Aircraft design projects - part 8 pps

“chap08” — 2003/3/10 — page 250 — #49 250 Aircraft Design Projects (E) Although point D above gives the max. sustained turn rate, the tangent of a radial from the origin to the zero SEP curve gives the smallest sustained turn radius. In the dry thrust case, the values are 7.5 ◦ /s at 200 kts with a radius of 2582 ft. The wet thrust intersection coincides with the minimum speed boundary. ∗ The minimum speed boundary in this calculation assumes that the high angle of attack required to achieve the max. C L value is controllable. It is likely that in our design, this may only be possible with a contribution from vectored thrust. The component of force from the thrust vectoring has not been included in the calculations because this would require more aircraft and propulsion details than are available at this stage. As we are relying on assistance from thrust vectoring for landing control, it may be possible to design the system to provide an integrated aerodynamic and propulsion control system in the turn manoeuvre without much additional complexity. We have easily met the instantaneous turn requirement so we will assume that the minimum speed boundary is achievable and not critical. Recommendations All of the specified manoeuvre and turn requirements have been easily met with the current design but a word of caution is appropriate. As the value of SEP at a particular flight condition is dependent on the difference between two relatively large numbers (thrust and drag), small percentage changes in either will result in l arge variations in SEP. At this stage in the design process, when only crude estimates have been made about aerodynamic and propulsion characteristics, this must concern us. For example, when considering flight at high manoeuvre load factors, the lift-induced drag becomes a significant component of drag. As this is dependent on the estimation of the induced drag factor, which is difficult to predict accurately for our planform, there could be uncertainty in the ‘high g’ performance. Also, the engine performance is affected by the detail layout and control mechanisms in the intake. A poor estimate of intake efficiency will significantly affect the net thrust available. For these reasons it is important to obtain better (higher confidence) estimates of these parameters in the next phase of the design process. 8.9.2 Mission analysis There are four cruise stages to be assessed: outbound supercruise, outbound dash, return dash and return supercruise. Each of these stages is to be flown at M1.6. It is necessary to determine the optimum (based on minimum fuel burn) cruise height for each stage. From the engine data it is possible to extract the performance (net thrust and sfc) versus altitude for the cruise speed of M1.6, see Figure 8.29. This provides the variation of sfc to be used to predict fuel burn by assuming that the thrust required equals the aircraft drag. From section 8.8.4, the aircraft drag polar at M1.6 is: C D = 0.0205 + 0.514C 2 L The dynamic pressure q (=0.5ρV 2 ) in the stratosphere (above 36 089 ft) where the speed of sound is constant at 986 kts, can be determined for each height using the ISA formula for relative density multiplied by the sea-level value (0.002378 slugs/cu. ft) as shown below: q = 0.5(0.2971e −x )0.002378(1.6 × 986) 2 “chap08” — 2003/3/10 — page 251 — #50 Project study: advanced deep interdiction aircraft 251 TSFC (1/hr) and net thrust (lb) for M1.6 0 5000 10 000 15 000 20 000 25 000 30 000 35 000 1.40 1.30 1.20 1.10 20 000 25 000 30 000 35 000 40 000 45 000 50 000 55 000 60 000 65 000 70 000 Altitude (ft) Net thrust (lb) TSFC/hr SFC/hr (a) (b) (c) (d) Net thrust per engine (lb) Fig. 8.29 Engine performance at M1.6 where x = (H − 36089)/20806.7 H = altitude (ft) For our aircraft, the reference area is 1430 sq. ft (133 sq. m). As we are unaware of the fuel burnt in each segment at this stage in the design process, it will be necessary to make some assumptions regarding the weight of the aircraft at the start of each stage, as shown below: (a) Outbound supercruise = 0.9 MTOW (b) Outbound dash = 0.8 MTOW (c) Return dash = 0.7 MTOM (d) Return supercruise = 0.6 MTOW And if the return stages follow the release of munitions: (e) Return dash = 0.7 MTOM – 8000 lb (f) Return supercruise = 0.6 MTOW – 8000 lb where, from section 8.8.2, MTOM = 114 082 lb (51 739 kg). The aircraft weight definesthe C L which in turn defines the C D from which the aircraft drag is calculated. This is multiplied by the engine sfc to obtain the fuel used per hour. This procedure is easily performed using a spreadsheet method. The results are shown in Figure 8.30. This clearly shows an optimum altitude for each stage. The optimum heights are cross-plotted against aircraft weight in Figure 8.31. The associated fuel consumption is also plotted on this graph. At this point it is possible to use the fuel consumption results to determine the overall fuel burnt on the mission (assuming that the fuel consumption in each stage is the average between the start and end values). The time spent on each stage is the stage distance divided by the aircraft speed. As the speed is constant (933.5 kt), the “chap08” — 2003/3/10 — page 252 — #51 252 Aircraft Design Projects 10 000 15 000 20 000 25 000 30 000 35 000 40 000 35 000 40 000 45 000 50 000 55 000 60 000 65 000 70 000 Cruise altitude (ft) (a) Outward supercruise  = 0.9 (b) Outward dash  =0 .8 (c) Return dash  =0 .7 (c) L e ss w e ap o n s (d ) R eturn dash  =0 .7 (d) L ess w eapo n s Fuel burn (lb/hr) Fig. 8.30 Aircraft fuel burn versus altitude Best alt 50 000 55 000 60 000 65 000 25 000 20 000 15 000 70 000 65 000 70 000 75 000 80 000 85 000 Aircraft weight (lb) 90 000 95 000 100 000 105 000 Optimum (min fuel burn) altitude Fuel burn optimum altitude Altitute (ft) Fuel burn (lb/hr) Fig. 8.31 Optimum cruise versus aircraft weight supercruise stages of 1000 nm take 1.07 hr and the dash stages of 750 nm take 0.80 hr. The analysis is shown in Table 8.8. This is much larger than originally estimated due to the lower lift to drag ratio (4.87 compared to 5.56 assumed earlier). This would mean that the aircraft MTOW “chap08” — 2003/3/10 — page 253 — #52 Project study: advanced deep interdiction aircraft 253 Table 8.8 Fuel burn per hour Stage Start End Average Time Fuel used (lb) (a) 25 521 22 742 24 131 1.07 25 820 (b) 22742 19 968 21 355 0.80 17 148 (c) 19 968 17 178 18 573 0.80 (14 914) (d) 17178 14 000 ∗ 15 589 1.07 (16 680) (e) 18 012 15 219 16 620 0.80 13 346 (f) 15 219 13 000 ∗ 14 109 1.07 15 097 71 411 lb ∗ Guessed values. 0 5000 10 000 15 000 20 000 25 000 15 000 20 000 25 000 30000 35 000 40 000 45 000 50000 55 000 60 000 65 000 Altitude (ft) Thrust and drag (lb) Thrust (single eng.) @ M0.9 Thrust (single eng.) @ M1.6 Aircraft drag (50%): (a) 1st supercruise (b) 1st dash (c) 2nd dash (d) 2nd supercruise a b c d Fig. 8.31 optimum Fig. 8.32 Engine thrust at M0.9 versus altitude must be increased. As the above mission assumed that only 8000 lb of weapon load would be dropped and about 13 000 lb was used in the mass statement to account for different missions, we could substitute 5000 lb into the fuel load. This would mean that the aircraft weight would need to be raised by 11 400 lb. However, some of this penalty could be set against potential improvements in engine design as mentioned in section 8.8.5. (For example, if the sfc could be reduced from 1.2 to 1.1 the fuel load would reduce by 6000 lb.) The cruise analysis predicted the drag. This can be compared to the available thrust which has been extracted from the engine data and shown on Figure 8.32. The analysis shows that the engine needs to be slightly more powerful to fly at optimum (minimum fuel burn) altitudes. However, as the aircraft L/D ratioislower than expected no change should be made to the engine until a more accurate estimation of aircraft drag is available. “chap08” — 2003/3/10 — page 254 — #53 254 Aircraft Design Projects Recommendations Assuming a 5 per cent reduction in engine sfc is possible from a new design, it is suggested that a fuel load of 68 000 lb (30 840 kg) should be provided in the next review of the aircraft mass. The review should also reduce the weapons load to 9400 lb (4263 kg). 