Aircraft Design Projects Episode 9 ppt

“chap08” — 2003/3/10 — page 221 — #20 Project study: advanced deep interdiction aircraft 221 10° α 0.5 1.0 1.5 20° 30° C L Theory E x p e rim e n t Source: McCormick 10 Aspect ratio 1.5 W ith o u t v o rte x lift Fig. 8.8 Section lift coefficient versus angle of attack Source: McCormick 10 Angle of attack V Spiral vortex sheet around leading edge Fully developed spiral vortex flow Fig. 8.9 Vortex induced flow lift capability is not appropriate in this situation. As shown on Figure 8.8, the max. lift coefficient will not be reached until exceptionally high nose angles have been pulled. When the aircraft is on or close to the ground (e.g. on take-off and landing) such high angles will cause the aircraft rear fuselage structure to scrape the runway. The geometry of the aircraft will limit the max. attitude to about 15 ◦ . At this angle Figure 8.8 shows that the C L is approximately 0.52. Therefore, the limit of lift generation on or close to the runway will be set by the aircraft tail scrape angle of 15 ◦ . For flight away from the ground, the max. lift coefficient will be set by the limit of controllable angle of attack. 8.7 Constraint analysis As the initial mass (weight) and aerodynamic estimates have now been made, it is possible to conduct a constraint analysis to determine if the original choice of thrust and wing loading values are reasonable. As these were derived from data on other aircraft, it is likely that a better selection can improve the design. This process will also indicate which of the constraints on the problem are most critical. “chap08” — 2003/3/10 — page 222 — #21 222 Aircraft Design Projects The equation below, as developed from the specific excess power relationship in Chapter 2, is the general form of the constraint function: (T SSL /W TO ) = (β/α)[{(q/β)(C DO /(W TO /S)}+{[k 1 · n 2 · (W TO /S)]/(q/β)} + (1/V ) · dh/dt + ([1/g]·[dV/dt]) In order to draw the constraint diagram (thrust versus wing loading) it is necessary to determine the values of the coefficients (etc.) to be used in the equation. These are defined below: List 1 T SSL = engine static sea-level thrust W TO = aircraft take-off weight S = aircraft reference wing area W = aircraft weight at the condition under investigation T = engine thrust at the condition under investigation List 2 β = aircraft weight fraction for the case under investigation = (W /W TO ) α = thrust lapse rate at the altitude and speed under investigation = (T /T SSL ) q = dynamic pressure at the altitude and speed under investigation = (0.5ρV 2 ) V = aircraft speed at the condition under investigation h = aircraft altitude at the case under investigation ρ = air density at height h C DO = aircraft zero-lift drag coefficient k 1 = aircraft-induced drag coefficient n = aircraft normal load factor = L/W dh/dt = aircraft rate of climb at the case under investigation g = standard gravitational acceleration = 32.2 ft/s 2 (or 9.81 m/s 2 ) dV /dt = aircraft acceleration at the case under investigation For each constraint case, the analysis requires all the values for the parameters in the second list above to be substituted into the equation for (T SSL /W TO ) above. Selected values of wing loading (W TO /S) are then used to determine corresponding values for thrust loading (T SSL /W TO ). These values are then plotted to indicate the constraint boundary for the case. This process is repeated for all constraints. In the design proposal, there are several performance requirements: • Take-off from 8000 ft (2440 m) runway, on standard day with icy runway. • Climb to optimum supercruise altitude. • Supercruise at optimum altitude at M1.6 for 1000 nm (less climb distance). • Dash at M1.6 at 50 000 ft (min.). • Manoeuvre with specific excess power (SEP), at specified weapon load and 50 per cent fuel: –at1g, M1.6, alt. = 50 000 ft with SEP = 0 ft/s with no afterburning –at1g, M1.6, alt. = 50 000 ft with SEP = 200 ft/s with afterburning –at2g, M1.6, alt. = 50 000 ft with SEP = 0 ft/s with afterburning • Land onto 8000 ft runway, on standard day with icy runway. Before the analysis can be made there are several assumptions that must be made: • Take-off from icy* conditions will be with afterburning (called maximum thrust). • Take-off in normal conditions will be with no afterburning (called military thrust). “chap08” — 2003/3/10 — page 223 — #22 Project study: advanced deep interdiction aircraft 223 • As some of the constraints are related to military thrust, it is necessary to define the increase in thrust from afterburning. We will initially assume (T max /T mil ) = 1.5. • Initial climb to supercruise with final rate of climb of 1000 fpm (our requirement). • Supercruise starts with 90 per cent MTOM. • Dash starts with 80 per cent MTOM. • Manoeuvres are at aircraft mass empty + crew + weapons + 50 per cent fuel (25 846 + 500 + 4000 + 15 180 = 45 526 kg (100 385 lb)). Basing all of the constraint analysis on our original mass estimate of 66 000 kg (145 530 lb) gives β manoeuvre = (W /W TO ) = 0.69. • Landing approach speed less than 160 kts (82 m/s) at 95 per cent MTOM. • Landing on an icy ∗ runway with fuel dumping and possibly emergency braking parachute. • Landing in normal conditions will be determined at 95 per cent MTOM with emergency braking (µ = 0.5). ( ∗ Operation from icy runways requires directional control that is not reliant on tyres.) Aerodynamic surfaces and engine thrust mechanisms are the only alternatives. Lateral thrust vectoring will be available for take-off but not for landing. Operation from icy runways may be difficult unless other solutions can be found. The last three constraints dictate maximum vales for (W/W TO ). The approach speed is only affected by the aircraft minimum speed. As we will not have a reverse-thrust capability on the aircraft, the landing distance calculations will be independent of engine thrust. The appropriate calculations are shown below and the results plotted in Figure 8.10. 0.3 0.4 0.5 0.6 0.7 0.8 Thrust loading (SSL) 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 700 0 Wing loading (N/sq. m) 0.9 1 1.1 F-117 F-16XL F-15EX F-14DX Normal landing Approach speed Original design point Revised design point B-2 Normal TO Concorde Supercruise Dash +Climb F-22 F-23 F-32 Manoeuvre 200 ft/s 100 ft/s 15 0ft/s Fig. 8.10 Constraint diagram “chap08” — 2003/3/10 — page 224 — #23 224 Aircraft Design Projects (a) For the approach speed (W TO /S) = 1/β{0.5ρ(V approach /1.2) 2 C Lapproach } where β = 0.95 ρ = 1.225 kg/m 3 V approach = 82 m/s C Lapproach = 0.52 We are assuming that the approach speed is 1.2V stall . This is slower than normal. This gives (W TO /S) = 1566 N/sq. m (max.) (For reference 1000 N/sq. m = 20.9 lb/sq. ft.) This is much too low. It will create a large wing area which will be inefficient in the cruise phases. (For reference, the initial estimate for wing loading is 3880 N/sq. m.) It will be necessary to generate more C L from the wing. The value used above was consistent with an unflapped delta wing limited to a maximum angle of attack of 15 ◦ . For an aircraft of our layout it may be possible to adopt a high angle of attack (HAA) approach (see Figure 8.11) as demonstrated by the X31 vector technology demonstrator. 9 This uses an HAA to provide a slow speed flight at low decent rate for most of the approach. Obviously, the aircraft must be stable in such a flight attitude and must be capable of maintaining its heading. As the aircraft gets near to the ground the angle of attack is raised to the maximum value to slow the aircraft. Just prior to the tail scrape, the incidence is rapidly reduced (nose-down). This will cause the aircraft to effectively have a controllable crash landing onto the runway threshold. This manoeuvre will demand an extra strong landing gear to withstand the high loads required to absorb the vertical energy. This flight profile requires automation as the pilots will not be capable of reacting to such a landing manoeuvre. (Most large civil aircraft landings are automatic these days, although not like