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I am a private pilot. I posed a question to instructors in the flight school and got differing answers. For a given aircraft, landing at the same speed, with everything the same except for landing weight, and using maximum braking without skidding the tires, which will stop in a shorter distance after the wheels touch down: the plane at max gross weight, or the same plane lightly loaded?

I submit that because the heavy plane has more energy to dissipate, it would take longer to stop. Some instructors said that because there is more weight on the wheels, the heavier plane would stop in a shorter distance. I think it is a simple energy conversion physics issue.

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    $\begingroup$ First "because the heavy plane has more energy to dissipate, it would take longer to stop" isn't necessarily true. More energy just means the brake gets hotter after the plane lands and nothing else. What actually makes a difference is aerodynamics, like air brakes, spoilers, reverse thrust. Also you made a very unrealistic requirement that airspeed being equal. Say you have an airshow and want to show off short field landing technique you would always unload the plane to nothing but minimum fuel and the thing just lands like feather. $\endgroup$ – user3528438 Oct 21 '17 at 23:21
  • $\begingroup$ Pretty sure you are right. That's one of the reasons why aircraft dump fuel before landing in an emergency return to field. $\endgroup$ – Trevor_G Oct 21 '17 at 23:39
  • $\begingroup$ Interesting question, but know that in the "real world" (with larger airplanes, where the differences are more exaggerated than a light airplane) landing distance goes down with a decrease in weight, but this is mainly attributed to the reduction in approach speed as the weight decreases. It may also be why you are getting different answers: some people may be missing the significance of the "at the same speed" point and answering just based on knowledge of actual airplane performance charts. $\endgroup$ – Lnafziger Apr 8 '18 at 23:56
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If you discount aerodynamic effects, and assume the same landing speed, on a theoretical level mass makes no difference. The additional kinetic energy to be dissipated is exactly cancelled by the additional normal force. The kinetic energy is given by the formula:

E = 1/2 mv², where

  • E is the kinetic energy
  • m is the vehicle's mass
  • v is the speed at the start of braking

The work done by braking is given by:

W = μmgd, where

  • W is work done
  • μ is the coefficient of friction between the road surface and the tires
  • g is the gravity of Earth
  • d is the distance travelled

The braking distance given an initial speed v is then found by putting W = E, from which it follows that

d = v²/2μg

  • d = minimum stopping distance
  • v = velocity when brakes were applied
  • μ = coeff. of friction between wheels and road
  • g = gravitational constant

Mass is removed from the equation. The kinetic energy is tranferred into heat, so the larger the mass the more heat is generated. But theoretically the stopping distance is unaffected.

In reality, though, the coefficient of friction of tires does not remain constant with changing normal force. As normal force increases the coefficient of friction (μ) of tires decreases.

Since the μ does not remain linear with mass a heavier plane will have an increase in kinetic energy which is higher than the corresponding increase in friction of the tires, thus a longer stopping distance.

All of this assumes that “maximum braking” means you have as much brake power available as the maximum friction force of the tires. Above a certain weight/speed this will not be the case for an airliner. They are designed for a particular amount of deceleration force at MTOW and that will be less than the amount of friction tires can provide on a dry runway. (See this question for more details on that.) That will also cause a heavier plane to roll farther.

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    $\begingroup$ Note that usually on dry surface the brakes exert much less force than the tires could handle. $\endgroup$ – Jan Hudec Oct 22 '17 at 19:28
  • $\begingroup$ @JanHudec Thanks. Added a paragraph about that $\endgroup$ – TomMcW Oct 23 '17 at 19:18
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Yes the maximum friction (and therefore braking) force is a linear function of weight. The higher braking force cancels out the higher kinetic energy, if we assume:

  • That the landing speed is equal at the higher weight.
  • That the lift dumpers are not operated.
  • That all the wheels have brakes.
  • That the brakes can get rid of all of the heat at the higher weight, without blowing the tyres.

Landing Speed. Braking distance is a quadratic function of touchdown speed. For a small weight increase, we can maintain the same landing speed without much safety implication. Sure airliners dump fuel when attempting a landing close to max gross weight, but that is mainly to limit the touchdown slam, the main gears are not dimensioned for doing this at TO weight. Making the observation at the same landing speed is a really important limitation though, since braking distance is equal to landing speed square.

Residual lift. Lift dumpers remove lift: just after touchdown, the wing still creates considerable lift which reduces the normal (and therefore braking) force. Since the landing speed does not change in our assumptions, the lift decrease is the same, regardless of weight: $ F_N = W - L + \Delta L$. If we take two gross weights and work out the percentage of down force, we get:

enter image description here

So the higher landing weight keeps a higher downforce percentage after deploying spoilers etc. If we consider residual lift, the higher weight causes a shorter braking distance.

Nose Wheel Brake. enter image description here

Usually, only the main wheels have brakes, but braking creates a forward moment which reduces the downforce on the main wheels. The magnitude of the effect depends on the height of the c.g. off of the ground, but if the nose wheel has no brakes, this effect increases braking distance at higher weight.

Heat buildup This is another factor that increases quadratically with landing velocity, and only linearly with weight. The brakes are dimensioned for emergency stop below V1 after engine fail, so they will cope with a normal landing at normal weights.

All in all, there is one effect that decreases landing distance as function of landing weight, and one that increases landing distance. We could say that they cancel each other out and that at the same landing speed and within normal limits, braking distance is independent of weight.

But the very shortest landing distance is at the very lightest weight: stay just above stall speed for the weight, even a small reduction is speed has a large effect on distance.

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As a practical matter, the aircraft you describe will not fly as you assume. You would have to land very differently at the lighter weight to get the plane on the ground at the same speed as at the heavier weight. With normal landing technique you will touch down a good bit slower at the lighter weight. This will result in a much shorter rollout, as kinetic energy varies with speed squared.

The other answers deal better with your theoretical impossibility.)

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