3.2.1 Cruise performance Cruise speed options When an aircraft is cruising, flying from point A to point B, the pilot has several options for setting cruise speed: One choice might be to get there as soon as possible, in which case the pilot would operate the engine at the maximum continuous power allowed by the engine designer. The recommended maximum continuous power is usually around 75% of the rated power of the engine and provides performance cruise. Another choice might be to get there using as little fuel as possible but in a reasonable time, in which case the pilot might choose a 55% power setting to provide an economy cruise airspeed. Or the pilot might choose any power setting, in the usual engine design range, 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 balance induced and parasite drag. Power is the rate of doing work, so power (in watts) 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 and multiply it by V we get: (Equation #1.3) Power required for level flight [watts] = CD × ½rV³ × S (note V cubed). The total drag curve can be converted into a 'power required' diagram — usually called the power curve — if you know the total drag at each airspeed between the minimum controllable speed and the maximum level flight speed. It is a different curve from that for total drag, because the power required is proportional to speed cubed rather than speed squared. This means that (ignoring the related CD change) if speed is doubled, drag is increased four-fold but power must be increased eight times — which indicates why increasing power output from, say, 75% power to full rated power, while holding level flight, doesn't provide a corresponding increase in airspeed. The diagram above is a typical level-flight power curve for a light aircraft. 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, the more power is needed, because of induced drag at a high angle of attack. The lowest possible speed for controlled flight is the stall speed, which is discussed in the 'Airspeed and properties of air' module. Two aerodynamic cruise 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 curve where drag force × speed has the lowest value). To maintain level flight at speeds less than or greater than the minimum power airspeed, power must be increased. Power available The engine provides power to the propeller. The propellers used in most light 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 the propeller and the forward speed of the aircraft all establish the angle of attack 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 at one combination of rpm and forward speed — this is usually in the mid-range between best rate of climb and the performance cruise airspeeds. Propeller blades are sometimes pitched to give the best efficiency near the best rate of climb speed (climb prop), or pitched for best efficiency at the performance cruise airspeed (cruise prop). The efficiency of all types of propellers falls off either side of the optimum; one with 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 any time. With the advent of higher-powered four-stroke light engines, such as the Jabiru 3300, there has been a corresponding increase in the availability of more advanced light-weight propeller systems, providing maximum effective power utilisation during all stages of flight. For more information refer to the 'Engine and propeller performance' module. Speed, power and altitude At sea-level, an aero-engine will deliver its rated power — provided it is in near-perfect ex-factory condition, properly warmed up and using fuel in appropriate condition. However, because air density decreases with increasing altitude, and an engine's performance depends on the weight of the charge delivered to the cylinders, then the full throttle power of a non-supercharged four-stroke engine will decrease with height. So, at about 6000–7000 feet above mean sea-level the maximum power available at full throttle may drop below 75% of rated power. At 12 000 feet full throttle power may be less than 55% of rated power. Thus, as altitude increases, the range of cruise power airspeeds decreases. For best engine performance, select a cruise altitude where the throttle is fully open and the engine is delivering 65% to 75% power. A couple of points to note from the speed-power diagram above: As air density, and consequently drag, decreases with height, then airspeed, from a particular power level, will increase with height; e.g. the airspeed attained with 65% power at sea-level is 90 knots increasing to 100 knots at 10 000 feet. At sea-level, an increase in power from 75% to 100% only results in an increase in airspeed from 100 to 110 knots. This is the norm with most light aircraft — that last 33% power increase to rated power only provides a 10% increase in airspeed. Power required vs power available In the 'power available' diagram at left, power available curves have been added to the earlier 'power required' diagram. The dashed red curve indicates the rated power — that is, the full throttle engine power delivered to the propeller over the range of level flight speeds at sea-level. The upper green curve — maximum thrust power, is that engine power converted by the propeller after allowing for 80% maximum propeller efficiency. The lower green curve is the propeller thrust power available with the engine throttled back to 75% power at sea-level, 