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Groundschool Theory of Flight
Aircraftmanoeuvring
forces
Revision 72 page content was
last changed 15 August 2012
The performance of an aircraft in the hands of a competent pilot at
a given altitude results from the sum of power, angle of attack,
attitude and configuration. Power provides thrust and consequently
contributes to acceleration, airspeed, lift, drag and radius of turn. The
angle of attack dictates the dimensions of the lift force and the
induced drag and contributes to airspeed; also the angle of attack is a
significant contributor to the aircraft attitude. Attitude is the angle the
aircraft longitudinal axis subtends above or below the horizon(usually called the 'pitch' which has another meaning associated with
propellers) plus the angle of bank and the degree ofslip. Attitude
dictates the direction of the lift, thrust and drag vectors and,
consequently, converts power into velocities and accelerations in the
three planes. Configuration relates to the deployment of lift/drag
changing devices.
Airspeed is dependent on power, angle of attack, configuration and
attitude under any given set of conditions and attitude in flight
is readily checked by reference to the horizon. The lift, thrust anddrag forces produce manoeuvring loads on the aircraft structure,
generally expressed in terms of 'g', that must be kept within defined
limits. There is a fourth performance factor energy management
which is an art that supplements attitude plus power plus height to
produce maximum aircraft performance. The epitome of such an art is
demonstrated by air-show pilots who produce extraordinary
performances from otherwise relatively mundane aircraft.
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Content
1.7 Cruise performance
Cruise speed optionsThe power required curvePower availableSpeed, power and altitudePower required vs power available
1.8 Forces in a climb
1.9 Forces in a descent
1.10 Turning forcesCentripetal forceTurn forces and bank angleManoeuvring loads
Increasing the lift force in a turnWing loading W/S
1.11 Limiting loads and ultimate loadsManoeuvring loads and gust induced loads
Airworthiness certification categories
1.12 Conserving aircraft energyEnergy available
Kinetic energy measurementMomentum conversion
Abridged trigonometrical tables
Things that are handy to know
1.7 Cruise performance
Cruise speed options
When an aircraft is cruising, flying from point A to point B, the pilot hasseveral options for setting cruise speed:
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One choice might be to get there as soon as possible, in whichcase the pilot would operate the engine at the maximumcontinuous power allowed by the engine designer. Therecommended maximum continuous power is usually around 75%
of the rated powerof the engine and provides performance
cruise.
Another choice might be to get there using as little fuel as possiblebut in a reasonable time, in which case the pilot might choose a
55% power setting to provide an economy cruise airspeed. Or thepilot might choose any power setting, in the usual engine designrange, between 55% and 75%; refer to cruise speeds in the'Airspeed and properties of air' module.
The power required curve
In level flight at constant speed thrust power is required to balanceinduced and parasite drag. Power is the rate of doing work, so power (inwatts) is force (newtons) distance (metres) / time (seconds).Distance/time is speed so power required is drag force (N) aircraft speed(m/s). Thus, if we use the expression for total drag from section 1.6 andmultiply it by V we get:
(Equation #1.3) Power required for level flight [watts] = CD rV S(note V cubed).
The total drag curve can beconverted into a 'power required'
diagram usually called the power
curve if you know the total dragat each airspeed between theminimum controllable speed and themaximum level flight speed. It is adifferent curve from that for totaldrag, because the power required isproportional to speed cubed ratherthan speed squared. This meansthat (ignoring the related CD change)if speed is doubled, drag is
increased four-fold but power must be increased eight times whichindicates why increasing power output from, say, 75% power to full ratedpower, while holding level fl ight, doesn't provide a corresponding increasein airspeed.
The diagram above is a typical level-flight power curve for a light aircraft.
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The part of the curve to the left of the minimum power airspeed is known
as the back of the power curve where the slower you want to fly, themore power is needed, because ofinduced drag at a high angle of attack.The lowest possible speed for controlled flight is the stall speed, which isdiscussed in the 'Airspeed and properties of air' module. Two aerodynamiccruise speeds are indicated the speed associated with minimum drag
(the point on the curve where the drag force factor has the lowest value)and the speed associated with minimum power (the point on the curvewhere drag force speed has the lowest value). To maintain level flight atspeeds less than or greater than the minimum power airspeed, powermust be increased.
Power available
The engine provides power to the propeller. The propellers used in mostlight aircraft have a maximum efficiency factor, in the conversion of engine
power to thrust power, of no more than 80%. (Thrust power= thrust forward speed.) The pitch of the blades, the speed of rotation of thepropeller and the forward speed of the aircraft all establish the angle ofattack of the blades and the thrust delivered. The in-flight pitch of ultralight
and light aircraft propeller blades is usually fixed (though many such types
are adjustable on the ground) so that the maximum efficiency will occur atone combination of rpm and forward speed this is usually in themid-range between best rate of climb and the performance cruiseairspeeds. Propeller blades are sometimes pitched to give the bestefficiency near the best rate of climb speed (climb prop), or pitched forbest efficiency at the performance cruise airspeed (cruise prop). The
efficiency of all types of propellers falls off either side of the optimum; onewith a too high pitch angle may have a very poor take-off performance,while one with a too low pitch may allow the engine to overspeed at anytime.
