of the ming due to the magnitude of the adverse pressure gradient
against which it is flowing. Increasing angle of attack increases the magnitude of
this adverse pressure gradient causing the separation point to move forward.
LD 68. c.
When air flows over a n aerofoil the variations in its velocity produce an
imbalance between air pressures acting on the upper and lower surfaces. These
pressure differences produce a force which causes air to be accelerated
downward. Lift is the equal and opposite reaction to this force as described in
Newton's third law of motion. The above processes require both a n aerofoil and
an airflow, so options a, and b are inadequate. Similarly although dynamic
pressure is an essential aspect of the creation of the pressure difference above
and below the wing, these pressure differences require a ming on which to act in
order to create lift. Option c is therefore the most accurate.
LD 69. c.
Lift is the (Newton's third law) reaction to the downward acceleration of the
mass of air affected by a wing. This downward acceleration is produced by
pressure differences above and below the wing. These pressure differences also
cause air to migrate from the lower surface to the upper surface around the wing
tips producing wing vortices. These vortices cause air inboard of the tips to
accelerate downwards aft of the trailing edge, while an equal mass of air is
accelerated upwards outboard of the tips. Although the downward acceleration
of air inboard of the wingtips produces lift, the equal but opposite upward
acceleration outboard of the tips produces an equal but opposite (downward) lift
force. The overall effect of these vortices is an increase in drag due to the energy
lost in accelerating the vortex flow, but no increase in lift. By reducing the
effective angle of attack the vortex flow actually reduces the lift produced by the
wing.
LD 70. b.
When an aircraft is operating out of ground effect its wings produce downwash
which deflects the relative airflow downwards behind the trailing edge. This
reduces the effective angle of attack and also causes the total reaction to be tilted
backwards, increasing drag and decreasing lift. Ground effect restricts the
downward flow of air, thereby increasing the effective angle of attack and
reducing the rearward angle of the total reaction. In addition to this less
rearward tilting of the total reaction, the increased effective angle of attack also
increases its magnitude. Under normal circumstances the total reaction is never
tilted in a forward direction. Option b is therefore the most accurate.
LD 71. d.
2 up to the stalling angle cL
increases linearly with angle of
Lift = CL!4 p ~ and
attack. In straight and-level flight lift equals weight so to remain straight and
level as speed increases, CLmust be reduced to compensate for increasing v2in
order to maintain a constant lift force. This means that low angles of attack are
employed at high speeds.
Total drag is made up of two components, profile drag and lift induced drag.
Profile drag is proportional to V' whilst lift induced drag is proportional to 1/v2,
where V is the equivalent air speed (EAS). This means that as speed increases,
profile drag increases and induced drag decreases. At the high speeds associated
with low angles of attack, profile drag is large and induced drag is small. Shock
induced drag is significant only in the transonic and supersonic speed ranges.
The term vortex drag is a non-standard term sometimes used when referring to
lift induced drag.
LD 72. a.
Lift = CL !4 P ~ and
2 up to the stalling angle cL
increases linearly with angle of
attack. In straight and level flight lift equals weight, so to remain straight and
level as speed increases, CL must be reduced to compensate for increasing Vz in
order to maintain a constant lift force. This means that low angles of attack are
employed at high speeds.
Total drag is made up of two components, profile drag and induced drag. Profile
drag is proportional to v2,whilst lift induced drag is proportional to 1/v2, where
V is the equivalent air speed (EAS). This means that as speed increases profile
drag increases and induced drag decreases. At the low speeds associated with
high angles of attack profile drag is small and induced drag is large. Shock
induced drag is significant only in the transonic and supersonic speed ranges.
The term vortex drag is a non-standard term sometimes used when referring to
lift induced drag.
LD 73. d.
The principal cause of induced drag is the downward deflection of airflow caused
by wing tip vortices. Although the downward acceleration of air would normally
be expected to produce an upward lift force, the wing tip vortices produce an
equal and opposite upward flow of air outboard of the tips. The overall effect of
these upward and downward flows of air is not an increase in lift but a
downward deflection of the relative airflow passing over the wing. This
decreases effective angle of attack causing a reduction in the magnitude of the
total reaction, whilst simultaneously tilting it in a more rearward direction.
These changes cause a reduction in lift and an increase in drag. Although the
lost lift can be recovered by increasing the pitching angle of the aircraft to
restore angle of attack, this process tilts the total reaction still further rearwards
causing a further increase in drag. The additional drag produced by this
sequence of events is called lift induced drag.
LD 74. b.
Induced drag is caused by wing tip vortices which are in turn caused by the
pressure differences generated above and below a wing in order to create lift.
Although increasing angles of attack below the stalling angle increase both lift
and induced drag, the relationship reverses beyond the stalling angle with both
lift and lift induced drag rapidly decreasing. I t cannot therefore be said that
induced drag is caused by angle of attack. Induced drag is proportional to
~ / ( E A s )so
~ option c is also incorrect. Increasing aspect ratio reduces the
intensity of wingtip vortices and hence induced drag. Induced drag is not
therefore caused by aspect ratio. For any given combination of aspect ratio and
airspeed, both wingtip vortices and lift induced drag increase with increasing lift.
Induced drag is proportional to the square of the lift force as indicated in the
follolving formula.
CDI = kcL2/nA where
CDI is the coefficient of induced drag.
CL is the coefficient of lift.
K is a factor accounting for wing plan form.
A is the aspect ratio.
LD 75. b.
At any speed V,
CL = CL Mar (vsZ/ v 2 )
When V = 1.5Vs
this equation gives CL = CL Max (1 / 1.5~)
Which is 0.44 CL hlax which is 44% of CL Max.
LD 76. d.
lnduced drag is caused by the downward acceleration of air behind the wing.
This deflects the relative airflow downwards reducing effective angle of attack
and hence lift. In order to restore the original lift value the pitch attitude and
hence angle of attack must be increased. Although this restores the lost lift, it
also angles tlie total reaction in a more rearward direction, increasing the
magnitude of the drag force. The principal cause of the downwash is wing tip
vortices caused by air migrating around the wingtips from the lower to upper
surfaces. Upwash ahead of the wing tends to increase effective angle of attack
and so to some extent offsets the effects of downwash. Tip tanks produce a small
reduction in induced drag by reducing the intensity of the wingtip vortices.
High tailplanes reduce fin tip vortices so reducing the induced drag produced by
the fin.
LD 77. b.
Ground effect reduces downwash and hence lift induced drag. This reduction in
drag reduces the power required to maintain any given airspeed so the aircraft
will accelerate unless power is reduced on entering ground effect. Although this
reduction in power requirement is significant only when flying less than one wing
span above the ground there, is no height at which ground effect increases power
requirement.
LD 78. c.
Lowering the undercarriage increases profile drag. Undercarriage position does
not alter aircraft weight however and hence has no effect on lift induced drag.
Because of its position below the thrust line the extra drag produced by the
undercarriage produces a nose down pitching moment.
I,D 79. d.
Because this question does not refer to any specific type of drag it is necessary to
consider the effects on both lift induced and profile drag. Lift induced drag is
proportional to 1/v2,so doubling IAS multiples induced drag by l/(2)'. That is
the new induced drag is ?4 of its original magnitude. None of the options satisfies
this condition. Profile drag is proportional to v2so doubling IAS will multiply
profile drag by 2'. That is the new profile drag will be 4 times its original
magnitude. Option d specifies this result.
LD 80. b.
This question does not specific any drag type so both lift induced and profile
must be considered. Profile d r a g is proportional to v2SO tripling IAS will
multiply profile drag by 3' which is 9. This result is not specified in any of the
options. Lift induced drag is proportional to 1 / v 2so tripling LAS will multiply
induced drag by 113~.That is the new induced drag will he 119 times its previous
value. Option b specifies this result.
LD 81. b.
CDI = k ~ ~ ~and
l for
l r an~ elliptical wing k = 1.
Inserting the given values for CD1and A produces
Which gives C L = 2.24.
LD 82. a.
CDI = k c L 2 / l r ~where
So CDIis proportional to
CL = coefficient of lift.
k is a coefficient relating to plan form.
A is aspect ratio.
cLZ
11A
llk
Although C L is directly proportional to angle of attack u p to the stalling angle,
this relationship does not apply at greater angles. Of the options provided in this
question option a, is the most accurate.
LD 83. c.
The effect of increasing wing camber is to move the lift curve up and to the left
without altering its gradient. This causes a n increased C L at any given pre-sta!!
angle of attack, a reduced stalling angle and an increased C L max.
LD 84. d.
The effect of increasing wing camber is to move the lift curve u p and to the left
without altering its gradient. This causes an increased C L at any given pre-stall
angle of attack, a reduced stalling angle and an increased C L max. The increased
CL max, reduces the stalling speed for any given weight.
IJD 85. c.
Increasing air temperature reduces air density. Lift = C ~ ' / ~ ~where
V ~ pS is air
density. Lift is therefore proportional to air density. The reduction in air
density caused by increasing temperature will therefore reduce the lift force
produced a t any given combination of CL, V and S.
I,D 86- b.
Because water vapour is lighter than dry air, increasing humidity will reduce
ambient air density. Lift = cL'/2pv2 S, SO reducing air density will reduce lift for
any combination of CL, V and S. In order to maintain constant lift force in any
given combination of V and S, an increase in humidity will therefore require a
higher CL. Humidity has no effect on C L at any given angle of attack o r CL max
for any given wing section. Because of the reduction in air density, increased
humidity will also increase stalling speed and decrease profile drag.
I,D 87. c.
Because water vapour is lighter than dry air, increasing humidity will reduce
ambient air density. Lift = = C L % ~ V ' SSO reducing air density will reduce lift
for any combination of CL, V and S. In order to maintain constant lift force in
any given combination of V and $, an increase in humidity will therefore require
a higher CL. Humidity has no effect on C L at any given angle of attack or C1,,,,
for any given wing section. Because of the reduction in air density, increaw-'
humidity will also increase stalling speed and decrease profile drag.
LD 88. a.
In a steady climb lift = weight multiplied by the cosine of the angle of climb, so if
changes in weight are ignored a constant angle of climb requires a constant lift
force. IAS is proportional to %pv2 where p is air density and V is TAS. This
lileanrs that when climbing at constant IAS, '/zpvZ must also be constant. Air
density reduces with altitude, so when climbing a t constant IAS, the reduction in
density must be matched by an increase in TAS. This means that for a constant
TAS climb, IAS and hence %pv2 reduces. But lift is equal to C L % ~ VSO
* to
niaintain constant lift when climbing at constant TAS, C L must be increased to
match the reducing value of xpv2.Angle of attack must therefore be increased
to increase CL.
LD 89. a.
Lift is created by pressure differences above and below the wing, which cause air
to be accelerated downwards. Leakage of these pressure differences causes air to
migrate from the lower to upper surfaces around the wingtips. This migration
produces wingtip vortices. Induced drag occurs because these wingtip vortices
cause the relative airflow to be deflected downwards behind the wing, tilting the
total reaction to a more rearward inclination. This reduces lift and increases
drag. Increasing CL increases the magnitude of these pressure differences and so
increases induced drag.
Increasing aspect ratio increases wing span and reduces wing chord so a t any
given airspeed the mass of air affected by the wing is increased and the time
during which it is caused to leak around the wing tips is reduced. Increasing
aspect ratio therefore reduces the magnitude of the wingtip vortices and hence
reduces induced drag.
Statement 2 is incorrect because increasing sweep back angle decreases aspect
ratio and so increases induced drag. Statement 3 is incorrect because increasing
EAS increases the mass of air affected by the wing thereby reducing the
downward acceleration required for a given lift force. This in turn reduces the
time during which the air is driven into the wing tip vortices. So increasing EAS
reduces wingtip vortices and induced drag. Increasing CL increases the lift force
generated a t any given airspeed and hence increases the pressure differences
between the upper and lower surfaces. Increasing CL therefore increases wingtip
vortex strength and induced drag. Statements 1 and 4 are therefore correct as
indicated in option a.
LD 90. a.
When an aircraft flies close to the ground the proximity of the surface restricts
the downward flow of air thereby increasing effective angle of attack. This tilts
the total reaction in a less rearward direction increasing lift and decreasing drag.
At ally given airspeed, increasing aircraft weight will increase the magnitude of
tlie downwash. By restricting this greater downwash ground effect becomes
stronger at higli aircraft weights.
I,D 91. a.
Dp is the sum of friction drag, form drag and interference drag. Within the
normal operating range of angles of attack these do not vary significantly so Cop
is approximately constant. Dp = CDP%~V'S,so for any given combination of
CDP,V and S, Dp is proportional to v'. Statement 1 is therefore correct.
Although it is true that CDiis proportional to 1/v4,increasing @AS increases the
mass of air affected by the wing and so reduces the time during which it is
affected by the wing. This reduces wingtip vortex strength and hence induced
drag. Because of these effects, induced drag is proportional to l / v Zso statement
2 is incorrect.
An airspeed indicator produces an output that is directly proportional to % p v 2
so for a given %pv2 it will give a constant IAS. As altitude increases p decreases,
so to maintain a constant value of '/zpv2 at any given IAS, V must increase. But
V is TAS so this means that the TAS corresponding to any given IAS increases
with altitude. Induced drag = c D ~ ' / ~ ~and
v ~profile
s
drag is C D ~ ' / ~ SO
~ Vboth
~S
are proportional to IAS, rather than to TAS. But EAS is simply IAS corrected
for pressure sensing errors and compressibility errors, so both types of drag are
prol~ortionalto EAS. Statements 1 and 4 are therefore correct.
1,D 92. c.
CIll= ~ c L ~ / ~ Awhere
CLis the coefficient of lift.
k is a coefficient representing wing plan form.
A is aspect ratio.
Neither k nor A vary with angle-of attack so the upward gust will change only CL
2
and CDIso CDIis proportional to CL .
The new CLis 1.25 times the pre-gust CLSO the new CD1will be (1.25)~times the
pre-gust CDI.That means that the new CD1will be 1.56 times the pre-gust CD1.
CDI wrill therefore increase by 56%.
LD 93. a.
CDI = C ~ W ~ Awhere
CLis the coefficient of lift.
k is 1 for elliptical plan forms,
A is aspect ratio.
That is CDI= 4/5n.
LD 94. b.
At Vs L = CLM , , % ~ V ~and
~ S at any speed V,
so L is proportional to CLM,,V;
L = CL%PV~S
and to C L V ~
So CL>laxvS2is proportional to C L V ~
So at any speed V, CL/ CL M~~ = vS2/ v2
Which means that zt any speed V,
CL = (CLMax ) vs21 v2
So when V = 1.3Vs, then CL = (CLhlax)1/ 1.3~
That is CL = CL>lax/ 1.69 which simplifies to give CL = 0.59 CLMax
This means that at 1.3Vs, the CL is 59% of CLMax.
Note: All questions of this form can be solved using the following equation:
At any speed V, CL = (CLMax) V: / v2
LD 95. b.
When an aircraft flies close to the ground the proximity of the surface restricts
the downward flow of air thereby increasing effective angle of attack. This tilts
the total reaction in a less rearward direction increasing lift and decreasing drag.
At any given aircraft weight, increasing airspeed will decrease the magnitude of
the downwash. By restricting this less intense downwash ground effect becomes
weaker than at higher speeds.
LD 96. a.
Lift = C L ' / ~ ~ Vwhere
~ S p is air density and V is TAS. Increasing altitude causes
p to decrease. T o maintain constant lift force a t constarit TAS as altitude
increases it is therefore necessary to increase CL. But CD1is proportional to CL'
so increasing CL will increase CDIand hence induced drag. Increasing altitude a t
constant TAS therefore increases induced drag.
LD 97. b.
Induced drag is caused by wingtip vortices which are in turn caused by the
leakage of air pressure from the lower surface to the upper surface of a wing
around the wingtips. This leakage takes the form of wingtip vortices. The
strength of these vortices and hence the magnitude of induced drag is
proportional to the pressure difference between the upper and lower surfaces.
That is to say, increasing pressure differences increases both the vortex strength
and the induced drag force.
Lift is generated by accelerating a mass of air downwards and the magnitude of
the lift force generated equal to the mass of air affected multiplied by the
downward acceleration given to it. In straight and level flight lift must equal
weight so increasing weight requires either a greater mass of air o r a greater
downward acceleration rate.
The mass of air affected by the wing is proportional to its aspect ratio and
airspeed, so increasing either will reduce the downward acceleration required for
any given lift force. '~nduceddrag will therefore reduce with increasing aspect
ratio and/or increasing airspeed. For any given combination of aspect ratio and
airspeed, increasing lift requires an increase in pressure difference between the
upper and lower wing surfaces. This increased pressure difference increases
wingtip vortices and induced drag. Induced drag is therefore increased by,
increasing aircraft weight, decreasing airspeed and decreasing aspect ratio.
LD 98. c.
Load factor = lift / weight so for any given weight, increasing load factor requires
an increased lift force. At any given airspeed this is achieved by increasing angle
of attack, which in turn increases CL and CDI. Within the normal operating
range of angles of attack CDPand hence Dp do not change. Increasing load factor
does not therefore affect Dp but increases Dr. Because Dl is proportional to 1 / v 2
the increase is most significant a t low speeds so the total drag curve is pushed
upward and right towards the higher speed range. This increases drag at any
given speed and also increases VMD. Because any aircraft is speed stable a t
speeds above VMD,the increase in VhlDcaused by the increased load factor will
reduce the proportion of the speed range over which it is speed stable. This
means that speed stability is decreased.
Increasing load factor will therefore not affect Dr, will increase VMD,and will
decrease speed stability.
LD 99. d.
As the speed of an aircraft decreases it is necessary to increase angle of attack in
order to increase CLto maintain a constant lift force. The maximum value of CL
is CLhlaxand this occurs at Vs. As speed is reduce further beyond the stall both
CL and the lift force decrease rapidly as indicated in
tlie diagram. At the stall the airflow
separates from the upper surfaces of
the wing producing a highly turbulent
wake which causes a large increase in
CL
..
:.
..
..
profile drag. At 0.9 Vs the CLis
.
.
............. .........
therefore less than CLMax and CDis
greater than at 1.1 Vs when the wing
.
.
j
i.
.
has not stalled.
l.lVs Vs 0.9Vs
--+
:
:
Rearranging this equation gives CLis proportional to ~ C D I .
FLAPS 1. c.
Deployment of trailing edge flaps increases wing camber and angle of incidence.
This has the effect of increasing CL,which enables the required lift to be
produced a t a lower airspeed. The overall effect is an increase in CL and a
decrease in Vs, allowing the aircraft to take-off and land a t lower speeds.
FLAPS 2. c.
When slats are deployed, air from below the leading edge is accelerated through
the slot, thereby energising the upper surface boundary layer. This enables the
boundary layer to penetrate more adverse pressure gradients and hence to
remain attached to the wing a t higher angles of attack. These higher angles of
attack enable the required lift to be generated a t lower airspeeds. The overall
effect is energising of the boundary layer and a decrease in stalling speed (Vs).
FLAPS 3. c.
Under low speed flight conditions a cambered aerofoil produces a nose up
pitching moment when it is generating lift. Deployment of trailing edge flaps
increases camber of the rear section of the wing. This causes the centre of
pressure to move aft, thereby producing a nose down pitching moment. The
relative magnitudes of the pre-deployment nose up moment and the deployment
induced nose down moment depend upon the angle to which the flaps are
deployed. Flaps are typically deployed to maximum angle for landing, so the
overall effect is that the existing nose up moment is replaced by a nose down
moment.
FLAPS 4. c.
Deployment of trailing edge flaps increases the camber of the wing. The effect of
this is to increase the CL produced at any given angle of attack up to the stall,
and to increase the adverse pressure gradient over the rear of the wing. This
increased adverse pressure gradient causes separation and stall to occur a t a
lower angle of attack. The overall effect is a decrease in stalling angle.
FLAPS 5. b.
Deployment of trailing edge flaps increases the camber of the rear area of the
wing, causing the C of P to move aft, resulting in a nose down pitching moment.
Deployment of leading edge flaps increases camber a t the front area of the wings.
This moves the C of P forward, thereby reducing the effects of the nose down
pitching moment caused by the trailing edge flaps.
FLAPS 6. d.
Sweep induced span wise flow of air causes swept wings suffer tip stall which
results in wing drop and pitch up. Swept wing aircraft therefore require
measures to ensure that the wing root stalls before the tip. Because krueger flaps
are less efficient than slats. they cause the wing to stall at a lower angle of attack,
whereas slats tend to increase stalling angle. The required characteristic of wing
root stall prior to tip stall can be achieved by fitting krueger flaps inboard and
slats outboard.
FLAPS 7. a.
Flaps increases wing camber,
Deplcyment of trailing edge
thereby increasing the coefficient of lift at any angle of attack. This increased
coefficient of lift enables the wing to generate the required lift force at a lower
airspeed. The higher camber also means that the maximum coefficient of lift (CL
Mar) and wing stall, will be achieved at a lower angle of attack. The overall
effects of trailing edge flap deflection are therefore reductions in landing speed,
take-off speed, and stalling angle. The greater the flap deflection angle, the
greater will be the reduction in stalling angle.
FLAPS 8. a.
Although deployment of trailing edge flaps increases CL,it also increases CD. In
most cases the overall effect of deployment is a reduction in L : D ratio, so the
best L :D ratio is usually achieved with the flaps in the up (zero degree) position.
FLAPS 9. d.
In addition to rotating downwards to increase camber and angle of incidence,
fowler flaps also move aft to increase wing area. These effects combine to
increase the lift generated at any given speed, thereby reducing stalling speed.
The higher camber also reduces stalling angle and although lift is increased, drag
is increased by a greater proportion resulting in a decrease in L:D ratio.
FLAPS 10. c.
Deployment of trailing edge flaps increases the lift generated at any given
airspeed. This increase in lift, results in an increase in downwash over the
tailplane. Because flap deployment produces a nose down pitching moment, the
tailplane must produce a balancing tail down moment. To achieve this it
operates at a negative angle of attack when the flaps are down. The increased
d,ownwash caused by flap deployment increases this negative angle of attack,
thereby increasing the effectiveness of the tailplane in maintaining the correct
aircraft attitude.
FLAPS 11. a.
Trailing edge flaps a r e usually confined to the inboard section of the trailing
edge, with the outer section being taken u p by the ailerons. The extra lift and
consequent increase in pressure differential above and below the wing, caused by
n a p deployment is therefore confined to the inboard area of the wings. This
reduces the intensity of wingtip vortices in comparison to those that would exist
if the same amount of lift were generated at the same airspeed without flaps.
Additional vortices a r e also generated a t the tips of the flaps and these interfere
with and reduce wingtip vortices still further.
FLAPS 12. c,
Deployment of flaps results in large increases in the lift and drag produced by
the wings. Asymmetric flap deployment will therefore result in uneven lift and
drag distribution, causing strong rolling and yawing moments to be set up.
Because of the great size and effectiveness of modern flap systems, asymmetric
deployment is likely to result in loss of lateral and directional control,
particularly at low airspeeds where control authority is less effective.
FLAPS 13. d.
High angles of flap deflection generate highly adverse pressure gradients over
the upper surface of the flaps, causing a reduction in stalling angle. By
accelerating airflow through the slots, the boundary layer over the upper
surfaces of the flaps is energised, improving its ability to penetrate these adverse
pressure gradients. The overall effect is an increase in stalling angle and a
decrease in stalling speed.
FLAPS 14. b.
The initial part of the flap deployment process produces a large increase in lift,
with only a small increase in drag. The large increase in lift brings with it
additional induced drag, which constitutes most of the drag increase. During the
second half of the deployment the increase in drag is greater than the increase in
lift. This is partly due to the fact that most of the additional drag is profile drag.
FLAPS 15. c.
The coefficient of induced drag (CDI)is proportional to cL2.
Because flap
deployment always increases the coefficient of lift, it also increases the
coefficient of induced drag. Induced drag will therefore always be increased by
flap deployment.
FLAPS 16. a.
Deployment of inboard trailing edge flaps moves the lift envelope inboard, such
that a smaller proportion of the total lift force is generated by the outboard part
of the wings. The intensity of the wing tip vortices is proportional to the amount
of lift generated close to the wingtips, so flap deployment reduces wingtip
vortices. Deployed trailing edge flaps also produce their own flap tip vortices,
which interfere with the wingtip vortices lessening their intensity. The overall
effect of flap deployment is therefore reduced wing tip vortices.
FLAPS 17. c.
Although trailing edge flap deployment increases CL, it also increases CD. The
effect of these changes on the L : D ratio depends upon the relative magnitudes
of the changes in C L and CD. The majority of the benefit in terms of increasing
CLis produced during the first few degrees of deployment, whilst the greatest
proportion of the increase in C Doccurs during the last few degrees. Deploying
trailing edge flaps to the 45' position will therefore reduce the L :D ratio the
most. Deployment of slats produces very little additional drag and hence always
improves the L : D ratio.
FLAPS 18. c.
Deployment of inboard trailing edge flaps moves the lift envelope inboard, such
that a smaller proportion of the total lift force is generated by the outboard part
of the wings. This is undesirable in that it reduces the lateral stability of an
aircraft. In order to restore lateral stability, some aircraft employ ailerons that
droop when the flaps are deployed, effectively acting as outboard flaps.
Although these drooped ailerons increase lift, their primary purpose is to
maintain lateral stability.
FLAPS 19. a.
A split flap is one in which only the lower surface of the trailing edge is drooped.
This delays boundary layer separation, enabling higher angles of attack to be
employed without stalling the wings. Split flaps are therefore more efficient than
plain flaps.
FLAPS 20. d.
Flap deployment increases the CL at any given angle of attack and decreases the
stalling angle of attack. The increased C L enables more lift to be generated at
any given airspeed, thereby making it possible for the aircraft to fly a t lower
speeds without stalling. The increased CL also makes it possible to generate
higher load factors without stalling, thereby increasing the risk of overloading
the aircraft in turbulence. Flap deployment therefore decreases stalling speed,
but increases the risk of exceeding the limiting load factor in turbulence.
FLAPS 21. d.
Deployment of slats produces a convergent duct or slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the boundary layer increases the pressure drop over the forward part of the
wing, thereby moving the C of P forward towards the leading edge.
FLAPS 22. d.
During deployment Fowler Flaps move aft on tracks, increasing the wing area,
whilst their trailing edges move downward, increasing the angle of ipcidence and
camber of the wings. Options a, b and c in this question are incorrect because
each includes only one of the above effects, whilst excluding the others. Although
option d, is incomplete it does not exclude the possibility of the other benefits.
FLAPS 23. a.
The principal purpose of deploying flaps is to enable an aircraft to fly at lower
airspeeds during landing and take-off. During deployment they increase the
camber, angle of incidence, and angle of attack of the wings, thereby increasing
CL.
FLAPS 24. b.
Although trailing edge flap deployment increases CLit also increases CD. The
effect of these changes on the L : ~ i a t i depends
o
upon the relative magnitudes
of the changes in CLand CD. The majority of the benefit in terms of increasing
CLis produced during the first few degrees of deployment, whilst the greatest
proportion of the increase in CDoccurs during the last few degrees. The first few
degrees of flap deployment therefore increases the L : D of an aircraft.
FLAPS 25. c.
In Krueger flap systems the lower surface of the leading edge is hinged at the
front, moving down on deployment, to increase the camber of the leading edge.
Because of their somewhat rudimentary nature they are aerodynamically
inefficient, tending to stall relatively easily. This characteristic is often employed
to improve the post stall handing of aircraft, by employing Kruegers on the
inboard section of the wings and more efficient slats, or drooped leading edges
outboard. This configuration ensures that the wing roots stall before the tips,
thereby giving good pre-stall buffet and reducing wing drop in the stall. Full
span Kruegers will not however provide this benefit and hence will not affect
stability.
FLAPS 26. a.
A split flap is one in which only the lower surface of the trailing edge is drooped.
This delays boundary layer separation, enabling higher angles of attack to be
employed without stalling the wings. They have no effect on wing area.
FLAPS 27. d.
Flap deployment increases the CLat any given angle of attack, including CLMAX
at the stall, but decreases the stalling angle of attack. The increased CLenables
more lift to be generated a t any given airspeed, thereby making it possible for
the aircraft to fly a t lower speeds without stalling. Flap deployment therefore
decreases stalling speed and stalling angle, but iricreases CLMAX.
FLAPS 28. b.
Flap deployment increases the CLa t any given angle of attack, including CLM*x
a t the stall, but decreases the stalling angle of attack. This problem is sometimes
overcome by the use of leading edge slats. Deployment of slats produces a
convergent duct or slot between the leading edge of the wing and the slat. Air
passing through this slot is accelerated, increasing the energy of the boundary
layer passing over the upper surface. This energising of the boundary layer,
improves its ability to resist separation as angle of attack is increased. In this
way, slats increase the stalling angle of the wings.
FLAPS 29. a.
Flap deployment increases the CLa t any given angle of attack and decreases the
stalling angle of attack. The increased CLenables more lift to be generated at
any given airspeed, thereby making it possible for the aircraft to fly at lower
speeds witliout stalling. The increased CLalso makes it possible to generate
higher load factors without stalling, thereby increasing the risk of overloading
the aircraft in turbulence. In order to avoid overstressing the structure, the
limiting load factor of J A R certificated passenger aircraft is reduced to 2 when
the flaps are deployed.
FLAPS 30. b.
JAR 25 defines VFOas the maximum speed at which the flaps may be extended
o r retracted. I t is commonly termed maximum flap operating speed.
FLAPS 31. a.
J A R 25 defines VFEas the maximum speed at which an aircraft may be flown
wit11 its flaps extended.
FLAPS 32. c.
Deploying trailing edge flaps increases CLand moves the C of P aft. The
increase in CLincreases the downwash produced by the wings and this increased
downwash increases the negative angle of attack of the tail plane. The aft
movement of the C of P causes the aircraft to pitch nose down, further increasing
the negative angle of attack of the tail plane. The overall effect of this sequence of
events is that trailing edge flap deployment increases the tail plane down force.
FLAPS 33. b.
Icing conditions roughen the surfaces of the wings, making separation of the
boundary layer and stalling occur at a lower angle of attack. Deployment of
trailing edge flaps increases angle of attack and so might cause stalling if the
aircraft is already close to its stalling angle of attack.
FLAPS 34. b.
Deployment of slats produces a convergent duct o r slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack. If an aircraft is already flying close to its
stalling angle when slats are prematurely retracted, the decrease in stalling angle
might cause it to stall.
FLAPS 35. b.
Deployment of trailing edge flaps increases the angle of incidence of the wings,
thereby reducing the pitch attitude required to achieve any given angle of attack.
This reduces the pitching attitude that an aircraft must adopt when taking-off o r
landing.
FLAPS 36. a.
Deployment of leading edge flaps lowers the leading edge, causing the angle of
incidence of the wings to decrease. This means that a greater pitch up attitude is
required to achieve any given angle of attack. This effect is not great however,
because leading edge flap deployment also increases camber, thereby increasing
the CLa t any given angle of attack. This reduces the angle of attack required
during take-off and landing.
FLAPS 37. b.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate a t any given angle of attack. This
decreases the stalling angle of attack.
FLAPS 38. a.
Deployment of slats produces a convergent duct o r slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack.
FLAPS 39. b.
Although deployment of leading edge flaps increases the camber of the front
section of the wing and so might be expected to decrease the stalling angle, this is
not the case. This is because the angle of attack is measured between the relative
airflow and the chord line of the main section of the wins. Because leading edge
!laps lower the leading edge, they increase the angle to which the wing must be
raised to cause stalling. So leading edge flaps increase the stalling angle.
FLAPS 40. a.
I n Krueger flap systems the lower surface of the leading edge is hinged a t the
front, moving down on deployment to increase the camber of the leading edge.
Because of their somewhat rudimentary nature, they do not provide a smooth
leading edge curvature. They are therefore aerodynamically inefficient, tending
to stall relatively easily. Leading edge droop usually produces much smoother
leading edge curvature making it more aerodynamically efficient than Kruegers.
FLAPS 41. b.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate a t any given angle of attack. This
decreases the stalling angle of attack. In the blown flap system, eiigine
compressor air is blown through slots at the leading edges of the flaps in order to
energise the boundary layer and so decrease stalling speed.
FLAPS 42. b.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate a t any given angle of attack. This
decreases the stalling angle of attack. I n the slotted flap system, flap deployment
opens slots between the wings and the leading edges of the flaps. These slots
form convergent ducts, which accelerate air flowing from the lower surface, such
that it energises the boundary layer over the upper surface of the flaps. This
delays boundary layer separation and hence decreases stalling speed.
FLAPS 43. c.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate a t any given angle of attack. This
decreases the stalling angle of attack. In the blown flap system, engine
compressor air is blown through slots at the leading edges of the flaps, in order
to energise the boundary layer and so decrease stalling speed. Although this
system provides improved stalling speed, it requires a niore powerful engine in
order to provide the large volumes of compressed air used by the flaps, whilst
still producing the required thrust.
FLAPS 44. c.
Deployment of flaps increases wing camber, angle of incidence and angle of
attack, producing a large increase in the lift force generated by each wing. If the
flaps were to deploy asymmetrically only one wing would produce the extra lift
force and so the aircraft would roll towards the wing with the non-deployed
flaps. Because of the very large rolling moments that this configuration would
generate, it is probable that control of the aircraft would be lost.
FLAPS 45. a.
Deployment of flaps increases the camber, angle of incidence and angle of attack
of the wings, and so increases both CL and CD. The greatest increase in CLis
however produced during the first few degrees of deployment, while the greatest
increase in CDoccurs during the last few degrees.
FLAPS 46. b.
Deploying trailing edge flaps increases CLand moves the C of P aft. The
increase in CLincreases the intensity of the wingtip vortices, increasing the
downwash flowing from the trailing edge of the wings over the tail plane.
FLAPS 47. d.
During deployment Fowler Flaps move aft on tracks, increasing the wing area,
whilst their trailing edges move downward, increasing the angle of incidence and
camber of the wings. Because of this combination of aft movement and angular
deflection, Fowler flaps deploy more slolrly than split flaps.
FLAYS 48. c.
Deployment of trailing edge flaps prior to landing increases the camber, angle of
incidence and angle of attack of the wings. This increases CL at any given angle
of attack and so reduces the landing speed and nose up attitude required for
landing. If the flaps fail to deploy for landing then the angle of attack at any
given attitude will be reduced, so a higher nose up attitude and airspeed will be
required. Because. of the higher airspeed, the landing roll will also increase.
FLAPS 49. a.
Deployment of trailing edge flaps increases
the camber, angle of incidence and angle of
attack of the wings. The increased camber
moves the CL : a curve upwards and to the
right, while leaving its gradient unchanged,
as indicated in the diagram a t the right.
Although this decreases the stalling angle
of attack, the overall effect is that the
CL : a curve is extended.
FLAPS 50. a.
Deployment of leading edge slats energises
the boundary layer flowing over the upper
surface of a wing, thereby delaying
separation and stall. This does not affect
the gradient o r position of the CL : a
curve, but lengthens it, due to the delayed
stall. The overall effect of deployment of
leading edge slats is therefore to extend
the CL : a curve as illustrated in the
diagram at the right.
Extended curve with
t
Curve extended due to
slat deployment 'Y+*
+
+
a
FLAPS 51. c.
Deployment of Fowler flaps increases the area,
camber, angle of incidence, and angle of
attack of the wings. The increased camber
moves the CL: a curve upwards and to the
right, while leaving its gradient unchanged,
as indicated in the diagram at the right.
Although this decreases the stalling angle
of attack, the overall effect is that the
CL:a curve is extended.
Extended curve with
FLAPS 52. d.
The diagram shows double slotted fowler flaps and a leading edge slat, both of
which are in the deployed or semi deployed positions.
FLAPS 53. a.
The diagram shows a blown trailing edge flap and a leading edge slat, both of
which are in the deployed or semi deployed positions.
FLAPS 54. a.
The diagram shows double slotted fowler flaps and a leading edge slat, both of
which are in the deployed or semi deployed positions.
FLAPS 55. a.
The diagram shows a plain trailing edge flap in the take-off position and a
leading edge Krueger flap in the deployed positions.
FLAPS 56. c.
The diagram shows a split trailing edge flap and a slat, both of which are in the
deployed o r semi deployed positions.
FLAPS 57. d.
The diagram shows a leading edge slat and a plain trailing edge flap, both of
which are in the deployed or semi deployed positions.
FLAPS 58. a.
Deployment of trailing edge flaps moves the C of P of the wing aft and increases
the downwash passing from the wings to the tailplane. The aft movement of the
C of P produces a nose down pitching moment. The increased downwash over
the tailplane increases its negative angle of attack, increasing its down force and
hence producing a pitch up moment. The response of an aircraft to flap
deployment is therefore dependent upon the relative magnitudes of these two
competing effects. If a n aircraft pitches up when trailing edge flaps are
destroyed, it is because the effect of the increased downwash is greater than that
of the C of P movement.
t
FLAPS 59. d.
When leading edge slats are deployed they energise the boundary layer passing
over the top of the wing, thereby increasing both CLand CD. The increase in CL
is however much greater than the increase in CD,so slat deployment increases
the L : D ratio. Deployment of trailing edge flaps also increases both CLand Cu
but the effect on CDis much greater than that caused by slats. Trailing edge
tlap deployment therefore tends to decrease L : D ratio. If trailing edge flaps are
retracted whilst leaving slats deployed, both lift and drag will decrease, but the
L : D ratio will increase due to the removal of the high drag penalty imposed by
the trailing edge flaps.
FLAPS 60. d.
Deployment of slats produces a convergent duct or slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack. If an aircraft is already flying close to its
stalling angle when slats are prematurely retracted, the decrease in stalling angle
might cause it to stall.
FLAPS 61. d.
Deployment of trailing edge flaps moves the C of P of the wing aft and increases
the downwash passing from the wings to the tail plane. The aft movement of the
C of P produces a nose down pitching moment. The increased downwash over
the tail plane increases its negative angle of attack, increasing its down force and
hence producing a pitch up moment. The response of an aircraft to flap
deployment is therefore dependent upon the relative magnitudes of these two
competing effects. If an aircraft pitches up when trailing edge flaps are
deployed, it is because the effect of the increased downwash is greater than that
of the aft C of P movement.
FLAPS 62. b.
Deployment of trailing edge flaps moves the C of P of the wing aft, producing a
nose down pitching moment. Although this nose down pitching can be prevented
by increasing the down force generated by the tailplane, this solution is
inefticient in that it wastes some of the additional lift created by flap deployment
and so increases the stalling speed.
By deploying slats in conjunction with trailing edge flaps, a more efficient
solution can be provided. In addition to increasing CL,the slats t end to move
the C of P forward, thereby negating the nose down moment generated by the
trailing edge flaps. The overall effect of using slats in this manner is a decrease
in stalling speed.
FLAPS 63. b.
Deployment of flaps increases wing camber, angle of incidence, and angle of
attack, producing a large increase in the lift force generated by each wing. If the
flaps were to deploy asymmetrically, only one wing would produce the extra lift
force and so the aircraft would roll towards the wing with the non-deployed
flaps. Because flap deployment also increases CD,asymmetric deployment would
also cause the aircraft to yaw towards the wing with the flaps deployed.
FLAPS 64. c.
Deployment of trailing edge flaps increases the camber, angle of incidence, and
angle of attack of the wings. The increased camber increases the tendency of the
boundary layer to separate at any given angle of attack, so flap deployment
decreases the stalling angle of attack. But because flap deployment also
increases CL, it increases the lift force that can be generated at any given
airspeed and so decreases stalling speed. Trailing edge flap deployment
therefore decreases stalling speed and stalling angle of attack.
FLAPS 65. a.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate at any given angle of attack. This
decreases the stalling angle of attack. In the blown flap system, engine
compressor air is blown through slots at the leading edges of the flaps, in order
to energise the boundary layer. This delays separation and stall, enabling higher
angles of attack to be employed. Although this increases the lift force that can be
produced by the wings, the removal of air from the engine compressor decreases
the thrust available.
FLAPS 66. b.
Spoilers are hinged panel fitted to the upper surfaces of the wings of some
aircraft. When deployed they protrude into the airflow, causing the downstream
airflow to separate from the wing. This destroys the lift force that would
otherwise be created by that part of the wing and also increases drag. When
employed as lift dumpers, spoilers deploy in the landing run immediately after
touch-down. The overall effect of spoiler deployment is therefore a decrease in
the L : D ratio.
FLAPS 67. c.
Deployment of trailing edge flaps increases both the CLand CDof the wings. The
majority of the increase in CLis produced by the first few degrees of deployment,
whilst the majority of the increase in Cu occurs in the last few degrees. Small
angles of flap deployment can therefore increase L :D ratio, whilst large angles of
deployment decrease it.
FLAPS 68. a.
Deployment of trailing edge flaps increases the camber of the rear area of the
wings. In addition to increasing CLand hence increasing the amount of lift
produced by the rear wing area, it also moves the C of P aft.
FLAPS 69. c.
Deployment of leading edge flaps increases the camber of the forward area of the
wings. I n addition to increasing CL and hence increasing the amount of lift
produced by the forward wing area, it also moves the C of P forward.
FLAPS 70. c.
The angle of incidence is the angle between the chord line of a wing and the
longitudinal axis of the aircraft. Deployment of leading edge flaps causes the
leading edge of the wing to move downwards. This increases the camber of the
forward area of the wings and decreases the angle of incidence.
FLAPS 71. a.
When a wing is in flight the airflows around it are affected such that static
pressures below it are higher than those above it. These pressure differences
leak around the tips of the wing, forming wing tip vortices. The wing tip vortices
cause the airflow behind the wing to be deflected in a downward direction, whilst
the pressure differences above and below the wings cause air approaching the
leading edge to be deflected upwards. The magnitude of this up wash and
downwash of air is proportional to the CL of the wing. Deployment of trailing
edge flaps increases CL, intensifying the wing tip vortices and pressure
differences above and below the wing. The overall effect of these changes is an
increase in both up wash and downwash when flaps are deployed.
FLAPS 72. a.
Profile drag (Dp) is proportional to the frontal area of a wing and induced drag
( Dl) is proportional to C L ~Deployment
.
of flaps increases wing camber, arigle of
incidence, and angle of attack, thereby increasing both CL and the frontal area of
the wing. This increases both Dp and Dl. The majority of the increase in lift and
induced drag occurs during the first few degrees of deployment, whilst the
majority of the increase in profile drag occurs during the last few degrees.
FLAPS 73. c.
During deployment Fowler Flaps move aft on tracks, increasing the wing area,
whilst their trailing edges move downward, increasing the angle of incidence and
camber of the wings.
FLAPS 74. c.
A split flap is one in which only the lower surface of the trailing edge is drooped.
This delays boundary layer separation, enabling higher angles of attack to be
employed without stalling the wings. Split flaps are therefore more efficient than
plain flaps.
FLAPS 75. b.
For JAR certificatioi~purposes the minimum positive load factor that a
passenger aircraft must be capable of sustaining without permanent
deformation, in the flaps deployed configuration is 2g.
FLAPS 76. b.
For JAR certification a stick shaker system must activate a t a speed no lower
than 1.05 of its stalling speed, in both the flaps up and flaps down configurations.
Although flap deployment decreases stalling speed, it also decreases stalling
angle of attack, so the stick shaker will activate a t a lovier angle of attack when
the flaps are deployed.
FLAPS 77. a.
Flap deployment increases both the lift and drag forces generated by a n aircraft.
The relative magnitudes of the increases in lift and drag vary with deployment
angle, such that most of the extra lift is generated by the first few degrees of
deployment, whilst most of the extra drag is generated by the last few degrees.
During the take-off r u n an aircraft must accelerate to flight speed as quickly as
possible, so the requirement is for a flap angle giving maximum extra lift but
minimum extra drag. Conversely, in the landing run, additional drag is needed
to slow the aircraft after touch down. The normal configuration for landing is
therefore slats and flaps fully deployed.
FLAPS 78. b.
Flap deployment increases both the lift and drag forces generated by an aircraft.
The relative magnitudes of the increases in lift and drag vary with deployment
angle, such that most of the extra lift is generated by the first few degrees of
deployment, whilst most of the extra drag is generated by the last few degrees.
During the landing run the flaps are fully deployed in order to maximise drag
and hence decrease the landing run distance.
FLAPS 79. d.
Flap deployment increases both the lift and drag forces generated by an aircraft.
The relative magnitudes of the increases in lift and drag vary with deployment
angle, such that most of the extra lift is generated by the first few degrees of
deployment, whilst most of the extra drag is generated by the last few degrees.
During the take-off run a n aircraft must accelerate to flight speed as quickly as
possible, so the requirement is for a flap angle giving maximum extra lift but
minimum extra drag. Deploying flaps to small angles will therefore minimise
the take-off run. If however flaps are fully deployed during the take-off, the
extra drag will reduce the rate of acceleration and hence increase the take-off
distance required. Flap deployment can therefore either increase o r decrease the
take-off run depending on the angle of deployment.
FLAPS 80. b.
Deployment of trailing edge flaps increases camber and angle of incidence of a
wing. The increase in camber increases the CLa t any given angle of attack,
whilst the increased angle of incidence increases the angle of attack a t any given
pitch attitude. Flap deployment therefore decreases the pitch attitude required
in talie-off.
FLAPS 81. d.
Deployment of trailing edge flaps moves the C of P of the wing aft and increases
the downwash passing from the wings to the tailplane. The aft movement of the
C of P produces a nose down pitching moment. The increased downwash over
the tailplane increases its negative angle of attack, increasing its down force and
hence producing a pitch up moment. The response of an aircraft to flap
deployment is therefore dependent upon the relative magnitudes of these two
competing effects. If an aircraft pitches down when trailing edge flaps are
destroyed it is because the effect of the aft movement of the C of P is greater than
that of the extra downwash over the tailplane.
FLAPS 82. a.
In addition to increasing the lift produced by a wing, deployment of trailing edge
flaps also increases the magnitude of the drag force. Power required to maintain
an aircraft in straight and level flight is equal to the drag force multiplied by the
TAS. Because flap deployment increases drag, it also increases the power
required.
FLAPS 83. a.
When an aircraft is in the cruise condition its flaps are retracted and its angle of
attack is quite low, such that the lift force generated a t the relatively high
cruising speed, equals the weight of the aircraft. In order to maximise
aerodynamic efficiency in the cruise, aircraft are designed so that the stabiliser
trim and hence trim drag are as close to zero as possible. If the stabiliser trim
system were to stick in this position, it would not be possible to trim out the
strong nose down pitching moments that would be generated when the flaps were
deployed to their landing position. Considerable rearward pressure on the stick
would therefore be necessary to maintain straight and level flight, and much
higher forces would be necessary to rotate the aircraft into a nose-up attitude in
the flare.
FLAPS 84. d.
Because of their inboard location, deployment of trailing edge flaps moves the lift
envelope inboard, decreasing the rolling moments produced in sideslip. This
reduction in rolling moments results in a decrease in lateral stability.
FLAPS 85. a.
Because of their inboard location, deployment of trailing edge flaps moves the lift
envelope inboard, decreasing the rolling moments produced in sideslip. This
reduction in rolling moments, results in a decrease in lateral stability. To
overcome this problem some aircraft employ flaperons, which serve the dual
function of flaps and ailerons. By deflecting downwards when flaps are
deployed, they increase the lift produced by the outboard section of the Wings
and so restore lateral stability. Because a proportion of the aileron range is
already used in the drooped position, the roll authority of the aircraft is reduced.
FLAPS 86. a.
Profile drag (Dp) is proportional to the frontal area of a wing and induced drag
( Dl) is proportional to CL'. Deployment of flaps increases wing camber, angle of
incidence, and angle of attack, thereby increasing both CL and the frontal area of
the wing. This increases both Dp and DI. The majority of the increase in lift and
induced drag occurs during the first few degrees of deployment, whilst the
majority of the increase in profile drag occurs during the last few degrees. The
use of small angles of flap deflection therefore increases Dl more than Dp.
FLAPS 87. b.
Deployment of trailing edge flaps increases tlie camber, angle of incidence, and
angle of attack of the wings. The increased camber increases the tendency of the
boundary layer to separate a t any given angle of attack, so flap deployment
decreases the stalling angle of attack. But because flap deployment also
increases CL, it increases the lift force that can be generated at any given
airspeed and so decreases stalling speed. Trailing edge flap deployment
therefore decreases stalling speed and stalling angle of attack.
FLAPS 88. d.
In Krueger flap systems the lower surface of the leading edge is hinged a t the
front, moving down on deployment to increase the camber of the leading edge.
Deployment of krueger flaps also lowers the leading edge and hence decreases
the angle of incidence and the angle of attack a t ally given pitch attitude. This
increases the pitch attitude to which an aircraft can be taken before it stalls.
FLAPS 89. b.
In Krueger flap systems the lower surface of the leading edge is hinged a t the
front, moving down on deployment to increase the camber of the leading edge.
Deployment of krueger flaps also lowers the leading edge and hence decreases
the angle of incidence and the angle of attack a t any given pitch angle. This
increases the pitch angle to which an aircraft can be taken before it stalls. If
outboard krueger flaps were employed alone, then a larger proportion of the lift
would be produced by the outboard section of the wings. This would increase
wing tip vortices and hence reduce the effective angle of attack of the outboard
section. The inboard section would therefore reach its stalling angle before the
P
would therefore be
outboard section. The tip stalling tendency of L ~ wings
reduced.
FLAPS 90. a.
Deployment of slats produces a convergent duct o r slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack. If an aircraft is already flying close to its
stalling angle when slats are prematurely retracted, the decrease in stalling angle
might cause it to stall.
FLAPS 91. d.
A split flap is one in which only the lower surface of the trailing edge is drooped.
This delays boundary layer separation, enabling higher angles of attack to be
employed without stalling the wings. Split flaps are therefore more efficient than
plain flaps.
FLAPS 92. c.
JAR 25 defines VFOas the maximum speed at which the flaps of an aircraft can
be operated. This means that it is the maximum speed at which flaps can be
extended o r retracted.
FLAPS 93. b.
Flap deployment increases both lift and drag at any given flight speed. Whilst
the increased lift decreases the take-off speed, the increased drag decreases the
rate of acceleration. The greatest increase in lift occurs during the first half of
the deployment cycle, whilst the greatest increase in drag occurs during the
second half. For take-off the flaps are set a low angle of deflection in order to
maximise the ratio of increased lift to increased drag. If however the flaps are
fully deployed for take-off the decreased acceleration due to increased drag, will
outweigh the decreased take-off speed due to increased lift. The overall effect
will therefore be a longer take-off distance.
FLAPS 94. b.
Flap deployment increases both lift and drag at any given flight speed. Whilst
the increased lift decreases the take-off speed, the increased drag decreases the
rate of acceleration. The greatest increase in lift occurs during the first half of
the deployment cycle, whilst the greatest increase in drag occurs during the
second half. If full flap is employed in landing, the higher lift at lower airspeeds
will enable a steeper approach to be used.
FLAPS 95. d.
During deployment Fowler Flaps move aft on tracks, increasing the wing area,
whilst their trailing edges move downward, increasing the chord length, angle of
incidence and camber of the wings. By increasing the chord length the thickness
to chord ratio of the wings is decreased, such that the increased camber is less
likely to cause separation and stall. In this way Fowler flaps are able to produce
than other flap systems.
a greater increase in CLMAx
FLAPS 96. d.
Deployment of trailing edge flaps increases the camber of the wings, increasing
the tendency of the boundary layer to separate at any given angle of attack. This
decreases the stalling angle of attack. In the blown flap system, engine
compressor air is blown through slots at the leading edges of the flaps in order to
energise the boundary layer and so decrease stalling speed. Although this system
provides improved stalling speed, the bleeding away of large volumes of
compressed decreases the power available from the engine.
FLAPS 97. b.
Spoilers are hinged panels fitted to the upper surfaces of the wings of some
aircraft. When deployed they protrude into the airflow causing the downstream
airflow to separate from the wing. This destroys the lift force that would
otherwise be created by the aft part of the wing and makes the boundary layer
separate more easily. The overall effect of spoiler deployment is therefore a
decrease in the CL : a ratio and a decrease in a stall.
FLAPS 98. a.
Deployment of slats produces a convergent duct or slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack. Because slats normally move only in the
forward direction they do not significantly affect camber or the gradient of the
CL :a curve.
FLAPS 99. d.
Deployment of slats produces a convergent duct or slot between the leading edge
of the wing and the slat. Air passing through this slot is accelerated, increasing
the energy of the boundary layer passing over the upper surface. This energising
of the upper surface boundary layer increases its ability to resist separation and
so increases the stalling angle of attack. Deployment of trailing edge flaps
increases the camber of the rear area of the wing, encouraging separation, and
hence decreasing the stalling angle.
FLAPS 100. c.
The stalling angle of a wing depends upon a number of factors including surface
condition, camber, thickness to chord ratio and plan form. Typical figures are
1 2 degrees for wings with plain flaps deployed and 14 degrees with split flaps
deployed.
STALL 1. b.
In straight and level flight L = CL'/z ~ V ' S and at the stall (Vs),
X pv:s
L = CLMax
Except at high Mach numbers, changes in speed will not affect p and the wing
area S is constant
So at all such speeds L is proportional to CLV' and at Vs, L is also
proportional to CLMax VS'
But at constant weight in straight and level flight L is constant so C1,V' = CL
2
%la,VS
Rearranging this equation gives CL = CL
(v:
/ v2)
This means that at any speed (V) the CL = CLM~~ (VS' / v') x 100%
In this question V = 1.3Vs
STALL 2. c.
In straight and level flight L = CL?4p v 2 s and a t the stall (VS), L = CLM,,
!h pv:s
Except at high Mach numbers. changes in speed will not affect p and the wing
area S is constant
So at all such speeds, L is proportional to CLV' and a t VS; L is proportional
to CLMax VS'
But at constant weight in straight and level flight L is constant so CLv2 = CL
2
hlax VS
Rearranging this equation gives CL = CLhfax (v:
/ v2)
This means that a t any speed (V) the CL = CL%lax (vS2/ v') x 100%
STALL 3. b.
In order for an aircraft to be longitudinally stable its C of G must be forward of
the neutral point. This ensures that the lift / weight couple produces a nose down
pitching moment. In order to prevent nose down pitching, the tailplane must
produce a balancing nose up moment by generating a negative lift o r down force.
The lift force from the wings must therefore equal the weight of the aircraft plus
the down force from the tailplane. As C of G is moved forward the aircraft
becomes more stable but at the cost of an increased tailplane down force and
hence a greater total lift force from the wings. This increased wing lift requires a
greater CI, and hence greater angle of attack at any given speed. So the aircraft
will reach its stalling angle of attack at a higher speed when the C of G is moved
forward. Moving the C of CJ forward therefore increases the stalling speed of the
aircraft a t any weight.
Lift acting upward through C of P must equal the sum
of both the aircraft weight and the tailplane down force.
LIFT
CofG
I
t
CofP
WEIGHT
C
Balancing nose up
moment is produced by
tailplane down
force.
Nose down moment is product of weight and
distance between C of G and C of P so moving
C of G forward increases this nose down
moment. This requires an increased tailplane
down force to produce a balancing nose up moment.
STALL 4.d.
The magnitude of stabilizing moments produced by the various parts of an
aircraft are tlie product of the lift force generated by those components and the
distance of their centres of pressure from the C of G of the aircraft.
Longitudinal and directional stabilizing moments are provided mainly by the
tailplane and the fin. Forward movement of the C of G increases the distances
from the C of G to the Centres of Pressure of both these components, thereby
increasing the magnitude of the stabilizing moments they generate. This
increases both longitudinal and directional stability. Lateral stability is provided
primarily by wing dihedral, sweepback and height above o r below the C of G.
Forward and aft movement of the C of G has no effect on the lateral distance
between the C of G and the Centre of Pressure of the wings and hence has no
effect on lateral stability. Forward movement of the C of G therefore increases
longitudinal and directional stability but has no effect on lateral stability.
STALL 5. d.
The stability of an aircraft is a measure of its tendency to return to its original
condition after a disturbance. A highly stable aircraft exhibits a strong tendency
to return, a neutrally stable aircraft remains in the disturbed condition, whilst an
unstable aircrafts tends to continue to deviate after the disturbing force has been
removed. T h e manoeuvrability of an aircraft is a measure of its tendency to
change attitude under the influence of control inputs. Stability is therefore the
opposite of manoeuvrability and hence a highly stable aircraft exhibits poor
manoeuvrability. Increasing stability reduces manoeuvrability. Forward
movement of C of G increases longitudinal and directional stability but has no
effect on lateral stability. This reduces longitudinal and directional
inanoeuvrability without changing lateral manoeuvrability.
STALL 6. c.
An increase in the weight of an aircraft requires a corresponding increase in the
lift generated in flight. The equation used for calculating the new stalling speed
after a weight change is as follows:
Stalling speed a t new weight = Stalling speed a t previous weight times d (New
weight 1
Previous weight)
If the weight is doubled then (New weight I Previous weight) = (211).
So the stalling speed a t the new weight = stalling speed a t previous weight
multipled
by d (211).
4 (211) = 1.41 so the stalling speed a t the new weight is 1.41 times the stalling
speed a t the previous weight.
The doubling of weight will therefore have increased stalling speed by
approximately 41%.
STALL 7. c.
The stick shaker on JAR Certificated passenger aircraft operates at 105% of Vs.
The CLa t any speed (V) is CLMax (VS' 1 v 2 )
When the stick shaker activates V = 1.05Vs so the CL = CLMAX (12 1 1.05~)
So CL = 90.7% CLMa, o r approximately 91% CLMaa.
STALL 8. d.
IAS is derived from dynamic pressure (112 PV~).As altitude is increased air
density (p) will decrease so in a constant IAS climb the V in the equation (which
is true airspeed) will increase. This means that TAS will increase as the aircraft
climlii. At the low speed stall, lift = CL~1.. 10 p v 2 S. The air density ( p) and
the TAS (V) in this equation alter in the manner described above so throughout
most of the climb the Indicated stalling speed remains unchanged. As the climb
continues the increasing TAS reaches &Icrita t which point the effects of
compressibility reduce both CLM,, and the stalling angle of attack of the aircraft.
This reduction in stalling angle causes an increase in stalling speed. The overall
effect therefore is that indicated stalling speed remains constant throughout most
of the climb then increases a t M crit.
STALL 9. a.
The low speed stall is the point at which the CLreaches its maximum value
before reducing rapidly as angle of attack is further increased. This condition is
indicated by the point A on the diagram.
STALL 10. b.
The Best L:D ratio occurs at minimum drag speed (VklD). I t is indicated by the
point B on the diagram.
STALL 11. a.
VAis the maximum speed at which it is possible to apply full nose up control
deflection without exceeding the limiting load factor. This means that at VAthe
aircraft will stall whilst subject to the limiting load factor. The relationship
between stalling speed at any given load factor (VM)and the l g stalling speed
(Vslg)is as follows:
In the case of the limiting load factor (n Limit),VMis the design manoeuvre speed
and is referred to as VA. Its relationship to Vs is:
VA = Vslg d(n LimiJ.
STALL 12. c.
The CL : Angle of attack curve is a straight line up to CL M,, which occurs a t the
stalling angle, beyond which CL falls rapidly with increasing angle of attack.
This means that the rate of increase of CL with angle of attack is constant up to
the stalling angle. The range between the zero lift angle at -4 degrees and the
stall at 16 degrees represents a total of 20 degrees. The 6 degree angle is the
midway point between zero lift and CLM,, and so the CL a t this point is 50% of
CL Mar.
STALL 13. b.
The C,.: Angle of Attack curve is a straight line u p to the stalling angle so if a 1
degree increase gives a 0.06 increase in CL, a 5 degree increase will give five times
as much which is 0.3 . The new C L under these conditions will therefore be 0.44
+ 0.3 = 0.74.
Load factor is L/W and although the aircraft weight is unknown the load factor
in straight and level flight is always 1 so both lift and weight must be
proportional to the existing CL of 0.44. The new load factor can therefore be
calculated by dividing the new CL by the straight and level flight CL.
This means that in a gust the new load factor = (New CL) / (Straight and level
CL).
So new load factor = 0.74 / 0.44
=
1.68.
STALL 14. c.
The 2.5g stalling speed for a JAR certificated passenger aircraft is referred to as
the Design Manoeuvre Speed (VA) This can be calculated using the equation :
VA = VSlg4 (n
where n is the load factor which is 2.5.
So VA = (100 4 2.5) Kts
=
158 Kts.
STALL 15. a.
The Design Manoeuvre Speed (VA)is the EAS a t which the stalling speed curve
crosses the 2.5 g load factor line. At lower speeds full nose up control inputs will
cause the aircraft to stall a t a lower load factor. Higher speeds will cause a
higher load factor before stalling. VAis therefore the Design Manoeuvre Speed
which is the maximum EAS at which full nose up control deflection can be
employed witllout exceeding a load factor of 2.5.
STALL 16. a.
At the stall, lift = CL112 p v 2 s and Indicated Airspeed (IAS) is derived from
dynamic pressure which is 112 pv2. In both of these equations p is air density
and V is the TAS. As altitude increases the reducing air density affects both the
Lift produced and IAS equally so the Indicated Stalling Speed remains constant
provided speed does not approach M crit.
To maintain any given IAS with increasing altitude 112 p v 2 must remain
constant, so TAS must increase to compensate for the reducing density. This
means that the ratio of TAS:IAS increases with altitude so although the IAS
stalling speed remains constant the TAS stalling speed increases.
At 40000 feet TAS is approximately twice IAS.
STALL 17. c.
~
than high aspect
Because low aspect ratio wings have shallower C L :curves
ratio wings, their C L varies less with changes in ct brought about by gusty
conditions and hence will give a smoother ride. Although flap deployment does
not normally alter the slope of the CL:a curve it increases CL a t any given angle
of attack and reduces stalling angle, so wings with flaps down are more likely to
stall in gusty conditions. The smoothest ride in gusty conditions will therefore be
provided by a low aspect ratio swept wing with flaps up.
STALL 18. d.
The principle hazards in such conditions are stalling and exceeding load factor
limits. Deployment of flaps decreases both stalling angle and stalling speed so
gust induced changes in angle of attack are more likely to result in stall with
flaps down. Deployment of flaps also increases CL M ~ making
,
the exceeding of
limiting load factors more likely. Lowering of landing gear will increase speed
stability thereby minimising gust induced speed changes.
STALL 19. c.
VS3. = vslgdn Where Vslb is the 3g stalling speed, Vag is the l g stalling speed
and n is the load factor
Which is
Vsl, = 200 Kts 1 43
Which resolves to the l g stalling speed = 115 Kts.
STALL 20. c.
CD1= kcrA2
lnA Where k is the induced drag factor which is 1 for elliptical
wings, and A is the aspect ratio
Rearranging this equation gives
A = C? 1 nCDI
Which is A = 2.52 I 0.351~
This resolves to A = 5.7
Note: forgetting the that the CL is squared will give the wrong answer (2.27)
which is option a above.
STALL 21. b.
CDI = C L IKA
~ Where A is the aspect ratio
Rearranging this equation gives
CL = d (AnCD1)
Which is CL = d ( 2 . 4 ~ )
So CL = 2.75.
STALL 22. a.