8.9.3 Field performance There are four operational issues to be investigated in this performance section: 1. Normal take-off distances to the point at which the aircraft achieves lift-off. 2. Balanced field length and the decision speed, for single engine operation. 3. Approach speed. 4. Landing distances from aircraft touchdown. The calculations for each of the cases above require an analysis of the forces on the aircraft (weight, lift, drag, thrust and ground friction). Our previous estimations of mass, aerodynamic and propulsion characteristics are sufficient to use as input to the analysis. The prediction of take-off and landing distances requires a step-by-step calculation which can be done using a spreadsheet application method. Normal take-off distances The take-off distance is the sum of the ground distance (s G ) and the rotation distance (s R ). The ground distance is that travelled along the runway up to the point at which the rotation speed is reached. The rotation distance is a nominal distance to account for the rotation of the aircraft to achieve the initial lift-off manoeuvre, prior to the climb from the runway. Take-off speed (V TO ) is defined as that reached at the point that the aircraft leaves the runway. To avoid inadvertent instabilities in the initial climb phase, this speed must be faster than that related to the lift coefficient at rotation in the take-off configuration. The allowance above stall on conventional aircraft is typically set at 15 to 20 per cent. As we are well away from the max. C L angle in our aircraft, we can either no factor is applied or that the factor is small: Take-off speed (V TO ) =[W TO /(0.5ρSC Lto )] 0.5 where for our aircraft: W TO = 51 739 kg (114 082 lb) ρ = ISA sea level air density S = reference area = 130 sq. m (1340 sq. ft) C Lto = maximum lift coefficient ∗ ∗ As the aircraft does not have any flaps, this is taken as the lift coefficient at the maximum aircraft tail-down attitude of 15 ◦ . From Figure 8.22 this is seen to be 0.52. Using the values above gives: V TO = 36.3 m/s (118 ft/s, 70 kt) The numerical integration of the ground distance covered is calculated in steps of aircraft speed from brake release (zero speed) to V TO . Although the aircraft would accelerate during the rotation phase, which would reduce the calculated distance, we “chap08” — 2003/3/10 — page 255 — #54 Project study: advanced deep interdiction aircraft 255 will concede this small advantage (inaccuracy) to make the calculation simpler. The formula to estimate ground distance is available in most textbooks and repeated below: s G = 0.5  (1/a)d(V 2 ) where a = aircraft acceleration =[T − D − µ(W − L)]/M T = take-off thrust. As there is only a small variation of thrust during the take-off speed change, we will assume that the thrust remains constant at the average energy speed (i.e. 0.707V TO ). From engine data, this relates to a thrust of 32 950 lb per engine. D = drag in the take-off configuration. This is calculated from the zero-lift drag coefficient estimated as 0.01148 in section 8.5.7, and the induced drag coefficient at subsonic speeds is assumed to be 0.15. With the wing at a 4 ◦ angle of attack on the ground, the lift coefficient (from Figure 8.22) is 0.15. Hence the aircraft drag coefficient is 0.01148 + (0.15 × 0.15 2 ) = 0.01486. µ = the coefficient of ground friction without braking. Design textbooks suggest this is 0.04 for dry runways and 0.02 for icy ones. (W − L) = the ground reaction force. Where W is the aircraft take-off weight (114 082 lb, 507.44 kN) and L is the lift generated with the lift coefficient of 0.15 mentioned above. M = aircraft mass = W /g. The ground distance, calculated by the step-by-step integration, is 583 ft for the dry runway and 563 ft for the icy one. In this case, the ice reduces ground friction and is therefore not critical except that the aircraft may be less directionally stable (see further comments in the landing section below). The time spent in the rotation phase is assumed to be 3 seconds. Hence, at the take-off speed of 118 m/s, the distance covered during rotation (s R ) = 354 ft. For normal take-off, at maximum take-off weight, the max. total take-off distance is 937 ft. Even if the usual 1.15 factor to account for pilot and atmospheric