The length of the runway available dictates the decision speed V1. If the airfield length is much longer, then a pilot may have a chance to land the aircraft if an engine failure occurs immediately after liftoff; it may be possible to stop within the available airfield length, which must have some clearway past the runway end.

This is compared with the minimum MILSPEC requirement of µ = 0.3. Equation 13.2 for average acceleration reduces to:

a¯ = 32.2 × [(−0.3) − (0.5 × 8.7)/58)(0.1 − 0.3)] = 32.2 × [−0.3 + (0.87/58)]

= 32.2 × (−0.3 + 0.015) = −9.18

Using Equation 13.3, the distance covered from VB to zero:

SG 0 = Vave × (V1 − V0)/a¯ ft = 87.3 × (174.6 − 0)/9.18 = 1, 660 ft

This value is on the high side. The minimum runway length for takeoff should be =

SG 1 + SG 1 B + SG 0.

The CRL = 912 + 524 + 1,660 = 3,096 ft is on the high side but within the specification of 3,600 ft. Therefore, the higher brake coefficient of 0.4 is used. This is not problematic because wheels with good brakes currently have a much higher friction coefficient µ.

A reduction of the decision speed to 140 ft/s (83 kts) reduces the SG 0 = 1,068 ft, decreasing the CRL to 2,504 ft.

Verifying the Climb Gradient at an 8-Deg Flap

Using AJT V2 = 206 ft/s (see Table 13.15) and W = 10,580 lb (4,800 kg) gives CL = (2 × 10,580)/(0.5 × 0.002378 × 183 × 2062) = 21,160/22,680 = 0.932.

The clean aircraft drag coefficient from Figure 9.16 gives CDclean = 0.1. AddCDflap = 0.012 and CD U/C = 0.022, giving CD = 0.1 + 0.012 + 0.022 = 0.134.

[1 + 0.0224] = (206 × 1,960 × 60)/(10,817) = 2,240 ft/min. This capability satisfies the military requirement of 500 ft/min (2.54 m/s). Readers may verify the 20-deg flap setting.

13.6.3 Landing Field Length (AJT)

Keeping a reserve fuel of 440 lb (200 kg), the landing weight of an AJT is 8,466 lb (wing-loading = 42.26 lb/ft2) and at full flap extended, CLmax = 2.2. Therefore:

Vappr = 1.2Vstall@land = 160 ft/s

VTD = 1.1Vstall@land = 146 ft/s

The average velocity from a 50-ft altitude to touchdown = 153 ft/s. The distance covered before brake application after 6 s from a 50-ft altitude:

SG TD = 6 × 153 = 918 ft

For an aircraft in full braking with µB = 0.45, all engines shut down, and average CL = 0.5, CD/CL = 0.1.

Equation 13.2 for average acceleration is based on 0.7 VTD = 107.1 ft/s. Then:

q = 0.5 × 0.002378 × 107.12 = 13.64

Deceleration, a¯ = 32.2 × [(−0.45) − (CLq/42.26)(0.1 − 0.45)] = 32.2 × [−0.45 + (0.15 × 13.64/42.26)] = 32.2 × [−0.45 + 0.0484] = −12.93 ft/s2.

The distance covered during braking, SG Land = (146 × 73)/12.93 = 824 ft. The landing distance SG Land = 918 + 824 = 1,742 ft. Multiplying by 1.667, the rated LFL = 1.667 × 1,742 = 2,904 ft and is expected to be less than the TOFL at an 8-deg flap setting (but not always). This is within the specification of 3,600 ft.

13.6.4 Climb Performance Requirements (AJT)

Military trainers should climb at a much higher rate of climb than civil aircraft. The requirement of 50 m/s (10,000 ft/min) at normal training configuration (NTC) is for an unaccelerated climb for comparison with accelerated climb. Unaccelerated rate of climb varies depending on the constant speed (i.e., EAS) climb, making a comparison difficult. This section presents calculations for both rates of climb.

This section checks only the enroute climb with a clean configuration. The unaccelerated climb Equation 13.7 is used. The MTOM at the NTC is 4,800 kg (10,582 lb). The wing area SW = 17 m2 (183 ft2).

