15.1.1 Section 15.13: Summarised Discussion of the DesignClasswork Content

At this stage of project design, only the substantiation of requirements/specifications are required for the sized aircraft configurations culminating in ‘concept finalisation’ as worked out in Chapter 14. Full aircraft performance for the full envelope is carried out after go‐ahead is obtained, as done in [1] by the same authors.

The readers will have to compute the following for their projects. This follows that the readers have made engine performance available as shown in Sections 13.4 and 13.5.

1. Substantiate the takeoff field length (TOFL). This has to demonstrate that the aircraft is capable of meeting the Federal Aviation Administration (FAA) second segment (see Figure 15.3) climb gradient requirement for the most critical case.
2. Substantiate the initial climb rate. This is not a FAA requirement but a market specification.
3. Substantiate the initial high speed cruise (HSC) capability. This is also not a FAA requirement but a market specification.
4. Substantiate the LFL. This has to demonstrate that the aircraft is capable of meeting the FAA requirement for the most critical case.
5. Substantiate the payload‐range capability. This is also not a FAA requirement but a market specification. Payload‐range computation requires the specific range, integrated climb and descent performance of the aircraft and the readers will have to compute these. Use of spreadsheets is recommended.

For the trainer aircraft, the following are the additional requirements.

6. Compute turn capability. This is not a Milspec requirement but a MoD (Ministry of Defence) specification.
7. Compute the training mission profile to substantiate the capability.

If any of the aircraft performances is not met then aircraft has to be configuration iterated (see Chapter 8 up to Chapter 14) until the requirements are met. The spreadsheet method proves convenient for iteration.

Airworthiness regulations differ from country to country. Military certification standards are different from the FAA. In the USA, this is guided by Milspecs. Certification authorities have mandatory requirements to ensure safety at takeoff and landing. For further details on respective certification regulations, the readers may refer to their official websites. Readers are recommended to read references 2–8 for supplementary coverage.

15.2.1 Aircraft Speed

Aircraft speed is a vital parameter in computing performance. It is measured from the difference between the total pressure, p_t and static pressure, p_s, expressed as (p_t − p_s). Static pressure is the ambient pressure in which the aircraft is flying and the total pressure, p_t, is aircraft speed dependent. The value of (p_t − p_s) gives the dynamic head, which depends on the ambient air density ρ and aircraft velocity (speed). Unlike automobile ground speed measured directly, aircraft ground speed needs to be computed from (p_t − p_s). The pilot reads the gauge reading converted from (p_t − p_s). The following are the various forms of aircraft airspeeds that engineers and pilots use. As can be seen, these require some computations – nowadays onboard computers do it all.

• V_i = Gauge reading is what the uninstalled bare instrument supplied by the manufacturer reads. The instrument includes standard adiabatic compressible flow corrections for high subsonic flight at a sea level STD day. But it still requires other corrections as follows:
• V_I = Indicated Airspeed (IAS). Manufactured instruments will have some built‐in instrumental errors, ΔV_i, (normally very small but an important consideration when close to stall speed). Instrument manufacturers supply the error chart for each instrument. The instrument is calibrated to read correct ground speed at sea level STD day with compressibility corrections. When corrected the instrument reads IAS.
  

• V_C = Calibrated Airspeed (CAS). Instrument manufacturers calibrate the uninstalled bare instrument for sea level conditions. Once installed on an aircraft, depending where it is installed, the aeroplane flow field will distort instrument readings. Therefore, it will require position error (ΔV_p) corrections by aircraft manufacturers.
  

• V_EAS = Equivalent Airspeed (EAS). Air density ρ changes with altitude – it goes down because atmospheric pressure reduces with gain in altitude. Therefore, at the same ground speed (also known as True Air Speed, TAS), the IAS would read lower values at higher altitudes. The mathematical relation between TAS and EAS reflecting the density changes with altitude can be derived as TAS = EAS/√σ, where σ is the density ratio (ρ/ρ₀) in terms of sea level value, ρ₀. A constant EAS has a dynamic head invariant. For high subsonic flights, it will require an adiabatic compressibility correction (ΔV_c) for the altitude changes.
  

V_EAS is what the pilot in the flight deck reads. In the past, the pilot/flight engineer had to manually correct the instrument and position errors. Today, these corrections are embedded in the aircraft microprocessors off‐loading manual effort.

Compressibility corrections for position errors is also there but at this stage such details can be avoided without any loss of conceptual design work undertaken in this book. Supersonic flight will require further adjustments to make the corrections resulting from the shock wave associated with it.

15.2.2 Some Prerequisite Information

The readers will have to find available engine performance data, as done in Chapter 13 in which the installed thrust and fuel flows for the various engines are worked out. The worked‐out examples in this chapter uses the following figures.

• Turbofan with a BPR (bypass ratio) ≈ 4 ± 1, (in the class of Honeywell TPE731 turbofan) – Bizjet application
• Figure 13.11 – Installed takeoff thrust and fuel flow per engine
• Figure 13.12 – Installed maximum climb thrust and fuel flow per engine
• Figure 13.13 – Installed maximum cruise thrust and fuel flow per engine
• Figure 13.14 – Installed idle engine thrust and fuel flow per engine
• Military Turbofan with a BPR < 1, (in the class of Rolls‐Royce/Turbomeca Adour) – AJT application
• Figure 13.15 – Installed maximum thrust and fuel flow for all the three ratings.
• Turboprop engine, in the class of Garrett TPT 331 – TPT application
• Figure 13.16 – Installed takeoff thrust and fuel flow per engine
• Figure 13.17 – Installed maximum climb thrust and fuel flow per engine
• Figure 13.18 – Installed maximum cruise thrust and fuel flow per engine

Using the engine and aircraft data obtained thus far during the conceptual design phase, this chapter checks whether the configured aircraft satisfies the airworthiness (FAR) requirements and the customer specifications in the takeoff/landing, the initial climb rates and the maximum initial cruise speed plus the payload‐range capability (civil aircraft).

15.2.3 Cabin Pressurisation

Aircraft payload‐range computation requires integrated climb and descent performances. It is topical to introduce the role of cabin pressurisation in assessing aircraft climb and descent performance. Ambient pressure reduces non‐linearly with increase in altitude as can be seen in Figure 1.8. At the tropopause it is 472.68 lb ft⁻² (3.28 lb in⁻²). For average human physiology, big jets maintain cabin pressure around 8000 ft altitude (ambient pressure of 1571.88 lb ft⁻² (10.92 lb in⁻²) resulting in a differential pressure of 7.63 lb in⁻²). Smaller Bizjets fly at higher altitudes to stay separate from big‐jet traffic, as high as 50 000 ft altitude (242.21 lb ft⁻², i.e. 1.68 lb in⁻²) when aircraft structural design considerations are made to maintain the inside cabin pressure at 10 000 ft altitude (1455.33 lb ft⁻², i.e. 10.11 lb in⁻², a differential pressure of 8.42 lb in⁻²). For big jets, a maximum of 8.9 lb ft⁻² (for Bizjets to 9.4 lb in⁻²) of differential pressure is used in practice. Cabin pressurisation is in use for certain types military aircraft with large fuselage volumes. Combat aircraft have different arrangements for higher altitude provision in the flight deck; pilots wear pressure suits and helmets with masks supplying them with oxygen.

The differential pressure between the outside and inside of the cabin is substantial and it cycles through every sortie flown. Stress and fatigue considerations of fuselage structural design are constrained by weight considerations. Aircraft cabins are built as sealed pressure vessels (very low leakage), kept airtight and provided with complex environment control systems (ECSs), which have pressure control valves to automatically regulate cabin pressure (and air‐conditioning). The aircraft crew can select the desired cabin pressure within the design limits. Unless there is a catastrophic failure, the sealed cabin can hold pressure to give enough time for pilots to descend fast enough in case there is a malfunction in pressurisation. If cabin pressure exceeds the limit then automatic safety valves can relieve cabin pressure to bring it down within safe limits.

The classwork example of the Bizjet has cabin pressure selectable from sea level to 10 000 ft altitude, limiting the differential pressure to within 9.4 lb ft⁻². During the climb the equivalent Bizjet cabin pressure rate of change is designed to 2600 ft min⁻¹. Bizjet type aircraft engine power is adjusted to have cabin pressurisation at the climb rate to a maximum of about 1800 ft min⁻¹ to stay within the ECS system capability, sufficient for the sealed cabin that starts with sea level pressure, and as a result en route climb rate is not required to be restricted. Section 15.10.5 works out the Bizjet climb performance. However, during descent, the cabin needs to gain pressure as it descends to lower altitudes. Average human physical tolerance is taken at 300 ft min⁻¹ descent and the ECS system pressurisation capability meets that.

The following equations are derived in this chapter.

1. TOFL equations
2. Climb and descent equations
3. Cruise equations
4. LFL
5. Range equations to establish the payload‐range capability
6. Aircraft turn equations

15.3.1 Civil Transport Aircraft Takeoff [FAR (14CFR) 25.103/107/109/149]

To ensure safety, the takeoff procedure must satisfy the airworthiness requirement so that the aircraft will be able to climb to safety or stop safely in a case where one critical engine has failed. FAR (14CFR) 1.1 defines the ‘critical engine’ whose failure would most adversely affect aircraft takeoff procedure. Evidently, the most outboard engine of a multi‐engine aircraft is the critical one. Designing for critical engine failure also covers the safety aspects of the failure of a non‐critical engine for an aircraft with more than two engines.

Certification authorities demand mandatory requirements to cater for taking off with one engine inoperative to clear a 35 ft height representing an obstacle and continue with the required climb gradient (Figure 15.3). One engine inoperative TOFL is computed by considering BFL(Section 15.3.2) when the stopping distance after an engine failure at the decision speed, V₁, is the same as the distance taken to clear the obstacle at MTOM.

FAR (14CFR) 25 requires takeoff substantiation for the both dry and wet runways. This book deals only with the dry runway considerations.

15.3.2 Balanced Field Length (BFL) – Civil Aircraft

The rated TOFL at MTOM is determined by the BFL in the event of an engine failure. Figure 15.2 shows the segments involved in computing BFL. It can be seen that taking off with one engine inoperative (failed) above the decision speed V₁ has three segments to clear 35 ft height. Designers must provide the decision speed V₁ to pilots below which, if an engine fails, then the takeoff has to be aborted for safety reasons. The BFL segments are defined next:

• Segment A. The distance covered by all engines operating a ground run until one engine fails at the decision speed V₁.
• Segment B. The distance covered by one engine inoperative acceleration from V₁ to V_LO.
• Segment C. Continue with the flare distance from lift‐off speed V_LO to clear the 35 ft obstacle height reaching aircraft speed V₂.