variability is applied to this figure, it is still within the 8000 ft specified in the design brief. Therefore, the all-engines take-off distance is shown to be not critical. Balanced field length If an engine fails during the take-off run, the pilot must make a decision either to continue the take-off with only one engine working, or to abort and bring the aircraft to rest further down the runway. If the failure occurs late in the take-off run he would naturally continue and vice versa if it happened earlier. The aircraft speed at which it is better to continue the take-off is called the decision speed. The pilot will be aware of this speed from the aircraft flight manual before starting the take-off manoeuvre. To determine this speed, it is necessary to calculate separately (for each of the possible speeds at which an engine might fail) the distances required to effect an ‘accelerate-go’ and an ‘accelerate-stop’ manoeuvre. For the accelerate-go case, the calculation includes 1 second of travel after the engine failure to recognise the fault and to take the necessary actions. After this period, the failed engine is assumed to be shut down and a ‘normal’ take-off performed with only “chap08” — 2003/3/10 — page 256 — #55 256 Aircraft Design Projects the remaining engine producing thrust. During this time, no changes to the aircraft configuration are allowed. An increase in drag is applied to account for the drag from the failed engine and the trim forces from the control surfaces required to stabilise the asymmetric flight condition. For the accelerate-stop case, again a 1 second delay is applied before any action is taken. After this time, a 3 second allowance is given to account for the application of brakes and the deployment of other drag devices (e.g. air brakes, reverse thrust, drag chutes). As our engine is relatively complex due to the reheat and vector thrust mechanisms, it is unlikely that thrust reversal will be available. For reasons of stealth and aerodynamic efficiency, the smooth wing profiles will not be disturbed by the installation of air brakes. Hence, the deployment of braking parachutes seems to be the preferred method of providing extra retardation at high speeds. Reference 12 provides a value for the drag area of parachutes. Using the figure of 1.4 times the canopy maximum area gives a C D of 0.076 for two 2 m (7 ft) diameter drag chutes. The results ofthe calculation using the previous aircraft characteristics and the opera- tional assumptions above for the accelerate-go and accelerate-stop cases, for dry and icy conditions, are shown in Figure 8.33. The intersection of the lines for the go and stop cases define the decision speed and the balanced field length. These distances are again substantially less than the required 8000 ft specified. In fact, the high thrust to weight ratio of the aircraft means that, if necessary, the take-off could be achieved with only one engine operating from the start (this is not a common feature on most aircraft). Calculations show that single-engine take-off can be achieved in 1688 ft for a dry runway and 1596 ft for the icy condition. Approach speed The approach speed is dependent on the value of the maximum C L in the approach condition and the maximum aircraft landing weight. Using the high angle of attack 0 250 500 750 1000 1250 1500 1750 2000 30 40 50 60 70 80 Decision speed (ft/s) 90 100 110 120 130 Dry runway: accel-go To distance (ft) Ice runway: accel-go Ice runway: accel-stop with brake-chute Ice Dry Dry runway: accel-stop Fig. 8.33 Balanced field length curves “chap08” — 2003/3/10 — page 257 — #56 Project study: advanced deep interdiction aircraft 257 on approach as described in section 8.7 and the lift data in Figure 8.22 at an assumed angle of attack of 30 ◦ , provides a C Lland of 1.4. The maximum landing weight is set by the operational requirements of the aircraft. If it is necessary to allow for a landing immediately following take-off (e.g. emergency due to system or engine failure) the landing weight could be up to 95 per cent of the take-off weight. If it was possible to burn or dump fuel before landing then a lower landing weight could be set. To avoid