With the advent of higher-powered four-stroke light engines, such as theJabiru 3300, there has been a corresponding increase in the availability ofmore advanced light-weight propeller systems, providing maximumeffective power utilisation during all stages of flight. For more informationrefer to the 'Engine and propeller performance' module.
Speed, power and altitude
At sea-level, an aero-engine willdeliver its rated power provided itis in near-perfect ex-factorycondition, properly warmed up andusing fuel in appropriate condition.
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However, because air densitydecreases with increasing altitude,and an engine's performancedepends on the weight of the chargedelivered to the cylinders, then thefull throttle power of a
non-supercharged four-strokeengine will decrease with height. So,at about 60007000 feet above meansea-level the maximum poweravailable at full throttle may drop
below 75% of rated power. At 12 000 feet full throttle power may be lessthan 55% of rated power. Thus, as altitude increases, the range of cruisepower airspeeds decreases. For best engine performance, select a cruisealtitude where the throttle is fully open and the engine is delivering 65% to75% power.
A couple of points to note from the speed-power diagram above:
As air density, and consequently drag, decreases with height, thenairspeed, from a particular power level, will increase with height;e.g. the airspeed attained with 65% power at sea-level is 90 knotsincreasing to 100 knots at 10 000 feet.
At sea-level, an increase in power from 75% to 100% only resultsin an increase in airspeed from 100 to 110 knots. This is the normwith most light aircraft that last 33% power increase to ratedpower only provides a 10% increase in airspeed.
Power required vs power available
In the 'power available' diagram atleft, power available curves havebeen added to the earlier 'powerrequired' diagram. The dashed redcurve indicates the rated power
that is, the full throttle engine powerdelivered to the propeller over therange of level flight speeds atsea-level. The upper green curve maximum thrust power, is thatengine power converted by thepropeller after allowing for 80%
maximum propeller efficiency. The lower green curve is the propeller thrustpower available with the engine throttled back to 75% power at sea-level,
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or if flying at an altitude such that full throttle opening will only deliver 75%of rated power. The intersection of those power available curves with thepower required curve indicates the maximum cruise speed in eachcondition.
The region between the maximum thrust power curve and the power
required (to maintain level flight) curve indicates the excess poweravailable at various cruise speeds this excess power is available forvarious manoeuvres if the throttle is fully opened. The simplest use wouldbe a straight unaccelerated climb, in which case the maximum rate ofclimb would be achieved at the airspeed where the two curves are furthestapart. It can be seen that the best rate of climb speed is around the sameairspeed as the minimum drag airspeed shown in the earlier poweredrequired diagram.
The rate of climb will decrease at any speed either side of the best rate ofclimb speed because the power available for climb decreases. The rate of
climb (metres/second) = excess power available (watts)/aircraft weight (N).
For example, lets assume the preceding diagram is representative ofan aircraft fitted with a 100 hp engine, and at the best rate of climbspeed the engine/propeller has 25 hp (18 600 watts) of excess thrustpower available. The aircraft weight is 4000 N so the rate of climb = 18600/4000 = 4.65 m/s. To convert metres/second to feet/minute, multiplyby 200 = 930 feet/minute as the maximum rate of climb.
One thing to bear in mind is that we have assumed the aircraft'saerodynamic shape its configuration is constant. However if theaircraft is fitted with flaps, high lift devices or spoilers the pilot is able to
change its configuration and consequently its performance. Thus
performance is dependent on power, plus attitude (pitch, bank,
sideslip and aoa) plus configuration.
1.8 Forces in a climb
When cruising, the difference between the current power requirement andpower available the excess power can be used to accelerate theaircraft or climb, to accelerate and climb, or perform any manoeuvre thatrequires additional power. For instance if the aircraft has potential poweravailable and the pilot opens the throttle, the thrust will exceed drag andthe pilot can utilise that extra thrust to accelerate to a higher speed whilemaintaining level flight. Alternatively the pilot can opt to maintain the
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existing speed but use the extra thrust to climb to a higher altitude. The
rate of climb (altitude gained per minute) depends on the amount ofavailable power utilised for climbing, which depends in part on theairspeed chosen for the climb. There are other choices than the best rateof climb speed available for the climb for example, the best angle ofclimb speed (which is around the same as the speed for minimum power)
or a combination enroute cruise/climb speed. The climb speed chosendepends on terrain, weather, cloud cover and other operating variables.