In the case of engine failure, past the decision speed V₁, there are two segments (which replace segments B + C), as follows.

• Segment D. The distance covered during the reaction time for the pilot to take braking action after an engine fails. (Typically, 3 s is taken as the pilot recognition time and brakes to act, spoiler deployment etc. At engine failure the thrust decay is gradual and within this reaction time before brake application there is a small speed gain shown in the diagram.) Typically, (V_B ≥ V₁) but can be taken as (V_B ≈ V₁).
• Segment E. Distance to stop from V_B to V₀ (maximum brake effort).
  15.1

15.3.2.1 Normal All‐Engine Operating (Takeoff) – Civil Aircraft

Normal takeoff with all engines (TOFL_{all_eng}) operating will take considerably lower field length than the one engine failed case (TOFL_1eng‐faled). To ensure safety, FAR (14CFR) requires a 15% allowance is added to TOFL_{all_eng} over the estimated value. Therefore, it must be at least equal to the greater of the two as shown next.

15.2

15.3.3 Civil Aircraft Takeoff Segments [FAR (14CFR) 25.107 – Subpart B]

A takeoff process with one engine inoperative (FAR requirement) can be divided into four segments (Figure 15.3) of climb procedure as follows:

undercarriage retraction starts as soon as a gradient is established and completes at a height above 35 ft, and can be seen within the first segment by the International Civil Aviation Organization (ICAO) requirement. FAR second segment climb is taken from 35 ft, a known height, as the undercarriage is retracted. The operational takeoff requirements by ICAO are in more detail than the design requirements governed by the FAA.

The capability at the end of the fourth segment is a customer requirement and not a FAR requirement. Tables 15.1 and 15.2 provide the aircraft configuration and power settings at the aircraft takeoff segment FAR requirement.

Note that some engines at their takeoff rating can have augmented power rating (APR) that could generate 5% higher thrust than the maximum takeoff thrust for short period of time. These types of engines are not considered in this book.

It may be noted that only V_stall is determined from aircraft weight and C_Lmax and FAR requires that V₂ ≥ 1.2 V_stall. All other speed schedules are interrelated with regulation requirements. At the conceptual design stage, the ratios are taken from the statistics of past designs and as the project progresses these are refined, initially from wind‐tunnel tests and finally from flight tests. Table 15.3 gives the important speed schedule ratios.

Performance engineers must know the sequence of the takeoff speed schedules stipulated by certification agencies as given in Table 15.4.

The higher the thrust loading (T/W), the higher the aircraft acceleration. For small changes, V_R/V_stall and V_LO/V_stall may be linearly decreased with an increase in T/W. The decision speed V₁ with an engine failed is established through iterations as shown in Section . In a family of derivate aircraft, the smaller variant can have V₁ approaching values close to V_R.

Military aircraft requirements (MIL‐C5011A) are slightly different from the civil requirements. The first segment clearing height is 50 ft instead of 35 ft. Many military aircraft have a single engine where the concept of BFL is not applicable. Military aircraft have to satisfy the Critical Field Length (CFL) as described in Section 15.11.2. The second segment rate of climb is to meet a minimum of 500 ft min⁻¹ for multi‐engine aircraft.

15.3.4 Derivation of Takeoff Equations

The derivation of takeoff performance equations is done at sea level STD day, an airfield with no gradient, zero wind conditions and thrust is aligned along the flight path (i.e. no thrust vectoring). The generalised equation for takeoff performance is derived in detail in [1].

15.3.4.1 Acceleration

Let the acceleration of the aircraft be denoted by a. The force balance equation resolved along aircraft velocity vector is given next (Figure 15.1).

Net force on the aircraft, (W/g)a = Σ_Applied forces along aircraft velocity vector = ΣF_applied

15.3

where, V and a are instantaneous velocity and acceleration of the aircraft

Rearranging Eq. 15.3,

15.4

In terms of coefficients and rearranging (all‐engine operating)

15.5

A simplified approach can be taken by taking average acceleration ā for aircraft at 0.707 V_LO (average of V_LO² from zero, that is, at √(V_LO²/2) of the following equation). Then

15.6

where S_W is the wing area and q = 0.5ρV² is the dynamic head.

The aircraft on the ground encounters rolling friction (coefficient μ = 0.025–0.03 for a paved, metalled runway). At braking friction, μ_B = 0.3–0.5 (for the Bizjet, it is taken as 0.4). Thrust loading (T/W) is obtained from the sizing exercise. Average acceleration, ā, is taken at 0.707 V_LO. FAR requires V₂ = 1.2 V_stall. This gives V₂² = [2 × 1.44 × (W/S_W)]/(ρC_Lstall). An aircraft stalls at C_Lmax.

15.3.4.2 Takeoff Field Length (TOFL) Estimation – Distance Covered from zero to V₂

Figures 15.1 and 15.2 show that aircraft is under all‐engine operation up to the decision speed, V₁, and, thereafter, if the decision is made, continues with the takeoff procedure: the aircraft is rotated at V_R to flare out and have lift‐off at V_LO to reach to V₂ to climb to a 35 ft altitude. It takes about 3–4 s time to reach V₂ from V_R. If aborted at V₁, then the aircraft is braked to stop.

Form Figure 15.1, let dS be the elemental ground run distance covered by the aircraft in infinitesimal time dt. If the instantaneous velocity of the aircraft at that point is V then.

15.7

Then the distance, S, covered from the initial velocity V_i = 0 to final velocity, V_f, is obtained by integrating Eq. . Aircraft acceleration and engine thrust varies with aircraft speed. At the conceptual stage of design, taking average values would prove convenient. Taking average acceleration as a constant and using Eq. 15.7, the equation for distance, S, can be written as follows.

15.8a

For initial velocity V_i ≠ 0 the ground run, Eq (15.8) can be re‐written in terms of the average condition as follows.

15.8b

where using Eq. 15.5, the average acceleration can be written in coefficient form as follows

ā = {g [(T/W – μ) – (C_LSq/W)(C_D/C_L – μ)]}_ave (in System International (SI) units ‘g’ drops out).

Equation (15.8) can now be written separately for each segment and then equated for the BFL. The average acceleration ā is of the segment. A simplified approach can be taken by taking average acceleration ā for aircraft at 0.707 ΔV (average of ΔV within the velocity interval, at √(ΔV²/2) of the following equation).

15.9

In case it is aborted, the distance, S_abort, covered is as follows.

15.10

For the BFL,

The proper choice of the decision speed V₁ will give the BFL. A number of iteration may be required to arrive at the proper V₁, as shown in the classroom example in Section 15.10.1.2.

(In later stages of the project, the computation could be done more accurately in smaller steps of speed increment within which average values of the variables are considered as constants. During takeoff, aircraft C_L and C_D vary with speed changes. In the worked‐out examples, average values are taken. Reference [1] describes the takeoff procedure in detail.)

During takeoff, FAR requires V₂/V_stall > 1.2. The speed gain continues during the rotational (V_R) through aircraft flaring out to V₂. In this phase, typically around 5% change occurs in aircraft velocity through acceleration at the most critical condition; it amounts to about a 5–7 kt speed increase. Therefore, only the proper choice of the decision speed V₁ will give the BFL. A number of iterations may be required to arrive at the proper V₁, as shown in the classroom example in Section 15.7.

Given next are the typical takeoff speeds V₂ (with high‐lift device deployed) for some of the larger jet transport currently in operation.

Boeing737: 150 mph	Boeing747: 180 mph	Airbus320: 170 mph
Boeing757: 160 mph	Concorde: 225 mph	Airbus340: 180 mph

15.4.1 Approach Climb and Landing Climb and Baulked Landing

In the case of a missed approach, the aircraft should be able to climb way above a 50 ft height to make the next attempt to land. The FAR (14CFR) 25.121 subpart B requirements for the approach climb gradient are given in Table 15.6. The aircraft is configured with full flaps, undercarriage retracted and the engine in full takeoff rating.

A discontinued approach takes place above 400 ft with the undercarriage retracted.

In the case of a baulked or called‐off landing with all engines operating, the aircraft should be able to climb away from below a 50 ft height to make the next attempt to land. The FAR (14CFR) 25.119 subpart B requirements for the landing climb gradient are 3.2%.

In both these cases, the aircraft is considerably lighter at the end of mission on account of consuming onboard fuel, so there is less reserve.

15.4.2 Derivation of Landing Performance Equations

15.4.2.1 Ground Distance During Glide, S_glide

Normally, at a steady approach the aircraft is kept throttled back to idle producing no thrust. Figure 15.4 gives the generalised force diagram. It shows thrust, T, assisted by a component of weight, Wsinγ, in the descent angle, γ. In equilibrium, lift, L, is balanced by the weight component, Wcosγ. Equating for equilibrium conditions:

and D = T + W sin γ

or γ = C_D/C_L – T/W

Figure 15.4 gives S_glide = 50/tanγ. For a small γ, S_LG = 50/γ.

Substituting the relation for γ, it gives

15.11

Velocity at touchdown = 1.15 V_{stall_land}.

However, as mentioned before, instead of using Eq. 15.11 the ground distance during glide, S_glide is computed as follows:

15.12

15.13

Ground distance from V_B to stop, V = 0 is formulated next.

To keep the derivation generalised, negative thrust of the thrust reverser is included. However, in the Bizjet example there is no thrust reverser. (Note: the average condition is at 0.707 V_TD).

15.14

where

15.15

15.16

LFL,

15.17

15.5.1 Derivation of Climb Performance Equations

From Figure 15.5, the force equilibrium gives (T – D) = mgsinγ + (m)dV/dt

15.18

Write dV/dt = (dV/dh) × (dh/dt), then;

15.19

By transposing and collecting dh/dt,

15.20a

For un‐accelerated rate of climb, Eq. becomes as follows.

