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I have long wondered whether its possible (in theory at least) for jetliners under typical loading scenarios (pax and cargo), landing at a normal airport such as LHR to come to a slow enough speed to exit the runway just by using spoilers, full flaps and normal friction (air and tyres) i.e. no brakes and no reverse thrust ? Are the runways at commercial airports long enough for this scenario?

I understand that busy airports expect aircraft to leave the runway as soon as possible but I would still like to know if its theoretically possible.

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it's possible for jetliners under typical loading scenarios... to come to a slow enough speed to exit the runway just by using spoilers, full flaps and normal friction (air and tyres) i.e. no brakes and no reverse thrust?

In this case, deceleration at landing is only due to aerodynamic drag and rolling resistance of the landing gears. As a first approximation we neglect the engine's thrust as well (which even at idle are anyway pushing the airplane, so a complete stop without brakes can be achieved only switching the engines off).

From Newton's second law $F=ma$ we simply get that the deceleration is $a=F/m$ where $F$ is the decelerating force due to the sum of the aerodynamic drag and the rolling resistance, as said.

  • For the drag we use the usual equation $D=½\rho V²SC_d$.
  • Rolling resistance is simply the weight times a coefficient $\mu$ which depends on the materials; $\mu=0.03$ is the standard value used for airplane tires on tarmac.

We get then:

$a=F/m=(½\rho V²SC_d + \mu mg)/m$

By definition, acceleration is the variation (derivative) of speed in time and that equation is therefore a differential equation which, without entering in mathematical details, possesses the solution:

$S_g=\frac{m}{\rho SC_d} \ln\left({(\mu+\frac{\rho SC_d V_{end}²}{2mg})}/{(\mu+\frac{\rho SC_d V_{td}²}{2mg})}\right)$

where:

  • $S_g$ is the ground roll length;
  • $m$ is the mass of the airplane at touchdown; in an emergency landing this can be as high as $0.85\times MTOW$;
  • $\rho$ is air density;
  • $V_{end}$ the final speed we want to reach;
  • $V_{td}$ the touchdown speed.

Let's do the math for an A380. Let's say that it touches down at some $m=330'000kg$, it reaches an end speed $V_{end}=0$ and has a touchdown speed $V_{td}=80m/s$ (150kts); for a jetliner in landing roll configuration a $C_d=0.03$ should be appropriate; as said typically $\mu=0.03$; the A380 as a wing area of $S=845m²$. Substituting we get:

$S_g=\frac{330'000}{1.125\times845\times 0.03} \ln\left((0.03+0)/(0.03+\frac{1.125\times 845 \times0.03\times 80²}{2 \times330'000\times 9.8q})\right)=7'667m$

A very long runway :⁠-⁠)

Obviously a lower weight, lower touchdown speed and/or higher rolling friction would reduce the ground roll: landing in wet grass ($\mu=0.08$) would for example reduce the A380's ground roll $S_g$ at less than 4'000m.

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  • $\begingroup$ thank you @Sophit thats a facinating answer and I like that you used a real life example to illustrate the point. Just to tidy up my understanding - (A) do your workings calculate the length of runway needed to bring the airplane to a complete stop or just a typical runway exit speed ? I guess if complete stop then this would require quite a bit longer runway than exit speed ? (B) What does Cd represent in your equation ? (C) do the equations take into account the drag from the Flaps and Spoilers being down ? $\endgroup$
    – robbie70
    Oct 17, 2023 at 20:24
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    $\begingroup$ @robbie70: thanks for your comment 🤗 (A) Yes, I considered a complete stop. If the end speed isn't zero then the equation changes a bit $\rightarrow$ I'll post the modified equation. (B)+(C) It's the drag coefficient. I've used 0.03 which is a reasonable number for a jetliner in landing roll configuration. I didn't specified it $\rightarrow$ I'll add it to the answer. $\endgroup$
    – sophit
    Oct 18, 2023 at 1:13
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    $\begingroup$ +1 for actually calculating the stopping distance! If you're bored, you could also calculate the remaining speed when reaching the end of a 3900m runway ;) $\endgroup$
    – Bianfable
    Oct 18, 2023 at 13:13
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    $\begingroup$ @sophit, 7667m? That's shorter than either runway 17/35 or runway 15/33 at Edwards Air Force Base, though since that's salt pan rather than tarmac, your number for rolling resistance is off. Edwards also has runways 07/25 and 18/36, at about 7000 meters each, sufficient if you're willing to do a rolling turnoff rather than a hard stop. $\endgroup$
    – Mark
    Oct 18, 2023 at 22:44
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    $\begingroup$ As I started reading, I thought, "here's a Peter Kämpf answer", but I was wrong! Yes, that's a compliment! Well done. $\endgroup$
    – FreeMan
    Oct 19, 2023 at 11:43
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No, in the general case it isn't even close to possible.

You land a 60-ton aircraft at 120 mph, and it has to slow to (let's say) 30 mph in the space of (being generous) 2 miles, with no braking? No way.