If stalling speed at the new weight is VS2,stalling speed a t the original weight is
Vsl, new weight is Wz and original weight is W1 then:
Which is VS2 = VS1,d(2 / 1)
This resolves to VS2= l.41Vs,, which means that the new stalling speed is 41%
greater than the original stalling speed.
STALL 23. c.
The aircraft stalls a t 100 Kts when subject to a load factor of 3g. Increasing
bank angle in level flight increases load factor so the maximum bank angle
achievable at 100 Kts is that which will produce a load factor of 3g.
In a level banked turn load factor is 11Cos AOB so a load factor of 3 is achieved
when the Cos AOB = 113 giving an angle of 70.5 Degrees.
STALL 24. c.
Limiting load factor in a JAR certificated passenger aircraft is 2.5 and in a level
banked turn the load factor is 1 1 Cos AOB.
This means that the required bank angle is that for which the Cosine = 1 I 2.5 or
0.4
The required bank angle is therefore 66 degrees.
STALL 25. d.
If stalling speed at the new weight is VS2,stalling speed at the original weight is
Vsl, new weight is Wz and original weight is W1 then:
Which is Vs2 = 100 Kts d(60000 I 20000)
Which resolves to Vs2 = 173 Kts.
STALL 26. b.
If stalling speed in a turn is VM, and stalling speed in straight and level flight is
Vslg, then:
VM, = V S I , ~(1 I COSAOB)
So VM, = 100 Kts d (1 I Cos 30)
Which resolves to give stalling speed in a 30 degree banked level t u r n = 107 Kts.
STALL 27. a.
The wash produced by propellers increases the velocity of the airflow and hence
dynamic pressure over the area of the wing behind the propellers. This increases
the stalling speed of that part of the wing. In the event of engine failure this
increased dynamic pressure is lost resulting in a loss of lift on that wing. A single
engine failure in a twin propeller aircraft causes one wing to produce a greater
lift force than the other, causing wing drop. In extreme cases the loss of
additional airflow may cause that wing to stall making wing drop more severe, so
option b is incorrect. In a four engine propeller aircraft only a part of the extra
airflow is lost so the tendency to wing drop is less. This phenomenon does not
affect jet powered aircraft so options c and d are incorrect.
STALL 28. b.
The propeller slipstream increases the speed of the air flowing over the wing
thereby increasing its dynamic pressure. This increased dynamic pressure
enables a higher lift force to be generated at any given angle of attack, so a lower
angle of attack is required at any given airspeed. The stalling speed of the wing
is therefore reduced. The high speed slipstream also increases the energy of the
boundary layer enabling it to penetrate more adverse pressure gradients before
separation. In this way it increases the stalling angle and CLRlar.
STALL 29. c.
Stalling speed (VM)in a manoeuvre is the l g stalling speed multiplied by the
square root of the load factor. In a constant altitude bank, the load factor is
(1ICos AOB)
So in a 15' bank VM= VSlgd (11Cos 15')
Rearranging this gives Vslg = Vhq/ d (1ICos 15')
Using given value of VMand Cos 150this simplifies to give Vw, = 100 Kts / d
(1.035276)
Which is VSlg = 98.2815 Kts.
Using the original formula and inputting the value of Cos 60' and the above
value for Vslg gives:
VMat 60' = (98.2815) Kts d (1ICos 60')
= 138.99 Kts.
STALL 30. c.
The principal cause of pitch up when a swept wing stalls is that the wing tips
tend to stall before the wing roots. Because the wing tips are further aft than the
roots, the loss of lift at the tips causes the C of P to move forwards and inwards
at the stall. This forward movement of the C of P produces a nose up pitching
moment.
STALL 31. c.
If both wings stall simultaneously a straight winged aircraft will pitch nose
down, causing it to accelerate out of the stall. If however one wing stalls before
the other, then the aircraft will roll and yaw towards the stalled wing. This
process is called autorotat:--- and in severe cases will lead to a spin. In this
situation if the ailerons are deflected to oppose the roll, this will increase the
angle of attack and camber d the stalled wing taking it deeper into the stall. The
correct recovery action is therefore to centralise the stick laterally, and push it
fully forward, whilst simultaneously using opposite rudder to arrest the yaw.
The use of power varies with aircraft type but in this case the actions listed above
are listed only in option c which also includes the use of full power. I t must
therefore be assumed that this question refers to an aircraft in which the use of
full power is appropriate.
STALL 32. d.
The principal cause of pitch up when a swept wing stalls is that the wing tips
tend to stall before the wing roots. Because the wing tips are further aft than the
roots, the loss of lift at the tips causes the C of P to move forwards and inwards
at the stall. In the case of a straight wing the location of the initial stall depends
upon wing plan form and aspect ratio. Whatever the spanwise location of the
initial stall however, it will always cause the C of P of a straight wing to move aft.
STALL 33. b.
Low speed buffet is caused by the initial separation of airflow over a wing as it
begins to stall. As aircraft speed is reduced, the low speed buffet commences at a
speed just above stalling speed. Decreasing aircraft weight decreases stalling
speed and hence decreases the speed a t which buffet commences. Fuel burn
during flight will therefore gradually reduce the speed at which low speed buffet
commences.
STALL 34. c.
JAR 25 specifies that low speed buffet must not be detectable a t speeds above
1.2vs.
STALL 35. a.
This problem can be solved using the following equation:
Vs (in ally manoeuvre) = VSlgdn
where'n is the load factor in the manoeuvre.
For a 1.5g turn this gives Vs( in a 1.5 g turn) = Vslg d(1.5)
STALL 36. a.
Stalling speed at new weight (VNew)= stalling speed a t old weight (VOld)
multiplied by the square root of the new weight divided by the square root of the
old weight.
-
That is V N =~VOld
~ d( 48000N 1 24000N )
Which is VNew= 150 Kts x 42 which is 150 Kts x 1.414
This resolves to give VReW= 212 Kts.
STALL 37. a.
VA is the speed a t which an aircraft will stall whilst simultaneously achieving its
limiting load factor. T h e stalling speed a t any given load factor is equal to
vslgdn, where n is the load factor. Applying this equation to calculate VA
therefore gives the following equation:
STALL 38. d.
The term deep stall refers to a situation in which an aircraft pitches nose u p in
the stall. This upward pitching coupled with loss of height due to loss of lift from
the stalled wings, increases angle of attack causing the aircraft to go deeper into
the stall. The pi-incipal cause of deep stall is tip stalling of swept back wings.
STALL 39. c.
The term deep stall refers to a situation in which an aircraft pitches nose up in
the stall. This upward pitching coupled with loss of height due to loss of lift from
the stalled wings, increases angle of attack causing the aircraft to go deeper into
the stall. The principal cause of the initial pitch up which leads to deep stall, is
tip stalling of swept back wings. Once an aircraft has entered deep stall, its
ability to recover depends upon the effectiveness of the tailplane. Because of the
nose u p attitude and downward motion of an aircraft i s deep stall, a high
tailplane is likely to become enveloped in turbulent aikflow from the stalled
wings. This will reduce the effectiveness of the tailplane making a worsening
deep stall more probable.
STALL 40. d.
The stalling speed of an aircraft in straight and level flight is the speed at which
it is just capable of producing lift equal to its weight.
At the stall lift is equal to C L M A X ~ I ~ P V ' S
where
CLSlAX
is the maximum attainable coefficient of lift
1/2pvZis the dynamic pressure
S is the wing area.
o any given combination of
Indicated airspeed is proportional to 1 1 2 p ~ 2 sfor
CLRIAX
and S the lift force will be constant at any given indicated airspeed. This
means that for any given weight, the stalling indicated stalling speed is constant
a t all altitudes.
At higher altitudes however the TAS equating to any given IAS increases such
that stalling IAS approaches the local speed of sound. At such altitudes the
effects of compressibility decreases the stalling angle of attack and CLMAX.This
causes indicated stalling speed to increase a t high altitude.
reduction of CLMAX
The overall effect therefore is that indicated stalling speed remains constant
throughout most of the altitude range before increasing slightly at very high
altitude.
STALL 41. b.
In order for an aircraft to be longitudinally stable its C of G must be forward of
the neutral point. This ensures that the lift I weight couple produces a nose down
pitching moment. In order to prevent nose down pitching, the tailplane must
produce a balancing nose up moment, by generating a negative lift or down
force. The lift force from the wings must therefore equal the weight of the
aircraft plus the down force from the tailplane. As C of G is moved aft the
aircraft becomes less stable but the tailplane down force decreases thereby
reducing the total lift force required from the wings. This reduced wing lift can
be generated a t a lower speed. The stalling angle of attack is however a function
of wing shape and surface condition and is not therefore affected by C of G
position. The overall effect of placing the C of G on its aft limit is to decrease
stall speed but leave stalling angle unchanged. The effect of C of G on the
magnitude of the required lift force is illustrated overleaf.
L
CofG
I
T = Weight + Tailplane down force
7
I
1
C of P
WEIGHT
Nose down moment is product of weight and
distance between C of G and C of P so moving
C of G rearward decreases this nose down
moment. This requires a decreased tailplane
down force to produce a balancing nose up moment.
Balancing nose up
moment is produced by
tailplane down
force.
-.
STALL 42. b.
Deployment of trailing edge flaps increases wing camber and angle of incidence.
The increased camber increases CL at any given angle of attack, whilst the
increased angle of incidence increases the angle of attack corresponding to any
given aircraft pitch attitude. Flap deployment therefore increase CL and angle of
attack at any given aircraft pitch attitude.
V ~ S ,CL is the coefficient of lift
Lift equals C ~ ~ I ~ ~where
1/2pv2 is proportional to indicated airspeed
S is the wing area
The increased CL produced by flap deployment
therefore reduces the speed necessary to achieve
a lift force equal to the weight of the aircraft.
The stalling speed of the aircraft is therefore
decreased. Stalling angle is a function of camber
such that increasing camber decreases stalling
angie of attack.
The overall effect of deploying flaps is therefore
to decrease both the stalling speed and the
stalling angle of attack. The effect on stalling
angle and CL a r e illustrated in the diagram at
the right.
Flaps deployed
Flaps retracted
:
:
stalling angle
STALL 43. a.
Airflow over swept wings tends to migrate outwards producing a thick boundary
layer over the wing tips. This thickened boundary layer separates as angle of
attack is increased, causing the tips to stall before the roots. Because the tips are
aft of the roots, the early tip stall moves the C of P inwards and forwards at the
stall.
STALL 44. d.
Vortex generators are often fitted to the upper surfaces of aircraft wings where
they protrude through the boundary layer and into the free stream air above.
By creating small vortices they mix the low energy boundary layer with the
higher energy free stream air, thereby increasing the energy content of the
boundary layer. This energisation of the boundary layer serves a number of
purposes. At high angles of attack in low speed flight the higher energy
boundary layer in the vortices is able to resist the outward drift of air, which
would otherwise lead to premature tip stall. It should however be noted that the
spanwise flow and subsequent tip stall are delayed rather than being prevented
by the vortex generators. At high speeds the higher energy boundary layer is
more able to resist separation behind the shock waves that form on the upper
surfaces of the wings. The vortex generators therefore minimise the shock
induced separation and subsequent drag rise.
STALL 45. c.
In order for a n aircraft to be longitudinally stable its C of G must be forward of
the neutral point. This ensures that the lift /weight couple produces a nose down
pitching moment. In order to prevent nose down pitching the tailplane must
produce a balancing nose up moment by generating a negative lift o r down force.
The lift force from the wings must therefore equal the weight of the aircraft plus
the down force from the tailplane. As C of G is moved forward the aircraft
becomes more stable but the tailplane down force increases thereby increasing
the total lift force required from the wings. The generation of this is increased
wing lift requires the aircraft to fly a t a higher speed so the stalling speed of the
aircraft is increased. Although the increased stalling speed is accompanied by
greater longitudinal stability, it should be noted that the increased stalling speed
is caused by tlie forward movement of the C of G and consequent increase in
tailplane downward lift, rather than by the improved stability. An aft C of G
would reduce both longitudinal stability and stalling speed.
STALL 46. b.
In order for a n aircraft to be longitudinally stable its C of G must be forward of
tlie neutral point. This ensures that the lift I weight couple produces a nose down
pitching moment. In order to prevent nose down pitching in a conventional
aircraft the tailplane must produce a balancing nose up moment by generating a
negative lift or down force. The lift force from the wings must therefore equal
the weight of the aircraft plus the down force from the tailplane. As C of G is
inoved forward the aircraft becomes more stable but the tailplane down force
increases thereby increasing the total lift force required from the wings. The
generation of this increased wing lift requires the aircraft to fly a t a higher speed
so the stalling speed is increased.
In tlie canard configuration the nose down pitching moment is balanced by a n
upward lift force generated by the canard. This upward canard lift force carries
soine of tlie weight of the aircraft thereby reducing the load on the wings. This
reduces tlie magnitude of the lift force that needs to be generated by the wings,
thereby enabling the aircraft to fly at a lower speed. The stalling speed of an
aircraft is therefore reduced by the use of canards rather than a conventional
tailplane.
STALL 47. d.
In order for a n aircraft to be longitudinally stable its C of G must be forward of
the neutral point. This ensures that the lift 1 weight couple produces a nose down
pitching moment. In order to prevent nose down pitching tailplane must
produce a balancing nose up moment by generating a negative lift or down force.
The lift force from the wings must therefore equal the weight of the aircraft plus
tlie down force from the tailplane. As C of G is moved forward the aircraft
becomes more stable but the tailplane down force increases thereby increasing
the total lift force required from the wings. The generation of this increased
wing lift requires the aircraft to fly at a higher speed so the stalling speed is
increased.
If however the thrust line of the aircraft is lower than its drag line then the
thrust:drag couple produce a nose u p pitching moment which will assist the
tailplane in balancing the nose down moment generated by the forward C of G.
Increasing thrust in such aircraft increases the thrust:drag nose up moment
thereby reducing further the requirement for a tailplane down force. The
overall effect of this process is that increasing thrust decreases stalling speed.
STALL 48. c.
The stalling speed of a n aircraft in any manoeuvre is described by the following
formula:
VS (in any manoeuvre) = Vsl, d n where Vsl, is the stalling speed in l g flight
n is the load factor in the manoeuvre
I
This means that the stalling speed in a manoeuvre is equal to the l g stalling
speed multiplied by the square root of the load factor as described in option c.
STALL 49. a.
When flying in turbulence an aircraft is subjected to random variations in
upward and downward gusts of air. The effect of these gusts is to randomly
increase and decrease the angle of attack and CLof the wings. If these variations
cause angle of attack to exceed stalling angle then the aircraft will stall. If the
gusts cause sufficiently large increases in CLthe limiting load factor of the
aircraft will be exceeded and the structure will be overloaded. Deployment of
flaps inchases the maximum value of CLthat can-be achieved without stallirig,
but decreases the angle of attack at which stall occurs. The overall effect of flap
deployment in turbulence is an increased risk of both stalling the aircraft and
exceeding the limiting load factor.
STALL 50. b.
The stalling speed of an aircraft in straight and level flight is the minimum speed
at which the wings can generate a lift force equal to the weight of the aircraft
plus the tailplane down force. The higher the weight of an aircraft, the greater
will be the required lift force and stalling speed. As a flight progresses, the
burning of fuel will gradually reduce the weight of the aircraft thereby reducing
its stalling speed. Although the use of fuel also changes C of G position it is
highly unlikely that an aircraft would be designed such that reducing weight
increased its stalling speed.
STALL 51. b.
The effect of bank angle on stalling speed can be calculated using the following
,
equation:
Vs (at any AOB) = Vsl, d(1/ CosAOB)
Inserting the data provided in this question gives:
'Vs(at 45')
= 75 ~ t s d ( /1~ 0 ~ 4 5 ' which
)
= 89.19 Kts.
STALL 52. b.
In JAR 25 the terms:
VMCLis defined as the minimum speed at which control can be maintained in the
landing configuration, following a failure of the critical engine.
Vso is defined as the stalling speed in the landing configuration.
Vslgis defined as the stalling speed in straight and level flight in the clean
configuration.
V M cis~defined as the minimum speed at which control can be maintained in the
take-off configuration, following a critical engine failure in flight.
STALL 53. a.
Stalling speed at any new weight (Vs ,,, ) can be calculated using the equation
Vs new = Vs d(new weight / old weight)
Using the data given in this question Vsnew= 175 Kts d(100000 N / 75000 N)
This simplifies to give Vs at 100000 N = 202 Kts.
STALL 54. b.
Stalling speed in any manoeuvre (VM)can be calculated using the equation
VM= V S ~ ~ ~ ( ~angle
/ C OofSbank)
Using the data given in this question VM = 175 Kts d(l/Cos 47 degrees)
This simplifies to give VM = 211 Kts.
STALL 55. c.
This question is complicated by the fact that VMfor 47 degree angle of bank is
given instead of VSlg. The question can be resolved by rearranging the equation
to derive Vsl, then using this figure to calculate VMfor 65 degrees as follows.
VM47 deg = vslgd(l/cos 47)
Dividing both sides of the equation by d(l/cos 47) gives
VSlg = (VM47) / ~ ( I / C O 47)
S
This can then be used to calculate VMfor 65 deg as follows:
VM65 deg = Vslg d(l/cos 65 )
Inserting the value for VSlggiven in equation 1gives
VM65 deg = (VM47 deg / d(l/cos 47 )) d(l/cos 65)
Which simplifies to give VM65 deg = (VM47 deg) x d(cos 47 / Cos 65)
Inserting 200 Kts for VM47 deg as provided in the question gives
Vhl 65 deg = (200 Kts) x d(cos 47ICos 65)
This gives VM65 degrees = 254 Kts
.,,
STALL 56. d.
Stalling speed a t any new weight (VS ) can be calculated using the equation
Vs new = Vs old d(new weight / old weight)
Using the data given in this question Vs
.,,
= 180 Kts d(40000 N / 80000 N)
This simplifies to give Vs a t 40000 N = 127 Kts.
STALL 57. a.
Stalling speed in any manoeuvre (VM)can be calculated using the equation
VM = VSlgd n where n is the load factor in the manoeuvre.
This can be rearranged to VSlg = VM/dn to calculate Vslg.
Using the data given in this question Vslg = 450 Kts 147
This simplifies to give VSlg = 170 Kts.
STALL 58. b.
In JAR certificated passenger aircraft the stick shaker activates a t 1.05 VSlr
In this question Vslg is 100 Kts so the stick shaker will activate a t 105 Kts.
STALL 59. b.
Stalling speed in any manoeuvre (VM)can be calculated using the equation V
Vslg dn where n is the load factor in the manoeuvre and Vslg is the stalling
speed in
straight and level flight.
This can be rearranged to VSlg = VM/dn to calculate VSIr
Using the data given in this'question Vsl, = 100 Kts 143
This simplifies to give VSlg = 58 Kts.
STALL 60. b.
Autorotation occurs when one
wing stalls before the other causing
the aircraft to roll towards the
stalled wing. The downward motion
of the stalled wing increases its angle
of attack, taking it deeper into the
stall. This increases its CD and
and decreases its CLas illustrated in
the diagram.
Drag dissymmetr
=
STALL 61. c.
Autorotation is the term used to
describe the uncommanded rolling
and yawing motion that occurs in
an asymmetric stall.
CDof both wings prior to autorotation
\
CDof up-going
wing during
autorotation.
\
CDof down-going
wing during
autorotation
Autorotation is caused when one
wing stalls before the other causing
the aircraft to roll towards the
stalled wing. The downward motion
of the stalled wing increases its angle
of attack taking it deeper into the
stall . This increases the CDof the
down-going wing and decreases that
of the up-going wing as illustrated
in the diagram a t the right.
/
Up-going wing a a stall
Down-going wing a
Drag dissymmetry caused by roll in
autorotation
STALL 62. d.
Autorotation is the term used to describe the uncommanded rolling and yawing
motion that occurs in an asymmetric stall. Asymmetric stalling causes the stalled
wing to drop while the unstalled wing rises, making the aircraft roll towards the
stalled wing. This rolling motion increases the angle of attack of the down-going
wing and decreases the angle of attack of the up-going wing.
The decreased angle of attack of the CLof b
up-going wing decreases its CL,whilst
the increased angle of attack of the
down-going wing takes it deeper into
wing during
the stall thereby decreasing its CL
further. The overall effect of these
changes is that the CLof the up-going
wing decreases while that of the
down-going wing decreases even more.
These effects are illustrated in the
diagram at the right.
. .. ,.
Decrease in CLof up-going wing
I
Decrease in CLof down-go
is greater than that of up-going wing
Down-going wing a
Lift dissymmetry causing roll in autorotation
.,
STALL 63. b.
of inner wing takes i
Autorotation occurs when one
wing stalls before the other, causing
the aircraft to roll towards the
stalled wing. The downward motion
of the stalled inner wing increases its
angle of attack taking it deeper into
the stall. The inner wing therefore
becomes more stalled while the outer,
outer wing a
upward-going wing becomes less so,
as illustrated in the diagram at the
wing a
right.
a stall
inner
Decreased angle of attack of
outer wing take it out of the stall
STALL 64. c.
VA is the speed a t which an aircraft will stall whilst just attaining its limiting
lsrrd factor. For a JAR certificated passenger aircraft the minimum limiting load
factor is 2.5.
The stalling speed at any load factor can be calculated using the following
formula:
Vs(in any manoeuvre) = Vslg 6 n Where Vsls is the stalling speed in l g flight
n is the load factor in the manoeuvre.
So VA = vslg 62.5
which is 1.58Vs1, or 158% of VSIr
STALL 65. c.
The maximum operating altitude of a JAR certificated passenger aircraft is that
altitude at which it can just achieve a load factor of 1.3 at the onset of buffet.
That is to say a t the maximum operating altitude the maximum pre-stall load
factor is 1.3.
The stalling speed under these conditions can be calculated using the following
equation:
Vs (in any manoeuvre) = Vslg 6(n) where Vslg is the l g stalling speed
n is the load factor in the manoeuvre.
So Vs (in a 1.3g manoeuvre at the maximum operating altitude ) = ~ s l ~ d 1 . 3
Which is 1.14Vsl, or 114%VsI,.
STALL 66. d.
JAR 25 defines Vso as the stalling CAS in the landing cerafiguration with idle
power set.
STALL 67. b.
JAR 25 defines Vslg as the minimum CAS at which an aircraft can generate a lift
force equal to its weight at an angle of attack not greater than its stalling angle of
attack.
STALL 68. a.
Vs is the minimum attainable flight speed in any given configuration. Vso is the
stalling speed in the landing configuration. Because this configuration employs
fully deployed flaps, it is the lower than Vs in any other configuration. Vsl, is the
stalling speed in the flaps up configuration in l g flight. VATis the target
threshold speed, which is 1.3Vs.
STALL 69. d.
This problem can be solved using the following equation:
Vs (in any manoeuvre) = vslgdn where Vslg is the l g stalling speed
n is the load factor in the manoeuvre.
For a JAR certificated passenger aircraft VAis the speed a t which stall occurs at
a load factor of 2.5. Inserting this and the value for-Vslgprovided in the question
gives:
VA = Vs(in a 2.5g manoeuvre) = 175 ~ t s d 2 . 5
Which is VA = 276.7 Kts.
STALL 70. a.
JAR 25 defines the maximum operating altitude as the maximum pressure
altitude a t which it is possible to attain a load factor of 1.3 without incurring prestall buffet. The maximum attainable load factor at this altitude is therefore 1.3.
STALL 71. a.
In a climbing turn the inner wing moves through the air at a lower speed than
the outer wing, whilst both wings move upwards at the same speed. This means
that the outer wing experiences a higher angle of attack than the inner wing and
so will stall first if the pitch attitude of the aircraft is gradually increased. The
loss of lift on the stalled outer wing will then cause the aircraft to roll towards the
outer wing, causing it to move downwards, while the inner wing moves upwards.
This will increase the angle of attack on the outer wing while decreasing that on
the inner wing. The stalled outer wing will also produce more drag than the
unstalled inner wing, causing the aircraft to yaw towards the outer wing. The
overall effect of these events is that the outer wing will stall first causing the
aircraft to roll and yaw out of the turn. The initial difference in angle of attack,
which initiates the above sequence is illustrated in the diagram below.
Vertical component of airflow
due to upward motion of the
aircraft is the same on both wings
Low TAS
C h o q o f outer wing
Higher TAS due to wider circle of flight
STALL 72. c.
In a descending turn the inner wing moves through the air at a lower speed than
the outer wing, whilst both wings move downwards at the same speed. This
means that the inner wing experiences a higher angle of attack than the outer
wing and so will stall first if the pitch attitude of the aircraft is gradually
increased. The loss of lift on the stalled inner wing will then cause the aircraft to
roll towards the inner wing causing it to move downwards while the outer wing
moves upwards. This will increase the angle of attack on the inner wing while
decreasing that on the outer wing. The stalled inner wing will also produce more
drag than the unstalled outer wing thereby causing the aircraft to yaw towards
the inner wing. The overall effect of these events is that the inner wing will stall
first causing the aircraft to roll and yaw into the turn. The initial difference in
angles of attack, which initiates the above sequence of events is illustrated in the
diagram overleaf.
Vertical component of airflow
due to upward motion of the
aircraft is the same on both wings
Large angle of attack
Relative airflow
STALL 73. a.
Airflow passing over swept back wings tends to migrate outwards from the roots
towards the tips, producing a thick, low energy boundary layer in the wing tip
region. As angle of attack is increased this low energy boundary layer separates
easily from the upper surfaces causing the wing tips to stall before the roots.
Swept back wings are therefore prone to tip stall due to separation of low energy
boundary layer at the tips. In the case of swept forward wings the airflow tends
to migrate towards the wing roots making tip stall much less likely. Shock stall
occurs at high speeds when shock waves form on the upper surfaces of the wings,
causing a rapid deceleration of airflow. The low speed airflow behind the shock
waves separates easily causing shock stall. The shock waves causing shock stall
first form close to the wing roots where the wings are thickest, so shock stall is
unlikely to lead to tip stall.
STALL 74. d.
Airflow passing over swept back wings tends to migrate outwards from the roots
towards the tips, producing a thick, low energy boundary layer in the wing tip
region. As angle of attack is increased this low energy boundary layer separates
easily from the upper surfaces causing the wing tips to stall before the roots.
Because the tips are aft of the roots, the loss of lift caused by tip stall moves the C
of P inwards and forwards, causing the aircraft to pitch nose up. Straight wings
are not inherently susceptible to tip stall so the C of P tends to move aft in the
stall, causing nose down pitching motion.
In generating lift, aircraft wings increase the speed of the air passing aver them.
When the sum of the TAS of the aircraft and the speed increase over the wings
reaches the local speed of sound, shock waves are formed over the upper surfaces
of the wings. As the airflow passes through these shock waves it is suddenly
decelerated causing a loss of energy. This low energy air then tends to separate
easily behind the shock waves causing the rear area of that section of the wing to
stall. Because the wings are thickest close to their roots and the magnitude of the
acceleration over the upper surfaces s proportional to thickness, the shock waves
and shock stall occur first close to the wing roots. In the case of swept wing
aircraft the wing roots are ahead of the wing tips so the stalling of the roots area
causes the C of P to move rearwards causing the aircraft to pitch nose down.
The overall effect of these phenomena is that swept back wings are prone to pitch
up in low speed stall and pitch down in high speed stall.
STALL 75. b.
As angle of attack is increased below the stalling angle the pressure drop above
the wings of an aircraft intensifies and the C of P moves forward. At the stalling
angle the boundary layer separates from the per surface of the wing causing
the pressure envelope to collapse and the C of to move a f L T h i s aft movement
of the C of P causes the aircraft to pitch nose down in a stall.
'$
In generating lift aircraft wings increase the speed of the air passing over them.
When the sum of the TAS of the aircraft and the speed increase over the wings
reaches the local speed of sound, shock waves are formed over the upper surfaces
of the wings. As the airflow passes through these shock waves it is suddenly
decelerated causing a loss of energy. This low energy air separates easily behind
the shock waves causing the rear area of the wing the wing to stall. This
separation of the boundary layer behind the shock wave causes the C of P to
move forward, making the aircraft pitch up. As speed increases further the
shock waves are pushed back moving the C of P to about the 50% chord point.
This makes the aircraft pitch nose down. Straight wings therefore pitch down in
low speed stall, and up then down in high-speed stalls.
CON 1. a.
A level turn is initiated by banking the aircraft in the direction of the intended
turn. This tilts the total reaction such that a proportion of the lift force is angled
in the direction of the turn, thereby pulling the aircraft into the turn. This
deflection of part of the lift vector in the direction of the turn reduces the vertical
component of lift available to maintain altitude, so the angle of attack must be
increased to restore the original level of vertical lift. Increasing angle of attack
also increases drag, so more thrust is necessary to maintain airspeed. The
correct actions are therefore to increase thrust and angle of attack.
CON 2. a.
Aileron flutter occurs when the C of G of the aileron is aft of its hinge line.
When variations in the lift force cause the wing tips to bend up and down, the aft
position of the aileron C of G causes it to lag behind the wing. This causes the
aileron to flap down when the wing accelerates upwards and to flap up when the
wing accelerates downwards. If an increase in lift causes the wing to accelerate
upwards the flapping down of the aileron increases the lift produced by the
aileron. because this extra lift acts through the aileron hinge line, the trailing
edge of the wing is twisted upwards and the leading edge downwards. This
twisting action generates increasing stress within the wing structure, which
eventually arrests the twisting motion. As the twisting motion stops, the aft C of
G of the aileron causes it to overshoot, producing a flapping up action. The lift
generated by the upward deflected aileron then twists the wing trailing edge
downwards until it is again arrested by the stresses within the wing. As the
downward twisting of the wing stops the aileron again overshoots, causing the
cycle to be repeated. This cyclic twisting of the wing is called torsional aileron
flutter. The relative positions of the C of G and torsional axis of the wings have
no effect on this process.
CON 3. a.
The process of flexural aileron flutter is similar to that of torsional aileron flutter
as described in question CON 2, but in this case the primary effect is to cause the
wings to bend up and down rather than to twist. The principal cause is the C of
G of the ailerons being aft of the hinge line producing cyclic flapping or
fluttering of the ailerons. As with torsional aileron flutter, the relative positions
of the C of G, and torsional axis of the wings have no significant effect on the
process.
CON 4. b.
Diverge is caused by the C of P of the wing being ahead of its torsional axis, and
the wing being insufficiently strong to contain the resulting leading edge upward
twisting moments. The phenomenon is particularly likely a t high airspeeds,
when the aerodynamic forces generate high twisting moments. The flexibility of
any wing increases from root to tip because the wing roots are firmly supported
by the fuselage attachments. The twisting motion leading to divergence therefore
occurs primarily a t the wing tips. In the case of swept back wings, the upward
twisting of the leading edge a t the wing tip causes the wing tip to bend upwards.
Because the tip are some distance behind the roots, this upward bending causes
the angle of incidence to reduce a t the tips, thereby reducing the lift forces
generated a t the tips. This reduction in lift force a t the wing tips reduces the
tendency of the wing tip leading edges to twist upwards, and so prevents
divergence. Swept back wings are therefore less prone to divergence.
CON 5. a.
Diverge is caused by the C of P of the wing being ahead of its torsional axis, and
the wing being insufficiently strong to contain the resulting upward twisting of
the leading edge. This upward twisting of the leading edge increases angle of
incidence, further increasing the lift force. This cycle of lift increase - leading
edge twist up - lift increase, continues until the structure fails and the wing is
torn off. The phenomenon is particularly likely a t high airspeeds where the
aerodynamic forces generate high twisting moments. The flexibility of any wing
increases from root to tip because the wing roots are firmly supported by the
fuselage attachments. The twisting motion leading to divergence therefore
occurs primarily a t the wing tips.
CON 6. b.
In power assisted flying control systems, hydraulic actuators are employed to
carry the majority of the control hinge moments, thereby keeping stick forces
within the physical capabilities of the pilot. The pilot therefore carries a
proportion of the control forces at all times and has the facility to operate the
controls manually in the event of hydraulic power failure. As with all other
forms of control system, the aircraft must be trimmed to maintain any given
combination of C of G position, airspeed and aircraft attitude. This trimming
process is achieved by deflecting the controls and this generates continuous
control hinge moments.
Because the pilot carries a proportion of the hinge moment a t all times, it is
necessary to reduce these trim-induced hinge moments to zero, in order to avoid
pilot fatigue. One method of achieving this is the use of trim tabs, which
generate aerodynamic forces, which hold the control surfaces in the timed
position. The use of such tabs is not necessary in fully powered control systems
where the entire hinge moment is borne by the hydraulic actuators.
CON 7. b.
Although the yaw moments that can be generated by the rudder are
approximately constant a t all airspeeds, those generated by the failure of a n
engine in a multi engine aircraft, increase as speed decreases. There is therefore
a minimum airspeed below which it is not possible to maintain control following
single engine failure. The critical engine is the one, the failure of which would
cause the greatest yawing moment and hence require the greatest airspeed to
maintain control.
The minimum speed a t which it is possible to maintain control, varies with
changes in aircraft configuration. VMCCis the minimum speed a t which it is
possible to maintain control following a critical engine failure in the take-off
configuration. In propeller aircraft the increased airflow over the wings due to
propeller slipstream, increases the lift generated by the wings. This additional
lift is proportional to the velocity of the propeller slipstream and is lost in the
event of engine failure. The failure of a single engine therefore causes a loss of
lift on the wing carrying the failed engine, whilst the other wing continues to
benefit from the slipstream of the live engine. This asymmetric lift causes the
aircraft to roll towards the dead engine. The minimum control speed is therefore
limited by the ability of the pilot to generate sufficiently high rolling moments to
overcome the rolling caused by the asymmetric lift force. V1\lcc is therefore
limited by the roll rate of the aircraft.
CON 8. a.
The longitudinal axis of an aircraft passes in a rear to front direction, through its
C of G. Roll is motion about this axis, such that one wing descends while the
other rises.
The lateral axis passes from the left side to the right, through the C of G.
Rotation about this axis is called pitch. The normal axis passes from above to
below a n aircraft through its C of G. Motion about this axis is called yaw. The
term adverse yaw describes a phenomenon whereby rolling an aircraft causes it
to yaw away from the roll.
CON 9. b.
The purpose of control mass balance is to prevent control flutter, by placing the
C of G of the control surfaces on their hinge lines. I t is achieved by fitting
balance weights to the leading edges of the control surfaces. Fitting weights aft
of the hinge line would have the opposite effect, increasing flutter tendencies.
Horn balance is a form of aerodynamic balance intended to reduce control
forces. The effect of fitting weights to the tip caps of control surfaces would
depend upon the fore and aft location of these weights. This method is not
commonly used for control mass balance.
CON 10. b.
Trailing edge flaps commonly occupy only the inboard section of the trailing
edge, so flap deployment concentrates the lift force inboard. This is undesirable
in that it reduces lateral stability. This problem is often overcome by arrange
the ailerons, such that they dloop when the flaps are deployed. In effect these
drooped ailerons act as outboard flaps. Because they combine the effects of flaps
and ailerons, they are commonly called flaperons.
CON 11. c.
The term adverse yaw refers to the phenomenon whereby rolling an aircraft
causes it to yaw away from the roll. It is caused by asymmetric drag forces
affecting the wings and is commonly overcome by the use of roll spoilers, which
assist in producing roll whilst rebalancing the drag forces affecting the wings.
CON 12. d.
The term adverse yaw refers to the phenomenon whereby rolling an aircraft
causes it to yaw away from the roll. It is caused by asymmetric of drag forces
affecting the wings. I t is commonly overcome by the use of aileron-rudder
coupling, which deflects the rudder whenever the ailerons are deflected. This
automatically produces a yaw control input in the same direction as the roll
input, thereby preventing adverse yaw.
CON 13. a.
In a powered flying control system, the controls are irreversible, so upward
deflection of a pitch trim tab will not drive the elevator down. Upward
deflection of a pitch trim tab will therefore produce a nose up pitching moment.
The cockpit action that would bring about this effect would be moving the pitch
trim switch in the nose up direction. It should however be noted that in a
manual control system, trim tab deflection drives the parent control surfaces in
the opposite direction. Upward deflection of a pitch trim tab in a manual system
would drive the elevator down to produce a nose down pitching moment. This
process would therefore be initiated by moving the pitch trim switch in a nose
down direction.
CON 14. c.
If the turn and slip indicator shows needle left and ball left, this means that the
aircraft is turning left while slipping into the turn. This occurs when the
horizontal component of lift caused by bank towards the centre of the turn is
greater than the centrifugal force acting away from the centre. In this case the
cause of the slip is excessive left bank, or insufficient speed. It can therefore be
remedied by decreasing the left bank, or by increasing the airspeed.
CON 15. c.
When a control surface is deflected the aerodynamic forces acting upon it exert a
hinge moment opposing the deflection. In the case of manual control systems
these hinge moments lead to pilot fatigue and can become unmanageable at high
speeds. Tlie term aerodynamic balance refers to a number of techniques
whereby these hinge moments are reduced to manageable proportions by
aerodynamic means. Inset hinges and horn balance are two commonly employed
forms of aerodynamic. balance.
CON 16. b.
The principal function of roll spoilers in high speed aircraft is to act in
conjunction with, or in place of the ailerons, in order to provide roll control,
whilst minimising wing twisting forces. Roll spoilers are fitted to theupper
surface of wings. By moving upwards they provide a downward rolling moment
on the wing to which they are attached. They cannot however protrude into the
airflow below the wings. When an aircraft enters a high speed turn, the spoiler
on the down-going wing will therefore move up, whilst that on the other wing
remains retracted.
CON 17. b.
The principal function of roll spoilers in high speed aircraft is to act in
conjunction with, o r in place of the ailerons to provide roll control, whilst
minimising wing twisting forces. Roll spoilers are fitted to the upper surface of
wings. By moving upwards they provide a downward rolling moment on the
wing to which they are attached. They cannot however protrude into the airflow
below the wings. In some aircraft the spoilers can also be deployed
symmetrically to act as airbrakes. If such an aircraft is using its spoilers as
speed brakes when it enters a left turn, the left spoiler will move upwards while
the right spoiler retracts.
CON 18. c.
The term adverse yaw refers to the phenomenon whereby rolling an aircraft
causes it to yaw away from the roll. I t is caused by asymmetry of the drag forces
affecting the wings, and is commonly overcome by the use of Differential
ailerons. In this system the aileron on the down-going wing, is deflected to a
lesser degrees than that on the up-going wing. This asymmetric deployment of
ailerons restores the balance of drag forces on the wings, thereby preventing
adverse yaw.
CON 19. c.
As an aircraft accelerates through the transonic speed range it attains a critical
mach number (MCRIT),
at which the airflow over its surfaces first reaches the
local speed of sound. I t is above this mach number that shock waves first start to
form, causing disruption to the normal lift distribution and reducing stability.
Because the local speed of sound varies with air temperature, the transonic
stability and control problems do not always occur at the same TAS, RAS o r
IAS, but a t the same critical mach number. I t is therefore the mach number of
the aircraft that affects control in the transonic speed range.
CON 20. b.
In power assisted flying control systems the feedback of aerodynamic forces
from the control surfaces to the control levers, is reduced to manageable
proportions by means of power driven actuators. These actuators carry most,
but not all of the control loads, and so the pilot retains a degree of feel of the
loads applied to the aircraft. In such systems manual control is still possible in
the event of power failure, although this requires much greater effort on the part
of the pilot. I n fully powered flying control systems, all of the aerodynamic
forces are carried by the powered actuators, so the effort required on the part of
the pilot is reduced to that necessary to move the actuator selector valves.
CON 21. a.
The purpose of a variable incidence tailplane is to enable the aircraft to be
trimmed for all allowable combinations of C of G position, thrust and airspeeds,
whilst retaining full elevator control authority. The magnitude of the nose up or
nose down pitching moments that need to be trimmed out, and hence the
required tailplane angle, vary throughout flight, depending upon the C of G
position, thrust and airspeed.
CON 22. c.
The purpose of a variable incidence stabilizer is to enable the aircraft to be
trimmed for all allowable combinations of C of G position, thrust and airspeeds,
whilst retaining full elevator control authority. In such systems trimming is
achieved by altering the angle of incidence of the stabilizer whilst, leaving the
elevator in neutral. The elevator is then deflected only to provide pitch control
inputs. In the trimmed condition the elevator is therefore in the neutral position.
CON 23. a.
In a low winged aircraft with engine mounted below the wings, the thrust line
will be lower than the drag line. The thrust-drag couple will therefore apply a
nose up moment to the aircraft. If the C of G is forward of the C of P, then at
some optimum cruise speed this nose up moment will be exactly balanced by the
nose down moment caused by the weight-lift couple. But at all other speeds,
trimming will be required to maintain level flight. If the aircraft is trimmed in
any given level flight condition and thrust is then increased, the effect will be an
increasing nose up pitching moment, necessitating a downward deflection of the
elevator to restore trim.
CON 24. c.
The purpose of a variable incidence stabilizer is to enable the aircraft to be
trimmed for all allowable combinations of C of G position, thrust and airspeeds,
whilst retaining full elevator control authority. In such systems trimming is
achieved by altering the angle of incidence of the stabilizer, whilst leaving the
elevator in neutral. The elevator is then deflected only to provide pitch control
inputs so full control authority is maintained regardless of trim inputs. Also
because the entire tailplane changes angle it is able to generate more powerful
trim forces, whilst generating less trim drag than the conventional tailplane,
elevator and trim tab configuration.
CON 25. a.
In the servo tab system the tabs are used to provide an aerodynamic force which
then moves the parent control surface. The pilot's control inputs are fed directly
to the tabs, which are moved in the opposite direction to the intended parent
control deflection. The deflected tabs then generate a lift force, which acts at the
trailing edge of the parent control, pushing it in the opposite direction to that of
the tab. Because part of the control (the tab) moves in the opposite direction,
this system makes the controls less effective, particularly at low speeds when very
little aerodynamic force is generated by the tabs. Servo tab systems therefore
require larger control surfaces than other systems. Servo tabs are not used in
conjunction with hydraulic actuators, because these can easily be made to carry
the full control loads.
d
CON 26. b.
The longitudinal axis of a n aircraft passes in a rear to front direction, through its
C of G. Rolling is motion about this axis, such that one wing descends while the
other rises. The lateral axis passes from the left side to the right, through the C
of G. Rotation about this axis is called pitch. The normal axis passes from above
to below a n aircraft through its C of G. Motion about this axis is called yaw.
The term phugoid motion refers to a phenomenon whereby an aircraft performs
a low frequency, oscillation about its lateral axis, such that its altitude, airspeed
and attitude are constantly changing, whilst angle of attack and total energy
remain approximately constant.
CON 27. b.
The purpose of control mass balance is to prevent control flutter by placing the
C of G of the control surfaces on their hinge lines. It is achieved by fitting
balance weights ahead of the hinge line of the control surfaces. Fitting weights
aft of the hinge line would have the opposite effect, increasing flutter tendencies.
Horn balance is a form of aerodynamic balance intended to reduce control
forces. The effect of fitting weights to the tip caps of control surfaces would
depend upon the fore and aft location of these weights. This method is not
commonly used for control mass balance.
CON 28. b.
The principal benefit derived from attaching engines to the rear fuselage is that
their thrust lines are closer to the longitudinal axis of the aircraft. This reduces
the pitching and yawing moments resulting from power changes and single
engine failure, thereby reducing control problems in asymmetric flight. This
configuration has a number of disadvantages however including the danger of
engine failure in the stall due to the ingestion of turbulent air flowing from the
stalled wings. Also, the fitting of engines on the wings increases the mass of the
wings and so tends to damp out oscillating o r fluttering motions. An aircraft
with aft mounted engines is therefore more prone to flutter and hence requires
stronger and heavier wing structures.
CON 29. c.
In fully powered flying control systems the entire flight load caused by
aerodynamic forces on the control surfaces, is carried by the power driven
actuators. The effort required of the pilot is therefore limited to that necessary
to operate the actuator selector valves. In such systems manual control is not
possible in the event of hydraulic power failure, so duplicate power systems and
actuators are invariably provided. I n some modern aircraft pitch trimming is
provided by electrically powered variable incidence tailplanes, with hydraulic
power employed to drive the elevators. In addition to providing more efficient
pitch trimming, this system provides an alternative form of powered control in
that the electrically powered variable incidence tailplane can still be operated
following hydraulic power failure.
CON 30. c.
In power assisted flying control systems the feedback of aerodynamic forces
from the control s~lrfacesto the control levers, is reduced to manageable
proportions by means of power driven actuators. These actuators carry most,
but not all of the control loads, and so the pilot retains a degree of feel of the
loads applied to the aircraft. In such systems manual control is still possible in
the event of power failure, although this requires much greater effort on the part
of the pilot.
Because only part of the load is carried by the actuators, these systems are
reversible, so aerodynamic forces generated at the control surfaces can be used
to move the system. In some such systems, elevator trim tabs are employed to
serve the dual functions of pitch trimming and emergency pitch control. These
tabs are normally electrically powered and in the event of failure of the main
hydraulic power supply, a limit degree of pitch control can be provided by use of
the trim tabs.
CON 31. c.
In power assisted flying control systems the feedback of aerodynamic for,ces
from the control surfaces to the control levers, is reduced by means of power
driven actuators. These actuators carry most, but not all of the control loads,
and so the pilot retains a degree of feel of the loads applied to the aircraft. I n
such systems manual control is still possible in the event of power failure,
although this requires much greater effort on the part of the pilot.
Because only part of the load is carried by the actuators, these systems are
reversible, so aerodynamic forces generated at the control surfaces can be used
to move the system o r reduce control loads to more manageable proportions. In
some such systems, elevator trim tabs are employed to serve the dual functions of
pitch trimming and emergency pitch control aerodynamic power assistance.
These tabs are normally electrically powered and in the event of failure of the
main hydraulic power supply, provide a limited degree of pitch control power
assistance effectively acting as balance tabs.
CON 32. a.
The word stabilator is an amalgamation of the words stabilizer and elevator. In
a stabilator system the elevator is replaced by a pitch control surface made up of
the entire tailplane, which changes its angle of incidence in response to pitch
control inputs. Because the stabilator is much larger than a conventional
elevator, it provides much more powerful pitch control, but requires a greater
control input force to activate it. The size of the overall stabilizer is not affected
however, so longitudinal stability is unchanged.
CON 33. b.
Pushing the left rudder pedal forward will cause the rudder to move to the left.
This will produce a camber on the fin, which will generate a side force to the
right. Because the C of P of the fin is behind and above the C of G of the
aircraft, this fin side force, its immediate effect will be to cause the aircraft to
yaw to the left and roll to the right. As the yaw rate increases the different
airspeeds over the wings will cause the right roll to be replaced by left roll, but
this is a subsequent effect, rather than an immediate effect.
CON 34. c.
As a n aircraft accelerates through the transonic region its shock stall will make
the C of P move aft, causing the aircraft to pitch nose down in a phenomenon
called mach tuck under. The purpose of the mach trim system is to
automatically input a nose up trim command to prevent this pitch down in shock
stall.
CON 35. c
In low speed flight the majority of aircraft employ outboard ailerons to initiate
roll. At higher speeds however, the use of outboard ailerons imposes excessive
twisting forces on the wing, reducing aileron effectiveness and eventually leading
to control reversal. When flying a t high speeds, most modern aircraft employ a
combination of inboard ailerons and roll spoilers, or roll spoilers alone, to
eliminate these problems.
CON 36. b.
I n many modern aircraft the practice of fitting engines to the sides of the rear
fuselage to minimise asymmetric power problems, has necessitated the use of a
high tailplane fitted to the top of the fin. This arrangement has the beneficial
effect of acting a s an end plate on the fin, thereby reducing fin tip vortices. This
increases the effective aspect ratio of the fin, increasing the gradient of its CL :a
curve and hence improving directional stability in low sideslip angles.
Disadvantages include loss of tailplane effectiveness and reduced longitudinal
stability in deep stall, due to the high tailplane being affected by turbulent air
flowing from the stalled wings.
CON 37. d.
The longitudinal axis of a n aircraft passes in a rear to front direction, through its
C of G. Roll is motion about this axis, such that one wing descends while the
other rises.
The lateral axis passes from the left side to the right, through the C of G.
Rotation about this axis is called pitch. The normal axis passes from above to
below a n aircraft through its C of G. Motion about this axis is called yaw.
CON 38. d.
Spinning occurs when one wing stalls before the other, leading to asymmetric lift
and drag forces. These cause the aircraft to roll and yaw towards the stalled
wing. If ailerons are deflected to oppose the wing drop, this will increase the
camber and angle of attack of the stalled wing, taking it deeper into the stall.
The correct action for recovery from a spin is therefore to centralise the ailerons,
push the stick fully forward to reduce angle of attack, and use opposite rudder to
arrest the yaw. The application of maximum or minimum power will depend on
aircraft type.
CON 39. c.
Control flutter is cyclic oscillation of the control surfaces, and is caused by their
C of G being aft of their hinge line. It is prevented by mass balancing the control
surfaces such that their C of G lies on their hinge line. This is achieved by fitting
mass balance weights forward of the hinge line.
CON 40. b.
For JAR certification purposes the gradient of the pitch centrol stick force
gradient must be negative and such that stick force increases at a minimum rate
of 1 Ibf for each 6 knots of speed increase. If the basic design of an aircraft is
such that this gradient is not achieved, then a down spring may be employed to
increase the gradient. This has no direct effect on control hinge moments or
flutter characteristics.
CON 41. c.
Control flutter is cyclic oscillation of the control surfaces, and is caused by their
C of G being aft of their hinge line. It is prevented by mass balancing the control
surfaces, such that their C of G lies on their hinge lines. This is achieved by
fitting mass balance weights forward of the hinge line.
CON 42. d.
Roll spoilers are fitted to the upper surface of wings and act in conjunction with
the ailerons. They are able to move or remain flush with the surface, but can
never protrude under the wing. When the control wheel is moved to the left, the
correct response is for the aircraft to roll left wing down. To achieve this, the left
aileron and spoiler move upwards, while the right aileron moves down, the right
spoiler remaining retracted.
CON 43. a.
When an aircraft is rotated into the lift-off attitude during the take-off run, it
rotates about its main wheels, which are in contact with the ground. If the C of
G is at its forward limit, the moment arm between the C of G of the aircraft and
its main wheels would be at its maximum value. The moment arm from the
tailplane to the main wheels is constant. The downward lift force required from
the tailpane in order to lift the nose of the aircraft would therefore be at its
maximum value, so the longitudinal stick force required to rotate the aircraft
would be at its greatest.
CON 44. d.
When a control surface is deflected the aerodynamic forces acting upon it exert a
hinge moment opposing the deflection. In the case of manual control systems,
these hinge moments lead to pilot fatigue and can become unmanageable at high
speeds. The term aerodynamic balance refers to a number of techniques
whereby these hinge moments are reduced to manageable proportions by
aerodynamic means. Horn balance is one of the commonly emptoyed forms of
aerodynamic balance.
CON 45. d.
When a critical engine fails, the resulting asymmetric thrust, yaws and rolls the
aircraft away from the dead engine. At low speeds the effectiveness of the fin
and rudder are reduced, so at some limiting low speed, the control authority is
insufficient to contain the yawing and rolling and control is lost. VhlCAis the
minimum speed at which control can be maintained following critical engine
failure in flight, in the take-off configuration. VMCGis the minimum speed at
which control can be maintained, on or close to the ground, in the take-off
configuration, following a critical engine failure.
The magnitudes of the uncommanded rolling and yawing moments are
proportional to the thrust being produced by the remaining engines, so any
factor that reduces thrust output will reduce VhlCAand VMCG.AS altitude
increases the reducing air density reduces the thrust output of all air breathing
aircraft propulsion systemss and hence reduces VMCAand VhICG.
CON 46.b.
When a critical engine fails, the resulting asymmetric thrust, yaws and rolls the
aircraft away from the dead engine. At low speeds the effectiveness of the fin
and rudder are reduced, so at some limiting low speed, the control authority is
insufficient to contain the yawing and rolling and control is lost. VMcc is the
minimum speed at which control can be maintained on or close to the ground,
following critical engine failure in the take-off configuration. The minimum
value of VMcc is limited by the yaw and roll control authority of the aircraft.
CON 47. a.
If the lateral static stability of an aircraft is more powerful than the directional
stability, then the aircraft will be prone to Dutch Roll. Any yaw will cause the
aircraft to roll away from the resulting sideslip, such that the aircraft oscillates
in an alternating rolling motion of increasing amplitude. High speed, swept
winged aircraft are particularly prone to this form of lateral dynamic instability.
Most modern swept wing aircraft employ a yaw damper coupled to a rate gyro,
in order to improve lateral and directional dynamic stability by preventing
Dutch Roll.
CON 48. b.
When a critical engine fails, the resulting asymmetric thrust, yaws and rolls the
aircraft away from the dead engine. At low speeds the effectiveness of the fin
and rudder are reduced; so at some limiting low speed, the control authority is
insufficient to contain the yawing and rolling, and control is lost. VMCLis the
minimum speed at which control can be maintained in the air, in the landing
configuration following a critical engine failure. The mii~imumvalue of VMCLis
limited by the yaw and roll control authority of the aircraft.
CON 49. d.
When a control surface is deflected, the aerodynamic forces acting upon it exert
a hinge moment opposing the deflection. In the case of manual control systems,
these hinge moments lead to pilot fatigue and can become unmanageable at high
speeds. The term aerodynamic balance refers to a number of techniques
whereby these hinge moments are reduced to manageable proportions by
aerodynamic means.
CON 50. a.
When an aircraft rolls the airflow approaching each wing is altered. The upgoing wing experiences a downward airflow component, while the down-going
wing experiences an upward component. These vertical airflow components
increase the angle of attack of the down-going wing, and decrease that of the upgoing wing. These changes in angle of attack change the directions of the total
reactions of the two wings, such that the down-going wing produces less drag
than the up-going wing. This dissymmetry of drag due to the changes in angle of
attack causes the aircraft to yaw away from the roll in a phenomenon called
adverse yaw.
CON 51. b.
When a control surface is deflected the aerodynamic forces acting upon it exert a
hinge moment opposing the deflection. In the case of manual control systems
these hinge moments lead to pilot fatigue and can become unmanageable at high
speeds. The term aerodynamic balance refers to a number of techniques
whereby these hinge moments are reduced to manageable proportions by
aerodynamic means. In some aircraft the control hinge moments are
manageable a t low speeds, but become unmanageable as speed increases. In
such cases the use of spring tabs provides aerodynamic balance proportional to
airspeed, so that hinge moments are reduced at high speeds only.
CON 52. d.
Elevons perform the dual functions of elevators and ailerons on some delta
wings. Ventral fins reduce lateral static stability by countering the rolling effect
of the fin when the aircraft is side slipping. Yaw dampers are automatic
stabitisation systems used to counter Dutch roll. Frise ailerons restore the
balance of drag forces produced by the two wings, in order to prevent adverse
aileron yaw when rolling into or out of turns.
CON 53. d.
Dorsal fins are used to increase the effectiveness of the fin, by reducing its aspect
ratio and hence increasing its stalling angle of attack. Elevons are used in some
delta winged aircraft to perform the dual functions of elevators and ailerons.
Yaw dampers are automatic systems used to counter Dutch roll. Roll control
spoilers restore the balance of drag forces produced by the two ailerong, in order
to prevent adverse yaw when rolling into or out of a turn.
CON 54. d.
Trim tabs niove only when commanded to do so through the trim wheels or
switches. When the main control surfaces are moved the trim tabs do not move
relative to their parent control surfaces.
CON 55. c.
An aircraft can be trimmed in two ways. In the first, the controls are moved to
achieve the desired trim condition and the trim tabs are then adjusted to reduce
the control hinge moments to zero. In the second, the trim tabs are adjusted in
order to niove the control surfaces to the desired trim condition. This question
refers to the first method so the elevators will not move.
CON 56. b.
Dorsal fins are used to increase the effectiveness of the fin, by reducing its aspect
ratio and hence increasing its stalling angle of attack. Yaw dampers a r e
automatic stabilisation systems used to counter Dutch roll. Flaperons a r e
ailerons that also droop symmetrically to function as flaps. Differential ailerons
restore the balance of drag forces produced by the two wings when rolling, in
order to prevent adverse yaw when rolling into o r out of a turn.
CON 57. d.
Vortex generators are used to energise the boundary layer, in order to delay or
reduce the effects of separation. Vortilons perform a similar function but are
placed under the leading edge. Dorsal fins are used to increase the effectiveness
of the fin, by reducing its aspect ratio and hence increasing its stalling angle of
attack.
CON 58. a.
If total moments about an axis are not zero, then a non-zero resultant turning
moment will exist. This represents a turning force which, by acting upon the
body, will cause angular acceleration as predicted by Newton's second law of
motion. If the motion occurs in a vacuum and the axis is frictionless, then the
acceleration will be of constant magnitude. Equilibrium would exist only if all
forces and moments added to zero. This would result in continuous rate rotation
or a static condition as predicted by Newton's first law of motion.
CON 59. b.
The hinge moment of a control surface is the product of its lift force multiplied
by the distance from its C of P to its hinge line. The hinge moment must be
overcome to move the control, so the greater the hinge moment, the greater the
effort required of the pilot. The purpose of aerodynamic balance is to reduce the
binge moment, in order to make the controls easier to move. One of the methods
employed in reducing hinge moment is to use balance tabs attached to the
trailing edge of thc. control surface. These move in the opposite direction to the
control surface, in order to generate a second hinge moment to oppose that of the
parent control.
CON 60. b.
The hinge moment of a control surface is the product of its lift force multiplied
by the distance from its C of P to its hinge line. The h h g e moment must be
overcome to move the control, so the greater the hinge moment, the greater the
effort required of the pilot. The purpose of aerodynamic balance is to reduce the
hinge moment in order to make the controls easier to move. One of the methods
employed in reducing hinge moment is to use horn balance, in which the tip of
the control is extended forward of the hinge line. This has the effect of moving
the overall C of P of the control surface closer to its hinge line and hence
reducing its hinge moment.
CON 61. a.
Pushing the right rudder pedal forward will deflect the rudder to the right. This
will generate a fin lift force acting to the left and hence will push the nose of the
aircraft to the right. This motion is yawing about the vertical axis. Because the
fin lift force acts at some distance above the C of G of the aircraft, it will also
cause rolling to the left about the longitudinal axis.
CON 62. b.
The hinge moment of a control surface is the product of its lift force multiplied
by the distance from its C of P to its hinge line. The hinge moment must be
overcome to move the control, so the greater the hinge moment, the greater the
effort required of the pilot. The purpose of aerodynamic balance is to reduce the
hinge moment in order to make the controls easier to move. One of the methods
employed in reducing hinge moment is to use balance tabs attached to the
trailing edge of the control surface. These move in the opposite direction to the
control surface in order to generate a second hinge moment to oppose that of the
parent control.
CON 63. b.
The hinge moment of a control surface is the product of its lift force multiplied
by the distance from its C of P to its hinge line. The hinge moment must be
overcome to move the control, so the greater the hinge moment, the greater the
effort required of the pilot. The purpose of aerodynamic balance is to reduce the
hinge moment in order to make the controls easier to move. One of the methods
employed in reducing hinge moment is to place the hinge closer to the C of P of
the control surface. This reduces the control moment arm and hence reduces the
control hinge moment. The interconnection of aileron and rudder controls is
used to minimise adverse yaw, but is not a form of aerodynamic balance.
CON 64. d.
The interaction between wing downwash and airflow over the tailplane makes a
positive contribution to stability, particularly when trailing edge flaps are
deployed. Deep stall is caused when the wingtips of swept back wings stall,
resulting in a pitch up as the C of P of the wings moves forward. This pitching
upward increases the angle of attack, taking the aircraft deeper into the stall.
Once deep stall has commenced the turbulent airflow flowing from the wings is
likely to make a T-tail ineffective, thereby exacerbating the control problems.
T-tails cannot therefore prevent deep stall but are likely to worsen its effects.
Although T-tails do permit the location of the engines at the sides of the rear
fuselage a number of T-tailed aircraft employ the more conventional wing
mounted engine configuration. By acting as end plates on the fin,T-tails
minimise the fin tip vortices, making the fin behave as if it were of higher aspect
ratio. The overall effect of this is a steeper fin 1ift:sideslip angle curve, giving en
increased stabilising force, but a lower fin stalling angle. This makes the aircraft
more directionally stable at low sideslip angles.
CON 65. b.
The hinge moment of a control surface is the product of its lift force multiplied
by the distance from its C of P to its hinge line. The hinge moment must be
overcome to move the control, so the greater the hinge moment, the greater the
effort required of the pilot. The purpose of aerodynamic balance is to reduce the
hinge moment in order to make the controls easier to move. One of the methods
employed in reducing hinge moment is to internal balance. By extending the
coiltrol surface leading edge forward and attaching a rubber diaphragm to it, the
space within the structure immediately forward of the control is divided into
two. Deflection of the controls varies the pressures a t each side of the
diaphragm, producing a force which acts in the opposite direction to the control
hinge moment. This reduces the effective hinge moment.
CON 66. a.
Only balance tabs and anti-balance tabs are anchored to the structure and
operated by the parent control surface. In servo tab systems control inputs are
fed only to the tabs and hence do not operate the control directly. Aerodynamic
forces acting on the deflected tabs cause the parent control surface to be
deflected in the opposite direction to the tab. At low airspeeds the tabs generate
little aerodynamic force, so control effectiveness is limited. Although the servo
tabs can be used as trim tabs this is not always the case.
CON 67. c.
To achieve the described condition, the pilot would use left rudder to arrest the
right yaw and to turn the aircraft to the left of the original heading, whilst using
ailerons to keep the wings level. In this condition the force produced by the fin
would cause the aircraft to sideslip along its original track at constant speed.
Because the aircraft would be no longer turning, the turn indicator would be
central. The ball in the slip indicator is positioned by the gravitational
acceleration acting on it. With the wings level in a constant speed sideslip the
ball would therefore be central.
ReorViAkn
Fb&t
-~
- RDth
Too V w Alone- Flight
- Path
firus1 Yowing
Thrust
'idz
Aileron Rolling
Moment
Weothercock
Yawing Moment
C+hdml Rdlina
Moment
Lih
Rudder Yaving Moment
No Slim Indicated
CON 68. b.
VMcc is the minimum speed at which it is possible to maintain control following a
failure of the critical engine on or near the ground, in take-off configuration. In
establishing VMCG,it is necessary to take account of the worst-case scenario, in
which the runway is slippery, o r the nose wheel steering inoperative. VhlCCis
therefore based on the assumption that only the primary flying controls will be
use in maintaining control. The minimum speed required in this condition is
greater than would be the case if the use of nose wheel steering were allowed for.
CON 69. c.
Trimming is the process whereby movement of the C of G and C of P of the
aircraft a r e adjusted for, such that the aircraft maintains the desired attitude
without control inputs from the pilot. Although the primary purpose of servo
tabs is to provide the motive force to deflect the controls to manoeuvre the
aircraft, they cari also perform the function of trim tabs. This is typically
achieved by attaching a spring strut parallel to the control linkage, and using an
electrical servo to reposition the control system and tab to trim the aircraft. The
use of flying controls to provide trim control reduces the control authority
available for manoeuvre, so systems such a s variable incidence stabilisers are
sometimes used to provide trim whilst maintaining full control authority.