this profile!) Touchdown rollout Runwa y threshold Rapid de-rotation and level-out Curved slow speed landing trajectory Max. HAA stabilised altitude HAA approach with low decent rate Transition to higher angle of attack Source: S. W. Kandebo 9 Fig. 8.11 High angle of attack approach profile “chap08” — 2003/3/10 — page 225 — #24 Project study: advanced deep interdiction aircraft 225 As we have already decided that artificial vision and automatic landing systems would be incorporated into the aircraft to avoid the forward cockpit profiling, the unusual aircraft attitude should not present a problem. Increasing the angle of attack to 35 ◦ would raise the C L to 1.3. The wing could be fitted with a leading edge (vortex) flap to increase C L to 1.5. Even when not deployed the additional mechanisms and systems needed to deploy the flaps would affect the stealth image of the aircraft so may not be desirable. Vortex flaps will not be included but will provide some insurance if the flight profile is seen in flight tests to require extra lift capability. We could also assume some fuel dumping or burn-off before landing. This would reduce β to 0.8. These changes would increase the maximum wing loading (N/sq. m) as shown below: (a) baseline = 1566, (b) with HAA profile = 3914, (c) HAA plus fuel dumping = 4648. The aircraft conditions to be adopted will be decided when all the constraints have been assessed. (b) For normal landing (W TO /S) = (s L · ρ · C Llanding · g · µ)/(1.69 · β) where s L = available runway length = 2440 m (8000 ft) C Llanding = 0.52 (see above) µ = 0.5 β = 0.95 This gives a maximum value of (W TO /S) = 4748 N/sq. m. Note that an approach speed of 1.3 times minimum speed has been assumed above (factor 1.69). This is typical of conventional aircraft to protect from stall due to sudden changes in atmospheric conditions. As the delta planform flying at 15 ◦ angle of attack is well away from the max. lift angle it may be argued that this factor could be ignored. If so, the maximum value of wing loading would be 8024 N/sq. m. (c) For icy landing The same formula as above is applicable if braking parachutes (etc.) are not used with input values of: s L = available runway length = 2440 m (8000 ft) C Lland = 0.52 µ = 0.1 β = 0.95 Giving: (W TO /S) = 950 N/sq. m Or without the factor = 1605 N/sq. m. These are obviously too low, therefore extra retardation is required. As we are likely to need thrust-vectoring and afterburning on the engine, it is unfeasible to expect thrust reversal to be available. Braking parachutes, air brakes, runway-retarding devices and ice removal offer some possibilities. As all of these devices complicate the analysis, it is not appropriate to get too involved in detail design at this early stage in the design of the aircraft. “chap08” — 2003/3/10 — page 226 — #25 226 Aircraft Design Projects (d) Normal take-off For take-off conditions the constraint equation reduces to: (T SSL /W TO ) =[(1.44 · β 2 )/(α · ρ · C Lto · g · s TO )]·(W TO /S) where s TO = available runway length = 2440 m (8000 ft) α and β = 1.0 C Lto = 0.52 As the equation is a straight line through the origin, it is only necessary to evaluate it for one value of wing loading. For (W TO /S) = 5000 N/sq. m, giving (T SSL /W TO ) = 0.472. The same argument as outlined above for landing can be made for the avoidance of the 1.44 factor in the take-off equation. In this case, the (T SSL /W TO ) reduces to 0.328. (e) For icy take-off The calculation requires the estimation of the balanced field length using the maxi- mum thrust for the flight condition but not for the braking condition to determine the decision speed. The braking part of the calculation involves the same difficulties as described in the icy landing description above. As with landing, it is too early in the design process to perform these calculations in sufficient detail. We will need to return to this subject later in the design process. (f) Supercruise at optimum altitude For a parabolic drag polar the condition for maximum range