penalising the aircraft for the exceptional emergency case we will assume the more conventional landing weight of MTOW less 50 per cent of fuel. For our aircraft, this definition makes the landing weight: W land = 114 082 − (0.5 × 55 000) = 86 582 lb (39 266 kg) For many conventional aircraft, the minimum approach speed is set at 1.3 times stall speed. As our aircraft must be fully automated for landing (due to the poor pilot visibility) and will have precision positioning systems we can assume this safety factor to be reduced to 1.2. In this landing case (as compared to the take-off), the aircraft is flying close to its maximum C L so a factor is still appropriate. Therefore: V land = 1.2[86 582/(0.5 × 0.002377 × 1340 × 1.4)] 0.5 = 236.5 ft/s (72.1 m/s, 140 kts) This seems reasonable compared to estimates of the approach speeds for similar mil- itary aircraft (F-14 = 134, F-117 = 144, Su-33 = 194!, B-2 = 140, B-52 = 140 kts). However, the analysis for our aircraft was based on assumptions for the landing weight and C Lmax for each aircraft which may be in error, so a sensitivity study was undertaken. The result is shown in Figure 8.34. 50 000 60 000 70 000 80 000 Aircraft weight (lb) Approach speed (ft /s) (=1.2V stall ) 1.0 1.2 1.4 90 000 10 0000 110 000 120 000 160 180 200 220 240 260 280 300 320 340 C Lmax Fig. 8.34 Approach speed versus aircraft weight “chap08” — 2003/3/10 — page 258 — #57 258 Aircraft Design Projects Landing distance Landing distance is computed in a similar method to that for take-off except that thrust is set to zero. To stabilise the aircraft on the ground and to apply maximum braking, a free-roll on touchdown of 3 seconds is assumed. In conventional landing procedures, the touch-down speed is lower than the approach speed due to the drag produced in the flare phase. In our design the high angle of attack on approach will be reduced prior to landing to avoid scraping the rear fuselage. This may suggest that the touchdown speed will be higher than the approach speed. However, to simplify the calculation we will assume that the touchdown speed is equal to the approach speed. The detailed landing calculation shows that, at the landing weight assumed above, the unfactored distance is 2535 ft (773 m) on a dry runway. On an icy runway, the distance increases to 9273 ft (2828 m). This is beyond the available runway length of 8000 ft specified in the project brief. It will therefore be necessary to use brake chutes to reduce the distance. Braking parachutes are particularly useful devices as they are most effective at higher speedswhen the aircraft brakes are less powerful (due totheunwanted lift reducing the ground reaction force). Using the two 7 ft diameter chutes described previously, the landing distance on an icy runway is reduced to 7047 ft (2150 m). This brings the distance within the available length. In fact the aircraft would be able to land at 95 per cent MTOW within the 8000 ft allowance. Figure 8.35 shows the variation of unfactored landing distance against landing weight. Although the results above look acceptable, it must be remembered that the landing manoeuvre may not be as precise as we have assumed in the analysis. For example, the approach speed may be higher than expected or the aircraft may overshoot the runway threshold due to gust disturbance just prior to touchdown. To guard against such possibilities it is common practice to apply a factor to the calculated landing distance. Typically, this is set at 1.67. Applying this to the dry distance of 2535 ft and the icy distance of 7074 ft increases them to 4233 ft and 11 768 ft. The normal, dry runway landing is still acceptable but clearly the icy one is still much too long. 2000 70 80 90 Dry MLW 60% Wet Ice Max. factored distance 100 Weight (lb×10 –3 ) 110 MTOW 120 3000 4000 6000 8000 Unfactored hourly distance (ft) Fig. 8.35 Landing distance versus aircraft weight “chap08” — 2003/3/10 — page 259 — #58 Project study: advanced deep interdiction aircraft 259 As military airfields are fully serviced, it is not unreasonable to expect that in icy conditions the runway surface will be treated to dissolve the ice (as on highways). Recalculating the landing distances using the accepted runway friction coefficient for wet surfaces (0.3) over the last 60 per cent of the runway length, instead of that for ice (0.1), reduces the landing distance to 4715 ft (unfactored) and 7874 ft (factored). This is within the allowable runway length. Treating the runway to avoid ice contamination will also avoid potential directional instabilities and skidding problems. 