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 the power required curve indicates the maximum cruise speed in each condition. The region between the maximum thrust power curve and the power required (to maintain level flight) curve indicates the excess power available at various cruise speeds — this excess power is available for various manoeuvres if the throttle is fully opened. The simplest use would be a straight unaccelerated climb, in which case the maximum rate of climb would be achieved at the airspeed where the two curves are furthest apart. It can be seen that the best rate of climb speed is around the same airspeed as the minimum drag airspeed shown in the earlier powered required diagram. The rate of climb will decrease at any speed either side of the best rate of climb 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 of an aircraft fitted with a 100 hp engine, and at the best rate of climb speed the engine/propeller has 25 hp (18 600 watts) of excess thrust power available. The aircraft weight is 4000 N so the rate of climb = 18 600/4000 = 4.65 m/s. To convert metres/second to feet/minute, multiply by 200 = 930 feet/minute as the maximum rate of climb. One thing to bear in mind is that we have assumed the aircraft's aerodynamic shape — its configuration is constant. However if the aircraft 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. 3.2.2 Forces in a climb When cruising, the difference between the current power requirement and power available — the excess power — can be used to accelerate the aircraft or climb, to accelerate and climb, or perform any manoeuvre that requires additional power. For instance if the aircraft has potential power available and the pilot opens the throttle, the thrust will exceed drag and the pilot can utilise that extra thrust to accelerate to a higher speed while maintaining level flight. Alternatively the pilot can opt to maintain the 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 of available power utilised for climbing, which depends in part on the airspeed chosen for the climb. There are other choices than the best rate of climb speed available for the climb — for example, the best angle of climb speed (which is around the same as the speed for minimum power) or a combination enroute cruise/climb speed. The climb speed chosen depends on terrain, weather, cloud cover and other operating variables. If an aircraft is maintained in a continuous full-throttle climb, at the best rate of climb airspeed, the rate of climb will be highest at sea-level; it will decrease with altitude, as engine power decreases. The aircraft will eventually arrive at an altitude where there is no excess power available for climb, then all the available power is needed to balance the drag in level flight and there will be only one airspeed at which level flight can be maintained. 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 the aircraft's service ceiling should appear in the aircraft's performance specification. The service ceiling is the altitude at which the rate of climb falls below 100 feet per minute; this is generally considered the minimum useful rate of climb. This diagram of forces in a climb and the subsequent mathematical expressions, have been simplified, aligning the angle of climb with the line of thrust. In fact the line of thrust will usually be 4 to 10° greater than the climb angle. The climb angle (c) is the angle between the flight path and the horizontal plane. The relationships in the triangle of forces shown are: Lift = weight × cosine c Thrust = 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 straight climb 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 It is power that provides a continuous rate of climb, but momentum may also be used to temporarily provide energy for climbing; see 'Conserving aircraft 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 airspeed and angle of climb is controlled with the attitude and particularly the included angle of attack. This is somewhat of a simplification, as the pilot employs both power and attitude in unison to achieve a particular angle and rate of climb or descent. The angle of attack in a climb is the pitch attitude minus the angle of climb being achieved plus the wing incidence. A very important consideration, particularly when manoeuvring at low level at normal speeds, is that the steeper the climb angle the more thrust is required to counter weight. For example, if you pulled the Jabiru up into a 30° 'zoom' climb the thrust required = drag + weight × sine 30° (= 0.5) so the 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 will lead to a dangerous situation if the pilot is slow in getting the nose down to an achievable attitude. Never be tempted to indulge in zoom climbs — they are killers at low levels. 