If an aircraft is maintained in a continuous full-throttle climb, at the bestrate of climb airspeed, the rate of climb will be highest at sea-level; it willdecrease with altitude, as engine power decreases. The aircraft willeventually arrive at an altitude where there is no excess power available forclimb, then all the available power is needed to balance the drag in levelflight and there will be only one airspeed at which level flight can bemaintained. Below this airspeed the aircraft will stall. This altitude is the
aircraft's absolute ceiling. However, unless trying for an altitude record,
there is no point in attempting to climb to the absolute ceiling so theaircraft's service ceiling should appear in the aircraft's performancespecification. The service ceiling is the altitude at which the rate of climbfalls below 100 feet per minute; this is generally considered the minimumuseful rate of climb.
This diagram of forces in a climb andthe subsequent mathematicalexpressions, have been simplified,aligning the angle of climb with theline of thrust. In fact the line of thrust
will usually be 4 to 10 greater thanthe climb angle. The climb angle (c)is the angle between the flight pathand the horizontal plane.
The relationships in the triangle offorces shown are:Lift = weight cosine cThrust = drag + (weight sine c)
In a constant climb the forces are again in equilibrium, but now thrust + lift
= drag + weight.
Probably the most surprising thing about the triangle of forces in a straightclimb is that lift is less than weight. For example, let's put the Jabiru into a
10 climb with weight = 4000 N. (There is an abridged trig. table at the end
of this page.)
Then, Lift = W cos c = 4000 0.985 = 3940 N
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It is power that provides a continuous rate of climb, but momentum mayalso be used to temporarily provide energy for climbing; see 'Conservingaircraft energy' below. It is evident from the above that in a steady climb,the rate of climb (and descent) is controlled with power, and the airspeedand angle of climb is controlled with the attitude and particularly theincluded angle of attack. This is somewhat of a simplification, as the pilot
employs both power and attitude in unison to achieve a particular angleand rate of climb or descent.
The angle of attack in a climb is the pitch attitude minus the angle of climbbeing achieved plus the wing incidence.
A very important consideration, particularly when manoeuvring at low levelat normal speeds, is that the steeper the climb angle the more thrust isrequired to counter weight. For example, if you pulled the Jabiru up into a30 'zoom' climb the thrust required = drag + weight sine 30 (= 0.5) sothe engine has to provide sufficient thrust to pull up half the weight plus
overcome the increased drag due to the increased aoa in the climb.Clearly, this is not possible, so the airspeed will fall off very rapidly and willlead to a dangerous situation if the pilot is slow in getting the nose down toan achievable attitude. Never be tempted to indulge in zoom climbs they are killers at low levels.
1.9 Forces in a descent
If an aircraft is cruising at, for instance, the maximum 75% power speedand the pilot reduces the throttle to 65% power, the drag now exceedsthrust and the pilot has two options maintain height, allowing theexcess drag to slow the aircraft to the level flight speed appropriate to 65%power; or maintain the existing speed and allow the aircraft to enter a
steady descent or sink. The rate of sink (a negative rate of climb, oraltitude lost per minute) depends on the difference between the 75%power required for level flight at that airspeed and the 65% power utilised.
This sink rate will remain constant as long as the thrust plus weight, which
are together acting forward and downward, are exactly balanced by the lift
plus drag, which are together acting upward and rearward. At a constantairspeed, the sink rate and the angle of descent will vary if thrust is varied.For example, if the pilot increased thrust but maintained constantairspeed, the rate of sink will decrease even becoming positive; i.e. arate of climb.
If the pilot pushed forward on the control column to a much steeper angleof descent, while maintaining the same throttle opening, the thrust plusweight resultant vector becomes greater, the aircraft accelerates with
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consequent increase in thrust power and the acceleration continues untilthe forces are again in equilibrium. Actually, it is difficult to hold a stableaircraft in such a fixed angle 'power dive' as the aircraft will want to climb but an unstable aircraft might want to 'tuck under'; i.e. increase theangle of dive, even past the vertical. We discuss the need for stability inthe 'Stability' module.
When the pilot closes the throttle completely,there is no thrust, the aircraft enters a glidingdescent and the forces are then as shown inthe diagram on the left. In the case ofdescent at a constant rate, the weight isexactly balanced by the resultant force of liftand drag.
From the dashed parallelogram of forcesshown, it can be seen that the tangent of the
angle of glide equals drag/lift. For example,assuming a glide angle of 10 (from theabridged trigonometrical table below, thetangent of 10 is 0.176), the ratio of drag/liftin this case is then 1:5.7 (1/0.176 =5.7).
Conversely, we can say that the angle of glide depends on the ratio oflift/drag [L/D]. The higher that ratio is, then the smaller the glide angle andconsequently the further the aircraft will glide from a given height.
For example, to calculate the optimum glide angle for an aircraft with a L/Dof 12:1.
Drag/lift equals 1/12, thus tangent = 0.08 and, from the trigonometricaltable, the glide angle = 5.