8.10 Cost estimations Estimating the costs of future aircraft has always been seen as an inexact science. Evidence from previous design programmes show that even the seasoned professionals in industry do not have a good track record at making such estimates. For students, and even faculty, in an academic environment it is impossible to predict the absolute costs associated with a new project. Too many of the factors that are needed are only available within a commercial organisation. Such factors relate to the accountancy practices used, the organisation of the company (or more likely the consortia of companies that are formed to share the design and manufacturing tasks), the interrelationship between government and industry, and many more non-technical issues. For military projects, the need to incorporate modern and advanced technologies is paramount. The timescales involved in the development of such technologies often overlaps the aircraft development period. This leads to uncertainties in the costs incurred. For our project there are at least six technological areas (e.g. stealth, propul- sion, aerodynamic design, structures and materials, and systems) which need to be matured before an exact cost can be assumed. Notwithstanding these difficulties, it is often financial parameters that are used to choose between different design options. It is therefore essential to be able to determine relative costs to create a framework for such decision making and to be able to compare our design with competitor aircraft. Fortunately, historical data shows that many of the cost parameters are related to aircraft design variables (e.g. aircraft empty weight, installed engine thrust, number of engines, aircraft operational speed, and the overall system complexity). Other factors are related directly to manufacturing variables (e.g. labour rates, number of aircraft produced and the production rate). Due to the variability of the value of a currency with time, it is always essential to ‘normalise’ the quoted cost numbers to a specific date (year). This means that inflation rates for the currency must be applied to any data used. Cost estimates must always state the year to which they are indexed. Several aircraft design textbooks provide details of cost estimation methods but in this study the method published by the Society of Allied Weight Engineers (SAWE) 13 is used. This paper describes fully all the details required to estimate the significant cost values at the preliminary design stage. It also provides a spreadsheet method and example. The method is based on regression of historical data from aircraft of specific types. As new designs will be more technically complex than older aircraft it is necessary to apply factors to account for the increase in costs associated with these new features. Our aircraft has many new technical features including new structural materials and construction processes, a sophisticated flight and weapon control system, vectoring engine nozzles, efficient high altitude and fast flight, and advanced stealth features. Each of the technical factors in the SAWE method will need to be set at high [...]