15.20b

The rate of climb is a point performance and is valid at any altitude. The term is a dimensionless term. It penalises unaccelerated rate (numerator of Eq. (15.20)) of climb depending on how fast the aircraft is accelerating during the climb. Part of the propulsive energy is consumed for speed gain instead of altitude gain. Military aircraft make an accelerated climb in the operational arena when the term reduces the rate of climb depending on how fast the aircraft is accelerating. On the other hand, a civil aircraft has no demand for high accelerated climb, instead it makes en route climbs to cruise altitude at a quasi‐steady state climb by holding aircraft climb speed at a constant EAS/Mach number (very slow rate of speed change as altitude increases). Constant EAS climb makes TAS increase with altitude gain, that is, also the Mach number. A constant indication of speed eases pilot workload. During a quasi‐steady state climb at constant EAS, the contribution by the term is low. The magnitude of the acceleration term reduces with altitude gain and becomes close to zero at the ceiling (defined when R/C_accl = 100 ft min⁻¹). (Remember: V = V_EAS/√σ and V_EAS = Ma√i].

15.5.2 Quasi‐Steady State Climb

Civil aircraft en route climb is executed at a constant EAS when the acceleration term,  ≈ 0, and dγ/dt =  ≈ 0. This makes the term in Eq. (15.20) small enough as derived follows.

(The other possibility is to have quasi‐level flight. This is when the climb angle, γ, is small (typically <10°) but the acceleration term, is not. sinγ ≈ γ radians, cosγ ≈ 1. It gives γ² < < 1. This is not dealt with in this book.)

15.5.2.1 Constant EAS Climb

The term can be worked out in terms of constant EAS as follows.

15.21

Below the troposphere (for air, γ = 1.4, R = 287 J kg K⁻¹, g = 9.81 m s⁻²)

In SI, Eq. (1.7a) gives in troposphere T = (288.16–0.0065 h) and

Substituting the values of Eq. (15.21)

15.22

Above the tropopause (for air, γ = 1.4, R = 287 J kg K⁻¹, g = 9.81 m s⁻², a = 295.07 m s⁻¹).

From 11 to 20 km the ambient temperature is held constant at 216.65 K. Eq. (1.8g) gives the density relation in this regime as ρ kg m⁻³ = where base pressure ρ₁ = is at 11 km.

That gives σ = ρ/ρ₀ = (ρ/ρ₁)(ρ₁/ρ₀) = (ρ/ρ₁)(0.36392/1.225) = 0.297 × (ρ/ρ₁) =

or σ = = 0.297 × e^{−[0.0001578](h − 11, 000)} = 0.297 × e^{(1.7355 − 0.0001578h)}

Then (dσ/dh) = 0.297 × e^{(1.7355 − 0.0001578h)}[−0.0001578]

Then Eq. can be written as follows.

15.23

15.5.2.2 Constant Mach Climb

The term can be worked out in terms of a constant Mach number climb as follows.

15.24

Below the tropopause (for air, γ = 1.4, R = 287 J kg K⁻¹, g = 9.81 m s⁻²)

From Eq. (1.7a), atmospheric temperature, T, can be expressed in terms of altitude, h, as follows:

T = (288–0.0065 h) where h is in metres. Substituting the values of γ, R and g, the following is obtained.

Substituting the value in Eq. ,

15.25

Above the tropopause (for air, γ = 1.4, R = 287 J kg K⁻¹, g = 9.81 m s⁻², a = 295.07 m s⁻¹)

In the similar manner, the relations above the tropopause can be obtained. Above the tropopause up to 25 km, the atmospheric temperature remains constant at 216.65 K, therefore the speed of sound remains invariant, that is (da/dh) = 0 making = 0.

Table 15.7 summarises the relation between and the constant speed climb schedules as derived before.

	Below tropopause	Above tropopause
At constant EAS	0.566 M2	0.7 M2
At constant Mach number	−0.133 M2	0 (Mach held constant)

To prepare the Pilot Manual, the airspeed needs to be in V_CAS and the conversion from V_EAS requires incorporating relevant errors associated with the airspeed indicator (see Section ). Readers may refer to [1] for details on compressibility correction. Nowadays, ready data is available in the Electronic Flight Information System (EFIS) memory.

With the loss of one engine at the second segment climb, an accelerated climb would penalise the rate of climb. Therefore, second segment climb with one engine inoperative is done at an unaccelerated climb speed, at a speed V₂ with the undercarriage retracted. The unaccelerated climb equation is obtained by dropping the acceleration term in Eq. 15.18, yielding the following equations.

15.26

The en route climb performance parameters vary with altitude. En route climb performance up to cruise altitude is computed at discrete steps of altitude (say in steps of 5000 ft altitude – see Figure 15.5) within which all parameters are considered invariant and taken as average value within the step altitudes. The engineering approach is to compute the integrated distance covered, time taken and fuel consumed to reach the cruise altitude in steps of small altitudes and then summed up. The procedure is shown next.

Infinitesimal time to climb is expressed as dt = dh/(R/C_accl). The integrated performance within the small steps of altitudes can be written as

15.27

15.28

The distance covered during climb can be expressed as

15.29

where V = average aircraft speed within the step altitude

Fuel consumed during climb can be expressed as

15.30

Military combat aircraft with high (TD) climb is performed in an accelerated climb. The equation for accelerated climb can be derived as follows (Figure 15.5). To keep it simpler, the subscript ∞ is dropped to represent aircraft velocity. This is not dealt with in this book.

• Summary
• Distance covered during climb, R_climb = ∑Δs is obtained by summing up the values obtained in the small steps of altitude gain.
• Fuel to climb, Fuel_climb = ∑Δ_fuel is obtained by summing up the values obtained in the small steps of altitude gain.
• Time to climb, time_climb = ∑Δt is obtained by summing up the values obtained in the small steps of altitude gain.

15.6.1 Derivation of Descent Performance Equations

Descent uses the same equations, except that the thrust is less drag the rate of descent (R/D_accl) is opposite to the rate of climb. The rate of descent is expressed as follows:

15.31

Unlike climb, gravity assists descent, hence it can be performed without any thrust (engine kept at idle rating producing zero thrust). However, passenger comfort and aircraft structural considerations require a controlled descent with maximum rate limited to a certain value depending on the design as explained in Section 15.2.3. Controlled descent is carried out at a part throttle setting. To obtain maximum range, the aircraft should ideally make its descent at the minimum rate. These adjustments will entail varying speed at each altitude. To ease pilot load, descent is made at a constant Mach number and when it has reached the V_EAS limit it adopts a constant V_EAS descent, in a quasi‐steady state, as is done for climb. During quasi‐steady state descent at constant EAS, the contribution by the term is low.

Some special situations may be pointed out as follows.

For unaccelerated descent Eq. gives

15.32

At a higher altitude, the prescribed speed schedule for descent is at a constant Mach, hence above the tropopause V_TAS is constant and descent is kept in unaccelerated flight.

At zero thrust, Eq. 15.32 becomes

15.33

It indicates that at constant V(L/D) the R/C_descent is the same for all weights.

As in climb, the other parameters of interest during descent are range covered (R_descent), fuel consumed (Fuel_descent) and time taken (time_descent) during descent. There are no FAR requirements for the descent schedule. Descent rate is limited by the cabin pressurisation schedule for passenger comfort. FAR requirements are enforced during approach and landing.

The descent velocity schedule is supplied to cater for the ECS capability (cabin descent rate = 300 ft min⁻¹, actual aircraft descent is higher than that, as explained in Section 15.2.3). This requires a shallow descent gradient, part throttle is required to maintain the gradient with the benefit of distance gained during descent.

Section 15.6 gives the worked example of the Bizjet. Integrated performances for climb to cruise altitude and the descent to sea level are computed and the values for distance covered, time taken and fuel consumed are estimated to obtain the aircraft payload‐range. Reference [1] be consulted for details of climb and descent performances on this Bizjet example.

15.9.1 Kinetics of a Coordinated Turn in Steady (Equilibrium) Flight

The equation of motion at an instant of a steady turn has a radius R of turn and aircraft velocity, V, tangential to the flight path. In elemental time Δt, the turning angle is Δψ.

Then, the instantaneous angular velocity can be written as (dψ/dt) in rad s⁻¹.

The instantaneous tangential velocity can be written as V = R(d/ψdt) in ft s⁻¹.

15.45

The thrust line is taken along the flight path. Turn is in a bank angle of ϕ with L′ as the lift force acting at the aircraft plane of symmetry normal to the instantaneous tangential velocity V (true airspeed).

In the horizontal plane, thrust = drag

In the lateral plane, weight, W = mg = L′cosϕ

15.46

where L′ − L = ΔL = W/cosϕ – W = W(1/cosϕ − 1)

15.47

15.48

where n = aircraft load factor acting in the plane of symmetry of the aircraft along L′.

Perpendicular to a flight path towards the instantaneous centre of turn, the force balance gives:

centrifugal force = centripetal force,

15.49

Therefore, the centripetal acceleration acting radial (normal to the flight path), a_n = V²/R

15.50

Combining Eqs. 15.46 and 15.49,

m = L′cosϕ/g = RL′sinϕ/V²

or V²/Rg = sinϕ/cosϕ = tanϕ

Substituting Eq. 15.45,

15.51

15.52

Eq. 15.49 gives,

15.53

Using Eq. 15.47 n = L′/W = L′/mg, Eq. 15.54 gives the radius of turn

15.54

• Load factor, n:

From Eq. 15.53 gives, R = mV²/L′sinϕ

15.55

where, C_L^′ is the lift coefficient while turning under load factor, n.

where L′ = nmg = 0.5ρV²S_WC_L^′

15.56

Substituting Eq. 15.56 into Eq. 15.55,

T = D = 0.5ρV²S_W[C_DPmin + k(2nmg/ρV²S_W)²]

or T = 0.5ρV²S_WC_DPmin + k(0.5ρV²S_W)(2nW/ρV²S_W)²]

or T = 0.5ρV²S_WC_DPmin + k(2n²W²/ρV²S_W)]

15.57

Solve Eq. 15.57 for load factor, n

or n² =

15.58

15.9.2 Maximum Conditions for a Turn in the Horizontal Plane

Differentiating Eq. 15.58 with respect to V setting it equal to zero gives the maximum conditions of V_{n_max} as follows.

15.59

Substituting in Eq. 15.18, the maximum load factor, n, can be obtained

(15.60).

15.10.1 Checking TOFL (Bizjet) – Specification Requirement 4400 ft

During take‐off ground run, aircraft accelerates from zero to lift‐off speed, V_{lift_off}, causing continuous changes in aircraft aerodynamics characteristic, e.g., lift and drag. An accurate TOFL estimation requires C_L and C_D variation with speed gain. Readers may refer to [1] for the methodology adopted to compute accurate TOFL using a graph showing the varying force parameters during the ground run.

