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This question already has an answer here:

I suppose they do both, but what percentage of the kinetic energy is absorbed by the wheel brakes and what percentage by air-friction (in typical cases)?

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marked as duplicate by ratchet freak, GdD, FreeMan, fooot, Simon Mar 16 '15 at 17:16

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  • $\begingroup$ Not a duplicate (probably) as the link specifically addresses large jets, and this question is asking about planes, which in the absence of any qualification includes everything from ultralights on up. $\endgroup$ – jamesqf Mar 17 '15 at 6:34
  • $\begingroup$ Also see spoilers which are used both to increase descent rate at constant speed in flight, and to remove residual lift and increase drag on the ground. $\endgroup$ – mins Mar 17 '15 at 23:41
  • $\begingroup$ @mins: Yes, especially in sailplanes. Or in light aircraft, if you're landing on grass or dirt, you might not want to use much braking (skidding, nosewheel digging in, etc), while on a short paved field you'd want full flaps & brakes... $\endgroup$ – jamesqf Mar 18 '15 at 1:30
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Depends on how long the runway is.

If you have a 12,000' runway and your terminal is at the far end of it, you can roll out with no wheel braking and as little (or no) reverse thrust as you're allowed, and you'll be close to taxi speed by the time you're ready to turn off. In that case, the brakes only absorb enough energy to get you from maybe 60-80 knots down to 20-30, since the drag devices lose most of their effectiveness by that point.

On the other hand, on a really short runway, you may be getting on the brakes immediately after touchdown, which means that the amount of time that you're dissipating energy due to drag (spoilers) and reverse thrust is reduced, and much less energy is lost that way. So in that case the brakes get a great deal of your kinetic energy.

The case beyond that would be if you didn't use reverse thrust, your brakes would get nearly all of your energy, with only a small amount dissipated due to aerodynamic drag.

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