Although there is no thrust associated with the power-off glide, the powerrequired curve is still relevant. The minimum drag airspeed shown in thatdiagram is roughly the airspeed for best glide angle and the speed forminimum power is roughly the airspeed for minimum rate of sink in a glide.This is examined further in the 'Airspeed and the properties of air' module.
It may be useful to know that in a glide, lift = weight cosine glide angleand drag = weight sine glide angle. There is further information on glide
angles and airspeeds in the lift/drag ratio section of module 4.
1.10 Turning forces
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Centripetal force
When an aircraft turns in any plane, an additional force must becontinuously applied to overcome inertia, particularly as an aircraft's
normal tendency is to continue in a straight line. This is achieved byapplying a force towards the centre of the curve or arc the centripetalforce which is the product of the aircraft mass and the accelerationrequired. Remember that acceleration is the rate of change of velocity either speed or direction, or both.
The acceleration, as you know from driving a car through an S curve,depends on the speed at which the vehicle is moving around the arc andthe radius of the turn. Slow speed and a sweeping turn involves very littleacceleration. But high speed and holding a small radius involves highacceleration, with consequent high radial g or centripetal force and
difficulty in holding the turn. Even when an aircraft enters a straight climbfrom cruising flight, there is a short transition period between the straightand level path and the straight and climbing path, during which the aircraftmust follow a curved path a partial turn in the vertical plane.
An aircraft turning at a constant rate turn is continuously acceleratingtowards the centre of the turn. The acceleration towards the centre of the
turn is V/rm/s. The centripetal force required to produce the turn is m
V/rnewtons, where m is the aircraft mass in kilograms and r is the turnradius in metres. Note this is aircraft mass, not weight.
Turn forces and bank angle
The diagram below shows the relationships between centripetal force,weight, lift and bank angle.
In a level turn, the vertical component of the lift (Lvc) balances the
aircraft weight and the horizontal component of lift (Lhc) provides the
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centripetal force.
(Note: in a right-angle triangle the tangent of an angle is the ratio of the
side opposite the angle to that adjacent to the angle. Thus, the tangent of
the bank angle is equal to the centripetal force [cf] divided by the weight
ortan = cf/W. Or, it can be expressed as tan = V/gr. In the diagram, I
have created a parallelogram of forces so that all horizontal lines representthe centripetal force or Lhcand all vertical lines represent the weight or
Lvc.)
Let's look at the Jabiru, of mass 400 kg, in a 250 m radius horizontalturn at a constant speed of 97 knots or 50 m/s:
Centripetal acceleration = V / r = 50 50 / 250 = 10 m/sCentripetal force required = mass V / r = mass 10 = 400 10 =4000 N
The centripetal force of 4000 N is provided by the horizontal component ofthe lift force produced by the wings when banked at an angle from thehorizontal. The correct bank angle depends on the airspeed and radius;think about a motorbike taking a curve in the road. During the level turn,the lift force must also have a vertical component to balance the aircraft'sweight, in this case it is also 4000 N. But the total required force is not thesum of 4000 N + 4000 N = 8000 N; it is less and we have to find the one and only one bank angle where Lvc is equal to the weight and Lhc is
equal to the required centripetal force.
What then will be the correct bank angle () for a balanced turn? Well, wecan calculate it easily if you have access to trigonometrical tables. If youhaven't then refer to the abridged version below.
So, in a level turn requiring 4000 N centripetal force with weight 4000N, the tangent of the bank angle = cf/W = 4000/4000 = 1.0, and thus(from the table) the angle = 45. Actually, the bank angle would be 45for any aircraft of any weight moving at 97 knots in a turn radius of 250
metres provided the aircraft can fly at that speed, of course. (Do the
sums with an aircraft of mass 2500 kg, thus weight = 25 000 N.).
Now, what total lift force will the wings need to provide in a level turn ifthe actual weight component (aircraft plus contents) is 4000 N and theradial component also 4000 N?
Resultant total lift force = actual weight divided by the cosine of the
bank angle orL = W / cos . Weight is 4000 N, cosine of 45 is 0.707 =
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4000/0.707 = 5660 N.
The load on the structure in the turn is 5660/4000 = 1.41 times normal,or 1.41g. Alternatively the 'load factor' = 1/cosine (bank angle); so,cosine 45 is 0.707 = 1/0.707 = 1.41g.
Manoeuvring loads
In aviation usage, the lowercase 'g' denotes the acceleration caused bythe force of gravity. When an aircraft is airborne maintaining a constantvelocity and altitude the total lift produced equals the aircraft's weightand that lift force is expressed as being equivalent to a '1g' load. Similarly,when the aircraft is parked on the ground, the load on the aircraft wheels
(its weight) is a 1g load.
Any time an aircraft's velocity is changed, there are positive or negativeacceleration forces applied to the aircraft and felt by its occupants. The
resultant manoeuvring load is normally expressed in terms ofg load,
which is the ratio of all the aerodynamic forces experienced during theacceleration to the aerodynamic forces existing at the normal 1g level flightstate.
