# Why is the minimum runway length non-linear in relation to the landing weight?

The graph below shows the landing runway length requirements for a Boeing 747-8.

(taken from the Boeing 747-8 Airplane Characteristics for Airport Planning)

Considering the fact that the kinetic energy and impulse are directly proportional to the mass, I would have expected this graph to be linear –­ but clearly that's not the case. What is causing the non-linearity here?

• Probably a function of the efficiency of the spoilers and brakes in dissipating the kinetic energy (${1\over 2}mv^2$ is non-linear with velocity itself). Could also be affected by the safety margins built into that landing distance, as those numbers do not represent the absolute minimum distance needed to stop the airplane. Commented Jul 15, 2014 at 20:57
• @casey Decreasing brake performance was my first idea too, but then I would have expected a more gradual change in slope. Commented Jul 15, 2014 at 21:09
• Kinetic energy is proportional to mass, but so is the braking force! (it's really proportional to weight, but for all practical purposes those two are proportional to each other). Commented Jul 16, 2014 at 7:36

## 1 Answer

While the kinetic energy is proportional to mass, it is not proportional to velocity.

The heavier the aircraft is, the higher the landing speed. So the stopping distance increases mainly because of the higher landing speed.

Comparing two landings, first at mass = $m_1$, the second at mass = $m_2 =1.1 \cdot m_1$, the landing speed will be $\sqrt{\frac{m_2}{m_1}}$ higher during the second landing.

The kinetic energy during the landing will scale proportional to the square of the mass.

• Ah, I didn't realize the landing speed was also affected by the mass, but obviously that has to be the case. Then the only remaining question is why the graph doesn't look like a parabola, but more like two straight lines spliced together. Commented Jul 16, 2014 at 3:25
• @Ventero I'm not sure a parabola would be expected. The aerodynamic drag and reverse thrust are non-linear with speed, but unaffected by aircraft mass. The wheel braking system is probably capable of providing deceleration independent of the mass (more mass = proportionally more grip = proportionally higher braking force = same deceleration) up to a mass where the maximum breaking power (heat generation) is reached and the deceleration will need to be lower. This could be the explanation for the two segment, however some curving is still expected. Simplification is probably the explanation. Commented Jul 16, 2014 at 6:17
• The stopping distance is not proportional to kinetic energy. Since friction is proportional to normal force (which is weight), the maximum deceleration the brakes can effect does not depend on mass at all. So the stopping distance should only depend on square of speed and since speed depends on square root of mass, one would still expect linear relationship. So this requires explaining what are the limiting factors for brakes. Commented Jul 16, 2014 at 7:54