The acceleration term in take‐off has a strong contribution to the determination of BFL. Ref [1] shows that TOFL estimation using average acceleration gives result within 1% of the accurate TOFL value for all engine operating case. It is convenient to use average acceleration, ā, term at this stage of study.

Equation gives average acceleration, ā, as follows.

The average acceleration, ā, can be applied for the segment with initial velocity, V_i, and final velocity, V_f, i.e., at 0.707V = 0.707 × (V_f − V_i) + V_i of the segment of operation

For the ground run, Eq. 15.8a gives the following.

15.8b

Table 15.5 may be used for Bizjet aircraft take‐off estimation at a sea level ISA day to prepare the spreadsheet for repeated computations. Table 15.8 gives the Bizjet aircraft data generated thus far. To make the best use of the available data all computations are done in the FPS system. The results can be subsequently converted to the SI system.

At V₂ speed, T@119.25 kt = 2860 lb per engine (20° flap)

Rolling friction coefficient on paved runway, μ = 0.025.

Several decision speeds are worked out to estimate takeoff capabilities. First, the all‐engine TOFL is estimated. Next, the BFL is estimated, for which the decision speed V₁ is to be determined in the inadvertent case of one critical engine failure.

15.10.1.1 All‐Engine Takeoff – 20° Flap

Table 15.9 gives the values for all‐engine TOFL with the Bizjet at 20° flap. Due to extended undercarriage and flap deflection, the aircraft lift to drag ratio degrades to a typical value of approximately 10 for the Bizjet.

			Vstall	VR
V, kt	80	90	100	110
V, ft s−1	135.04	151.92	168.8	185.68
0.5ρSW	0.384	0.384	0.384	0.384
0.707V	95.47	107.41	119.34	131.28
āavg	8.91	8.85	8.793	8.73
ΔV	135.04	151.92	168.8	185.68
Vavg	67.52	75.96	84.4	92.84
SG, ft	1024	1304	1620	1976

15.10.1.1.1 Flare from V_R to V₂

The aircraft is under manoeuvre load through and in climb, and still accelerates at a lower rate. Aircraft flaring through rotation is a complex physics and is described in detail in [1]. Typically, in this class of aircraft, it takes 3 s. Here the computation is simplified by taking the average velocity from V_R to V₂ to cover field length distance in 3 s.

Taking flare time as 3 s for the Bizjet, the average V_{LO to V2} = (192.43 + 202.56)/2 = 197.496 ft s⁻¹

Computed all‐engine operating field length = 1976 + 592.3 ≈ 2568 ft

To ensure safety, the FAR requires an all‐engine operating TOFL at a 15% higher margin than the computed value and it may exceed the value obtained for a one engine inoperative BFL value. In that case, the higher value of the two is used.

FAR all‐engine operating field length for a Bizjet at 20° flap, TOFL_{all_eng} = 1.15 × 2568 = 2953 ft

To compare with the simplified method to compute , take average acceleration at

compared to 1976.3 ft in Table 15.9. This shows excellent agreement.

15.10.1.2 One Engine Inoperative – Balanced Field Takeoff (BFL) Inoperative

Section 15.3.2 gives the BFL takeoff to make sure that the aircraft can stop on the airfield in case an engine fails. It requires a decision speed V₁, below which the aircraft stops and above which the aircraft continues with the takeoff operation with power available to maintain the FAR stipulated minimum climb gradient up to the second segment and then return to land immediately. At first, V₁ is estimated and subsequently establishes the BFL for takeoff. Table 15.9 showing all engine takeoff values can be used up to a speed of V₁.

• Segment A – all engines operating up to the decision speed V₁ – BFL run

To determine the decision speed V₁, estimate three speeds, for example, 80, 90 and 100 kt. The all‐engine operating case up to these speeds can be taken from Table 15.9.

• Segment B – one engine inoperative acceleration from V₁ to lift‐off speed, V_R

This phase is a transient one; the aircraft leaves the ground to become airborne. Its lift and drag characteristics are changing fast. With the aircraft in a high drag configuration with one engine inoperative, drag estimation is quite difficult. Therefore, a simplified approach is taken to compute distance covered and time taken from V₁ to lift‐off speed, V_LO. The simplification gives a reasonable result and conveys the physics involved. Industries using computers and more accurate lift and drag data adopt detailed calculations.

In flight, the lift to drag ratio in such a dirty configuration will be low, in this class in the order of around 9.5. On the ground below rotation speed, the lift is low and so is drag is low; the simplification takes the same value of (L/D) ≈ 9.5. Therefore, the average value of (C_D/C_L) ≈ 0.105 is taken. A typical average C_L ≈ 0.8 is taken, the weight on the wheels is lighter on account of some lift generated on the wing. Because one engine is inoperative there is a loss of power by half [(T/W)_avg = 0.154)]. The simplified method is used because the difference will be small.

The velocity that would give average acceleration is V_0.707 = 0.7 × (V_R − V₁) + V₁.

One engine inoperative ā for the Bizjet after decision V₁ is as follows

20° flap

Braking takes place after an engine fails when it has the all‐engine acceleration. With the loss of thrust aircraft on account of engine failure the acceleration does not suddenly drop to low levels in a discrete step down, but the aircraft gradually retards to a lower level of acceleration. Therefore, for the interval of computation, the average value has to be taken. This is an important consideration that is often overlooked. ā_ave = 0.5 × (ā_{all_engine} + ā_{one_engine_failed})

Section gives the ground distance covered, S_G = V_ave × (V_f − V_i)/ā

Table 15.10 computes Segment B, the ground distance covered from V₁ to V_LO.

• Segment C – Flaring distance with one engine inoperative from V_R to V₂

			Vstall	VR
V, kt	80	90	100	110
V, ft s−1	135.04	151.92	168.8	185.68
VLO, ft s−1	192.432	192.432	192.432	192.432
ΔV	57.392	40.512	23.632	6.752
Vavg	163.74	172.18	180.62	189.06
0.707V	175.62	180.56	185.51	190.45
Ā	3.416	3.374	3.331	3.286
āave	6.162	6.114	6.062	6.006
SG, ft	1525	1141	704	213

This is the flaring distance to reach V₂ from V_R. From the statistics, time taken to flare is 2 s. Tables 15.10–15.12 compute the ground distance covered from V_LO to V₂ with one engine inoperative for the flap settings. In this segment the aircraft is airborne, hence there is no rolling friction. By taking the average velocity between V₂ and V_R the distance covered during flare is given.

The next step is to compute the stopping distance with the maximum application of brakes.

			Vstall	VR	VLO	V2
V (kt)	80	90	100	110	114	120
Segment (B + C)	2118	1734	1297	781	593	593

• Segment D – Engine failed (recognition time)

Table 5.12 computes the ground distance covered from V₁ to V_B.

VR at 1.11Vstall f/s	184.61
VLO at 1.15Vstall f/s	193.00
V2 at 1.2Vstall f/s	201.39
Vave f/s	197.20
SGflair (3 s), ft	591.6

The distance is covered in 1 s due to pilot recognition time and 2 s for the brakes to act from V₁ to V_B (flap settings are of little consequence).

Table 15.12 shows the Segment D engine failure recognition distance.

• Segment E – braking distance from V_B to zero velocity (Flap settings are of little consequence)

V – knots	80	90	100	114
V ft s−1	135.04	151.92	168.8	192.432
Distance in 3 s at V1, SG_B – ft	405.12	456	506.4	577.3

Reaction time to apply the brake after the decision speed, V₁ is 3 s. The aircraft continues to accelerate a little in the 3 s but speed returns to V_B – this is ignored. Table 15.13 computes the ground distance covered from V_B to stop.

V, kn	80	90	100
V, ft s−1	135.04	151.92	168.8
ΔV, ft	135.04	151.92	168.8
Vavg, ft	67.52	75.96	84.4
0.707V	95.47	107.41	119.34
Ā	−12.08	−11.86	−11.62
SG, ft	755	973	1197
(D + E)	1160	1429	1732

Aircraft in full brake mode with μ_B = 0.4, all engines shut down and average C_L = 0.5,

Using Eq. 15.6, the average acceleration based on 0.707 V_B (≈ 0.707 V₁) reduces to

Select the distance when segments (B + C) equals segment (D + E) as shown in Figure 15.8. With 20° flap, the decision speed V₁ is 94 kt covering a distance of 1570 ft. Then TOFL becomes 1430 + 1570 = 3000 ft. This is summarised in Table 15.14 and Figure 15.8. Both satisfy the specified TOFL requirement of 4400 ft.

	BFL
Flap, degrees	20
Decision speed, V1, knots	94
Stall speed, Vstall, knots	99.3
Rotation speed, VR, knots	109.3
Lift‐off speed, VLO, knots	114
V2 speed, V2, knots	119.3
Distance, ft at (B + C) = (D + E), ft	1570
All‐engine distance at V1 speed, ft	1430
BFL, ft	3000
second segment climb gradient (computed in next chapter)
All‐engine distance at V2 speed, ft	2568 ft
All‐engine increase by 15%	2953 ft

The TOFL requirement is 4400 ft, recommended takeoff procedure uses a 20° flap (Figure 15.9). At lower takeoff weight and/or a longer runway length (4400 ft), the pilot may choose an 8° flap that gives a better second segment climb gradient. With the air‐conditioning switched on and at an eight flap setting, the BFL should come close to 4000 ft. The readers may compute for an 8° flap decision speed.

• Discussion for takeoff analysis

Increase of flap setting improves the BFL capability at the expense of a loss of climb gradient (this will be shown in the next chapter). With one engine inoperative, the percentage loss of thrust for a two engine aircraft is the highest (50% lost). With one engine failed, the aircraft acceleration suffers and the ground run taken from V₁ to lift‐off is higher. Table 15.15 summarises the takeoff performance and the associated speed schedules for the two flap settings. The ratio of speed schedules can be made to vary for pilot ease, as long as it satisfies FAR requirements. This is an example of the procedure.

	20		8
Flap setting, degrees	knots	ft s−1	knots	ft s−1
Vstall @ 20 680 lb	99.38	167.83	106	179
bVmc at 0.94V1	84.4 b	142.8	91.2 b	153.9
V1 decision speed	90.00	151.92	96.5	163.74
VR = 1.05Vstall	109.32	184.61	116.60	196.92
VLO at 1.15Vstall	114.28	193.00	121.90	205.8
aV2 = 1.2 Vstall at 35 ft altitude	119.26	201.39	127.2	214.82
Vmu at 1.02 VR (lower than VLO)	115.50	188.30	119.00	200.85
BFL, ft	2940		3220
All‐engine takeoff time, s	≈31 s

^aIf required, V₂ can be higher than 1.2 V_stall.