You will come across terms such as '2g turn' or 'pulling 2g'. What is beingimplied is that during a particular manoeuvre the lift force is doubled and a
radial acceleration is applied to the airframe for the Jabiru a 2g load =400 kg 20 m/s = 8000 N. The occupants will also feel they weigh twiceas much. This is centripetal force and 'radial g'; it applies whether theaircraft is changing direction in the horizontal plane, the vertical plane oranything between.
You may also come across mention of 'negative g'. It is conventional todescribe g as positive when the l ift produced is in the normal directionrelative to the aircraft. When the lift direction is reversed, it is described asnegative g. Reduced g and negative g can occur momentarily in
turbulence. An aircraft experiencing a sustained 1g negative loading isflying in equilibrium, but upside down. It is also possible for somehigh-powered aerobatic aircraft to fly an 'outside' loop; i.e. the pilot's headis on the outside of the loop rather than the inside, and the aircraft (and itsvery uncomfortable occupants), will be experiencing various negative gvalues all the way around the manoeuvre.
It can be a little misleading when using terms such as 2g. For instance,
let's say that a lightly loaded Jabiru has a mass of 340 kg, and if you again
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do the precedingcentripetal force calculation in a 45 banked turn using
340 kg mass you will find that the centripetal acceleration is 10 m/s,
centripetal force is 3400 N, weight is 3400 N and total lift = 4800 N. The
total lifting force is 15% less than in the 400 kg mass calculation but it is
still a 1.41g turn; i.e. the ratio 4800/3400 = 1.41.
Rather than thinking in terms of ratios, it may be appropriate to considerthe actual loads being applied to the aircraft structures. The norm is to usethe lift load produced by the wing as a primary structural load reference. Inthe 400 kg mass calculation the load produced is 5660/8 = 707 N/m,compared to the 500 N/m load in normal cruise. However, even if the totalweight of the aircraft changes, the forces experienced individual structuralitems the engine mountings for example will vary according to the gforce produced by the wings.
Increasing the lift force in a turn
You might wonder how does the Jabiru increase the lift (or more correctly,the aerodynamic force) if it maintains the same cruise speed in the level
turn? Well, the only value in the equation lift = CL rV S that canthen be changed is the lift coefficient. This must be increased by the pilot
increasing the angle of attack. (Conversely ifCL the angle of attack is
increased during a constant speed manoeuvre the lift and consequently
the aerodynamic load factor must increase.) Increasing aoa will alsoincrease induced drag, so that the pilot must also increase thrust tomaintain the same airspeed. Thus, the maximum rate of turn for an aircraftwill also be l imited by the amount of additional power available toovercome induced drag.
The radius of turn = V/g tan metres. For a level turn, the slowestpossible speed and the steepest possible bank angle will provide both thesmallest radius and the fastest rate of turn. However there are severallimitations:
When the steepest bank angle and slowest speed is appliedthe necessary centripetal force for the turn is provided by theextra aerodynamic force gained by increasing the angle ofattack ( orCL ) to a very high value. Also due to the lowerairspeed a larger portion of the total lift is provided by CL ratherthan V. Consequently the induced drag will increasesubstantially requiring increased thrust power and there willbe a bank angle beyond which the engine/propeller will not beable to supply sufficient thrust to maintain the required lift, andthus height in the turn.
1.
All aircraft that are not certificated under the utility or aerobatic2.
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categories are limited to bank angles not exceeding 60. Abank angle of 75 in a level turn would induce a 3.8g loadfactor the load limit for a normal category certification.Similarly a level turn bank angle of 77 would induce the 4.4gload limit for an utility category aircraft.
The stall speed increases with bank angle, or more correctlywith load factor, thus the lowest possible flight speed increasesas bank in a level turn increases.
3.
Turns at high bank angles, near the accelerated stall speed,with maximum power applied, leaves the aircraft with nothing inreserve. Any mishandling or turbulence may result in a violentwing and nose drop with substantial loss of height.
4.
(For more information on turn physics see 'Turning back procedure and
dynamics'.)
If you consider an aerobatic aircraft weighing 10 000 N and making a turnin the vertical plane such as a loop and imagine that the centripetalacceleration is 2g; what will be the load factor at various points of the turn?
Actually, the centripetal acceleration varies all the way around because the
airspeed and radius must vary. For simplicity we will ignore this and say
that it is 2g all around. If the acceleration is 2g then the centripetal forcemust be 20 000 N all the way around.
A turn in the vertical plane differs from a horizontal turn in that, at both
sides of the loop, the wings do not have to provide any lift component tocounter weight, only lift for the centripetal force so the total load atthose points is 20 000 N or 2g. At the top, with the aircraft inverted, theweight is directed towards the centre of the turn and provides 10 000 N ofthe centripetal force while the wings need to provide only 10 000 N. Thus,the total load is only 10 000 N or 1g, whereas at the bottom of a continuingturn the wings provide all the centripetal force plus counter the weight so the load there is 30 000 N or 3g.