3.2.3 Forces in a descent If an aircraft is cruising at, for instance, the maximum 75% power speed and the pilot reduces the throttle to 65% power, the drag now exceeds thrust and the pilot has two options — maintain height, allowing the excess 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, or altitude 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 constant airspeed, the sink rate and the angle of descent will vary if thrust is varied. For example, if the pilot increased thrust but maintained constant airspeed, the rate of sink will decrease — even becoming positive; i.e. a rate of climb. If the pilot pushed forward on the control column to a much steeper angle of descent, while maintaining the same throttle opening, the thrust plus weight resultant vector becomes greater, the aircraft accelerates with consequent increase in thrust power and the acceleration continues until the forces are again in equilibrium. Actually, it is difficult to hold a stable aircraft 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 the angle of dive, even past the vertical. We discuss the need for stability in the 'Stability' module. When the pilot closes the throttle completely, there is no thrust, the aircraft enters a gliding descent and the forces are then as shown in the diagram on the left. In the case of descent at a constant rate, the weight is exactly balanced by the resultant force of lift and drag. From the dashed parallelogram of forces shown, it can be seen that the tangent of the angle of glide equals drag/lift. For example, assuming a glide angle of 10° (from the abridged trigonometrical table below, the tangent of 10° is 0.176), the ratio of drag/lift in 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 of lift/drag [L/D]. The higher that ratio is, then the smaller the glide angle and consequently the further the aircraft will glide from a given height. For example, to calculate the optimum glide angle for an aircraft with a L/D of 12:1. Drag/lift equals 1/12, thus tangent = 0.08 and, from the trigonometrical table, the glide angle = 5°. Although there is no thrust associated with the power-off glide, the power required curve is still relevant. The minimum drag airspeed shown in that diagram is roughly the airspeed for best glide angle and the speed for minimum 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 angle and drag = weight × sine glide angle. There is further information on glide angles and airspeeds in the lift/drag ratio section of module 4. 3.2.4 Turning forces Centripetal force When an aircraft turns in any plane, an additional force must be continuously applied to overcome inertia, particularly as an aircraft's normal tendency is to continue in a straight line. This is achieved by applying a force towards the centre of the curve or arc — the centripetal force — which is the product of the aircraft mass and the acceleration required. 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 and the radius of the turn. Slow speed and a sweeping turn involves very little acceleration. But high speed and holding a small radius involves high acceleration, with consequent high radial g or centripetal force and difficulty in holding the turn. Even when an aircraft enters a straight climb from cruising flight, there is a short transition period between the straight and level path and the straight and climbing path, during which the aircraft must follow a curved path — a partial turn in the vertical plane. An aircraft turning at a constant rate turn is continuously accelerating towards the centre of the turn. The acceleration towards the centre of the turn is V²/r m/s². The centripetal force required to produce the turn is m × V²/r newtons, where m is the aircraft mass in kilograms and r is the turn radius 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 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 — or tan ø = cf/W. Or, it can be expressed as tan ø = V²/gr . In the diagram, I have created a parallelogram of forces so that all horizontal lines represent the centripetal force or Lhc and all vertical lines represent the weight or Lvc.) Let's look at the Jabiru, of mass 400 kg, in a 250 m radius horizontal turn at a constant speed of 97 knots or 50 m/s: Centripetal acceleration = V² / r = 50 × 50 / 250 = 10 m/s² Centripetal force required = mass × V² / r = mass × 10 = 400 × 10 = 4000 N The centripetal force of 4000 N is provided by the horizontal component of the lift force produced by the wings when banked at an angle from the horizontal. 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's weight, in this case it is also 4000 N. But the total required force is not the sum 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 (ø) for a balanced turn? Well, we can calculate it easily if you have access to trigonometrical tables. If you haven't then refer to the abridged version below. So, in a level turn requiring 4000 N centripetal force with weight 4000 N, 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 45° for 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 if the actual weight component (aircraft plus contents) is 4000 N and the radial component also 4000 N? Resultant total lift force = actual weight divided by the cosine of the bank angle or L = W / cos ø. Weight is 4000 N, cosine of 45° is 0.707 = 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 by the force of gravity. When an aircraft is airborne maintaining a constant velocity and altitude — the total lift produced equals the aircraft's weight and 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 negative acceleration forces applied to the aircraft and felt by its occupants. The resultant manoeuvring load is normally expressed in terms of g load, which is the ratio of all the aerodynamic forces experienced during the acceleration to the aerodynamic forces existing at the normal 1g level flight state. You will come across terms such as '2g turn' or 'pulling 2g'. What is being implied 