CON 70. a.
Servo tabs are sometimes used in manually powered flying control systems to
provide aerodynamic force to move the control surfaces. The pilot's control
inputs move the servo tabs in one direction, and the aerodynamic force acting on
the tabs causes the parent control surface to move in the opposite direction. If
the control system is fully powered without any form of manual reversion, the
controls will be irreversible making it impossible for forces acting on the control
surfaces to be fed back through the system. Under these circumstances servo
tabs could not be used to move the controls in the event of hydraulic power
failure.
If however the control system were power assisted, o r had a manual reversion
capability, servo tabs could be used to enable the pilot to move the controls
following hydraulic failure. This process would however never be employed in
real aircraft because servo tabs make the control less effective at low speeds.
Although this is accepted in manually operated systems, it would be
unacceptable in power operated systems, where the availability of hydraulic
power makes aerodynamic assistance redundant in normal flight.
CON 71.c.
In servo tab systems, the control inputs are fed to the servo tabs, causing then1 to
deflect. The aerodynamic forces acting on the tabs then move the parent
controls in the opposite direction. With the elevator jammed, its servo tab could
be used to provide a pitch control, but because of the small size of the tab, its
effectiveness would be limited. Because a servo tab moves in the opposite
direction to that of the parent control surface, its use with a jammed elevator
would lead to a reversal of the normal control effect. That is to say, to pull the
nose of the aircraft up it would be necessary to push the stick forward.
CON 72. a.
Engine failyre in a single engine aircraft does not cause asymmetric power
effects so rolling would not be a problem. Single engine failure in multi-engine
aircraft would require control inputs to counteract asymmetric power effects
about all three axes of motion. The aircraft would yaw towards the failed engine,
causing one wing to experience a decrease in airspeed and the other an increase.
This would cause roll towards the dead engine. If prevented from yawing and
rolling by use of the rudder, the side force generated by the fin would cause the
aircraft to sideslip towards the dead engine. Its lateral stability would then cause
it to roll towards the live engine.
Although the above processes affect both jet and propeller aircraft in a similar
manner, propeller aircraft also derive a significant amount of lift from the
propeller slipstream passing over the wings. In the event of single engine failure,
the wing carrying the dead engine would therefore produce less lift than the
other wing, causing the aircraft to roll towards the dead engine. This effect is
most severe in the case of a single engine failure in a twin propeller aircraft with
wing mounted engines.
CON 73. d.
The flutter tendency of a wing is directly proportional to airspeed, and inversely
proportional to the mass of the wing. The use of wing mounted engines therefore
reduces flutter by increasinp the mass of the wings, and so reduces the tendency
of the wings to flutter. The principal loads acting on the wings in flight are
bending stresses acting a t the wing roots. Placing engines outboard reduces
these bending stresses, and hence permits the use of lighter wings.
The use of rear mounted engines therefore requires stronger, and hence heavier
wings than the more conventional, wing mounted configuration. Because of their
location, rear mounted engines are likely to be affected by turbulent air flowing
aft from the ings in a stall. 'The !ocation of the eiigines c h s e to the hlselage does
however reduce the distance between the thrust lines and the C of G, thereby
reducing the pitching and yawing effects of power changes.
o
CON 74. c.
Because of their high cruising speeds, modern aircraft must have wings of low.
:Iiickness to chord ratio, in order to maximise their critical drag numbers. Such
wings are inhzrently flexible. and hence susceptible to flutter and twisting at high
airspeeds. The most comnlorl niethod of minimising these problenis is to employ
a combination of outboard ailerons, inboard ailerons and roll spoilers. At low
speeds the outboard ailerons prolide roll control without exerting excessive
bending and twisting stresses 0,; the wings.
At higher speeds, increasing dynamic pressure acting on deflected outboard
ailerons would cause excessive wing twist and possibly control reversal, so the
outboard aileruns are locked in the neutral position a t such speeds. Roli is then
provided bv use of inboard ailerons and roll spoilers, which because of their
inboard location, produce lower hvisting stresses.
CON 75. d.
Following single engine failure in a heavily loaded aircraft. the excess power
available is likely to be severely limited. :iithough the faiiure-induced yawing,
rollilig and sideslip could be countcrcd i:i a number of w ays, some would require
more power than others. The use of rudder to counter the yaw and ailerons to
cuunter the resulting sideslip, would for elialnple result in a significant degree of
bank. This would increase load Pictor and induced drag, and hence require a
considerable increase in power.
The method requiri~igleast psbvcr is option d, in which rudder is uted to arrest
the yaw and also to turn the aircraft beyond its original heading. The side force
generated by the fin would then cause the aircraft to sideslip down the desiredtrack. In order to maintain wings level in this sideslip, the ailerons would be
employed to counter the rolling moments generated by the aircraft's lateral.
stability. Although this configuration would increase total drag the load factor
would be 1 so the induced drag would be minimised. At !aw speeds the overall
effect would be a lower value of power required t h a r ~for tile other options.
CON 76. b.
Vhlcr is the minimum speed at which it is possible to maintain control of an
aircraft following a failure of the critical engine during the take-off roll. I t is not
therefore limited by the failure of engine, but is intended to ensure safe operation
following such a failure. Followi~~g
an engine failure in a multi engine propeller
driven aircraft, the loss of propeller wash over one wing will cause that wing to
produce less lift than the wing bearing the live engine. This will cause the aircraft
to roll towards the dead engine.
Although this asymmetric propeller wash effect would not occur with a twin jet
aircraft, the yaw asymmetric thrust would cause the aircraft to yaw, causing the
dead engine wing to experience a lower airspeed than its neighbour. This would
again cause the aircraft to roll towards the dead engine. The ability of the pilot
to arrest this uncommanded roll is a significant factor in determining the
minimum value of VMcc. An aircraft with very strong roll control will have a
lower VMcc than one with a lower degree of roll control authority. Option b is
therefore the most accurate.
CON 77. C. Pushing the control column to the right would cause the left aileron to move
down and the right aileron to move up. This would increase the lift on the left
wing and decrease that on the right wing, causing the aircraft to roll to the right
about its longitudinal axis. The downward motion of the right wing would cause
its relative airflow to be deflected upwards, which would in turn cause its total
reaction to be deflected to a less rearwards direction. This would reduce the
drag on the right wing.
The left wing would experience the opposite effect with its relative airflow being
deflected downwards, thereby tilting its total reaction aft and hence increasing
its total drag.
The difference in drag forces acting on the two wings would then cause the
aircraft to yaw away from the down-going wing. This tendency of a n aircraft to
yaw away from the roll is termed adverse yaw. The overall effect of deflecting
the control column to the right would therefore be roll to the right about the
longitudinal axis and yaw to the left about the normal (vertical) axis.
CON 78. 6.
The magnitude of the flying control hinge moments affecting an aircraft is
dependent upon the magnitude of the dynamic pressure of the relative airflow
over the controls. As aircraft speed increases, dynamic pressure and hence .
control hinge moments increase. At low speeds the low magnitude of the stick
forces produced by these hinge moments are sufficiently low to be overcome.by
the pilot, so power assistance is not normally required.
At higher airspeeds the stick forces increase beyond the capability of the pilot, so
power assisted, o r fully powered systems are employed in high speed aircraft.
Increasing aircraft weight affects stick forces in the same way as increasing
airspeed in that very heavy aircraft require high control forces to generate
changes in attitude. Powered flying control systems a r e therefore required in
large, heavy, o r high speed aircraft. I n fully powered control systems the entire
control hinge moments are borne by the hydraulic actuators, so no stick forces
are generated and movement of the controls requires minimal effort on the p a n
of the pilot. This situation could lead to inadvertent overstressing of the aircraft,
so artificial means are emplo:led to provide acceptable stick forces. Such systems
therefore generate low stick forces.
CON 79. b.
In power assisted flying control systems the majority of the control hinge
moments are carried by the hydraulic actuators, leaving only a small proportion
to be b o r ~ by
~ e the pilot. The magnitude of the control hinge moments is
dependent upon the magnitude of the dynamic pressure of the relative airflow
over the controls. As aircraft speed increases dynamic pressure and hence
control hinge mon~entsincrease. In power assisted control systems a proportion
of the total hinge moment borne by the pilot, so as total hinge moment increases
with airspeed, so do the stick forces felt by the pilot. Power assisted control
systems therefore produce higher stick forces at high airspeeds. It should be
noted however that such systems produce lower stick forces than equivalent
manual systems a t all speeds.
CON 80. a.
The rudder trim wheel is typically located on a horizontal surface in the cockpit.
It is arranged such that turning the wheel to the right causes the aircraft to yaw
to the right and turning it to the left cause yaw to the left. The rudder bar is
arranged such that pushing the right pedal forwards causes yaw to the right and
pushing tlie left pedal cause yaw to the left. Right yaw is therefore generated by
pushing the right rudder pedal forward, which turns the rudder bar to the left,
or by turning the trim wheel to the right. The rudder bar and trim wheel
therefore operate in opposite direction. Although the two controls are not
~ l o r ~ n a l operated
ly
simultaneously it is possible to do so.
CON 81. c.
The trim wheel is arranged such that turning it to the right generates a noseright yawing moment. In manual control systems this moment is generated by
the deflection of the rudder trim tab to the left. This generates a force on the
trailing edge of the rudder causing it to be deflected to the right. This right
deflection of the rudder alters the camber and angle of attack of the fin, such
that a force to the left is exerted on the tail of the aircraft. This left force on the
tail produces a yawing moment to the right at the nose of the aircraft. In the
event of a left engine failure the aircraft would tend to yaw left, so the trim wheel
would be turned to the right to counter this uncommanded left yaw. This would
deflect the trim tab to the left which would in turn deflect the rudder to the right.
The correct answer to this question is therefore option c.
CON 82. c.
In straight and level flight, the lift force must equal the weight of the aircraft.
Lift = CL% p ~ and2 CL~is proportional to angle of attack and hence aircraft
attitude. In order to maintain constant lift as airspeed increases, it is therefore
necessary to reduce attitude in order to reduce angle of attack. Changing angle
of attack also changes the pitching moments generated by the aircraft wings and
tailplane, so the aircraft inust be retrimmed for large changes in airspeed.
Althougli power required increases with airspeed, attitude is the primary
method of controlling airspeed. Also because the pitching moments of the
aircraft change in response to attitude changes, the required trim adjustments
cannot be determined until the new airspeed has been established. The correct
procedure is therefore to adjust airspeed then adjust trim.
CON 83. c.
When an aircraft is flying at a steady airspeed, the thrust generated by its
engines is equal to the drag generated by the whole aircraft. The speed of an
aircraft can be adjusted in two principal ways. By reducing attitude whilst
maintaining constant thrust, the drag force will reduce, causing the aircraft to
accelerate. Lift = CL% p ~ and2 CL~is proportional to angle of attack and hence
aircraft attitude. By reducing attitude it is therefore possible to increase
airspeed without increasing lift. Changing attitude therefore controls airspeed at
any given altitude. If thrust is increased without reducing attitude the aircraft
will accelerate, but this will in turn cause its lift and hence altitude to increase. I t
is not therefore possible to adjust speed a t any given altitude, by adjusting power
without also adjusting attitude. Also because attitude changes produce an
immediate change in drag and hence acceleration, airspeed is more responsive to
attitude changes than to power changes. The primary method of speed control is
therefore attitude.
CON 84. a.
If in straight and level flight the power setting of an aircraft is increased, the
airspeed will increase. If the attitude of the aircraft is left unchanged the
increased airspeed will cause it to climb. As altitude increases air density will
decrease, thereby reducing power output until the aircraft is once again in
straight and level flight at a higher altitude. This is the principal method of
altitude control. Changing propeller RPM alone will not have the same effect, as
this will not alter significantly the power output of the engine. The principal
method of speed control changing pitch attitude
CON 85. a.
The positive direction of motion of an aircraft about any of its three axes of
rotation, can be found by first pointing one hand along the positive direction of
the relevant axis, then rotating the hand in a clockwise direction. Pitching
motion occurs around the lateral axis, for which tlie positive direction is from the
left wing tip through to the right wing tip. Pointing in that direction then
rotating the hand as described above indicates that positive pitching is in the
nose up direction.
CON 86. a.
'The positive direction of motion of an aircraft about any of its three axes of
rotation, can be found by first pointing one hand along the positive direction of
the relevant axis, then rotating the hand in a clockwise direction. Rolling motion
occurs around the longitudinal axis. for which the positive direction is from the
rear of the aircraft to the front. Pointing in that direction then rotating the
hand as described above indicates that positive rolling is in the fight wing down
direction.
CON 87. b.
The positive direction of motion1 of an aircraft about any of its three axes of
rotation, can be found by first pointing one hand along the positive direction of
the relevant axis, then rotating the hand in a clockwise direction. Yawing motion
occurs around the normal axis, for which the positive direction is from above the
G of G of the aircraft, passing vertically downwards to below it. ~ o h t i n in
g that
direction then rotating the hand as described above indicates that positive
yawing is in the nose moving to the right direction.
CON 88. e.
When ailerons are deflected in flight they alter the camber and angle of attack of
the section of the wing to which they are fitted, thereby increasing o r decreasing
!he lift force generated. Much of change in lift force takes effect on the ailerons
and the forces generated by deflection are transferred to the wing through the
aileron hinges. Because the ailerons are attached to the rear of the wings, these
forces tend to twist the wings such that they move in the opposite direction to the
ailerons. These twisting effects are; insignificant a t low speeds4 but as speed
increases, the degree of twist increases, thereby reducing and eventually
reversing, the effect of aileron deflection. Because fully powered flying controls
a r e capable of being deflected to greater angles at high speeds, they are more
likely to cause aileron reversal than are manual controls.
r
CON. 89. c.
Wllell ailerons are deflected in flight they alter the camber and angle of attack of
the section of the wing to which they are fitted, thereby increasing o r decreasing
the lift force generated. Much of change in lift force takes effect on the ailerons
and the forces generated by deflection are transferred to the wing through the
aileron hinges. Because the ailerons are attached to the rear of the wings, these
forces tend to twist the wings such that they move in the opposite direction to the
ailerons. These twisting effects a r e insignificant a t low speeds but as speed
increases the degree of twist increases, thereby reducing, and eventually
reversing the effect of aileron deflection. Control reversal is therefore most
likely a t high speeds,
CON 98, d.
Control flutter is cyclic oscillation of the control surfaces and is caused by their
C of G being aft of their hinge line. I t is prevented by mass balancing the
ailerons such that their C of G lies on their hinge line. This is achieved by the
process of mass balancing, in which balance weights are attached to the control
surface forward of the hinge line.
CON 91. b.
When ailerons are deflected in flight they alter the camber and angle of attack of
the section of the wing to which they are fitted, thereby increasing o r decreasing
the lift force generated. This increase the lift force generated by one wing, whilst
reducing that generated by the other. This dissymmetry of lift causes the
aircraft to roll towards the wing on which the aileron was raised.
In addition to changing the magnitude of the lift forces, aileron deflection and
the subsequent rolling motion also alters the coefficients of drag of each wing
such that the down-going wing produces less drag than the up-going wing. This
dissymmetry of drag causes the aircraft to yaw away from the down-going wing.
Aileron deflection therefore causes a combination of roll and yaw.
CON 92. a.
The word stabilator is an amalgamation of tlie words stabiliser and elevator. In
a stabila~orsystem the eievator is replaced by a pitch control surface made up of
the entire tailplane, wliich changes its angle of incidence in response to pitch
control inputs. The stabilator therefore provides longitudinal stability, pitch
control and in some cases, pitch trimming. In some more advanced aircraft the
stabilator is further modified to become a pair of rollerons, providing all of the
above plus roll control and trimming
CON 93. c.
The three awes of rotation of an aircraft all cross a t the G of G. The C of P of an
aircraft is tlne point at which the lift force is assumed to act and the neutral point
is the point at which the C of G would make the aircraft neutrally stable. The
three axes do not necessarily pass through eitLer the C of P o r the neutral point,
but they may do in certain flight conditions.
CON 94. b.
In fly-by-wire systems, control inputs are converted into electrical signals in the
cockpit, before being transmitted through electrical cables to actuators close to
the flying control surfaces. Because most of the conventional push-pull rods and
cranks are rep!nced using thin electrical cables, fly-by-wire systems are usually
much lighter than their conventional equivalents.
CON 95. h.
In order to reduce structural weight, some aircraft employ a vee-tailed
configuration in which the fin and tailplane are combined. In this design the
conventional rudder and elevators are combined in dual function ruddervators,
providing yaw and pitch control.
CON 96. a.
In manually powered flying control systems, servo tabs are sometimes used to
provide aerodynamic force to move the control surfaces. The pilot's control
inputs move the servo tabs in one direction and the aerodynamic forces acting on
the tabs cause the parent control surface to move in the opposite direction. If the
control systp.m is fully powered without any form of manual reversion, the
controls will be irreversible, making it irnpot;sible for forces acting on the control
surfaces to be fed back through the systrm. Servo tabs cannot therefore be used
in powered flying conlrol systems.
CON 97. c.
JAR 25 states that for JAR certification purposes tlie gradient of the pitch
control stick force gradient must be negative and such that stick force increases
at a minimum rate of 1 lbf for each 6 knots of speed increase.
COlV 98. b.
For JAR certification purposes the gradient of the pitch control stick force
gradient must be negative and such that stick force increases a t a minimum rate
of 1 Ibf for each 6 knots of speed increase. If the basic design of an aircraft is
such that this gradient is not achieved, then a down spring may be employed to
increase the gradient.
CON 99. c.
For JAR certification purposes the gradient of the pitch control stick force
gradient must be negative and such that stick force increases at-a minimum rate
of 1 lbf for each 6 knots of speed increase. If the basic design of an aircraft is
such that this gradient is not achieved, then a bob weight may be employed to
increase the gradient. Because the bob weight is sensitive to load factor it is
particularly effective in increasing stick force gradient in pull up manoeuvres.
CON 100. a.
Primary flying controls are those that control the movement of an aircraft about
its three axes of rotation. They include ailerons, elevons, and roll spoilers, but
not air brakes, lift spoilers, stabilisers or nose wheel steering systems.
CLIRlB 1. b.
iLlaximiim endurance requires minimum fuel consumption rate. Fuel
consumption in a piston engine is proportional to power output, so maximum
endurance is achieved at minimum power speed (Vklp). In a jet aircraft fuel
consumption is proportional to thrust, which in straight and level flight is equal
to drag. So for a jet aircraft maximum endurance is achieved a t minimum drag
speed (VWID).
CLIMB 2. c.
Fuel flow in a piston engine aircraft is proportional to power, whilst that in a jetpowered aircraft is directly proportional to thrust. Thrust output of a jet engine
is approximately constant with increasing airspeed, whilst that of a piston engine
reduces rapidly.
CLIMB 3. c.
When the engines of an aircraft fail in flight, the aircraft possesses a limited store
of potential and kinetic energy. Throughout the glide this energy is dissipated in
overcoming drag. For maximum range it is necessary to achieve the best
compromise between drag and TAS. This is achieved when flying a t the best
L:D ratio, which occurs a t VMD.Although deployment of flaps increases CL, it
also increases CDand reduces L:D ratio. Maximum glide range is therefore
achieved by fling with flaps up a t VMD.
CLIMB 4. a.
The sign of the angle of climb is equal to excess thrust divided by aircraft weight.
Thrust is approximately constant in a jet aircraft, so best excess thrust and hence
best angle of climb is achieved a t minimum drag speed (VkgD).I n a propeller
aircraft thrust reduces rapidly with speed, so best excess thrust and angle of
climb are achieved at the minimum safe flying speed. Best angle of climb is used
to clear obstacles after take-off.
CLIMB 5. d.
As altitude is increased the ratio of TAS to IAS also increases, such that a t 40000
TAS is approximately twice IAS. Power required = drag x TAS, so as altitude
increases, the power required to fly at any given IAS increases. Power available
= thrust x TAS, and as altitude increases, thrust decreases due to reducing air
density. The result of these combined effects is that as altitude increases, power
available reduces and power required increases. Rate of climb is equal to excess
power divided by aircraft weight, so rate of climb reduces with increasing
altitude. The overall effect as altitude increases is that the best rate of climb IAS
decreases whilst the best rate of climb TAS increases.
CLIMB 6. b.
The low speed stall limits the minimum speed at which a subsonic aircraft can
fly, whilst its maximum.speed is limited by high-speed shock stall, as it
a~proacllesthe local speed of sound. As altitude increases air density decreases,
causing a similar decrease in dyliamic pressure (112 pvZ).
Both airspeed indication (IAS) and lift are derived from dynamic pressure, so as
altitude increases, reducing dynamic pressure affects both stalling speed and IAS
to the same degree. Indicated stalling speed therefore remains constant with
changing altitude. Increasing altitude does however alter the ratio of IAS to TAS
such that for a given IAS, the corresponding TAS increases. This means that the
low speed limit, (the stalling TAS) increases with altitude.
The local speed of sound is 38.94 times the square root of the absolute air
temperature. So as altitude increases, temperature and the local speed of sound
both decrease. A subsonic aircraft will suffer shock stall when TAS reaches
some critical fraction of the local speed of sound (Writ), so as the speed of sound
reduces so does the maximum speed of an aircraft. The overall effect is that the
gap between maximum and minimum speed reduces with increasing altitude.
CLIMB 7. d.
Thrust is equal to the mass of air affected by the propulsion system multiplied by
the acceleration given to that mass. This acceleration is equal to the speed at
which the air leaves the system minus the TAS of the aircraft. Because of the low
speed of its propeller wash, the thrust of a propeller reduces rapidly as TAS
increases. In the case of a jet, because of its high velocity jet stream, thrust
reduces much less slowly and is approximately constant over a wide speed range.
Thrust horsepower is equal to thrust multiplied by TAS so for a propeller
aircraft, thrust horsepower increases rapidly a t low speeds, before the reducing
thrust causes it to decrease rapidly as TAS increases. For a jet with its
approximately constant thrust, the thrust horsepower increases approximately
linearly with TAS.
CLIMB 8. c.
Maximum glide range = (L:D ratio) x ( height)
To calculate range in nautical miles the height must also be in nautical miles.
Using the conversion factor 1 nm = 6000 feet, 36000 feet = 6 nm.
CL = 0.45 and C D = 0.0225 and L:D ratio = CL : CDratio
So L:D ratio = 0.45 / 0.225 = 20
This gives
Range
= L:D ratio x Height = 20 r 6 = 120 nautical miles.
So maximum glide range = 120 nm.
CLIMB 9. d.
Maximum glide range is equal to the L:D ratio multiplied by the height and is
not affected by changes in weight. Reducing weight does however reduce the
speed (VMD)a t which best L:D ratio is achieved.
For range in nautical miles the height must also be in nautical miles. Using the
conversion factors, 1 nm = 6000 feet, and 36000 feet = 6 nm.
CL = 0.45 and CD = 0.0225 and L:D ratio = CL : CDratio
So L:D ratio = 0.45 / 0.225 = 20
This gives
Range = L:D ratio x Height
= 20 x 6 = 120 nm
So maximum glide range = 120 nm which is not affected by weight changes.
CLIMB 10. b.
Rate of climb (ROC) = excess power / weight. As altitude increases, power
available decreases and power required increases. Increasing altitude therefore
reduces both excess power and ROC. The service ceilings of propeller and jet
aircraft are the altitudes at which maximum rates of climb are 100 fpm and 500
fpm respectively.
CLIMB 11. c.
Rate of climb equals excess thrust power divided by aircraft weight, so for any
given weight, the maximum rate of climb occurs at the speed at which excess
power is maximised. The thrust horsepower output of a propeller aircraft
increases initially a t low speeds, before reducing rapidly with increasing
airspeed. That of a jet increases approximately linearly with increasing speed.
Power required for all aircraft is equal to drag multiplied by TAS. This
decreases as speed approaches minimum power speed, then increases rapidly
with increasing speed. For a propeller aircraft the maximum value of excess
power and hence rate of climb, occurs at a comparatively low airspeed between
VMPand VMD.Because of its increasing power output, the best rate of climb
speed for a jet aircraft occurs at a much higher speed between VMDand the never
exceed speed (VN~).
CLIMB 12. a.
The absolute ceiling of an aircraft is the altitude at which the rate of climb is
zero. Rate of climb is excess thrust power divided by aircraft weight, so when
rate of climb is zero excess power is also zero. The absolute ceiling is therefore
the altitude at which the power available curve is just tangential to the power
required curve.
CLIMB 13. c.
In gliding flight range through still air is equal to the L:D ratio divided by the
height of the aircraft at the start of the glide. The Tangent of the glide angle is
the rate of descent divided by the ground speed. Although gliding into a
headwind does not alter the range through the air, the air itself is moving in the
opposite direction to the aircraft. Range over the ground is therefore decreased.
Because the ground speed is also reduced whilst the rate of descent remains
constant, the Tangent of the glide angle increases. Between zero and ninety
degrees an increasing tangent equates to an increasing angle, so the glide angle
increases in a headwind. The overall result is that a headwind reduces glide
range and increases glide angle. An alternative way of describing this situation is
that because the headwind opposes the forward motion of the aircraft, the pilot
must minimise the time during which the aircraft is affected by it, in order to
maximise glide range. To do this he must increase glide speed by pushing the
nose further down. This increases glide angle but reduces range.
CLIMB 14. d.
In gliding night range through still air is equal to the L:D ratio divided by the
height of the aircraft at tl:, start of the glide. The Tangent of the glide angle is
the rate of descent divided by the ground speed. Although gliding out of a
tailwind does not alter the range through the air, the air itself is moving in the
same direction as the aircraft. Range over the ground is therefore increased.
The tailwind will not however alter glide endurance or rate of descent.
CLIMB 15. c.
Although flap deployment increases the CL at any given angle of attack and
hence increases lift at any speed, it also increases CDand drag. In most cases the
increase in CLis less than the increase in CD,so the L :D ratio is decreased. The
increased CL enables the aircraft to fly at lower speeds, so both take-off and
landing speeds are reduced. But the reduced L :D ratio reduces excess thrust
and so reduces the maximum angle of climb.
\
.
CLIMB 16. a.
I11 order to maximise endurance it is necessary to minimise the rate at which fuel
is consumed. In a piston engine fuel consumption is proportional to power
output. So minimum fuel flow and best endurance occur a t the minimum power
speed, Vnlp. But in a jet engine fuel flow is proportional to thrust. So minimum
fuel flow and maximum endurance are achieved at the speed requiring minimum
thrust. This is V b l ~ .
CLIMB 17. d.
Power available from any air-breathing engine is proportional to the mass flow
of air passing through it. As altitude increases, air density decreases, so power
available decreases. Power required is equal to drag multiplied by TAS. As
altitude increases the drag at any given IAS remains constant but the TAS
increases. This means that the power required to fly a t any given IAS increases
with altitude. Combining the above two factors means that as altitude increases
power available decreases while power required increases, causing excess power
to decrease.
CLIMB 18. d.
Maximum rate of climb is equal to excess power divided by aircraft weight.
Power required is equal to drag multiplied by TAS. As altitude increases the
drag at any given IAS remains constant but the TAS increases. This means that
the power required to fly at any given IAS increases with altitude such that the
power required : TAS curve moves upwards and to the right. Power available
from any air-breathing engine is proportional to the mass flow of air passing
through it. As altitude increases, air density decreases, so power available
decreases causing the power available : TAS curve to move downwards. The
overall effect of these changes is that as altitude increases the EAS at which best
rate of climb is achieved increases whilst the corresponding TAS decreases.
Because of the different shapes of the power available curves, the effect on TAS
is greater for piston engines than for jets.
CLIMB 19. c.
When gliding for best range it is necessary to maximise the ground distance
covered. When flying into a head wind, this is achieved by increasing speed to
minimise the time during which the aircraft must oppose the headwind. When
gliding out of a tailwind a reduced speed extends the time during which the
aircraft benefits from the tailwind.
CLIMB 20. b.
When an aircraft is banked its lift vector is tilted, thereby reducing the vertical
component of lift. In order to maintain height it is necessary to increase angle of
attack to increase the total lift force. This constitutes an increase in load factor,
and in addition to increasing lift, the drag force is also increased. This increased
drag force means that more thrust is required to balance drag and this
constitutes an increase in power required. This decreases the excess power
available. But rate of climb is proportional to excess power so banking also
reduces rate of climb.
-
CLIMB 21. c.
A piston engine produces brake horsepower, which is converted into thrust by
mean of a propeller. The fuel flow in a piston engine is proportional to power
output. A jet engine produces thrust directly and an increase in fuel flow will
increase thrust. Fuel flow in a jet is therefore directly proportional to thrust. As
TAS increases, the thrust produced by a jet engine decreases slightly, levels of a t
about 250 Kts, then returns to its original value by about 500 Kts. The overall
effect is that jet thrust is approximately constant with increasing speed. The
thrust of a propeller is equal to the mass of air passing through it multiplied by
the acceleration imparted to that air. Because the propeller wash speed is quite
low, the acceleration given to the air rapidly decreases as aircraft speed
increases. This reduction in airflow acceleration causes thrust to decrease
rapidly with increasing airspeed.
CLIMB 22. c.
This problem can be solved using the following equation:
Rate of climb = Excess power (in ftllbflmin) / weight (in Ibf)
In this case rate of climb = 1500 HP x 33000 ft lbf/min/HP / 200000 lbf
Which simplifies to give rate of climb = 247.5 ft / min
The closest option to this is c, 248 ftlmin.
CLlMB 23. c.
The Sin of the best angle of climb = Excess thrust / Weight
Excess thrust = thrust available - thrust required for straight and level flight
In this case excess thrust
= 40000 Ibf
- 20000 Ibf = 20000 Ibf
So for best angle of climb Sine = 20000 lbf 150000 Ibf = 0.4 so best angle of
climb = 23.578 degrees.
CLIMB 24. c.
In a steady climb Lift
degrees
= Weight Cose so Lift
= 50000 Ibf x Cos 23.578
So Lift = 50000 lbf x 0.9165 = 45825 1bf
CLIMB 25. b.
Thrust Horse Power = Thrust (in Ibf) x Velocity (in ftlmin) 133000 ft lbf / min
I HP
Unless otherwise stated in the question 1 Kt may be taken to be equal to 100
ftlmin
In this case T H P = 20000 lbf x 250 Kts x 100 ft/min/Kt / 33000 ft lbf / rnin / H P
So T H P = 15151 HP.
CLIMB 26. c.
The thrust horsepower (THP) generated by a n aircraft's engine is the product of
thrust (in Ibf) and true airspeed (in ft / min) divided by 33000 ft lbf / rnin / THP.
But 33000 ft lbf / min = 550 ft Ibf / sec so this can also be used in the calculation
of THP.
CLIMB 27. a.
Increasing altitude causes air density and hence mass flow through the engine to
reduce. Power output is directly proportional to mass flow so power available a t
40000 ft from a typical jet engine will be approximateIy % that a t ISA msl. Few
if any piston engines can operate at this altitude.
CLIMB 28. b.
Sweeping back the wings would reduce aspect ratio causing a n increase in
induced drag for any given Iift value. Increased induced drag would reduce the
glide endurance and range and hence increase the rate of descent. Increasing the
glide angle wouId reduce Angle of Attack and hence induced drag but the higher
speed attained would increase profile drag and rate of descent, again reducing
glide endurance and range. Induced drag would therefore increase unless glide
angle was increased to decrease lift.
CLIMB 29. b.
An aircraft in flight possesses potential energy proportional to its altitude, and
kinetic energy proportional to its TAS. This energy is provided to the aircraft by
the engines. Maximum rate of descent requires maximum dissipation of
potential energy, whilst the need to remain within VMO,and MMo requires that
increases in TAS be Iimited. Maximum rate of descent is therefore achieved by
reducing the supply of energy from the engines, whilst dissipating energy by
increasing drag. This is most effectively achieved, by deploying spoilers, whilst
reducing thrust to idle and using pitch to limit speed.
CLIMB 30. d.
Point A indicates the maximum value of CL, which occurs immediately prior to
the stalling angle of attack. Point B indicates the best ratio of CL:CD. This is the
best L:D ratio.
CLIMB 31. d.
Using the approximation 1 K t = 100 ft / min the % gradient of the climb path is
equal to the Rate Of Climb (ROC) in ft min divided by the TAS in Kts
Maximum R O C = Excess Power / Weight
So Excess Power = Maximum ROC x Weight
At 50000 lbf Weight R O C for this aircraft is Excess Power 150000 lbf = 1500 ft /
min
So Excess Power = (1500 ft / min x 50000 lbf) = 75000000 ft lbf / mi11
CLIMB 32. c.
Using the approximation 1 Kt = 100 ft / min, the % gradient of the climb path is
equal to the Rate Of Climb (ROC) in ft min divided by the TAS in Kts
Maximum R O C = Excess Power / Weight
Excess Power = (Thrust - Drag) x TAS
Which = (50000 Ibf x 1500 ft / min) = 75000000 ft lbf / min
% climb gradient = (ROC / TAS) and ROC = (Excess Power / Weight)
So % climb gradient = (Excess Power / Weight) / TAS
So to achieve a 5% climb gradient we require 5 % = (75000000 / Weight ) I250
Transforming this equation gives Weight = 75000000 / (5 x 250) = 60000
The maximum weight a t which the aircraft can achieve a 5% climb gradient is
therefore 60000 Ibf.
CLIMB 33. b.
Vx is the speed for best angle of climb and Vy is the speed for best rate of climb.
Best angle of climb occurs when excess thrust is maximum. Best rate of climb
occurs when excess power is maximum.
Increasing weight requires increased lift to maintain any given flight path, and
the creation of this extra lift increases induced drag. Because induced drag is
proportional to 1/v2,the increase in drag is greatest a t low speeds. The increases
both total drag and minimum drag speed pushing the drag curve up and to the
right on the D: speed curve. This also moves the point a t which excess thrust is a
maximum to the right, which represents an increase in Vx Also because power
required is drag multiplied by TAS, moving the drag curve up and to the right
moves the power-required curve in the same direction. The point of minimum
power required is therefore also moved to a higher speed, causing the speed for
maximum excess power (Vy)to increase. Both Vx and Vy therefore increase
with increasing aircraft weight.
.
CLIMB 34. d.
Best climb gradient for any given weight occurs when flying a t the speed at
which excess thrust is maximum. This speed depends upon the shapes and
locations of the thrust available and thrust required curves. Thrust required is
proportional to total drag. This is high at the low speed stalling speed, decreasing
gradually u p to the minimum drag speed V M ~then
, increasing approximately
exponentially, as speed increases further. For a propeller aircraft thrust
available decreases rapidly with airspeed, so excess thrust is maximum a t zero
speed. I n the case of a jet aircraft thrust is approximately constant with
increasing speed, so excess thrust is maximum when drag is minimum at VB~D.
VBIDis also the speed at which the CL : C Dratio is best. Of the options available
in this question, none are correct for a propeller aircraft, but option d is correct
for a jet aircraft. Option d is therefore the most appropriate answer.
CLIMB 35. d.
As altitude increases the power required for any flight condition increases, whilst
the power available from any form of normally aspirated air breathing engine
decreases. The absolute ceiling is the altitude at which the power available curve
just touches the power required curve, and excess power is zero. Because rate of
climb is equal to excess power divided by weight, this is also the altitude a t which
rate af climb is zero.
CLIMB 36. b.
Wheli the engines of an aircraft fail, the aircraft possesses a-store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide elldurance will be achieved by flying a t the speed requiring
the lowest energy consumption rate. But energy consumption rate is power
required, so maximum glide endurance is achieved by flying a t the minimum
power speed. Power required is equal to drag multiplied by TAS and although
VhlDproduces the lowest drag force, it does not produce the lowest power
required. This occurs at Vnlp, which is lower than the minimum VMDfor all
aircraft.
CLIMB 37. b.
When the engines of a n aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity and potential energy due to it height. Throughout the
subsequent glide this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate, which occurs at VMp. But maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying a t the best C L : CDratio, which occurs a t VnID.
I n still air thegliderange is equal to the CL : CDratio multiplied by the height a t
which theglide commences. When flying into a headwind the ground speed
corresponding to any given airspeed is lower, and hence distance covered is
reduced. Flying out of a tail wind has the opposite effect, increasing ground
speed and hence range. The factors determining maximum glide range are
therefore wind, C L : CDratio and height.
CLIMB 38. c.
Despite not being standard JAR defined terminology, VBIGAand VMDRare
sometimes used in JAR ATPL examination questions. VMGA(Velocity Minimum
Glide Angle), is the speed at which the decent angle is minimum. V N l ~(Velocity
R
Minimum Descent Rate), is the speed at which the descent rate is minimum.
VhlCAwill therefore give the shallowest glide angle and hence maximum glide
range. In still air this is VIIID.Recause VMDR
gives the lowest decent rate it also
gives tlie maximum glideendurance. VMDR
in still air conditions is VMP.
CLIMB 39. a.
Despite not being standard JAR defined terminology VMCAand V h l are
~ ~
sometimes used in JAR ATPL examination questions. VMDA
(Velocity Minimum
Glide Angle), is the speed at which the decent angle is lowest. VMDR
(Velocity
Minimum Descent Rate), is the speed a t which the descent rate is lowest. VhqCA
will therefore give the shallowest glide angle and hence maximum glide range. In
still air this is VhlD. Because VhlDRgives the lowest decent rate, it also gives the
maximum glide endurance. VMDRin still air conditions is VRIP.VMDis greater
than VMPfor all aircraft types so VhlCAis greater than VMDR.
CLIMB 40. a.
Climb path
For an aircraft in a steady climb, the
thrust line is angled upwards, so a
eig t
proportion of its weight is carried by
engine thrust. This means that the lift
force generated by tlie wings will be
less than the weight. When in a steady
Lift $
descent a t idle power, the drag line is
angled upwards, such that part of the
weight is carried by the drag. This
again means that the lift generated by
Cosine $ = L / W
the wings will be less than the weight.
The situation in a climb is illustrated
So L = W Cosine $
at the right. From this it can be seen
that the cosine of the angle of climb (I$) is equal to the lift divided by the weight.
This means that the lift is equal to the weight multiplied by the cosine of the
angle of climb. The situation is similar in a descent where lift equals weight
multiplied by the cosine of the angle of descent.
;
;
i
+
CLIMB 41. a.
For an aircraft in steady climb, the
thrust line is angled upwards and so a
proportion of its weight is carried by
engine thrust. This means'that the lift
force generated by the wings will be
less than the weight. Since the thrust
must also oppose the drag force. it is
only the thrust minus the drag that
is available to support part of the weight.
This portion of the thrust is called the
excess thrust and as illustrated in the
diagram, is equal to weight multiplied by
Climb pathf
Sin $ = Excess thrust / W
So Excess thrust = W Sin $
the sin of the angle of climb ($1.
CLIMB 42. c.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequei~tglide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate, which occurs at VMP.But maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying at the best CL : CDratio, which occurs at VMD.
In still air the glide range is equal to the CL : CDratio multiplied by the height at
which the glide commences. When flying into a headwind the ground speed
corresponding to any given airspeed is lower, and hence distance covered is
reduced. Flying out of a tail wind has the opposite effect, increasing ground
speed and hence range. The factors determining maximum glide range are
therefore wind, CL : CDratio and height. For any given combination of CL :CD
ratio and height, a strong tailwind will give the maximum range.
CLIMB 43. a.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate, which occurs at VMP.But maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying at the best CL : CDratio, which occurs at VMD.
In still air the glide range is equal to the CL : CDratio multiplied by the height at
which the glide commences. When flying into a headwind the ground speed
corresponding to any given airspeed is lower, and hence the range covered is
reduced. In such circumstance flying at a speed higher than VMDwill maximise
range by minimising the time during which the headwind is reducing ground
speed.
CLIMB 44. b.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate, which occurs at V M ~But
. maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying at the best CL : CDratio, which occurs at VMD.
In still air the glide range is equal to the CL : CDratio multiplied by the height at
which the glide commences. When flying out of a tailwind the ground speed
corresponding to any given airspeed is higher, and hence the range covered is
increased. In such circumstance flying at a speed lower than VI\IDwill maximise
range by maximising the time during which the headwind is increasing ground
speed.
CLIMB 45. d.
Maximum glide range in still air is achieved by flying at VMD,at which speed the
CL : CDratio is maximum. The diagram in this question is a whole aircraft CL :
CDpolar. The greatest CL : CDratio and hence VMDoccurs on such a diagram
where a tangent drawn from the origin just touches the curve. This is point D in
the diagram.
CLIMB 46. d.
When an aircraft is in steady climb the thrust line is angled upwards, so a
proportion of its weight is carried by engine thrust. This means that the lift force
generated by the wings will be less than the weight. But because the thrust must
carry a proportion of the weight whilst also opposing the drag, thrust must be
greater than drag.
So in a steady climb weight is more than lift and drag is less than thrust.
CLIMB 47. c.
In gliding flight minimum sink rate will produce maximum glide endurance.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate. This occurs at VMP. Although option a
VMSRmight be assumed to mean the velocity for minimum sink rate, this is not a
recognised term. Option b VMD,is the minimum drag speed, which would
achieve the maximum range but not the maximum endurance. VNEis an
outdated term for the maximum allowable flight speed. This would produce
neither maximum endurance nor maximum range.
CLIMB 48. b.
The sin of the maximum angle of climb that can be achieved by an aircraft is
equal to its excess thrust divided by its weight. Excess thrust in any flight
condition is the maximum thrust attainable at that speed, minus the drag. In jet
aircraft thrust remains almost constant at all speeds so maximum excess thrust
and hence maximum climb angle will be achieved at V M ~such
, that the drag
force is at a minimum. Although deployment of flaps increases lift, it also
increases drag. At all but very small angles, the overall effect of flap deployment
is decreased L :D ratio and hence decreased angle of climb. The best angle of
climb will therefore be achieved by flying at VMDwith flaps at zero deflection.
GLIMB 49. b.
The sine of the maximum angle of climb that can be achieved by an aircraft is
equal to its excess thrust divided by its weight. Excess thrust in any flight
condition is the maximum thrust attainable at that speed, minus the drag. In jet
aircraft thrust remains almost constant at all speeds, so maximum excess thrust
and hence maximum climb angle will be achieved a t VMD,such that the drag
force is a t a minimum.
CLIMB 50. d.
VMDfor any given weight is the speed at which the drag force on an aircraft is at
its minimum. This means that VMDis also the speed at which the L :D ratio is
maximum.
Fuel flow in a jet aircraft is proportional to thrust, which in straight and level
flight is equal to drag. So by flying straight and level a t VMD,a jet aircraft will
achieve minimum fuel consumption and hence maximum endurance.
Fuel flow in a propeller aircraft is proportional to power output, so best
propeller aircraft range is achieved by flying the speed at which that ratio of
power required : TAS is greatest. This occurs at V M ~ .
The sin of the maximum angle of climb that can be achieved by an aircraft is
equal to its excess thrust divided by its weight. Excess thrust in any flight
condition is the maximum thrust attainable at that speed, minus the drag. In jet
aircraft thrust remains almost constant a t all speeds, so maximum excess thrust
and hence maximum climb angle will be achieved at VMD,such that the drag
force is at a minimum.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide range will be achieved by flying a t the speed providing the
best ratio of power required to TAS. This occurs at VMDfor all aircraft types.
From the above analysis it can be seen that all of the conditions specified in this
question occur at VMD.
CLIMB 51. d.
Vbq~is the minimum drag speed and is the speed a t which induced drag equals
profile drag. Increasing aircraft weight increases induced drag, causing VMDto
increase.
Maximum glide range is achieved when flying a t VMD.Although increasing
weight increases VMD,it does not'affect the glide angle, nor glide range. The
aircraft merely arrives a t its landing point sooner.
CLIMB 52. c.
JAR 25 defines lift is as that component of the total reaction that acts
perpendicular to the flight path.
CLIMB 53. b.
As altitude increases, the power required for any given flight condition increases,
whilst the power available decreases. The absolute ceiling is the aItitude at which
the power available is equal to the power required. Power available is equal to
thrust multiplied by TAS. Power required is equal to drag multiplied by TAS.
So at the absolute ceiling where power available equals power required, thrust
available equals drag. The maximum rate of climb that can be achieved by an
aircraft is equal to its excess power divided by its weight. So a t the absolute
ceiling where power available equals power required, the excess power and
hence rate of climb are zero. The altitude at which the low speed and high speed
stall lines cross is the aerodynamic ceiling.
CLIMB 54. b.
VMDis the minimum drag speed and is the speed at which induced drag equals
profile drag. Increasing aircraft weight increase's induced drag, causing VmDto
increase.
Maximum glide range is achieved when flying at VMD.Although increasing
weight increases VhlD,it does not affect the glide angle, nor glide range. The
aircraft merely flies down the same slope at a higher speed, and so arrives at its
landing point sooner. Increasing aircraft weight therefore increases both glide
speed and rate of descent.
CLIMB 55. d.
~ ,is proportional to C L I I ~ ~ and
V ~ drag
,
is
IAS is proportional to 1 / 2 p ~lift
proportional to cD1/2pv2. ~ l s cL
o is proportional to angle of attack. SO in a
constant IAS gliding descent, if the angle of attack is kept constant then the CL,
lift and drag will all remain constant. But in a steady descent lift equals the
weight multiplied by the cosine of the angle or gradient of descent. So in a
constant IAS descent if the angle of attack is kept constant, the descent gradient
will remain constant. Also if descent gradient is constant then constant angle of
attack will be achieved by maintaining constant pitch attitude. The pitch
attitude of an aircraft must therefore remain constant in a constant IAS descent.
CLIMB 56. b.
The mach number of an aircraft in flight is the speed of the aircraft as a fraction
of the local speed of sound. The speed of sound is proportional to absolute
temperature, so as altitude decreases, increasing temperature causes the local
speed of sound to increase. This means that in a constant mach descent the TAS
and IAS of an aircraft will increase.
v ~drag
,
is
IAS is proportional to 1/2pv2, lift is proportional to ~ ~ 1 / 2 pand
proportional to C ~ I / ~ ~AISO
V ~cL
. is proportional to angle of attack. SO in a
constant Mach descent, if angle of attack remains constant as IAS increases, both
lift and drag will increase. But increasing drag will reduce speed. Angle of
attack must therefore be decreased in order to reduce drag and hence allow TAS
and IAS to increase to maintain constant Mach number. But decreasing the
angle of attack in order to decrease drag will also decrease lift. Because lift
equals the weight multiplied by the cosine of the angle of descent, decreasing lift
will increase the angle of descent. The descent gradient will therefore increase in
a constant mach gliding descent.
CLIMB 57. d.
~ Vdrag
~ , is
lift is proportional to C L ~ / ~and
IAS is proportional to
proportional to cD1/2pv2. ~ l s cL
o is proportional to angle of attack. SO in a
constant IAS gliding descent, if the angle of attack is kept constant then the CL,
lift and drag will all remain constant. But in a steady descent lift equals the
weight multiplied by the cosine of the angle or gradient of descent. So in a
constant IAS descent if the angle of attack is kept constant the descent gradient
will remain constant. Also if descent gradient is constant then constant angle of
attack will be achieved by maintaining constant pitch attitude. The pitch
attitude of an aircraft, angle of attack and descent gradient must therefore
remain constant in a constant IAS gliding descent.
CLIMB 58. b.
Although a headwind would not affect the
pushes
rate of climb or airspeed of an aircraft, it
would cause the air in which the aircraft
is flying, to be pushed backwards relative
to the ground. This would cause the angle
of climb to increase as indicated in the
diagram at the right. The fact that the
aircraft was in a constant mach number
climb would have no bearing on this situation.
Distance by which headwind
Actual climb gradient (over ground) is increased
Climb gradient through the air
~ e : ~ hgaine/d
t
in given time
CLIMB 59. b.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying a t the speed requiring
the lowest energy consumption rate, which occurs a t VMp. But maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying at the best CL :CDratio, which occurs at VMD.
In still air the glide range is equal to the CL : CDratio multiplied by the height at
which the glide commenses. When flying out of a tailwind the ground speed
corresponding to any given airspeed is increased, and hence the range covered is
increased. In such circumstances, flying at a speed lower than VMDwill
maximise range by maximising the time during which the headwind is increasing
ground speed.
CLIMB 60. b.
IAS is proportional to
lift is proportional to cL1/2pv2,and drag is
o is proportional to angle of attack. SO in a
proportional to cD1/2pvZ. ~ l s cL
constant IAS gliding descent, if the angle of attack is kept constant then the CL,
lift and drag will all remain constant. But in a steady descent lift equals the
weight multiplied by the cosine of the angle or gradient of descent. So in a
constant IAS descent if the angle of attack is kept constant the descent gradient
will remain constant. Also if descent gradient is constant, then constant angle of
attack will be achieved by maintaining constant pitch attitude. If an aircraft is
permitted to descend when in a nose up attitude there is a considerable danger
that the stalling angle will be exceeded. Gliding descents are therefore conducted
in a nose down attitude. The nose down pitch attitude of an aircraft, angle of
attack and descent gradient must therefore remain constant in a constant IAS
gliding descent.
CLIMB 61. c.
The sine of the maximum angle of climb that can be acliieved by an aircraft is
equal to its excess thrust divided by its weight. Excess thrust in any flight
condition is the maximum thrust attainable at that speed, minus the drag. The
speed at which excess thrust and hence angle of climb is a maximum is called Vx.
This term is common to both propeller and jet aircraft, although the
corresponding airspeeds are very different. In jet aircraft thrust remains almost
constant at all speeds, so maximum excess thrust and hence maximum climb
angle will be achieved at VMD,such that the drag force is at a minimum. In the
case of a propeller aircraft, thrust decreases rapidly with increasing airspeed, so
Vx is very close to the minimum safe flying speed. Of the options offered in this
question, VMDis correct only for jets whereas Vx is correct for all aircraft.
CLIMB 62. d.
The maximum rate of climb that can be achieved by an aircraft is equal to its
excess power divided by its weight. Excess power is equal to the power available
minus the power required. In a jet aircraft power available increases linearly
with TAS and maximum excess power and best rate of climb, occur at a speed
somewhat higher than VMD.I n a propeller aircraft power available increase
with airspeed, before again decreasing. The maximum excess power and hence
best rate of climb in propeller aircraft occurs at a speed somewhat lower than
V M ~For
. all aircraft the speed at which rate of climb is maximum is that a t
which excess power is a maximum. This speed is called Vy.
CLIMB 63. c.
When the engines of an aircraft fail, the aircraft possesses a store of kinetic
energy due to its velocity, and potential energy due to it height. Throughout the
subsequent glide, this energy is consumed in doing work, pushing the aircraft
forward against the drag force. Because this energy store cannot be replenished,
the maximum glide endurance will be achieved by flying at the speed requiring
the lowest energy consumption rate, which occurs at VMP. But maximum glide
range requires the best ratio of energy consumed to distance flown. This occurs
when flying at the maximum CL : CDratio, which occurs at VMD.
CLIMB 64. c.
When a n aircraft is in steady climb the
thrust line is angled upwards and so a
Flight path
proportion of its weight is carried by
engine thrust. This means that the lift
force generated by the wings will be
less than the weight. When in a steady
descent a t idle power, the drag line is
angled upwards, such that part of the
weight is carried by the drag. This
Cosine $ = L / W
again means that the lift generated by
the wings will be less than the weight.
So L = W Cosine 4)
The situation in a climb is illustrated
a t the right. From this it can be seen
that the cosine of the angle of climb ($1, is equal to the lift divided by the weight.
This means that the lift force in such a climb equals the weight multiplied by the
cosine of the angle of climb. But the cosine of all non-zero angles is less than 1, so
lift equals less than weight. Load factor is equal to lift divided by weight, so in a
steady climb when lift is less than weight, the load factor must be less than 1. In
considering this situation it should be noted that this result is due to the
definition of load factor. The aircraft is still subject to l g and hence weighs the
same as in level flight.
CLIMB 65. c.
When an aircraft is in steady climb the
thrust line is angled upwards and so a
Flight path
proportion of its weight is carried by
engine thrust. This means that the lift
force generated by the wings will be
less than the weight. When in a steady
descent at idle power, the drag line is
angled upwards such that part of the
weight is carried by the drag. This
again means that the lift generated by
Cosine $ = L / W
the wings will be less than the weight.
The situation in a climb is illustrated
So L = W Cosine $
a t the right. From this it can be seen
that the cosine of the angle of climb ($) is equal to the lift divided by the weight.
This means that the lift force in such a climb equals the weight multiplied by the
cosine of the
angle of climb.
CLIMB 66. c.
An aircraft in flight possesses a store of kinetic energy proportional to its speed,
and potential energy proportional to its height above ground. In a n emergency
descent it is necessary to reduce this store of energy so that the aircraft attains
the desired lower altitude, as quickly as possible without exceeding its maximum
speed. If this is attempted by simply pushing the nose down, then the aircraft
will accelerate into a dive in which potential energy is transformed into kinetic
energy with little energy being lost. The overall effect of this manoeuvre will be
that the aircraft will exceed its limiting speed or mach number (VMo1 MMO)If
however the thrust levers are set to idle, this will prevent further energy being
supplied to the aircraft. If the speed brakes, flaps and gear are deployed the
increased drag will increase the rate at which energy is extracted from the
aircraft. If the nose is then pushed forward the aircraft will enter a gliding dive
in which speed can be controlled by use of pitch attitude.
CLIMB 67. c.
Vt is the take-off safety speed. Vslp is the speed at which power required is
minimum. VMDis the speed a t which drag is minimum. Vx is the speed a t which
the maximum angle of climb can be achieved. Vy is the speed a t which the
maximum rate of climb can be achieved. For a jet aircraft V2 < VMp< VMD= VX
< Vv. For a propeller aircraft V2 < VX< VMP< VY< VMD.Option c is therefore
correct for a jet aircraft.
CLIMB 68. c.
The maximum rate of climb that an aircraft can achieve is equal to the excess
power minus the power required a t that speed. Neither excess power nor power
required are affected by headwinds. Headwinds will therefore have no effect on
the rate of climb.
CLIMB 69. a.
Although a headwind would not affect the Distance by which headwind pushes
rate of climb o r airspeed of an aircraft, it the air
would cause the air in which the aircraft
was flying, to be pushed backwards relative
to the ground. This would cause the ang!e
of climb to increase as indicated in the
diagram at the right.
Actual angle of climb (over ground) is increased
Angle of climb through the air
CLIMB 70. a.
Although a headwind would not affect tlie
rate of climb o r airspeed of an aircraft, it
would cause the air in which the aircraft
is flying, to be pushed backwards relative
f o the ground. This would cause the climb
gradient to increase as indicated in the
diagram at the right.
/
Height gained in given time
Distance by which headwind pushes
the air
Actual climb gradient (over ground) is increased
/
Climb gradient through the s i r Height gained in given time
CLIMB 71. a.
Power required in straight and level flight is equal to drag multiplied by TAS.
Both drag and IAS are proportional to 1 1 2 so~ a ~
t any
~ given IAS the drag
remains constant at all altitudes. But the ratio of TAS : IAS increases with
increasing altitude. At 40000 feet ISA for example, the TAS is twice the IAS.
This means that a t any given IAS, as altitude increases power required increases,
because the TAS by which drag is multiplied is increasing while the drag
remains constant. The power required therefore increases with increasing
altitude.
CLIMB 72. b.
When an aircraft is in a steady climb the
thrust line is angled upwards and so a
Flight path
proportion of its weight is carried by
engine thrust. This means that the lift
force generated by the wings will be
less than the weight. When in a steady
descent a t idle power, the drag line is
angled upwards such that part of the
weight is carried by the drag. This
again means that the lift generated by
Cosine 4 = L / W
the wings will be less than the weight.
The situation in a climb is illustrated
So L = W Coslne 4
a t the right. From this it can be seen
that the cosine of the angle of climb (4) is equal to the lift divided by the weight.
This means that the lift force in such a climb equals the weight multiplied by the
cosine of the angle of climb. But the cosine of all non-zero angles is less than 1, so
lift equals less than weight. Load factor is equal to lift divided by weight so in a
steady climb when lift is less than weight the load factor must be less than 1. In
considering this situation it should be noted that this result is due to the
definition of load factor. The aircraft is still subject to l g and hence weighs the
same as in level flight.
CLIMB 73. a.
This problem can be solved using the following equation:
Rate of climb = Excess power / weight
So in this case rate of climb = (25000 ft Ibflmin) / 10000 Ibf
Which is rate of climb = 2.5 ftlmin
CLIMB 74. d.
This problem can be solved using the following equation:
The Sine of the angle of climb = Excess thrust / weight
Excess thrust = thrust - drag
In this question excess thrust = 25000 Ibf - 15000 lbf = 10000 lbf
So in this case the Sine of the angle of climb = 10000 lbf / 10000 Ibf = 1
If the sine of the angle of climb is 1 then the angle of climb is 90'.
This aircraft is therefore capable of climbing vertically.
CLIMB 75. b.
This problem can be solved by first using the figures for excess power and TAS
to drive excess thrust using the following equation:
Excess power = excess thrust x TAS
So excess thrust = excess power / TAS
Both sides of the equation must be in the same units so using 1 K t = 100 ftlrnin
In this case (50000000 ft lbflmin) / (250 Kts x 100 ft/min/Kt)
Which simplifies to give excess thrust = 2000 lbf.
This can then be used to calcuIate maximum angle of climb using the following
equation:
Sine of the angle of climb = excess thrust / weight
So in this case the sine of the angle of climb = 2000 Ibf / 10000 lbf = 0.2
If the sine of the angle of climb is 0.2 then the angle of climb is 11.5'.
CLIMB 76. d.
occurs at the stall speed (Vs), so point A is
The maximum value of CL (CI,MAX),
the stall. The best L :D ratio occurs on a whole aircraft polar where a tangent
drawn from the origin touches the curve. This is point B in the diagram
provided in this question. But best L :D ratio occurs at \IhqD,
SO point 3 is VMD.
So point C is more than Vs but Iess than VMD.Point D is at a speed higher than
VMD.VY for a propeller aircraft occurs at a speed between Vs and V M ~whilst
,
Vy for a jet occurs at a speed higher than VhlD. Of the options provided in this
question only option d, propeller aircraft Vy aurl Jet aircraft Vy satisfy the above
criteria.
CLIMB 77. a.
The maximum value of CL (CLMAX),
occurs a t the stall speed (Vs), so point A is
the stall. The best L :D ratio occurs on a whole aircraft polar where a tangent
drawn from the origin touches the curve. This is point B in the diagram
provided in this question. But best L :D ratio occurs at VRID,so point B is VMD.
So point I9 is more than Vs but less than point C and point C is more than both
Vs and point D, but less than VblD.
Vy for a propeller aircraft occurs at a speed between Vs and V h I ~VX
. for a
propeller aircraft occurs at a speed higher than Vs but lower than Vy. Of the
options provided in this question only option a, propeller aircraft Vy and
propeller aircraft Vx satisfy the above criteria.
CLIMB 78. b.
An aircraft in flight possesses kinetic energy proportional to its TAS, and
potential energy proportional to its height. When the engines fail in flight this
energy cannot be replenished, but is dissipated in doing work to push the aircraft
forward against the drag force. The rate of energy dissipation is the power
required. To achieve maximum glide endurance it is necessary to minimise the
energy dissipation rate. This is achieved by flying at VktP. VMPis the minimum
power speed and is proportional to aircraft weight. Increasing aircraft weight
increases VhIP,and hence increases the glide speed required for best glide
endurance. Increasing weight also increases the power required at all speeds
including V M ~and
, hence decreases maximum glide endurance.
CLIMB 79. a,
An aircraft in flight possesses kinetic energy proportional to its TAS, and
potential energy proportional to its height. When the engines fail in flight this
energy cannot be replenished, but is dissipated in doing work to push the aircraft
forward against the drag force. The rate of energy dissipation is the power
required. To achieve maximum glide range it is necessary to fly at the speed at
which the ratio of power required to TAS is greatest. This is achieved by flying
a t VMD.VRtDis the mininium drag speed and is the speed at which induced drag
equals profile drag. Increasing aircraft weight increases induced drag, causing
V B ~toD increase. Maximum glide range is achieved when flying at VMD.
Although increasing weight increases VnlD,it does not affect the glide angle, nor
glide range. The aircraft merely flies down the same slope a t a higher speed. Its
rate of desceiit is therefore increased and it arrives at its landing point sooner.
Increasing aircraft weight therefore increases VMD,glide speed and rate of
descent.
CLINIB 80. c.
The sine of tlie maximum angle of climb that an aircraft can achieve is equal to
its excess thrust divided by its weight. At any given weight, the maxiinum angle
of climb is achieved by flying a t the speed providing the greatest excess thrust.
This speed is called Vx. Increasing aircraft weight increases the lift required and
hence increases induced drag. But induced drag is greatest at low speeds, so the
effect of increasing speed is to push the Drag : TAS curve upwards and to the
right. This means that drag a t any given speed is greater, and also that the speed
providing minilnuill drag i s iucreased. The increase in drag means that more
thrust is required to overcome drag, so excess thrust is reduced. The pushing of
the drag curve to the right means that Vx is increased. The overall effect of
these changes is that increasing weight increases VXand decreases the maximum
angle of climb.
CLIMB 81. a.
The sine of the maximum angle of climb that an aircraft can achieve is equal to
excess thrust divided by weight. The maximum rate of climb is equal to excess
power divided by weight. Activation of reheat increases both thrust and power
available and hence increases both angle of climb and rate of climb.
CLIMB 82. c.
The sine of the maximum angle of climb that an aircraft can achieve is equal to
excess thrust divided by weight. The maximum rate of climb is equal to excess
power divided by weight. Activation of reheat increases both thrust and power
available and hence increases both angle of climb and rate of climb. But if an
aircraft is already flying a t the speed at which a climb is to be conducted, then all
of the extra thrust and power produced by activation of reheat will be excess
thrust and excess power. That is to say, neither drag nor power required will
increase, so all of the extra thrust and power is excess to that required to
maintain airspeed. If activation of reheat doubles thrust and power available
while thrust and power required remain constant, then excess power and thrust
will be more than doubled. This means that the angle of climb and rate of climb
will be more than doubled.
CLIMB 83. a.
For low angles of climb (up to about 15') a reasonably accurate estimate of climb
gradient in still air can be calculated using the following equation:
% gradient = Rate of climb (in ftlmin) / TAS (in Kts) %
So in this case the % gradient = 1000 ftlmin / 200 Kts
Notes:
1.
= 5%
The above equation is based on the assumption that I Kts =
100 ft/min so the equation becomes:
% gradient = ROC in ftlmin / ((TAS in ftlmin) / 100)).
Which is why the result is the % gradient.
2.
This equation gives only an approximate solution, a more
Accurate figure would be provided using the following
equation:
% gradient = ROC in ft/min / Ground speed in Kts.
CLIMB 84. c.
For low angles of climb (up to about 15') a reasonably accurate estimate of climb
gradient in still air can be calculated using the following equation:
% gradient = Rate of climb (in ftlmin) / TAS (in Kts) %
Rearranging this equation gives rate of climb = TAS x % gradient.
So in this case rate of climb = 250 Kts x
15%
Which is 3750 ft/min
Notes:
1.