^bAircraft control surfaces (mainly rudder) are sized to get a V_mc below V₁ for the pilot to have the speed margin. Here, it arbitrarily chosen to at 75 kt – this needs to be computed (a design task beyond the scope of this book). This is because the lowest V₁ is when the aircraft is lightest (say at the landing weight of 15 800 lb) and at the highest flap deployment (for an inadvertent case, it can be at 40° flap, otherwise at 20° flap). This point is often overlooked in aircraft design coursework. Figure 15.9 gives the Bizjet takeoff speed schedules.

Higher flap settings would give more time between decision speed V₁ and the rotation speed V_R. However, it is not a problem if V₁ is close to V_R, so long there is pilot reaction time is available if one engine fails; if not then in a very short time the rotation speed V_R is reached and the aircraft takes off (typically, BFL is considerably less than the available airfield length). Also V_mu is close to V_R, hence there is little chance for tail dragging. If the pilot makes an early rotation then V_mu may not be sufficient for a lift‐off and the aircraft will tail drag until it gains sufficient speed for the lift‐off. If the engine fails early enough then the pilot has sufficient time to recognise it and act to abort takeoff.

With more than two engines, the decision speed V₁ is further away from the rotation speed V_R. The pilot must remain alert as the aircraft speed approaches the decision speed V₁ and must react quickly if an engine fails.

The readers should compute for other weights to prepare graphs for ISA day and ISA + 20 °C.

15.10.2 Checking Landing Field Length (Bizjet) – Specification Requirement 4400 ft

Landing weight of the Bizjet is 15 800 lb (wing loading = 48.92 lb ft⁻²) and at full flap 40° extended C_Lmax = 2.2.

Average velocity from 50 ft height to touch down = 168 ft s⁻¹.

Distance covered before brake application after 5 s (may differ) from 50 ft height,

Aircraft in full brake with μ_B = 0.4, all engines shut down and average C_L = 0.4, C_D/C_L = 0.11.

Section 15.10.1 for average acceleration based on 0.707 V_TD = 110.6 ft s⁻¹.

Then q = 0.5 × 0.002378 × 110.6² = 14.54

The deceleration equation becomes,

Deceleration, ā = 32.2 × [(−0.4) + (0.119) × (0.29)] = −11.8 ft s⁻²

Distance covered during braking, S_{G_0Land} = (158 × 79)/11.8 = 1058 ft

Landing distance S_{G_Land} = 840 + 1058 = 1898 ft

Multiplying by 1.667 (dry runway), the rated LFL = 1.667 × 1898 = 3164 ft, within requirement of 4400 ft. Typically, BFL and LFL are close to one another.

Designers must also check the capability of go‐around in case missed approach (approach climb) or baulked landing (landing climb) as described in Section 15.4.1. These cases are not worked out in this book. Generally, airfield is sufficiently long and with robust design ensures that this situation can handled by pilots. Safety can never be compromised.

15.10.3 Checking Takeoff Climb Performance Requirements (Bizjet)

There are three requirements to be substantiated for the climb performance. The three requirements are given here for the two‐engine aircraft. The first two are the FAR requirements. Table 5.3 gives aircraft configuration and FAR requirements for the first and second segment climb. The four takeoff climb segments are shown in Figure 15.3.

1. Check that the FAR first segment climb requirement of positive gradient is maintained.
2. Check that the FAR second segment climb gradient requirement exceeds 2.4%.
3. Check the market specification of the initial en route climb rate equals or exceeds 2600 ft min⁻¹. The cabin pressurisation system should cope with the rate of climb. This is a customer specification and not a FAR requirement.

The second segment starts at 35 ft altitude with flaps extended and undercarriage retracted (one engine inoperative) and ends at 400 ft altitude. Aircraft is maintained at V₂ speed to keep the best gradient. A loss of 50% thrust does not favour accelerated climb, which will be low in the case. Engine is at takeoff rating. The available one engine installed thrust is taken from Figure 13.11. The thrust is kept invariant at takeoff rating through first segment and second segment climb. Table 15.16 summarises the first and second segment climbs for both 8° and 20° flap settings.

15.10.4 Checking Initial En Route Rate of Climb – Specification Requirement is 2600 ft min⁻¹

Initial en route climb starts at 1000 ft altitude (ρ = 0.0023 slug ft⁻³, σ = 0.9672), all engines throttled back to maximum climb rating and the aircraft has a clean configuration. The aircraft makes an accelerated climb from V₂ to reach 250 KEAS, which is kept constant in a quasi‐steady state climb until it reaches Mach 0.7 at about 32 000 ft altitude, from where the Mach number is held constant in the continued quasi‐steady state climb until it reaches the cruise altitude. Fuel consumed during second segment climb is small and taken empirically (from statistics) to be 80 lb (see Table 15.17). Therefore, aircraft weight at the beginning of en route climb, W = 20 600 lb.

At 250 kt (422 ft s⁻¹, Mach 0.35) the aircraft lift coefficient C_L = M/qS_W = 20 600/(0.5 × 0.0023 × 422² × 323)= 20 600/66150 = 0.311.

Clean aircraft drag coefficient from Figure 11.2 at C_L = 0.311 gives C_Dclean = 0.0242.

Clean aircraft drag, D = 0.0242 × (0.5 × 0.0023 × 422² × 323) = 0.0242 × 66 150 = 1600 lb

Available all‐engine installed thrust at maximum climb rating from Figure 13.12 at Mach 0.378 is T = 2 × 2450 = 4900 lb

From Eq. 15.20a, quasi‐steady state rate of climb is given by,

At quasi‐steady state climb, Table 15.7 gives

The capability satisfies the market requirement of 2600 ft min⁻¹ (15.2 m s⁻¹).

(Civil aircraft rate of climb is limited by the cabin pressurisation schedule. The aircraft is limited to 2600 ft min⁻¹ at an altitude where the cabin pressurisation rate reaches its maximum capability. Naturally, at a low altitude of 1000 ft, this limit is not applicable.)

15.10.5 Integrated Climb Performance (Bizjet)

Integrated climb performance is not a specification for substantiation – it is used to obtain aircraft payload‐range capability, which is a requirement to be substantiated. It is reasoned in Section 15.1 that only the results of the integrated climb performances in graphical form are given as shown in Figures 15.10 and 15.11.

It is convenient to first establish the climb velocity schedule (Figure 15.10a) and the point performances of rate of climb (Figure 15.10b) up to ceiling altitudes and for at least three weights for interpolation. Bizjet carries out quasi‐steady state climb at a constant V_EAS = 250 kt until it reaches Mach 0.7 and thereafter continues at constant Mach until it reaches the ceiling (rate of climb = 100 ft min⁻¹).

The computations for the integrated climb performance are carried out in steps of approximately 5000 ft altitude (as convenient – at higher altitudes in smaller steps) in which the variables are kept invariant using their mean values (see Figure 15.5). Eqs. 15.27–15.29 are used to compute the integrated performances. Figure 15.11 gives the integrated performances of fuel consumed, distance covered and time taken to climb at the desired altitude.

The detailed computations to obtain all the graphs given in this section are shown in the companion book [1]. Classroom instructors may assist the computational work to obtain similar performance graphs for their projects.

15.10.6 Checking Initial High‐Speed Cruise (Bizjet) – Specification Requirement of High‐Speed Cruise Mach 0.75 at 41 000 ft Altitude

Aircraft at high speed initial cruise is at Mach 0.75 (716.4 ft s⁻¹) at 41 000 ft altitude (ρ = 0.00055 slug ft⁻³). Fuel burned to climb is computed (not shown) to be 700 lb. Aircraft weight at initial cruise is 20 000 lb.

At Mach 0.75 the aircraft lift coefficient, C_L = MTOM/qS_W = 20 000/(0.5 × 0.00055 × 716.4² × 323) = 20 000/45627 = 0.438.

Clean aircraft drag coefficient from Figure 11.2 at C_L = 0.438 gives C_Dclean = 0.0305.

Clean aircraft drag, D = 0.0305 × (0.5 × 0.00055 × 716.4² × 323) = 0.0305 × 45 627 = 1392 lb.

Available all‐engine installed thrust at the maximum cruise rating at the speed and altitude from Figure 13.13 at Mach 0.75 is T = 2 × 750 = 1500 lb (adequate).

The capability satisfies the market requirement of Mach 0.75 at HSC.

15.10.7 Specific Range (Bizjet)

Specific range is a convenient way to present cruise performance. Using Eq. 15.42 the specific range, Sp.Rn is computed. It is needed to compute the cruise segment performances (fuel, distance and time). Figure 15.12 gives the specific range for the Bizjet (classroom example).

Cruise segment distance cover is follows.

Cruise range = 2000 − (Distance to climb to 43 000 ft) − (distance to descend from 45 000 ft altitude).

From the Sp.Rn values the distance covered is worked out, which in turn gives the time taken for the distance.

15.10.8 Descent Performance (Bizjet) – Limitation Maximum Descent Rate of 1800 ft min⁻¹

It is also reasoned in Section 15.1 that only the results of the integrated descend performance in graphical form is supplied as shown in Figures 15.13 and 15.14. It is convenient to first establish the descent velocity schedule (Figure 15.13) and the point performances of the of descent (Figure 15.14) up to sea level (valid for all weights – the difference between the weights is ignored). While redoing the calculations by the readers, there could be small differences in the results.

The related governing equations are explained in Section 15.6.1. It also mentions that descent rate is restricted by rate of cabin pressurisation schedule to ensure passenger comfort. Two of the difficulties in computing descent performance are to have the part throttle engine performance and the ECS pressurisation capabilities that dictate the rate of descent, which in turn stipulates the descent velocity schedule. These are not supplied in the book. Classroom instructors are to assist to establish these two graphs. In absence of any information these two may be used. The following simplifications could prove useful.

The first simplification is in obtaining part throttle engine performance as follows.

1. Descent is carried out anywhere at 50–70% of maximum power rating.
2. Zero thrust at idle rpm at about 40% of the maximum rated power/thrust.

The second simplification is that the descent velocity schedule is supplied to cater for the ECS capability (cabin descent rate = 300 ft min⁻¹, actual aircraft descent is higher than that – Figure 15.13b). This is explained in Section 15.2.3.