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/s² = 8000 N. The occupants will also feel they weigh twice as much. This is centripetal force and 'radial g'; it applies whether the aircraft is changing direction in the horizontal plane, the vertical plane or anything between. You may also come across mention of 'negative g'. It is conventional to describe g as positive when the lift produced is in the normal direction relative to the aircraft. When the lift direction is reversed, it is described as negative g. Reduced g and negative g can occur momentarily in turbulence. An aircraft experiencing a sustained 1g negative loading is flying in equilibrium, but upside down. It is also possible for some high-powered aerobatic aircraft to fly an 'outside' loop; i.e. the pilot's head is on the outside of the loop rather than the inside, and the aircraft (and its very uncomfortable occupants), will be experiencing various negative g values 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 do the preceding centripetal force calculation in a 45° banked turn using 340 kg mass you will find that the centripetal acceleration is 10 m/s², 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 consider the actual loads being applied to the aircraft structures. The norm is to use the lift load produced by the wing as a primary structural load reference. In the 400 kg mass calculation the load produced is 5660/8 = 707 N/m², compared to the 500 N/m² load in normal cruise. However, even if the total weight of the aircraft changes, the forces experienced individual structural items — the engine mountings for example — will vary according to the g force 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 × ½rV² × S — that can then be changed is the lift coefficient. This must be increased by the pilot increasing the angle of attack. (Conversely if CL — the angle of attack — is increased during a constant speed manoeuvre the lift — and consequently the aerodynamic load factor — must increase.) Increasing aoa will also increase induced drag, so that the pilot must also increase thrust to maintain the same airspeed. Thus, the maximum rate of turn for an aircraft will also be limited by the amount of additional power available to overcome induced drag. The radius of turn = V²/g tan ø metres. For a level turn, the slowest possible speed and the steepest possible bank angle will provide both the smallest radius and the fastest rate of turn. However there are several limitations: When the steepest bank angle and slowest speed is applied the necessary centripetal force for the turn is provided by the extra aerodynamic force gained by increasing the angle of attack ( or CL ) to a very high value. Also due to the lower airspeed a larger portion of the total lift is provided by CL rather than V². Consequently the induced drag will increase substantially — requiring increased thrust power and there will be a bank angle beyond which the engine/propeller will not be able to supply sufficient thrust to maintain the required lift, and thus height in the turn. All aircraft that are not certificated under the utility or aerobatic categories are limited to bank angles not exceeding 60°. A bank angle of 75° in a level turn would induce a 3.8g load factor — the load limit for a normal category certification. Similarly a level turn bank angle of 77° would induce the 4.4g load limit for an utility category aircraft. The stall speed increases with bank angle, or more correctly with load factor, thus the lowest possible flight speed increases as bank in a level turn increases. Turns at high bank angles, near the accelerated stall speed, with maximum power applied, leaves the aircraft with nothing in reserve. Any mishandling or turbulence may result in a violent wing and nose drop with substantial loss of height. (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 turn in the vertical plane —such as a loop — and imagine that the centripetal acceleration 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 force must 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 to counter weight, only lift for the centripetal force — so the total load at those points is 20 000 N or 2g. At the top, with the aircraft inverted, the weight is directed towards the centre of the turn and provides 10 000 N of the 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 continuing turn 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 reinforced by 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 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 . (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 with low W/S have lower stall speeds than aircraft with higher W/S — so consequently have shorter take-off and landing distances. High W/S aircraft are less affected by atmospheric turbulence. W/S is expressed in pounds per square foot [psf] or kilograms per square metre [kg/m²]. The second wing loading connotation is as the operating wing loading; if the aircraft takes off at a gross weight lower than the designer's maximum, then the operating wing loading — in level unaccelerated flight — will also be lower than the design W/S, as will its stall speed. The third is the load applied by the pilot in manoeuvring flight. As we saw above, 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 in an overloaded aircraft (i.e. the aircraft's weight exceeds the design MTOW) and conducts a 2g steep turn, then that manoeuvring wing loading will be greater than the designer's expectations. 