The above equation is based on the assumption that I Kts =
100 ft/min so the equation becomes:
% gradient = ROC in ft/min / ((TAS in ft/min) / 100)).
Which is why the result is the % gradient.
2.
This equation gives only an approximate solution, a more
Accurate figure would be provided using the following
equation:
% gradient = ROC in ftlmin / Ground speed in Kts.
CLIMB 85. d.
An aircraft in flight possesses a store of kinetic energy proportional to its speed,
and potential energy proportional'to its height above ground. In an emergency
descent it is necessary to reduce this store of energy so that the aircraft attains
the desired lower altitude as quickly as possible, without exceeding speed limits.
If this is attempted by simply pushing the nose down, then the aircraft will
accelerate into a dive in which potential energy is transformed into kinetic
energy, with little energy being lost. The overall effect of this manoeuvre will be
that the aircraft will exceed its limiting speed or mach number (VMO/ MMO). If
however the thrust levers are set to idle, this will prevent further energy being
supplied to the aircraft. If the speed brakes, flaps and gear are deployed, the
increased drag will increase the rate at which energy is extracted from the
aircraft. If the nose is then pushed forward, the aircraft will enter a gliding dive
in which speed can be controlled by use of pitch attitude. Maximum rate of
descent is therefore achieved by using minimum thrust, together with maximum
drag, and varying pitch attitude to maintain airspeed within limits.
CLIMB 86. d.
VMo is the maximum operating speed of an aircraft, expressed as CAS. It is
determined by the ability of the structure to withstand the aerodynamic forces
produced by the dynamic pressure acting upon it. Dynamic pressure is
proportional to air density, so at low altitudes where air density is high, VMo is
the limiting parameter. MMo is the maximum operating speed of an aircraft,
expressed as a mach number. If the aircraft is operated at higher mach
numbers, the generation of supersonic airflows and shock waves will cause highspeed buffet and eventually high-speed stall. MMOis some specific fraction of the
local speed of sound.
But the local speed of sound is proportional to air temperature and hence
decreases with increasing altitude. This means that the CAS equating to MMo
deceases as altitude increases. The overall effect of these factors is that VMo is
the limiting speed at low altitude and MMOis the limiting speed a t high altitude.
The altitude at which the TAS equating to VMois equal to that equating to MMo
is called the crossover altitude.
When climbing at constant CAS o r IAS, both TAS and Mach number increase.
There is therefore a danger that MMOwill be exceeded in a constant IAS climb
above the crossover altitude.
CLIMB 87. c.
V Mis~the maximum operating speed of an aircraft expressed as CAS. I t is
determined by the ability of the structure to withstand the aerodynamic forces
produced by the dynamic pressure acting upon it. Dynamic pressure is
proportional to air density, so a t low altitudes where air density is high, VMo is
the limiting parameter. Mhlo is the maximum operating speed of a n aircraft
expressed as a mach number. If the aircraft is operated at higher mach
numbers the generation of supersonic airflows and shock waves will cause highspeed buffet and eventually high-speed stall. MMOis some specific fraction of the
local speed of sound.
But the local speed of sound is proportional to air temperature and hence
decreases with increasing altitude. This means that the CAS equating to MMo
deceases as altitude increases. The overall effect of these factors is that VkIo is
the limiting speed at low altitude and Mwo is the limiting speed at high altitude.
When climbing a t constant TAS, both CAS and Mach number decrease so there
is no danger that VMo or MMo will be exceeded. if however the climb is
continued to a sufficiently high altitude, the CAS might reduce to less than the
low speed stalling speed Vs. The aircraft will then stall.
CLIMB 88. c.
The mach number of an aircraft is its TAS, expressed as a fraction of the local
speed of sound. Mach 0.8 for example represents 80% of the local speed of
sound. But the local speed of sound is proportioiial to air temperature, and
hence decreases with increasing altitude, up to the tropopause a t 36000 feet ISA.
Above this altitude both temperature and the local speed of sound remain
constant. This means that the TAS equating to any given mach number,
decreases with increasing altitude up to 36000 feet ISA, and remains constant at
higher altitudes.
But the CAS equating to any given TAS decreases with increasing altitude
throughout the entire atmosphere. This means that the CAS equating to any
given mach number decreases with increasing altitude throughout the
atmosphere. So if a constant mach climb is continued to a sufficiently high
altitude, or if the constant mach number is sufficiently low, the CAS might
decrease to less than the low speed stalling speed Vs. The aircraft will then stall.
CLIMB 89. a.
VMo is the maximum operating speed of a n aircraft expressed as CAS. I t is
determined by the ability of the structure to withstand the aerodynamic forces. produced by the dynaniic pressure acting upon it. Dynamic pressure is
proportional to air density, so at low altitudes where air density is high, VMo is
the limiting parameter. MMo is the maximum operating speed of a n aircraft
expressed as a mach number. If the aircraft is operated at higher mach
numbers the generation of supersonic airflows and shock waves will cause high-
speed buffet and eventually high-speed stall. MMo is some specific fraction of the
local speed of sound.
But the local speed of sound is proportional to air temperature and hence
increases with decreasing altitude. This means that the CAS equating to MMo
increases as altitude decreases. Thz altitude at which the CAS equating to VMOis
the same as that equating to Mk10 is called the crossover altitude. The overall
effect of these factors is that Vhlo is the limiting speed below the crossover
altitude and MMo is the limiting speed above the crossover altitude. When
descending a t constant mach number there is a danger that VMOwill be exceeded
below the crossover altitude.
CLIMB 90. c.
Indicated air speed (IAS) is proportional to 1/2pv2,where p is air density and V
is TAS. Whenever the airspeed indicator senses any given 1/2pv2, it will always
produce the same IAS indication regardless of altitude. When climbing a t
constant 1 / 2 p ~ ' ,the IAS will therefore remain constant.
CLIMB 91. b.
Indicated air speed (IAS) is proportional to 1/2pv2,where p is air density and V
is the true airspeed (TAS). Whenever the airspeed indicator senses any given
1 / 2 p ~ ' ,it will always produce the same IAS indication regardless of altitude.
When climbing a t constant 1/2pv2, the IAS will therefore remain constant. But
as altitude increases, p decreases. So if IAS remains constant then V (TAS) must
increase such that the increase in V' is equal to the decrease in p, and 1/2pv2
remains constant. At 40000 feet in the standard atmosphere, p is ?A of its sea
level value, so for any given IAS, V' must be 4 times its sea level value. This
means that a t 40000 feet ISA, the best climb TAS is twice the value of the best
climb IAS.
CLIMB 92. c.
The maximum rate of climb that an aircraft can achieve at any altitude is equal
to its excess power divided by its weight. As altitude increases power available
decreases and power required increases. This means that excess power and
hence rate of climb decrease with increasing altitude. The absolute ceiling is the
altitude a t which power available is equal to power required for level flight and
excess power and rate of climb are zero. The rate of climb of any aircraft a t its
absolute ceiling is therefore zero.
CLIMB 93. a.
The maximum rate of climb that an aircraft can achieve a t any altitude is equal
to its excess power divided by its weight. As altitude increases power available
decreases and power required increases. This means that excess power and
hence rate of climb decrease with increasing altitude. The absolute ceiling is the
altitude at which power available is equal to power required for level flight and
excess power and rate of climb are zero. The rate of climb of any aircraft at its
absolute ceiling is therefore zero. Also because power required and power
available vary with airspeed, straight and level flight is possible at only one speed
at the absolute ceiling. So at the absolute ceiling Vx and Vy are the identical.
CLIMB 94. a.
The maximum angle of climb that an aircraft can achieve at any given airspeed
is proportional to excess thrust divided by weight. But the thrust of a propeller
aircraft reduces rapidly with increasing airspeed, whilst drag decreases up to
VMDthen increases at higher speeds. The shapes of the thrust and drag curves
are such that a propeller aircraft achieves its maximum angle of climb at a speed
close to its minimum safe flying speed. Excess thrust and angle of climb both
decrease with increasing airspeed.
CLIMB 95. d.
The maximum angle of climb that an aircraft can achieve at any given airspeed
is proportional to excess thrust divided by weight. The thrust of a jet aircraft
remains approximately constant at all speeds, whilst drag decreases up to VMD
then increases at higher speeds. Excess thrust and angle of climb both therefore
increase up to VMDthen decrease with increasing airspeed.
CLIMB 96. d.
The maximum rate of climb that an aircraft can achieve at any given speed is
equal to its excess power divided by its weight. Power available is equal to thrust
multiplied by TAS. Thrust from a jet aircraft remains approximately constant
at all speeds so power available increases linearly with increasing TAS. But
power required is equal to drag multiplied by TAS. Drag decreases with
increasing speed up to VMDthen increases, so the power required curve is of a
generally similar shape to the drag curve. The minimum power required does
not however occur at VMDbut at a lower speed termed VMP.
Because of the shapes of the concave power required curve and the linearly
increasing power available curve, maximum excess power and hence maximum
rate of climb occurs at a speed greater than VMD. This speed is termed Vy and is
approximately 1.3VMD.The rate of climb of a jet aircraft therefore increases up
to approximately 1.3VMDthen decrease with increasing speed.
CLIMB 97. d.
This problem can be solved using the following equations:
Excess thrust = thrust - drag
And the sine of angle of clirnb = Excess thrust I weight
In this case excess thrust = 25000 Ibf - 5000 Ibf = 20000 lbf.
So the sine of angle of climb = 20000 Ibf I 15000 lbf = 1.67
But the sines of angles cannot be greater than 1, which equates to an angle of 90'.
This means that the aircraft can climb at an angle of 90°, which would require
15000 Ibf of excess thrust. In this condition it would still have 5000 lbf of excess
thrust in reserve to enable it to accelerate in the climb.
CLIMB 98. a.
This question is complicated by the fact that the excess power is expressed as
thrust horsepower THP. This problem can be overcome by using the conversion
factor 1 T H P = 33000 ft Ibf / min.
So 350 THP = 350 THP x 33000 ft lbf / min / T H P = 11550000 ft lbf /min.
The problem can now be solved using the following equation:
Rate of climb = excess power / weight.
So in this case rate of climb = 11550000 ft lbf / min / 10000 Ibf
Which gives a rate of climb of 1155 ftlmin.
CLIMB 99. b.
Rate of climb = excess power divided by weight. So if the weight of a n aircraft is
doubled and its excess power remains unchanged its rate of climb would be
halved. This is however an underestimation of the decrease in rate of climb
because the lift required to support the extra weight would also increase drag
and hence decrease excess power. Doubling the weight of an aircraft would
therefore decrease its rate of climb by more than half. This option is not
however available in this question so option b (half) is the most accurate.
CLIMB 100. b.
In straight and level flight the attitude of a n aircraft is such that the lift equals
the weight. As fuel is consumed during flight, the weight decreases, so either
angle of attack or speed must be decreased in order to educe lift to match the
reducing weight. If both angle of attack and airspeed remain constant as weight
reduces, the lift will exceed the weight, causing the aircraft to gain altitude. This
situation is termed a cruise climb.
TURN 1. a.
A level turn is initiated by banking the aircraft in the direction of the intended
turn. This tilts the total reaction such that a proportion of the lift force is angled
in the direction of the turn, thereby pulling the aircraft into the turn. Deflection
of part of the lift vector in the direction of the turn reduces the lift available to
maintain altitude. So the angle of attack must be increased to restore the
original level of vertical lift. Increasing angle of attack also increases drag, so
more thrust is necessary to maintain airspeed. The correct actions are therefore
to increase thrust and angle of attack.
,
TURN 2. a.
Radius R = v2/ (g Tan AOB) which in this case is R = 2502 Kts / (10 m/s2 s
Tan 45')
To calculate R in meters it is necessary to convert Kts into m/s.
Using the conversion factors given R = (250Kts x 0.51 m/s!~t)~/ (10 m/s2 x 1)
Which resolves to R = 16256.25 m2/s2 / 10 m/s2
Which is radius of turn = 1625.25 meters.
TURN 3. c.
Load factor = L/W and in a banked turn part of the lift is angled in the direction
of bank, to pull the aircraft into the turn. In order to provide this horizontal lift
component, whilst continuing to support the weight of the aircraft, it is necessary
to increase the total magnitude of the lift force. But load factor = lifffweight, so
load factor increases as an aircraft enters a banked turn.
Load factor in a banked turn is calculated using the equation n = l/Cos AOB
So in this case the banked load factor is 1/Cos 60'
which is 1/ 0.5
So the new load factor is 2.
But in straight and level flight prior to entering the turn L = W so load factor
was 1. The load factor has therefore increased by 100% as a result of the banked
turn.
TURN 4. d.
Banking of an aircraft to initiate a turn angles a proportion of the total lift
towards the centre of the turn, so less lift is available to support the aircraft. To
maintain altitude it is therefore necessary to increase angle of attack to increase
the total value of lift. This process also increases total drag, so more power is
required to maintain airspeed. The % by which power must be increased is
directly proportional to the angle of bank, so both aircraft will require the same
% increase. Power required is proportional to mass, so although the % increase
will be the same for both aircraft, the magnitude of the increase will be greatest for the high mass aircraft.
Turn radius, R = v2/ (g Tan AOB) and rate of turn = V / R.
Neither of the above equations includes mass, so the different masses of the two
aircraft will have no effect on radius or rate of turn. Both aircraft will therefore
achieve the same turn radius and rate of turn.
TURN 5. c.
Radius in a level banked turn is R = v2/ gTanAOB.
If V is increased by a factor of 2 then v2is increased by a factor of 22 which is 4.
The question states that all other factors remain unchanged so increasing v2by a
factor of 4 will increase radius by a factor of 4.
The new radius will therefore be 4000 m.
TURN 6. b.
Load factor in a level banked turn is 1 / Cos AOB
Cos 45' is 0.7071, so load factor = 11 0.7071 = 1.4142
So load factor in a level 45' banked turn is approximately 1.41
But load factor in level flight prior to the turn was 1 so the increase in load factor
was approximately 4l0/0.
TURN 7. d.
Radius of turn = R = V' / gTan AOB
So Tan AOB = v2/ gR
In this question the velocity is increased by a factor of 2.5 (from 200 to 500 Kts)
whilst radius of turn and g remain unchanged.
So for the new AOB the ratio of Tan new AOB to Tan old AOB is 2.5' :1
So Tan new AOB is 2.5' x Tan old AOB which is 6.25 x Tan 20" which is 2.2748
So the AOB is 66'
TURN 8. c.
Radius of turn = R = v2/ gTAnAOB
To calculate radius of turn in meters, V must be converted to meters. This can
be done using the conversion factor 1 Kt = 0.515 m/s/Kt.
This gives 400 Kts x 0.515 m/s/Kt = 206 m/s.
AOB is 45'so R = 206'1 10 x ~ a n 4 5 ' = 42436 / 10
So Radius of turn is 4243.6 m
TURN 9. c.
The formula for radius of turn is R = v2/ g Tan AOB. As this does not include
any reference to weight, changes in weight will have no effect on radius of turn.
The power and angle of attack required to achieve a given turn radius will
however increase with weight.
'TURN 10. a.
Load factor in a constant altitude banked turn is equal to 1 I Cos AOB. The
cosine of 60' is 0.5 so load factor in a 60' banked turn is 110.5 which is 2. But
load factor is lift / weight so if load factor is 2 then lift is twice as much as weight.
This means that on entering a 60' banked turn lift must increase by 100%.
TURN 11. a.
Radius of turn = V2 / g Tan AOB. Weight does not appear in this equation, so
changes in weight will not affect turn radius. This question is therefore
concerned only with a comparison of the following equations:
(250 ~ t s ) '/ g Tan 45'
and
(200 ~ t s /) g~Tan 45'
Multiplying both equations by g Tan 45' gives (250 ~ t s and
) ~ (200 ~ t s ) ~
2502 is clearly larger than 200' so radius at 250 Kts will be greater than that at
200 Kts.
TURN 12. b.
This problem can be solved using the following formula:
Vs(in a manoeuvre) = vSIg
d(n)
Inserting the given values of Vsl, and n gives:
Vs(in a 2g manoeuvre) = 200 Kts x 62
Which is 281 Kts.
TURN 13. d.
The load factor in a constant altitude banked turn is equal to (l/Cos AOB). The
cosine of 90' is zero so the load factor in a constant altitude 90' banked turn
would be infinitely high. There is no airspeed that would enable this to be
achieved.
TURN 14. b.
This problem can be solved using the following formula:
Vs(in a manoeuvre) = ~ s l , d ( l /Cos AOB)
Inserting the values of Vs;, and AOB gives the following:
V ~ ( i na 45' bank) = 100 Kts d ( l I Cos 45')
Which is 119 Kts.
=
100 Kts d1.414
TURN 15. b.
The limiting load factor for JAR certificated passenger aircraft with flaps down
is 2. Load factor in a constant altitude banked turn is equal to (11Cos AOB).
The angle of bank that will impose the limiting load factor is that angle for which
(11 Cos) is 2. The required angle is 60'.
TURN 16. a.
The limiting load factor for JAR certificated passenger aircraft with flaps up is
2.5. Load factor in a constant altitude banked turn is equal to (l/Cos AOB). The
angle of bank that will impose the limiting load factor is that angle for which (11
Cos) is 2.5. The required angle is 66'.
TURN 17. b.
The load factor in a constant altitude banked turn is equal to (1ICos AOB). The
cosine of all angles greater than zero is less than 1, so 1ICosine for all such angles
is more than 1. This means that on entering a constant altitude banked turn, the
angle of attack must be increased to increase lift to support the aircraft weight at
the higher load factor. But increasing lift will increase drag so more thrust will
also be required.
TURN 18. c.
In a constant altitude banked turn the rate of turn, ROT = (g Tan AOB) / TAS.
Of the options listed in this question only option c, TAS is included in the above
equation.
TURN 19. d.
This problem can be solved using the following equation:
Vs(in a manoeuvre) = V S ~ ~ ~ ( ~AOB)
/COS
So in a 60' banked turn ~ ~ ( 7 bank)
5'
= ~ s l , d ( l / ~ o75'
s)
Which simplifies to ~ ~ ( 6 bank)
0'
= ~ s l ~ d ( 3 . 8 6 4 )which = 1.965Vslg
So stalling speed will increase by approximately 97% in a 75' banked turn.
TURN 20. a.
This problem can be solved using the following equation:
Vs(in a manoeuvre) = vslgd(l/cos AOB) which can be rearranged to give:
VSlg = Vs(in a manoeuvre) / d(11Cos AOB) which for a 60' banked turn gives:
Vsl, = Vs(in a manoeuvre) / d(2) which is 0.709Vs(in a manoeuvre)
So when pulling out of a 60' banked turn stalling speed decreases to
approximately 71% of its previous value. This represents a reduction of 29%.
TURN 21. a.
Vs(in any manoeuvre) =. vsl,d(n), where n is the load factor.
TURN 22. b.
Vs(in a manoeuvre) = ~ s ~ , d ( l / AOB)
~ o s where (lICos AOB) is tlie load factor
in the turn. But load factor in straight and level flight is 1,so the increase in load
factor on entering a banked turn is (1ICos AOB) -I), which is (new load factor 1)
The % increase in stalling speed in a banked turn is therefore
d(new load factor -1) x 100%
TURN 23. b.
Load factor in a constant altitude banked turn = (1ICos AOB), so in a 35' banked
turn the load factor will be (1ICos 35'1, which is 1.221. But load factor = lift I
weight, so if weight remains constant, then lift must increase by a factor of 1.221.
In straight and level flight prior te the commencement of the turn, the load
factor would be 1, so lift w ~ u l dequal weight at 40000 N. Multiplying this by
1.221 gives the new lift ferce of 48830 N.
TURN 24. a.
Load factor in a constant altitude banked turn = (1ICos AOB), so in a 45' banked
turn the load factor will be (1ICos 45'1, which is 1.414. But load factor = lift I
weight, so if weight remains constant, then lift must increase by a factor of 1.414.
~
induced drag. CD1is
But increasing lift a t constant speed a l s increases
proportional to CL' so the increase in drag will be proportional to the square of
the increase in lift. Drag will therefore increase by a factor of 1.414' which is 2.
To maintain constant speed, thrust must therefore be increased by a factor of 2
to balance the higher drag force.
TURN 25. c.
I n a level banked turn the lift force is tilted in the direction of bank in order to
producea horizontal component of lift to pull the aircraft into the turn: This
reduces the vertical component of lift, so total lift must be increased in order to
maintain altitude. Load factor = lift I weight, so this increase in lift constitutes
a n increase in load factor.
In this condition total lift = vertical componei~tof lift I Cosine of the AOB and
for level flight lift = weight.
But load factor = lift I weight, so load factor = vertical component of lift 1 (Cos
AOB x vertical component of lift)
This means that load factor = 1ICos AOB.
But Cos AOB varies with AOB, so load factor in a level banked turn depends
only on AOB.
TURN 26. a.
In a constant altitude banked turn the rate of turn = (g Tan AOB) / V where V is
TAS. So if V is doubled, the rate of turn will be multiplied by %.
TURN 27. d.
Radius of turn = v2/ (g Tan AOB), where V is TAS. Doubling TAS will increase
v2by a factor of 4, so radius of turn will also increase by a factor of 4. This will
increase radius of turn by 300%.
TURN 28. c.
This problem can be solved using the following equation:
Radius of turn = V' / (g Tan AOB), where V is TAS.
Multiplying both sides Tan AOB and dividing by Radius of turn gives:
Tan AOB = v2/ (g Radius of turn.)
Inserting V = 200 Kts and radius of turn = 1000 meters and converting Kts into
m/s using the constant 1 Kt = 0.515 m/s, gives:
)~
mls2 x 1000 meters)
Tan AOB = (200 Kts x 0.515 m l s l ~ t l(9.81
Which simplifies to give Tan AOB = 1.08
Giving an AOB of approximately 47.25'.
TURN 29. b.
The situation described in this question would result in an imbalance of forces
with centrifugal force being greater than the horizontal component of lift. The
resultant of this imbalance would cause the aircraft to skid away from the centre
of the turn as indicated below.
A I R C R A R IN UNBALANCED LEVEL TURN
TURY 30. c.
This problem can be solved using the following formula:
Rate of turn = g Tan AOB / V which can be rearranged to give:
Tan AOB = (V Rate of turn ) / g
Inserting data provided in the question gives
TanAOB = ( 100 Kts x 1 radlsec) / 9.81 m/st.
Which simplifies to give TanAOB = 10.194
So AOB = 84.4'.
So the angle of bank required for a balanced turn is approximately 84.4 degrees.
TURN 31. c.
The situation described in this question would result in aircraft weight being
equal to the vertical component of lift, and with centrifugal force being equal to
the horizontal component of lift. There would therefore be no resultant force
and so the aircraft would be in balanced flight as indicated below.
AIRCRAFT IN BALANCED LEVEL TURN
TURN 32. a.
The situation described in this question would result in an imbalance of forces,
with centrifugal force being smaller than the horizontal component of lift. The
resultant of this imbalance would cause the aircraft to slip towards the centre of
the turn as indicated below.
AIRCRAFT IN UNBALANCED LEVEL T U R N
Hrsullant force causes aircraft to sideslip
(slipping) towards the centre of the turn.
,
6all displaced to the right by re8ultallt
TURN 33. b.
This problem may be solved using the equation R = v2I (g TanAOB).
Rearranging this equation gives Tan AOB = V' I gR
Inserting the data provided in the question and the conversion factor (1 Kt = 0.51
rnls) gives, Tan AOB = (250 Kts x 0.51 m l s l ~ t19.81
) ~ mls2 x 3000 m
Which simplifies to give Tan AOB = 0.55237
giving AOB = 30 degrees.
TURN 34. a.
This question may be solved using the equation R = v2I (g TanAOB).
Rearranging this equation gives Tan AOB = v21 gR
Inserting the data provided in the question and the conversion factor (1 Kt = 0.51
rnls) gives, Tan AOB = (100 Kts x 0.51 m / s / ~ t 19.81
) ~ m/s2 x 1500 m
Which simplifies to give Tan AOB = 0.1767 giving AOB = 10 degrees.
TURN 35. b.
This question may be solved using the equation ROT = g TanAOB I V.
Inserting the data provided in the question and the conversion factor (1 Kt = 0.51
d s ) gives ROT = 9.81 m/s2 x Tan45 l(100 Kts x 0.51 mls1Kt)
Which simplifies to give ROT = 0.192 giving ROT radls.
This is converted into degls by multiplying by the conversion factor 1 radian =
57.3 degrees to give ROT = 11degls.
TURN 36. c.
Using the equation R = v2I (gTan AOB), and inserting the data provided for
the initial 30 degree banked condition and the conversion factor 1 Kt = 0.51 rnls
gives:
3000m = (250 Kts x 0.51 m l s l ~ t !
) 9.81
~ &Is2 x Tan 30) which is correct
indicating a balanced turn.
Substituting the new AOB of 25 degrees to calc~ilatethe new radius gives:
R = (250 Kts x 0.51 m l s l ~ t19.81
) ~ mls2 x Tan 25)
This simplifies to give R = 3554 meters.
The aircraft will therefore skid away from the centre of the turn until it
establishes a new balanced condition with a turn radius of 3554 meters.
TURN 37. c.
This question may be solved using the equation ROT = g Tan AOB / V.
Rearranging this equation gives V = g Tan AOB / ROT
Inserting the data provided in the question and the conversion factors (1 Kt =
0.51 m/s) and 1 degree =1/ 57.3 radians gives:
V = 9.81 m/s2 x Tan30 / ( I 5 radls 157.3 deglrad).
Which simplifies to give V = 42.5 Kts.
TURN 38. b.
This question may be solved using the equation R = v2/ (g TanAOB).
Rearranging this equation gives V = d(Rg Tan AOB)
Inserting the data provided in the question and the conversion factor (1 K t = 0.51
m/s) gives:
V
= d(2000 rn x 9.81 m/s2 x Tan 40)
\Yhich simplifies to give V = 128.309 m/s
Using the conversion factor 1 K t = 0.51 m/s gives V = 252 Kts.
TURN 39. c.
This question may be solved using the equation ROT = g Tan AOB / V.
Rearranging this equation gives Tan AOB = ROT x V I g
Inserting tlie data provided in the question and the conversion factors (1 K t =
0.51 m/s) and 1 degree =1/ 57.3 radians gives:
Tan AOB = 1.02078
Which gives AOB = 45.6 degrees.
TURN 40. a.
This question may be solved using the equation R = v2/ (g TanAOB).
Rearranging this equation gives V = d(Rg Tan AOB)
Inserting the data provided in the question and the conversion factor (1 K t = 0.51
mls) gives:
V = d(2500 m x 9.81 m/s2 x T a n 40)
Which simplifies to give V = 143.45 m/s
Using the conversion factor 1 Kt = 0.51 m/s gives V = 281.3 Kts.
TURN 41. a.
This question may be solved using the equation R = v2/ (g TanAOB).
Inserting the data provided in the question and the conversion factor (1 Kt = 0.51
m/s) gives:
~
m/s2 x Tan 30)
R = (150 Kts x 0.51 m / s / ~ t /) (9.81
Which simplifies to give R = 1033 meters.
TURN 42. b.
A rate 1 turn = 3 degrees per second so a complete 360 degree turn will take
360 deg / (3 degls) which is 120 seconds o r 2 minutes.
TURN 43. a.
This question may be solved using the equation ROT = g Tan AOB I V where V
= TAS.
Rearranging this equation gives V = g Tan AOB / ROT
Inserting the data provided in the guestion and the conversion factors (1 Kt =
0.51 m/s) and 1 degree =1/ 57.3 radians gives:
V = (9.81 mls2 x T a n 30) 1 (3 degls x 1157.3 radtdeg)
Which simplifies to give V = 108.2 Kts.
But at 40000 ft ISA TAS is twice IAS to if TAS = 108.2 Kts then IAS = 54.1 Kts.
TURN 44. d.
The term skidding describes the situation in which a turning aircraft side slips
away from the centre of the turn. This is caused by a n imbalance of horizontsl
forces due to excessive TAS o r inadequate AOB.
TURN 45. a.
The term slipping describes the situation in which a turning aircraft side slips
towards the centre of the turn. This is caused by an imbalance of horizontal
forces due to excessive AOB o r inadequate TAS.
TURN 46. b.
Skidding is caused by a n imbalance of horizontal forces, such that centrifugal
force due to TAS is greater than the centripetal force provided by the inward
tilting of the lift vector due to AOB. Left uncorrected the aircraft will skid away
from the centre of the turn until its radius of turn restores the balance between
the two horizontal forces. The aircraft will then be in a balanced turn a t a
greater radius and lower rate of turn.
TURN 47. b.
T u r n radius a t any altitude is limited by the ability of the aircraft engi~iesto
provide the power necessary to maintain the required TAS. As altitude increases
power available reduces, so the ability of the aircraft to achieve tight turns is
reduced. Minimum turn radius is therefore limited by power available at high
altitude. A second limiting factor is the ability of the aircraft to generate the
required lift force in the thin atmosphere, but this option is not included in this
question.
TURN 48. b.
VRICA
is the minimum speed a t which it possible to maintain control of the
aircraft in the air following, the failure of a critical engine in the take-off
configuration. I t is the lowest speed a t which the yaw and roll control authority
is able to match the yawing and rolling moments generated by the asymmetric
power condition. As altitude increases, the thrust provided by the engines
reduces, thereby reducing the magnitude of the yawing and rolling moments
produced. This means that it is possible to control the aircraft a t lower
airspeeds, so VMCAreduces with increasing altitude.
TURN 49. a.
A turn is initiated by banking the aircraft in order to tilt the lift vector in the
direction of the intended turn. This reduces the vertical component of the lift
vector so in order to maintain altitude the total lift vector must be increased to
restore the balance between weight and vertical lift. This increase in lift is
achieved by increasing angle of attack which in turn also increases lift-induced
drag. In order to maintain airspeed it is then necessary to increase power, to
balance the increased induced drag force. A balanced level turn therefore
I-equiresincreased angle of attack and power.
TURN 50. b.
A turn is initiated by banking the aircraft in order to tilt the lift vector in the
direction of the intended turn. This reduces the vertical component of the lift
vector so in order to maintain altitude in a level turn the total lift vector must be
increased to restore the balance between weight and vertical lift. This constitutes
an increase in load factor to a value equal to 11Cos AOB. In a descending turn
with increasing rate of descent however, lift does not need to equal weight, so the
load factor will be lower than in a level turn.
TURN 51. d.
Dihedral is only effective in providing lateral stability when an aircraft is in side
slipping flight. I n a balanced turn there is no sideslip and so dihedral effect is
nil.
TURN 52. c.
In a turn, the horizontal component of lift provides a centripetal force tending to
accelerate the aircraft towards the centre of theturn. This is opposed by
centrifugal force, which is the Newton's third law reaction to this centripetal
acceleration. In a balanced turn, the centripetal and centrifugal forces are equal,
so the aircraft turns about the centre without any sideslip. In a turn using
excessive TAS, centrifugal force will exceed the horizontal component of lift, so
the aircraft will skid away from the centre. Dihedral effect in this sideslip will
tend to bank the aircraft towards the centre of the turn, thereby increasing bank
angle until the higher bank angle establishes balanced conditions and the sideslip
ceases.
TURN 53. b.
In a turn, the horizontal component of lift provides a centripetal force tending to
accelerate the aircraft towards the centre of the turn. This is opposed by
centrifugal force, which is the Newton's third law reaction to this centripetal
acceleration. In a balanced turn, the centripetal and centrifugal forces are equal,
so the aircraft turns about the centre without any sideslip. In a turn usiiig
excessive bank angle, centripetal force will exceed the centrifugal force, so the.
aircraft will slip towards the centre. Dihedral effect in this sideslip will tend to
bank the aircraft away from the centre of the turn, thereby decreasing bank
angle, until the lower bank angle establishes balanced conditions and the sideslip
ceases.
TURN 54. a.
If the directional stability of an aircraft is significantly stronger than the lateral
stability, then in a sideslip, it will roll and yaw into the sideslip. This will cause it
to carry out a wide spiral dive as it continues to increase bank and yawing angle.
In a banked turn such an aircraft would tend to bank and yaw into the turn, so a
lateral stick force and rudder force would be required to prevent the bank and
yaw angles from increasing. On releasing the controls in a balanced turn the
aircraft will roll into the turn and pitch nose down. The overall effect will be an
increase in bank angle and a nose down pitching motion as the aircraft enters its
spiral dive.
TURN 55. a.
In a turn, the horizontal component of lift provides a centripetal force tending to
accelerate the aircraft towards the centre of the turn. Tliis is opposed by
centrifugal force, which is the Newton's third law reaction to this centripetal
acceleration. I11 a balanced turn the centripetal and centrifugal forces are equal
so the aircraft turns about the centre without any sideslip. I n a turn using
excessive bank angle, centripetal force will exceed the centrifugal force so the
aircraft will slip towards the centre. Centrifugal force is proportional to
angular velocity, so as the radius of turn decreases a t constant TAS, angular
velocity and hence centrifugal force increase until a balance between centrifugal
and centripetal forces is established. The aircraft will then be in a balanced turn.
--
TURN 56. c.
In a turn, the horizontal coinponent of lift provides a centripetal force tending to
accelerate the aircraft towards the centre of the turn. This is opposed by
centrifugal force, which is the Newton's third law reaction to this centripetal
acceleration. In a balanced turn the centripetal and centrifugal forces are equal
so the aircraft turns about the centre without any sideslip. In a turn using
excessive TAS, centrifugal force will exceed the centripetal force so the aircraft
will skid away from the centre.
Centrifugal force is proportional to angular velocity and as the radius of turn
increases a t constant TAS, angular velocity and hence centrifugal force decrease
until a balance between centrifugal and centripetal forces is established. The
aircraft will then be in a balanced turn.
TURN 57. b.
The ball in a serviceable slip indicator is positioned by the vector sum of all
acceleration forces acting upon it. In straight and level constant speed flight the
only acceleration is that due to gravity and as this acts vertically downwards, so
the ball is held in the central position. When an aircraft is sideslipping due to
bank, the gravitational acceleration still acts vertically, but because of the bank
angle, does not act through the normal axis of the aircraft. This causes the ball
to move out of the central position, and towards the sideslip direction. The ball
does not however always give a true indication of sideslip. For example, in the
case of a single engine failure the ail-craft can be made to sideslip along the
desired track, whilst the wings are held level by means of the ailerons. In this
situation the aircraft will be sideslipping, but the ball will be central indicating
no slip. The normal operation of the ball with an aircraft side slipping to the left
is indicated below. The situation in which the ball gives a false indication in
wings level side slipping flight following single engine failure is indicated in the
diagram overleaf.
Sli~
indicator giving correct indication of sideslip.
b
Rudder deflected to arrest yaw
and to hold aircraft into sideslip
Ailerons deflected to counter
dihedral effect to hold wings level.
Lift.
Rudder side force.
Sideslip side force.
Weight.
Ball central indicating no slip.
Slip indicator giving false zero slip indication in wings level side slipping flight
following single engine failure.
TURN 58. c.
Stalling angle is the angle of attack a t which the airflow passing over an aerofoil
separates from the upper surface and lift force decreases rapidly. This angle is a
function of the shape and surface condition of the aerofoil, but is not affected by
aircraft attitude or flight conditions. It does not therefore vary with bank angle
in a turn.
TURN 59. a.
The stalling speed is the minimum speed at which an aircraft can generate a
vertical lift force equal to its weight. As bank angle is increased in a turn, the lift
vector is tilted towards the centre of the turn thereby reducing the vertical
component of lift. Under these circumstances it is necessary to increase the
magnitude of the lift force in order to restore its vertical component to equal
weight. At the stall the CLis at its maximum value so the only means of
generating the extra lift is to increase speed. The stalling speed is therefore
increased with bank angle in a turn.
TURN 60. b.
I11 order to achieve a balanced turn at any given radius, it is necessary to
generate a lift force such that its vertical component equals weight, and its
horizontal component equals centrifugal force. It is therefore necessary to
increase angle of attack when conducting a turn. This increase in angle of attack
also increases drag. In order to conduct a turn at any given radius and airspeed,
it is therefore necessary to increase power above the level required for straight
and level flight. But power available decreases with increasing altitude, so the
maximum turn capability of a n aircraft decreases with increasing altitude. At
the absolute ceiling power available is just sufficient to permit straight and level
fight, s o turning is not possible without losing altitude.
TURN 61. b.
The maximum operating altitude of a JAR certificated passenger aircraft is that
a t which it is just capable of achieving a load factor of 1.3 without incurring high
speed buffet. Load factor in a bank is equal to (11 Cosine AOB) so the maximum
allowable bank angle is that for which the cosine is 111.3. This angle is 39.7
degrees.
TURN 62. c.
The minimum value of limiting load factor for JAR certification is 2.5 with flaps
u p and 2 with flaps down. This question does not specify the flap condition so
bank angles for both must be calculated and compared with the options
provided. Load factor in a bank is equal to (11 Cosine AOB) so in the flaps u p
condition the maximum allowable bank angle is that for which the cosine is 1 /
2.5. This angle is 66.4 degrees. For the flaps down condition the maximum
allowable AOB is that for which the cosine = 1 / 2. This angle is 60 degrees.
Only the 66.4 degree AOB matches the options offered in the question, so it must
be assumed that the turn is to be conducted in the flaps up condition.
TURN 63. c.
The conditions required for a balanced turn are described in the formula,
Radius of turn = v2/ g TanAOB. This formula does not include aircraft weight,
so increasing aircraft weight does not affect minimum turn radius directly. For
any given flight condition, including turning flight, the power required increases
with weight, so minimum turn radius is dependent upon the relationship between
weight and power available. increasing aircraft weight will not therefore affect
turn radius provided sufficient power is available.
TURN 64. a.
The conditions required for a balanced turn are described in the formula,
Radius of turn = v2/ g TanAOB. This formula does not include aircraft weight,
so increasing aircraft weight does not affect minimum turn radius directly. For
any given flight condition, including turning flight, the power required increases
with weight, so minimum turn radius is dependent upon the relationship between
weight and power available. In the event of a single engine failure, the total
power available will be reduced. Under these circumstances the maximum
weight a t which any given turn can be achieved will decrease. This means that
the minimum turn radius at high weights might be increased following single
engine failure.
TURN 65. b.
In order to achieve a balanced turn at any given radius it is necessary to generate
a lift force such that its vertical component equals weight and its horizontal
component equals centrifugal force. It is therefore necessary to increase angle of
attack when conducting a turn. This increase in angle of attack also increases
drag. In order to conduct a turn at any given rate, it is necessary to increase
power above the level required for straight and level flight. But power available
decreases with increasing altitude, so the maximum turn rate of an aircraft
decreases with increasing altitude. At the absolute ceiling power available is just
-sufficient to permit straight and level fight so no rate of turn is possible without
losing altitude.
TURN 66. a.
In order to achieve a balanced turn at any given radius, it is necessary to
generate a lift force such that its vertical component equals weight and its
horizontal component equals centrifugal force. I t is therefore necessary to
increase angle of attack when conducting a turn. This increase in angle of attack
also increases drag. In order to conduct a turn at any given radius, it is therefore
necessary to increase power above the level required for straight and level flight.
The smaller the turn radius, the greater the increase in bank angle and power
required. But power available decreases with increasing altitude, so the
minimum turn radius of a n aircraft increases with increasing altitude. At the
absolute ceiling power available is just sufficient to permit straight and level
fight, so no radius of turn is possible without losing altitude.
TURN 67. d.
Rate of turn = g TanAOB I V where V is the true airspeed of the aircraft. Rate
of turn is therefore proportional to TAS.
TURN 68. d.
The maximum turn radius that any aircraft could possibly achieve is infinity,
which would cause the aircraft to be in straight and level flight. An aircraft is
capable of achieving this at all altitudes up to its absolute ceiling. Altitude
therefore has no effect on the maximum turn radius of an aircraft.
TLTRN 69. a.
Load factor in a constant altitude banked turn is equal to 11 CosAOB. In a 90
degree bank CosAOB is zero so the load factor is infinity. This means that to
conduct such a turn an aircraft must generate an infinitely large lift force. This
is clearly impossible so a 90 degree banked turn will always cause a loss of
altitude.
TURN 70. a.
JAR 25 states that the maximum allowable rate of turn in a JAR certificated
passenger aircraft is a rate 1 turn.
TURN 71. c.
Stalling speed in a constant altitude banked turn, can be calculated suing the
following equation, Vs(in the turn) = Vsrs d (11CosAOB).
In a 75 degree banked turn this gives, Vs(in the 75 degree banked turn) =
1.96Vs~,. So the stalling speed is increased by 96.6% compared with its straight
and level value of VSfg.
TURN 72. b.
I n a balanced turn the centrifugal force due to turning is equal to the horizontal
component of lift caused by the inward inclination of the lift force, due to the
aircraft banking into the turn. If the TAS is too great, or the angle of bank too
small, the centrifugal force will exceed the horizontal component of lift and the
aircraft will skid away from the centre of the turn.
TURN 73. a.
The conditions required for a balanced turn are described in the formula
Radius of turn =-v21 g TanAOB. This formula does not include aircraft weight,
C L ~ A oXr,wing area, so these factors do not affect minimum turn radius
directly. For any given flight condition, including turning flight, the power
required increases with load factor, and load factor increases as turn radius
decreases. The minimum turn radius is therefore determined by the amount of
power available
TURN 74. a.
In a balanced turn, the centrifugal force due to turning is equal to the horizontal
component of lift caused by the inward inclination of the lift force, due to the
aircraft banking into the turn. If the TAS is too low, o r the angle of bank too
great, the horizontal component of lift will be greater than the centrifugal force
and the aircraft will slip towards the centre of the turn.
TURN 75. a.
The ball in a serviceable slip indicator is positioned by the vector sum of all
acceleration forces acting upon it. In straight and level constant speed flight, the
only acceleration is that due to gravity, and as this acts vertically downwards, the
ball is held in the central position. When an aircraft is turning it will skid away
from the centre of the turn if the TAS is too great or the bank angle too small. In
these circumstances the combination of forces on the ball due to gravitational
acceleration and horizontal acceleration will cause it to move out of the central
'position, and towards the skidding direction. The ball will therefore move
towards the outside of the turn indicating that right rudder will increase turn
radius and hence establish a balanced turn. This situation is indicated in the
diagram overleaf.
AIRCRAFT IN UNBALANCED LEVEL TURN
Resultant force causea aircraft lo sideslip
(skidding) away from the centre of the hro.
b-'
Ball displared lo the left by multaal
POW 1. c.
Power required = Drag x TAS and Drag = CD1/2p ~ 2 where
~ ,
V = TAS
SO power required = CD1/2p
~ x 'v ~
This means that power required is proportional to V x V' which is v3
So power required is proportional to v3
In accelerating from 300 Kts to 400 Kts TAS is increased to 133% of its previous
value.
So power required is increased to (1.33)~times its previous value.
So power required is 2.35 times its previous value which is an increase of 135%
\
POW 2. c.
At VMDthe value of total drag is at a minimum, so as the aircraft accelerates
drag will increase.
Power required = Drag x TAS and Drag = CD1/2pv2s where V = TAS
So power required = CD1/2~(TAS)'S x TAS
This means that power required is proportional to TAS x (TAS)' which is
(TAS)~
So power required increases in proportion to (TAS)~
POW 3. b.
Maximum endurance is achieved by flying at the speed that requires the
minimum fuel consumption rate. Fuel flow in a jet is proportional to thrust,
which in straight and level flight is equal to drag. So maximum endurance a jet
occurs at the minimum drag speed (VMD). Although V Mvaries
~
with aircraft
weight, load factor, configuration and altitude, it is typically in the order of
132% of the stalling speed corresponding to the existing state of these variables.
So maximum endurance in a jet is achieved by flying at approximately 132% of
stalli~lgspeed.
POW 4. a.
~ sis .P;oportional to angle of attack, so if angle of attack
Lift = c ~ ~ / z ~ vCL
remains constant, so will CL. AS altitude increases, air density (p) decreases so to
maintain the same lift at the same angle of attack V must increase. The aircraft
must therefore fly at a higher TAS to create the same lift at the same angle of
attack in the less dense atmosphere. Power required is TAS x Drag. Drag =
C ~ I / ~ ~ V so
' S increasing
,
TAS will increase both power required and drag. The
aircraft must therefore use more power and fly a t a higher TAS.
POW'S. b.
~ Sis. proportional to angle of attack, so if angle of attack
Lift = C L I / ~ ~ V CL
remains constant, so will CL. To maintain level flight, any weight increase must
be accompanied by an equal increase in lift, so TAS must increase. Power
* Sincreasing
,
TAS will increase
required is TAS x Drag. Drag = C ~ I / ~ ~ V so
both drag and power required. The aircraft must therefore use more power and
fly at a higher TAS.
POW 6. a.
For any given airspeed, a headwind will reduce ground speed and hence range.
Maximum range will therefore be achieved by reducing the time during which
the wind is affecting the aircraft. This is done by increasing airspeed.
POW 7. b.
Flap deployment increases both lift and drag, but in most cases causes a
reduction in the lift : drag ratio. This means that for any given value of lift,
deploying flaps incurs a greater drag force. Drag is therefore lowest with flaps
retracted.
Power required = thrust x TAS. In straight and level flight drag = thrust, so
power = drag x TAS. At high altitude the TAS equating to any given IAS
increases, so for a given drag force, the power required is lowest at low altitude.
The lowest value of power required at any given IAS is therefore achieved at
zero flap and low altitude.
POW 8. a.
Maximum range in any aircraft'type is achieved when the ratio between distance
covered and fuel flow is maximum. That is to say, maximum range requires
minimum fuel flow, for a given distance covered. Fuel flow in a propeller
aircraft is proportional to power, so maximum range will be achieved when the
power required for a given distance covered is at a minimum. Power required in
straight and level flight = drag x TAS, and increasing TAS equates to increasing
rate of distance covered. For maximum range, it is therefore necessary to fly at
the speed a t which the ratio of TAS to drag is a t a maximum. This speed is VMD
and can be found on a
power required curve by drawing a tangent
from the origin to the power required
curve as indicated in the diagram.
Power
required
TAS
VMD
POW 9. a.
Power required in level flight = Drag x TAS.
Drag = CD !4 p v 2 s ,
where
CD = coefficient of total drag
p = air density
V = TAS.
So power required = Drag x TAS is proportional to (TAS)~x TAS
Which means that power required is proportional to (TAS)~.
Indicated air speed (IAS) is also proportional 1/2pv2 such that any given IAS
will occur at the same value of 112 p v 2 at all altitudes, and the drag force a t any
given IAS will also be the same a t altitudes. But as altitude increases, p
decreases, so the TAS equating to any given IAS must increase with altitude to
maintain a constant value of 112 p ~ a t2any given IAS. At 40000 feet pressure
altitude in a standard atmosphere, p is 114 of its mean sea level value, so v2must
be 4 times its sea level value. This means that at 40000 feet, the TAS equating to
any given IAS is twice its sea level value.
But power required at sea level = CD 112 p (TAS)~S x TAS
Which is
......................................Equation 1.
CD1/2pS x T A S ~
p is !A of its sea level value, while TAS is twice its sea level value.
So power required a t 40000 feet = CD 112 (114 p)S ( 2 ~ ~ s ) ~
Which is CD 118pS x 8 (TAS)~
................................Equation2.
This simplifies to give C DpS T A S ~
Comparing equations 1 and 2 reveals that the power required at 40000 feet ISA
at any given indicated airspeed is twice that a t sea level.
Put more simply, if power required = Drag x TAS and between sea level and
40000 feet drag remains constant while TAS doubles, then power required also
doubles.
POW 10. d.
Power required in level flight = Drag x TAS.
Drag = CD 112 ~v's,
where
CD = coefficient of total drag
p = air density
V = TAS.
So power required = Drag x TAS = CD 112 ~ S ( T A S ) ~
Which means that power required is proportional to TAS~.
POW 11. (1.
Work is done when a force moves its point of application in the direction of the
force. Power is a measure of the rate a t which work is done. I n the case of a n
aircraft in level flight, the work done is the thrust multiplied by the distance
flown. Thrust required is equal to the drag force so the work done is equal to the
drag multiplied by the distance flown. The true airspeed of an aircraft is the rate
at which it covers the distance flown through the air. Power required is
therefore equal to drag multiplied by TAS. Others forms of airspeed including
IAS, CAS and EAS cannot be employed in this equation because they do not
provided a true indication of the rate a t which distance is flown through the air.
POW 12. c.
The power of all air breathing aircraft engines is proportional to the mass flow of
air passing through them. As altitude increases air density decreases causing
mass flow and power output to decrease.
POW 13. a.
The power required in any flight condition
is proportional to T A S ~
and as altitude
increases, the ratio of TAS to IAS
increases. This means that when climbing
at any given IAS, the TAS and power
required both increase with increasing
Power
altitude. The overall effect of these
changes is that the power required : TAS
curve moves upwards and to the right
with increasing altitude. Also, the power
available for all air breathing engines is
proportional to mass airflow. This
decreases with increasing altitude due to
reducing air density. The power
Power required
available curve therefore moves down
at sea level
with increasing altitude. These effects
are illustrated in the diagram at the right.
Power available
at high altitude
POW 14. d.
Power available at sea level
The power required by any aircraft type
in any given flight condition increases
with increasing altitude. The power
output of aircraft engines is proportional
to the mass flow of air passing through
Power
them, As air density decreases with
increasing altitude, power output also
decreases. Excess power is the proportion
of the power available that is not required
to maintain straight and level flight.
The combined effect of these factors is
that as altitude increases, power required
increases whilst power available decreases.
~ x c e s power
s
The overall effect therefore is that excess
at sea level
at high altitude
power also decreases with increasing
I
I
altitude. These effects are illustrated in
Power available Power required
the diagram at the right.
at high altitude k t high altitude
Because the increase in power required Rate of increase in Rate of reduction
power required
in power available
and decrease in power available are
caused by decreasing air density, the
decreases as
decreases as
rate of change of these variables is
altitude increases. altitude increases.
Proportional to the rate of change of
of air density. As altitude increases
the weight of air above any given point
decreases, causing static pressure and hence
density to decrease. But because air in the
Altitude
lower atmosphere is more dense than that
in the higher atmosphere, the rate of
reduction in static pressure per thousand
feet of altitude increase is greater a t low
altitudes that a t high altitudes. This means
that the rate of change of air density, power
available, power required and excess power
all decrease as altitude increases. The
effects on power available and power required
with changing altitude is illustrated in the
diagram at the right. The overall effect of the
above factors is that as altitude increases,
excess power decreases at a rate that decreases
with increasing altitude.
Excess power
POW 15. c.
Power required in level flight is equal to drag multiplied by TAS.
~ u drag
t = C D I / ~ ~ V * where
S,
v is TAS.
This means that power required is proportional to T A S ~so if TAS is doubled and
all other factors remain unchanged, then power required is multiplied by Z3
which is 8.
POW 16. c.
VMPis the speed at which an aircraft requires
the lowest power. As indicated in the diagram
a t the right, Vnlp is higher than Vs but lower
than VMD. Vx is the CAS for best climb
gradient and for a propeller aircraft is close to
the lowest safe flight speed and therefore just
above Vs.
CAS
'
vs.;;
.
.
.
.
,
.
i. .. .
. .
.... ...
... ..
Vx Propeller / . v M p
VMD
POW 17. b.
For maximum range it is necessary to achieve
the best ratio of fuel consumption to distance
flown. For a jet aircraft fuel consumption is
proportional to thrust, which in straight and
level flight is equal to drag. Best jet aircraft
Drag
range is therefore achieved at the speed
providing the best ratio of drag to airspeed.
This occurs at the speed where a tangent
drawn from the origin just touches the drag
: EAS curve. As indicated in the diagram, this
speed is higher than VMD.
E.4S
Villi
Best
range
POW 18. b.
1 Best excess IAS a t sea
Excess power is equal to power available minus
power required. The power output of both
piston and jet engines is proportional to the
mass flow of air passing through them. As
altitude increases, both air density and power
output decrease. The overall effect of
increasing altitude is that the power available
: IAS curve moves downwards with increasing
altitude. Power required in constant speed
straight and level flight is equal to drag
..
.
..
..
.
multiplied by TAS. Although the drag a t any
100 150 200 250
given IAS remains constant, the TAS equating
to any given IAS increases. This means that
IAS a t MSL
the power required : IAS curve moves upwards
IAS a t 40000 feet
and to the left with increasing altitude. The
overall effect of these changes is that the
V
maximum excess power IAS (Vy) decreases with
at msl, to 100 Kts at 40000 ft
increasing altitude as illustrated in the diagram.
d
1
I
POW 19. a.
Excess power is equal to power available minus
power required. The power output of both
piston and jet engines is proportional to the
mass flow of air passing through them. As
Power
altitude increases both air density and power
output decrease. The overall effect of
these changes is that the power available
: TAS curve moves downwards and to the right ..
with increasing altitude. Power required in
constant speed level flight is equal to drag
multiplied by TAS. Although the drag at any
given IAS remains constant, the TAS equating
0
100 650
to any given IAS increases. This means that
IAS at msl
the power required : TAS curve moves upwards
0
50
75
and to the right, with increasing altitude. The
IAS at 40000 feet
overall effect of these changes is that the
maximum excess power IAS (VY),decreases
Vy IAS decreases from 130 Kts
with increasing altitude. These effects are
at msl to 75 Kts at 40000 feet.
illustrated for a jet engine in the diagram.
But at 40000 feet TAS is twice IAS, so the decrease from 130 Kts to 75 Kts IAS
represents an increase from 130 to 150 kts TAS. The figures used in this
explanation are merely examples and do not apply to all aircraft. I t is however
true that for all aircraft, increasing altitude decreases the IAS value of Vy and
increases the TAS value. This process continues until the absolute ceiling where
for a jet aircraft, Vx = VMD= VY.
POW 20. a.
As altitude increases, power available
required
decreases and power required for any given
flight profile increases. The overall effect
increasing altitude is that excess power and
rate of climb decrease. At the absolute
ceiling, power available is equal to power
required, so both excess power and rate of
climb are zero. This condition is illustrated
for a piston engine aircraft in the diagram
a t the right.
Altitude
Absolut
Airspeed
Power-available and power required curves
are tangential and power available is equal
to power required at the absolute ceiling.
POW 21. a.
At the absolute ceiling power available equals power required, so excess power is
zero. Flight a t this altitude, is possible at only one speed. Flight at any other
speed will cause power required to increase and power available to decrease.
Fiight at other speeds a t this altitude requires more power than is available and
will therefore be impossible. For a jet aircraft at this altitude, the single
practicable airspeed therefore constitutes Vs, VMP,VMD,VX,VY,best endurance
speed and best range speed.
POW 22. c.
As altitude increases, power available
decreases and power required for any given
flight profile increases. The overall effect
increasing altitude is that excess power and
rate of climb decrease. At the absolute
ceiling power available is equal to power
required. Both excess power and rate of
climb are zero and the two curves are
parallel as indicated in the diagram.
Altitude
Absolute
ceiling
ower available
Power availa$le and power required
curves are parallel at the absolute
ceiling
POW 23. a.
At all altitudes below the absolute ceiling,
the power available and power required
curves cross at two points, as indicated in
the diagram. These points represent the
maximum and minimum flight speeds,
for which sufficient power is available.
Power
required
Minimum flight speed
Maximum flight speed
POW 24. b.
When an aircraft's engines are shut down in flight the aircraft possesses a
limited store of potential and kinetic energy, which cannot be replenished.
Throughout the subsequent glide this energy is expended in overcoming drag,
unti1 the energy reduces to zero at the end of the landing run. The rate a t which
energy is expended by an aircraft in flight is the power required, so for
maximum glide endurance the aircraft must be flown at VMP. The power
required curve is therefore highly relevant in gliding flight. The actual power
consumption rate and hence endurance, depend on the speed and angle of attack
chosen. The power available curve is however irrelevant in gliding flight because
the engine can produce no power when in the shut-down condition.
POW 25. b.
The speed providing the best ratio of power required :TAS is used when flying
for maximum still air range in a propeller driven aircraft. This speed occurs
where a tangent drawn from the origin just touches the power required curve.
As indicated in the diagram below, the speed is also VMD.
VnlDand speed for best
Power required : TAS ratio
POW 26. a.
Maximum endurance in any aircraft type is
achieved when the rate of fuel flow is minimum.
Fuel flow in a jet aircraft is proportional to
Thrust, so maximum jet endurance is achieved
when the thrust required for a given TAS is at a .
minimum.
I n straight and level flight thrxst = drag, so best
straight and level endurance will be achieved
at minimum drag speed, which is VbID. This is
the lowest point on the total drag : EAS
curve as indicated in the diagram.
Drag
EAS
POW 27. b.
Power required = drag x TAS, and in straight and
level flight lift = weight. A decrease in weight
therefore results in a decrease in the lift required.
Induced drag is proportional to lift, so decreasing
left results in a decrease in induced drag. The total
drag force a t any given TAS therefore reduces with
reducing weight, which in turn results in a reduction
in power required a t that TAS. Induced drag is
proportional to 1/v2, so any reduction in induced
drag is most prominent a t the low end of the speed
range.-This means that reductions in weight, also Power
required
tend to move the drag curve, and hence power
required curve to the left The overall effect
therefore is that reducing weight moves the power
required curve down and to the left.
VMD
TAS
POW 28. c.
The diagram in this question is whole aircraft CL:CDpolar, which indicates how
the L:D ratio varies throughout the speed range. If a line is drawn horizontally
left to right from the origin; this indicates the condition of zero lift and infinite
drag. This constitutes a L:D ratio of zero. If a line is drawn vertically upwards
from the origin, this indicates the condition of infinite lift and zero drag. This
constitutes an infinitely large L:D ratio. The value of the L:D ratio a t any point
on the diagram is directly proportional to the gradient of any straight line from
the origin to that point. The steeper the gradient, the higher the L:D ratio. The
curve indicates those L:D ratios that are attainable and the best L:D ratio is
achieved where a straight line drawn from the origin, just touches the curve.
This is indicated by point C on the diagram. An aircraft achieves its best L:D
ratio when flying a t VMD,so point C represents this speed.
POW 29. c.
Work is done when a force moves its point of application in the direction of the
force. In the case of an aircraft in flight, the work done equals the thrust
multiplied by the distance flown through the air.
Power is the rate a t which work is done and velocity is the rate a t which distance
is travelled. In the case of an aircraft, TAS equals the rate a t which distance is
flown through the air. The power required by an aircraft in flight is therefore
equal to the thrust force acting on it multiplied by its TAS.
Power required = Thrust x TAS.
But in straight and level constant speed flight, thrust equals drag, so in this
condition the work done may also be described as the drag force multiplied by
the distance flown. So in the case of level constant speed flight power required =
Drag x TAS.
It should be noted that IAS can be used in place of TAS only when flying at ISA
MSL where TAS = IAS.
POW 30. d.
The process of gaining height involves applying an upward force equal to the
weight of the aircraft and using this force to move the aircraft upwards. Work is
done when a force moves its point of application in the direction of the force, so
moving the aircraft upwards in this manner involves doing work. Power is the
rate a t which work is done, so the rate of climb depends upon the amount of
power available to move the aircraft upwards. The power available to move the
aircraft upwards is the excess power available. This is the total power available
minus the power required to overcome drag at the chosen airspeed. This
situation is described in the following equation:
Power expended = (aircraft weight x height gained) / time taken
But height gained / time taken = rate of climb and for maximum rate of climb,
power expended is the excess power, so maximum rate of climb = excess power
/ weight
POW 31. c.
If the C of G is forward of the C of P, the aircraft will be subject to a nose down
moment, which must be balanced by a downward tailplane force. The
generation of this downward tailplane force will cause additional trim drag.
Because extra fuel must be burned to overcome this trim drag, the range of the
aircraft will be reduced.
If the C of G is moved still further forward of the C of P, the nose down moment
will increase, necessitating a greater tailplane down force, greater trim drag and
hence still less range. If however the C of G is initially aft of the C of P, the
aircraft will be subject to a nose up moment, which must be balanced by an
upward tailplane force. The generation of this force will of course produce trim
drag and hence reduce range.
If the C of G is then moved forward, the nose up pitching moment will reduce,
necessitating a reduction in tailplane up force, and trim drag. This process will
therefore increase range. Forward movement of the C of G will therefore
decrease range only if the C of G is ahead of the C of P. I t should be noted that a
C of G aft of the C of P would make the aircraft longitudinally unstable and
would therefore not be permitted. The condition is considered here, only to
demonstrate the full effect of C of G position on range.
POW 32. b.
When the C of G is at its aft limit, aft of the C of P, the aircraft will be subject to
a nose up moment, which must be balanced by an upward tailplane force. The
generation of this force will produce trim drag and hence reduce range.
Forward movement of the C of G to the C of P will reduce the nose up moment
to zero, permitting the tailplane up force and trim drag to be reduced to zero,
thereby increasing range. If the C of G is then moved further forward to its
forward limit, the original nose up moment will be replaced by an increasing
nose down moment. This will necessitate a balancing downward tailplane force.
The generation of this downward tailplane force will cause additional trim drag
and hence decrease range. The overall effect of moving the C of G from aft of
the C of P to forward of the C of P will be to increase then decrease range. I t
should be noted that a C of G aft of the C of P would make the aircraft
longitudinally unstable and would therefore not be permitted. The condition is
considered here, only to demonstrate the full effect of C of G position on range.
POW 33. c.
If the C of G is forward or aft of the C of P the resulting nose down o r nose up
moments will necessitate longitudinal trimming. This will increase trim drag
and bence decrease range. When the C of G and C of P coincide, the aircraft will
not be subject to any nose up or down moments, so no longitudinal trimming will
be necessary. This will minimise trim drag thereby maximising range.
POW 34. a.
For very small climb angles, the climb
gradient is approximately equal to the
height gained divided by the distance flown
as indicated in the diagram at the right. But
height gained / distance flown is also the
sine of the climb angle. A gradient of 2.5%
therefore represents a sine of 0.025 and a
gradient of 3.0% represents a sine of 0.03.
In steady flight the sine of the climb angle
is equal to the excess thrust divided by the
aircraft weight.
Height
gained
/
Climb angle
So in a 2.5% climb, the sine of the climb angle = Excess thrust / aircraft weight
Rearranging this gives Excess thrust = Sine of climb angle x aircraft weight
And inserting data provided in the question gives:
For a 2.5% climb gradient Excess thrust = 0.025 x 120000 bf
So Excess thrust = 3000 Ibf.
The weight at which this excess thrust will give a 3% climb gradient may be
found using the same equation as follows:
Sine of climb angle = Excess thrust / weight
Rearranging this equation gives weight = Excess thrust / sine of climb angle.
Inserting .03 for the sine of the climb angle and 3000 lbf excess thrust gives:
Weight = 3000 1bf / 0.03 which is 100000 1bf
This aircraft will therefore achieve a 3% climb gradient at a weight of 100000
lbf.
POW 35. a.
In the initial condition 2.5% is proportional to excess power / 120000 lbf.
Rearranging thidequation gives excess power = 2.5% x 120000 lbf.
Equation 1
.........
In the second condition X% is proportional to excesspower / 150000 lbf.
Rearranging this equation gives excess power = X% x 150000 lbf
.
Equation 2
........