In industry, the exact installed engine performance at each part throttle condition is computed from the engine deck supplied by the manufacturer. Also, the ECS manufacturer supplies the cabin pressurisation capability from which aircraft designers work out the velocity schedule.

The inside cabin pressurisation is restricted to an equivalent rate of 300 ft min⁻¹ at sea level to ensure passenger comfort. The aircraft rate of descent is then limited to the pneumatic capability of the ECS. The Bizjet is restricted to a maximum of 1800 ft min⁻¹ at any time (for higher performance, at lower altitudes it can be raised up to 2500 ft min⁻¹). Descent speed schedule is to continue at Mach 0.6 from cruise altitude until it reaches the approach height when it changes to constant V_EAS = 250 kt until the end (for higher performance, it can be raised to Mach 0.7 and V_EAS = 300 kt). The longest range can be achieved at minimum rate of descent. Evidently, these will require throttle dependent descents to stay within the various limits.

It is convenient to establish first the point performances of velocity schedule (Figure 15.13a) and rate of descent (Figure 15.13b – all weights – variation is small). Descent is performed within the limits of human comfort level. However, in an emergency, rapid descent becomes necessary to compensate for loss of pressure and recover oxygen.

Integrated descent performance is computed in the same way as for climb. The integrated descent performance is computed in steps of approximately 5000 ft altitude (or as convenient) in which the variables are kept invariant. No computational work is shown here.

Figure 15.14 plots the fuel consumed, time taken and covered during descent from ceiling altitude to sea level. While redoing the integrated descent performances by the readers, there could be small differences in the results.

15.10.9 Checking out the Payload‐Range Capability – Requirement of 2000 nm

A typical transport aircraft mission profile is shown Figure 15.15.

Equations 15.34–15.35 give the mission range and fuel consumption expressions as:

where R_climb = ∑Δs_climb and R_descent = ∑Δs_descent computed from steps of heights.

where Fuel_climb = ∑Δfuel_climb and Fuel_descent = ∑Δfuel_descent computed from steps of heights.

Minimum reserve fuel is computed for an aircraft holding an 5000 ft altitude at around Mach 0.35–0.4 at about 50% of the maximum rating for 45 min or an 100 nm diversion cruising at 0.5 Mach and 25 000 ft altitude plus 20 min. The amount of reserve fuel has to be decided by the operator suited to the region of operation. The worked‐out example takes the first option.

Fuel is consumed during taxiing, takeoff and landing without any range contrition and this fuel is to be added to the mission fuel with the total, known as block fuel. The time taken from the start of the engine at the beginning of the mission to the stop of the engine at the end of the mission is known as block time, in which a small part of time is not contributing to gain in range. Taken from the statistics of operation, these additional fuel burn and time consumed without contributing to range are given in Table 15.17.

Reserve fuel: At 5000 ft altitude (ρ = 0.00204 lb ft⁻³) and 0.35 Mach (384 ft s⁻¹) it gives C_L = 0.323 resulting in C_D = 0.025 (Figure 9.2). Equating thrust to drag, T/engine = 610 lb. Corresponding fuel flow rate at this speed altitude is 210 lb/h per engine; holding for 45 min consumes 2 × 0.75 × 210 = 315 lb. It is safer to keep more in reserve if a 100 nm diversion is required, which is not a stipulation if 45 min hold is taken. Therefore, it is decided to carry 600 lb reserve fuel for holding/diversion around landing airfield. The range performance can be improved with a step‐climb from 43 000 to 45 000 ft as the aircraft becomes lighter with fuel consumed.

The cruise altitude of the Bizjet starts at 43 000 ft and ends at 45 000 ft (the design range is long to make a step cruise). Taking the average value of cruising at 44 000 ft (ρ = 0.00048 slug ft⁻³), The methods to compute R_climb and R_descent are discussed in Sections 15.5 and 15.6. Using Figures 15.11 and 15.14, the required values are given as R_climb = 170 nm, Fuel_climb = 800 lb, Time_climb = 25 minutes, R_descent = 160 nm and Fuel_descent = 340 lb and Time_descent = 30 minutes (in part throttle gliding descent). Table 15.17 gives the segment details of aircraft weight, distant covered, fuel burned and time taken. The aircraft is at LRC schedule operating at Mach 0.7.

Table 15.17 gives the following.

MTOM = 20 680 lb (9400 kg)

Landing weight = 20 680–5500 (onboard fuel) + 600 (reserve fuel, not consumed) = 15 780 lb (rounded to 15 800 lb).

Initial and final cruise weights can then be computed from the climb and descent performance data (Figures 15.11 and 15.14, respectively) = 800 lb and 340 lb, respectively.

Initial cruise weight W_icr = 20 680 (MTOM) + 80 (taxi) + 340 (takeoff) − 800 (climb) = 19 680 lb

Final cruise weight W_fcr = 15 800 (landing weight) + 340 (descent) + 80 (taxi‐in) = 16 220 lb

Available cruise fuel = 19 680–16 220 = 3460.

Split the fuel load to 1460 lb cruising at 43 000 ft altitude and climb to 45 000 ft to use the 2000 lb fuel.

In operational practice, it is not done in this way. This example is to demonstrate the capability and logic of the operation. It can be seen that the step‐climb part has not been computed. It assumes that it is a discrete jump without losing fidelity to get the extent of possible range. Industries meticulously plan stage to make the climb to get the best range. Today, microprocessor‐based FBW operation makes the optimised cruise operation that is in a continuous shallow climb. Aircraft Performance Management (APM – see Section 16.5.2) is integral to flight and fleet management. There are several ways to climb, cruise and descend, treated extensively in Ref. [1].

In case, the specific range graphs are not available, the cruise segment range can also be obtained as shown next. First, compute the thrust required. Then from the sfc of the engine the fuel flow rate can be obtained and using the time required for the sector length, fuel required is computed.

From Table 15.17, the mid‐cruise weight is (19 700 + 16 240)/2 = 17 970 lb.

LRC is at Mach 0.7 (677.7 ft s⁻¹, 401.5 kt). Engine power setting is below maximum cruise rating.

Aircraft lift coefficient, C_L = 17 970/(0.5 × 0.00046 × 677.7² × 323) = 17 970/34120 = 0.527.

From Figure 11.2 the clean aircraft drag coefficient, C_D = 0.033.

Aircraft drag, D = 0.033 × (0.5 × 0.00046 × 677.7² × 323) = 0.033 × 34 120 = 1126 lb.

Thrust required per engine is 1126/2 = 563 lb. Figure 13.13 is meant for maximum cruise rating to achieve HSC. Therefore, at LRC at Mach 0.7 it is throttled back with reduced fuel flow, sfc stays the same. From Figure 13.4, the sfc at the speed and altitude is 0.72 lb h⁻¹ lb⁻¹ Fuel flow for two engines = 0.72 × 1126 = 811 lb h⁻¹ at 43 000 ft.

Mission range satisfies the requirement of 2000 nm. Block time for the mission is 5.45 h and block fuel consumed is 4923 lb (2233 kg). On landing, the return taxi‐in fuel of 20 lb is taken from the reserve fuel of 600 lb (45 min holding or diversion). Total onboard fuel carried is therefore: 4880 + 600 = 5480 ≈ 5500 lb (2500 kg).

This checks the Bizjet weights breakdown given in Table 15.17.

The payload‐range graph is given in Figure 15.16. A summarised discussion of the Bizjet design is given in Section 15.13.1.

The fuel tank has a larger capacity than what is required for the design payload‐range. Payload could be traded to increase range until the tanks fill up by trading the payload. Further reduction of payload would make the aircraft lighter, thereby increasing the range.

h3

15.11.1.5. Mission Profile

Fuel load and management depends on the type of mission. Military mission profiles are varied. Figure 15.17 gives a typical normal training profile to gain airmanship and navigational skills in an advanced aircraft.

The training profile segment breakdown is given in Table 15.18.

	Engine rating = % rpm
Taxi and takeoff	60% (idle)
takeoff and climb to 6 km altitude	TO @ 100% then @ 95%
Four turns/stalls	1 min @ 95% + 3 min @ 60%
Climb from 5 to 8 km altitude	95%
Four turn spins	60%
Climb from 5 to 8 km altitude	95%
Four turn spins	60%
Four turns/stalls	1 min @ 95% + 3 min @ 60%
Climb from 5 to 6 km altitude	95%
Aerobatics practise	95%
Descent and practice force landing	2 min @ 95% + 6 min @ 60%
Three circuits for landing practise	Average 80%
Approach, land return taxi	60%
Trainee pilot allowance	95%

Block fuel and block time are computed by adding fuel and time consumed in each segment. However, distance covered is meaningless as the aircraft returns to the base. Armament training practise closely follows the combat mission profile. Combat missions depend on target range and the expected adversary defence capability. Two typical missions are shown in Figure 15.18. Air defence would require continuous intelligence information feedback.

The study of a combat mission requires complex analysis by specialists. Defence organisations conduct these studies and are understandably kept confidential in nature. Game theory, twin dome combat simulations and suchlike are some of the tools for such analysis. Actual combat may yet prove quite different as not all is known about the adversary's tactics and capabilities. Detailed study is beyond the scope of this book and possibly of most academies.

15.11.2 Checking TOFL (AJT) – Specification Requirement 3600 ft (1100 m)

Being single engine, there is no question regarding BFL. Military aircraft TOFL must satisfy the CFL. The CFL can also have a decision speed V₁, which is determined by whether the pilot can stop within the distance available, or otherwise the pilot needs to eject.

To obtain the CFL for a single engine aircraft, the decision speed is at the critical time just before the pilot initiates rotation at V_R. Then the decision speed V₁ is worked out from the V_R by giving 1 s as the recognition time. Engine failure occurring before this V₁ leads the pilot to stop the aircraft by applying full brakes and deploying any other retarding facilities available, for example a brake‐parachute. There should be sufficient runway available for the pilot to stop the aircraft from this V₁. If there is not enough runway available, then a new decision speed V₁, considerably lower than the V_R, has to be worked out in a similar manner as done for the Bizjet. Engine failure occurring after the V₁ means the pilot will have little option other than to eject (if there is not enough runway/clearway available to stop). For multi‐engine aircraft, determining the CFL follows the same procedure as the civil aircraft computation but complying with Milspecs.

In this book, the following three (CFL) requirements for the AJT are worked out.