3.2.5 Limiting loads and ultimate loads Manoeuvring loads and gust induced loads To receive type certification the design of a general or recreational aviation factory-built aircraft must conform to certain airworthiness standards among which the in-flight manoeuvring loads and the loads induced by atmospheric turbulence, that the structure must be able to withstand, are specified. The turbulence loads are called the gust-induced loads. The U.S. Federal airworthiness standards FAR Part 23 are the recognised world standards for light aircraft certification and the following are extracts [emphasis added]: Three seconds is not much time, so any inflight excursion above the ultimate load will probably result in rapid structural failure. The safety factor of 1.5 applies to fairly new aircraft in good condition; as very light aircraft age aerodynamic stresses, corrosion, hard landings and inadequate maintenance contribute to reduction of that safety factor. Airworthiness certification categories Light aeroplanes can be certificated in one or more of four airworthiness categories — 'normal', 'utility', 'acrobatic' and 'light sport aircraft' (LSA). The minimum positive limit flight load factor that an aircraft in the normal certification category (at maximum gross weight) must be designed to withstand is 3.8g positive. The LSA category minimum positive limit load is 4g. The negative limit flight load factor is –1.5g for the normal category and –2g for the LSA category. Recreational aviation aeroplanes, which are limited to banked turns not exceeding 60°, generally fit into either the normal category or the LSA category. The ultimate loads for the normal category 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 limiting load 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. aircraft designed to perform aerobatics) limit loads are +6g and –3g. Sailplanes and powered sailplanes are generally certificated in the utility or acrobatic categories of the European Joint Airworthiness Requirements JAR-22, which is the world standard for sailplanes; aerobatic sailplanes have limit loads of +7g and -5g. The 'flight load factor' calculation is defined as the component of the aerodynamic force acting normal (i.e. at right angles) to the aircraft's longitudinal axis, divided by the aircraft weight. A positive load factor is one in 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 flight envelope. It should not be thought that aircraft structures are significantly weaker in the negative g direction. The normal level flight load is +1g so with a +3.8g limit then an additional positive 2.8g acceleration can be applied while with a –1.5g limit an additional negative 2.5g acceleration can be applied. The manufacturer of a particular aircraft type may opt to have the aircraft certificated within more than one category, in which case there will be different maximum take-off weights and centre of gravity limitations for each 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 been exposed 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. 3.2.6 Conserving aircraft energy Energy available An aircraft in straight and level flight has: linear momentum — m × v [kg·m/s] kinetic energy (the energy of a body due to its motion) — ½mv² [joules or newton metres (N·m)]; remembering that 'm' in the ½mv² 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 this guide, it is more helpful to express the kinetic energy in the N·m form rather than joules — the N·m and the joule are dimensionally equivalent) gravitational potential energy — in this case, the product of weight in newtons and height gained in metres chemical potential energy in the form of fuel in the tanks air resistance that dissipates some kinetic energy as heat or 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 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 = 8 million N·m • kinetic energy = ½mv² = ½ × 400 × 50 × 50 = 500 000 N·m • momentum = mass × v = 400 × 50 = 20 000 kg·m/s • chemical energy = 50 litres @ 7.5 million joules = 375 million joules. Because it is the accumulation of the work done to raise the aircraft 6500 feet, the potential energy is 16 times the kinetic energy, and is obviously an asset that you don't want to dissipate. It is equivalent to 2% of your fuel. It is always wise to balance a shortage of potential energy with an excess of kinetic energy, and vice versa. For example, if you don't have much height then have some extra speed up your sleeve for manoeuvring or to provide extra time for action in case of engine or wind shear problems. Or if kinetic energy is low (because of flying at lower speeds than normal) make sure you have ample height or, if approaching to land, hold height for as long as possible. The only time to be 'low and slow' is when you are about to touch down. However, during take-off it is not possible to have an excess of either potential or kinetic energy; thus, take-off is the most critical 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. Kinetic energy measurement Kinetic energy is a scalar quantity equal to ½mv² joules if the aircraft is not turning. The velocity must be measured in relation to some frame of reference, and when we discuss in-flight energy management, the aircraft velocity chosen is that which is relative to the