can be shown 4 to be: C Do = (3 · k 1 · C 2 L ) For our aircraft: C Do = 0.01996 and k 1 = 0.3 Hence, C L for max. range is 0.149 Using the definition of lift: L = W = 0.5 · ρ · V 2 · S · C L With W = 0.9 · 66 000 · 9.81, V = M1.6 = 1.6 · 295 = 472 above 11 000 m, S = 170 sq. m, gives ρ = 0.2295. From ISA tables this density occurs at 14 000 m (46 000 ft) altitude, this is the initial supercruise height. This calculation involves the initial guess for the wing loading (i.e. 3808 N/sq. m). The equation above can be solved in terms of other values for wing loading to indicate the sensitivity of (W TO /S) against initial optimum altitude: Wing loading (N/sq. m) 3 000 4 000 5 000 Optimum altitude (m) 16 000 13 000 11 600 Optimum altitude (ft) 52 500 42 000 38 030 As fuel is used and the aircraft gets lighter the wing loading will reduce and the optimum cruise height will rise. On the return supercruise phase (and for the dash manoeuvres) when the aircraft is lighter the cruise height will be increased providing that the engine thrust is large enough to reach these altitudes. “chap08” — 2003/3/10 — page 227 — #26 Project study: advanced deep interdiction aircraft 227 Artificially fixing the supercruise height for the initial calculation at 14 000 m it is possible to determine the relationship of thrust to wing loading using the constraint equation above. The result is shown below (assuming β = 0.9): Wing loading (N/sq. m) 2000 3000 4000 5000 6000 7000 Thrust loading (T SSL /W TO ) 0.905 0.656 0.548 0.496 0.473 0.465 (g) Initial climb to supercruise altitude The required thrust to achieve the supercruise condition, as calculated above, must include sufficient climbing ability at the start of cruise. The minimum for this type of aircraft is 1000 ft/min (5.08 m/s). The thrust loading to give this rate of climb is calculated by the climb term [(1/V)·dh/dt] in the constraint equation, suitably adjusted to the take-off condition (i.e. multiplied by β/α). Hence, (T SSL /W TO ) =[(1/V ) · dh/dt]·β/α = (1/472) · 5.08 · (0.9/0.3) = 0.0323 (h) Dash at 50 000 ft altitude This is similar to the supercruise case except that the starting mass will be lower due to the fuel used in the previous sector. We will assume β = 0.8. The calcula- tion is performed with and without the climb requirement. The results are shown in Figure 8.10. (i) Manoeuvres There are three separate manoeuvres that have to be investigated (as described in section 8.3 as cases (a) to (c)). The constraint equation is used with the afterburning thrust ratio (1.5) for cases (b) and (c). Case (a) is similar to the initial dash phase described above except that the aircraft weight is lower (β = 0.69). This will make it uncritical and therefore not worth investigating for the constraint analysis. Case (b) is very critical as shown by the results plotted in Figure 8.10. This require- ment overpowers all other constraints and will solely dictate the aircraft layout. For aircraft design this is an undesirable situation and calls into question the validity of this requirement. The specified climb rate of 200 ft/s (12 000 fpm) at the high altitude and high weight may be desirable for avoidance of threats but seems excessive in view of the stealth characteristics of the aircraft. It would be sensible to discuss this prob- lem with the originators of the RFP to establish how ‘firm’ they are on retaining the requirement. Requirements often fall into two categories: ‘demands’ and ‘wishes’. Part of the constraint analysis is concerned with distinguishing between these two types for the critical design requirements. To assist with the discussion it is worth showing the sensitivity of the climb requirement by performing the analysis for different values; in this case, for 100 and 150 ft/s. These extra cases are shown on Figure 8.10. The 100 ft/s case seems to offer the most ‘balanced’ design and still provide a respectable 6000 fpm climbing ability. Case (c) is similar to case (a) but with the normal acceleration value (n) increased to 2, and with afterburning applied. As seen on the constraint diagram the case fits well with the other requirements. 