1. (i) To meet the takeoff distance of 3600 ft (1100 m) to clear 15 m
2. (ii) To meet the initial rate of climb (unaccelerated): 10 000 ft min⁻¹ (50 m s⁻¹)
3. (iii) To meet the maximum cruise speed of Mach 0.85 at 30 000 ft altitude

MIL‐C501A requires a rolling coefficient μ = 0.025 and minimum braking coefficient μ_B = 0.3. Training aircraft will have a good yearly utilisation operated by relatively inexperienced pilots (with about 200 h of flying experience) carrying out a very large number of take‐offs and landings on relatively shorter runways. AJT brakes are generally more robust in design giving a brake coefficient, μ_B = 0.45, which is much higher than the minimum Milspec requirement. Figure 15.19 gives the speed schedules of single engine military type aircraft.

Both the Bizjet and the AJT have the same class of aerofoil and same types of high‐lift devices. Therefore, there is a strong similarity in the wing aerodynamic characteristics. Table 15.19 gives the AJT data pertinent to takeoff performance.

Equation 15.6 gives average acceleration as ā = g [(T/W – μ) – (C_LSq/W)(C_D/C_L – μ)]

Using the values given in Table 15.19,

15.11.2.1 Takeoff with 8° Flap

Refer to Figure 15.19 showing an AJT takeoff profile.

• Distance covered from zero to the decision speed V₁

The decision speed V₁ is established as the speed at 1 s before the rotation speed V_R. Estimate the acceleration of 16 ft s⁻² (can be computed first and then iterate without estimation).

Aircraft velocity at 0.707 V₁ = 122.22 ft s⁻¹.

Up to V₁, the average C_L = 0.5 (still at low incidence)

Using Eq. 15.8b, the distance covered until lift‐off speed is reached,

However, provision must be made if the engine fails at the decision speed, V₁, when braking has to be applied to stop the aircraft. The designers must make sure that the operating runway length is adequate to stop the aircraft.

If the engine is operating, then the AJT continues with the takeoff procedure when lift‐off takes place after rotation is executed at V_R.

• Distance covered from zero to lift‐off speed V_LO

Using the same equation for distance covered up to the decision speed V₁ with the change to lift‐off speed, V_LO = 192.5 ft s⁻¹. Aircraft velocity at 0.707 V_LO = 134.75 ft s⁻¹.

Up to V_LO, the average C_L = 0.5 (still at low incidence)

Then ā = 32.2 × (0.51 − C_Lq/773.33) = 32.2 × (0.51–4.35/773.33) = 32.2 × 0.5004 = 16.24 ft s⁻².

Using Eq. , the distance covered until lift‐off speed is reached,

• Distance covered from V_LO to V₂

The flaring distance is required to reach V₂ clearing the 50 ft obstacle height from V_LO. From statistics; the time taken to flare is 3 s with at 8° flap and at V₂ = 206 ft s⁻¹.

In 3 s the aircraft would cover S_{G_LO_V2} = 3 × 206 = 618 ft.

• Total takeoff distance

The takeoff length is S_{G_LO} + S_{G_LO_V2} = (1140 + 618) = 1758 ft, much less than the required TOFL is 3500 ft to cater for the full weight 6800 kg (≈15 000 lb) for armament training. In that case, a higher flap setting of 20° may be required. At a higher flap setting, the aircraft will have a shorter CFL for the same weight.

• Stopping distance and the CFL

The next step is to check what is required to stop the aircraft if an engine has failed at V₁. Stopping distance has three segments. CFL is normally longer than TOFL.

1. Distance covered from zero to the decision speed V₁, S_{G_1}. It is computed as before to be 912 ft.
2. Distance covered from V₁ to braking speed V_B, S_{G_1_B}.
3. Braking distance from V_B (≈V₁) to zero velocity, S_{G_B_0}.

The other two are estimated here:

• Distance covered from V₁ to braking speed V_B

At engine failure, assume 3 s for pilot recognition and taking action. Therefore, the distance covered from V₁ to V_B is S_{G_1_B} = 3 × 174.6 = 524 ft

• Braking distance from V_B (≈V₁) to zero velocity (flap settings are of little consequence)

The most critical moment of brake failure is at the decision speed V₁ when the aircraft is still on the ground. With full brake coefficient μ = 0.45 and average C_L = 0.5, V₁ = 174.6 ft s⁻¹

Aircraft velocity at 0.707 V₁ = 122.22 ft s⁻¹.

The equation for average acceleration reduces to

Using Eq. , the distance covered till lift‐off speed is reached,

Therefore, the minimum runway length (CFL) for takeoff should be = S_{G_1} + S_{G_1_B} + S_{G_0}

TOFL = 940 + 524 + 1119 = 2583 ft. It is still within the specified requirement of 3600 ft.

The takeoff length is thus 1746 ft, much less than the CFL of 2511 ft computed before. The required TOFL is 3500 ft to cater for the full weight 6800 kg (≈15 000 lb) for armament training. In that case, a higher flap setting of 20° may be required. At a higher flap setting the aircraft will have a shorter CFL for the same weight.

Evidently, the length of the runway available will dictate the decision speed V₁. If the airfield length is much longer then the pilot may have a chance to set down the aircraft if an engine failure occurs immediately after lift‐off and it may be possible to stop within the available airfield length, which must have some clearway past the runway end.

Compare with the minimum Milspec requirement of breaking friction, μ = 0.3.

The equation for average acceleration reduces to

Using Eq. 18b, the distance covered from V_B to zero

This value is on the high side.

The minimum runway length for takeoff should be = S_{G_1} + S_{G_1_B} + S_{G_0}

The CFL (critical field length) = 912 + 524 + 1660 = 3096 ft is on the high side but within the specification of 3600 ft. It is for this reason that a higher brake coefficient of 0.45 is taken. This is not a problem as nowadays wheels with good brakes have much a higher friction coefficient μ.

A reduction of the decision speed to 140 ft s⁻¹ (83 kt) would reduce the S_{G_0} = 1068 ft bringing down the CFL to 2504 ft.

15.11.3 Checking the Second Segment Climb Gradient at 8° Flap

V₂ = 206 ft s⁻¹ (see Table 15.19) and W = 10 580 lb (4800 kg)

Clean aircraft drag coefficient from Figure 11.12 gives C_Dclean = 0.098.

Available thrust is 5100 lb (from Figure 13.15).

At V₂ speed to clear 50 ft, the climb is at an unaccelerated rate but thereafter it will be at an accelerated rate. From Eq. steady state rate of climb is given by,

Hence, R/C_accl = {[206 × (5100–2995) × 60]/10 580} = (206 × 2005 × 60)/(10 580) = 2388 ft min⁻¹

The capability satisfies the military requirement of 500 ft min⁻¹ (2.54 m s⁻¹). The reader may check for the 20° flap setting.

15.11.4 Checking Landing Field Length (AJT) – Specification Requirement 3600 ft

Keeping a reserve fuel of 440 lb (200 kg), the landing weight of the AJT is 8466 lb (wing loading = 42.26 lb ft⁻²) and at full flap extended C_Lmax = 2.2.

V_appr = 1.2V_stall@land = 160 ft s⁻¹

V_TD = 1.1V_stall@land = 146 ft s⁻¹

Average velocity from 50 ft height to touch down = 153 ft s⁻¹.

Distance covered before brake application after five seconds from 50 ft height,

S_{G_TD} = 5 × 153 = 765 ft.

Aircraft in full brake with μ_B = 0.45, all engines shut down and average C_L = 0.5, C_D/C_L = 0.1.

Average speed at 0.707 V_TD = 107.1 ft s⁻¹.

Then q = 0.5 × 0.002378 × 107.1² = 15.64

Distance covered during braking, S_{G_0Land} = (146 × 73)/12.93 = 824 ft

Landing distance S_{G_Land} = 765 + 824 = 1742 ft

Multiplying by 1.667, the rated LFL = 1.667 × 1742 = 2649 ≈ 2650 ft. It is within the specification of 3600 ft, the margin allowing for higher MTOM = 15 210 lb loaded with armament.

15.11.5 Checking the Initial Climb Performance – Requirement 50 m s⁻¹ (10 000 ft min⁻¹) at Normal Training Configuration (NTC)

Military trainers should climb at a much higher rate of climb than civil aircraft. The requirement of 50 m s⁻¹ (10 000 ft min⁻¹) at NTC is for unaccelerated climb as a capability for comparison. Unaccelerated rate of climb varies depending on the constant speed (EAS) climb, making comparison difficult. This section shows calculations for both the rates of climb.

This section will check only the en route climb with clean configuration. Unaccelerated climb Eq. is used. MTOM at NTC is 4800 kg (10 582 lb). Wing area S_W = 17 m² (183 ft²).

At en route climb, the aircraft has a clean configuration. Under maximum takeoff power, it makes an accelerated climb to 800 ft (ρ = 0.00232 slug ft⁻³, σ = 0.9756) from the second segment velocity of V₂ to reach a 350 KEAS climb speed schedule to start the en route climb. At en route climb the engine throttle is retarded to maximum climb rating. The quasi‐steady state climb schedule holds to 350 KEAS and the aircraft accelerates with altitude gain at a rate of dV/dh until it reaches Mach 0.8 around 25 000 ft, from where the Mach number is held constant until it reaches the cruise altitude. Consider that 100 kg of fuel is consumed to taxi and climb to 800 ft altitude where the aircraft mass is 4700 kg (10 362 lb).

At 350 kt (590.8 ft s⁻¹, Mach 0.49) the aircraft lift coefficient

Clean aircraft drag coefficient from Figure 11.17 at C_L = 0.141 gives C_Dclean = 0.0225

Available engine installed thrust at maximum continuous rating (95% of maximum thrust given in Figure 13.15) at Mach 0.49 (459.8 ft s⁻¹) is T = 0.95 × 5200 = 5016 lb.

From Eq. , the accelerated rate of climb as follows.

At quasi‐steady state climb, Table 15.5 gives = 0.56 M² = 0.56 × 0.49² = 0.1345

On substitution, rate of climb,

Therefore, an unaccelerated rate of climb, R/C = 11 395 ft min⁻¹.

Aircraft specification is based on a constant EAS climb rate of 10 000 ft min⁻¹, which is just met. There are many ways to climb that are dealt with in detail in [1]. (Here, the cabin area is small and the pressurised pilot suit allows a high rate of climb.)