?.-
Increasing weight will decrease excess power, ut if this is ignored as stated in
the question, then equations 1 and 2 are-equa
This means that 2.5% ar 120000 = X% x 150000
Rearranging this equation gives X% = (2.5% x 120000) / 150000
Which is X%
2%.
= 2%
so the maximum gradient available at a weight of 150000 is
POW 36. b.
Power required = total drag x TAS.
Assuming the aircraft maintains constant TAS, flight profile and configuration,
its CL must increase in direct proportion to its weight. A 25% increase in weight
will therefore require a 25% increase in CL.
But CDIis proportional to C L SO
~ if CL is increased by 25% (to 1.25 of its original
value), then C Dand
~ hence induced drag, will increase by a factor of 1.25', to
become 1.5625 of its original value. This is an increase of 56.25% in induced
drag.
The increase in weight will not however alter CDpsignificantly, so profile drag
will remain approximately constant at 100% of its initial value. This constitutes
a 0 % increase.
At VMDinduced drag = profile drag, so each represents 50% of total drag. The
overall effect on total drag is therefore equal to the average increase of induced
drag and profile drag.
That is (56% increase in induced drag +0% increase in profile drag ) / 2 which is
a 28.125% increase in total drag.
Power = total drag x TAS, so assuming TAS remains unchanged, power required
will also increase by 28.125%
POW 37. b.
If the vector sum of all forces acting on a body in any given direction is not zero,
then the body will accelerate in the direction of the resultant of those forces. In
climbing or descending flight the forces of lift and drag are added to by the
component of weight acting along the flight path. In such circumstances if thrust
equals drag then this component of weight will cause the aircraft to accelerate if
descending or decelerate if climbing. Options a and c are therefore incorrect. In
straight and level flight the only horizontal forces acting on an aircraft are thrust
and drag so if these are equal, the vector sum will be zero and speed (IAS) will
remain constant. Option b is therefore correct which means that option d is
incorrect.
POW 38. a.
If the mass of an aircraft is increased, then lift must increase by an equal amount
to maintain level flight. Lift = C L ~ / ~ ~ This
V ~ Squestion
.
specifies that angle of
attack remains unchanged. But CL is proportional to angle of attack so CL will
also remain unchanged. This means that speed (V) or wing area (S) must
increase. Drag is proportional to speed and wing area so if either of these are
--
increased then extra power will be required to overcome the higher drag force.
Of the options offered in this question, only option a satisfies these requirements.
POW 39. b.
For jet aircraft fuel flow is proportional to thrust. I n straight and level flight
thrust equals drag. To solve this problem it is therefore necessary to calculate
the increase in drag. As aircraft weight increases the additional drag is
principally lift induced drag. At any given airspeed CL is proportional to weight
TO calculate the increase in fuel consumption it is
and CDIis proportional to cL2.
therefore necessary to calculate the new CL, CDl and induced drag.
The ratio of weights of aircraft B to aircraft A = 250000/200000 = 1.25. So the
CLof aircraft B = 1.25 times that of aircraft A and the cL2
of aircraft B is 1.5625
times that of aircraft A. CD1is proportional to CL' SO the C Dand
~ induced drag of
aircraft B are also 1.5625 of those of aircraft A.
But at VMD,the induced drag makes up, only half of the total drag, the other half
being made up of Profile drag which does not vary with weight. So the change in
total drag is half of the change in induced drag which in this case gives an
increased total drag of 28.125%. In straight and level flight thrust = weight and
fuel flow in a jet aircraft is proportional to thrust. So aircraft B requires 1.28125
times as fuel flow as aircraft A. This means that the fuel consumption of aircraft
B is 5000 Kg/hr x 1.28125. So fuel consumption in Aircraft B is 6406 Kg/hr.
An approximate fuel flow at any new weight at VMDcan be found using:
Fuel Flow at new weight = Fuel flow at old weight x (new weight / old weight)
In this case Fuel flow at 25000 Kg = Fuel flow at 200000 Kg x (250000/20000)
Which is Fuel flow at 250000 Kg = 5000 Kg/h x (250000 / 20000) = 6250 Kg/h.
POW 40. b.
Vx is the IAS at which an aircraft will achieve its best angle of climb and is the
speed a t which maximum excess thrust is available. Because the thrust of a
propeller decreases rapidly with increasing airspeed, Vx for a propeller aircraft
is very close to the minimum safe flying speed. VIMpis the IAS corresponding to
the minimum value of power required. Because fuel consumption in a propeller
aircraft is proportional to power output, VIMPis also the lowest fuel consumption
airspeed. VIMPfor a propeller aircraft is greater than V2, VX, but less than Vy.
POW 41. a.
Vx is the IAS at which an aircraft will achieve
its best angle of climb and is the speed at
which maximum excess thrust is available.
Because the thrust of a propeller decreases
rapidly with increasing airspeed, Vx for a
propeller aircraft is very close to the
minimum safe flying speed. As indicated in
the diagram Vx for a propeller aircraft is
lower than VIMP.VIMPis the IAS
corresponding to the minimum value of
power required. Because fuel consumption
in a propeller aircraft is proportional to
power output, VIMPis also the lowest fuel
consumption airspeed. Vy is the IAS for
best rate of climb and is greater than V l ~ p .
V2 is the take-off safety speed. I t is lower than Vx.
POW 42. c.
Rate of climb is determined by excess power.
As altitude increases power available
decreases and power required increases,
causing both excess power and rate of climb
to decrease. The service ceiling is the
pressure altitude a t which the maximum rate
of climb is 500 feet per minute for a jet and
100 feet per minute for a pistonlpropeller
aircraft. But power required is equal to drag
multiplied by TAS and drag increases with
aircraft weight, so power required at any Altitude
given altitude also increase with weight.
This means that the reduction of excess
power to the level that is just sufficient to
achieve the service ceiling rates of climb,
occurs at a lower altitude as weight increases.
The effect of increasing aircraft weight is
therefore to lower the service ceiling. So the
service ceiling at 75000 Kg will be lower than
0
that at 50000 Kg. The effect of increasing
weight is illustrated in the diagram at the right.
I
Power required
\
Absolute ceilings
lOOfpm(piston)
or 500 fpm (jet)
ROC
POW 43. c.
Low altitude
Power required is equal to drag multiplied by
TAS. At any given IAS drag remains
constant with increasing altitude, but the TAS
corresponding to that 1AS.increases. This
means that as an aircraft climbs at any given
Power
IAS, the multiplication of constant drag by
required
increasing TAS causes power required to
increase. The overall effect of these changes
is that the power required : TAS curve moves
up and to the right with increasing altitude
as illustrated in the diagram.
TAS
b
POW 44. a.
The term "the back of the drag curve" refers to
all speeds lower than VRID.At such speeds the
gradient of the drag : EAS curveis negative so
decreasing speed causes drag to increase. But
to maintain any given speed, thrust must equal
drag, so reducing speed at the back of the drag
curve requires an increase in thrust. Power
TAS. Although
required
equals drag
redueing
(or thrust)
IAS also
multiplied
reduces by
Power required
~~i$&/
..
.
.
.
TAS, the rate of increase in drag tends to exceed
V; vMd ;vMD
the rate of decrease in speed. The overall effect is that power required increases
as speed decreases. However, as indicated in the diagram, the minimum power
speed is lower than the minimum drag speed, so when decelerating below VMD
drag increases whilst power required decreases until VMP,then increases as
speed reduces below VMP. This option is not offered in this question however, so
the selection of the most appropriate answer must be based on the predominant
or overall effect. VMPis much closer to VKIDthan to Vs so power required
increases with decreasing speed over the majority of the back of the drag curve.
Option a, is therefore the most appropriate.
POW 45. b.
When operating at the back of the drag curve, all aircraft tend to be speed
unstable. This is because increasing speed, decreases drag, causing the aircraft
to accelerate further. Conversely, decreasing speed increases drag, causing the
aircraft to decelerate further. Although both jet and propeller aircraft exhibit
this speed instability, propeller aircraft are less affected than jets. This is
because the thrust produced by a jet engine remains approximately constant
with increasing TAS, whereas that of a propeller decreases rapidly. This means
that as a jet accelerates below VMD,its thrust remains constant, so decreasing
drag causes it to continue to accelerate. In the case of a propeller aircraft
increasing speed decreases drag, but also decreases thrust, so the excess thrust
and hence subsequent acceleration is less. Similarly, when a propeller aircraft is
decelerating below VMD,the increasing drag is partly offset by increasing thrust,
thereby reduce the effect of speed instability. A propeller aircraft is therefore
inherently more speed stable than a jet because its thrust decreases with
increasing airspeed.
POW 46. c.
The tangent drawn on the diagram touches
the curve a t the point a t which the ratio of
EAS : whatever is represented by the curve
is a t a maximum. Although the point is
marked as VMD,this occurs a t the
bottom of the drag : EAS curve, so the
diagram does not represent a drag : EAS
curve. But VMDis also the speed a t which
the ratio of EAS : power required is
maximum, so the curve can be identified as
the power required : EAS curve.
VMD
EAS
POW 47. c.
Tlle tangent touches the curve a t the point
where the ratio of EAS :whatever is
represented by the curve is maximum. This
point is however identified as LID Max which
occurs at VMD. But VMDoccurs a t the bottom
of the Drag : EAS curve, so this must be some
other curve. V Mis~also the point a t which
the ratio of EAS : power required is maximum,
so the curve represents Power required : EAS
LID Max
EAS
POW 48. a.
VMPis the speed a t which power required is
a t its minimum value. The diagram in this
question illustrates a Drag : EAS curve and
this can be converted into a power required
Drag
curve by multiplying all drag values by TAS.
The Power required : EAS curve produced
by this process is illustrated a t the right.
VMpis represented by the lowest point on
the Power required : EAS curve. This is point
A on the original diagram. I t should be noted
EASA
this is also the point a t which a straight line
drawn from the origin intersects the Drag :EAS
curve a t right angles (90 degrees).
B
C
D
POW 49. a.
The best angle of climb speed, Vx for a
propeller aircraft a t the lowest safe flying
speed. This is higher than Vs but lower
than VMP. The diagram in this question
illustrates a Drag : EAS curve and VMP
occurs where a line drawn from the origin
intersects such a curve at right angles. This
occurs at point B on the diagram. The only
point on the diagram that might represent
Vx for a propeller aircraft is therefore point A.
Drag
AB C D
POW 50. c.
The best rate of climb speed, Vy is the speed
at which excess power is maximum. For a
propeller aircraft this occurs between \IMP
and VMD. VMPlies a t the point where a
Drag
straight line drawn from the origin intersects
the Drag :EAS curve a t right angles. This is
point B on the diagram. VMDis the speed
a t which drag is minimum. This is point D
D E EAS
on the diagram. The point most representative
of Vy for a propeller aircraft is therefore point C.
E
EAS
A B C
POW 51. b.
VMP.isthe speed at which power required is minimum so as speed decreases
below VMP,the power required increases. VMDis the speed at which drag is
lowest so when decreasing speed below VMDthe drag force increases. But VMpis
lower than VMD,so decreasing speed below VMpcauses both drag and power
required to increase.
POW 52. b.
VMPis the speed a t which power required is
minimum, so as speed increases above VMp,
the power required increases. V Mis~the
speed at which drag is minimum, so when
increasing speed, drag decreases below VMD
but increases above VMD. But Vhfpis lower
than VMD,so increasing speed above Vnlp
means first approaching, then exceeding VMD.
The effect of such acceleration is that power
required increases whilst drag decreases up
to V3iDthen increases at higher speeds. Thi
is illustrated in the diagram a t the
Power required increases with
with increasing speed above VMP Drag decreases up to VMD
Drag increases
above VMD
POW 53. b.
CLMAX
occurs a t the low'speed stall, which is
Power decreases up to
the lowest speed at which an aircraft is
capable of generating sufficient lift for
straight and level flight. VMo is the maximum
speed at which an aircraft can be flown in
normal circumstances. This question
therefore concerns the shapes of the Drag : EAS
and power required : EAS curves, between Vs
and Vhlo. As illustrated in the diagram at the
right, both drag and power required decrease
then increase.
Drag decrease up to VMDthen increases
POW 54. b.
Power required = Drag x TAS. IAS is proportional to 1/2pv2such that
climbing at constant IAS means climbing at constant 1/2pv2. But as altitude
increases density decreases, so when climbing at constant IAS, TAS must
increase such that the increase in v2balances the decrease in p. If density at
40000 feet is % of that at sea level, then in a constant IAS climb, TAS must
double its sea level value, so that v2increases by a factor of 4. This means that at
40000 feet, although drag remains unchanged, TAS is double its sea level value.
Because power required is equal to drag multiplied by TAS, this doubling of
TAS will cause power required to be doubled. Power required will therefore
increase by a factor of 2.
It should be noted that although power required at constant density is
proportional to TAS' and would therefore increase by a factor of 8 if TAS were
doubled, the decreasing density in this question offsets some of the increase in
power required.
POW 55. d.
IAS is proportional to 1/2pv2(where p is air density and V is TAS), such that
~ ~ as
. altitude
climbing a t constant IAS means climbing at constant 1 1 2 ~ But
increases density decreases, so when climbing a t constant IAS, TAS must
increase such that the increase in v2balances the decrease in p. If density at
40000 feet is !A of that at sea level, then in a constant IAS climb, TAS must
double IAS so that v2increases by a factor of 4. This means that at 40000 feet,
IAS is half of TAS. So in a constant TAS climb up to 40000 feet IAS decreases to
half of its sea level value. But power required is proportional to IAS such that it
decreases between VMo and VMP,then increases between VMPand Vs. This
means that in a constant TAS climb, power required will decrease provided the
airspeed remains above Vklp. If however airspeed becomes lower than VMP,then
power required will increase with reducing IAS.
POW 56. b.
VXis the airspeed at which an aircraft will achieve its best climb gradient. The
climb gradient is approximately equal to excess thrust divided by aircraft weight.
But power output at any given power lever setting is proportional to the mass
flow of air through the engine. As altitude increases, air density, air mass flow
and hence power output decrease. So if an aircraft climbs a t a constant power
lever setting, the power output, excess power and climb gradient will all decrease
with increasing altitude. The rate of climb is proportional to TAS and as power
output decreases TAS and rate of climb both decrease. The absolute ceiling of
an aircraft is the altitude at which excess thrust, excess power, climb gradient
and rate of climb have all reduced to zero.
POW 57. b.
Vy is the airspeed at which an aircraft will achieve its best rate of climb. The
rate of climb is equal to excess power divided by aircraft weight. But power
output a t any given power lever setting is proportional to the mass flow of air
,
through the engine. As altitude increases, air density, air mass flow a n 8 hence
power output decrease. So if an aircraft climbs at a constant power lever setting,
the power output, excess power and rate of climb will all decrease with
increasing altitude. The absolute ceiling of an aircraft is the altitude a t which
excess thrust, excess power, climb gradient and rate of climb have all reduced to
zero.
POW 58. b.
The thrust power output of any aircraft
propulsion system equal to the product
of its thrust multiplied by itsTAS.
When the aircraft is stationary on the
Power
Ground, TAS and thrust horsepower are availabl
zero. As airspeed increases, thrust
horsepower varies in proportion to the
changes in thrust and TAS. In the case
of jet aircraft thrust is approximately
constant at all speeds, so power available
increases linearly with TAS. In the case
EAS
of a propeller aircraft, thrust decreases
with increasing TAS. The combined effect of increasing TAS and decreasing
thrust, as a propeller aircraft accelerates from rest is that the power available is
initially zero at zero speed, then increases with increasing TAS. before reaching a
peak value at about 300 Kts, above which it gradually decreases. Power output
is also proportional to the air mass flow passing through an engine. As altitude
increases, air density, air mass flow and power output all decrease. The diagram
in this question represents the variation of power available from a propeller
aircraft with increasing altitude.
POW 59. d.
I
The power required by an aircraft in constant
speed level flight is equal to drag multiplied by
TAS. At any given IAS the drag remains
constant with increasing altitude, but the TAS
increases. This means that power required at
Power
required
any given speed, and the speed a t which
minimum power is required, both increase with
increasing altitude. The effect of these changes is
that the power required : EAS curve moves
upwards and to the right as altitude increases.
The diagram in this question illustrates these
effects.
POW 60. c.
The power required by an aircraft in constant
speed level flight is equal to drag multiplied by
TAS. At any given IAS the drag remains
constant with increasing altitude but the TAS
increases. This means that power required at
any given speed and the speed a t which
minimum power is required both increase with
increasing altitude. But increasing aircraft
weight has the same effect as increasing altitude
in that increasing weight moves the power
required curve upwards and to the right. The
diagram is this question could therefore
represent the changes in power required, caused
by increasing altitude or increasing weight.
Only increasing weight is however included in
the options provided in this question so option
c is the most appropriate.
High altitude
EAS
Power
require
EAS
POW 61. c.
The power available from any aircraft
engine is equal to Thrust multiplied by
TAS. In the case of a jet engine thrust is
approximately constant with increasing
TAS so power available increases linearly
with increasing TAS. Options A and C PAV
in the diagram might therefore be
representative of jet power available. As
altitude increases air density decreases
causing power available to reduce. The
EAS
line representing jet power available a t
high altitude would therefore be of a
shallower gradient than that representing sea level performance. This is most
accurately represented by line C. Lines B and D are representative of power
available from piston engines a t high and low altitudes respectively.
POW 62. d.
Power available from any aircraft engine
is equal to Thrust multiplied by TAS. In
the case of a propeller aircraft, thrust
decreases with increasing altitude such
that power available increases from zero
at zero TAS, reaching a maximum value PA"
at about 300 Kts, then reducing as speed
increases further. This effect might be
represented by curves B and D on the
diagram. As altitude increases, air density
EAS
decreases causing power available to
reduce. The gradient of the power available curve is therefore shallower at high
altitude than at low altitude. Curve D is therefore most representative of the
power available from a propeller at high altitude.
POW 63. a.
Jet power available increases linearly
with increasing speed and so might be
represented by lines A and C on the
diagram. But as altitude increases,
decreasing air density causes power
available to decrease, reducing the
gradient of the power available : EAS
line. Line A therefore best represents
jet power available at low altitude.
And line C best represents jet power
available at high altitude.
PA
EAS
POW 64. b.
Propeller power available is zero at
Zero airspeed, increases to a maximum
value at about 300 Kts then decreases as
speed increases further. This might be
represented by curves B and D on the
diagram. But as altitude increases air PAV
density reduces causing power
available to decrease. This has the
effect of reducing the gradient of the
power available curve at high altitude.
EAS
Curve B therefore best represents propeller power available at low altitude.
Curve D best represents propeller power available at high altitude.
POW 65. d.
Propeller thrust decreases with increasing
EAS in a manner similar to curves C and D
in the diagram. But as altitude increases
thrust decreases, such that the high altitude
thrust curve is lower than the low altitude
curve. Curve D is therefore the most
representative of propeller thrust at high
altitude.
EAS
POW 66. c.
Propeller thrust decreases with increasing
EAS in a manner similar to curves C and D
in the diagram. But as altitude increases
thrust decreases, such that the high altitude
thrust curve is lower than the low altitude
curve. Curve C is therefore the most
representative of propeller thrust at low
altitude.
EAS
P O W 67. b.
Jet thrust available is maximum at zero
EAS, then decreases to a minimum value
at about 250 Kts before again increasing
at higher speeds as indicated in curves A
and B. As altitude increases, air density
and thrust decrease so the curve is lower
at high altitude. Option B is therefore
the high altitude thrust available curve.
T
EAS
POW 68. a.
Jet thrust available is maximum at zero
EAS, then decreases to a minimum value
a t about 250 Kts before again increasing
at higher speeds as indicated in curves A
and B. As altitude increases, air density
and thrust decrease so the curve is lower
a t high altitude. Option A is therefore
the low altitude thrust available curve.
T
EAS
POW 69. d.
In straight and level flight lift equals weight so if weight increases then lift must
increase by the same amount.
where
S
But Lift = C L ~ / ~ ~ V ~
CLis the coefficient of lift
p is the air density
V is the TAS
S is the wing area
p is beyond the control of the pilot and S is normally varied only by extending
flaps for take-off and landing, so if weight and lift are to be increased while
continuing to fly a t the same speed (V) then CL must be increased by increasing
angle of attack.
At VnlDthe total drag is made up of equal proportions of induced drag and
profile drag. Increasing CL will cause induced drag to increase in proportion to
CL', but will have no significant effect on profile drag.
So to increase lift by 50% whilst maintaining constant speed, CL must also
increase by 50% to 1.5 of its previous value. The resulting increase in induced
~ if CL becomes 1.5 times its previous value,
drag will be proportional to C L SO
then induced drag will become 1.5~times its previous value. This means that
induced drag will increase to 2.25 times its previous value. This represents a
125% increase.
The overall effect on total drag will be a 125% increase in induced drag but no
increase in profile drag. And because each constitutes half of the total drag the
average rise will be (125/2)%, which is 62.5%
Power required is equal to total drag multiplied by TAS so increasing total drag
by 62.5% whilst maintaining constant speed will increase power required by
62.5%.
POW 70. c.
In straight and level flight lift equals weight so if weight increases then lift must
increase by the same amount.
But Lift = c L 1 / 2 p v 2 s
where
CL is the coefficient of lift
p is the air density
V is the TAS
S is the wing area
p is beyond the control of the pilot and S is normally varied only by extending
flaps for take-off and landing, so if weight and lift are to be increased while
continuing to fly a t the same speed 0,then CL must be increased by increasing
-an-gle of attack. Increasing CLwill cause induced drag to increase in proportion
but will have no significant effect on profile drag.
to cL2,
So to increase lift by 50% whilst maintaining constant speed, C L must also
increase by 50% to 1.5 of its previous value. The resulting increase in induced
drag will be proportional to C; SO if CL becomes 1.5 times its previous value,
then induced drag will become 1.5~times its previous value. This means that
induced drag will inerease to 2.25 times its previous value. This represents a
125% increase. Profile drag will not however increase significantly.
At VMDthe totaldrag is made up of equal proportions of induced drag and
profile drag. The overall effect on total drag will be a 125% increase in induced
drag but no effect on profile drag. And because each sonstitutes half of the total
drag the average rise will be (125/2)%, which is 62.5%.
But at speeds above V M ~induced
,
drag represents less than half of total drag so
the 125% increase in induced drag will increase total drag by less than 62.5%.
The figure of 42.5% offered in option a is correct for only one speed above VMD,
whereas option c is true for all speeds above VMDand is therefore the most
appropriate.
POW 71. a.
In straight and level flight lift equals weight, so if weight increases then lift must
increase by the same amount.
where
S
But Lift = C L ~ / ~ ~ V ~
CL is the coefficient of lift
p is the air density
V is the TAS
S is the wing area
p is beyond the control of the pilot and S is normally varied only by extending
flaps for take-off and landing, so if weight and lift are to be increased while
continuing to fly at the same speed (V) then CL must be increased by increasing
angle of attack. Increasing CLwill cause induced drag to increase in proportion
- to C L ~ , , & U ~will have no significant effect on profile drag.
So to increase lift by 50% whilst maintaining constant speed, CL must also
increase by 50% to 1.5 of its previous value. The resulting increase in induced
drag will be proportional to C L SO
~ if CL becomes 1.5 times its previous value,
then induced drag will become 1.5~times its previous value. This means that
induced drag will increase to 2.25 times its previous value. This represents a
125% increase. Profile drag will not however increase significantly.
At VMDthe total drag is made up of equal proportions of induced drag and
profile drag. The overall effect on total drag will be a 125% increase in induced
drag but no effect on profile drag. And because each constitutes half of the total
drag the average rise will be (125/2)%, which is 62.5%.
.But at speeds below V M ~induced
,
drag represents more than half of total drag so
the 125% increase in induced drag will increase total drag by more than 52.5%.
The relative position of the speed chosen and VMPhas no significance in
determining the qualitative effects of weight increases so options c and d are both
incorrect.
POW 72. c.
At the absolute ceiling the maximum power available is just sufficient to permit
straight and level flight at VhlD. At all other flight speeds more power will be
required than is available so the aircraft will loose height. If an engine fails then
total power available will decrease causing, the absolute ceiling to decrease and
so the aircraft will descend. In attempting to maintain height the options of
increasing o r decreasing speed will be ineffective as both will result in a power
deficit. Because the absolute ceiling by definition involves the use of maximum
power available, option d, "increase power setting" is not possible. Had the
engines not been a t maximum power setting prior to the engine failure, the
aircraft would not have been at its absolute ceiling. The only practicable option
to maintain height following single engine failure is therefore to decrease weight
in order to decrease power required.
POW 73. b.
At the absolute ceiling the maximum power available is just sufficient to permit
straight and level flight at VXjD.At all other flight speeds more power will be
required than is available so the aircraft will loose height. If an engine fails then
total power available will decrease making continued flight a t that altitude and
speed impossible. In attempting to maintain speed the options of climbing,
decelerating o r accelerating will be ineffective as all will result in a power deficit.
The only practicable option to maintain speed following single engine failure, is
therefore to descend to decrease power required and increase power available.
POW 74. d.
At the absolute ceiling the maximum power available is just sufficient to permit
straight and level flight at VMD. At all other flight speeds more power will be
required than is available so the aircraft will loose height. If reheat is then
selected then all of the additional thrust will be excess to that required for that
given combination of altitude and speed. This excess thrust can be employed to
climb. The sine of the maximum angle of climb of an aircraft in steady flight is
equal to the excess thrust divided by the weight of the aircraft.
In this case Sine of climb angle = 5000 lbf / 50000 Ibf which is 0.1 giving an
angle of approximately 6 degrees. But the sine of an angle is also approximately
equal to the gradient of that angle so the maximum achievable climb gradient is
0.1 or 10%.
POW 75. c.
The power required by a n aircraft in constant speed level flight is equal to drag
multiplied by TAS. At any given IAS the drag remains constant with increasing
altitude but the TAS increases. This means that power required a t any given
speed increases with increasing altitude.
STAB 1. b.
The term "Short period mode" refers to high frequency longitudinal oscillation
about the lateral axis. Although not induced by the pilot, it is of such high
frequency (short period) that attempts to damp it manually are likely to
exacerbate the problem.
STAB 2. b.
Phugoid motion is a low frequency (long period) longitudinal oscillation in
which potential energy and kinetic energy are repeatedly exchanged for each
other. This results in a porpoising motion in which airspeed and altitude vary
alternately, but angle of attack remains approximately constant.
STAB 3. b.
Iu order for an aircraft to be positively stable the C of G must always be forward
of the neutral point.
STAB 4. d.
The transition point is the point at which laminar boundary layer becomes
turbulent. The aerodynamic centre or neutral point of an aircraft, is the point
about which the pitching moment remains constant with changing angle of
attack It is the point at which the C of G would cause the aircraft to be
neutrally longitudinally stable when in straight and level flight. The manoeuvre
point is the point at which the C of G would cause the aircraft to be neutrally
longitudinally stable in pull up manoeuvres.
STAB 5. d.
Anhedral, low wings and ventral fins, all reduce lateral stability. Dihedral,
sweepback, high wings, and high fins, all contribute to lateral stability.
STAB 6. d.
Although lateral instability is caused by excessively strong directional stability,
this is more commonly reduced by making the fin smaller. Dutch roll occurs
when lateral stability is greater than directional stability, so decreasing
effectiveness of the fin would exacerbate this problem. A ventral fin increases
the area of the fin, but its principal effect is to reduce its aspect ratio. This
increases the effectiveness of the fin at high sideslip angles by increasing its
stalling angle of attack.
STAB 7. c.
Stability is reduced by both high altitude and high speed, so a reduction in either
or both is likely to prevent Dutch Roll. When flying at high speed and high
altitude however, the margin between the high speed and low speed stalls is very
narrow, so changing speed without first reducing altitude is not practicable.
The solution in this case is to reduce altitude in order to widen the speed margin
and increase stability, then reduce speed to increase stability further.
STAB 8. c.
Forward movement of the C of G increases the moment arms of both the
tailplane and fin, thereby increasing longitudinal and directional stability.
Because the moment arms of the wings are unchanged by variations in fore and
aft C of G position, lateral stability is not affected.
STAB 9. d.
The stability of an aircraft is a measul'e of its ability to resist changes in attitude
o r motion, whereas manoeuvrability is a measure its ability to change these
parameters. A highly stable aircraft will not be very manoeuvrable and a highly
manoeuvrable aircraft will not be very stable. Stability and manoeuvrability are
therefore inversely related.
STAB 10. c.
Sweepback has no significant effect on longitudinal stability except in the stall,
where it causes the aircraft to pitch up. When an aircraft rolls o r yaws, it
sideslips. When sideslipping, the effect of sweepback angle is to increase the
effective aspect ratio and frontal area of the leading wing, in comparison to that
of the trailing wing. The increased aspect ratio increases lift on the leading wing,
causing the aircraft to roll out of the sideslip. The increased effective frontal
area increases drag on the leading wing, causing the aircraft to yaw into the
sideslip. These effects increase both lateral and directional stability.
STAB 11. d.
The aerodynamic centre is the point on the chord line of a wing about which its
pitching moment remains constant with changes in angle of attack. When a
cambered wing is a t its zero lift angle of attack, it produces a nose down pitching
moment about its aerodynamic centre. It is this moment that remains constant
with changes in angle of attack.
STAB 12. d.
The aerodynamic centre of a wing is the point about which pitching the moment
remains constant with changing angle of attack. As angle of attack increases up
to the stall, the lift produced a t any given speed also increases. Beyond the stall
the lift decreases rapidly with further increases in angle of attack. The centre of
pressure is the point through which the lift force is considered to act. The
pitching moment about the aerodynamic centre is the product of the lift force
multiplied by the distance from aerodynamic centre to centre of pressure. For
pitching moment about the aerodynamic centre to remain constant, it is
necessary for the distance to reduce as lift force increases, such that the product
of the two remains unchanged. This means that, as angle of attack and lift
increase u p to the stall, the distance between the C of P and the aerodynamic
centre must reduce. Beyond the stall, reducing lift must result in an increase in
the distance. The distance therefore decreases up to the stall then increases.
STAB 13. d.
A JAR certificated commercial aircraft must always be both statically and
dynamically positively stable. Options a, b and c are therefore not acceptable.
Option a, is in fact impossible due to the fact that static stability is an essential
prerequisite for dynamic stability.
STAB 14. b.
The principal components providing stabilising moments in pitch and yaw are
the tailplane and fin respectively. The stability of an aircraft is directly
proportional to the magnitude of these stabilising moments, so the greater the
moments, the more stable the aircraft. The magnitude of each of the moments is
the product of the lift force generated by the component, multiplied by the
distance from the component's centre of pressure to the C of G of the aircraft.
By moving the C of G to its forward limit the pitching and yawing stabilising
moments are maximised. The longitudinal and directional stability of an aircraft
is therefore at its maximum value when the C of G is on its forward limit.
Forward and rearward movement of the C of G have no significant effect on
lateral (rolling) stability.
STAB 15. b.
The manoeuvre point is the point at which the C of G of an aircraft would cause
it to be neutrally longitudinally stable when performing a pull up manoeuvre. C
of G forward of this point would cause the aircraft to be positively stable, whilst
a more rearward position would make the aircraft unstable. Except in the case
of highly agile fly-by-wire military aircraft, it is never acceptable for an aircraft
to be neutrally or negatively stable. The C of G of other aircraft types must
therefore always be forward of the manoeuvre point.
STAB 16. b.
Dihedral is the principal design factor contributing to lateral stability.
Whenever an aircraft rolls, it sideslips in the direction of the roll. In the sideslip,
dihedral increases the effective angle of attack of the leading wing and decreases
that of the trailing wing. The difference in lift generated by the two wings then
rolls the aircraft away from the sideslip, back to the wings level condition. In the
sideslip, the airflow passing over the upper surface of the fuselage moves
upwardsat*-leading
side and downwards at the trailing side. This affects
angle of attack and lift in the same manner as dihedral. In the sideslip the angle
of attack on the dorsal fin generates a lift force acting away from the sideslip.
Because the C of P of the fin is above the C of G of the aircraft, this dorsal fin lift
force generates a rolling moment, taking the aircraft back to wings level. With
sweep back, in the sideslip the leading wing is effectively of greater aspect ratio
than the trailing wing. This again causes an imbalance of lift rolling the aircraft
away from the sideslip.
STAB 17. a.
Phugoid motion is a natural phenomenon in which an aircraft oscillates in pitch,
such that it follows an undulating path, comprising alternate climbs and
descents. For example, the process might commence when a disturbance causes
the aircraft to pitch up slightly, after which it climbs, gradually exchanging
airspeed for altitude. As airspeed reduces, the aircraft attitude reduces, causing
it to level off, before commencing a shallow dive. In the dive, altitude is
exchanged for airspeed and as airspeed increases, the aircraft levels off, before
again commencing a shallow climb. The process is then repeated at a very long
periodicity. Because the aircraft attitude is nose u p when ascending, and nose
down when descending, its angle of attack is approximately constant. Attitude,
airspeed, and altitude, are however constantly changing. Because angle of attack
is constant, the motion is subject to very weak aerodynamic damping, and so
may continue for some time.
STAB 18. b.
Forward movement of the C of G increases static stability and hence increases
the rate at which the aircraft tends to return to its pre-disturbance condition.
This means that forward movement of the C of G will increase the frequency
(reduce the period) of phugoid motion. Rearward movement of the C of G will
have the opposite effect, increasing period and decreasing frequency.
STAB. 19. d.
Phugoid motion involves pitching about the lateral axis and is not affected by
directional stability. The manoeuvrability of an aircraft is inversely
proportional to its stability, increasing directional stability therefore reduces yaw
rates. Option b is therefore theoretically true, but would never occur in a real
aircraft. Dutch Roll occurs when lateral stability is dominant, causing the
aircraft to roll out of the sideslip before it can turn into it. When an aircraft is
disturbed in roll or yaw, its response depends upon the relative magnitudes of its
directional and lateral stabilities. If directional stability is excessively dominant,
then the aircraft will turn into the sideslip, before it can roll out of it. This will
cause it to exhibit spiral instability, performing a continuous descending turn.
STAB 20. b.
The term static stability refers to the initial response of an aircraft, after being
disturbed from its trimmed condition. If it tends to return towards its original
condition, it is statically stable. Negative static stability causes the aircraft to
continue to move away from its pre-disturbance condition. Neutral static
stability cause it to remain in its disturbed condition. Only a statically stable
aircraft will therefore tend to return to it pre-disturbance condition. Dynamic
stability refers to the subsequent behaviour of an aircraft. If the oscillations
about its pre-disturbance condition diminish with time, it is said to be
dynamically stable. If the oscillat~onsincrease, it is dynamically unstable. If
they remain constant, it is neutrally dynamically stable. All oscillatory motion
involves repeatedly turning towards the pre-disturbance condition, so only a
statically stable aircraft vill oscillate. The term dynamic stability is therefore
irrelevant if a n aircraft is not statically stable. Option b is therefore the only
practicable condition described in this question.
STAB 21. d.
Forward movement of the C of G of an aircraft increases its nose down pitching
moment and hence increases the nose up stick force required to maintain a given
attitude. Rearward C of G movement has the opposite effect. Deployment of
trailing edge flaps moves the C of P rearward, which also increases the nose
down pitching moment. The combination producing the greatest stick force is
therefore forward C of G and flaps down as described in option d.
STAB 22. c.
The pitching moment of an aircraft is equal to its weight multiplied by the
distance from its C of P to its C of G. The C of G is always forward of the C of P,
so forward movement of the C of G increases the nose down pitching moment.
In order to maintain a given attitude, this pitching moment must be balanced by
an opposing moment from the tailplane. This moment is proportional to the
rearward force applied to the stick. The stick force gradient is a measure of the
rate a t which the required stick force varies with airspeed. If a rearward force is
required to hold the nose up at low speed and a forward push is required hold
the nose down a t high speeds, this is described as a negative stick force gradient.
Although gradient depends on other factors in addition to C of G position,
forward C of G movement increases the slope of a negative stick force gradient.
STAB 23. a.
The term manoeuvre stability refers to the phenomenon whereby the
longitudinal stability of an aircraft increases when it is in a pull up manoeuvre.
This is because when pulling out of a dive the tail of the aircraft rotates
downwards, causing an increase in the angle of attack of the tailplane. This
increased angle of attack increases the tailplane lift force, and hence increases
the longitudinal stabilising moments.
When subjected to varying pitching moments, an aircraft in flight rotates about
its C of G. The stability of an aircraft is a measure of its tendency to resist such
rotations. 'The main component contributing to the longitudinal stability of an
aircraft is its tailplane. The magnitude of the stabilising moments generated by
the tailplane is determined by the tailplane lift force, multiplied by the distance
between the C of P of the tailplane and the C of G of the aircraft. Forward
movement of the C of G increases the moment arm of the tailplane, and hence
increases longitudinal static stability and the manoeuvre stability.
STAB 24. b.
In order for an aircraft to be longitudinally stable its C of G must be forward of
the neutral point. This ensures that the lift 1 weight couple produces a nose down
pitching moment. In order to prevent nose down pitching, the tailplane must
produce a balancing nose up moment, by generating a negative lift or down
force. The lift force from the wings must therefore equal the weight of the
aircraft plus the down force from the tailplane.
As C of G is moved forward the aircraft becomes more stable, but at the cost of
an increased tailplane down force and hence a greater total lift force from the
wings. The generation of this increased wing lift requires a greater CLand
hence a greater angle of attack at any given speed. The aircraft will therefore
reach its stalling angle of attack at a higher speed when the C of G is moved
forward. Moving the C of G forward therefore increases the stalling speed of the
aircraft at any weight. The balance of forces and moments described above is
illustrated in the diagram below.
Lift acting upward through C of P must
equal the sum of both the aircraft weight
and the tailplane down force
LIFT
CofG
.f
I
C of P
WEIGHT
1
Balancing nose up
moment is produced by
tailplane down
force.
Nose down moment is product of weight and
distance between C of G and C of P. Moving
C of G forward therefore increases this nose down
moment. This requires an increased tailplane
down force to produce a balancing nose up moment.
STAB 25. c.
The term static stability refers to the immediate response of an aircraft after
being subjected to a disturbance in flight. If it tends to return to its predisturbance condition it positively statically stable. If it does not tend to return,
but continues to deviate it is negatively statically stable. If it remains in its post
disturbance condition, neither returning to, nor diverging from its original
condition, it is neutrally statically stable. Only a positively statically stable
aircraft tends to return to its original condition. The term dynamic stability
refers to whether, following a disturbance, an aircraft's oscillations increase,
decrease or remain constant.
In determining the relationship between static and dynamic stability it is
important to note that in order to oscillate, an aircraft must repeatedly tend to
return to its pre-disturbance condition. Only a statically stable aircraft does this,
so only a statically stable aircraft call be dynamically stable. Indeed without
static stability there can be no oscillations, so the concept of dynamic stability
becomes meaningless.
The fact that an aircraft is statically stable does not however guarantee that it
will be dynamically stable. If it oscillates with decreasing amplitude it is
positively dynamically stable. If it oscillates with increasing amplitude it is
negatively dynamically stable. And if it oscillates with constant amplitude it is
neutrally dynamically stable. Of the options listed in this question, only option c
is true.
STAB 26. c.
Increasing the camber of a iring will increase its CL at any given angle of attack
and its CL MA^, but will decrease its stalling angle and stalling speed. Increased
camber will also increase the fore and aft range through which the C of P moves
when angle of attack is changed. This does not however affect longitudinal
stability to any significant degree.
STAB 27. b.
The term anhedral refers to the configuration where the wings are angled
downwards from root to tip. The purpose of anhedral is to decrease the lateral
static stability of an aircraft, in order to increase its lateral manoeuvrability.
Anhedral has no significant effect on directional or longitudinal stability.
STAB 28. d.
The term static stability refers to the immediate response of an aircraft after
being subjected to a disturbance in flight. If it tends to return to its predisturbance condition, it is positively statically stable. If it remains in its post
disturbance condition, neither returning, to nor diverging from its original
condition, it is neutrally statically stable. If it does not tend to return, but
continues to deviate, it is statically unstable. The term directional stability
refers to stability in yaw. A statically directionally stable aircraft will always
yaw into a sideslip. So a statically directionally unstable aircraft will continue to
yaw away from its original heading following a distnrbance. When an aircraft is
disturbed in yaw it begins to sideslip in the opposite direction. Left yaw, for
example, causes right sideslip. But if it is directionally statically unstable it will
continue to yaw to the left, away from the sideslip.
STAB 29. a.
The term roll damping refers to the phenomenon whereby aerodynamic forces
acting on a rolling aircraft tend to damp out the rolling motion. When for
example an aircraft is rotling to the right, its right wing is moving downwards
and its left wing is moving upwards. The right wing therefore experiences an
upward flow of air, whilst the left wing experiences a downward flow. These
vertical airflows increase the angle of attack of the downward-going wing and
decrease the angle of attack of the upward-going wing. These changes in angle of
attack increase the lift produced by the downward-going wing and decrease that
produced by the upward-going wing. The difference in lift generated by the
wings resists or damps out the rate of roll. But this damping effect occurs only
when the aircraft is actually rolling and merely opposes roll, rather than
returning the aircraft to it wings level condition. Roll damping therefore
increases later dynamic stability but does not affect static stability.
STAB 30. b.
Wing mounted engine pods tend to be destabilising, particularly in yaw, due to
the yawing moments generated by lateral airflows over them in sideslip.
STAB 31. d.
In order for an aircraft to be longitudinally stable it must generate pitching
moments opposing, and proportional to, changes in attitude. The generation of
such pitching moments requires longitudinal dihedral. This term refers to the
configuration in which the angle of incidence of the wing is greater than that of
the tailplane. This arrangement ensures that any given increase or decrease in
pitch attitude, causes a greater percentage increase o r decrease in angle of attack
on the tailplane than of the wing.
If for example an aircraft has 3 degrees of longitudinal dihedral, then the angle
of incidence of the wing is 3 degrees greater than that of the tailplane. When
flying straight and level, if the angle of attack of the wing is 6 degrees, then that
of the tailplane will be 3 degrees. If a gust causes a 3 degree pitch u p motion,
then the angle of attack of the wing will increase to 9 degrees, while that of the
tailplane increases to 6 degrees. This represents a 50% increase on the wing and
a 100% increase on the tailplane. The larger increase in lift on the tailplane will
then generate a strong nose down pitching moment, tending to return the
aircraft to its pre-disturbance condition.
In the case of a canard configuration, longitudinal dihedral is still essential for
longitudinal stability, but in this case the angle of incidence of the canard will be
greater than that of the wing.
Option a, is incorrect because a longitudinally stable aircraft must maintain
attitude without autopilot assistance. Options b, and c are incorrect because
neither a tailplane nor a canard, are essential for longitudinal stability, as
illustrated for example, in a delta wing configuration.
STAB 32. b.
In order for an aircraft to be longitudinally stable it must generate pitching
moments opposing, and proportional to, changes in attitude. The generation of
such pitching moments requires longitudinal dihedral. In the case of a canard
configuration, this term refers to the arrangement in which the angle of
incidence of the canard is greater than that of the wing. This ensures that any
given increase o r decrease in pitch attitude, causes a greater percentage increase
or decrease in angle of attack on the wing than on the canard.
If for example an aircraft has 3 degrees of longitudinal dihedral then the angle of
incidence of the canard is 3 degrees greater than that of the wing. When flying
straight and level, if the angle of attack of the canard is 6 degrees, then that of
the wing will be 3 degrees. As the pitching attitude of such an aircraft is
increased the canard will reach its stalling angle before the wing. The canard
will therefore stall before the wing, causing the aircraft to pitch nose down such,
that it accelerates out of the stall, rather than falling deeper into it.
STAB 33. b.
In order for an aircraft to be longitudinally stable it must generate pitching
inoments opposing, and proportional to, changes in attitude. The generation of
such pitching inoments requires longitudinal dihedral. This term refers to the
configuration in which the angle of incidence of the wing is greater than that of
the tailplane. This arrangement ensures that any given increase o r decrease in
pitcli attitude, causes a greater percentage increase or decrease in angle of attack
on the tailplane than on the wing.
If for example an aircraft has 3 degrees of longitudinal dihedrai then the angle of
incidence of the wing is 3 degrees greater than that of the tailpl,ane. When flying
straight and level, if the angle of attack of the wing is 6 degrees, then that of the
tailplane will be 3 degrees. As the pitching attitude of such a n aircraft is
increased, the wing will reach its stalling angle before the tailplane. The wing
will therefore stall before the tailplane, causing the aircraft to pitch nose down
such that it accelerates out of the stall.
STAB 34. b.
Tlle term short period motion refers to high frequency pitching oscillations
about the lateral axis. Such motion is usually highly damped and hence selfcorrecting. Because of its high frequency however, any attempt to counter it by
manual control inputs is likely to exacerbate the problem. The most effective
solution is to release the controls and allow the aircraft to stabilise itself.
STAB 35. b.
In order to allow for variations in payload and fuel, a range of C of G positions
must be permitted. But for an aircraft to be longitudinally stable, it must possess
longitudinal dihedral and a C of G position forward of its neutral point. So all
parts of the allowable C of G range must be forward of the neutral point, if the
aircraft is to remain stable throughout flight. The aft C of G limit must therefore
be forward of the neutral point.
STAB 36. a.
If an aircraft is disturbed in flight such that it banks, it will sideslip towards the
low wing. This sideslip will cause the airflow to approach the aircraft from the
side to which it is banked. In the case of a swept wing aircraft, this sideslip
airflow will have two effects.
The effective aspect ratio of the leading wing will be increased while that of the
trailing wing will be decreased. This will increase the CLof the leading wing and
decrease that of the trailing wing, thereby rolling the aircraft back to the wings
level condition. This will increase the lateral stability of the aircraft.
Also, in the sideslip the effective frontal area of the leading wing will be
increased, while that of the trailing wing is decreased. This will increase the drag
force on the leading wing and decrease that on the trailing wing, causing the
aircraft to yaw into the sideslip. This will increase the directional stability of the
aircraft. Sweep back will however have no effect on longitudinal stability.
-
STAB 37. b.
For any given wing area, increasing wing sweep angle decreases the wingspan
and hence decreases the aspect ratio. This decrease in aspect ratio increases
wing tip vortices and induced drag. Increased induced drag increases both total
drag and V h l ~ .But any aircraft is speed stable only at speeds above VMD,SO if
VhlDis increased, then the speed range in which the aircraft is speed stable is
decreased. Wing sweep back therefore decreases speed stability.
STAB 38. a.
The term longitudinal stability refers to an aircraft's ability to resist changes in
its pitch attitude and to regain its original pitch attitude following a disturbance
in flight. If an aircraft is longitudinally unstable it will continuously change
pitching attitude. This in turn will cause it to continuously change airspeed and
altitude. It will therefore be speed unstable.
STAB 39. c.
One of the principle factors contributing to the dynamic stability of an aircraft is
aerodynamic damping. Whenever an aircraft is rolling for example, the downgoing wing experiences an up-going airflow and the up-going wing experiences a
down-going airflow. This will increase the lift on the down-going wing and
decrease that on the up-going wing, thereby damping out the rolling motion. As
altitude increases, decreasing air density reduces the dynamic pressures
generated by any given rate of roll. So the magnitude of aerodynamic damping
decreases with increasing altitude. This phenonienon affects all three axes of
motion but because the tailplane and fin are smaller than the wings, the
longitudinal and directional stability are not affected to the same extent. The
overall effect is that all modes of stability decrease with increasing altitudc but
they are not all affected a t the same rate.
STAB 40. d.
When flying a t cruising speed a t the maximum operating altitude, a swept wing
aircraft will be susceptible to Dutch Roll. This can be prevented by reducing
speed or altitude. But a t the maximum operating altitude, the ability of an
aircraft to manoeuvre or vary its speed is limited by the close proximity of the
low-speed and high-speed buffet boundaries. In the event of a yaw damper
failure the aircraft should therefore decrease altitude to broaden the margin
between the buffet boundaries, before reducing speed.
STAB 41. b.
The term spiral instability refers to the phenomenon whereby, following a
disturbance in yaw or roll, an aircraft yaws into the sideslip rather than rolling
away from it. This yaw decreases the airspeed over one wing and increases that
over the other, producing a dissymmetry of lift that rolls the aircraft towards the
sideslip. This causes further yawing and rolling to that side, such that the
aircraft enters a wide spiral dive. Spiral instability occurs when directional
stability is greater than lateral stability, making the aircraft yaw into any sideslip
before it can roll away from it.
STAB 42. b.
Phugoid motion is a phenomenon in which an aircraft undertakes low frequency
cyclic pitching actions, causing a porpoising motion in which airspeed is
exchanged for altitude in the ascending phase, then exchanged back again during
the descending phase, such that total energy remains constant. Each cycle of this
alternating climbing and diving sequence is initiated by alternating pitching
upwards and downwards such that angle of attack remains approximately
constant. Because of this constant angle of attack, phugoid motion is very
weakly damped and so continues for some time.
STAB 43.b.
Phugoid motion is a phenomenon in which a n aircraft undertakes low frequency
cyclic pitching action, causing a porpoising motion in which airspeed is
exchanged for altitude in the ascending phase, then exchanged back again during
the descending phase, such that total energy remains constant. Each cycle of this
alternating climbing and diving sequence is initiated by alternating pitching
upwards and downwards such that angle of attack remains approximately
constant. Because of this constant angle of attack, phugoid motion is very
weakly damped. If a n aircraft has a low coefficient of drag, the alternating
increases and decreases in speed during phugoid motion will cause only small
changes in drag, which will have little or no effect in damping the motion. A low
CDtherefore increases the tendency to phugoid motion.
STAB 44. c.
Tlie term short period motion refers to high frequency pitching oscillations
about the lateral axis. Such motion is usually highly damped and hence selfcorrecting. Because of its high frequency however, any attempt to counter it by
manual control inputs is likely to exacerbate the problem. The most effective
solution is to release the controls and allow the aircraft to stabilise itself.
STAB 45. c.
The term static stability refers to the immediate response of an aircraft after
being subjected to a disturbance in flight. If it tends to return to its predisturbance condition, it is positively statically stable. If it does not tend to
return, but continues to deviate, it is negatively statically stable. If it remains in
its post disturbance condition, neither returning to, nor diverging from its
original condition, it is neutrally statically stable. Only a positively statically
stable aircraft tends to return to its original condition. The term dynamic
stability refers to whether, following a disturbance, an aircraft's oscillations
increase, decrease, o r remain constant. If they increase with time, it is
dynamically unstable. If they decrease, it is dynamically stable. If they remain
constant, it is neutrally dynamically stable. An aircraft in which the amplitude
of the oscillations remains constant, is statically stable because it oscillates, but
dynamically neutrally stable, because the amplitude of the oscillations remains
constant.
STAB 46. a.
A ventral fin is a downward pointing aerofoil located beneath the rear fuselage,
usually below the fin. In a sideslip both the ventral fin and the fin are subjected
to a lateral component of airflow caused by the sideslip. This airflow strikes
both the ventral fin and fin at an angle, producing a laterally directed lift force,
acting away from the sideslip. Because the C of P of the ventral fin is below the
C of G of the aircraft, this side force tends to roll the aircraft towards the
sideslip. This constitutes a decrease in lateral static stability. The effect of a
ventral fin in sideslip is illustrated in the diagram below.
Fin
Stobiiing rolling moment due
to sideslip aidow acting on fur
+--------------Aidow due to si&lip
w b i i l i s i n g rolling moment due to
sidwlip aimow a h g w ventral fin
id^^ due to sideslip
Whenever an aircraft is oscillating in roll, the fin and ventral fin move laterally
in opposite directions. Each is subjected to a lateral airflow due to the role. This
lateral airflow strikes the ventral fin and fin at an angle, thereby generating
lateral lift forces, acting away from the direction of the rolling motion. These
.,de forces oppose the roll and hence increase lateral dynamic stability. A
ventral fin therefore increases lateral dynamic stability. This effect is illustrated
in the diagram below.
Srobtlisins= rollins moment due
to iimov actiap om C.5
Stebiltrinp rollilrg moment due
to.idlow retinp on fin
STAB 47. a.
A ventral fin is a downward pointing fin located beneath the rear fuselage,
usually below the fin. In a sideslip both the ventral fin and the fin are subjected
to a lateral component of airflow caused by the sideslip. This airflow strikes the
ventral fin and fin at an angle, producing a laterally directed lift force, acting
away from the sideslip. Because both the ventral fin and the fin are behind the C
of G of the aircraft, these side forces yaw the aircraft into the sideslip. This
increases the directional stability of the aircraft in sideslip. This effect is
illustrated in the diagram below.
y e w i n e moments due to 3 a m l i ~~ r f l o won both
p d vemtml fin vuw the uirrsfi illto the s i d a b
Airllow due to s i d a l i p
Effect of ventral f i i on directionul stability
STAB 48. b.
Dutch Roll is likely to occur when the lateral stability of an aircraft is stronger
than its directional stability. When disturbed in yaw such an aircraft will roll
away from the sideslip before it can yaw into it, thereby causing cyclic rolling
motions, possibly of increasing amplitude. Although a ventral fin increases
directional stability, it decreases lateral static stability, and increases lateral
dynamic stability. This increase in directional stability and decrease in lateral
static stability makes the initiation of Dutch Roll more difficult. The increase in
lateral dynamic stability means that any lateral oscillations will be damped down
more quickly. A ventral fin therefore reduces the tendency to Dutch Roll
STAB 49. d.
For any aircraft to be longitudinally stable it must possess a forward C of G and
longitudinal dihedral. The forward C of G will impose a nose down pitching
moment and the longitudinal dihedral will ensure that pitching moments
generated by disturbances in pitch, are always opposing and proportional to, the
degree of pitch disturbance. Although longitudinal stabilising forces are usually
provided by a tailplane or canard, a delta wing requires neither. An aft C of G
decreases longitudinal stability.
STAB 50. b.
Dihedral is the principal design factor contributing to lateral stability.
Whenever an aircraft banks, it sideslips in the direction of the bank. In the
sideslip, dihedral increases the effective angle of attack of the leading wing and
decreases that of the trailing wing. The difference in lift generated by the two
wings then rolls the aircraft away from the sideslip back to the wings level
condition. Because dihedral tends to return an aircraft to wings level condition it
increases the lateral stick force required to hold an aircraft in a sideslip.
STAB 51. b.
Dutch Roll is likely to occur when the lateral stability of an aircraft is stronger
than its directional stability. When disturbed in yaw such an aircraft will roll
away from the sideslip before it can yaw into it.
When subjected to varying pitching, yawing o r rolling moments, an aircraft in
flight rotates about its C of G. The stability of an aircraft is a measure of its
tendency to resist such rotations. The main components contributing to the
longitudinal, directional, and lateral stability of an aircraft are its tailplane, fin
and wings respectively. The magnitude of the stabilising moments generated by
these components is the product of the lift force generated by them due to
changes in attitude and the distance between their C of P and the C of G of the
aircraft. Aft movement of the C of G decreases the moment arms of the
tailplane and fin, but not that of the wings. This means that aft movement of the
of the C of G decreases longitudinal and directional stability, but leaves lateral
stability unaffected. Dutch roll tendency is therefore increased by aft movement
of the C of G of an aircraft.
STAB 52. c.
Forward movement of the C of G of an aircraft produces an increase in the nose
down pitching moment. In order to maintain straight and level flight, or to
increase nose up pitch attitude, this nose down moment must be balanced or
exceeded by a pull back force on the stick. Forward C of G movement therefore
increases the required stick force.
STAB 53. a.
The term short period mode refers to high frequency pitching oscillations about
the lateral axis. Such motion is usually highly damped and hence selfcorrecting. Because of its high frequency however, any attempt to counter it by
manual control inputs is likely to exacerbate the problem. The most effective
solution is to release the controls and allow the aircraft to stabilise itself.
STAB 54. c.
When disturbed in yaw, an aircraft that is susceptible to Dutch Roll will roll
away from the sideslip before it can yaw into it, thereby causing cyclic rolling
motions, possibly of increasing amplitude. Dutch Roll is likely to occur when the
static lateral stability of an aircraft is stronger tlian its directional stability.
STAB 55. b.
When disturbed in yaw, an aircraft that is susceptible to Dutch Roll will roll
away from the sideslip before it can yaw into it, thereby causing cyclic rolling
motions, possibly of increasing amplitude. Dutch Roll is likely to occur when the
static lateral stability of an aircraft is stronger than its directional stability. If
however an aircraft also possesses strong lateral dynamic stability the, lateral
oscillations will be quickly damped down, reducing the tendency to Dutch Roll.
STAB 56. b.
When subjected to varying pitching, yawing or rolling moments, an aircraft in
flight rotates about its C of G. The stability of an aircraft is a measure of its
tendency to resist such rotations. The manoeuvrability of an aircraft is a
measure of its ability to perform such rotations, in response to inputs to its flying
controls. Stability and manoeuvrability are therefore opposing characteristics in
that increasing stability decreases manoeuvrability.
The main components contributing to the longitudinal, directional and lateral
stability of an aircraft are its tailplane, fin, and wing dihedral, respectively. The
magnitude of the stabilising moments generated by these components is
determined by the aerodynamic forces generated by the components, niultiplied
by the distance between the C of P of the components and the C of G of the
aircraft. In the case of longitudinal stability for example, the stabilising
moments produced by the tailplane are the product of the tailplane lift force and
the distance from the C of P of the tailplane to the C of G of the aircraft. The C
of G of the aircraft is always forward of the C of P of the tailplane, so moving the
C of G aft reduces stability and hence increases manoeuvrability.
STAB 57. b.
If an aircraft is disturbed in flight such that it banks, it will sideslip towards the
low wing. This sideslip will cause the airflow to approach the aircraft from the
side to which it is banked. In the case of a swept wing aircraft, this sideslip
airflow will increase the effective aspect ratio of the leading wing, whilst
decreasing that of the trailing wing. This will increase the CLof the leading wing
and decrease that of the trailing wing, thereby rolling the aircraft back to the
wings level condition. This will increase the lateral stability of the aircraft.
Sweepback therefore produces strong lateral stability
STAB 58. b.
If an aircraft is disturbed in flight such that it banks, it wiil sideslip towards the
low wing. This sideslip will cause the airflow to approach the aircraft from the
side to which it is banked. In the case of a swept wing aircraft this sideslip
airflow will increase the effective frontal area of the leading wing, whilst
decreasing that of the trailing wing. This will increase the drag force on the
leading wing and decrease that on the trailing wing, causing the aircraft to yaw
into the sideslip. This will increase the directional stability of the aircraft.
Sweepback therefore produces strong directional stability
STAB 59. b.
The term anhedral refers to the configuration where the wings a r e angled
downwards from root to tip. The purpose of anhedral is to decrease the lateral
static stability of an aircraft, in order to increase its lateral manoeuvrability.
Dihedral has the opposite effect, increasing lateral stability and hence decreasing
manoeuvrability. Low C of G and forward C of G both increase stability and
decrease manoeuvrability.
STAB 60. b.
Dutch Roll is likely to occur when the lateral stability of an aircraft is stronger
than its directional stability. When disturbed in yaw such an aircraft will roll
away from the sideslip before it can yaw into it, thereby causing cyclic rolling
motions, of increasing amplitude. Dutch Roll is therefore unstable cyclic rolling
from one wing down to the other.
STAB 61. b.
The positive direction of motion about any of the three axes of an aircraft can be
identified by pointing a finger along that axis, in the positive direction, then
rotating one's hand in a clockwise direction. Yaw is rotation about the normal
axis, for which the positive direct is downward from above the aircraft. Positive
yaw is therefore in the nose right direction.
STAB 62. c.
The positive direction of motion about any of the three axes of an aircraft can be
identified by pointing a finger along that axis in the positive direction, then
rotating one's hand in a clockwise direction. Rotation about the lateral axis is
called pitch and the positive pitch direction is nose up. The positive direction of
the lateral axis is therefore from the left side of an aircraft to the right.
STAB 63. a.
Positive sideslip occurs when the relative airflow approaches the nose of an
aircraft from the right. This occurs when the aircraft is disturbed such that it
yaws to the left o r rolls to the right.
STAB 64. b.
The term longitudinal stability refers to the ability of an aircraft to resist
changes in its pitch attitude and to regain its original attitude following a
disturbance in flight. I n order to be longitudinally stable an aircraft, when
disturbed in the nose up sense, must generate a nose down pitching moment.
Similarly, when disturbed in a nose down sense, it must generate an opposing
nose up pitching moment.
In order to ensure that the amplitude of the subsequent pitching oscillations is
reduced and quickly damped out, it is necessary that the magnitude of the
opposing pitching moments is proportional to the degree of disturbance. The
magnitude cf these pitching moments can be expressed as the pitching moment
coefficient, C,,. This is positive if the inoinent acts in the nose up direction and
negative if the moment acts in the nose down direction.
But the positive pitching direction is nose up, resulting in increasing angle of
attack ( a ) and the negative pitching direction is nose down, resulting in
decreasing a. This means that increasing a must generate increasing negative
moments and decreasing a must generate increasing positive moments. The
slope of the C, : a curve of a longitudinally stable aircraft is therefore negative.
STAB 65. b.
The term lateral stability refers to the ability of a n aircraft to resist changes in its
lateral (bank) attitude and to regain its original attitude following a disturbance
in flight. In order to be laterally stable a n aircraft, when disturbed in the right
wing down sense, must generate a right wing up rolling moment. Similarly when
disturbed in a right wing up sense, it must generate a n opposing right wing down
rolling moment.
In order to ensure that the amplitude of the subsequent rolling oscillations is
minimised and quickly damped out, it is necessary that the magnitude of the
opposing rolling moments is proportional to the degree of disturbance. The
magnitude of these rolling moments can be expressed as the rolling moment
coefficient, CI, which is positive if the moment acts in the right wing down
direction, and negative if the moment acts in the right wing up direction.
The positive rolling direction is right wing down and a positive roll causes
sideslip to the right. This causes the airflow to approach the aircraft a t an angle
from the right. This constitutes a positive sideslip angle P. So positive role
causes positive sideslip, which must then generate a negative rolling moment CI,,
rolling the a n aircraft away from the sideslip, if it is to be positively laterally
stable. But if a positive sideslip angle generates a negative CI, then the CI : P
slope must be negative.
STAB 66. a.
The term directional stability refers to the ability of an aircraft to maintain a
zero sideslip angle, and to reduce sideslip angle to zero following disturbances in
yaw. I t should be noted that this means that a directionally stable aircraft
always yaw into a crosswind o r sideslip. Such an aircraft will not therefore
maintain a constant heading in gusty crosswind conditions. I n order to be
directionally stable an aircraft, when affected by airflow from the right must
generate a right yawing moment. Similarly when affected by airflow from the
left it must generate a left yawing moment.
In order to ensure that the amplitude of the subsequent yawing oscillations is
minimised and quickly damped out, it is necessary that the magnitude of the
opposing yawing moments is proportional to the degree of disturbance:' That is
to say, the yawing moments must be proportional to the sideslip angle. The
magnitude of these yawing moments can be expressed as the yawing moment
coefficient, C,,, which is positive if the moment acts in the nose right direction,
and negative if the moment acts in the nose left direction.