... M E., Aircraft Performance – Theory and Practice, Butterworth-Heinemann and AIAA Education Series, 2000, ISBN 1-5 634 7-2 5 0-3 and 1-5 634 7-3 9 8- 4 8 AIAA Aerospace Design Engineers Guide, 19 98, ISBN 1-5 634 7-2 83 -X 1 9 ESTOL aircraft landing profile for the X31 demonstrator (www.aviationnow.com) 10 McCormick, B W., Aerodynamics, Aeronautics and Flight Mechanics, Wiley and Sons, 1979, ISBN 0-4 7 1-0 303 2-5 11... Engineering, AIAA Education Series, 1995, ISBN 1-5 634 7-1 0 5-1 4 Raymer, D P., Aircraft Design: A Conceptual Approach, AIAA Education Series, 1999, ISBN 1-5 634 7-2 8 1-0 5 Brandt, S A et al., Introduction to Aeronautics: A Design Perspective, AIAA Education Series, 1997, ISBN 1-5 634 7-2 5 0-3 “chap 08 — 2003/3/10 — page 2 68 — #67 Project study: advanced deep interdiction aircraft 6 Aviation Week Source Book, published... observable signature Design limits +7/−3g, VD = M2.0 and max dynamic pressure = 2133 lb/sq ft (equivalent to 80 0 kt at SL) “chap 08 — 2003/3/10 — page 265 — #64 265 266 Aircraft Design Projects Weapons: Common racking for combination weapon loads as defined below: (4) Mk -8 4 LDGP + AIM-120 (4) GBU-27 + AIM-120 (4) 2000 lb JDAM + AIM-120 (4) AGM-154 JSOW + AIM-120 (16) 250 lb small smart bomb Aircraft data Dimensions:... Jet Aircraft Design, AIAA Education Series and Butterworth-Heinemann, 1999, ISBN 1-5 634 7-3 50-X and 0-3 4 0-7 4152-X 15 Southampton University, UK, ‘AIAA undergraduate team design competition – Group report’, June 2002 16 Loughborough University, UK, ‘AIAA undergraduate team design competition – Group report’, June 2002 “chap 08 — 2003/3/10 — page 269 — # 68 269 9 Project study: high-altitude, long-endurance... to give a design value of 0.53 Using the above values in the initial aircraft take-off mass equation gives: MTO = 80 0/(1 − 0. 38 − 0.53) = 88 88 kg (19 600 lb) To provide for some design flexibility in the subsequent work a design (max.) mass of 9200 kg (20 280 lb) will be assumed “chap09” — 2003/3/10 — page 284 — #15 High-altitude, long-endurance (HALE) uninhabited aerial surveillance vehicle (UASV) 9.7.2... length 23 m 8. 4 m 75.3 ft 27.5 ft “chap09” — 2003/3/10 — page 277 — #8 277 2 78 Aircraft Design Projects Wing area (ref gross) Wing aspect ratio Empty mass TO mass Cruise speed Lift/drag ratio 50 @ 18. 7 sq m 28. 2 670 kg 980 kg 46 m/s 30 m/s 200 sq ft 1477 lb 2161 lb 89 kt 58 kt Powered glider derivatives employ a power-off ‘glide down’ technique to save fuel and extend flight duration Another aircraft project... low-observable, deep interdiction aircraft Design features: Mid-wing, diamond planform, blended body, tailless, twin-engine layout All weapons stored internally in a central bomb bay below the engine and equipment compartments Side-by-side, high mounted, low-bypass engines with 2D variable geometry, under-wing intakes positioned close to the “chap 08 — 2003/3/10 — page 263 — #62 263 264 Aircraft Design. .. Mattingly, J D., Aircraft Engine Design, AIAA Education Series, 1 987 , ISBN 0-9 3040 3-2 3-1 12 Nicolai, L M., Fundamentals of Aircraft Design, METS Inc., San Jose, California 95120, USA, 1 984 13 Society of Allied Weight Engineers Inc., J Wayne Burns, Aircraft cost estimation methodology and value of a pound derivation for preliminary design development applications’, SAWE Paper No 22 28 Cat No 29, May... in Figures 8. 37, 8. 38 and 8. 39 At this stage in the design process, it is advisable to compile a detailed description of the aircraft so that the work that follows (often by different specialists) will have a common basis The section below is typical of the detail that should be included in such a description 8. 12.1 Final baseline aircraft description Aircraft description Aircraft type: Two-seat, high... Wing aspect ratio Empty mass 80 74 kg TO mass 18 145 kg Cruise speed 192 m/s Ceiling (operational) 22.4 km Ceiling (absolute) 27.4 km Range 380 0 nm Duration 12–15 hours Structural limit 2.5g Engine: F 188 -GE-101 turbofan Max thrust (total) 84 .5 kN 111 ft 63 ft ∗ sq ft 17 80 0 lb 40 010 lb 373 to 470 kt 70 to 73 400 ft 90 000 ft 19 000 lb (∗ As some of the details of the new U-2S aircraft are unavailable, . Butterworth-Heinemann and AIAA Education Series, 2000, ISBN 1-5 634 7-2 5 0-3 and 1-5 634 7-3 9 8- 4 . 8 AIAA Aerospace Design Engineers Guide, 19 98, ISBN 1-5 634 7-2 83 -X 1. 9 ESTOL aircraft landing profile for the X31 demonstrator. 22 742 24 131 1.07 25 82 0 (b) 22742 19 9 68 21 355 0 .80 17 1 48 (c) 19 9 68 17 1 78 18 573 0 .80 (14 914) (d) 171 78 14 000 ∗ 15 589 1.07 (16 680 ) (e) 18 012 15 219 16 620 0 .80 13 346 (f) 15 219 13. and Sons, 1979, ISBN 0-4 7 1-0 303 2-5 . 11 Mattingly, J. D., Aircraft Engine Design, AIAA Education Series, 1 987 , ISBN 0-9 3040 3-2 3-1 . 12 Nicolai, L. M., Fundamentals of Aircraft Design, METS Inc., San