This highlights an important point: when acceleration loads are reinforcedby the acceleration of gravity, the total load can be very high.
If you have difficulty in conceiving the centripetal force loading on the
wings, think about it in terms of the reaction momentum, centrifugal force
which, from within the aircraft, is seen as a force pushing the vehicle and
its occupants to the outside of the turn and the lift (centripetal force) is
counteracting it. Centrifugal force is always expressed as g multiples.
Wing loading W/S
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The term 'wing loading' has three connotations. The prime connotation is
the standard expression design W/S (usually just 'W/S', pronounced'w-over-s') which is the ratio of the aircraft designer's maximum
allowable take-off weight [W] to the gross wing area [S]. (There are some
complications when national regulations specify a maximum allowable
weight for an aircraft category that is lower than the design weight of a
particular aircraft type; see the 'Weight and balance'module.) Aircraft withlow W/S have lower stall speeds than aircraft with higher W/S soconsequently have shorter take-off and landing distances. High W/Saircraft are less affected by atmospheric turbulence. W/S is expressed inpounds per square foot [psf] or kilograms per square metre [kg/m].
The second wing loading connotation is as the operating wing loading; ifthe aircraft takes off at a gross weight lower than the designer's maximum,
then the operating wing loading in level unaccelerated fl ight willalso be lower than the design W/S, as will its stall speed.
The third is the load appl ied by the pilot in manoeuvring flight. As we sawabove, pulling 2g in a steep turn will produce a manoeuvring wing
loading that is double the operating wing loading. So, if a pilot takes off inan overloaded aircraft (i.e. the aircraft's weight exceeds the design MTOW)and conducts a 2g steep turn, then that manoeuvring wing loading will begreater than the designer's expectations.
1.11 Limiting loads and ultimate loads
Manoeuvring loads and gust induced loads
To receive type certification the design of a general or recreational aviationfactory-built aircraft must conform to certain airworthiness standards
among which the in-flight manoeuvring loads and the loads induced byatmospheric turbulence, that the structure must be able to withstand, are
specified. The turbulence loads are called the gust-induced loads. TheU.S. Federal airworthiness standards FAR Part 23 are the recognised
world standards for l ight aircraft certification and the following are extracts[emphasis added]:
"... limit loads ... [are] the maximum loads to be expected in
service (i.e. the highest load expected in normal operations)
andultimate loads ... [are] limit loads multiplied by [a safety
factor of 1.5 1.75 for sailplanes]."
"The structure must be able to support limit loads without
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detrimental,permanent deformation. At any load up to limit
loads, the deformation may not interfere with safe operation ...
The structure must be able to support ultimate loads without
failure for at least three seconds ..."
Three seconds is not much time, so any inflight excursion above theultimate load will probably result in structural failure.
Airworthiness certification categories
Light aeroplanes can be certificated in one or more of four airworthinesscategories 'normal', 'utility', 'acrobatic' and 'l ight sport aircraft' (LSA).
The minimum positive limit flight load factorthat an aircraft in the
normal certification category (at maximum gross weight) must be designedto withstand is 3.8g positive. The LSA category minimum positive limit loadis 4g. The negative limit flight load factor is 1.5g for the normal categoryand 2g for the LSA category. Recreational aviation aeroplanes, which arelimited to banked turns not exceeding 60, generally fit into either thenormal category or the LSA category. The ultimate loads for the normalcategory are +5.7g and 2.25g and, for the LSA category, +6g and 4g.
Amateur builders should aim to meet the same minimum values for limitingload and ultimate load factors.
The 'utility' category (which includes training aircraft with spin certification)
limit loads are +4.4g and 2.2g while the 'acrobatic' category (i.e. aircraftdesigned to perform aerobatics) limit loads are +6g and 3g. Sailplanesand powered sailplanes are generally certificated in the utility or acrobaticcategories of the European Joint Airworthiness Requirements JAR-22,which is the world standard for sailplanes; aerobatic sailplanes have limitloads of +7g and -5g.
The 'flight load factor' calculation is defined as the component of theaerodynamic force acting normal (i.e. at right angles) to the aircraft'slongitudinal axis, divided by the aircraft weight. A positive load factor is onein which the force acts upward, with respect to the aircraft; a negative load
factor acts downward. The inflight load factor is a function of wing loading,dynamic pressure and the aoa, i.e. lift coefficient, but see the flightenvelope.
It should not be thought that aircraft structures are significantly weaker inthe negative g direction. The normal level flight load is +1g so with a +3.8glimit then an additional positive 2.8g acceleration can be applied while witha 1.5g limit an additional negative 2.5g acceleration can be applied.