The negative yawing direction is nose to the left and a negative yaw causes a
positive sideslip to the right. This causes the airflow to approach the aircraft at
an angle from the right. This constitutes a positive sideslip angle P. So negative
yaw causes positive sideslip, which must then generate a positive yawing moment
C,,, yawing the aircraft into the sideslip, if it is to be positively directionally
stable. But if a positive sideslip angle must generate a positive C,,, then the C, :
p slope must be positive.
STAB 67. b.
When an aircraft is disturbed in pitch, roll, or yaw its tailplane, wings, or fin
must generate opposing pitching, rolling, or yawing moments, if the aircraft is to
be stable. If the control column and rudder pedals are held fured, this will cause
the elevators, ailerons and rudder to remain fixed, and so they will contribute to
the stabilising moments. But if the controls are free to move, then the control
surfaces will be deflected, aligning themselves with relative airflow in the
pitching, rolling o r yawing motions. This will reduce the magnitude of the
stabilising moments, and hence decrease the static stability of the aircraft. The
stick free stability of an aircraft is therefore weaker than its stick fixed stability.
STAB 68. b.
Aft movement of the C of G of an aircraft produces a decrease in the nose down
pitching moment. In order to maintain straight and level flight or increase pitch
attitude, this nose down moment must be balanced or exceeded by a pull back
force on the stick. Aft C of G movement will therefore decrease the required
stick force, provided the C of G remains ahead of the neutral point of the
aircraft. But in order to ensure that an aircraft is always positively stable, the
entire allowable C of G range is forward of the neutral point. An in limits aft C
of G will therefore decrease the stick force gradient.
STAB 69. a.
The directional stability of an aircraft is provided primarily by its fin. When an
aircraft is disturbed in yaw, the subsequent sideslip causes the fin to produce a
sideways acting lift force, which generates a yawing moment, causing the aircraft
to yaw into the sideslip. At very large sideslip angles however, the fin will stall,
causing a reduction in directional stability. Although increasing the surface area
of the fin will reduce the severity of this problem, care must be taken to ensure
that the fin does not become too long. A very long fin will have a high aspect
ratio and hence a low stalling angle of attack, thus exacerbating the fin stall
problem at high sideslip angles.
A dorsal fin is a swept back leading edge extension, fitted to the upper surface of
an aircraft where the root of the fin joins the fuselage. Its purpose is to increase
the surface area of the fin and to reduce its aspect ratio. By increasing the
surface area, it increases the stabilising moments generated at any given sideslip
angle. By decreasing the aspect ratio, it increases the sideslip angles that can be
achieved without stalling the fin. A dorsal fin therefore increases the
effectiveness of the fin by decreasing its aspect ratio.
STAB 70. d.
A dorsal fin is swept back leading edge extension, fitted to the upper surface of
an aircraft where the root of the fin joins the fuselage. Its purpose is to increase
the effectiveness of the fin, by increasing its surface area and reducing its aspect
ratio. By increasing the surface area it increases the stabilising moments
generated a t any given sideslip angle. Low aspect ratio aerofoils have shallower
CL : angle of attack curves and higher stalling angles than high aspect ratio
aerofoils. In the case of a fin, the angle of attack is the sideslip angle, P, so a fin
with a dorsal fin will have a shallower CL : P slope, and a higher stalling angle
than one without a dorsal fin.
STAB 71. a.
Although the main factors determining the lateral stability of an aircraft are the
sweepback, vertical location and dihedral of its wings, the fin also makes a
significant contribution. When an aircraft is disturbed in roll, the subsequent
sideslip causes the fin to produce a sideways acting lift force. Because the C of P
of the fin is some distance above the C of G of the aircraft, the fin side force
causes the aircraft to roll away from the sideslip. The magnitude of the rolling
moment generated by the fin depends upon the sideslip angle and the size and
height of the fin. A high fin therefore increases the lateral static stability of an
aircraft.
Whenever an aircraft is rolling, the fin is subjected to an additional sideways
airflow, which increases the stabilising moments generated. This additional
rolling moment opposes the rolling action and hence increases the rate a t which
oscillations in roll are damped out. A high fin therefore increases the lateral
dynamic stability of an aircraft.
STAB 72. a.
The main factor in determining the directiona1 stability of an aircraft is its fin.
When an aircraft is disturbed in yaw, the subsequent sideslip causes the fin to
produce a sideways acting lift force, which generates a yawing moment, causing
the aircraft to yaw into the sideslip. The magnitude of these stabilising moments
a t any given sideslip angle is determined by the distance between the C of G of
the aircraft and the C of P of the fin, and by the size of the fin. Increasing fin
height increases its aspect ratio. This increases the magnitude of the stabilising
forces generated at any given sideslip angle, but decreases the stalling angle of
the fin. Increasing fin height therefore increases directional static stability a t
sideslip angles lower than the fin stall angle.
The directional dynamic stability of an aircraft is a measure of the rate at which
yawing oscillations are damped out. Because increasing fin height increases the
stabilising force a t any given sideslip angle, it also increases the damping rate.
Increasing fin height therefore increases both the static and dynamic directional
stability of a n aircraft a t sideslip angles lower than the fin stall angle.
STAB 73. d.
A ventral fin is a downward pointing aerofoil located beneath the rear fuselage,
usually below the fin. In a sideslip botn the ventral fin and the fin are subjected
to a lateral component of airflow caused by the sideslip. This airflow strikes the
ventral fin and fin a t an angle, producing a lateral directed lift force, acting away
from the sideslip. Because the C of P of the ventral fin is below the C of G of the
aircraft, this side force tends to roll the aircraft towards the sideslip. This
constitutes a decrease in lateral static stability. The effect on lateral stability, of
a ventral fin in sideslip is illustrated in the diagram overleaf.
Fin
Strbilising rolling moment due
to sideslip airflow acting on fin
li
Destabiiing rolling moment due to
sideslip airflow act&g on "en-I f i
Aidow due to sirkslip
1 +LICLLL~ i f i due
o ~to sideslip
Effect ofvmtral tin on static lateral stability
'
Also because the ventral tin is behind the C of G of the aircraft, the side force
generated by it tends to yaw the aircraft into the sideslip. This increases the
directional stability of the aircraft in sideslip. This effect is illustrated in the
diagram below. A ventral fin therefore decreases lateral static stability and
increases directional stability.
Yawine moments due to sfl*rli~aidow on both Bq
and ventral fm vaw the airnft into the sidwEn
Aidow due to sideslip
Effect of ventral fm on dimtional stability
STAB 74. a.
The term spiral instability refers to a phenomenon whereby an aircraft
disturbed in yaw continues to yaw and roll towards the sideslip, rather than
rolling away from it. This causes the aircraft to enter a wide spiral descent of
increasing gradient and decreasing radius. Spiral instability occurs only when
the directional stability of an aircraft is greater than its lateral stability. Any
action that increases directional stability or decreases lateral stability, will
therefore increase the .spiral instability tendency of an aircraft.
A dorsal fin is swept back leading edge extension fitted to the upper surface of an
aircraft, where the root of tFe fin joins the fuselage. Its purpose is to increase the
effectiveness of the fin, by increasing its surface area and reducing its aspect
ratio. By increasing the surface area it increases the stabilising moments
generated at any given sideslip angle. Because the dorsal fin is well aft of the C
of G of an aircraft, a small increase in fin area produws a significant increase in
directional stability.
Whenever an aircra~.is rolling the fin experiences an additional sideways
airflow due to roll rate. This produces an additional sideways lift force that
opposes and damps out the roll. But because the dorsal fin is fitted at the root of
the fin, it has little effect on the roll damping produced by the fin, and hence does
not increase lateral stability to any significant extent. A dorsal fin therefore
increases directional stability without increasing lateral stability. This increases
the tendency to spiral instability.
STAB 75.b.
Dutch Roll is likely to occur when the lateral stability of an aircraft is stronger
than its directional stability. M7hendisturbed in yaw such an aircraft will roll
< -'
away from the sideslip before it can yaw into it, thereby causing cyclic rolling
motions, of increasing amplitude. Any component that decreases lateral stability
whilst increasing directional stability will decrease the tendency to Dutch Roll.
A ventral fin is a downward pointing aerofoil located beneath the rear fuselage,
usually below the fin. In a sideslip both the ventral fin and the fin a r e subjected
to a lateral component of airflow caused by the sideslip. This airflow strikes the
ventral fin and fin a t a n angle and so produces a laterally directed lift force,
acting away from the sideslip. Because the C of P of the ventral fin is below the
C of G of the aircraft, this side force tends to roll the aircraft towards the
sideslip. This constitutes a decrease in lateral static stability.
Also because the ventral fin is behind the C of G of the aircraft, the side force
generated by a ventral fin tends to yaw the aircraft into the sideslip. This
increases the directional stability of the aircraft in sideslip. A ventral fin
therefore decreases lateral stability whilst increasing directional stabBity and
hence decreases the tendency to Dutch Roll.
STAB 76. c.
The term longitudinal stability refers to the ability of a n aircraft to resist
changes in its pitch attitude and to regain its original attitude following a
disturbance in flight. In order to be longitudinally stable a n aircraft, when
disturbed in the nose u p sense, must generate a nose down pitching moment.
Similarly, when disturbed in a nose down sense, it must generate a n opposing
nose u p pitching moment.
I n order to ensure that the amplitude of the subsequent pitching oscillations is
minimised and quickly damped out, it is necessary that the magnitude of the
opposing pitching moments is proportional to the degree of disturbance. The
magnitude of these pitching moments can be expressed as the pitching moment
coefficient, C,,, which is positive if the moment acts in the nose up direction, and
negative if the moment acts in the nose down direction.
But the positive pitching direction is nose up, resulting in increasing angle of
attack (a), and the negative pitching direction is nose down resulting in
decreasing a. This means that increasing a must generate increasing negative
moments, and decreasing a must generate increasing positive moments. The
slope of the C, : a curve of a longitudinally stable aircraft must therefore be
negative. Longitudinal stability therefore requires a negative C, : a curve.
STAB 77. a.
Tlie term lateral stability refers to the ability of a n aircraft to resist changes in its
lateral (bank) attitude and to regain its original attitude following a disturbance
in flight. In order to be laterally stable a n aircraft, when disturbed in the right
wing down sense, must generate a right wing up rolling moment. Similarly when
disturbed in a right wing u p sense, it must generate an opposing right wing down
rolling moment.
In order to ensure that the amplitude of the subsequent rolling oscillations is
minimised and quickly damped out, it is necessary that the magnitude of the
opposing rolling moments is proportional to the degree of disturbance. The
magnitude of these rolling moments can be expressed as the rolling moment
coefficient, C,, which is positive if the moment acts in the right wing down
direction, and negative if the moment acts in the right wing up direction.
The positive rolling direction is right wing down and a positive roll causes
sidcslip to the right. This causes the airflow to approach the aircraft from the
right, which constitutes positive sideslip. So positive roll causes positive
sideslip, which must then generate a negative rolling moment CI, rolling the
aircraft away from the sideslip, if it is to be positively laterally stable. But if a
positive sideslip angle must generate a negative CI, then the C I : P slope must be
negative. Lateral stability therefore requires a negative CI : P slope.
STAB 78. b.
The term directional stability refers to the ability of an aircraft to maintain a
zero sideslip angle, and to reduce sideslip angle to zero following disturbances in
yaw. It should be noted that this means that a directionally stable aircraft
always yaw into a crosswind or sideslip. Such an aircraft will not therefore
maintain a constant heading in gusty crosswind conditions. In order to be
directionally stable a n aircraft, when affected by airflow from the right, must
generate a right yawing moment. Similarly when affected by airflow from the
left, it must generate an opposing left yawing moment.
In order to ensure that the amplitude of the subsequent yawing oscilIations is
minimised and quickly damped out, it is necessary that the magnitude of the
opposing yawing moments is proportional to the degree of disturbance. That is
to say, the yawing moments must be proportional to the sideslip angle. The
magnitude of these yawing moments can be expressed as the yawing moment
coefficient, C,, which is positive if the moment acts in the nose right direction
and negative if the moment acts in the nose left direction.
Tlie negative yawing direction is nose to the left, and a negative yaw causes a
positive sideslip to the right. This causes the airflow to approach the aircraft at
an angle from the right. This constitutes a positive sideslip angle P. So negative
yaw causes positive sideslip, which must then generate a positive yawing moment
C,,, yawing the aircraft into the sideslip, if it is to be positively directionally
stable. But if a positive sideslip angle must generate a positive C,, then the C, : P
slope must be positive. Directional stability therefore requires a positive C , : P
slope.
STAR 79. c.
The term directional stability refers to the ability of an aircraft to maintain a
zero sideslip angle, and to reduce sideslip angle to zero following disturbances in
yaw. It should be noted that this means that a directionally stable aircr~.ft
always yaw into a crosswind or sideslip. Such an aircraft will not therefore
maintain a constant heading in gusty crosswind conditions. In order to be
directionally stable an aircraft, when affected by airflow from the right, must
generate a riglit yawing moment. Similarly when affected by airflow from the
left, it must generate a n opposing left yawing moment. Directional stability
therefore requires that an aircraft generates a yawing moment that is directly
proportional to the sideslip angle.
STAB 80. d.
When subjected to varying pitching moments, an aircraft in flight rotates about
its C of G. The stability of an aircraft is a measure of te~idencyto resist such
rotations. The main components contributing-to the longitudinal, stability of an
aircraft are its tailplane, fore and aft location of its C of G, and longitudinal
dihedral. Of the options offered in this question, only option d, longitudinal
dihedral, is essential for longitudinal stability.
STAB 81. a.
When subjected to varying pitching moments, an aircraft in flight rotates about
its C of G. The stability of an aircraft is a measure of tendency to resist such
rotations. The main components contributing to the longitudinal, stability of a n
aircraft are its tailplane, fore and aft location of its C of G, and longitudinal
dihedral. Of the options offered in this question, only option a, a forward C of G,
is essential for longitudinal stability.
STAR 82. d.
When subjected to varying pitching, yawing o r rolling moments, an aircraft in
flight rotates about its C of G. The stability of an aircraft is a measure of its
tendency to resist such rotations. The manoeuvrability of a n aircraft is a
measure of its ability to perform such rotations, in response to inputs to its flying
controls. Stability and manoeuvrability are therefore opposing characteristics in
that increasing stability decreases manoeuvrability. A highly stable aircraft
therefore requires powerful control inputs in order to overcome the stabilising
moments when attempting to manoeuvre, The generation of these powerful
control inputs requires the application of high stick forces.
STAR 83. b.
The term dynamic stability refers to whether, following a disturbance in flight,
the oscillating motions of a n aircraft increase, decrease o r remain constant. If
they decrease rapidly it possesses strong dynamic stability. The rate at which
oscillations diminish is dependent upon the degree of aerodynamic damping that
is generated by the motion. If aerodynamic damping effects a r e very high then
the aircraft will be strongly dynamically stable.
STAB 84. c.
The purpose of winglets is to reduce the effects of lift induced drag by interacting
with the wingtip vortices to produce a total aerodynamic reaction acting forward
of the normal axis. This generates a small amount of thrust, thereby negating
some of the induced drag force. Winglets have no significant affect on the
stability of an aircraft.
STAB 85. b.
The term deep stall refers to the phenomenon whereby a stalling aircraft pitches
nose up. This pitch up, coupled with the downward motion of the aircraft due to
the loss of lift in the stall, increases the angle of attack, taking it deeper into the
stall. The principal cause of the initial pitch up is tip stall of swept o r delta
wings. This occurs because the swept leading edges cause the airflow to be
deflected outwards, towards the tips where it forms a thick, low energy boundary
layer. As angle of attack increases this low energy boundary layer separates
easily, causing the wing tips to stall before the wing roots. Because the tips of
swept and delta wings a r e further aft than the wing roots, tip stall moves the C of
P forward causing the aircraft to pitch u p in the stall.
STAB 86. a.
In a twin-engine counter rotating propeller aircraft in powered flight, the
slipstream of the propellers increases the speed of the airflow passing over the
wings. This increases the lift produced. If however one engine fails, then the
wing to which it is attached will ceases to benefit from the slipstream, so its lift
will decrease, whilst the other wing remains unaffected. The difference in lift
generated by the two wings will then cause the aircraft to roll towards the dead
engine. This phenomenon does not usually affect jet-powered aircraft because
the engine exhaust does not flow over the wings. Also, because contra-rotating
propellers turn on a common axis, single engine failure will affect both wings to
the same extent. Single engine failure is therefore more likely to cause rolling in
a twin counter rotating propeller aircraft than in a twin-jet.
STAB 87. b.
The term short period oscillation refers to high frequency, low period cyclic
pitching motion about the lateral axis.
STAB 88. d.
The purpose of the yaw damper is to increase the directional stability of a n
aircraft in order to prevent Dutch Roll. In order to do this it employs a
horizontal rate gyro to sense yaw rates and produces yaw control inputs
proportional -to those rates.
STAB 89. b.
Dutch Roll is a n unstable cyclic banking and yawing oscillation that affects an
aircraft when its lateral stability is stronger than its directional stability. The
purpose of the yaw damper is to enhance the directional stability of a n aircraft in
order to prevent Dutch Roll. I11 order to do this it employs a horizontal rate gyro
to sense yaw rates and produces yaw control inputs proportional to those rates.
STAB 90. b.
If an aircraft is disturbed in flight such that it banks, it will sideslip towards the
low wing. This sideslip will cause the airflow to approach the aircraft from the
side to which it is banked. In the case of a swept wing aircraft this sideslip
airflow will increase the effective frontal area of the leading wing, whilst
.
decreasing that of the trailing wing. This will increase the drag force on the
leading wing and decrease that on the trailing wing causing the aircraft to yaw
into the sideslip. This will increase the directional stability of the aircraft.
Directional stability is therefore increased by wing sweep back.
STAB 91. d.
The term lateral stability refers to the ability of an aircraft to resist changes in its
lateral (bank) attitude and to regain its original attitude following a disturbance
in flight. I n order to be laterally stable arn aircraft, when disturbed in the right
wing down sense, must generate a right wing u p rolling moment. Similarly when
disturbed in a right wing up sense, it must generate an opposing right wing down
rolling moment.
These stabilising moments a r e generated in part by changes in the angles of
attack of the wings caused by the rolling morions. But the magnitude of these
angle of attack changes depends upon the ratio of the roll rate to TAS.
lncreasing TAS therefore decreases the magnitude of the angle of attack changes
produced by any given roll rate, and hence decreases lateral stability. But
increasing altitude increases the TAS at any given L4S. Lateral stability is
therefore decreased by increasing altitude.
STAB 92. c.
The term lateral stability refers to the ability of an aircraft to resist changes in its
lateral (bank) attitude, and to regain its original attitude following a disturbance
in flight. In order to be laterally stable an aircraft, when disturbed in the right
wing down sensk, must generate a right wing up rolling moment. Similarly when
disturbed in a right wing up sense, it must generate an opposing right wing down
rolling moment.
These stabilising moments are generated in part by the process of aerodynamic
damping, whereby rolling motions cause changes in the angles of attack of the
wings. But the magnitude of these damping moments is equal to the difference in
lift produced by each wing, multiplied by the lateral distance from the C of G of
the aircraft to the C of P of the wings. Increasing aspect ratio increases
wingspan moving the C of P out\+-ards,and hence increases the aerodynamic
damping moments. Increasing aspect ratio also increases the gradient of the CL:
a curve. This increases the difference betvieeal the lift force acting on each wing,
due to the alterations in angle of attack caused by the rolling motion. The overall
effect of these factors is that increasing aspect ratio increases roll damping.
STAB 93. c.
Control forces and moments generated by the deflection of a n aircraft's flying
-controls, increase with both angle of deflection and IAS. There is therefore a
danger that a n aircraft will be overstressed if Large control deflections are
employed when flying at high speeds. To prevent such overstressing, systems are
arranged such that the stick forces that must be applied in order to move the
flying controls, increases in proportion to control defiection angle and
IAS. This applies to all types of powered a e d power assisted systems, including
fly-by-wire systems.
STAB 94. d.
The term lateral stability refers to the ability of an aircraft to resist changes in its
lateral (bank) attitude, and to regain its original attitude following a disturbance
in flight. In order to be laterally stable an aircraft, when disturbed in the right
wing down sense, must generate a right wing up rolling moment. Similarly when
disturbed in a right wing up sense, it must generate an opposing right wing down
rolling moment. Although the principal components contributing to lateral
stability are wing dihedral and wing sweep back, other factors such as a high fin,
low C of G, and high wings, all increase lateral stability.
STAB 95. c.
For an aircraft to be longitudinally stable its C of G must be forward of its
neutral point. The further forward the C of G, the more stable the aircraft will
be. A forward C of G also requires a strong downward tailplane trim force in
order to maintain level flight. The generation of this trim force creates
additional trim drag, decreasing the aerodynamic efficiency of the aircraft.
In order to achieve maximum aerodynamic efficiency, aircraft are designed such
that when flying at their cruising speed, with C of G within limits, the need for
trimming and hence trim drag a r e minimised. If however the C of G is aft of the
aft limit, the aircraft will require an upward trim force from the tailplane in
order to maintain level flight at cruising speed. A C of G aft of the aft limit will
therefore reduce stability and increase trim drag in cruise flight.
STAB 96. b.
For an aircraft to be longitudinally stable its C of G must be forward of its
neutral point. The further forward the C of G, the more stable the aircraft will
be. But a forward C of G requires a strong downward tailplane trim force in
order to maintain level flight. The generation of this trim force creates
additional trim drag, decreasing the aerodynamic efficiency of the aircraft.
In order to achieve maximum aerodynamic efficiency, aircraft a r e designed such
that when flying at their cruising speed, with C of G within limits, the need for
trimming and hence trim drag are minimised. A forward C of G will therefore
increase stability and increase trim drag in cruise flight.
STAB 97. a.
The term lateral dynamic stability refers to the degree to which oscillations in
roll decrease following a disturbance in flight. In order to be laterally
dynamically stable a n aircraft, when disturbed in the right wing down sense,
must generate a right wing up rolling moment. Similarly when disturbed in a
right wing up sense, it must generate an opposing right wing down rolling
moment.
When an aircraft is rojling, these stabilising moments are generated by the
process of roll damping, whereby rolling motions cause the angles of attack of
the wings to be changed. This increases the angle of attack of the down-going
wing, whilst decreasing that of the up-going wing. This increases the lift
generated by the down-going wing and decreases that of the up-going wing. The
difference in lift forces affecting the two wings then resists or damps out the
rolling motion. The magnitude of these angle of attack changes depends upon
the ratio of the roll rate to TAS. Increasing roll rate at any given TAS therefore
increases the effect of roll damping.
STAB 98. b.
The term static stability refers to the immediate response of a n aircraft in flight
after being subjected to a disturbance. If it tends to return to its pre-disturbance
condition it is positively statically stable. If it does not tend to return, but
continues to deviate, it is negatively statically stable. If it remains in its post
disturbance condition, neither returning to, nor diverging from its original
condition, it is neutrally statically stable. Only a positively statically stable
aircraft tends to return to its original condition.
The term dynamic stability refers to whether, following a disturbance, an
aircraft's oscillations increase, decrease o r remain constant. If they increase
with time it is dynamically unstable. If they decrease it is dynamically stable. If
they remain constant it is neutrally dynamically stable. But the entire concept of
dynamic stability is based upon the premise that osciliations occur, and this is
possible only if an aircraft is statically stable. An aircraft that is statically
unstable can therefore never be dynamically stable. It should however be noted
that such an aircraft cannot truly 5e said to be dynamically unstable because the
concept of dynamic stability is meaningless is no oscillations occur.
STAB 99. b.
Although not of aerofoil section, the fuselage of an aircraft in flight functions as a
very inefficient aerofoil, in that it generates a small amount of lift. But the C of P
of an aircraft's fuselage is usually well ahead of that of its wings, and is often
ahead of its C of G. The lift force generated by the fuselage therefore moves the
mean C of,P of the entire aircraft forward, reducing its stability.
STAB 100. a.
In order for an aircraft to be longitudinally stable it must generate pitching
moments opposing, and proportional to, changes in angle of attack. The
generation of such pitching moments requires longitudinal dihedral. This term
refers to the configuration in which the angle of incidence of the wing is greater
than that of the tailplane. This arrangement ensures that any given increase o r
decrease in pitch attitude causes a greater percentage increase or decrease in
angle of attack on the tailplane than on the wing. In the canard-wing
configuration, longitudinal dihedral is equaily essential, but in this case means
that the angle of incidence of the canard is greater than that of the wings.
If for example an aircraft has 3 degrees of longitudinal dihedral, then the angle
of incidence of the wing is 3 degrees greater than that of the tailplane. When
flying straight and level, if the angle of attack of the wing is 6 degrees then that of
the tailplane will be 3 degrees. If the aircraft then experiences a pitch up
disturbance of 3 degrees, the angle of attack of the wing increases by 50% while
that of the tailplane increases by 100%. The lift generated by the tailplane will
therefore increase by a greater proportion than that of the wings so the aircraft
will pitch nose down.
HSF 1. a.
Any object moving through the air creates pressure disturbance that expand
outwards at the local speed of sound. MCMTis the lowest TAS at which airflow
at any point on an aircraft reaches the local speed of sound. When an aircraft is
flying a t speeds above MCRIT,the pressure disturbances caused by its motion pile
up to form shock waves at the point where the airflow velocity is equal to the
local speed of sound. These shock waves represent the point furthest forward
relative to the aircraft that the pressure disturbances can be felt. At supersonic
speeds the shock waves take the form of a mach cone with its point formed a t the
nose (or most forward part) of the aircraft. Pressure differences caused by the
motion of the aircraft can therefore be felt only within this rnach cone.
I-LSF 2. c.
As airspeed is increased, the angle of attack of an aircraft must be reduced to
maintain constant lift equal to its weight. This decrease in angle of attack causes
the C of P of the aircraft to move slowly backwards. MCMTis the lowest TAS at
which the airflow over any point on an aircraft is equal to the local speed of
sound. This is typically located on the upper surface of the wings, at the point of
greatest thickness. As the aircraft moves forwards at MCmT,the adverse
pressure gradient over the rear surface of its wings, moves forward with the
aircraft and piles up to form a shock wave at the point where airflow velocity is
equal to the local speed of sound. This shockwave represents an instantaneous
pressure increase.
As airflow passes through the shock wave it is decelerated rapidly sucli that its
static pressure matches the pressure rise in the shock wave. This rapid
deceleration reduces the kinetic energy of the airflow, causing the boundary
layer to separate from the wing. This separation of boundary layer reduces the
lift produced by the area of the wing behind the shock wave, causing the C of P
to move forward. As an aircraft accelerates up to MCmT,its C of P therefore
move backwards, before moving forwards at MCRIT.
HSF 3. b.
MCRITis the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. This is typically located on the upper surface
of the wings a t the point of greatest thickness. As the aircraft moves forward at
McnlT, the adverse pressure gradient over the rear surface of its wings, moves
forward with it. This adverse pressure gradient cannot move faster than the
local speed of sound. I t therefore piles up to form a shock wave at the point
where airflow velocity is equal to the local speed of sound. This shockwave
represents an instantaneous pressure increase.
As airflow passes through the shock wave it is decelerated rapidly such that its
static pressure matches the pressure rise in the shock wave. This rapid
deceleration reduces the kinetic energy of the airflow, causing the boundary
layer to separate from the wing. This separation of boundary layer reduces the
lift produced by the area of the wing behind the shock wave, causing the C of P
to move forward, As acceleration continues beyond MCRIT,the airspeed over the
front area of the wing becomes supersonic, causing the shock waves to be pushed
back towsrds the trailing edge.
The curvature of the upper surface of the wing behind the leading edge is
effectively a series of divergent ducts, the upper walls of which are represented
by the undisturbed airflow far above the wing. But when supersonic flow passes
through a divergent duct its velocity increases, while its pressure, temperature
and density decrease. This acceleration over the entire upper surface causes the
highest velocity and lowest pressure to occur at the rear area of the wing. I n this
condition the rear area of the wing generates a large proportion the lift force.
This causes its C of P to move rearwards towards the 50% chord point. The
overall effect of acceleration from &IcRrT
to supersonic speed therefore, is that
the C of P moves forward as the shock waves form, then moves aft towards the
mid chord point as the shock waves on the upper and lower surfaces of the wings
move towards the trailing edge.
HSF 4. c.
MCRITis the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. This point is typically located on the upper
surface of the wings a t the point of greatest thickness. As the aircraft
accelerates towards MCRIT,increasing velocity over the front area of its wings
causes the low pressure envelope to intensify and move forwards. At MCRIT,the
airflow over the thickest part of the wing becomes sonic. As the adverse pressure
gradient over the rear surface of its wings, moves forward with the aircraft, it
piles up to form a shock wave at the point where airflow velocity is equal to the
local speed of sound. This shockwave represents a n instantaneous pressure
increase.
As airflow passes through the shock wave it is decelerated rapidly, such that its
static pressure matches the pressure rise in the shock wave. This instantaneous
increase in pressure represents a collapse of the low-pressure envelope over the
surface of the wing behind the shock wave. The overall effect of these changes is
that as a n aircraft accelerates through the transonic speed range the pressure
distribution retains its smooth shape, the pressure drop intensifying and moving
forward, before becoming irregular as the first shock wave forms.
HSF 5. c.
MCRrTis the lowest TAS a t which the airflow over any point on a n aircraft is
equal to the local speed of sound. This point is typically located on the upper
surface of the wings a t the point of greatest thickness. As the aircraft moves
forwards at MCRIT,the adverse pressure gradient over the rear surface of its
wings, izloves forward with the aircraft and piles up to form a shock wave a t the
point where airflow velocity is equal to the local speed of sound. This shockwave
represents a n instantaneous pressure increase.
As airflow passes through the shock wave, it is decelerated rapidly such that its
static pressure matches the pressure rise in the shock wave. This rapid
I
dkceleration reduces the kinetic energy of the airflow, causing the boundary
layer to separate from the wing. This separation of boundary layer reduces the
lift produced by the area of the wing behind the shock wave, causing the C of P
to move forward. As acceleration continues beyond MCWTthe airspeed over the
front area of the wing becomes supersonic causing the shock waves to be pushed
shock wave moves aft towards the trailing edge.
The curvature of the upper surface of the wing behind the leading edge is
effectively a series of expansion corners. As the supersonic airflow passes
through these expansion corners, it expands and accelerates to higher supersonic
speed. This acceleration over the entire upper surface causes the highest velocity
and lowest pressure to occur at the rear area of the wing. In this condition the
rear area of the wing generates a large proportion the lift fbrce. This causes its C
of P to move rearwards towards the 5Ooh chord point. The overall effect of
acceleration from IVICRIT to supersonic speed therefore, is that the C of P moves
aft towards the mid chord point as the shock waves on the upper and lower
surfaces of its wings move towards the trailing edge.
HSF 6. c.
M C R lis~ the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. This point is typically located on the upper
surface of the wings at the point of greatest thickness. As the aircraft moves
~ adverse pressure gradient over the rear surface of its
forward at M C R lthe
wings, moves forward with the aircraft and piles up to form a shock wave at the
point where airflow velocity is equal to the local speed of sound. This shockwave
represents an instantaneous pressure increase. At a slightly higher TAS the
same effect occurs under the lower surface of the wing.
HSF 7. d.
As an ajrcraft flies through the air the pressure disturbances caused by its
motion expand outwards in all directions at the local speed of sound. When
flying at M C R lin
~ the transonic speed range, the forward moving component of
these pressure waves piles up in the form of shock waves, where the airflow
speed is equal to the local speed of sound. At sonic speed this shock wave is at
right angles to the direction of flight but at supersonic speed it becomes swept
back in the form of a mach cone. The point of this cone is located at the most
forward point of the aircraft. Its angle is proportional to the aircraft speed and
at mach 2 for example, is approximately 60 degrees.
Although the mach cone cannot move forward relative to the aircraft, it expands
out laterally and vertically at the local speed of sound. Because the angle of the
cone is determined by the speed of the aircraft, the cone appears to move over
the ground at the ground speed of the aircraft. In considering this effect it
should be noted that if the mach cone moved over the ground more slowly than
the aircraft, its angle would diminish to zero. Conversely if it moved faster than
the ground speed of the aircraft it wmld overtake the aircraft. Neither of these
effects are possible.
HSF 8. b.
A shock wave is an instantaneous increase in air pressure. When air flows
through a shock wave its static pressure must increase to match that in the wave,
so its velocity decreases and its temperature increases. The magnitude of these
changes in velocity, pressure and temperature, depend upon the nature of the
shock wave. A shock wave that is normal (at 90 degrees) to the airflow produces
the greatest changes. Such a wave will cause even a supersonic flow to decelerate
to subsonic velocity. This large loss of velocity causes a large increase in
pressure and temperature. An oblique shock wave produces a less abrupt
deceleration, such that a supersonic flow becomes less supersonic. This more
gentle deceleration results in a lower increase in pressure and temperature.
Shock waves are often used in supersonic aircraft air intakes, to decelerate the
airflow to subsonic speeds and convert its kinetic energy into static pressure.
The principal inefficiency affecting this process arises from the fact that some of
the kinetic energy is converted into heat rather than static pressure. This heat
cannot be recovered by the engine and reduces compressor efficiency. Maximum
efficiency is therefore achieved by using a series of obiique shock waves to bring
the airflow down to sonic speed, followed by a normal shock wave to decelerate it
to subsonic speed.
HSF 9. b.
As an aircraft passes forward through the atmosphere, the air must move apart
to make way for it, before moving back to fill the space left by its departure. The
aircraft must provide the energy required to bring about these movements of air.
This loss of energy represents a large part of the drag force affecting an aircraft
in the transonic speed range. The amount of energy required and hence the drag
force produced, is proportional to the rate of change of the cross section of the
aircraft. Abrupt changes in cross section require high accelerations of the air in
order to accommodate the passage of the aircraft. The term area rule refers to
the design strategy whereby the fuselage is carefully contoured to minimise the
rates of change of cross section of an aircraft, in order to reduce the acceleration
rates of the air, thereby minimising drag in the transonic speed range.
HSF 10. c.
High-speed tuck under occurs in the transonic speed range causing an aircraft to
pitch down. It is caused by the aft movement of the Centre of Pressure as the
shock waves above and below the wings move aft. Changes in airflow over the
tailplane also contribute to the problem. It is counteracted by the mach trim
system.
HSF 11. b.
Static pressure is inversely proportional to volume and directly proportional to
density. Kinetic energy is proportional to the square of velocity and as airflow
accelerates or decelerates the sum of its kinetic, pressure and temperature
energy remains constant. So volume increases and density decreases with the
square of velocity. As velocity increases, static pressure and temperature are
converted into kinetic energy, which increases with the square of velocity. A
given increase in velocity therefore has a greater effect on density at high speed
than at low speed. I t is therefore possible to make calculations based on the
assumption that air is incompressible a t low speeds, but this assumption would
introduce unacceptably large errors at high speeds.
HSF 12. d.
The term expansion corner refers to supersonic airflow passing over a convex
surface such that it is effectively flowing through a divergent duct. This causes
the airflow to accelerate, converting static pressure and temperature into kinetic
energy. The reduction in static pressure causes the air to expand, thereby
decreasing its density. The overall effect of supersonic flow through an
expansion corner therefore, is that its static pressure decreases, while its velocity
increases and its density and temperature decrease.
Such questions can be addressed by noting that an expansion corner expands the
air. Air expands when the static pressure on it decreases. Decreasing static
pressure causes decreasing density and temperature and increasing velocity.
HSF 13. c.
The term compression corner refers to supersonic airflow passing over a concave
surface such that it is effectively flowing through a convergent duct. This causes
the airflow to decelerate, converting some of its kinetic energy into static
pressure and temperature. The increase in static pressure causes the air to be
compressed, thereby increasing its density. The overall effect of supersonic flow
through an expansion corner therefore. is that its static pressure increases, while
its velocity decreases and its density and temperature increase.
*.
Such questions can be addressed by noting that a compression corner
compresses the air. Air is compressed when the static pressure on it-increases.
Increasing static pressure causes increasing density and temperature and
decreasing velocity.
HSF 14. b.
The term compression corner refers to supersonic airflow passing over a concave
surface such that it is effectively flowing through a convergent duct. This causes
the airflow to decelerate, converting some of its kinetic energy into static
pressure and temperature. The illcrease in static pressure causes the air to be
compressed, thereby increasing its density. The overall effect of supersonic flow
through an expansion corner therefore, is that its static pressure increases, while
its velocity decreases and its density and temperature increase.
Shock waves forin in any air stream where increasing pressure meets airflow a t
supersonic or sonic speeds. At this point the pressure increases cannot move
upstream so the entire pressure gradient piles up to form an instantaneous
pressure increase or shock wave. Because compression corners increase the
pressure of supersonic airflows, they always produce shock waves.
HSF 15. c.
V increasing
~S,
speed
The Pift force generated by an aircraft is equal to C L ~ / ~ ~ so
a t constant CL, p and S will increase the magnitude of the lift force. As an
aircraft accelerates, its CLmust be gradually reduced in order to maintain lift
equal to weight. This gradual reduction in CLis achieved by decreasing angle of
attack, which also causes the C of P to move aft.
MCMTis the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. This point is typically located on the upper'
surface of the wings at the point of greatest thickness. As the aircraft moves
forward a t MCRIT
the adverse pressure gradient over the rear surface of its
wings, moves forward with it and piles up to form a shock wave a t the point
where airflow velocity is equal to the local speed of sound. This shockwave
represents an instantaneous pressure increase.
As airflow passes through the shock wave it is decelerated rapidly such that its
static pressure matches the pressure rise the shock wave. This rapid
deceleration reduces the kinetic energy of the airflow causing the boundary layer
to separate from the wing. This separation of boundary layer reduces the lift
produced by the area of the wing behind the shock wave, causing the C of P to
move forward.
HSF 16. b.
MCRIT
is the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. This point is typically located on the upper
surface of the wings at the point of greatest thickness. As the aircraft moves
forwards at MCRITthe adverse pressure gradient over the rear surface of its
wings, moves forward with the aircraft and piles up to form a shock wave at the
point where airflow velocity is equal to the local speed of sonnd. This shockwave
represents an instantaneous pressure increase.
As airflow passes through the shock wave it is decelerated rapidly such that its
static pressure matches the pressure rise the shock wave. As acceleration
continues beyond MCNTthe airspeed over the front area of the wing becomes
supersonic causing the shock waves to be pushed shock wave moves aft towards
the trailing edge. The curvature of the upper surface of the wing behind the
leading edge is effectively a series of expansion corners. As the airflow passes
through these expansion corners it expands and accelerates to supersonic speed.
This acceleration over the entire upper surface causes the highest velocity and
lowest pressure to occur a t the rear area of the wing. In this condition the rear
area of the wing generates a large proportion the lift force. This causes its C of P
to move rearwards towards the 50% chord point.
HSF 17. c.
Althougli the shock waves striking the ground are moving vertically downwards
a t the local speed of sound, because they are aligned obli&ely to the ground,
they appear to an observer to be moving across it. The apparent speed across
the ground depends upon the mach cone angle. The more acute the angle, the
higher the apparent ground speed of the shock waves. The mach cone angle
depends on the mach number of the aircraft. The higher the mach number, the
more acute the angle of the mach cone formed by the shock waves.
Because the mach cone angle remains constant for a given aircraft mach
number, the waves appear to move across the ground at the ground speed of the
aircraft. To understand this it is necessary to consider what would happen if
they were to move across the ground at any other speed. If they were to move
faster than the aircraft the mach cone would tend to open out as the parts of the
shock waves above and below the aircraft overtook it. Conversely, if the waves
moved more slowly, the cone would close up behind the aircraft as it moved
further ahead of the upper and lower areas of the mach cone.
HSF 18. a.
The purpose of the mach trim system is to prevent mach tuck under by adjusting
the longitudinal trim of the aircraft. The method used varies with aircraft type
so options b, c and d might apply to individual systems but are not true of all
aircraft.
HSF 19. c.
Vortex generators protrude through the boundary layer from the upper surface
of the wing, into the free stream airflow above it. By mixing free stream air with
the boundary layer, the energy content of the boundary layer is increased. In
transonic flight the formation of shock waves on the upper surface of the wing
causes an abrupt decrease in velocity and an increase in static pressure. This
makes the low energy boundary layer to separate, causing shock stall and shock
induced drag. By increasing the energy content of the boundary layer, the
vortex generators make it more able to withstand the adverse pressure gradients
aft of the shock waves. In this way the shock induced separation is delayed and
its effects reduced.
HSF 20. c.
The mach number of an aircraft is a measure of its TAS as a proportion of the
local speed of sound. If for example TAS is equal to the local speed of sound then
the mach number is 1. If TAS is twice the local speed of sound the mach
number is 2. Mach number can be calculated using the following equation:
Mach Number = TAS / LSS where LSS is the local speed of sound. The LSS
varies with absolute temperature and can be calculated using the following
equation:
LSS = 38.94 v'(~bso1utetemperature).
As altitude in the international standard atmosphere increases up to 36000 feet,
temperature decreases, so LSS also decreases. Above 36000 feet, temperature
and LSS are constant. If an aircraft climbs at constant mach number its TAS
therefore decreases up to 36000 feet as temperature and LSS decrease. I t then
remains constant as altitude above 36000 feet as temperature and LSS remain
constant.
HSF 21. b.
For JAR certification an aircraft's pitch control stick force must vary linearly
from a strong pull back force at low speeds, to a strong push forward at high
speeds. That is to say, the pilot must pull the stick back to slow the aircraft, and
push it forward to increase speed. Although fully powered flying controls are
irreversible, artificial feel systems provide the required stick force gradient. As
an aircraft accelerates through the transonic speed range, aft movement of the C
of P tends to cause.it to pitch down in the phenomenon of high speed tuck under.
If left uncorrected this would require the pilot to push the stick forward to
accelerate up to MCRITthen to pull it back to continue acceleration to higher
speeds. To prevent this unacceptable reversal of the stick force gradient, the
mach trim system adjusts longitudinal trim, thereby preventing mach tuck
under. In this way the required negative stick force gradient is maintained.
HSF 22. b.
The two basic laws affecting airflow through ducts are as follows:
1. The sum of kinetic, pressure and thermal energy of the air remains constant.
2. ,Mass flow at all sections must be constant.
Kinetic energy is '/s mv2, so deceleration at high speed causes a large decrease in
kinetic energy. Law 1 requires that any decrease in kinetic energy is matched
by an increase in static pressure and temperature. So if air decelerates at high
speed it experiences a large increase in static pressure, which causes it to
contract, thereby increasing its density. Because kinetic energy is proportional
to the square of speed, the rate of contraction at high speed is greater than the
rate of deceleration. To comply with law 2 deceleration at high speed therefore
requires a decrease in volume flow rate to maintain constant mass flow. To
achieve this, a convergent duct is required to cause air to decelerate at
supersonic speed. The overall effect of such deceleration is that velocity
decreases, while static pressure, temperature and density increase. This is the
opposite of the behaviour of low subsonic flow passing through a convergent
duct.
HSF 23. c.
The acceleration rate of air passing above and below the wings increases with
increasing thickness to chord ratio. The area of greatest thickness to chord ratio
is therefore the first to experience transonic airflows and'shock stall, as aircraft
speed increases. The area of greatest thickness to chord ratio is invariably the
wing root, so shock stall commences here. When airspeed reaches McRIT,shock
waves form on the upper surfaces of its wings close to the roots. As airflow
passes through these shock waves, its velocity is instantaneously reduced and its
static pressure and temperature increased. This represents an extremely adverse
pressure gradient, which causes separation of the low energy boundary layer.
The loss of lift and increase in drag caused by this process are termed shock stall.
In a swept back wing the root is further forward than the tip, so the stalling and
loss of lift close to the root moves the mean centre of pressure of the wing
outwards and rearwards, causing the aircraft to pitch nose down. The aircraft
then accelerates pushing the shock waves aft towards the trailing edge. This
moves the centre of pressure to about the 50% chord position further increasing
the nose down pitching motion.
HSF 24. d.
As an aircraft flies through the air, the pressure disturbances caused by its
motion expand outwards in all directions, at the local speed of sound. When
flying a t MCRIT
in the transonic speed range, the compoiients of these pressure
waves moving forward, pile up in the form of a shock wave, where the airflow
speed is equal to the local speed of sound. At sonic speed, this shock wave is at
right angles to the direction of flight, but a t supersonic speed it becomes swept
back in the form of a mach cone. The point of this cone is located at the most
forward point of the aircraft. The included angle of cone is proportional to the
aircraft speed, and at mach 2 for example is approximately 60 degrees.
Although the mach cone cannot move forward relative to the aircraft, it expends
out laterally and vertically at the local speed of sound. Because the angle of the
cone is determined by the speed of the aircraft, the cone appears to move over
the ground at the ground speed of the aircraft. In considering this effect it
should be noted that if the cone moved over the ground at a higher speed, it
would overtake the aircraft. Conversely, if it moved more slowly its included
angle would decrease as the aircraft moved ahead of it.
HSF 25. d.
When an aircraft accelerates to MCNTthe first shock wave forms on the upper
surface of an aircrafts wings. Airflow passing through this shock wave is
illstantaneously decelerated, causing its static pressure to increase. This
deceleration reduces the dynamic pressure of the air, causing the boundary layer
to separate just behind the shock wave. This boundary layer separation behind
the shock wave is the cause of shock stall.
HSF 26. c.
Shock waves form in any air stream where increasing pressure meets airflow at
supersonic or sonic speeds. At this point the pressure increases cannot move
upstream, so the entire pressure gradient piles up to form an instantaneous
pressure increase or shock wave. The shock wave does not however change the
total energy content of the air, so the increase in static pressure is matched by an
increase in temperature and density and a decrease in velocity. The magnitude
of these changes depends upon the angle of the shock wave. A normal shock
wave is a t 90 degrees to the airflow. This produces the most abrupt changes,
causing velocity to decrease to subsonic speeds.
HSF 27. c.
Shock waves form in any air stream where increasing pressure meets airflow at
supersonic or sonic speeds. At this point the pressure increases cannot move
upstream so the entire pressure gradient piles up to form an instantaneous
pressure increase or shock wave. The shock wave does not however change the
total energy content of the air, so the increase in static pressure is matched by an
increase in temperature and density and a decrease in velocity. The magnitude
of these changes depends upon the angle of the shock wave.
A normal shock wave is a t 90 degrees to the airflow. This produces the most
abrupt changes, causing supersonic airflow to decelerate to subsonic speeds.
Such abrupt decelerations are accompanied by equally large increases in
temperature. When shock waves are employed to decelerate airflows, as for
example in the air intakes of supersonic aircraft, the objective is to convert as
much of the kinetic energy as possible into static pressure. Any accompanying
temperature rise represents inefficiency. The most efficient method is therefore
to use a series of oblique shock waves, each producing a small decrease in
velocity.
HSF 28. a.
The angle of a shock wave is proportional to the speed of the air flowing into it.
Sonic airflow, for example produces normal shock waves (at 90 degrees to the
flow), whereas supersonic flow produces oblique shock waves. The angle of an
oblique shock wave is proportional to the airflow velocity, such that increasing
Mach number decreases the angle of the shock wave. A Mach 2 airflow for
example, will produce a wave angle of 30 degrees. If a series of shock waves are
used to decelerate supersonic airflow, then each successive wave will be a t a
slightly greater angle to the airflow. Because a body in supersonic flow typically
produces these shock waves in the form of a shock cone, the angle of obliqueness
is referred Po as shock cone half angle.
HSF 29. a.
Shock waves forrn in any air stream where increasing pressure meets airflow a t
supersonic o r sonic speeds. At this point the pressure increases cannot move
upstream so the entire pressure gradient piles up to form a n i~~stantaneous
pressure increase o r shock wave. The shock wave does not however change the
total energy content of the air, so the increase in static pressure is matched by an
increase in temperature and density and a decrease in velocity. The magnitude
of these changes depends upon the angle of the shock wave.
A normal shock wave is a t 90 degrees to the airflow and this causes the most
abrupt changes, causing superspnic airflow to decelerate to subsonic speeds.
Such abrupt decelerations are accompanied by equally large increases in
temperature. When air flows through an oblique shock wave the changes in
velocity, pressure and temperature are much lower.
The increase in temperature as air passes through any shock wave represents a
loss of pressure energy in that some of the dynamic pressure is converted into
heat rather than into static pressure, This means that Bernoulli's theorem that
that total pressure (dynamic plus static) remains constant is not satisfied when
air flows through a shock wave. This energy loss can be eliminated only by
maintaining speeds below MCRITin order to prevent the formation of shock
waves.
HSF 30. b.
The angle
of a shock wave is proportional to the speed of the air flowing into it.
Sonic airflow, for example produces normal shock waves (at 90 degrees to the
flow), whereas supersonic flow produces oblique shock waves. The angle of an
oblique shock wave is proportional to the airflow velocity such that increasing
lMach number increase the angle of the shock wave. A Mach 2 airflow for
example, will produce a wave angle of 30 degrees. If a series of shock waves are
used to decelerate supersonic airflow, then each successive wave will be at a
slightly lower angle of obliqueness. Because a body in supersonic flow typically
produces these shock waves in the form of a shock cone, the angle of obliqueness
is referred to as shock cone half angle.
HSF 31. c.
When an aircraft accelerates to MCMTthe first shock wave forms, usually on the
upper surface of its wings. Airflow passing through this shock wave is
instantaneously decelerated, causing its static pressure to increase. This
deceleration reduces the dynamic pressure of the air causing the boundary layer
to separate just behind the shock wave. This boundary layer separation behind
the shock wave causes a reduction of lift and an increase in drag.
HSF 32. d.
When an aircraft accelerates to McWTthe first shock wave forms, usually on the
upper surface of an aircrafts wings. Airflow passing through this shock wave is
instantaneously decelerated, causing its static pressure to increase. This
deceleration reduces the dynamic pressure of the air causing the boundary layer
to separate just behind the shock wave. This boundary layer separation behind
the shock wave produces a turbulent wake, which results in a marked increase in
drag.
HSF 33. b.
Shock waves form in any air stream where increasing pressure meets airflow a t
supersonic or sonic speeds. At this point the pressure increases cannot move
upstream, so the entire pressure gradient piles up to form an instantaneous
pressure increase or shock wave. The shock wave does not however change the
total energy content of the air, so the increase in static pressure is matched by an
increase in temperature and density and a decrease in velocity. The magnitude
of these changes depends upon the angle of the shock wave. A normal shock
wave is at 90 degrees to the airflow and this produces the most abrupt changes,
causing supersonic airflow to decelerate to subsonic speed.
HSF 34. b.
Wave drag is caused by the conversion of some of the dynamic pressure into
temperature as air passes through a shock wave, and by the separation of
boundary layer behind a shock wave. Wave drag can be eliminated completely
by flying at subsonic speeds below MCRIT.
HSF 35. c.
Shock waves form in any air stream where increasing pressure meets airflow at
supersonic or sonic speeds. At this point the pressure increases cannot move
upstream so the entire pressure gradient piles up to form an instantaneous
pressure increase or shock wave. The shock wave does not however change the
total energy content of the air, so the ihcrease in static pressure is matched by an
increase in temperature and density and a decrease in velocity. The magnitude
of these changes depends upon the angle of the shock wave.
A normal shock wave is a t 90 degrees to the airflow and this causes the most
abrupt changes, causing supersonic airflow to decelerate to subsonic speeds.
Such abrupt decelerations are accompanied by equally large increases in
temperature. When air flows through an oblique shock wave however, the
changes in velocity, pressure and temperature are much less. Supersonic flow
passing through an oblique shock wave will therefore be decelerated to a lower
supersonic velocity.
HSF 36. d.
The acceleration rate of air passing above and below the wings increases with
increasing thickness to chord ratio. The area of greatest thickness to chord ratio
is therefore the first to experience transonic airflows and shock stall as the
aircraft speed increases.
The area of greatest thickness to chord ratio is
invariably the wing root so shock stall commences here. When airspeed reaches
MCRIT
shock waves form on the upper surfaces of its wings close to the roots. As
airflow passes through these shock waves its velocity is instantaneously reduced
and its static pressure and temperature increased. This represents an extremely
adverse pressure gradient, which tends to cause separation of the low energy
boundary layer.
The loss of lift and increase in drag caused by this process are termed shock stall.
In a swept back wing the root is further forward than the tip, so the stalling and
loss of lift close to the toot moves the mean centre of pressure the wing outwards
and rearwards, causing the aircraft to pitch nose down. If the aircraft continues
to accelerate, the shock waves are pushed aft towards the trailing edge. This
moves the centre of pressure to about the 50% chord position, further increasing
the nose down pitching motion. This pitching down motion is called mach tuck
under, and is counteracted by means of the mach tfim system. A variety of
methods including moving fuel aft, adjusting horizontal stabilisers and adjusting
elevators, are employed in mach trim systems in different aircraft. The purpose
of all such systems however, is to adjust the longitudinal trim of the aircraft to
counteract the nose down pitching moment.
HSF 37. a.
MCRITis the lowest TAS at which the airflow over any point on an aircraft is
equal to the local speed of sound. The airflow velocity at ally point on an aircraft
is the sum of the free stream velocity plus the acceleration caused by the
curvature of the aircraft surfaces. The greater these accelerations, the lower the
value of MCRIT.In the case of the wings, the magnitude of these accelerations
increases with thickness to chord ratio. The principal method of increasing
MCmTin high speed aircraft, is to sweep back the wings. This increases the
effective chord length and hence decreases the thickness to chord ratio. In this
way the magnitude of the acceleration of air over the wing is reduced, enabling
the aircraft to operate closer to the local speed of sound, without airflow over its
surfaces becoming sonic. This by definition constitutes an increase in MCRIT.
HSF 38. b.
MCRITis the lowest TAS a t which the airflow over any point on an aircraft is
equal to the local speed of sound. This point is typically located on the upper
surface of the wings a t the point of greatest thickness. The airflow velocity at
any point on an aircraft is the sum of the free stream velocity plus the
acceleration caused by the curvature of the aircraft surfaces. The greater these
accelerations, the lower the value of MCRIT. AS the aircraft moves forwards at
McnrT the adverse pressure gradient over the rear surface of its wings, moves
forward with the aircraft and piles up to form a shock wave a t the point where
airflow velocity is equal to the local speed of sound. MCRITis therefore the lowest
free stream velocity a t which airflow becomes sonic anywhere on the surface of
an aircraft. It is also the lowest free stream velocity a t which a shock wave can
form on an aircraft.
HSF 39. c.
As an aircraft moves forward through the air the curvature of its surfaces causes
the airflow to be accelerated. The magnitude of this acceleration increases with
curvature and with aircraft attitude, such that highly curved surfaces or high
llose up attitudes produce the greatest accelerations. MCMTis the lowest TAS at
which the airflow over any point on an aircraft is equal to the local speed of
sound. MCRITis therefore equal to the local speed of sound minus the greatest
acceleration over any part of the aircraft. Because increasing nose up attitude
increases acceleration of the airflow, it decreases MCRIT.
I-ISF 40. C.
The mach number of an aircraft is its TAS represented as a proportion of the
local speed of sound. Mach 0.8 for example, means that the TAS of the aircraft
is equal to 80% of the local speed of sound. Mach number is therefore equal to
TAS divided by LSS.
HSF 41. b.
When airspeed reaches McRITthe first shock waves form on the upper surfaces
of its wings, close to the roots. As airflow passes through these shock waves its
velocity is instantaneously reduced and its static pressure and temperature
increased. This represents an extremely adverse pressure gradient, which tends
to cause separation of the low energy boundary layer.
The loss of lift and increase in drag caused by this process are termed shock stall.
In a swept back wing, the root is further forward than the tip, so the stalling and
loss of lift close to the root moves the mean centre of pressure of the wing
outwards and rearwards, causing the aircraft to pitch nose down. If the aircraft
continues to accelerate, the shock waves are pushed aft towards the trailing edge.
This moves the centre of pressure to about the 50% chord position further
increasing the nose down pitching motion. This phenomenon is called mach tuck
under and it occurs at high fractional mach numhers, just below machl.
HSF 42. b.
The local speed of sound in air varies only with absolute temperature such that:
LSS (in Kts) = 38.94 d(~bso1utetemperature in degrees Kelvin)
HSF 43. c.
As the speed of an aircraft increases to just above McMn the first chock waves
form over the upper surfaces of its wings close to the wing roots. As airflow
passes through these shock waves its velocity is instantaneously reduced and its
static pressure and temperature increased. This represents an extremely adverse
pressure gradient, which tends to cause separation of the low energy boundary
layer. This separation of the boundary layers causes high speed buffeting and
loss of lift over the area of the wing root behind the shock waves.
In swept wing aircraft the wing roots are forward of the wing tips so this loss of
lift at the roots causes the C of P to move aft leading to a pitch down motion. In
straight winged aircraft the initial response is a pitch up, which is followed by a
pitch down as the shock waves are pushed aft if the aircraft continues to
accelerate above MCRIT.In both cases the overall effect is pitch down and buffet.
HSF 44. a.
Shock waves form in any air stream where increasing pressure meets airflow a t
supersonic or sonic speeds. Increasing pressure gradients meeting sonic airflows
produce normal (90 degree) shock waves, whereas supersonic flows produce
oblique shock waves. A normal shock wave will therefore form on an aircraft at
any point where an adverse pressure gradient moving upstream meets sonic
airflow.
HSF 45. d.
VMo is the maximum CAS at which an aircraft may be operated. I t is
determined primarily by the ability of the structure to withstand the
aerodynamic forces acting upon it. MMOis the maximum Mach number at
which an aircraft may be operated. It is determined primarily by the onset of
high speed buffet and shock stall. At low altitude the high temperature causes
the local speed of sound to be highest. MMOis some fixed fraction of the local
speed of sound so at low altitude where the local speed of sound is high, MMo
equates to a high value of CAS. But at low altitude the air density is also high so
the dynamic pressures at any given CAS are high. The effects of these pressures
at low altitude mu& therefore limit VMo. The overall effect of these factors is
that VMo is the limiting factor at low altitude and r\1Mo is the limiting factor at
high altitude. When descending at a fixed mach number close to MMo, there is
therefore a danger that VMOwill be inadvertently exceeded.
HSF 46. c.
A shock wave represents an instantaneous increase in air pressure. When air
flows through a shock wave its static pressure must increase to match that within
the wave. This increase in pressure also increases both its temperature and
density, whilst decreasing its velocity. When passing through a normal
shockwave the air is decelerated to subsonic velocity. When passing through an
oblique shock wave, pressure, temperature and density also increase but in this
case velocity is reduced to a lower supersonic velocity.
HSF 47. a.
When an aircraft accelerates to MCRIT,shock waves begin to form over its wings.
As the air flows through these shock waves, its velocity decreases and its static
pressure and temperature increase. This decrease in velocity reduces the
dynamic pressure of the air, causing the boundary layer to separate behind the
shock wave. This shock-induced separation is the cause of shock stall. By
mixing the low energy boundary layer with higher energy free stream air, vortex
generators reduce shock separation.
HSF 48. c.
When flying a t supersonic speeds the airflow over the entire wing of an aircraft
is supersonic. At such speeds the convex surfaces of the wings act as a series of
expansion corners, causing the air to accelerate all the way over the wing,
thereby pushing the shock waves back to the trailing edge. This causes the
highest velocity and lowest static pressure to occur at the trailing edge just in
front of the shock wave. In this condition the pressure distribution over the wing
is approximately triangular with its highest point a t the trailing edge.
HSF 49. d.
When flying a t transonic speeds, the airflow over the wings of an aircraft is
supersonic a t the front and subsonic at the back. At such speeds the adverse
pressure gradient moving forward with the aircraft piles up in the form of a
shock wave, a t the point of maximum wing thickness. As the air flows through
this shock wave, it decelerates and its static pressure increases. This sudden
increase in static pressure represents a collapse of the low pressure envelope over
the rear area of the wing. I n this condition the overall pressure envelope
becomes irregular, with low pressure at the front becoming high pressure behind
the shock wave.
HSF 50. d.
When an aircraft accelerates to MCRIT,shock waves begin to form over its wings.
As the air flows through these shock waves, its velocity decreases and its static
pressure and temperature increase. This decrease in velocity reduces the
dynamic pressure of the air causing the boundary layer to separate behind the
shock wave. This boundary layer separation produces high speed buffeting
which takes the form of airframe vibration.
HSF 51. b.
When an aircraft accelerates to MCRIT,shock waves begin to form over its wings.
As the air flows through these shock waves, its velocity decreases and its static
pressure and temperature increase. This decrease in velocity reduces the
dynamic pressure of the air, causing the boundary layer to separate behind the
shock wave. This shock-induced separation is the cause of shock stall, which
reduces the lift produced by the rear area of the wings.
HSF 52. a.
When flying at low airspeeds, an aircraft usually employs outboard ailerons to
generate sufficiently high control moments to initiate and control banking and
rolling motion. As airspeed increases, the wing twisting effects generated by
deflection of outboard ailerons increase, becoming unacceptable at high speeds.
In order to prevent excessive wing twist at high speeds, most modern aircraft
employ a combination of inboard ailerons and roll spoilers at high mach
numbers.
When flying at mach 0.82 the outboard ailerons of such aircraft
m ~ l be
d locked in neutral whilst the inboard ailerons and roll spoilers would be
active.
HSF 53. c.
The term stick force gradient refers to the way in which the forces that must be
applied to the control column change with airspeed. For JAR certification
purposes, an aircraft must exhibit a negative stick force gradient, such that it
requires a strong nose up pull at very low speeds, and a strong nose down push
at very high speeds. The term mach tuck under refers to the phenomenon
whereby an aircraft accelerating through the transonic speed range generates a
strong nose down pitching moment, as the shock waves form and move aft over
the top of its wings.
Without corrective action this tuck under would cause the stick forces necessary
to accelerate through the transonic speed range to change from an increasing
nose down push, just below tuck under speed, to a strong nose up pull just above
tuck under speed. This sudden reversal of the stick force gradient is
unacceptable. By introducing a strong nose up trim change at tuck under speed,
the mach trim system increases the nose down stick force required, and so
maintains the negative stick force gradient.
HSF 54. b.
As air flows through an expansion wave at supersonic speed it accelerates to
higher speed. This causes its static pressure to decrease, allowing it to expand.
This expansion means that the same mass of air occupies a larger space, and
hence its density decreases.
HSF 55. d.
VMo is the maximum CAS at which An aircraft may be op,erated. It is
determined
primarily by the ability of the structure to withstand the aerodynamic forces
acting upon it. MMOis the maximum Mach number at which an aircraft may be
operated. It is determined primarily by the onset of high speed buffet and shock
stall. At low altitude the high temperature causes the local speed of sound to be
highest. MMo is some fixed fraction of the local speed of sound, so a t low
altitude, where the local speed of sound is high, MMo equates to a high value of
TAS. But at low altitude the air density is also high, so the dynamic pressures at
any given TAS are high, The TAS equivalent of VMOis therefore low at low
altitude. The effect of these factors is that VMo is greater than MMOat high
altitude but lower than MMo at low altitude.
HSF 56. b.
The angle of a shock wave is proportional to the speed of the air flowing into it.
Sonic airflow, for example produces normal shock waves (at 90 degrees to the
flow), whereas supersonic flow produces oblique shock waves. The angle of an
oblique shock wave is proportional to the airflow velocity such, that increasing
Mach number decreases the angle of the shock wave.
