# Takeoff Field Length (AJT)

With a single engine, there is no question about the BFL. The military aircraft TOFL must satisfy the CFL. The CFL also can have a decision speed V1 that is determined by whether a pilot can stop within the distance available; otherwise, the pilot ejects.

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To obtain the CFL for a single-engine aircraft, the decision speed is at the critical time just before a pilot initiates rotation at VR. Then, the decision speed V1 is worked out from the VR by allowing 1 s for the recognition time. An engine failure occurring before this Vi leads a pilot to stop the aircraft by applying full brakes and deploying any other retarding facilities available (e. g., brake-parachute). There should be sufficient runway available for a pilot to stop the aircraft from this V1.

Figure 13.21. AJT takeoff |

Engine failure occurring after the V1 means that a pilot will have no option other than to eject (i. e., if there is not enough clearway available to stop). For a multiengine aircraft, determining the CFL follows the same procedure as in the civil – aircraft computation but it complies with MILSPECS.

Figure 13.21 shows the takeoff speed schedule of a single-engine military type aircraft.

In this book, the following three requirements for the AJT are worked out:

1. Meet the takeoff distance of 3,600 ft (1,100 m) to clear 15 m.

2. Meet the initial rate of climb (unaccelerated) of 10,000 ft/min (50 m/s).

3. Meet the maximum cruise speed of Mach 0.85 at a 30,000-ft altitude.

MIL-C501A requires a rolling coefficient, д = 0.025, and a minimum braking coefficient, дв = 0.3. Training aircraft have a good yearly utilization operated by relatively inexperienced pilots (with about 200 hours of flying experience) carrying out numerous takeoffs and landings on relatively shorter runways. The AJT brakes are generally more robust in design, resulting in a brake coefficient, дв = 0.45, which is much higher than the minimum MILSPEC requirement.

The Bizjet and the AJT have the same class of aerofoil and types of high-lift devices. Therefore, there is a strong similarity in the wing aerodynamic characteristics. Table 13.19 lists the AJT data pertinent to takeoff performance.

Equation 13.2 gives the average acceleration as:

a = g[(T/ W – д) – (SbSq/ W)(Cd/Cl – д)]

Using the values given in Table 13.15:

a = 32.2 x [(0.55 – 0.025) – (0*/58)(0.1 – 0.025)]

= 32.2 x [0.525 – (Ctq/773.33)]

Refer to Figure 13.21, which shows the AJT takeoff profile with an 8-deg flap. 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 VR. The acceleration of 16 ft/s2 (which can be computed first and then iterated without estimating) is estimated as follows:

Table 13.19. AJT takeoff distance, Sw = 17 m2 (183 ft2)
* Takeoff at 8- and 20-deg flaps. Landing at 35- to 40-deg flaps, engines at idle, and Vstall at aircraft landing weight of 8,466 lb. |

V1 = 190.6 – 16 = 174.6 ft/s. Aircraft velocity at 0.7 V1 = 122.22 ft/s. q = (at 0.7V1) = 0.5 x 0.002378 x 0.49V/ = 0.000583 x V/ = 8.7. Up to V1, the average CL = 0.5 (still at low incidence). Then:

a = 32.2 x (0.525 – Cl q/773.33) = 32.2 x (0.525 – 4.35/773.33)

= 32.2 x 0.519 = 16.72 ft/s2

Using Equation 13.3, the distance covered until the liftoff speed is reached:

Sg – 1 = Vave x (V1 – V))/a ft = 87.3 x (174.6 – 0)/16.72 = 912 ft

However, provision must be made for engine failure at the decision speed, V1, when braking must be applied to stop the aircraft. Designers must ensure that the operating runway length is adequate to stop the aircraft.

If the engine is operating, then an AJT continues with the takeoff procedure when liftoff occurs after rotation is executed at VR.

Distance Covered from Zero to Liftoff Speed VLO

Using the same equation for distance covered up to the decision speed V1 with the change to liftoff speed, VLO = 192.5 ft/s. Aircraft velocity at 0.7VLO = 134.75 ft/s, q = (at 0.7Vi) = 0.5 x 0.002378 x 0.49VLO = 10.586. Up to VLO, the average CL = 0.5 (still at low incidence). Then:

a = 32.2 x (0.525 – Cl q/773.33) = 32.2 x (0.525 – 4.35/773.33)

= 32.2 x 0.519 = 16.72 ft/s2

Using Equation 13.3, the distance covered until the liftoff speed is reached:

Sglo = Vave x (V1 – VO/a ft = 96.25 x (192.5 – 0)/16.42 = 1,128 ft

Distance Covered from VLO to V2

The flaring distance is required to reach V2, clearing a 50-ft obstacle height from VLO. From statistics, the time to flare is 3 s with an 8-deg flap and at V2 = 206 ft/s. In 3 s, the aircraft covers SG_LO_V2 = 3 x 206 = 618 ft.

