A Jet Transport With A Landing Speed Of 200 Km/H Reduces Its

A Jet Transport With A Landing Speed Of 200 Kmh Reduces Its Speed To

A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with a constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At the lower speed aerodynamic forces on the aircraft are small and may be neglected.

Paper For Above instruction

In this technical analysis, we examine the forces acting on a jet transport aircraft during the deceleration phase initiated by reverse thrusts. The primary objective is to determine the reaction force (N) under the nose wheel (B) just before mechanical braking begins, after the aircraft has reduced its speed from 200 km/h to 60 km/h over a distance of 425 meters. This scenario assumes that aerodynamic forces are negligible at the lower speed, which simplifies the calculations to focus mainly on the forces caused by the thrust reversers and the aircraft's weight distribution.

Understanding the dynamics of aircraft deceleration involves the principles of physics and mechanics, specifically Newton’s second law (F = ma). The initial and final velocities are given in km/h, which we convert to m/s for consistency:

  • Initial velocity, \( V_i = 200\, km/h = \frac{200 \times 1000}{3600} \approx 55.56\, m/s \)
  • Final velocity, \( V_f = 60\, km/h = \frac{60 \times 1000}{3600} \approx 16.67\, m/s \)

The total distance over which deceleration occurs is \( s = 425\, m \). The mass of the aircraft \( m = 140\, Mg = 140,000\, kg \) (since 1 Mg = 1000 kg). Under constant deceleration, we employ the kinematic equation:

V_f^2 = V_i^2 + 2as

where \( a \) is the deceleration. Rearranging for \( a \):

a = \frac{V_f^2 - V_i^2}{2s}

Plugging in the values:

a = \frac{(16.67)^2 - (55.56)^2}{2 \times 425} = \frac{277.78 - 3086.42}{850} \approx -3.20\, m/s^2

The negative sign indicates deceleration. The net force causing deceleration is given by Newton’s second law:

F_{net} = m \times a = 140,000\, kg \times (-3.20\, m/s^2) = -448,000\, N

This net force results from the thrust reversers providing a backward force R, counteracted by the aircraft's weight and the distribution of forces on the landing gear. Since aerodynamic forces are negligible at the lower speed, the primary forces acting vertically are the weight \( W = m \times g \), where \( g = 9.81\, m/s^2 \), and the normal reactions at the landing gear.

The aircraft's weight is:

W = 140,000\, kg \times 9.81\, m/s^2 \approx 1,373,400\, N

Assuming the mass center is at G and the nose wheel is at B, the reaction force \( N \) under the nose wheel is influenced by the vertical load distribution and the longitudinal deceleration. Mechanical braking is not yet applied, so the reaction is primarily due to the weight of the aircraft balanced by the forces during deceleration.

Considering the aircraft as a rigid body on a horizontal surface, and assuming the deceleration force acts horizontally at the center of mass, the reaction \( N \) under the nose wheel can be derived from the equilibrium of vertical forces, typically involving the weight and the distribution of forces due to deceleration. Since the aircraft decelerates uniformly and aerodynamic effects are neglected, the reaction under the nose wheel just before braking can be approximated by the vertical component balancing the weight, adjusted for any longitudinal forces acting on the aircraft.

Final calculations involve resolving the forces at the point of contact and considering the position of the center of gravity relative to the landing gear. If the center of gravity is located at a distance \( L_{CG} \) from the nose wheel and the aircraft's length is \( L \), then the reaction \( N \) at the nose wheel can be approximated by considering moments about the center of gravity, taking into account the deceleration force R due to thrust reversers acting at a certain point along the aircraft's longitudinal axis. Simplifying assumptions lead to an approximate estimate, often used in aircraft ground handling analyses.

Conclusion

Through the above calculations, we find that the total decelerating force acting on the aircraft is approximately 448 kN. The vertical reaction force under the nose wheel prior to mechanical braking is primarily determined by the aircraft’s weight, which is about 1.37 MN. Considering the aircraft's geometry and the placement of the center of gravity, the reaction force \( N \) under the nose wheel can be approximated to sustain the load due to the weight distribution, typically close to this magnitude. Exact numerical determination may require detailed aircraft geometry data, but this analysis provides a comprehensive understanding of the forces at play during the deceleration process.

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