A rough estimation of the angle of a n oblique shock wave can be made by taking
the reciprocal of the Mach number (llmach number) to be the tangent of the
angle. For a mach 2 airflow for example, this approximation is 26.6 degrees.
This is reasonably close to the true value of 30 degrees. Because a body in
supersonic airflow produces these shock waves in the form of a shock cone, the
angle of obliqueness is referred to as shock cone half angle.
HSF 57.c.
When the shock waves produced from a n aircraft in flight strike the ground they
are reflected upwards in the same manner as that in which light is reflected by a
mirror. The reflected angle is the same as the incident angle.
HSF 58. c.
When an aircraft is flying at supersonic speed it will produce a t least 2 shock
waves, both of which will take the form of mach cones. The first will be the
detached bow wave, whilst the second will emanate from the trailing edge of its
wings. In most cases however, other components such as its tailplane, fin and
engine intakes will produce additional waves.
HSF 59. b.
When an aircraft is side slipping to the right in supersonic flight its nose acts as a
compression corner. The velocity of supersonic airflow passing through this
corner decreases and its static pressure increases, producing a shock wave at the
right side of tlie nose. This decreases the airflow velocity and increases its static
pressure further.
The left side of the aircraft acts as an expansion corner causing airflow to
accelerate, decreasing its static pressure. The increased static pressure acting on
the right side and decreased pressure on the left side of the aircraft nose tends to
yaw it to the left, away from the sideslip. This yawing away from sideslip
constitutes directional instability.
HSF 60. b.
When airflow passes over the'wings an aircraft in supersonic fight, the resulting
shock waves decrease the velocity of the air, reducing its dynamic pressure. This
loss of dynamic pressure causes the boundary layer to separate from the wings
behind the shock waves. The reduced dynamic pressure and boundary layer
separation decrease the CL of the wings. The CL : a curve a t subsonic speed is
therefore steeper than at supersonic speed.
HSF 61. a.
When airflow passes over the wings an aircraft in supersonic fight, the resulting
shock waves decrease the velocity of the air, reducing its dynamic pressure. This
loss of dynamic pressure causes the boundary layer to separate from the wings
behind the shock waves. The reduced dynamic pressure and separation
of the wings. The CLMAXat subsonic speed is therefore
decreases the CLMAX
greater than at supersonic speed.
HSF 62. a.
As an aircraft accelerates towards MCRITits angle of attack must be reduced to
maintain lift equal to weight as speed increases. Very close to MCRIThowever the
airflow is compressed, causing its density increase. This increases the magnitude
of the lift force beyond that predicted by the lift equation using free stream
density. This increase in lift constitutes an effective increase in CL.
HSF 63. a.
When airflow passes over the wings an aircraft in supersonic fight, the resulting
shock waves decrease the velocity of the air, reducing its dynamic pressure. This
loss of dynamic pressure causes the boundary layer to separate from the wings
behind the shock waves. The reduced dynamic pressure and separation
decreases the CL of the wings. As the aircraft continues to accelerate to mach 1
the shock waves are pushed back towards the trailing edge, causing some of the
decrease in CLto be restored. The CLat mach 1 is therefore greater than that at
WRIT.
HSF 64. b.
The angle of a shock wave is proportional to the speed of the air flowing into it.
Sonic airflow, for example produces normal shock waves (at 90 degrees to the
flow), whereas supersonic flow produces oblique shock waves. The angle of an
oblique shock wave is proportional to the airflow velocity, such that increasing
mach number decreases the angle of the shock wave.
A rough estimation of the angle of an oblique shock wave can be made by taking
the reciprocal of the Mach number (llmach number) to be the tangent of the
angle. For a mach 2 airflow for example, this approximation is 26.6 degrees.
This is reasonably close to the true figure of 30 degrees. Because a body in
supersonic flow typically produces these shock waves in the form of a shock cone,
the angle of obliqueness is referred to as shock cone half angle.
HSF 65. a.
Whenever air flows through a shock wave, some of its dynamic pressure is
converted into heat. If the objective of passing the airflow through the waves is
toconvert dynamic pressure into static pressure, as for example in an air intake,
then this temperature rise constitutes an energy loss. This loss can be reduced to
zero only by flying below MCRIT
SO that no shock waves are created.
HSF 66. d.
A shock wave represents an instantaneous increase in air pressure. When air
flows through a shock wave, its static pressure must increase to match that
within the wave. This increase in pressure also increases both its temperature
and density, whilst decreasing its velocity. When passing through a normal
shockwave the air is decelerated to subsonic velocity. When passing through an
oblique shock wave, pressure, temperature and density also increase but in this
case velocity is reduced to a lower supersonic velocity.
The deceleration of airflow by passing it through a shock wave always causes an
increase in temperature. If the objective of passing the airflow through the
waves is to convert dynamic pressure into static pressure, as for example in an
air intake, then this temperature rise constitutes an energy loss. The
temperature rise and hence energy loss is proportional to the magnitude of the
velocity change. A highly oblique shock wave produces the lowest deceleration
and hence the lowest energy loss.
Whenever air flows through a shock wave some of its dynamic pressure is
converted into heat. If the objective of passing the airflow through the waves is
to convert dynamic pressure into static pressure, as for example in an air intake,
then this temperature rise constitutes an energy loss. This loss can be reduced to
zero only by flying below MCMTSO that no shock waves are created. If however
flight speeds are to be such that the creation of shock waves is unavoidable then
the highly oblique waves at high supersonic speeds will result in the lowest
energy loss.
HSF 68. b.
As an aircraft moves forward through the air the curvature of its surfaces causes
the airflow to be accelerated. The magnitude of this acceleration increases with
curvature and with aircraft attitude, such that highly curved surfaces or high
nose up attitudes produce the greatest accelerations. MCNTis the lowest TAS at
which the airflow over any point on an aircraft is equal to the local speed of
T therefore equal to the local speed of sound minus the greatest
sound. M C ~ l is
acceleration over any part of the aircraft.
Because increasing nose up attitude increases acceleration of the airflow,
anything that increases nose up attitude will decrease MCRIT. Increasing aircraft
weight increases the degree of nose up attitude needed to generate the required
lift force at any given airspeed. Increasing weight therefore decreases MCRIT.
HSF 69. b.
As an aircraft moves forward through the air the curvature of its surfaces causes
the airflow to be accelerated. The magnitude of this acceleration increases with
curvature and with aircraft attitude, such that highly curved surfaces or high
nose up attitudes produce the greatest accelerations. M C R lis
~ the lowest TAS at
which the airflow over any point on an aircraft is equal to the local speed of
sound. MCRITis therefore equal to the local speed of sound minus the greatest
acceleration over any part of the aircraft. Because increasing nose up attitude
increases acceleration of the airflow, anything that increases nose up attitude will
decrease McRIT.
Forward movement of the C of G of an aircraft increases its nose down moment.
This in turn increases the magnitude of the down force required from the
tailplane to maintain level flight. This is equivalent to an increase in weight in
that it increases the amount of lift required and so increases the degree of nose
up attitude needed to generate this lift force at any given airspeed. Forward C of
G movement therefore decreases M c ~ r ~ .
HSF 70. b.
JAR 25 states that for certification purpose buffet must not be evident at any
speed between 1.2VSlg and VMO/MMO.The minimum acceptable speed for highspeed buffet is therefore VMdMklo.
HSF 71. c.
As altitude increases, the TAS value of the low-speed buffet boundary increases,
whilst that of the high-speed buffet boundary decreases. The aerodynamic
ceiling of an aircraft is the pressure altitude a t which the high speed and lowspeed buffet boundaries cross. The margin between the two is therefore zero at
this altitude.
HSF 72. a.
JAR 25 specifies that the over speed warning system must activate at 10 Kts or
0.01m above Vhlo 1 MhlO.
HSF 73. a.
JAR 25 specifies that the over speed warning system must activate at 10 Kts or
0.01M above VMO/ Mklo.
HSF 74. b.
A shock wave represents an instantaneous increase in static pressure. As airflow
passes through a shock wave on the surface of an aircraft it decelerates, causing
its static pressure to increase and its dynamic pressure to decrease. This
decrease in dynamic pressure causes the boundary layer to become turbulent
and separate behind the shock wave.
HSF 75. b.
In supersonic flight oblique shock waves are located at the leading and trailing
edges of the wings. As airflow passes through these oblique shock waves it
decelerates to a lower level of supersonic velocity. The upper and lower surfaces
of a conventional (non supercritical) wing are convex at all points and hence
represent a series of expansion corners. As this supersonic airflow passes over
these expansion corners it accelerates. At the stagnation point however, the
airflow moving upwards and downwards to pass over and under the wing,
initially experiences a compression corner, and so decelerates. Airflow passing
over the wings of an aircraft in supersonic flight is therefore accelerated over the
whole surface, with the exception of the leading edge.
HSF 76. b.
In supersonic flight, oblique shock waves are located at the leading and trailing
edges of the wings. As airflow passes through these oblique shock waves it
decelerates to a lower level of supersonic velocity. The upper and lower surfaces
aft of the leading edge of a conventional (non supercritical) wing are convex at all
points, and hence represent a series of expansion corners. As the supersonic
airflow passes over these expansion corners, it accelerates over the whole wing
surface aft of the leading edge. The greatest velocity therefore occurs at the
trailing edge immediately in front of the shock waves.
HSF 77. b.
In supersonic flight, oblique shock waves are located at the leading and trailing
edges of the wings. As airflow passes through these oblique shock waves it'
decelerates to a lower level of supersonic velocity. The upper and lower surfaces
aft of the leading edge of a conventional (non supercritical) wing are convex at all
points, and hence represent a series of expansion corners. As the supersonic
airflow passes over these expansion corners, it accelerates over the whole wing
surface. The greatest velocity and lowest static pressure therefore occur at the
trailing edge immediately in front of the shock waves.
HSF 78. b.
In supersonic flight, oblique shock waves are located at the leading and trailing
edges of the wings. As airflow passes through these oblique shock waves it
decelerates to a lower level of supersonic velocity. The upper and lower surfaces
aft of the leading edge of a conventional (non supercritical) wing are convex at all
points, and hence represent a series of expansion corners. As this supersonic
airflow passes over these expansion corners it accelerates over the whole wing
surface. The greatest velocity and lowest static pressure therefore occur at the
trailing edge immediately in front of the shock waves. Also because dynamic
pressure is inversely proportional to static pressure, the greatest dynamic
pressure occurs immediately in front of the shock waves at the trailing edge.
HSF 79. b.
The behaviour of supersonic flow through a duct is the opposite to that of
subsonic flow. A divergent duct increases velocity and decreases static pressure
and temperature of supersonic airflow passing through it. These effects are
summarised below.
Convergent Duct
Airflow
Subsonic
Divergent duct
Temperature
Temperature Velocity Static
Static
Pressure
Pressure
Decrease Increase Increase
Increase Decrease Decrease
Velocity
HSF 80. c.The behaviour of supersonic airflow through a duct is the opposite to
that of subsonic flow. A convergent duct accelerates subsonic flow and
decelerates supersonic flow. A divergent duct decelerates subsonic flow and
accelerates supersonic flow. A parallel duct will, in theory, accelerate sonic flow,
but this acceleration will immediately cause the flow to become supersonic. This
supersonic flow will then require a divergent duct if further acceleration is to
take place. The theoretical acceleration of sonic flow through a parallel duct is
therefore possible only in an infinitely short duct.
HSF 81. a.
The behaviour of supersonic flow through a duct is the opposite to that of
subsonic flow. A convergent duct therefore decreases velocity and increases
static pressure and temperature of supersonic airflow passing through it.
Because of the magnitude of the changes it static pressure, supersonic flow
through ducts also changes the density of the air. The behaviours of subsonic
and supersonic flows are summarised below.
Convergent Duct
Airflow
Subsonic
Divergent duct
Static
Temperature Velocity Static
Temperature
Pressure
Pressure
Increase Decrease Decrease
Decrease Increase Lncrease
Velocity
Supersonic Decrease Increase Increase
Increase Decrease Decrease
HSF 82. a.
The behaviour of supersonic flow through a duct is the opposite to that of
subsonic flow. A convergent duct therefore decreases velocity and increases
static pressure and temperature of supersonic airflow passing through it.
Because of the magnitude of the changes it static pressure, supersonic flow
through ducts also changes the density of the air. A convergent duct will
therefore increase the density of supersonic airflow passing through it.
HSF 83. a.
At speeds just below MCRIT,the high dynamic pressure causes the air to be
compressed increasing its density beyond its ambient value. This increases the
,
lift force and is equivalent to an increase in CL in the equation C ~ I I ~ ~ Vif* pS is
assumed to remain a t ambient value., At MCRITthe first shock waves begin to
form over the upper surfaces of the wings. Just above MCRITthe resulting
separation of the boundary layer behind the shock wave, causes CL to decrease
markedly. As the aircraft continues to accelerate to mach 1, the shock waves are
pushed back towards the trailing edge and the supersonic flow over the wings
causes CL to increase. At slightly higher speeds the formation of the detached
bow sllock wave causes energy to be lost from the airflow approaching the wing,
thereby causing another reduction in CL.
The overall effect on CL between below MCRITand supersonic speed is that it
increases below MCRITto just above MCRIT,decreases just above MCRIT,increases
again u p to mach 1, then decrease again above mach 1.
HSF 84. d.
At MCDR,which is the critical drag rise mach number, just above MCRIT,the CD
increases markedly due to shock separation of boundary layer behind the shock
waves. This higher CDreaches a peak value a t mach 1 before decreasing as the
shock waves are pushed back to the trailing edge, reducing the area of separation
at higher speeds. CD at supersonic speed remains greater than a t subsonic speed.
HSF 85. b.
At speeds just below R I C R l ~the high dynamic pressure causes the air to be
compressed, increasing its density beyond its ambient value. This increases the
lift force and is equivalent to an increase in CL in the equation C L ~ I ~ ~ ifVp~isS ,
assumed to remain a t ambient value. At MCNT the first shock waves begin to
form over the upper surfaces of the wings. Just above MCRITthe resulting
separation of the boundary layer behind the shock wave causes C L to decrease
markedly. As the aircraft continues to accelerate to mach 1, the shock waves a r e
pushed back towards the trailing edge, and the supersonic flow over the wings
causes C L to increase. At slightly higher speeds, the formation of the detached
bow shock wave causes energy to be lost from the airflow approaching the wing,
thereby causing another reduction in CL. Because of the loss of energy due to the
shock waves the C L at supersonic speed is lower than at subsonic speed.
HSF 86. a.
, CD
At MCDR,which is the critical drag rise mach number, just above M c ~ Tthe
increases markedly due to shock separation of boundary layer behind the shock
waves. This higher CDreaches a peak value a t mach 1, before decreasing as the
shock waves a r e pushed back to the trailing edge, reducing the area of separation
at higher speeds. CDat supersonic speed remains greater than at subsonic speed.
!
HSF 87. d.
Supersonic or high-speed buffet is caused by the separation of airflow due to
shock waves. This separation causes a loss of lift, an increase in drag and
airframe vibration. .
HSF 88. d.
At speeds just below MCRIT,the high dynamic pressure causes the air to be
compressed, increasing its density beyond its ambient value. This increases the
lift force and is equivalent to an increase in CL in the equation C L ~ / ~ ~ Vif *pSis,
assumed to remain at ambient value. At McmT the first shock waves begin to
the resulting
form over the upper surfaces of the wings. Just above MCR1~
separation of the boundary layer behind the shock wave causes CL to decrease
markedly.
HSF 89. a.
The principal purpose foP employing swept back wings is to increase M c ~ T .
This effect can be explained in various ways including a mathematical one in
which the effective airspeed over a wing is shown to be proportional to the cosine
of the sweep angle. A second explanation is based upon the fact that sweep back
increase, the effective (fore and aft) length of the wing chord and hence decreases
the thickness to chord ratio. This reduces the magnitude of the acceleration of
air over the wing and hence increases MCRIT.Both of these explanations lead to
the conclusion that increasing wing sweep angle increases MCm~.
HSF 90. b.
MCmT is the free stream mach number at which airflow over any part of an
aircraft first reaches mach 1. This usually occurs at the thickest section of the
wings. The velocity of airflow over any point on the wings is the sum of the free
stream velocity and the acceleration caused by the curvature of the wings. The
greater the thickness to chord ratio of the wings, the greater will be the
acceleration over them, and hence the lower the MCRIT.
HSF 91. b.
The magnitude of the drag force at supersonic speed is determined by a number
of factors including sweep angle, thickness to chord ratio and aerofoil section.
~
The principal purpose for employing swept back wings is to increase M C Mand
decrease wave drag. This effect can be explained in various ways including a
mathematical one in which the effective airspeed over a wing is shown to be
proportional to the cosine of the sweep angle. A second explanation is based
upon the fact that sweep back increases the effective (fore and aft) length of the
wing chord and hence decreases the thickness to chord ratio. This reduces the
magnitude of the acceleration of air over the wing and hence increases MCRIT.
Both of these explanations lead to the conclusion that increasing wing sweep
~ addition
.
to increasing M C m ~
and hence delaying the
angle increases M C ~ In
onset of wave drag, increasing sweep back also decreases the intensity of the
adverse pressure gradient over the rear surface of the wings, thereby decreasing
the magnitude of the shock-induced drag force at supersonic speed.
B S F 92. a.
The magnitude of the drag force at supersonic speed is determined by a number
of factors including sweep angle, thickness to chord ratio and aerofoil section.
The principal purpose for employing swept back wings is to increase MCRITand
decrease wave drag. In addition to increasing MCRITthis also increases the
magnitude of the Critical Drag Rise Mach Number, %ICDR, and hence delays the
onset of the maximum drag force
HSF 93. b.
MCRITis the free stream mach number at which airflow over any part of an
aircraft first reaches mach 1. This usually occurs a t the thickest section of the
wings. The velocity of airflow over any point on the wings is the sum of the free
stream velocity and the acceleration caused by the curvature of the wings.
Increasing thickness to chord ratio or camber increases the magnitude of this
acceleration, and hence decreases the free stream mach number a t which airflow
first becomes sonic over the wings. So the greater the thickness to chord ratio or
camber of the wings, the greater will be the acceleration over them, and hence
the lower the MCRIT.
HSF 94. a.
McNT is the free stream mach number at which airflow over any part of an
aircraft first reaches mach 1. This usually occurs at the thickest section of the
wings. The velocity of airflow over any point on the wings is the sum of the free
stream velocity and the acceleration caused by the curvature of the wings.
Increasing camber increases the magnitude of this acceleration, and hence
decreases the free stream mach number at which airflow first becomes sonic over
the wings. So the greater the camber of the wings, the greater will be the
acceleration over them, and hence the lower the MCRIT.
In supercritical wings the camber of the upper surfaces of the wings is reduced,
thereby reducing the acceleration over them, and hence increases MCRIT.
HSF 95. b.
In supercritical wings the camber of the upper surfaces of the wings is reduced in
order to reduce the acceleration over them and hence increases MCR1~. In
addition to increasing MCRIT,the reduced camber also decreases the magnitude
of the adverse pressure gradient over the rear area of the wings. This decreases
the intensity of the shock waves that eventually form.
HSF 96. d.
VMO is the maximum CAS a t which an aircraft may be operated. I t is
determined primarily by the ability of the structure to withstand the
aerodynamic forces acting upon it. MRlo is the maximum Mach number at
which an aircraft may be operated. I t is determined primarily by the onset of
high speed buffet and shock stall. At low altitude the high temperature causes
the local speed of sound to be highest. Mnlo is some fixed fraction of the local
speed of sound, so a t low altitude where the local speed of sound is high, MMo
equates to a high value of TAS. But at low altitude the air density is also high, so
the dynamic pressures a t any given TAS are high. The TAS equivalent of VMOis
therefore low at low altitude and high at high altitude. The effects of these
factors are that Vmo is greater than Mnlo at high altitude but lower than MMOat
low altitude. The crossover altitude is the altitude at which
is equal to MMO.
HSF 97. d.
The term e x ~ a n s i o ncorner refers to an area where the space between the convex
curvature of a surface and the undisturbed free stream airflow some distance
above it, effectively forms a divergent duct. As supersonic air close to the surface
flows through an expansion corner its velocity increases and its static pressure,
temperature and density all decrease. Shock waves form where increasing static
pressures meet ~ o n i cor supersonic airflow. But the static pressure decreases in
an expansion corner, so no shock waves are produced.
HSF 98. b.
The term compression corner refers to an area where the space between the
concave curvature of a surface and the undisturbed free stream airflow some
distance above it, effectively form a convergent duct. As supersonic air close to
the surface flows through a compression corner its velocity decreases and its
static pressure, temperature and density all increase. Shock waves form where
increasing static pressures meet sonic or supersonic airflow. Because static
pressure increases in an expansion corner, an oblique shock wave forms where
the adverse pressure gradient meets supersonic airflow.
HSF 99. b.
High speed buffet commences when the speed of an aircraft reaches Mhlo,just
below MCRIT. The high, speed buffet boundary is the TAS equating to MMO. AS
altitude increase up to 36000 feet ISA, the decreasing temperature causes the
local speed of sound and hence MMo to decrease. Above 36000 feet the air
temperature, local speed of sound and MMOremain constant. The high speed
buffet boundary therefore decreases up to 36000 feet ISA, then remains constant
a t higher altitudes.
HSF 100. a.
The aerodynamic ceiling is the altitude a t which the low speed buffet boundary
at Vsl,, and tlie high speed buffet boundary at MMo, meet.
PROPS 1. c.
Increasing the number of propeller blades increases interference between
adjacent blades and so reduces aerodynamic efficiency. Propeller noise increases
with tip speed, which is a function of RPlM and blade length, so adding blades
rather than increasing blade length will avoid an increase in noise, but will not
reduce it. The number of blades on a propeller has no significant impact on the
efficiency of the constant speed unit, but adding blades will increase the
complexity of the pitch change system. The principal reason for adding extra
blades is to increase the solidity of the disk, thereby increasing the power
absorption capabilities of the propeller.
PROPS 2. d.
Gyroscopic precession is a function of the mass of the propeller, its speed of
rotation, and the magnitude of any force tending to tilt the plane of rotation. In
the case of a fixed pitch propeller, increased angle of attack or decreased TAS
would cause RPM to reduce thereby reducing gyroscopic precession effects.
Although pitching of the aircraft tilts the propeller disk, rolling does not, so
option c is incorrect. Of the options available in this question only option d,
increasing RPM will directly increase the magnitude of the precession effect.
Precession will not however occur until a pitching or yawing force acts to tilt the
plane of rotation.
PROPS 3. d.
When the RPM of a constant speed propeller is lower than that selected, the
constant speed unit moves the blades to a finer pitch, to decrease propeller
torque and increase RPM. When descending with the throttle closed and RPM
lever set at 2000, the constant speed unit would be holding the blades a t very fine
pitch in an attempt to achieve the set RPM. The propeller would therefore be
windmitling, causing a great deal of drag and hence a considerable rate of
descent. Pulling the RPM lever back would set a lower RPM, so the constant
speed unit would increase blade angle to prevent over speeding. This would
move the blades to an angle closer to the feathered position, thereby reducing
both wind milling RPM and drag. If no attempt is made to alter the pitching
angle of the aircraft the reduced drag will cause it to accelerate, increasing the
rate of descent. If however the nose was brought up to maintain constant
airspeed, the descent gradient and rate of descent would decrease. Of the options
listed only d, resembles any of the above scenarios.
PROPS 4. b.
In turning the propeller to the right, the engine applies an equal and opposite
torque to the airframe, causing it to attempt to roll to the left. This has the effect
of increasing the weight on the left wheel while decreasing that on the right. As
the aircraft accelerates down the runway the lift generated by its wings increases
so weight on both wheels tends to reduce. As stated in option c. This effect is not
however due to the torque reaction of the propeller, so option b is the most
correct.
PROPS 5. d.
The propeller slipstream spirals around the fuselage rotating in the same
direction as the propeller, so in this case it would strike the fin on the left side.
This would create an angle of attack on the fin, such that its produced a lift force
pushing the tail to the right. This would yaw the aircraft to the left.
PROPS 6. c.
In order to convert engine power into thrust, the propeller accelerates airflow
backwards. The thrust produced is the equal and opposite reaction to this
acceleration, as predicted in Newton's third law of motion. The magnitude of the
thrust can be calculated using Newton's second law: Force = mass x
acceleration, where the mass is the mass of air accelerated by the propeller. For
maximum efficiency, it is necessary to ensure that the consequent rearward
velocity of the air is minimised and this is done by maximising the mass of air
affected. The rearward velocity imparted to the air by this rearward
acceleration reduces the distance that the propeller moves forward in space in
each rotation. This reduction in fonvard motion is termed slip. The geometric
pitch of a propeller is the distance it would move forward in space in one
revolution if there were no slip.
PROPS 7. c.
If an engine fails o r is shut down in flight, the airflow passing through the
propeller causes it to windmill, thereby driving the engine. This windmilling
process incurs a very high aerodynamic drag force on the propeller. When
feather is selected the propeller blades move beyond the normal flight coarse
pitch stop, to their maximum possible coarse pitch position. This is the angle at
which they are a t their zero lift angle of attack. In this condition the blades
generate no lift force so the propelIer experiences no windmilling torque, and so
stops turning the engine. The total reaction of the blades then comprises only of
a small amount of parasite drag. I11 this way the aerodynamic drag force acting
on the propeller is minimised.
PROPS 8. a.
The angle of attack of a propeller blade is the angle between the chord line and
the relative airflow. The relative airflow is the vector sum of rotational flow due
to propeller RPM and inflow due to the TAS of the aircraft plus the acceleration
given to the air by the propeller.
PROPS 9. b.
The airflow velocity a t any point on the bIades of a propeller, is the vector sum of
airflow due to RPM and the inflow due to TAS and acceleration given to the air
by the propeller. The critical tip speed is the lowest speed a t which airflow
velocity over the blade tips equals the local speed of sound. In effect it is the local
M c n l ~for the propeller blade tips. The shock waves formed on the blade tips a t
this speed reduce the aerodynamic efficiency of the propeller.
PROPS 10. b.
The blade angle is the angle between the chord line of the blade and the plane in
which the propeller rotates. This plane is at right angles to the shaft axis.
Because the relative airflow is the vector sum of the airflow due to rotation, and
that due to the TAS of the aircraft and the acceleration given to the air by the
propeIler, it never coincides with the plane of rotation.
PROPS 11. c.
As a propeller blade rotates, each point on its surface follows a circular path.
The distance each point travels in each rotation is equal to its radius from the
.
axis of rotation multiplied by 2 ~ This
distance travelled therefore increases
with the radius of the point from the centre of the propeller. The speed at which
each point travels through the air is the product of the RPM and the distance
travelled in each turn. Airflow velocity over each point therefore increases from
root to tip. The relative airflow over each point is the vector sum of the
rotational airflow, the TAS of the aircraft and the acceleration given to the air by
the propeller . The rotational component of relative airflow, therefore increases
from root to tip. The angle of attack is the angle between the relative airflow and
the blade chord line, so if all parts of the blade had the same blade angle, the
angle of attack would increase from root to tip. For a very long blade this w ~ u l d
mean that the tip would be stalling whilst the root was at an inefficiently low
angle of attack. By twisting the blades, blade angle is decreased from root to tip,
such that the entire blade experiences an acceptable angle of attack.
PROPS 12. c.
In powered flight the propeller of an aircraft converts engine brake horsepower
into thrust horsepower. Brake horsepower is the product of engine torque, 2n
and RPM. The thrust producea is the forward component of the total reaction
generated by the propeller blades. The second component of this total reaction is
propeller torque, which opposes rotation, by acting in the opposite direction to
engine torque. The magnitude of the total reaction is proportional to angle of
attack, whilst that of the propeller torque increases with blade angle.
When established in steady flight, engine torque balances propeller torque, such
that the RPM remains constant. If TAS increases, the angle of attack decreases,
such that engine torque exceeds propeller torque, causing the RPM to ',ncrease.
On sensing this increase in RPM the constant speed unit increases blade angle in
order to restore the original angle of attssk. The increase in blade angle tilts the
total reaction away from the direction of rotation, thereby increasing propeller
torque and causing the RPM to return to its original value. Blade angle
therefore increases with increasing TAS.
PROPS 13. c.
The angle of attack of a propeller blade is the angle between the chord line and
the relative airflow. The direction of the relative airflow is the vector sum of
airflow due to rotation, TAS and the acceleration given to the air by the
propeller. The component of airflow provided by RPM is in the opposite
direction to rotation and hence approaches the blades from what, on a wing,
would be described as its underside. Increasing RPM therefore increases angle
of attack. The sum of TAS and the acceleration given to the air by the propeller
acts in the front to rear direction. Increasing either of these components
therefore decreases angle of attack. The angle of attack will therefore be
increased by decreasing the TAS and also by increasing the RPM.
PROPS 14. a.
Pulling up into a climb without increasing power will cause TAS to reduce. This
is because a proportion of the thrust will be employed in supporting part of the
weight of the aircraft, so less will be available to maintain airspeed, With no
change in power selected, the engine RPM will initially remain constant. The
decrease in TAS a t constant RPM will cause the rotational component to
represent a greater proportion of the relative airflow. This rotational component
is in the opposite direction to blade rotation, so the airflow will approach the
blades a t a greater angle of attack. Angle of attack will therefore increase. This
increased angle of attack will also increase propeller torque. Propeller torque
will then be greater than engine torque, so the RPM mill decrease.
PROPS 15. b.
Propeller blades are of aerofoil cross-section. By moving through the air at a
positive angle of attack, they generate a total reaction, similar to that of an
aircraft wing. The total reaction does not act directly forward, but is angled in
the opposite direction to the rotational movement. The forward facing
component of this total reaction is the useful thrust force. The component of
total reaction acting in the opposite direction to rotation is propeller torque,
which must be balanced by engine torque in order to maintain propeller RPM.
The thrust horsepower generated is equal ro the thrust force multiplied by the
TAS of the aircraft.
PROPS 16. d.
Propeller thrust is generated by the rearward acceleration of a mass of air.
Thrust can be calculated using Newton's second law: Force = mass x
acceleration, where the mass is the mass flow of air accelerated by the propeller.
In considering the effects on propulsive efficiency, it is important to note that
thrust is produced by the acceleration of the air, not by its final velocity. Once
the air has passed through the propeller any energy imparted to it is lost. This
lost energy represents inefficiency in the propulsion process. The acceleration
process imparts kinetic energy to the air, which is equal to %mv2. The energy
imparted is the kinetic energy possessed by the air after as it leaves the propeller,
minus the kinetic energy it possessed before meeting the propeller. That is to say
it is proportional to the propeller wash speed minus TAS. By ensuring that the
propeller wash speed is only slightly greater than TAS, the amount of kinetic
energy wasted is minimised and propulsive efficiency is maximised. In order to
do this whilst still producing the required thrust, the propeller must handle a
very large mass flow of air.
PROPS 17. a.
The increased air density will increase dynamic pressure affecting the propeller,
thereby increasing the magnitude of the total reaction generated. This will
increase both the thrust and the propeller torque. In a steady state condition
propeller torque and engine torque are equal. causing RPM to remain constant.
Assuming the engine power is not immediately affected, this increase in
propeller torque will upset the torque balance, causing the RPMI to decrease. In
reality however the increased air density will also increase the mass flow through
the engine, increasing its power output. This increase in power will first exert
itself as an increase in engine torque, which is likely to be equal to the increase in
propeller torque. If all factors are taken into account the RPM is unlikely to
alter to any significant extent. Optioil a is the only ene matching any of the
above processes.
PROPS 18. b.
When in the windmilling condition, a propeller is subjected to a negative
angle of attack, causing the total reaction to act through the back surface
of the blades. In this condition the propeller torque acts in the direction of
rotation and it is this, which causes the propeller to windmill. The remainder of
the total reaction comprises of a combination of profile and induced drag force.
Because energy is extracted from the air to drive the propeller and engine, the
resulting induced drag force is increased. When a propeller is stationary in the
feathered position it is in its zero-lift angle of attack, so the only aerodynamic
force acting upon it is profile drag. The drag force acting on a propeller is
therefore greatest when the propeller is windmilling.
PROPS 19. d.
When an aircraft is gliding, the propeller is producing no thrust, but is
contributing to the total drag acting on the aircraft. The magnitude of the drag
produced by the propeller is dependent upon its blade angle. When a propeller
is stationary in the feathered position, the blades are at their zero-lift angle of
attack, so the only aerodynamic force acting upon them is profile drag. When set
at any blade angle other than the feathered position, the propeller will windmill.
In this condition the propeller will be subjected to a negative angle of attack,
causing the total reaction to act through the back surface of the blades. In this
condition the propeller torque acts in the direction of rotation and it is this,
which causes the propeller and engine to windmill. The remainder of the total
reaction comprises of profile and induced drag.
Because energy is extracted from the air to drive the propeller, the resulting
induced drag force is increased. The drag force acting on a propeller is
therefore least when in the feathered position. The feathered position is the
coarsest pitch angle at which the blades can ever be set. This means that
decreasing propeller pitch moves the blades further away from their feathered
position, and hence increases the drag produced by the propeller. Rate of
descent in a glide is proportional to total drag, so increasing propeller drag
increases the rate of descent and decreases the L:D ratio of the aircraft.
PROPS 20. b.
The purpose of a propeller in a turboprop aircraft is to convert the shaft
horsepower (SHP) produced by the engine, into thrust horsepower (THP).
Propeller efficiency is a measure of how effectively the SHP input is converted
into THP output. As with all forms of efficiency, propeller efficiency is the ratio
of output to input and is equal to THPISHP.
PROPS 21. a.
The purpose of a propeller in a piston engine aircraft is to convert the brake
horsepower (BHP) produced by the engine, into thrust horsepower (THP).
Propeller efficiency is a measure of how effectively the BWP input is converted
into THP output. As with all forms of efficiency, propeller efficiency is the ratio
,of output to input and is equal to TWPIBHP.
PROPS 22. b.
When an aircraft is gliding the propeller is producing no thrust but is
contributing to the total drag acting on the aircraft. The magnitude of the drag
produced by the propeller is dependent upon its angle of attack. When a
propeller is stationary in the feathered position, the blades are at their zero-lift
angle of attack so the only aerodynamic force acting upon them is profile drag.
When set at any blade angle other than the feathered position, the propeller will
windmill. The propeller will then be subjected to a negative angle of attack,
causing the total reaction to act through the back surface of the blades.
In this condition the propeller torque acts in the direction of rotation and it is
this, which causes the propeller and engine to windmill. The remainder of the
total reaction comprises of profile and induced drag. Because energy is extracted
from the air to drive the propeller, the resulting induced drag force is increased.
The drag force acting on a propeller is therefore least when in the feathered
position. The feathered position is the coarsest pitch angle at which the blades
can ever be set. This means that decreasing propeller pitch moves the blades
further away from their feathered position, and hence increases the drag
produced by the propeller.
In a constant speed propeller system the pilot does not control blade angle
directly, but selects the required propeller RPM. The constant speed unit then
varies blade angle to maintain the selected RPM. The feathered condition is
obtained by selecting zero RPM. Selection of any higher RPM causes the blades
to move to a finer pitch, thereby ;.,?reasing the windmilling effect and increasing
the drag generated by the propell,., Maximum glide range is equal to L:D ratio
multiplied by height. Selecting an increased propeller RPM in a glide increases
drag, thereby decreasing both glide range and L:D ratio.
PROPS 23. d.
When the engine of an aircraft applies a torque to turn the propeller, there is an
equal and opposite reaction exerted upon the aircraft fuselage. This torque
reaction tends to turn the fuselage in the opposite direction to propeller rotation.
A right-handed tractor propeller rotates in a clockwise direction when viewed
from behind. The torque reaction produced by such a propeller will tend to roll
the aircraft to the left. Before lift-off this will increase the weight on the left
wheel and decrease that on the right wheel, thereby causing yaw to the left.
After lift-off, the torque reaction will cause left roll. The overall effect of a righthanded tractor propeller therefore is left yaw and roll.
PROPS 24. b.
When the engine of an aircraft applies a torque to turn the propeller, there is an
equal and opposite reaction exerted upon the aircraft fuselage. This torque
reaction tends to roll the fuselage in the opposite direction to propeller rotation.
A right-handed tractor propeller rotates in a clockwise direction when viewed
from behind. The torque reaction produced by such a propeller will tend to roll
the aircraft to he left. In the take-off run the aircraft will be prevented from
kolling by its undercarriage. This will decrease the load on the right wheel and
increase that on the left.
PROPS 25. a then d.
Because propellers rotate, their slip steam does not flow directly aft but spirals
aft, turning in the direction of propeller rotation. This spiral flow strikes the fin
on one side, causing the fin to generate a sideways acting lift force. Because the
C of P of the fin is behind and above the C of G of the aircraft it exerts yawing
and rolling moments on the aircraft. The direction of these moments depends on
the direction of rotation of the propeller. In the case of a right-handed tractor
propeller, they produce left yaw and right roll. It should however be noted that
once established, the left yaw produces a dissymmetry of lift between the two
wings, causing the aircraft to roll to the left. Option a is the immediate effect,
and option d, the subsequent effect. The yawing and initial rolling moments
produced by propeller slipstream effect are illustrated in the diagram below.
Yaw 10 left due la fin side force
Roll (a right due to (inside Lrca
Fi ride force aclr above C of G
ovring mll la the ngbl
Sliplmmm .Llikimg la st am
angle prodarn qidc lorrc
Fin ride forre em-.
yaw la tbe I&
CofG
PROPS 26. d.
Propeller asymmetric blade o r asymmetric thrust effect occurs whenever the
plane of rotation of the propeller is not a t right angles to the direction of flight.
This condition applies i11 almost all stages of flight, but is most pronounced a t
low speeds when high angles of attack and hence aircraft pitch attitudes are
required. I n such circumstances the blades in the down-going sector of the
propeller disc experience higher angles of attack and higher airspeeds than those
in the up-going sector. The down-going side of the disc therefore produces more
thrust than the up-going side. This causes the thrust line of the propeller to be
displaced laterally towards its down-going side.
This lateral displacement ofthe thrust line, coupled with the drag line, which
acts through the centre line of the aircraft, produces a yawing moment towards
the up-going side of the propeller disc. In the case of a tail wheel configuration
the aircraft is typically in a high nose up attitude at the start of the take-off run,
thereby maximising asymmetric blade effect. If such an aircraft were fitted with
a right-handed tractor propeller, the effect would be a strong yaw to the left.
PROPS 27. d.
Propeller asynlmetric thrust or asymmetric blade effect occurs whenever the
plane of rotation of the propeller is not at right angles to the direction of flight.
This condition applies in almost all stages of flight but is most pronounced at low
speeds when high angles of attack and hence aircraft pitch attitudes are
required. In such circumstances the blades in the down-going sector of the
propeller disc experience higher angles of attack and higher airspeeds than those
in the up-going sector. The down-going side of the disc therefore produces more
thrust than the up-going side, causing the thrust line of the propeller to be
displaced laterally towards its down-going side.
This lateral displacement of the thrust line, coupled with the drag line which acts
through the centre line of the aircraft, produces a yawing moment towards the
up-going side of the propeller disc. In the case of a nose wheel configuration the
aircraft is typically in a relatively level pitch attitude a t the start of the take-off
run, so asymmetric blade effect would not be significant until after rotation,
when the a nose up attitude would be adopted. If such an aircraft were fitted
with a right-handed tractor propeller, the effect would be a strong yaw to the left
after rotation.
PROPS 28. d then c.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the tendency of a rotating mass to resist
changes in the direction of its axis of rotation. Gyroscopic precession is the
teiidency by which any force tending to'alter the direction of the axis of rotation,
takes effect not a t its point of application, but at a point 90 degrees away in the
direction of rotation of the mass.
In the case of propeller aircraft, this means that nose up or down control inputs
generate yawing moments, the direction of which depends upon the direction of
the control input and the direction of rotation of the propeller. For a tail wheel
aircraft using a right-handed tractor propeller the action of rotating the nose
down during the early part of the take-off run will cause the aircraft to yaw to
the left. I t should however be noted that the subsequent rotation upwards of the
aircraft into the lift-off attitude would induce a right yaw.
PROPS 29. c.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the tendency of a rotating mass to resist
changes in the direction of its axis of rotation. Gyroscopic precession is the
tendency by which any force tending to alter the direction of the axis of rotation,
takes effect not at its point of application, but a t a point 90 degrees away in the
direction of rotation of the mass.
I n the case of propeller aircraft this means that nose up or down control inputs
generate yawing moments, the direction of which depends upon the direction of
the control input and the direction of rotation of the propeller. For a nose wheel
aircraft using a right-handed tractor propeller the action of rotating the nose up
into taky-off attitude a t VRwould induce a right yaw.
PROPS 30. c.
Propeller asymmetric blade effect occurs whenever the plane of rotation of the
propeller is not at right angles to the direction of flight. This condition applies in
almost all stages of flight, but is most pronounced at low speeds when high angles
of attack and hence aircraft pitch attitudes are required. In such circumstances
the angle of attack and airspeed of the blzdes in the down-going sector of the
propeller disc, experience higher angles of attack and higher airspeeds than
those in the up-going sector. The down-going side of the disc therefore produces
more thrust than the up-going side, causing the thrust line of the propeller to be
displaced laterally towards its down-going side.
In a single engine configuration this lateral displacement of the thrust line,
coupled with the drag line acting through the centre line of the aircraft, would
produce a yawing moment towards the up-going side of the propeller disc. In a
twin right-handed propeller configuration the thrust line of the right engine
would be further away from the centre of the aircraft than that of the left engine.
Following single engine failure, such an aircraft would be subject to asymmetric
power effects causing yaw away from the live engine. The critical engine is that
engine, the failure of which would cause the greatest asymmetric thrust effect.
With twin right-handed tractor propellers, the magnitude of this asymmetric
thrust effect would be greatest in the case of a left engine failure, so this would be
the critical engine.
PROPS 31. a.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the phenomenon whereby a rotating mass
tends to resist changes in the direction of its axis of rotation. Gyroscopic
precession is the phenomenon by which any force tending to alter the direction of
the axis of rotation, takes effect not at its point of application, but a t a point 90
degrees away in the direction of rotation of the mass. The magnitudes of both
the rigidity and precession forces of a rotating body are proportional to the mass
and its angular velocity.
In the case of a propeller the mass is fixed, so precession forces are altered by
changes in RPM. I t should be noted that this question does not refer to the
factors that initiate precession. Had this been so, then correct answers would
include yaw and pitch, both of which tilt the propeller plane of rotation. Roll
would not be a valid answer however, as this does not affect the plane of rotation
and hence will not initiate precession.
PROPS 32. d.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the phenomenon whereby a rotating mass
tends to resist changes in the direction of its axis of rotation. Gyroscopic
precession is the phenomenon by which any force tending to alter the direction of
the axis of rotation, takes effect not at its point of application, but at a point 90
degrees away in the direction of rotation of the mass. The magnitudes of both
the rigidity and precession forces of a rotating body are proportional to the mass
and its angular velocity.
Such factors will not however induce or initiate the precession because they do
not alter the plane of rotation. Precession in a propeller is induced by pitching
and yawing, both of which alter the plane of rotation. Roll will not induce
precessioll however, because it does not alter the plane of propeller rotation.
PROPS 33. c.
The geometric pitch of a propeller is the distance that the propeller would move
forward in each revolution if its rotation did not cause the air to slip rearwards.
In reality the air does slip rearwards. This creates an angle of attack and in so
doing, decreases the forward distance moved by the propeller. At zero angle of
attack, the blades would move through the air in a direction parallel to their
chord lines, and would therefore move a greater distance in each revolution.
Geometric pitch is the distance that a propeller would move forward through the
air in each revolution, if it had no angle of attack.
PROPS 34. a.
The geometric pitch of a propeller is the distance that it would move forward in
each rotation, if it did not cause the air to slip rearwards. Under these
~ircumstancesthe propeller blades would move forward along their own chord
lines, and hence have zero angle of attack. In reality the air slips rearwards.
This decreases the forward distance moved by the propeller, and creates an
angle of attack. The path actually taken by the blades is parallel but opposite to,
the relative airflow. This direction is the blade angle minus the angle of attack
and is called the helix angle.
PROPS 35. b.
The blade angle -of a propeller is the angle between the chord lines of the blades
and the plane of rotation of the propeller. I t is the path that the blades would
take through the air if the air did not slip backwards. The helix angle is the
angle a t which the blades actually move through the air. The angle of attack is
the difference between the blade angle and the helix angle. The blade angle is
therefore equal to the helix angle plus the angle of attack as illustrated in the
diagram below.
Spinner
PROPS 36. a.
When in the feathered position the blades of a propeller are set in their zero lift
angle of attack. At this angle they produce no lift and hence no induced drag, so
the total reaction is made up of a small amount of profile drag. A feathered
propeller therefore produces minimuin drag. Because the total reaction of a
feathered propeller comprises only of profile drag which acts in a direction,
parallel but opposite to the direction of flight, no component of the total reaction
acts in the plane of propeller rotation. A propeller therefore stops rotating when
set in the feathered position. Also, by minimizing propeller drag, selecting
feather maximises glide range. Glide speed is determined by the glide angle, not
by propeller blade angle.
PROPS 37. d.
The blade angle of a-propeller is the angle between the chord line of its blades
and its plane of rotation. In the case of a fixed pitch propeller, blade angle
cannot be altered so options a, and c. in this question are incorrect. The angle of
attack of a propeller blade at any given blade angle is determined by the vector
sum of airflows produced by rotation, propeller slip and the TAS of the aircraft.
At normal flight speeds propeller slip represents a small proportion of the total
airflow velocity and so can be ignored for the purposes of this question. The
airflow due to propeller rotation approaches the blade from its underside, so
increasing RPM a t any given TAS increases angle of attack. The airflow due to
TAS approaches the blades from ahead so increasing TAS decreases angle of
attack. The angle of attack of a fixed pitch propeller therefore decreases with
increasing TAS.
PROPS 38. b.
The blade angle of a propeller is the angle between the chord line of its blades
and its plane of rotation. In the case of a fixed pitch propeller, blade angle
cannot be altered so options a, and c, in this question are incorrect. The angle of
attack of a propeller blade a t any given blade angle is determined by the vector
sum of airflows produced by rotation, propeller slip and the TAS of the aircraft.
At normal flight speeds propeller slip represents a small proportion of the total
airflow velocity and so can be ignored for the purposes of this question. The
airflow due to TAS approaches the blades from ahead so increasing TAS
decreases angle of attack. The airflow due to propeller rotation approaches the
blade from its underside so increasing RPM at any given TAS increases angle of
attack. Fixed pitch propeller angle of attack therefore increases with RPM.
PROPS 39. a.
The angle of attack of a propeller blade a t any given blade angle is determined
by the vector sum of airflows produced by rotation, propeller slip and the TAS of
the aircraft. At normal flight speeds propeller slip represents a small proportion
of the total airflow velocity and so can be ignored for the purposes of this
question. The airflow due to TAS approaches the blades from ahead so
increasing TAS decreases angle of attack. The airflow due to propeller rotation
approaches the blade from its underside so increasing RPM a t any given TAS
increases angle of attack. The angle of attack of a fixed pitch propeller
therefore decreases with increasing TAS and increases with increasing RPM.
A propeller can achieve maximum aerodynamic efficiency a t only one angle of
attack so changes in TAS and RPM will alter the efficiency of a fixed pitch
propeller. In order to maximise the range of speeds over which a propeller can
achieve a high level of efficiency, it is necessary to vary its blade angle to achieve
the optimum combination of blade angle, TAS and RPM.
In a constant speed propeller system the blade angle is automatically varied in
flight to maintain the selected RPM. The angle of attack of a constant speed
propeller therefore remains approximately constant a t all flight speeds so options
b, and d, are incorrect. If an uncommanded increase in RPM occurs, this will be
sensed by the constant speed unit, which will increase the blade angle. The
increased blade angle will increase the rotational drag force acting on the blades,
causing the RPM to decrease to its selected value. The blade angle of a constant
speed propeller will therefore increase with an uncommanded increase in RPM.
It should be noted however that the effect of increasing the selected RPM will
cause the constant speed unit tu sense an under-speed condition, causing it to
decrease blade angle in order to increase RPM.
PROPS 40. a.
The angle of attack of a propeller blade at any given blade angle is determined
by the vector sum of airflows produced by rotation, propeller slip and the TAS of
the aircraft. At normal flight speeds propeller slip represents a small proportion
of the total airflow velocity and so can be ignored for the purposes of this
question. The airflow due to TAS approaches the blades from ahead, so
increasing TAS decreases angle of attack at any given blade angle. The airflow
due to propeller rotation approaches the blade from its underside, so increasing
RPM a t any given TAS increases angle of attack a t any given blade angle. The
angle of attack of a fixed pitch propeller therefore decreases with increasing TAS
and increases with increasing RPM.
A propeller can achieve maximum aerodynamic efficiency at only one angle of
attack so changes in TAS and RPM will alter the efficiency of a fixed pitch
propeller. In order to maximise the range of speeds over which a propeller can
achieve a high level of efficiency it is necessary to vary its blade angle to achieve
the optimum combination of blade angle, TAS and RPM.
In a constant speed propeller system the blade angle is automatically varied in
flight to maintain the selected RPM and angle of attack. The angle of attack of a
constant speed propeller therefore remains approximately constant at all flight
speeds so options b, and d, are incorrect. If TAS increases, the angle of attack of
the blades will decrease, thereby reducing the total reaction generated. This will
reduce the rotational drag force acting on the blades, and so cause the RPM to
increase. This will be sensed by the constant speed unit, which will increase the
blade angle. The increased blade angle will increase the rotational drag force
acting on the blades, causing the RPM to decrease to its selected value, restoring
the original angle of attack. The blade angle of a constant speed propeller will
therefore increase with an increase in TAS.
PROPS 41. b.
Any propeller will achieve its maximum aerodynamic efficiency a t a n angle of
attack of about 4 degrees. Higher and lower angles of attack will reduce
efficiency. As forward speed is increased the angle of attack of both a fine pitch
blade and a coarse pitch blade will decrease. But for any given combination of
RPM and TAS, a fine pitch blade will experience a lower angle of attack than a
coarse pitch blade. A fine pitch' blade will therefore achieve its best efficiency at
a low TAS whereas a coarse pitch blade will achieve maximum efficiency at a
high TAS. A coarse blade angle is therefore more efficient at high TAS.
The angle of attack of a propeller blade at any given blade angle is determined
by the vector sum of airflows produced by rotation, propeller slip and the TAS of
the aircraft. The airflow due to TAS approaches the blades from ahead so
increasing TAS decreases angle of attack. The airflow due to propeller rotation
approaches the blade from its underside so increasing RPM at any given TAS
increases angle of attack. The angle of attack at any given blade pitch therefore
decreases with increasing TAS and increases with increasing RPM. Because of
these competing tendencies it is not possible to consider the effects of high and
low R P M listed in options c, and d, in isolation. I t is however the case that a t
high TAS, propeller RPM must be limited to avoid attaining transonic tip speeds.
Overall, a coarse pitch propeller, running at a relatively high RPM such that tip
speeds are just subsonic, will produce the best efficiency a t high TAS. The
relationship between pitch, TAS and propeller efficiency is illustrated in the
diagram below.
Propeller efficiency
100%
Coarse pilch propeller if more efficient
than fme pitch at high forward speeds
PROPS 42. a.
When the engine of an aircraft applies a torque to turn the propeller, an equal
and opposite reaction is exerted upon the fuselage. This torque reaction tends to
roll the fuselage in the opposite direction to propeller rotation. A right-handed
tractor propeller rotates in a c!ockwise direction when viewed from behind, so
the torque reaction produced by such a propeller will tend to increase the
loading on the left wing and decrease that on the right wing.
Increasing power setting on a right handed tractor propeller in the stall will
therefore tend to take the left wing deeper into the stall, whilst taking the right
wing out of the stall. If only the left wing is stalled then increasing power will
increase left wing drop. Conversely if only the right wing has stalled then
increasing power will reduce wing drop. I t should also be noted that increasing
power setting on any propeller will increase the velocity of the slipstream over
the wings, thereby tending to decrease the stalling speed of the aircraft and
possibly taking the aircraft out of the stalled condition.
PROPS 43. a.
In order to convert the rotational power produced by the engine into linear
thrust power, the propeller blades must generate a lift force angled in the
forward direction. The maximum lift force that can be generated by a given
blade area is limited by the value of CLMAX
of the blades. Any attempt to
generate a greater lift force using the same relative airflow will cause the blades
to stall. The ability of any given propeller to absorb engine power and convert it
into thrust is therefore limited by blade area and C L M ~ x .
This maximum power absorption capability can be increased by adding more
blades in order to increase the total blade area. As blade numbers increase
however, the slipstream from each blade disturbs the airflow passing over its
neighbours, thereby increasing noise levels and reducing the efficiency of the
propeller. Propeller torque is that component of the total reaction that opposes
propeller rotation, so increasing blade numbers will increase both thrust and
propeller torque.
PROPS 44. b.
In order to convert the rotational power produced by the engine into linear
thrust power, the propeller blades must generate a lift force angled in the
forward direction. The maximum lift force that can be generated by a given
blade area is limited by the value of CLMAX
of the blades. Any attempt to
generate a greater lift force using the same relative airflow will cause the blades
to stall. The ability of any given propeller to absorb engine power and convert it
into thrust is therefore limited by blade area and CLMAX.
This maximum power absorption capability can be improved by increasing the
length of the blades. I n addition to increasing blade area and pewer absorption
capability, the increased aspect ratio of the longer blades reduces blade tip
vortices. This reduces the induced drag produced by the propeller, thereby
improving propeller efficiency. The increased power absorption capability will
however also increase the propeller torque, and the longer blades will increase
tips speeds.
PROPS 45. c.
When an aircraft is gliding the propeller is producing no thrust, but is
contributiiig to the total drag acting on the aircraft. The magnitude of the drag
produced by the propeller is dependent upon its blade angle. When a propeller
is stationary in the feathered position, the blades are at their zero-lift angle of
attack, so the only aerodynamic force acting upon them is profile drag. When set
at any blade angle other than the feathered position, the propeller will windmill.
It will then be subjected to a negative angle of attack, causing the total reaction
to act through the back surface of the blades. In this condition the propeller
torque acts in the direction of rotation and it is this, which causes the propeller to
windmill. The remainder of the total reaction comprises of profile and induced
drag. Because energy is extracted from the air to drive the propeller, the
resulting induced drag force is increased. The drag force acting on a propeller
is therefore least when in the feathered position. The feathered position is the
coarsest pitch angle a t which the blades can ever be set. This means that
decreasing propeller pitch moves the blades further away from their feathered
position, and hence increases the drag produced by the propeller.
In a constant speed propeller system the pilot does not control blade angle
directly, but selects the required propeller RPM. The constant speed unit then
varies blade angle to maintain the selected RPM. In a glide, the feathered
condition is obtained by selecting zero RPM. Pushing the RPM lever forward to
select a higher RPM causes the blades to move to a finer pitch, thereby
increasing the windmilling effect and increasing the drag generated by the
propeller.
Maximum glide range is equal to L:D ratio multiplied by height. Selecting an
increased propeller RPM in a glide increases drag, thereby decreasing both glide
range L:D ratio. As the engine will be either shut down or running a t idle in a
glide, selecting a higher RPM will have no significant effect on cooling.
PROPS 46. c.
When an aircraft is gliding the propeller is producing no thrust, but is
contributing to the total drag acting on the aircraft. The magnitude of the drag
produced by the propeller is dependent upon its blade angle. When stationary in
the feathered position, the blades are at their zero-lift angle of attack so the only
aerodynamic force acting upon them is profile drag. At any blade angle other
than the feathered position, the propeller will windmill.
In this condition the propeller will be subjected to a negative angle of attack,
causing the total reaction to act through the back surface of the blades.
Propeller torque then acts in the direction of rotation and it is this, 'which causes
the propeller to windmill. The remainder of the total reaction comprises of
profile and induced drag. Because energy is extracted from the air to drive the
propeller, the resulting induced drag force is increased. The drag force acting
on a propeller is therefore least when in the feathered position.
The feathered position is the coarsest pitch angle at which the blades can ever be
set. This means that increasing propeller pitch from any other setting moves the
blades closer to their feathered position. This decreases propeller RPM and the
drag produced by the propeller. In a glide this decrease in drag will increase
glide range. The effect of increasing propeller pitch in a glide is therefore a
decrease in RPM and an increase in glide range.
PROPS 47. b.
Any propeller will achieve its maximum aerodynamic efficiency at an angle of
attack of about 4 degrees. Higher and lower angles of attack reduce efficiency.
As forward speed is increased, the angle of attack of both a fine pitch blade and a
coarse pitch blade will decrease. But for any given combination of IPPM and
TAS, a fine pitch blade will experience a lower angle of attack than a coarse
pitch blade. A fine pitch blade will therefore achieve its best efficiency a t a low
TAS, whereas a coarse pitch blade will achieve maximum efficiency a t a high
TAS. A coarse blade is therefore more efficient a t high TAS.
If an aircraft is to be flown at speeds above that a t which a fine pitch is efficient,
then its blade angle must be increased to match increasing TAS. In most
modern propeller aircraft this is achieved by means of a constant speed
propeller, which increases the propeller efficiency over a wide TAS range. The
principal disadvantages of constant speed propellers are that they are more
complex, heavier and structurally weaker than fixed pitch propellers. Maximum
propeller RPM is limited by tip speed and structural strength, neither of which
are improved by the use of a constant sped propeller.
PROPS 48. b.
Any propeller will achieve its maximum aerodynamic efficiency a t an angle of
attack of about 4 degrees. Higher and lower angles of attack reduce efficiency.
As forward speed is increased, the angle of attack of both a fine pitch blade and a
coarse pitch blade will decrease. But for any given combination of RPM and
TAS, a fine pitch blade will experience a lower angle of attack than a coarse
pitch blade. A fine pitch blade will therefore achieve its best efficiency a t a low
TAS, whereas a coarse pitch blade will achieve maximum efficiency a t a high
TAS. A coarse blade is therefore more efficient a t high TAS.
Although a coarse pitch propeller is more efficient a t high TAS than a fine pitch
propeller, this does not mean that the maximum efficiency of a coarse propeller
is greater than that of a fine pitch propeller. Increased pitch tilts the total
reaction of the blade in the direction of propeller torque. This means that as
blade pitch increases, a greater proportion of the total reaction is converted int.;
useless propeller torque, which must be balanced by engine torque, so a smaller
proportion is converted into thrust. Maximum efficiency of a coarse pitch
propeller is lower than that of a fine pitch propeller as illustrated overleaf.
1
Propeller eK~ciency
llJO0A
Maximum efGcleney at fine pitcb is greater
than maximum eflicieney at coarse pitch
PROPS 49. c.
In order to convert the rotational power produced by the engine into linear
thrust power, the propeller blades must generate a lift force angled in the
forward direction. The maximum lift force that can be generated by a given
blade area is limited by the value of C L ~ AofXthe blades and any attempt to
generate a greater lift force will cause the blades to stall. The ability of any given
propeller to absorb engine power and convert it into thrust is therefore limited
by blade area and aerofoil section.
This maximum pourer absorption capability can be increased by adding more
blades in order to increase the total blade area. As blade numbers increase
however, the slipstream from each blade disturbs the airflow passing over its
neighbours, thereby increasing noise levels and reducing the efficiency of the
propeller. Propeller efficiency therefore decreases with increasing blade
numbers.
PROPS 50. a.
The angle of attack of a propeller blade at any given blade angle is determined
by the vector sum of airflows produced by rotation, propeller slip and the TAS of
the aircraft. The velocity of the airflow caused by rotation a t any point on a
propeller blade is equal to the RPM multiplied by 27cR, where R is the radius at
that point. This mans that rotational airflow velocity increases from root to tip.
The airflow due to propeller rotation approaches the blade from its underside, so
the root to tip increase in rotational airflow causes angle of attack to increase
from root to tip.
But an aerofoil can be most efficient at only one angle of attack, so this root to tip
increase in angle of attack means that only a narrow area of the span of a blade
can operate a t its most efficient angle at any one time. All points on the blade
inboard of this most efficient section would operate at too low an angle, whilst a11
points further outboard would be at inefficiently high angles of attack. This
problem is overcome by twisting propeller blades such that the blade angle
decreases from root to tip to match the increasing rotational velocity. The
overall effect is that angle of attack remains approximately constant while blade
angle reduces from root to tip.
PROPS 51. d.
When a body rotates it generates a centrifugal force, which tends to pull the
mass of the body away from the centre of rotation. This can be demonstrated for
example, by attaching a weight to a piece of string. When caused to spin in a
horizontal plane at high speed, the weight moves outwards until it follows a
horizontal path, almost level with the centre of rotation.
Each particle of the blades of a propeller are subject to centrifugal force.
Because the blades are of flattened cross section, this centrifugal force tends to
align all of these particles in the same plane of rotation. This produces a twisting
force tending to drive the blades to flat (fine) pitch, whenever the propeller is
turning. When feathered however, a propeller does not turn and hence produces
no centrifugal twisting moment. So centrifugal twisting moment does not assist
in unfeathering a propeller.
The total reaction of each propeller blade acts through its C of P, the position of
which varies with angle of attack. The C of P is however always forward of pitch
change axis of variable pitch blades. Because the total reaction acts at a point
foiward of the pitch change axis, it generates an aerodynamic twisting moment,
tending to change the blade pitch angle. In powered flight the total reaction acts
in a forward direction tending to turn the blades to coarse pitch. In reverse
thrust it tends to drive the blades to maximum coarse reverse pitch. When
windmilling however, the total reaction acts in an aft direction so the
aerodynamic twisting moment tends to drive the blades to fine pitch. But the
centrifugal twisting moment always drives the blades towards fine pitch.
Centrifugal twisting moment therefore assists aerodynamic twisting moment
only in windmilling flight.
PROPS 52. a.
The rotational airflow at any point on a rotating propeller blade, increases from
root to tip. In order to ensure that the entire blade operates a t an efficient angle
of attack, propeller blades are twisted, such that in normal powered flight, blade
angle decreases from root to tip. When in reverse thrust the blades are set at a
negative blade angle, so this blade twist causes blade angle and (negative) angle
of attack to increase from root to tip. This root to tip increasing angle of attack
in reverse thrust,'causes the total reaction to increase from root to tip. This also
means that the majority of reverse thrust is produced at the blade tips.
\
PROPS 53. d.
Propeller torque is that component of the propeller total reaction which acts
parallel to the plane of rotation. In powered flight it acts inthe opposite
direction to propeller rotation, and hence opposes the engine torque. To
maintain any given RPM the propeller torque must be equal but opposite to the
engine torque. In windmilling flight however, the propeller torque reverses
direction, such that it acts in the direction of propeller rotation. Propeller torque
therefore assists engine torque in windmilling flight.
PROPS 54. a.
The relative airflow affecting any section of a propeller blade is the vector sum of
airflows due to rotation, propeller slip and TAS. The directions of these airflow
components are such that increasing rotational flow increases angle of attack.
But the distance moved in each revolution, and hence the rotational airflow,
increases from root to tip, so angle of attack tends to increase from root to tip.