8.7.1 Conclusion The constraint analysis has shown that, in general, the aircraft requirements are well balanced. The exceptions to this optimism are concerned with the manoeuvre climb “chap08” — 2003/3/10 — page 228 — #27 228 Aircraft Design Projects requirement and the airfield performance onto icy runways. Both of these present prob- lems for the design. As discussed above, the climb requirement should be reduced to 100 ft/s. In the following work we will assume that this concession has been made by the customer. Operation from and onto icy runways is not avoidable so some extra retardation systems will have to be introduced or some other possibilities consid- ered. Incorporating reversed thrust into the already complex, engine-nozzle system appears to be unfeasible. Braking parachutes will have to be used together with a reduction in the touchdown speed to lower the energy to be dissipated. The best solu- tion would be the installation of an arrester-hook on the aircraft and some form of wire pick-up on the runway for those airstrips that are susceptible to icing. Such a concept is outside the remit for our design. It should be remembered that constraint analysis is a very crude process. It is based on potentially inaccurate data that has been generated from the initial ‘guesstimates’ of mass, aerodynamic and propulsion values and characteristics. Nevertheless, it offers the first tests of the initial layout and provides a direction to first revision of the aircraft geometry. 8.8 Revised baseline layout The most efficient aircraft layouts on the constraint diagram are those with lower values for thrust loading and higher values for wing loading. The original design point was set at a wing loading of 3826 N/sq. m and thrust loading of 0.6. From Figure 8.10, it is seen that this point violates the modified manoeuvre constraint. Moving to 4500 N/sq. m and 0.58 brings the design into the feasible region. This reduces the wing area by about 15 per cent and the engine by about 4 per cent. This should result in a reduction of the aircraft MTOM. Using the detailed mass estimate calculated earlier (section 8.6.1) and assuming a saving of 2000 kg in empty mass (about 8 per cent) to reflect the new aircraft parame- ters above, provides an initial value of MTOM of 60 806 kg (134 077 lb). This makes the new wing area = (60 806 × 9.81)/4500 = 133 sq. m approx. (i.e. 1425 sq. ft) The static sea-level military thrust (both engines) = (60 806 × 9.81) × 0.58 = 346 kN (77 782 lb) SSL thrust per engine = 173 kN (38 900 lb) It is now possible to modify the original aircraft general arrangement drawing and to make some detailed estimates for the aircraft mass, aerodynamic characteristics and engine performance. 8.8.1 General arrangement The new general arrangement drawing of the aircraft forms the basis of the input to the detailed technical analysis for the next stage of the design process. The basic layout of the aircraft will not be changed from that devised previously but some of the principal dimensions will be different. We now realise the importance of reducing the aircraft maximum cross-sectional area and lengthening the ‘fuselage’. This will be achieved by stretching the planform. “chap08” — 2003/3/10 — page 229 — #28 Project study: advanced deep interdiction aircraft 229 Table 8.3 TE sweep (forward) LE sweep 30 25 20 15 10 60 15.18 ∗ 15.56 15.93 16.31 16.70 17.53 17.10 16.70 16.31 15.93 65 13.98 14.27 ∗ 14.56 14.85 15.14 19.03 18.63 18.27 17.91 17.57 70 12.66 12.87 13.08 ∗ 13.28 13.49 21.03 20.67 20.34 20.03 19.72 30° 30° 20° 10° LE sweep 60° LE sweep 70° 17.53 m 15.18 m 21.03 m 12.66 m 20.34 m 13.08 m 19.72 m 13.49 m Centre line chord 16.70 m span 15.93 15.93 m 16.70 20° 10° Fig. 8.12 Planform shapes (a) Wing For a diamond wing planform of a specified wing area (133 sq. m) it is possible to apply simple geometry to determine the span and centre line chord for various leading and trailing