During an enroute climb, the aircraft has a clean configuration. Under maximum takeoff power, it makes an accelerated climb to 800 ft (ρ = 0.00232 slug/ft3, σ = 0.9756) from the second-segment velocity of V2 to reach a 350-KEAS speed schedule to start the enroute climb. During enroute climb, the engine throttle is retarded to the maximum climb rating. The quasi-steady-state climb schedule maintains 350 KEAS and the aircraft accelerates with an altitude gain at a rate of dV/dh until it reaches Mach 0.8 at around 25,000 ft. From there, the Mach number is held constant until it reaches the cruise altitude. We assume that 100 kg of fuel is consumed to taxi and climb to an 800-ft altitude, where the aircraft mass is 4,700 kg (10,362 lb). At 350 kts (590.8 ft/s, Mach 0.49), the aircraft lift coefficient is:

CL = MTOM/q SW = 10,362/(0.5 × 0.00232 × 590.82 × 183)

= 10,582/74,905 = 0.138

The clean aircraft drag coefficient from (see Figure 9.19) at CL = 0.141 gives CDclean = 0.023. The clean aircraft drag, D = 0.023 × (0.5 × 0.002378 × 590.82 × 183) = 0.023 × 74,905 = 1,723 lb. The available engine-installed thrust at a maximum continuous rating (95% of maximum thrust, as given in Figure 13.4) at Mach 0.49 (459.8 ft/s) is T = 0.95 × 5,000 = 4,750 lb. From Equation 13.10, the accelerated rate of climb is as follows:

V[(T − D)/ W]

R/Caacl = 1 + (V/g)(dV/dh)

458 Aircraft Performance

At a quasi-steady-state-climb, Table 13.5 gives:

V dV g dh

From Equation 13.5, the rate of climb is:

R/Caccl = {[590.8 × (4,750 − 1,723) × 60]/10,362}/]1 + 0.1345] = 10, 355/1.1345

= 9, 127 ft/min

Therefore, the unaccelerated rate of climb, R/C = 10,355 ft/min. The aircraft specification is based on an unaccelerated climb of 10,000 ft/min, which is just met. (Here, the cabin area is small and the pressurization limit is high.)

13.6.5 Maximum Speed Requirements (AJT)

An aircraft at HSC is at Mach 0.85 (845.5 ft/s) at a 30,000-ft altitude (ρ = 0.00088 slug/ft3). The fuel burned to climb is computed (but not shown) as 582 lb. The aircraft weight at the altitude is 10,000 lb.

At Mach 0.85, the aircraft lift coefficient CL = MTOM/qSW = 10,000/

the maximum rating) is from Figure 13.4 at Mach 0.85, and at a 30,000-ft altitude is T = 0.85 × 2,000 = 1,700 lb. (In the industry, the thrust is computed.)

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

13.6.6 Fuel Requirements (AJT)

Other than a ferry flight, military aircraft are not dictated only by the cruise sector, unlike in a civil aircraft mission. A short combat time at the maximum engine rating, mostly at low altitudes, is responsible for a suitable part of the fuel consumed. However, the range to the target area dictates the fuel required. A long-distance ferry flight and combat arena require additional fuel to be carried by drop tanks. Immediately before combat, the drop tanks (they are empty) by then can be jettisoned to gain aircraft performance capability. The CAS variant has this type of mission profile.

A training mission has a varied engine demand and it returns to its own base covering no range, as shown in Figure 13.19. Mission fuel is computed sector by sector of fuel burn, as shown as follows for the coursework example. To compute the fuel requirement, climb and specific-range graphs for the AJT at NTC are required (Figures 13.22 and 13.23). To compute the varied engine demand of a trainingmission profile, Figure 13.4 is used to establish the fuel-flow rate for the throttle settings. The graph is valid for 75% rpm to 100% ratings. Typically, it has the approximate following values:

at idle (50% rpm) ≈ 8 kg/minat 75% rpm ≈ 11 kg/min

at 95% rpm ≈ 16.5 kg/min