Given that the nose wheels of both the Boeing 747-8 and Airbus A380, as is true for almost all aircraft, are not equipped with brakes, do they encounter more weight during braking?

If so, how much more at maximum safe landing weight?

If they were equipped with brakes, would they encounter more weight during braking and what other effects would ensue?

  • 8
    $\begingroup$ Braking puts more weight on the front wheel(s) of any vehicle - physics demands it. $\endgroup$
    – FreeMan
    Jan 17, 2018 at 19:25
  • $\begingroup$ @FreeMan Beat me to it, how much more depends on how much it weighs, how hard it is braking, and the distance between the main wheels and the nose wheel. $\endgroup$
    – Ron Beyer
    Jan 17, 2018 at 19:26
  • 3
    $\begingroup$ @FreeMan be careful with the word any. What if the the vehicle's centre of gravity is below the wheels? What if aerodynamic braking is used? Or speed brakes? $\endgroup$
    – DeltaLima
    Jan 17, 2018 at 19:59
  • 2
    $\begingroup$ @DeltaLima good points. I would postulate, though, that weight will still shift forward for the suspended trolley, though that might produce a lifting force on the front wheels. Popping a chute behind the plane would cause a different weight shift. Original statement valid for any conventional, wheel-braked vehicle (that I can think of right now). $\endgroup$
    – FreeMan
    Jan 17, 2018 at 20:04
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    $\begingroup$ @ymb1 yes, because when going in reverse, the weight shifts forward relative to the motion of the vehicle, putting more weight on the wheel(s) that are now in the front. So there, smart alek! :D $\endgroup$
    – FreeMan
    Jan 17, 2018 at 21:40

1 Answer 1


Yes of course. And even with nose wheel brakes the main braking force would still be generated by the main wheels, so a substantial pitching moment is generated.

Now for a crude first-order approximation. We have an answer about the braking forces here, and using those 235 kN per wheel (the 747 has 16 main wheels) means that the center of gravity, which I guess to be at least 6 m above the ground, experiences inertial loads of 3,76 MN. This needs to be reacted in a moment around the main wheels with the lever arm of the wheel base. This is conveniently displayed in the drawing below and is 25.6 m. Thus, the reaction moment around the main wheels requires a nose wheel load of 881 kN in addition to the regular load of the nose wheel (which is maybe 5% of the aircraft's gross weight or 196 kN).

enter image description here

Boeing 747 side view (picture source)

Since the highest friction coefficient is reached at low speed, wing lift or tail downforce will not play a significant role. Braking to a full stop after an aborted take-off at full gross weight will load the nose wheels with a multiple of the static load of more than four.

In a normal landing, the forces involved are much lower and the multiple is close to one. This is really a quick back-of-the-envelope calculation, but the magnitude should still be right.

  • 1
    $\begingroup$ I would also point out that with regard to the extra load on the nose wheel, for a given deceleration it makes no difference how much of the braking force is provided by the nose wheel: since all wheels are the same distance below the line along which the aircraft's center of gravity travels, every newton of braking force on any wheel produces the same moment around the COG, and as this answer points out, that moment is what determines the load on the nose wheel. $\endgroup$
    – David K
    Jan 19, 2018 at 2:31

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