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The manufacturer of a particular aircraft type may opt to have the aircraftcertificated within more than one category, in which case there will bedifferent maximum take-off weights and centre of gravity limitations foreach operational category. See weight/cg position limitations.
The sustainable load factors only relate to a new factory-built aircraft.
The repairs, ageing and poor maintenance that any aircraft has beenexposed to since leaving the factory may decrease the strength of
individual structural members considerably. Read the current
airworthiness notices issued by the RA-Aus technical manager.
1.12 Conserving aircraft energy
Energy available
An aircraft in straight and level flight has:
linear momentum m v [kgm/s]kinetic energy (the energy of a body due to its motion) mv [joules or newton metres (Nm)]; remembering that'm' in the mv term represents mass(Note: normally, the newton metre the SI unit of moment of force
is not used as the measure of work or energy; however throughout thisguide, it is more helpful to express the kinetic energy in the Nm form
rather than joules the Nm and the joule are dimensionally
equivalent)
gravitational potential energy in this case, the product ofweight in newtons and height gained in metreschemical potential energy in the form of fuel in the tanksair resistance that dissipates some kinetic energy as heator atmospheric turbulence.
To simplify the text from here on, we will refer to 'gravitational
potential energy' as potential energy and 'chemical potential energy'as chemical energy.
We can calculate the energy available to the Jabiru
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cruising:
at a height of 6500 feet (2000 m)
and (air distance flown over time)= 97 knots (50 m/s) with mass = 400 kg, thus weight = 4000 N fuel = 50 litres.
Then:
potential energy = weight height = 4000 2000 = 8million Nm kinetic energy = mv = 400 50 50 = 500 000Nm momentum = mass v = 400 50 = 20 000 kgm/s
chemical energy = 50 litres @ 7.5 million joules = 375million joules.
Because it is the accumulation of the work done to raise the aircraft6500 feet, the potential energy is 16 times the kinetic energy, andis obviously an asset that you don't want to dissipate. It isequivalent to 2% of your fuel.
It is always wise to balance a shortage of potential energy with anexcess of kinetic energy, and vice versa. For example, if you don'thave much height then have some extra speed up your sleeve formanoeuvring or to provide extra time for action in case of engine orwind shearproblems. Or if kinetic energy is low (because of flyingat lower speeds than normal) make sure you have ample height or,if approaching to land, hold height for as long as possible. Theonly time to be 'low and slow' is when you are about to touchdown.
However, during take-offit is not possible to have an excess
of either potential or kinetic energy; thus, take-off is the mostcritical phase of flight, closely followed by the go-around
following an aborted landing approach. Ensure that a safe
climb speed is achieved as quickly as possible after becoming
airborne or commencing a go-around and before the
climb-out is actually commenced; see take-off procedure.
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Kinetic energy measurement
Kinetic energy is a scalar quantity equal to mv joules if theaircraft is not turning. The velocity must be measured in relation tosome frame of reference, and when we discuss in-fl ight energymanagement, the aircraft velocity chosen is that which is relative tothe air; i.e. the true airspeed. For a landborne (or about to belandborne) aircraft we are generally concerned with either the workto be done to get the aircraft airborne or the (impact) energyinvolved in bringing the aircraft to a halt. So, the velocity used isthat which is relative to the ground. Ground speed represents thehorizontal component of that velocity, and rate of climb/sinkrepresents the vertical component.
Kinetic energy, gravitational potential energy and energyconservation are complex subjects. If you wish to go further,google the search terms 'kinetic energy' and 'reference frame'.
Momentum conversion
Let's look at momentum conversion. Consider the Jabiru, weighing4000 N and cruising at 97 knots (50 m/s) and the pilot decides toreduce the cruise speed to 88 knots (45 m/s). This could be
accomplished by reducing thrust below that needed for 88knots allowing drag to dissipate the excess kinetic energy thenincreasing power for 88 knots. However, if traffic conditions allow,the excess kinetic energy can be converted to potential energy byreducing power, but only to that needed to maintain 88 knotscruise, and at the same time pulling up thus reducing airspeedbut still utilising momentum then pushing over into level flight
just before the 88 knot airspeed is acquired.
How much height would be gained?
Consider this:
kinetic energy at 97 knots = mv = 400 50 50= 500 000 Nm
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kinetic energy at 88 knots = mv = 400 45 45= 405 000 Nm
kinetic energy available = 95 000 Nm
but potential energy [Nm] = weight height thus height (gained) = energy available divided byweight
= 95 000 Nm / 4000 N = 24 metres = 78 feet, or 9 feetgained per knot of speed converted.
If we recalculate the preceding figures doubling the initial (100m/s) and final velocities (90 m/s) the height gained will increasefourfold to 96 metres, or about 18 feet per knot. Conversely, if wehalve the initial velocity to about 50 knots, the height gained perknot converted is halved, to about 4 feet. Note that as massappears in both the kinetic energy and the weight expressions, itcan be ignored; thus the figures are the same for any mass.Sometimes momentum (mass velocity) is confused with inertia (aparticular quality of mass).