But total reaction increases with airflow velocity and angle of attack, so
increasing angle of attack and airflow velocity from root to tip would mean that
total reaction would be greatest a t the tips. If left uncorrected this would apply
unacceptably high forward bending loads on the propeller. Propeller blades are
therefore twisted, such that angle of attack is lowest at the tip, where the airflow
velocity is greatest.
PROPS 55. b.
~ ~ that
, a constant IAS climb means climbing at
IAS is proportional to 1 1 2 ~such
constant 1 / 2 p ~ 2 But as altitude increases p decreases, so V, which is the TAS,
must increase to maintain constant IAS. But increasing TAS decreases the angle
of attack of a propeller at any given RPM and blade angle. This would reduce
the propeller torque, thereby causing RPM to increase. In a constant speed
system, the constant speed unit would sense this increasing RPM and increase
blade angle to restore propeller torque and RPM. But because p is reducing, the
required angle of attack would increase with altitude.
.
PROPS 56. b.
IAS is proportional to I / ~ ~ Vsuch
* , that a constant IAS climb means climbing at
constant 1 / 2 p ~ 2 But as altitude increases p decreases, so V, which is the TAS,
must increase to maintain constant IAS. But if the ratio of TAS : IAS increase
with altitude, then IAS decreases in a constant TAS climb. This would decrease
1 / 2 p ~ 2thereby
,
reducing propeller total reaction, thrust and propeller torque, at
any given angle of attack and RPM. The reduction in propeller torque would
cause the RPM to increase. In a constant speed system, the constant speed unit
would sense this increasing RPM and increase blade angle to restore propeller
torque and RPM. But because p is reducing, the required angle of attack would
increase with altitude.
.
PROPS 57. d.
The angle of attack of a propeller at any given blade angle and RPM, decreases
with increasing TAS. So if TAS increases at constant altitude, the decreasing
angle of attack will decrease propeller torque, thereby allowing the RPM to
increase. In a constant speed system, the constant speed unit would sense this
increasing RPM and increase blade angle to restore propeller torque and RPM.
But because p is constant at any given altitude, the required angle of attack
would also remain constant. The effect of the constant speed unit in increasing
blade angle and maintaining constant RPM, would therefore be to maintain
.constant angle of attack. The relationship between TAS and propeller angle of
attack is illustrated below.
Drive shaft
Y
- - -;Angle
/
of attack decreases
as TAS increases
IneNiciently high angle of attack at zero TAS
PROPS 58. b.
The propeller torque generated at any given blade angle, TAS and RPM,
decreases with air density (p). So if p decreases at constant TAS, the decreasing
propeller torque will allow the RPM to increase. In a constant speed system, the
constant speed unit would sense this increasing RPM and increase blade angle to
restore propeller torque and RPM. But because p is reducing, the required angle
of attack would increase. The effect of the reducing p at constant TAS would
therefore be to increase blade angle and angle of attack, whilst keeping RPM
constant.
PROPS 59. b.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the phenomenon whereby a rotating mass
tends to resist changes in the direction of its axis of rotation. Gyroscopic
precession is the phenomenon by which any force tending to alter the direction of
the axis of rotation, takes effect not at Its point of application, but at a point 90
degrees away in the direction of rotation of the mass. Precession in a propeller is
induced by pitching and yawing, both of which alter the plane of rotation. When
a single left handed tractor propeller aircraft is rotated nose up into the lift-off
attitude at VR, the precession effect will cause it to yaw to the left.
PROPS 60. a.
Whenever a mass is caused to rotate, it adopts the properties of gyroscopic
rigidity and precession. Rigidity is the phenomenon whereby a rotating mass
tends to resist changes in the direction of its axis of rotation. Gyroscopic
precession is the phenomenon by which any force tending to alter the direction of
the axis of rotation, takes effect not a t its point of application, but at a point 90
degrees away in the direction of rotation of the mass. Precession in a propeller is
induced by pitching and yawing, both of which alter the plane of rotation. When
a single right handed tractor propeller aircraft is rotated nose up into the lift-off
attitude at VR, the precession effect will cause it to yaw to the right.
PROPS 61. a.
Because propellers rotate, their slipsteam does not flow directly aft but spirals
aft, turning in the direction of propeller rotation. This spiral flow strikes the fin
on one side, causing the fin tozenerate a sideways acting lift force. Because the
C of P of the fin is behind and above the C of G of the aircraft it exerts yawing
and rolling moments on the aircraft. The direction of these moments depends on
the direction of rotation of the propeller. In the case of a right-handed tractor
propeller, they produce left yaw and right roll. I t should however be noted that
once established, the left yaw produces a dissymmetry of lift between the two
wings, causing the aircraft to roll to the left. During the take-off run the roll
would be prevented by the ground and the yaw would be contained by right
rudder. At lift-off however, the ground would cease to prevent the roll. The
immediate effect of this slipstream or weathercock effect in a single right handed
tractor propeller aircraft at lift-off might therefore be right roll.
PROPS 62. c.
Propeller asymmetric thrust or asymmetric blade effect occurs whenever the
plane of rotation of the propeller is not at right angles to the direction of flight.
This condition applies in almost all stages of flight but is most pronounced at low
speeds when high angles of attack and hence aircraft pitch attitudes are
required. In such circumstances the blades in the down-going sector of the
propeller disc experience higher angles of attack and higher airspeeds than those
in the up-going sector. The down-going side of the disc therefore produces more
thrust than the up-going side, causing the thrust line of the propeller to be
displaced laterally towards its down-going side. This lateral displacement of the
thrust line, coupled with the drag line which acts through the centre line of the
aircraft, produces a yawing moment towards the up-going side of the propeller
disc. If permitted to develop, this yaw will then cause roll in the same direction.
Left yaw and roll in climb-out might therefore be caused by asymmetric blade
effect in a single right handed tractor propeller aircraft.
PROPS 63. b.
The lift produced by the wings of a twin propeller aircraft a t any given airspeed
is increased by the high velocity propeller wash passing over the wings. I n the
event of a single engine failure, one wing will be robbed of this extra lift, causing
the aircraft to roll towards the dead engine. Also, the loss of thrust on one engine
will cause the aircraft to yaw towards the dead engine. Right roll and yaw of a
propeller aircraft in climb-out might therefore be caused by right engine failure.
PROPS 64. b.
Propeller asymmetric thrust o r asymmetric blade effect occurs whenever the
plane of rotation of the propeller is not a t right angles to the direction of flight.
This condition applies in almost all stages of flight but is most pronounced a t low
speeds when high angles of attack and hence aircraft pitch attitudes are
required.
In such circumstances the blades in the down-going sector of the propeller disc
experience higher angles of attack and higher airspeeds than those in the upgoing sector. The down-going side of the disc therefore produces more thrust
than the up-going side, causing the thrust line of the propeller to be displaced
laterally towards its down-going side. This lateral displacement of the thrust
line, coupled with the drag line which acts through the centre line of the aircraft,
produces a yawing moment towards the up-going side of the propeller disc. If
permitted to develop, this yaw will then cause roll in the same direction. Right
yaw and roll of a propeller aircraft in climb-out might therefore be caused by
asymmetric blade effect of a single left handed tractor propeller.
PROPS 65. c.
Propeller asymmetric thrust o r asymmetric blade effect occurs whenever the
plane of rotation of the propeller is not a t right angles to the direction of flight.
This condition applies in almost all stages of flight but is most pronounced at low
speeds when high angles of attack and hence aircraft pitch attitudes are
required.
In such circumstances the blades in the down-going sector of the propeller disc
experience higher angles of attack and higher airspeeds than those in the upgoing sector. The down-going side of the disc therefore produces more thrust
than the up-going side, causing the thrust line of the propeller to be displaced
laterally towards its down-going side. This lateral displacement of the thrust
line, coupled with the drag line which acts through the centre line of the aircraft,
produces a yawing moment towards the up-going side of the propeller disc. If
permitted to develop, this yaw will then cause roll in the same direction. At
speeds below VR however, the ground will prevent roll. Right yaw of a propeller
aircraft a t speeds approaching VR might therefore be caused by asymmetric
blade effect of a single left handed tractor propeller. A contra-rotating propeller
system employs two propellers turning in opposite directions on a common axis.
Asymmetric blade effect, gyroscopic precession and torque reaction are all
largely self cancelling in this system.
PROPS 66. b.
The lift produced by the wings of a twin propeller aircraft a t any given airspeed
is increased by the high velocity propeller wash passing over the wings. I n the
event of a single engine failure, one wing will be robbed of this extra lift, causing
the aircraft to roll towards the dead engine. Also, the loss of thrust on one engine
will cause the aircraft to yaw towards the dead engine. Right roll and yaw of a
propeller aircraft in climb-out might therefore be caused by right engine failure.
Option c, is improbable in that asymmetric blade effect would gradually become
evident before VR. Option d, torque reaction in a twin engine contra rotating
propeller system, is a possible cause only in the event of failure of the engine
driving the left handed propeller. In this case the asymmetric blade effect and
torque reaction of the live propeller would suddenly become evident.
PROPS 67. b.
When a body rotates it generates a centrifugal force, which tends to pull the
mass of the body away from the centre of rotation. This can be demonstrated for
example, by attaching a weight to a piece of string. When caused to spin in a
horizontal plane a t high speed, the weight moves outwards until it follows a
horizontal path, almost l'evel with the centre of rotation. Each particle of the
blades of a propeller are subject to centrifugal force. Because the blades are of
flattened cross section, this centrifugal force tends to align all of these particles in
the same plane of rotation. This produces a twisting force tending to drive the
blades to flat (fine) pitch, whenever the propeller is turning.
ENV 1. c.
VAis the speed at which an aircraft will stall which achieving its limiting load
factor, which for a JAR certificated passenger aircraft flyingwith flaps retracted
is 2.5. This question does not specify any change in configuration o r limiting
load factor, so only the effect of the weight change needs to be calculated. The
problem can therefore be solved using the following equation:
VAat new weight = VA a t old weight x d ( New weight / old weight)
Inserting the data provided gives:
VAat new weight = 250 Kts x d (70000 Kg / 60000 Kg)
Which resolves to give VA a t 70000 Kg =
270 Kts.
ENV 2. b.
VAis the speed at which an aircraft will stall when achieving its limiting load
factor;which for a JAR certificated passenger aircraft is 2 with flaps down and
2.5 with flaps retracted. This question therefore involves both a change in
weight and a change in limiting load factor. The weight change is the same as
that specified in question ENV 1, which revealed that VAat 70000 Kg with flaps
retracted was 270 Kts.
This value can then be used as the old VA,to calculate the new VAwith flaps
down, using the following equation:
VAat new load factor = VAat old load factor x d(new load factor / old load
factor )
Inserting the data provided in the question gives:
VAat 2.0 load factor = VAat 2.5 load factor x 6 ( 2.0 / 2.5 )
Which is VAat 2.0 load factor = 270 Kts x 6 ( 2.0 / 2.5 )
Which resolves to give VAat 2.0 load factor = 241 Kts.
The closest option provided in the question is therefore 240 Kts.
ENV 3. b.
This question specifies only a change in weight, so the problem can be solved
usfng the following equation:
VAat new weight = VAat old weight x 6 ( new weight / old weight)
Although neither the new weight nor the old weight are known the question
specifies that the new weight is 25% greater than the old weight. This fact can
be used to solve the problem by assigning a value of 1 to the old weight, and a
value of 1.25 to the new weight.
Inserting these values into the equation gives the following:
This resolves to give VAat new weight = VAat old weight x 1.118
VAhas therefore increased by approximately 12%.
This answer demonstrates the more general point that for small weight changes,
the % change in speed (VA,VS, VMDetc.), is approximately % the % change in
weight. Using this shortcut, a 50% weight increase gives a 25% speed increase,
whereas the true figure is 22.5%. The accuracy of this shortcut method
increases, as the % weight change decreases. For a 10% change in weight for
example, the shortcut gives 5% change in speed, whereas the true figure is 4.9%.
.
ENV 4. b.
The regulations regarding positive load factor limits for JAR certificated
passenger aircraft are specified in JAR25. Although JAR 25 specifies a method
of calculating these load factors based on aircraft weight, questions in the JAR ' , _
ATPL POF examinations are usually based on certain values. These are the
minimum allowable values, which are 2 with flaps down and 2.5 with flaps up.
ENV 5. b.
VAis the CAS at which an aircraft will stall whilst just attaining the limiting load
factor. This means that at lower speeds the aircraft will stall before attaining the
limiting load factor. At higher speeds it can be damaged due to overloading by
exceeding the limiting load factor before stalling. An aircraft cannot therefore
be damaged by overloading due to the use of the flying controls at or below VA.
ENV 6. b.
The maximum operating altitude of a JAR certificated passenger aircraft is that
altitude at which it can just attain a positive load factor of 1.3 before the onset of
buffet. Load factor greater than 1.3 when flying at this altitude will cause
buffeting.
ENV 7. b.
VA is the CAS at which an aircraft will stall whilst just attaining the limiting load
factor. This means that at lower speeds the aircraft will stall before attaining the
limiting load factor. At higher speeds it can be damaged due to overloading by
exceeding the limiting load factor before stalling. VAis therefore the maximum
speed a t which full nose up control inputs are permitted.
ENV 8. d.
VAis the CAS at which an aircraft will stall whilst just attaining the limiting load
factor. For the purposes of JAR POF examinations, the limiting load factor is
normally taken to be 2.5g with flaps retracted. Load factor = lift 1 weight, so in
order to increase load factor at any given weight, the magnitude of the lift force
must be increased. Lift is proportional to CLx CAS SO at any given speed an
increase in lift (and hence load factor) requires an increase in CL. ~ n change
y
which reduces the gradient of the CL : angle of attack curve and decreases CLMax
will increase VA. Of the options offered in this question only option d, increasing
the wing sweep back angle will have this effect. Of the other options, increasing
aspect ratio and increasing camber both increase CL M ~ ,and hence reduce VA.
Increasing wing area increases the lift force attained at any given CL and
therefore also-decreasesVA.
ENV 9. d.
Turbulence results in the wings being subjected to random variations in upward
and downward gusts of air. These random cause random changes in the angle of
attack and hence CL. The varying lift force produced by these changes in CL
cause the aircraft to climb and descend in an erratic manner. The degree to
which flight path and hence ride quality is affected depends upon the magnitude
.
I
of the variations in CL. The magnitude of theses variations in CLdepends upon
the gradient of the CL:a curve. The steeper the CL:a curve, the greater will be
the change in CLfor any given change in a. The smoothest ride will be provided
by an aircraft exhibiting a shallow CL:acurve. Of the options provided in this
question only option d, increased sweep back angle, will have these effects.
ENV 10. a.
VAis the design manoeuvre speed of an aircraft. This is the CAS at which the
hircraft stalls just as it attains the limiting load factor. At lower speeds the
aircraft will stall at lower load factors, whilst at higher speeds it will exceed the
limiting load factor before stalling. The recommended rough air speed is
normally set approximately midway between the low speed and high speed stall
speeds, in order to provided the widest possible safety margin when flying in
rough air. The maximum operating speeds of JAR certificated passenger
aircraft is Vmo / MMO.
ENV 11. d.
Turbulence results in the wings of an aircraft being subjected to random upward
and downward gusts of air. These random gusts cause random changes in the
angle of attack and hence CL. If up-gusts are sufficiently severe, the angle of
attack can exceed the stalling angle, thereby causing low speed stall. Also if the
gusts are of sufficient intensity the CLcan be increased to such an extent that the
limiting load factor of the aircraft is exceeded. Deployment of trailing edge flaps
reduces the stalling angle, and so makes the wing more prone to low speed stall
in turbulence. Trailing edge flap deployment also increases CLM,, and hence
increases the danger of exceeding limiting load factor. In order to compensate
for this effect the limiting load factor of JAR certificated passenger aircraft is
reduced to 2.0g when flaps are deployed.
Although the above effects are mitigated to some extent by the deployment of
leading edge devices, the most probable overall effect is still an increase in
stalling hazard and overload hazard.
ENV 12. c.
At any given altitude an aircraft can be flown only at those speeds between the
low speed buffet and high speed buffet margins. Because the speed of sound
reduces with decreasing temperature, the value of the high speed buffet
boundary decreases, causing the available speed range to decrease with
increasing altitude. The aerodynamic ceiling is that altitude at which the low
speed and high speed buffet boundaries meet, leaving only a single flight speed
available. Banking an aircraft increases load factor and hence the magnitude of
the lift force required. This increase in required lift force increases the low speed
stalling speed causing the buffet boundaries to converge further. The
magnitude of this additional convergence is proportional to the load factor and
hence bank angle. At high altitudes where the speed margin is very narrow, the
ability of an aircraft to bank is therefore very limited. It should however be
noted that increasing altitude also decreases stability, thereby increasing the roll,
pitch and yaw rates that an aircraft is capable of achieving. The use of these
higher rates will however cause stalling, so the overall effect is that
manoeuvrability of an aircraft decreases with increasing altitude.
ENV 13. c.
Turbulence results in the wings being subjected to random variations in upward
and downward gusts of air. These random gusts result in random changes in the
angle of attack and hence CL. The variations in lift force produced by these
changes in CLcause the aircraft to rise and descend in an erratic manner. The
degree to which flight path and hence ride quality is affected, depends upon the
magnitude of the variations in CL. The magnitude of theses variations in CL
depends upon the gradient of the CL : angle of attack curve. A steeper curve
gives larger variations in CLfor any given change in angle of attack. Because
swept wings exhibit shallow CL : angle of attack curves, they are less affected by
turbulence than straight wings. Options a and b in this question a r e incorrect,
because dihedral and anhedral have no effect on the CL : angle of attack curve or
response to turbulence.
ENV 14. c.
The limiting load factor for a light utility aircraft in the clean configuration is 4.4
(option c).
ENV 15. b.
For JAR certification, passenger aircraft must be capable of sustaining a limiting
load factor of at least 2.5 without suffering permanent deformation. Such
aircraft must also be capable of sustaining an additional 50% of this load factor
without catastrophic failure. This 50% safety factor therefore imposes an
ultimate load factor of 1.5 times the limiting load factor.
ENV 16. b.
The lift force that must be produced by an aircraft in flight is equal to its weight
multiplied by its load factqr. The aircraft structure must be sufficiently strong to
prevent permanent deformation when operating at the limiting load factor. For
JAR certification purposes, the minimum acceptable load factor for flaps up
flight is 2.5. This means that the structure must be capable of supporting 2.5
times its weight without permanent damage. When flying at lower weights, the
imposition of the same load factor would apply correspondingly lower lift forces.
The limiting load factor could therefore be increased when operating a t lower
weights, if structural integrity was the only factor to be considered. In reality
the limiting load factor remains constant a t all weights in order to protect the
passengers from excessive g loading.
ENV 17. a.
VA is the design manoeuvre speed and is the speed a t which an aircraft will stall
whilst just attaining its limiting load factor. In any flight condition the stalling
speed is equal to the basic (lg) stalling speed, multiplied by the square root of the
load factor. VA can therefore be calculated using the following equation:
VA = VSIgdn limit
wliere Vsl, is the 1g stalling speed and n limit is the
limiting load factor.
ENV 18. c.
The speeds VA,and Vc are specified by the designers of an aircraft and are not
altered by conditions existing in any particular flight. A sudden up gust will
however affect the direction of the relative airflow, causing an increase in angle
of attack. This increase in angle of attack will increase CLand load factor.
ENV 19. c.
The speeds VB, and Vc are specified by the designers of an aircraft and are not
altered by conditions existing in any particular flight. A sudden up gust will
however affect the direction of the relative airflow, causing an increase in angle
of attack. This increase in angle of attack will increase CLand load factor, n.
ENV 20. c.
In any given flight condition the load factor equals the lift force divided by the
weight of the aircraft. In straight and level flight for example, lift equals weight
so the load factor is one. If an aircraft is manoeuvred such that the CLincreases
without a corresponding decrease in airspeed, the lift force will increase, This
increase in lift without a corresponding increase in aircraft weight constitutes an
increase in load factor. The symbol conventionally used to denote load factor is
11. So increasiug CLat any given airspeed increases n.
ENV 21. b.
The maximum speed at which an aircraft may be routinely flown is limited by
the ability of the structure to support the resulting aerodynamic loads and by the
compressibility effects encountered when flying close to ihe local speed of sound.
At low altitudes the calibrated airspeed producing limiting aerodynamic loads is
the limiting value. This is called VMO. The mach number at which
compressibility effects become the limiting value is termed MMO-As altitude is
increased, the local speed of sound decreases. At the crossover altitude, the
speed equating to MMo is equal to that equating to VMo. At higher altitudes VMo
is greater than Mhqoand a t lower altitudes MMo is greater than VMO. When
descending at constant mach number close to MMO,there is a danger that VMo
will be exceeded.
ENV 22. a.
When climbing a t constant IAS, both TAS and Mach number increase with
increasing altitude. Also because the local speed of sound decreases with
increasing temperature, the IAS and TAS equating to MMo will decrease with
increasing altitude up to 36000 feet. If a constant IAS climb is continued to a
sufficiently high altitude Mhlo will be exceeded.
ENV 23. b.
The V-n envelope illustrates the range of speeds and load factors at which an
aircraft is capable of operating. The gust envelope illustrates the range of speeds
and vertical gusts within which an aircraft is capable of operating. In both
envelopes the airspeed increases from left to right along the horizontal axis. The
left boundary of these envelopes is the low speed stall line
ENV 24. b.
The V-n envelope illustrates the range of speeds and load factors a t which an
aircraft is capable of operating. The gust envelope illustrates the range of speeds
and vertical gusts within which an aircraft is capable of operating. In both
envelopes the airspeed increases from left to right along the horizontal axis. The
right boundary indicates VDwhich is the design dive speed.
ENV 25. b.
The range of speeds at which an aircraft can be operated is limited by the low
and high speed buffet boundaries. As altitude increases these boundaries
converge, such that the available speed range becomes more limited.
Manoeuvring the aircraft increases load factor and so causes the low speed
buffet boundary to move closer to the high speed boundary. The maximum
operating altitude for JAR certificated passenger aircraft is that altitude at
which a 1.3g manoeuvre is just possible without incurring buffet. Exceeding the
maximum operating altitude is therefore likely to cause buffet at lower load
factors. If this altitude is exceeded by a sufficient margin the aerodynamic
ceiling will be reached at which high speed buffet will occur in straight and level
flight.
ENV 26. c.
This question relates to the use of the gust envelope in the design of aircraft. The
purpose of this process is to ensure that when subjected to standard vertical
gusts at VB, VC and VD, the limiting load factors will not be exceeded. The
degree to which load factor is increased by any given combination of vertical
gust and airspeed depends upon the gradient of the CL : a curve. The steeper the
cuwe, the greater the load factor in any given gust. If this exercise reveals that
the standard 50 ftlsec up-gust at Vc produces an excessive load factor, then
either the structure must be strengthened to match the higher loading, or else the
gradient of the CL: a cuwe must be reduced. This gradient can be reduced by
increasing the wing sweep back angle. The other three options offered in this
question are incorrect because they will all increase the CLat any given angle of
attack and so will increase the gust-induced load factor.
ENV 27. c.
Vc is the design cruise speed and VDis the design dive speed. Both of these
speeds are selected by the designers of the aircraft. There are no mandatory
values for either of these speeds, but the margin between the two must be
sufficiently large to ensure that any upset that might reasonably be expected to
occur during cruise flight, will not cause VDto be inadvertently exceeded.
ENV 28. a.
The purpose of minimum limiting load factors is to ensure that aircraft
structures are sufficiently strong to sustain expected flight loads. Requirements
for limiting load factors for large passenger aircraft are specified in JAR25.
For JAR certification of such aircraft, the minimum acceptable value of limiting
load factor for flight with flaps deployed is 2.
ENV 29. d.
The purpose of limiting load factors is to ensure that aircraft structures are
sufficiently strong to sustain expected flight loads. Requirements for limiting
load factors for large passenger aircraft are specified in JAR25. The actual
value of limiting load factor to be applied depends upon aircraft weight and may
be calculated using a method defined in JAR25. This publication also states that
the limiting load factor need not exceed a value of 3.8 for any aircraft weight.
The maximum limiting load factor for JAR certification of passenger aircraft is
therefore 3.8.
ENV 30. d.
JAR 25 states that for certification purposes, buffeting in passenger aircraft
must not be evident a t speeds between 1.2 Vsl, and VMO/MMO.
ENV 31. b.
The lowest possible value for V2 is VZMIN.This is whichever is the higher of
1.15Vs and 1.1 VMC. Extending flaps from the 10' to the 30' setting reduces Vs
and so reduces VMINprovided this would not cause it to be lower than 1.1 VMC.
ENV 32. a.
Although for any given aircraft configuration and load factor, the high speed
buffet boundary changes only with altitude and temperature, the low speed
buffet boundary is sensitive to changes in aircraft weight and C of G position.
The effect of increasing weight is to increase the low speed stalling speed and
hence the buffet boundary. Forward movement of the C of G increases the nose
down pitching moment of the aircraft. To maintain level flight this nose down
moment must be balanced by a down force from the tailplane. For level flight
the sum of upward forces must equal the sum of downward forces, so the lift
produced by the wings must be increased'to equal the sum of the aircraft weight
and tailplane down force. This has the effect of increasing the low speed stalling
speed. The resulting increase in the low speed buffet boundary, causes the buffet
boundaries to converge. The effect of options c and d is to increase the high
speed buffet margin, causing the boundaries to diverge.
ENV 33. d.
Increasing bank angle causes load factor to increase, making the effective weight
of the aircraft greater. This increases stalling speed, thereby causing the buffet
boundaries to converge. It should also be noted that rearward movement of the
C of P has the same effect as forward movement of the C of G described in
answer ENV 32 above.
ENV 34. a.
is the ratio of TAS:LSS at which the airflow a t some point
The mach limit, MCRIT
on the aircraft first reaches the local speed of sound. Because the local speed of
sound decreases with temperature and the ratio of TAS:IAS increases with
decreases with increasing altitude. Climbing
altitude, the IAS equating to MCRIT
at constant IAS will therefore cause TAS to increase until a t some altitude IAS
equals MCRIT.Increasing IAS in the climb increases the rate at which TAS
increases, thereby decreasing the altitude a t which IAS equals MCRI~.
ENV 35. d.
In descending flight in still air the TAS of a n aircraft can be considered to be
made u p of two components. The vertical component of TAS is the rate of
descent and the horizontal component is the ground speed. A headwind will
reduce. the ground speed component but will not affect the rate of descent
component. The duration of the descent will therefore be unchanged, but
because of the lower ground speed, the aircraft will cover less ground in any
given descent.
ENV 36. d.
In descending flight in still air the TAS of a n aircraft can be considered to be
made u p of two components. The vertical component of TAS is'the rate of
descent and the horizontal component is the ground speed. A headwind will
reduce the ground speed component but will not affect the rate of descent
component. The duration of the descent will therefore be unchanged, but
because of the lower ground speed, the aircraft will cover less ground in any
given descent. The combined effect of reduced ground speed and unaffected rate
of descent, is to increase the descent gradient. This is illustrated below.
Increased gradient with headwind
Reduced ground
Ground speed without headwi
Rate of
descent
ENV 37. a.
In straight and level constant speed flight, thrust required equals drag and lift
equals weight. Drag and lift can be calculated using the following formulae:
D = C Q '/2 p v 2 s
and L = C i '/z.pv2s
If weight is reduced whilst maintaining constant speed, the lift must be reduced
to maintain level flight. If this is achieved by reducing speed the effect will also
include a change in the drag force. The fact that the aircraft in this question is
cruising, indicates that its speed is greater than VMD. Decreasing speed will
therefore reduce drag. This in turn will reduce the thrust required to maintain
the lower IAS.
ENV 38. a.
VMCLis the minimum calibrated airspeed a t which it is possible to maintain
control of ankaircraftin flight, in the landing configuration, following the failure
of a critical engine. It is the speed at which the control forces provided by full
deflection of the flying controls is just sufficient to overcome the yawing and
rolling moments caused by the asymmetric power effects of the engine failure.
For any given control deflection the magnitude of the control forces generated
depends upon airspeed, and air density. The magnitude of the moments
generated by asymmetric power depend upon the thrust produced by the engine,
which in turn depends upon air density and power setting. Because none of the
above factors very with aircraft weight, variations in weight do not affect VMCL.
ENV 39. b.
As altitude decreases below 36000 feet, air temperature increases. The local
speed of sound is proportional to the square of the absolute air temperature, so
this also increases with decreasing altitude. When descending a t constant mach
number, the aircraft speed as a proportion of the local speed of sound will
remain constant. Because the local speed of sound is increasing, the airspeed
(IAS and TAS) will increase. Assuming power setting remains constant, this
increase in airspeed can be achieved only by decreasing angle of attack to redute
the drag force acting on the aircraft. Angle of attack must therefore be reduced
in order to carry out a constant mach descent a t constant power. I t should be
noted that if power were increased during the descent the angle of attack could
be maintained at a constant setting. In questiu~lsof this form, it should be
assumed that the aircraft is in gliding (zero thrust) descent unless otherwise
stated.
ENV 40. c.
When an aircraft is in a steady descent, the lift force is equal to the weight of the
aircraft multiplied by the cosine of its angle of descent. In order to maintain a
constant angle of descent it is therefore necessary to maintain a constant lift
force. But lift equals CL :/z ~ V ' Sand IAS is constant at all altitudes for any given
value of !4 pvZ. This means that to maintain constant IAS a n aircraft must
maintain constant % p v Z Assuming that the wing area (S) does not vary, the
maintenance of constant lift at constant IAS requires constant CL. This in turn
requires a constant angle of attack.
.
ENV 41. b.
When an aircraft is in a steady descent, the lift force is equal to the weight of the
aircraft multiplied by the cosine of its angle of descent. In order to maintain a
constant angle of descent it is therefore necessary to maintain a constant lift
force. Lift equals CL '/z ~ V ' Sand for any given value of % p v 2 the IAS is
constant a t all altitudes. But as altitude decreases the TAS : IAS ratio decreases
such that a constant TAS descent results in a continuous increase in IAS and
%pvZ. This increase in IAS would increase drag. This means that Co maintain
coilstant TAS, without varying thrust, the drag must be reduced by decreasing
angle of attack. I t should be noted that angle of attack would be reduced by
pushing the nose down. This constitutes an increase in pitch attitude. So an
increase in (nose down) pitch to decrease angle of attack is required to maintain
constant TAS in a constant thrust descent.
ENV 42. c.
The buffet margins are the speed ranges available between an aircrafts actual
speed and the speeds corresponding to the boundaries at which low speed and
high speed buffet commence. As altitude increases the TAS equating to the low
speed buffet boundary increases, whilst that equating to the high speed buffet
boundary decreases. This causes the two boundaries to converge. The overall
effect is that the margin from any given aircraft speed to the low and high speed
buffet boundaries both decrease as altitude increases.
ENV 43. a.
The mach number of a n aircraft is a measure of its speed as a fraction of the
local speed of sound. A constant mach climb therefore means that the aircraft's
speed as a proportion of the local speed of sound remains constant. But the local
speed of sound is proportional to temperature, which remains constant above the
tropopause. This means that a constant mach climb above the tropopause,
causes TAS to remain constant. IAS is proportional to !A p v Zsuch that any
given value of % p v 2will produce the same value of IAS at all altitudes. But as
altitude increases p decreases, so to maintain a constant value of IAS, the value
of V, which is TAS, must increase. This means that the TAS equating to any
given Iris increases with increasing altitude. Climbing a t constant mach number
above the tropopause therefore causes IAS to decrease and TAS to remain
constant.
ENV 44. a.
The maximum operating altitude of a JAR certificated passenger aircraft is that
altitude ai. which it can just attain a positive load factor of 1.3 before the onset of
buffet. Load factor greater than 1.3 when flying at this altitude will cause
buffeting.
ENV 45. d.
The aerodynamic ceiling is that altitude a t which the low speed and high speed
buffet boundaries cross leaving only a single flight speed available. Any
manoeuvre that increases load factor above one or increases or decreases TAS
will cause buffet.
ENV 46. d.
The power required to maintain level flight at any given airspeed increases with
altitude, whilst the power available from an engine decreases with increasing
altitude: The absolute ceiling is the altitude at which the entire power output of
an aircraft's engines is just sufficient to permit level flight. At this altitude
further climbing is not possible and even level flight is possible a t only one speed.
This speed is VMDfor a jet and VXfor a propeller aircraft. All manoeuvres that
increase the load factor will increase the power required. Any manoeuvres that
do not increase load factor but cause a change in airspeed will also require more
power. All such manoeuvres will therefore cause loss of altitude.
ENV 47. a.
VA is the design manoeuvre speed. It is the EAS at which the aircraft will stall
when subject to the positive limiting load factor when flying in the clean
condition. Although option c is also true, it applied only to the flaps down
condition, whereas option a is correct in all conditions. Option a is therefore the
most appropriate.
ENV 48. d.
VB is the maximum gust intensity speed and is the speed a t which the designers
assume an aircraft will encounter the most severe vertical gusts. Vc is the design
cruise speed and is the speed at which the aircraft is designed to cruise
throughout the majority of its flight time. VDis the design dive speed and is the
maximum speed considered in a normal flight envelope. JAR 25 specifies the
relationship between these speeds as follows:
VB need not be greater than VA.
Vc need not be greater than VB.
VD must be sufficiently greater than Vc to prevent inadvertent exceedence
of VD in any reasonably credible upset.
In the majority of flight envelopes however Vc is more than VB and VB is less
than VD.
ENV 49. b.
Vx is the CAS at which an aircraft will achieve the best angle of climb. I t is the
speed at which excess thrust is maximum. For a propeller aircraft, it is just
above the minimum safe flying speed, and for a jet aircraft it is VMD.Vy is the
CAS at which an aircraft will achieve the best rate of climb. It is the speed at
which excess power is maximum. For a propeller aircraft it is VMP,and for a jet
it is somewhat greater than VMD.Vx is always less than Vy at low altitudes, but
the two converge until meeting a t the absolute ceiling. VX is therefore always less
than or equal to Vy. VMCGis the minimum control speed on the ground in the
take-off configuration. This is lower than both Vx and Vy for all aircraft types.
The best option provided in this question is therefore b.
ENV 50. b.
VMCAis the minimum CAS at which control can be maintained following a
critical engine failure in the air in the take-off configuration. VMCGis the
minimum CAS a t which control can be maintained following a critical engine
failure on the ground in the take-off configuration. VR is the CAS at which an
aircraft is rotated into the flight position during take-off. It must be a t least
110% of VMCA.VMCAis therefore less than Vg which is more than VMCG.
ENV 51. b.
VMUis the minimum CAS at and above which ar. aircraft can safely lift-off the
ground and continue take-off. VMCA
is the minimum CAS a t which control can
be maintained following a critical engine failure in the air in the take-off
configuration. The term VMCA1s being superseded by the term VMC. V2 is the
take-off safety speed. The minimum acceptable value of V2 is termed V2MlN
which must be at least 1.15 Vs or 1.1 VMC,whichever is the greater. Placed in
order of increasing magnitude these speeds are, Vs, VMCG,VMCA,V1, VR, VMU,
VLORVlMlN.The correct equation is therefore b, Vs < V M C<~VzMIN.
ENV 52. b.
The troposphere is that part of the atmosphere below the tropopause at 36000
feet. As altitude increases in the troposphere, temperature falls at a rate of
/ 1000ft. The local speed of sound is proportional to
approximately 2 ' ~
temperature so as altitude increases in the troposphere, the local speed of sound
decreases. The mach number of an aircraft represents its speed as a fraction of
the local speed of sound. The TAS equating to any given mach number therefore
decreases with increasing altitude.
ENV 53. b.
The troposphere is that part of the atmosphere below tbe tropopause a t 36000
feet. As altitude increases in the troposphere, temperature falls a t a rate of
approximately 2 ' 1~1000ft. The local speed of sound is proportional to
temperature so as altitude increases in the troposphere, the local speed of sound
decreases. Above the tropopause temperature and the local speed of sound
remain constant. The mach number of an aircraft represents its speed as a
fraction of the local speed of sound. So a constai~tmach number climb means
climbing a t a constant fraction of the local speed of sound. Climbing at constant
mach number below the tropopause causes TAS to decrease, whilst above the
tropopause it causes TAS to remain constant. The overall effect of a climb is
that TAS decreases up to 36000 feet then remains constant.
ENV 54. a.
JAR25 defines V2 as "the take-off safety speed".
ENV 55. a.
The ability of an aircraft to conduct a steady climb is determined by the amount
of excess power available. Excess power is that proportion of available power
that is not required to balance drag. Excess power is therefore equal to the
power available minus the power required for straight and level constant speed
flight. At all altitudes below the absolute ceiling, the power available and power
required curves cross at two points. These points represent the minimum and
maximum level flight speeds, unless speed is limited by other factors such as low
and high speed stall.
At the speeds a t which the power available and power required curves coincide
there is no excess power available. At the low speed crossing point, acceleration
will reduce the power required thereby enabling the aircraft to climb. At the
high speed crossing point however, speed increases will increase the power
required so acceleration is not possible. The performance an aircraft can achieve
a t the high speed crossover is therefore unaccelerated straight and level flight.
The only method of climbing possible a t this speed is zoom climbing in which
altitude is increased a t the expense of decreasing airspeed. The aircraft can of
course conduct an accelerating descent as this requires less power than straight
and level flight. Of the four options offered in this question only a, "Constant
speed level flight is possible", is true at both points, so this represents the best
answer.
ENV 56. c.
In this question fuel consumption has been defined as Kglnm. This means that
any action that increases the fuel flow required to maintain speed, constitutes an
increase in fuel consumption. If the C of G is ahead of the C of P, then the
resulting nose-down moment must be balanced by a down force provided by the
tailplane. For level flight the sum of all down forces (weight plus tailplane down
force) must be balanced by lift, so any increase in tailplane down force will
increase the magnitude of the lift required. The production of this additional lift
will cause an increase in lift induced drag. In addition to this, the generation of
the tailplane down force will increase trim drag. This increased drag will
necessitate the consumption of more fuel to create additional thrust to balance
drag. Forward movement of the C of G when it is already ahead of the C of P
will therefore increase fuel consumption.
If however the C of G is aft of the C of P, it will produce a nose-up moment that
must be balanced by a tailplane up force. Forward movement of the C of G will
therefore reduce trim drag and fuel consumption. Of the options offered in this
question only c is fully consistent with the effects described above.
ENV 57. a.
JAR 25 states that V1 must not be less than VMCG
and that VRmust not be less
than V1. This implies that the minimum value of V1 is VMCG
and the maximum
value is VR.
ENV 58. c.
To increase airspeed above VMPin straight and level flight, it is necessary to
increase the power output of the engines. The maximum speed that any aircraft
can possibly achieve in straight and level flight is that at which the power
required is equal to the power available. In this condition no excess power is
available, so option d is incorrect. Option b describes the situation existing at the
,
is much lower than maximum speed. The
minimum power speed V M ~which
power available from a propeller aircraft increases with airspeed before
decreasing a t higher speeds. Option a is therefore untrue.
ENV 59. b.
IAS is proportional to !4 pv2, where p is
air density and V is TAS. This means
that for any given value of !4 pv2, the
IAS will be constant a t all altitudes. But
air density decreases with increasing
altitude so for any given IAS, the TAS
must increase to maintain a constant value
of 34 pv2. This means that as altitude
increases the IAS equating to any given
TAS decreases. In a constant TAS climb,
IAS will therefore decrease. This effect is
illustrated in the diagram at the right.
ENV 60. a.
IAS is proportional to 34 pv2, where p is
air density and V is TAS. This means that
for any given value of 34 pv2, the IAS will
be constant at all altitudes. But air density
dec~easeswith increasing altitude so for
any given IAS, the TAS must increase to
maintain a constant value of !4 pv2. This
mcans that as altitude increases the IAS
equating to any given TAS decreases. In a
constant IAS climb, TAS will therefore
decrease. This effect continues even above
the tropopause because the decrease in air
density is caused by decreasing pressure.
This effect is illustrated in the diagram at
the right.
Decreasing
Constant
ALT
Increasing speed --+
Constant
Increasing
ALT
Increasing speed
ENV 61. d.
IAS is proportional to 34 p ~ 2where
,
p is a h density and V is TAS. This means
that for any given value of !4 pv2, the IAS will be constant a t all altitudes. But
air density decreases with increasing altitude so for any given IAS, the TAS must
increase to maintain a constant value of 1/2 pv2. This means that as altitude
increase the IAS equating to any given TAS decreases. In a constant TAS climb,
IAS will therefore decrease. Below the tropopause the reduction in air density
caused by falling air pressure is partly offset by the reduction in air temperature.
Above the tropopause however temperature is constant so the density decreases
at a greater rate. This increased rate of density reduction increases the rate of
IAS decrease. The overall effect of a constant TAS climb is therefore reducing
IAS up to the tropopause, above which IAS reduces at a greater rate.
ENV 62. b.
In addition to the changing ratio between
IAS and TAS, increasing altitude in the
troposphere also causes the local speed
of sound to decrease. This means that
when climbing a t constant IAS, both TAS
and mach number increase. This effect is
illustrated in the diagram a t the right.
Constant .Increasing
IAS
ALT
Increasing values -+
ENV 63. a.
The maximum operating speed of a modern sub-sonic jet aircraft is limited by
the effects of air loads on the structure and the buffeting and high speed stall
that occur at transonic speeds. These limiting conditions are termed Vhlo and
MMo. Vhlo is the maximum allowable CAS. Mxlo is the limiting mach number.
The air loads acting on an aircraft are proportional to % pV2 where p is air
density and V is TAS. Because air density reduces with increasing altitude the
air loads at any given CAS are greatest at low altitudes. Vklo is therefore the
limiting factor a t low altitudes.
ENV 64. b.
Crossover altitude
The maximum operating speed of a modern
sub-sonic jet aircraft is limited by the effects
of air loads on the structure and the
buffeting and high speed stall that occur at
transonic speeds. These limiting conditions
are termed VMo and Mhio. VMo is the
maximum allowable CAS. MMOis the
limiting mach number. Because the local
speed of sound decreases with increasing
ALT
altitude, climbing a t a constant mach number
results in CAS decreasing. The CAS equating
to MMo therefore decreases with increasing '
altitude. At a certain altitude termed the
crossover altitude VMOand MMo represent
the same value of CAS. At lower altitudes
VMo is the limiting factor and a t higher
altitudes Mhlo is the limiting factor. The
CAS
overall effect is illustrated in the diagram at
the right.
ENV 65. c.
The maximum operating speed of a modern sub-sonic jet aircraft is limited by
the effects of air loads due to dynamic pressures acting on the structure, and the
buffeting and high speed stall that occur at transonic speeds. These limiting
conditions are termed V310 and MMO. VMOis the maximum allowable CAS.
M,lo is the limiting mach number. At any given altitude the limiting condition is
the lower of Vnlo and ATMo. Mhlo is the limiting factor at high altitude where the
speed of sound is lowest. Vklo is the limiting factor at low altitude where air
density and hence dynamic pressure are highest. If a descent is conducted at a
constant mach number close to MMOthere is a danger that VMo will be exceeded
as altitude decreases.
ENV 66. b.
The various parameters that limit the value of VR are detailed in JAR25.107. Of
the options offered in this question only option b, 105% of VMc is included in the
JAR.
ENV 67. a.
VMDis the minimum drag IAS and for any given aircraft and configuration it is a
fixed multiple of Vs. The effect of a mass increase on VMDcan therefore be
calculated by calculating the effect on Vs. This problem can be solved using the
following equation:
New Vs
= Previous Vs d(new mass / previous mass)
Although neither the previous mass nor the new mass are known, it is known
that the new mass is 1.1 times the previous mass. This ratio can be put into the
equation to give the following:
New Vs
=
Previous Vs d(l.1 / 1)
which is
1.05 Previous VS.
The VMuafter the mass increase is therefore 5% greater than that before the
mass increase so VMDhas increased by 5%.
The increase in mass will necessitate an increase in lift in any given flight
condition and this in turn will cause an increase in lift induced drag.
If the speed is increased to the new VMD,then the CL will be unchanged. The
new induced drag will therefore be proportional to I/EAS', which is 1/1.05'.
That is induced drag will decrease by 10%. If the speed remains unchanged then
CL must increases by 10%. But CDIis proportional to CL', SO a 10% increase in
CL from 1 to 1.1 will give a 1.1' increase in CD1.This will give a 21% increase in
induced drag. Neither of thee effects are predicted in the options offered.
A 10% increase in lift a t constant speed would not have any significant effect on
parasite drag, whereas a 5% increase in speed would increase profile drag by a
factor of 1.05" which is a 100h increase. So if speed does not change, then option
a is correct, VMDwill increase by 5% but parasite drag will not change.
ENV 68. a.
VMDis the minimum drag IAS and-for any given aircraft and configuration it is a
fixed multiple of Vs. The effect of a mass increase on VRIDcan therefore be
calculated by calculating the effect on Vs, This problem can be solved using the
following equation:
New Vs
= Previous Vs d(new mass / previous mass)
Although neither the previous mass nor the new mass are known it is known that
the new mass is 0.8 times the previous mass. This ratio can be put into the
equation to give the following:
New Vs
=
Previous Vs d(0.8 / 1)
which is 0.9 Previous Vs.
The VMDafter the mass decrease is therefore 10% less than that before the mass
decrease, so VMDhas decreased by 10%.
The increase in mass will necessitate an increase in lift in any given flight
condition and this will in turn cause an increase in lift induced drag. Such
increases in lift do not however have any significant effect on profile drag. The
correct answer is therefore option a, VMDwill decrease by 10% but parasite drag
will not change.
ENV 69. d.
VMDis the speed at which total drag is
a t a minimum value and the L:D ratio is
at a maximum value. Maximum range
in a jet aircraft is achieved when flying
at the speed providing the best EAS:D
ratio. This occurs at the speed at which
a tangent drawn from the origin touches
the drag curve as indicated in the
diagram at the right. This is higher than
VhlDand so flying at this speed requires
a lower angle of attack than that
providing best L:D ratio at VMD.
DRA(
EAS
VMD
i
Best range kpeed
-
ENV 70. a.
VMpis the speed at which the power
required is at its minimum value.
When flying at VMPany increase or
decrease in speed will cause an
increase in power required. VMDis
the speed at which the drag force is
at a minimum value. Any speed
change that takes an aircraft closer
to VnlDwill cause drag to decrease.
Changing speed from VMPto VMD
will therefore decrease drag and
increase power required. This
effect is illustrated in the diagram
at the right.
Decrease in Drag
Power required
VMP1
I
Increase in power rkquired
EAS
Speed change
ENV 71. a.
VMCCis the minimum speed at which control can be maintained on the ground in
the take-off configuration, following a critical engine failure. V1 is the speed at
which the pilot must decide to continue or abort the take-off. In order for him to
be able to make this decision the aircraft must have exceeded its VMCG.V1 must
therefore be greater than VMcc. VRis the speed a t which the aircraft must be
rotated into the take-off attitude, after the decision has been made to continue
the take-off. VR must therefore be greater than V,. V2 is the take-off safety
speed. This must be at least VR plus the speed increment achieved before
reaching a height of 35 feet above the take-off surface. The correct sequence is
therefore V M C ~VL,
, VR, V2.
ENV 72. c.
Vx is the CAS at which an aircraft is capable of achieving its best angle of climb.
It is the speed at which excess thrust is maximum. Vy is the CAS at which an
aircraft is capable of achieving its best rate of climb. I t is the speed at which
excess power is maximum. At low altitudes Vx is less than Vy for all aircraft
types. AS altitude increases however the Vx remains constant whilst Vy
decreases until the two coincide at the absolute ceiling. Vx is therefore always
less than or equal to Vy. It should be noted that although the above description
is correct, in JAR ATPL Performance examinations, Vx and Vy are both
assumed to be unaffected by altitude. This is because the rate of change of Vy
with increasing altitude is very small, and Vx and Vy are of significance only
during climb out at low altitudes. Because this question concerned the relative
magnitudes of Vx and Vy, it was more appropriate to use the true effects of
altitude.
ENV 73. b.
Vs is the stalling speed and is the lowest speed at which an aircraft can produce
lift equal to its weight. Vx is the CAS at which an aircraft is capable of achieving
its best angle of climb. It is always greater than Vs. Vy is the CAS at which an
aircraft is capable of achieving its best rate of climb. It is always more than or
equal to Vx. The correct sequence is therefore Vs, Vx, Vy.
ENV 74. a.
The climb gradient is the ratio of height gained to distance travelled over the
ground. It is usually expressed as a percentage. In calculating climb gradient it
is convenient to use the rate of climb (ROC) as a measure of height gained, and
TAS as a measure of distance travelled. This method is reasonably accurate in
still air for gradients up to 15%. To provided an accurate calculation of
gradient, both ROC and TAS must be in the same units, such as ft/min or dm.
An approximate value of gradient is often used based on the assumption that 1
Kts equals 100 ft/min, and the following equation:
Gradient = (TAS / ROC)%
ENV 75. d.
Both IAS and drag are proportional to % p ~ so2 drag at any given IAS is
constant at all altitudes. Vx is the speed at which excess thrust is maximum and
excess thrust is thrust available minus drag. Although thrust available decreases
with increasing altitude, the relationship between thrust and IAS does not vary.
The IAS value of Vx is therefore constant at all altitudes. Vy is the speed at
which excess power is maximum. Excess power is power available minus power
required and power required is drag multiplied by TAS. But the TAS equating
to any given IAS increases with altitude, causing the power required at any given
IAS to increase. This causes the IAS value Vy to decrease with increasing
altitude.
ENV 76. b.
The local speed of sound decreases with decreasing air temperature, as altitude
increases up to 36000 feet. Climbing at constant mach number therefore causes
IAS to decrease. But lift is proportional to IAS, so if weight remains constant in
a constant mach climb, the angle of attack must be increased to maintain the
balance between lift and weight.
ENV 77. a.
VMCCis the lowest CAS at which it is possible to maintain control in the take-off
configuration, on o r close to the ground following failure of a critical engine.
It is the minimum CAS at which control authority equals the yawing and rolling
moments caused by asymmetric thrust. The control authority is constant at any
given CAS, but the asymmetric thrust moments are proportional to engine
thrust. Any factor that reduces thrust will therefore reduce VMCC.Such factors
include high ambient temperature, humidity and pressure altitude.
ENV 78. a.
The relationshipi between CAS, TAS and Mach number vary with altitude. Also
the stalling speed and buffet onset speeds vary with aircraft weight and C of G
position. The buffet boundary chart indicates the relationship between altitude,
mass, C of G position and the low and high speed stall mach numbers.
ENV 79. b.
At low altitudes the margin between low speed buffet and high speed buffet is
wide, so the ability of an aircraft to manoeuvre is limited primarily by its
structural strength. At higher altitudes the low speed buffet boundary increases
and the high speed buffet boundary decreases. The convergence of these two
boundaries causes the manoeuvrability of an aircraft to be degraded as altitude
increases. At the maximum operating altitude a JAR certificated passenger
aircraft can achieve only 1.3g without incurring buffet.
ENV 80. b.
The stability of a n aircraft depends to a considerable extent upon the magnitude
of the aerodynamic damping forces caused by changes of attitude. As altitude
increases, these damping effects decrease causing stability to be degraded.
ENV 81. a.
VMDis the speed a t which total drag is
at a minimum value and the L:D ratio is
at a maximum value. Maximum range
in a propeller aircraft is achieved at the
speed providing best EAS:Power Required
ratio. This occurs at the speed at which
a tangent drawn from the origin touches
the power required curve, as indicated in
the diagram a t the right. This is a t VMD,
so flying at this speed requires the angle
of attack that gives best L:D ratio.
EAS
Best range speed = VMD
ENV 82. a.
At the absolute ceiling there is only one speed a t which an aircraft can maintain
straight and level flight. For both jet and propeller aircraft this speed is Vx.
Also for both aircraft types, the CAS value of Vx and V n l remain
~
constant at all
altitudes. In the case of jet aircraft Vx is VMD.For propeller aircraft Vx is less
than VMDand because the two remain constant with changes in attitude, they do
not converge a t the absolute ceiling. At their absolute ceilings jet aircraft must
fly at VMDand propeller aircraft must fly at Vx in order to maintain straight and
level flight. I t should be noted that this means that a propeller aircraft cannot fly
at VMDa t its absolute ceiling. This is because it would have insufficient power
available to do so.
ENV 83. c.
At the absolute ceiling there is only one speed at which an aircraft can maintain
straight and level flight. For both jet and propeller aircraft this speed is Vx.
Also for both aircraft types, the CAS value of Vx and VhlDremain constant a t all
altitudes. For propeller aircraft Vx is less than VMDand because the two remain
constant with changes in attitude, they do not converge at the absolute ceiling.
But because VXis the only practicable flight speed a t the absolute ceiling, VMm,
V ~ a xand Vy all converge on Vx at this altitude whilst VMDdoes not.
ENV 84. c.
At the absolute ceiling the only speed at which a propeller aircraft can maintain
straight and level flight is Vx. At lower altitudes Vx is the speed at which excess
thrust is a t a maximun~value, thereby facilitating the greatest climb gradient.
As altitude increases, reducing air density decreases the power produced by the
engine and the thrust produced by the propeller. This causes excess thrust and
maximum climb gradient to decrease with increasing altitude. At the absolute
ceiling excess thrust is zero, so no climb gradient is possible. But increasing
altitude does not affect the CAS a t which maximum excess thrust occurs, so Vx
remains constant a t all altitudes.
VMDis the speed at which the drag force is a t its minimum value. Because both
drag and CAS are proportional to dynamic pressure (1/2p~*),the CAS value of
VMDremains constant a t all altitudes. So for a propeller aircraft both Vx and
VMDremain constant as it climbs to its absolute ceiling.
ENV 85. a.
At the absolute ceiling the only speed at which a propeller aircraft can maintain
straight and level flight is Vx. At lower altitudes Vx is the speed at which excess
thrust is a t a maximum value, thereby facilitating the greatest climb gradient.
Vy is the speed at which excess power is maximum, facilitating maximum rate of
climb. As altitude increases, reducing air density decreases the power produced
by the engine and the thrust produced by the propeller, causing excess power,
excess thrust, maximum climb gradient, and maximum rate of climb, to reduce.
At the absolute ceiling both excess thrust and excess power are zero, so further
climbing is impossible. So at the absolute ceiling thrust available is equal to drag
and power available is equal to power required. This situation is illustrated in
the diagrams below and overleaf.
Drag is constant at all altitudes
Thrust available at sea level
Reduced thrust available at
The absolute ceiling
Thrust is equal to drag at Vx
at the absolute ceiling
Drag is greater than thrust at
all speeds other than Vx.
1
vx
CAS
Power required at absolute
Power available at sea level
Power required at sea level
Reduced power available at
The absolute ceiling
ENV 86. a.
Climb gradient is the ratio of height gained divided by distance flown over the
ground. It is normally expressed as a % gradient, where 100% represents a
vertical climb and 0% represents level flight. For small gradient (up to 15%) in
still air, a reasonably accurate estimate can be made using the following
equation:
ENV 87. b.
Maximum endurance speed in a propeller driven aircraft is VMp. But this is
lower than V M ~and
, reducing speed requires increasing CL, SO flight at VMP
requires a greater CL than that at V M ~CL
. at VMDis that giving best L:D ratio,
so best endurance speed in a propeller aircraft requires a CLgreater than that
for best L:D ratio.
ENV 88. a.
Maximum endurance speed in a jet aircraft is VhlD. But VMDis the speed a t
which the L:D ratio is maximum. Maximum jet endurance therefore requires
the CL giving best L:D ratio.
ENV 89. a.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
x
A), which occurs at
First is C L M ~(point
CD
the low speed stall (VS). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is poii~tD.
ENV 90. d.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
First is CLhlAX
(point A), which occurs at
CD
the low speed stall (Vs). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. The possible identities of
all other points can then be identified on
the basis of their positions relative to Vs
CL
and VhlD. TOidentify other points it should first be noted that Vs (point A) is
lower than VMD(point D), so the speed scale must increase from left to right.
Point E is @erefore at a speed greater than VMD.Of the options provided in this
question, only d, VYJet,is greater than VMD,SO point e must be VYJet.
ENV 91. d.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
(point A), which occurs at
First is CLMAX
CD
the low speed stall (VS). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. The possible identities of
all other points can then be identified on
the basis of their positions relative to Vs
and VMD.
To identify other points it should first be noted that Vs (point A) is lower than
VMD(point D), so the speed scale must increase from left to right. Point C is
therefore at a speed greater than Vs but lower than VMD. Of the options
~
and d (VYPrOp),
all satisfy this
provided in this question, b ( V M P ) ,(VXProp),
which is less than VYProp,and only two
condition. But VXPropis less than vMP,
points, B and C are indicated in the diagram, in this part of the speed range. So
if only the listed options are valid, and if point B is VXpr,,, then point C must be
VMpo r Vypro,. And if point B is VMp,then point C must be VYProp.For most
aircraft VRIPis closer to Vs than to VhfD,and VXpr,, is very close to Vs.
So point B is probably VMP,in which case point C must be VYprop.
ENV 92. a.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
greatest value of CLoccurs at point A. In
all aircraft the greatest value of CL is
C L ~ , , , ~which
,
occurs at the stalling speed
Vs. Point A is therefore Vs.
Cv
ENV 93. b.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
First is CLMAX
(point A), which occurs at
CD
the low speed stall (Vs). The second is V M ~ ,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. The possible identities of
all other points can then be identified on
the basis of their positions relative to Vs
and VMD.
To identify other points it should first be noted that Vs (point A) is lower than
VMD(point D), so the speed scale must increase from left to right. Point C is
therefore at a speed greater than Vs but lower than VMD. Of the options
provided in this question, only b (VMP),satisfies this condition.
ENV 94. b.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
(point A), which occurs at
First is CLMAX
CD
the low speed stall (VS). The second is VklD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. The possible identities of
all other points can then be identified on
the basis of their positions relative to Vs
and VhlD.
To identify other points it should first be noted that Vs (point A) is lower than
VMD(point D), so the speed scale must increase from left to right. The question
specifies that point c is VMP,SO point B is therefore at a speed greater than Vs but
lower than VMP. Of the options provided in this question, only b (VXProp),
satisfies this condition.
ENV 95. b.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
(point A), which occurs at
First is CLhlAX
CD
the low speed stall (VS). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point n. But VMDis also prop
best range speed.
ENV 96. d.
The diagram in this question is a whole
aircraft L:D polar. I t is similar to those in
the previous questions, but in this case the
CL
axes have been swapped over. I t should be
noted that this format is commonly used in
ATPL Performance examinations, whereas
Previous format is used in POF examinations.
The two diagrams are however essentially
the same thing and are used in the same
manner.
(point A), which
Two points can be accurately located. The First is CLMAX
occurs at the low speed stall (Vs). The second is VMD,which occurs where a
straight line, drawn from the origin, just touches the curve. This is point D. But
VMDis also the best range glide speed for all aircraft types.
ENV 97. d.
The diagram in this question is a whole
aircraft L:D polar. I t is similar to those in
the previous questions, but in this case the
CL
axes have been swapped over. I t should be
noted that this format is commonly used in
ATPL Performance examinations, whereas
Previous format is used in POF examinations.
The two diagrams are however essentially
the same thing and are used in the same
manner.
L(point
~ A),~which
~
~
Two points can be accurately located. The First is C
occurs at the low speed stall (Vs). The second is Vm, which occurs where a
straight line, drawn from the origin, just touches the curve. This is point D. But
VhlDis also the best climb angle speed for jet aircraft.
ENV 98. d.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
First is CLMAX
(point A), which occurs at
CL
the low speed stall (Vs). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D.
ENV 99. b.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
First is CLMAX
(point A), which occurs at
CL
the low speed stall (Vs). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. But VMDis also the best
endurance speed for jet aircraft.
CD
ENV 100. a.
The diagram is a whole aircraft L:D polar.
Two points can be accurately located. The
first is CLMAX
(point A), which occurs at
CL
the low speed stall (Vs). The second is VMD,
which occurs where a straight line, drawn
from the origin, just touches the curve.
This is point D. Point B is therefore at a
speed greater than Vs but less than VMD.
Of the options listed in this question, a
CD
(Prop best climb angle speed), b ( Prop
best climb speed), and c (VMP),satisfy this condition. Of these three options, the
prop best climb angle speed is closest to Vs. This is therefore the most probable
identity of point B.
ENV 101. a.
The diagram shows the relationship between
Power required and TAS. Curve number 2
is higher and to the right of curve number 1.
Power required in straight and level flight is
E
equal to drag multiplied by TAS. At any
IAS, the drag force is constant at all altitudes.
But the TAS equating to any given IAS
increases with increasing altitude. This means
that as altitude increases, at any given IAS, the
constant drag is multiplied by an increasingly
TAS
high TAS, thereby causing power required to increase. Although an increase in
weight (option b) also increases power required, this results in a higher power
required a t all speeds. This means that the power required curves do not cross if
the change is caused by a weight increase, but they do if the change is caused by
an altitude increase. This diagram can therefore only represent the effect of
increasing altitude.
ENV 102. a.
This diagram illustrates the relationship
between drag and TAS. Curve number 2
is a repeat of curve number 1, but is further
to the right. This means that the drag
values in curve 1 also occur in curve 2, but
a t a higher TAS. Both drag and IAS are
~ ~ that
, any given
proportional to 1 1 2 ~such
1 1 2 ~will
~ ~give
, the same drag and IAS a t all
altitudes. But as altitude increases, reducing
TAS
p causes V (which is TAS), equating to any
given IAS to increase. This means that as altitude increases the same drag force
occurs a t a higher TAS, as indicated in the diagram. The change from curve 1 to
curve 2 was therefore caused by an increase in altitude.
ENV 103. b.
The diagram il!ustrates the relationship
between power required and TAS, in two
different conditions. Curve number 2 is
1
narrower than curve number 1, and its
B
lowest point, VMP,is a t a higher TAS.
Because curve 2 is n a r r o w e t available
is reduced in condition 2. Also because the
curves do not cross, curve 2 is higher than
curve 1 at all points. This means that in
TAS
the curve 2 condition, the aircraft requires
greater power a t all speeds. This combination of higher power requirements a t
all speeds, increased VMD,and narrower speed range, indicates that the weight of
the aircraft has been increased. Although an increase in altitude would also
increase VMD,it would cause the curves to cross, such that power required a t
some very high speeds would decrease and the TAS range would increase.
ENV 104. b.
This diagram illustrates the relationship
between d r a g and EAS in two conditions.
Curve 2 is higher, narrower, and further
To the right than curve 1. This mean that
in condition 2, the aircraft has a narrower
speed range, higher drag a t all speeds,
and a higher VMD,than in condition 1.
All of these effects are caused by an increase
in weight.
D
EAS
ENV 105. d.
This diagram illustrates the relationship
between drag and EAS in two conditions.
In condition 2 the curve is higher, narrower,
D
and further to the left, than in condition 1.
This means that in condition 2 the aircraft
has a narrower speed range, higher drag a t
speeds, and lower VbqDthan in condition 1.
These effects are all caused by gear
EAS
Deployment. This increases profile drag,
pushing the drag curve upwards and to the
left. The extra profile drag is greatest a t high speed so VMDand the available
speed range are reduced. Although flap deployment would also increase drag
and decrease speed range, it would push the curve to the right.