Total Takeoff Distance

The takeoff length is thus SGLO + SGLOV2 = (1,128 + 618) = 1,746 ft, much less than the required TOFL of 3,500 ft for the full weight of 6,800 kg («15,000 lb) for armament training. In this case, a higher flap setting of 20 deg may be required. At a higher flap setting, an aircraft has a shorter CFL for the same weight.

Stopping Distance and the CFL

• distance covered from zero to the decision speed V2, SG2; it was previously computed as 912 ft

• distance covered from V2 to braking speed VB, SG2B

• braking distance from VB («V/) to zero velocity, SgB0

The next step is to verify what is required to stop the aircraft if an engine fails at V2. The stopping distance has the following three segments. (The CFL is normally longer than the TDFL.)

Distance Covered from Vl to Braking Speed VB

At engine failure, 3 s is assumed for a pilot’s recognition and taking action. Therefore, the distance covered from V2 to VB is SG2B = 3 x 174.6 = 524 ft.

Braking Distance from VB (« Vl) to Zero Velocity (Flap Settings Are of Minor Consequence)

The most critical moment of brake failure is at the decision speed V2, when the aircraft is still on the ground. With the full brake coefficient, д = 0.45, and the average CL = 0.5, V2 = 174.6 ft/s. With aircraft velocity at 0.7, Vi = 122.22 ft/s, q = (at0.7V1) = 0.5 x 0.002378 x 0.49 V2 = 0.000583 x V2 = 8.7.

Equation 13.2 for average acceleration reduces to:

a = 32.2 x [(-0.45) – (0.5 x 8.7)/58)(0.1 – 0.45)]

= 32.2 x [-0.45 + (1.52/58)]

= 32.2 x (-0.45 + 0.026) = -13.65

Using Equation 13.3, the distance covered with the liftoff speed is reached:

Sg_0 = Vave x (V1 – V0)/a ft = 87.3 x (174.6 – 0)/13.65 = 1,119 ft

Therefore, the minimum runway length (CFL) for takeoff should be = SG2 + SG2B + SG0. The TOFL = 912 + 524 +1,119 = 2,555 ft, which is still within the specified requirement of 3,600 ft.

The takeoff length is thus 1,746 ft, much less than the CFL of 2,555 ft computed previously. The required TOFL is 3,500 ft (computation not shown) for the full weight of 6,800 kg («15,000 lb) for armament training. In this case, a higher flap setting of 20 deg may be required. At a higher flap setting, an aircraft has a shorter CFL for the same weight.

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 д = 0.3. Equation 13.2 for average acceleration reduces to:

a = 32.2 x [(-0.3) – (0.5 x 8.7)/58)(0.1 – 0.3)] = 32.2 x [-0.3 + (0.87/58)]

= 32.2 x (-0.3 + 0.015) = -9.18

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

SG-0 = Vave X (V1 – V))/a ft = 87.3 x (174.6 – 0)/9.18 = 1, 660 ft

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

Sqj + Sqj_b + Sq_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 д.

A reduction of the decision speed to 140 ft/s (83 kts) reduces the Sg_o = 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 x 10,580)/(0.5 x 0.002378 x 183 x 2062) = 21,160/22,680 = 0.932.

The clean aircraft drag coefficient from Figure 9.16 gives CDclean = 0.1. Add ACoflap = 0.012 and ACd_u/c = 0.022, giving CD = 0.1 + 0.012 + 0.022 = 0.134. Therefore, drag, D = 0.134 x (0.5 x 0.002378 x 2062 x 183) = 0.134 x 22,680 = 3,040 lb. The available thrust is 5,000 lb (see Figure 13.4).

From Equation 13.5, the quasi-steady-state rate of climb is shown by:

V[(T – D)/W]

1 + (V/g)(dV/dh)

At the quasi-steady-state climb, Table 13.5 shows V (dV) = 0.566 m2 = 0.56 x 0.22 = 0.0224. Hence, R/Cacci = {[206 x (5,000 – 3,040) x 60]/10,580}/ [1 + 0.0224] = (206 x 1,960 x 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.

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