You will come across the expression 'low inertia / high drag'applied to some recreational light aircraft. This means thatalthough all recreational light aircraft are low-inertia aircraft,
compared to other recreational l ight aircraft this minimum aircrafthas a relatively low inertial mass combined with a relatively highparasite drag profile; thus if the thrust is reduced or fails, the dragreduces the airspeed very rapidly. This is exacerbated if the aircraftis climbing. An aluminium tube and sailcloth aircraft at one end ofthe spectrum may be termed 'low momentum' or 'draggy', while anepoxy composite aircraft at the other end may be termed 'slippery';some are very slippery indeed. The standing world speed recordfor an aircraft under 300 kg is 213 miles per hour; that amateur-designed and amateur-built aircraft was powered by only a 65 hptwo-stroke Rotax. The handling characteristics for a low inertia/low
drag aircraft differ considerably from those of a low inertia/highdrag (low momentum) aircraft.
( The next section in the airmanship and safety sequence iscontained within section 12.2 'Factors affecting
safe landing performance' )
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The next module in this Flight Theory Guide examines aspects ofairspeed and air properties, but you may first wish to read thenotes below.
Abridged trigonometrical table
Relationship between an angle within a right angle triangle and the sides:
Tangent of angle=opposite side/adjacentSine of angle=opposite/hypotenuseCosine of angle=adjacent/hypotenuse
Degrees Sine Cosine Tangent Degrees Sine Cosine Tangent
1 0.017 0.999 0.017 50 0.766 0.643 1.192
5 0.087 0.996 0.087 55 0.819 0.574 1.428
10 0.173 0.985 0.176 60 0.866 0.500 1.732
15 0.259 0.966 0.268 65 0.910 0.423 2.145
20 0.342 0.939 0.364 70 0.939 0.342 2.747
30 0.500 0.866 0.577 75 0.966 0.259 3.732
40 0.643 0.766 0.839 80 0.985 0.173 5.672
45 0.707 0.707 1.000 90 1.000 0 infinity
Things that are handy to know
Rated poweris the brake horsepower delivered at the propellershaft of a direct drive engine, operating at maximum design rpm
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and best power fuel/air mixture, in standard sea-level air densityconditions. (In a regulatory sense the definition is a little morecomplex.) An engine is only operated at its rated capacity for shortperiods during flight, usually during take-off and the initial climb.Rated power for small aero-engines is usually expressed as brakehorsepower rather than the SI unit of kilowatts. Further discussionis provided in the 'Engine and propeller performance' module.
To convert horsepower to watts multiply by 745.7; or to calculatekilowatts, multiply by 0.75.
Design W/S is usually between 11 and 22 psf for GA aircraft, and 4
and 12 psf for ultralights. Gross wing area includes a notionalextension of each monoplane wing up to the fuselage centrelinebut excludes any fairings at the wing/fuselage junction. For multi-engined aircraft, with the engines enclosed in wing nacelles, thewing area would also include the area occupied by the nacelles.
Stuffyou don't need to know
High-performance military aircraft can achieve an aoa exceeding45.
Aerobatic pilots and combat pilots use a value termed
specific energy orenergy height, He. It is the potential energy
plus the kinetic energy per kg of aircraft weight; i.e.
He = mgh/W + mv/WAs W = mg, then the equation can be re-arranged as He = h +
V/2gwhere h = height.
What it expresses is the height that could be achieved if all kineticenergy were transferred to potential energy, but it is of little interestto recreational aviation.
The thermal energy content of one litre of avgas is 30 millionjoules. With good engine handling by the pilot, that litre can
provide 10 million joules of mechanical energy to the propellershaft of most engines. The propeller of the Jabiru is maybe 70%efficient at cruise speed and provides 7.0 million joules, or Nm, ofenergy from the litre of fuel. Roughly how far will that take theJabiru cruising at 97 knots? Easy! Drag is 540 N, so 7 000 000 /540 = 12 965 m or 7.0 air nautical miles. We specify air nauticalmiles because wind will affect the distance travelled over theground.
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Groundschool Flight Theory Guide modules
| Flight theory contents | 1. Basic forces | (1b. Manoeuvring forces) |
| 2. Airspeed & air properties | 3. Altitude & altimeters | 4. Aerofoils & wings |
| 5. Engine & propeller performance | 6. Tailplane surfaces | 7. Stability | 8. Control |
| 9. Weight & balance | 10. Weight shift control | 11. Take-off considerations |
| 12. Circuit & landing | 13. Flight at excessive speed | 14. Safety: control loss in turns |
Supplementary documents
| Operations at non-controlled airfields | Safety during take-off & landing |
Copyright 20002012 John Brandon (contact information)
Page edited by RA-Aus member Dave Gardinerwww.redlettuce.com.au
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