15.11.6 Checking the Maximum Speed – Requirement Mach 0.85 at 30 000 ft Altitude at NTC

Aircraft at HSC is at Mach 0.85 (845.5 ft s⁻¹) at 30 000 ft altitude (ρ = 0.00088 slug ft⁻³). Fuel burned to climb is computed (not shown) to be 582 lb. Aircraft weight at the altitude is 10 000 lb.

At Mach 0.85 the aircraft lift coefficient C_L = MTOM/qS_W = 10 000/(0.5 × 0.00088 × 845.5² × 183) = 10 000/57561.4 = 0.174.

Clean aircraft drag coefficient from Figure 11.17 at C_L = 0.174 gives C_Dclean = 0.025 (high speed)

Clean aircraft drag, D = 0.025 × (0.5 × 0.00088 × 858.5² × 183) = 0.025 × 57 561.4 = 1440 lb.

Available engine installed thrust at maximum cruise rating (90% of the maximum rating) is taken from Figure 13.15 at Mach 0.85 and 30 000 ft altitude as T = 0.9 × 2000 = 1800 lb.

Therefore, the AJT satisfies the customer requirement of Mach 0.85 at HSC.

15.11.7 Compute the Fuel Requirement (AJT)

Other than ferry flight, a military aircraft is not merely dictated by the cruise sector unlike a civil aircraft mission. A short combat time at maximum engine rating, mostly at low altitudes, is a good part of fuel consumed. However, the range to the target area would dictate the fuel required. Long distance ferry flight and the combat arena will require additional fuel carried by drop tanks. Just before combat, the drop tanks (by that time they are empty) could be jettisoned to gain aircraft performance capability. The CAS variant has these types of mission profiles.

A training mission has varied engine demand and it returns to its own base covering no range as can be seen in Figure 15.20. Mission fuel is computed sector by sector of fuel burn as shown next for the classroom example. To compute fuel requirement, climb and specific range graphs for the AJT at NTC are required (Figures 15.20 and 15.21). To compute the varied engine demand of a training mission profile, use sfc and Figure 13.15 to establish the fuel flow rate for the throttle settings. The graph is valid for 75% rpm to 100% ratings. Typically, it will have approximately the following values.

At idle (60% rpm) ≈ 8 kg min⁻¹ At 75% rpm ≈ 11 kg min⁻¹ At 95% rpm ≈ 16.5 kg min⁻¹.

Fuel and time consumed for the normal training profile of the AJT is given in Table 15.20.

15.11.8 Turn Capability – Check n_max at the Turn (AJT)

Specified g‐level: 7 to −3.5. The specification of n = 7 is the maximum permitted g‐level (n_max) in pitch plane. The specified practice tight turns has n = 4 at sea level.

Turning Eq. 15.60 gives n_max = .

To demonstrate the AJT turn capability, let the tight turn maneuver is executed at Mach 0.5 to 0.55 at mid‐sortie weight of about 90% of NTC weight at sea level. For the AJT, Figure 14.5 gives the installed thrust loading (T_SLS/W) = 0.55. Tight turning is a short duration maneuver executed that can be carried out at full throttle, but in this carried out at maximum continuous rating at 0.9 × T_SLS. There operating (T/W) = (0.9 × 0.55)/(0.9 × NTC weight) ≈ 0.55.

On substitution, becomes it becomes.

n_max = 0.5 × 0.55 × √[1/(0.07 × 0.0212)] = 0.275 × √ 674 = 7.1, i.e. have excess capacity to carry out practice tight turns at around 10,000 ft altitude at n = 4.

Thrust at 10000 ft turning at a speed Mach 0.5 to 0.55 gives thrust at maximum continuous rating, T/W = 0.9 × 0.692 × (T_SLS/W) = 0.623 × (T_SLS/W).

On substitution, n_max at 10000 ft is as follows.

n_max = 0.5 × 0.623 × 0.55 ×√[1/(0.07 × 0.0212)] = 0.171 ×√674 = 4.43 with a healthy margin over n = 4.

Tight turning at full throttle permits n = 4 at higher altitudes.

15.13.1 The Bizjet

The sizing point in Figure 14.3 gives a wing loading, W/S_W = 64 lb ft⁻² and a thrust loading, T/W = 0.34. It may be noted that there is little margin given for the landing requirement. The maximum landing mass for this design is at 95% of MTOM. If for any reason aircraft OEW (now OEM, Operating Empty Mass) grows then it would be better if the sizing point for the W/S_W is taken a little lower than 64 lb ft⁻², say at 62 lb ft⁻². A quick iteration would resolve the problem. But this choice is not exercised to keep wing area as small as possible. Instead the aircraft could be allowed to approach to land at a slightly higher speed as LFL is generally about the same length as the TOFL. This is easily achievable as commonality of undercarriages for all variants would start with the design for the heaviest, that is, for the growth variant and then its bulky components are shaved to lighten for the smaller weights. The middle variant is taken as the baseline version. Its undercarriage can be made to accept MTOM growth as a result of OEW growth instead of making the wing larger.

It should be borne in mind that it is recommended that civil aircraft should come in a family of variants to cover wider market demand to maximise sale. Truly speaking, none of the three variants are optimised although the baseline is carefully sized in the middle to accept one larger and one smaller variant. Even when the development cost is front loaded, the cost of development of the variant aircraft is low by sharing the component commonality. The low cost is then translated to lowering aircraft price that can absorb the operating costs of the slightly non‐optimised designs.

It is interesting to examine the design philosophy of the Boeing737 family and the Airbus320 family of aircraft variants in the same market arena. Together, more than 10 000 aircraft have been sold in the world market. This is a no small achievement in engineering. The cost of these aircraft is about $50 million apiece (2005). For airlines with deregulated fare structures, making a profit involves complex dynamics of design and operation. The cost and operational scenario changes from time to time, for example, rise of fuel cost, terrorist threat and so on.

As early as the 1960s, Boeing saw the potential for keeping component commonality in offering new designs. The B707 was one of the earliest commercial jet transport aircraft carrying passengers. It followed by a shorter version B720. Strictly speaking, the B707 fuselage relied on the KC135 tanker design of the 1950s. From the four‐engine B707 came the three‐engine B727 and then the two‐engine B737, both retaining considerable fuselage commonality. This was one of the earliest attempts to utilise the benefits of maintaining component commonality. Subsequently, the B737 started to emerge in different sizes of variants maximising component commonality. The original B737‐100 was the baseline design and all other variants that came later, up to the B737‐900, are larger aircraft. Although B737‐100 morphed to a new aircraft as B737‐900, retaining commonality in many areas of external geometries, it posed certain constraints, especially on the undercarriage length. On the other hand, the A320 serving as the baseline design was in the middle of the family, its growth variant is the A321 and shrunk variants are the A319 and A318. Figure 7.9 gives a good study on how the OEW is affected by the two examples of family of variants. A baseline aircraft starting at the middle of the family would be better optimised and hence in principle would offer a better opportunity to lower production cost of the variants in the family.

Failure of two engines simultaneously is extremely rare. If it happens after the decision speed and if there is not enough clearway available, then it will prove catastrophic. If the climb gradient is not in conflict with the terrain of operation then it is better to takeoff with higher flap settings. If a longer runway is available then a lower flap setting could be used. Takeoff speed schedules can slightly exceed the FAR requirements that stipulate the minimum values. There have been cases of all‐engine failures occurring in cruise on account of volcanic ash in atmosphere (also in the rare event of fuel starvation). Fortunately, the engines were restarted just before the aircraft would have hit the surface. All‐engine failure by bird strike also occurred in 2009 – all survived after the pilot ditched in the Hudson River.

15.13.2 The AJT

Military aircraft serves only one customer, the Ministry of Defence (MoD)/Department of Defense (DoD) of the nation that designed the aircraft. Frontline combat aircraft incorporate the newest technologies at the cutting edge to stay ahead of potential adversaries. Development costs are high and only a few countries can afford to produce advanced designs. International political scenarios indicate a strong demand for combat aircraft even for developing nations who must purchase from abroad. This makes military aircraft design philosophy different from civil aircraft design. Here, the designers/scientists have a strong voice unlike in civil design where the users dictate terms. Selling combat aircraft to restricted foreign countries is a way to recover some finance.

Once the combat aircraft performance is well understood over its years of operation, consequent modifications follow capability improvements. Subsequently, a new design replaces the older design – there is a generation gap between them. Military modifications for the derivative design are substantial. Derivative designs primarily come from revised combat capabilities with newer types of armament along with all round performance gains. There is also a need for modifications, seen as variants rather than derivatives, to sell to foreign customers. These are quite different to civil aircraft variants, which are simultaneously produced for some time, serving different customers, some operating all the variants.

Advanced trainer aircraft designs have variants serving as combat aircraft in a close air support (CAS) role. Advanced Jet Trainers (AJTs) are less critical in design philosophy in comparison with front line combat aircraft, but bear some similarity in design considerations. Typically, AJTs will have one variant in the CAS role produced simultaneously. There is less restriction to export these kinds of aircraft.

The military infrastructure layout influences the aircraft design and here the life cycle cost (LCC) is the prime economic consideration. For military trainer aircraft design, it is better to have a training base close to the armament practise arena, saving time. A dedicated training base may not have as long a runway as major civil runways. These aspects are reflected in the user specifications needed to start a conceptual study. The training mission includes aerobatics and flying with onboard instruments for navigation. Therefore, the training base should be far away from the airline corridors.

The AJT sizing point in Figure 14.5 gives a wing loading, W/S_W = 58 lb ft⁻² and a thrust loading, T/W = 0.55. It may be noted that there is a large margin, especially for the landing requirement. The AJT can achieve maximum level speed over Mach 0.88, but this is not demanded as a requirement. Mission weight for the AJT varies substantially; the NTC is at 4800 kg and for armament practise it is loaded to 6600 kg. The margin in the sizing graph covers some increase in loadings (the specification taken in the book is for the NTC only). There is a big demand for higher power for the CAS variant. The choice for having an up‐rated engine or to have an afterburner depends on the choice of engine and the mission profile.

Competition for military aircraft sales is not as critical compared to the civil aviation sector. The national demand would support the national product for producing a tailor made design with under manageable economics. But the trainer aircraft market can have competition, unfortunately sometimes influenced by other factors that may fail to bring out a national product even if the nation is capable of doing so. The Brazilian design Tucano was re‐engineered and underwent massive modifications by Short Brothers plc of Belfast for the Royal Air Force (RAF), UK. The BAE Hawk (UK) underwent made major modifications in the USA for domestic use.