edge sweep angles (degrees) (Table 8.3). In Table 8.3, the upper value is the wing span and the lower the centre line chord. Both values are in metres. The ∗ values represent a wing tip angle of 90 ◦ .Itisalways wise to visualise geometric data. To appreciate the wing planform shapes the options in the table are drawn in Figure 8.12. There are several considerations to take into account in making a choice of planfor m: • To reduce wave drag a long centre line chord is desirable. • Less TE sweep makes the trailing edge controls more effective. “chap08” — 2003/3/10 — page 230 — #29 230 Aircraft Design Projects • Less TE sweep pushes the centre of lift further aft. This could lead to aircraft balance and trim problems and demand larger control surfaces. • A long centre line chord will give a deeper wing profile (for a given profile thickness ratio). • A long centre line chord will reduce wing span and thereby reduce roll control but improve roll inertia. Such concerns require a number of design compromises to be made. At this stage in the design process, the sensitivity of the control responsiveness and the aircraft balance issues are unknown. As our previous analysis has highlighted the need to reduce wave drag, our choice will be towards this aspect. We will select the 70 ◦ sweep with the 90 ◦ wing tip angle. Rounding the exact values shown above gives: Wing span = 13 m, Centre line chord = 20.5 m, Wing area = 133.2 sq. m Wing span = 42.6 ft, Centre line chord = 67.2 ft, Wing area = 1432 sq. ft (b) Weapons In order to arrange the fuselage shape it is necessary to identify the requirements for weapon storage. All the weapons are carried internally to reduce drag and radar returns. The design brief defined the type and combination of weapons to be carried. There are two categories of weapon listed: air-to-surface munitions and an air-to-air defence missile. The latter is the only form of self-defence on the aircraft. There are five different types of munitions specified: • general purpose (GP) guided bombs, • cluster bomb units (CBU), • direct attack penetrators (JDAM), • stand-off weapons (JSOW), • small smart bombs (SSB). The air-to-air missile is the AIM-120/AMRAAM which is commonly carried on other US military aircraft. Descriptions and dimensions for the weapons are easily found in the aeronautical press 6 and web sites. Most of these weapons are already used on other aircraft but some are externally mounted, therefore some detailed modifications for internal storage will be required. The largest weapon in the list (Table 8.4) is the GBU. This will define the required weapon bay dimensions. The internal measurements will depend on the choice between four abreast or two abreast layouts (m/ft): Layout Length Width Depth 4 abreast 4.7/15.4 3.1/10.2 0.76/2.5 2 abreast 9.5/31.1 1.6/5.25 0.76/2.5 (c) Layout Making some assumptions with regard to the engine size and installation, it is now possible to draw the revised baseline general arrangement (GA). This is shown in Figure 8.13. To avoid the possibility of unstable flow conditions at the apex of the delta wing planfor m, a fuselage extension has been added (agreeing with the assumed length increase assumed in section 8.6.2). This will provide a separation of the airflow at the nose of the aircraft, it will add length which will reduce wave drag, and it will [...]... — #32 233 234 Aircraft Design Projects Table 8.5 Component lb kg 6 157 832 5 210 2 835 868 2 633 2 800 378 2 368 1 2 89 394 1 197 18 535 8 406 16.2 15 600 7 69 1 054 7 091 3 49 478 13.7 0.7 0 .9 PROPULSION 17 423 8 728 15.3 18.8 Fuel system and tanks Aircraft systems Avionics Cockpit systems Weapon systems 1 723 1 546 2 370 1 440 1 500 783 701 1 077 653 682 1.5 1.4 2.1 1.3 1.3 19. 5 14.0 9. 5 10.0 17.0 Wing... Max take-off 7.6 1 100 13 448 500 6 113 1.0 11.8 59 082 27 623 51 .9 25 000 48.1 51 7 39 18.0 23.5 16.5 19. 7 6.7 13.5 39. 1 114 082 ZERO FUEL 3 891 21 025 55 000 Crew and op items Weapons 8 5 79 44 537 EMPTY Arm (m) 100.0 10.0 17.5 18.0 (central) 20.5 (wing) estimated positions of the component masses With the aid of a spreadsheet, various combinations of aircraft loading can be assessed The main results... 