air; i.e. the true airspeed. For a landborne (or about to be landborne) aircraft we are generally concerned with either the work to be done to get the aircraft airborne or the (impact) energy involved in bringing the aircraft to a halt. So, the velocity used is that which is relative to the ground. Ground speed represents the horizontal component of that velocity, and rate of climb/sink represents the vertical component. Kinetic energy, gravitational potential energy and energy conservation 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, weighing 4000 N and cruising at 97 knots (50 m/s) and the pilot decides to reduce the cruise speed to 88 knots (45 m/s). This could be accomplished by reducing thrust — below that needed for 88 knots — allowing drag to dissipate the excess kinetic energy then increasing power for 88 knots. However, if traffic conditions allow, the excess kinetic energy can be converted to potential energy by reducing power, but only to that needed to maintain 88 knots cruise, and at the same time pulling up — thus reducing airspeed but 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 = ½mv² = ½ × 400 × 50 × 50 = 500 000 N·m • kinetic energy at 88 knots = ½mv² = ½ × 400 × 45 × 45 = 405 000 N·m • kinetic energy available = 95 000 N·m • but potential energy [N·m] = weight × height • thus height (gained) = energy available divided by weight • = 95 000 N·m / 4000 N = 24 metres = 78 feet, or 9 feet gained per knot of speed converted. If we recalculate the preceding figures — doubling the initial (100 m/s) and final velocities (90 m/s) — the height gained will increase fourfold to 96 metres, or about 18 feet per knot. Conversely, if we halve the initial velocity to about 50 knots, the height gained per knot converted is halved, to about 4 feet. Note that as mass appears in both the kinetic energy and the weight expressions, it can be ignored; thus the figures are the same for any mass. Sometimes momentum (mass × velocity) is confused with inertia (a particular quality of mass). You will come across the expression 'low inertia / high drag' applied to some recreational light aircraft. This means that although all recreational light aircraft are low-inertia aircraft, compared to other recreational light aircraft this minimum aircraft has a relatively low inertial mass combined with a relatively high parasite drag profile; thus if the thrust is reduced or fails, the drag reduces the airspeed very rapidly. This is exacerbated if the aircraft is climbing. An aluminium tube and sailcloth aircraft at one end of the spectrum may be termed 'low momentum' or 'draggy', while an epoxy composite aircraft at the other end may be termed 'slippery'; some are very slippery indeed. The standing world speed record for an aircraft under 300 kg is 213 miles per hour; that amateur-designed and amateur-built aircraft was powered by only a 65 hp two-stroke Rotax. The handling characteristics for a low inertia/low drag aircraft differ considerably from those of a low inertia/high drag (low momentum) aircraft. Abridged trigonometrical table Relationship between an angle within a right angle triangle and the sides: Tangent of angle=opposite side/adjacent Sine of angle=opposite/hypotenuse Cosine 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 power is the brake horsepower delivered at the propeller shaft of a direct drive engine, operating at maximum design rpm and best power fuel/air mixture, in standard sea-level air density conditions. (In a regulatory sense the definition is a little more complex.) An engine is only operated at its rated capacity for short periods during flight, usually during take-off and the initial climb. Rated power for small aero-engines is usually expressed as brake horsepower rather than the SI unit of kilowatts. Further discussion is provided in the 'Engine and propeller performance' module. To convert horsepower to watts multiply by 745.7; or to calculate kilowatts, 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 notional extension of each monoplane wing up to the fuselage centreline but excludes any fairings at the wing/fuselage junction. For multi-engined aircraft, with the engines enclosed in wing nacelles, the wing area would also include the area occupied by the nacelles. Stuff you don't need to know High-performance military aircraft can achieve an aoa exceeding 45°. Aerobatic pilots — and combat pilots — use a value termed specific energy, E or energy height, He. It is the potential energy plus the kinetic energy per kg of aircraft weight; i.e. He = mgh/W + ½mv²/W As W = mg, then the equation can be re-arranged as He = h + V²/2g where h = height. What it expresses is the height that could be achieved if all kinetic energy were transferred to potential energy, but it is of little interest to recreational aviation. The thermal energy content of one litre of avgas is 30 million joules. With good engine handling by the pilot, that litre can provide 10 million joules of mechanical energy to the propeller shaft of most engines. The propeller of the Jabiru is maybe 70% efficient at cruise speed and provides 7.0 million joules, or N·m, of energy from the litre of fuel. Roughly how far will that take the Jabiru 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 nautical miles because wind will affect the distance travelled over the ground. STRICT COPYRIGHT JOHN BRANDON AND RECREATIONAL FLYING (.com)
