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In a comment left on my answer here Jan Hudec says:

Note that usually on dry surface the brakes exert much less force than the tires could handle.

This got me wondering how much force large aircraft brakes can provide.

The limiting factor in brake force for a vehicle is the friction between the tire and the surface. On my car, even on dry pavement, the brake force available is always greater than the tire/ground friction, so I can always lock the brakes and set off the ABS. I may be misunderstanding it, but Jan’s comment makes me think that maybe large aircraft don’t have that much brake force available. If you commanded maximum braking on a dry runway would the brakes try to lock? Is the ABS really only necessary on a contaminated runway?

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  • $\begingroup$ Yes, they can lock the brakes up, and some (many large/commercial) aircraft have anti skid systems much like vehicle ABS. $\endgroup$
    – Ron Beyer
    Commented Oct 22, 2017 at 23:19
  • $\begingroup$ Related: This question. In @Federico 's answer, there is a link to this presentation. See page 16 and next. $\endgroup$
    – mins
    Commented Oct 22, 2017 at 23:23
  • $\begingroup$ Under normal use case I'd say it's indeed pretty far. Say dry surface with u=0.7 then you only need less than 450 meters to stop at 150 knots, and 700 meters for wet surface whose u=0.4, but usually for a plane that lands at that speed would have ~2000 meters to use. Even for a plane with 215 knot touch down speed(the worsts, like space shuttle, SR71,XB70,B58 are all around 175-185 knot range), with u=0.4 it would still be able to stop within 1500 meters. I don't know what kind of worst case ABS is planning for but for dry surface normal operation I don't think it is ever triggered. $\endgroup$
    – user3528438
    Commented Oct 23, 2017 at 2:00
  • $\begingroup$ However jet fighters seems to break traction limit regularly: even if they can lock the wheels, the engine can still push the fighter forward at full thrust (engine can provide close to 1g of acceleration while tires can only neutralize at most 0.7g of it). When they take off from sky-jump carriers there are wheel blockers that keep the plane still before taking off rather than relying brakes. $\endgroup$
    – user3528438
    Commented Oct 23, 2017 at 2:06
  • $\begingroup$ @RonBeyer: And which predate automobile ABS by a decade or so, and are the direct ancestors of same. $\endgroup$
    – Vikki
    Commented Oct 11, 2019 at 5:36

3 Answers 3

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This has been answered before, but in a different context. Therefore, I copied the essential part here:

Emergency procedures at take-off weight will produce the biggest braking loads. I will calculate the highest possible braking loads for a Boeing 747-400 that weighs up to 400 tons on take-off. For this I need this plot of a polynominal for the braking coefficient, which is the ratio between vertical and horizontal forces before the tire skids. I do not know the source; I collected it somewhere in the past and never found a reason to doubt its validity. tire friction coefficient

Shortly before the aircraft comes to a stop, the highest friction coefficient is reached, and then little lift is produced by the wings, so the vertical tire loads are those of the static case. The braked wheels are the 16 main wheels; I assume that the unbraked noswheels carry 4% of the total weight. At 96% of 400 tons acting on those 16 wheels, this is 24 tons = 235,344 N = 52,907 lbs of downward force per wheel. Since the friction coefficient is 1 at low speed, the same load is transferred horizontally from each wheel to the ground and into the gear struts.

And, yes, brakes will lock on a dry runway if the disk brakes are fully engaged.

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I read Jan's comment as stating that the tyres will remain intact when maximum braking force is applied on a dry runway: tyres are dimensioned such that they don't fail structurally from maximum braking effort.

They can fail thermally, after a blocked tyre has skidded at high speed for a while, and the resulting high temperature at the contact area eats through the tyre rubber. This is of course one of the reasons why anti-skid systems are implemented in aircraft, the other reason being that rolling friction is higher than skidding friction. This report describes what goes on when tyres skid, the high temperature causes part meltdown and gasification of the contact area, which provides a form of lubrication and therefore a lower resulting friction coefficient.

Without an anti-skid system, the brakes can most certainly lock up the wheels, and in fact car ABS systems were implemented decades after they had been installed in aircraft, as mentioned in the wiki. As to the maximum braking force acting on the aircraft mass, Torenbeek page 585 lists the maximum deceleration values as typically:

  • 0.55g: dry runway, maximum effort, ignoring passenger tolerance
  • 0.35g: wet runway, modern braking with anti-skid, lift dumpers, reverse thrust
  • 0.15g: wet runway, simple braking or flooded runway, reverse thrust
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  • $\begingroup$ That’s the way I am reading it too, that the brakes are not designed to have more braking force than the tires on a dry runway. On a wet runway they can, therefore you need anti-skid. I just wasn’t sure if they were usually designed that way or not. $\endgroup$
    – TomMcW
    Commented Oct 23, 2017 at 18:53
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The amount of force required (in many cases they are designed to deliver more force but this provides a minimum) is based on the aircrafts weights and speeds, you can find most of the info in

§ 25.735 Brakes and braking systems.

The most relevant section is that you need to be able to provide enough force to decelerate within limits

...The energy absorption rate derived from the airplane manufacturer's braking requirements must be achieved. The mean deceleration must not be less than 10 fps2.

Also as a matter of certification the brakes must be able to provide enough force to hold the aircraft in place on the ground under full thrust from the engines.

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  • $\begingroup$ It would be helpful to provide an estimate of the force based on these requirements. $\endgroup$
    – fooot
    Commented Nov 22, 2017 at 22:35

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