247 — #46 0 .90 0 247 248 Aircraft Design Projects As the radius of the turn equals the turn rate divided by the aircraft speed, it is possible to construct a set of straight lines on the graph that represent specific turn radii The turn diagram is made specific to a particular aircraft by introducing three boundaries: 1 Maximum positive structural normal acceleration factor (nmax ) For our design, this... turn rate for M0 .9 is limited by the nmax line at a value of 13.4◦ /s (the associated turn radius is 40 69 ft) The turn rate is in excess of the requirement of 8.0◦ /s Table 8.6 Altitude (ft) V (ft/s) 1340 1443 15 59 1835 2 190 SL 5 000 10 000 20 000 30 000 Mach no 1.2 1. 29 1.40 1.64 1 .96 Table 8.7 n Vmin (ft/s) Mach no 1 2 3 4 5 6 7 234 331 405 467 523 573 618 0.221 0.313 0.383 0.442 0. 494 0.524 0.585... rate For the dry engine the rate is 8◦ /s at 407 kts (note how flat the sustained rate against speed is on this aircraft) The corresponding values for the wet thrust are 12◦ /s at M0.35 ( 492 1 ft radius) This curve is also very flat “chap08” — 2003/3/10 — page 2 49 — #48 2 49 250 Aircraft Design Projects (E) Although point D above gives the max sustained turn rate, the tangent of a radial from the origin... 8.8.2 Mass evaluation With the aircraft geometry determined in the general arrangement drawing it is now possible to make a detailed assessment of the aircraft mass Using empirical formulae in design textbooks slightly modified to suit the particular features of our aircraft, each aircraft component can be evaluated The results are shown in Table 8.5 The blended profile of our aircraft layout makes it difficult... interdiction aircraft 0.025 65 k 0.020 SL 0.015 CDo Clean aircraft 65 k 20 k 0.010 SL 0.005 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Mach number Fig 8.20 Zero-lift drag versus Mach number Engine C A/c C Engine C Engine C Thrust vectors Chord line Fig 8.21 Modified engine nozzle geometry “chap08” — 2003/3/10 — page 2 39 — #38 2 39 240 Aircraft Design Projects Simple methods, however, may lead to substantial... M0.8 = 39% MAC, M1.0 = 50% MAC, M1.6 = 48% MAC This data will be useful to enable us to balance the aircraft for each of the flight and loading (weight) cases “chap08” — 2003/3/10 — page 240 — # 39 Project study: advanced deep interdiction aircraft 1.4 With vortex contribution Aircraft lift coefficient 1.2 1.0 Without vortex flow 0.8 0.6 0.4 0.2 0 5 10 15 20 25 30 35 Angel of attack (α) Fig 8.22 Aircraft. .. configuration and part of the engine depth This produces a blended body shape for the aircraft For stealth and to avoid flow problems at high angles of “chap08” — 2003/3/10 — page 231 — #30 231 232 Aircraft Design Projects attack, the intakes have been positioned on the underside of the wing profile Without details of the aircraft centre of gravity position but knowing that the centre of lift is slightly... presumptuous, so we will continue the design process without altering our revised baseline details and accepting the above mass values 8.8.3 Aircraft balance Although the aircraft control limits have not yet been determined, it is still worthwhile to check if the aircraft configuration provides a sensible location for the centre of gravity excursions Using the aircraft GA, Table 8.5 can be extended . our aircraft: C Do = 0.0 199 6 and k 1 = 0.3 Hence, C L for max. range is 0.1 49 Using the definition of lift: L = W = 0.5 · ρ · V 2 · S · C L With W = 0 .9 · 66 000 · 9. 81, V = M1.6 = 1.6 · 295 =. interdiction aircraft 2 29 Table 8.3 TE sweep (forward) LE sweep 30 25 20 15 10 60 15.18 ∗ 15.56 15 .93 16.31 16.70 17.53 17.10 16.70 16.31 15 .93 65 13 .98 14.27 ∗ 14.56 14.85 15.14 19. 03 18.63 18.27 17 .91 . appropriate to get too involved in detail design at this early stage in the design of the aircraft. “chap08” — 2003/3/10 — page 226 — #25 226 Aircraft Design Projects (d) Normal take-off For take-off