It's possible to stop an aircraft in most cases without the use of reverse thrust, and the difference between having or not having reverse thrust available is several hundreds of feet difference. The rest of the stopping force comes from the wheel brakes, and without those, it would take miles & miles to slow down that much.

Take as an example, landing an airliner on runway 22 in El Paso (KELP), which at 12,000' long is one of the longest runways at commercial airports in the US. If you touch down normally & plan to exit near the end, it's common to stay off the brakes, and let the aircraft roll for a distance, before applying the brakes to slow down & turn off on the high-speed taxiway that takes you to the terminal. During that time rolling, you're pretty much in the suggested state: idle thrust, no reverse, no brakes. And from personal experience, I'll tell you that the aircraft isn't slowing very much. I have no doubt at all that if a pilot were to never apply the brakes, the aircraft would roll off the far end of the runway, probably at 80+ knots (and that's after having initially applied reverse thrust for a while, to get from landing speed down to something close to 80 knots).

Is it possible to imagine a creative scenario involving immense up-slope on the runway, combined with strong headwinds, and maybe some other factor, that might result in getting the aircraft stopped? Probably; a 50-knot headwind would work wonders for you in that sort of an exercise. But how often do we land with that sort of a headwind? Almost never.

The main function of deploying spoilers on landing isn't the drag, although that's a secondary benefit. The main purpose is to kill the creation of lift from the wings, so that the entire weight of the aircraft is on the wheels, allowing them to brake harder without (or, before) skidding. If you had 60 tons of mass, being stopped with wheels that only have 10 tons of weight on them, the wheels would skid (or the anti-skid system would release the brakes to prevent the skid) before you could generate much in the way of stopping force. But if you have the whole weight of the aircraft on them, then you can apply the brakes much more firmly. That's why deploying spoilers is important on landing, much more so than the aerodynamic drag that they provide.

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    $\begingroup$ Common belief is that reverse thrust is needed, or that it's used to reduce brake wear. What I have read is that as you said in most cases on a dry runway reverse thrust isn't needed. Apparently it's mainly used to avoid brake overheating which can increase turnaround time waiting for the brakes to cool off. Apparently the wear on brakes is less expensive than the fuel burned and wear on the engines using reverse thrust, so some airlines prefer using idle reverse thrust when possible, which contributes very little to braking but at least there's no forward thrust. $\endgroup$ Oct 17, 2023 at 22:10
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    $\begingroup$ Note that a strong headwind would help you in two ways. It slows you down once you've landed, of course; but also your groundspeed can be correspondingly lower at touchdown, giving you less speed to get rid of on the runway. $\endgroup$ Oct 18, 2023 at 14:34
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    $\begingroup$ @KurtFitzner: If a headwind were to start just as the plane touched down, it would improve the effectiveness of aerodynamic drag mechanisms. Likewise if a headwind dissipated just as a plane touched down, the effectiveness of aerodynamic drag mechanisms would be decreased, but the amount of kinetic energy to be dissipated would still have been reduced prior to touchdown. $\endgroup$
    – supercat
    Oct 18, 2023 at 22:49
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    $\begingroup$ Not only is thrust reverse not needed, @StevePemberton, testing and certification of the braking system is done without thrust reverse at all. The brakes are required to be able to stop the aircraft without any engine assistance. That is how the landing rollout tables are built. $\endgroup$
    – FreeMan
    Oct 19, 2023 at 11:46
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    $\begingroup$ @KurtFitzner why would it not be able to? As supercat pointed out, the two effects occur at different times. There's no double counting. When flying you're in the reference frame of the air, so the headwind reduces your groundspeed when you're at touchdown airspeed. When you're on the ground, you're in the earth's reference frame, and the wind continues to oppose the remaining velocity. Really not sure what double counting you think is happening. $\endgroup$
    – llama
    Oct 20, 2023 at 22:22
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It's theoretically possible but in most cases not practically possible due to the limited runway length available. If the aircraft was very light (low threshold speed) and there was a strong headwind with an up slope then on a very long runway you could get down to a fast taxi speed. The problem at light weights is that your thrust at idle will tend to prevent slowing down to and maintaining taxi speeds without using the brakes or idle reverse. I've done it when the conditions were right on a very long runway. Not exactly SOP though!

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To complement the other answers with a summary:

No. Aerodynamic braking doesn't work well at low speeds. On a very long runway, it could at best slow the plane down to a non-lethal crash. Spoilers also don't produce as much drag as the huge brakes that jet fighters use - and fighters still need brakes or a parachute for a full stop.

However, transport category aircraft are required to be able to stop using brakes alone, no reverse thrust. They are also generally designed to be able to stop within the length of a runway using reverse thrust alone, no brakes.

This is for a runway. In a rough field, the stopping power from uneven terrain is stronger and sufficient for a full stop.

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Yes, it can be done. A medium jet slows down easily on light to moderate headwinds on a 4000m runway without reverse and wheel brakes to vacate via highspeed exit at the end of the runway.

And before someone comments, yes one should open reverse cowls (ie. idle reverse) and try out the brakes at safe distance remaining just to make sure they are there if required. And it is polite to make sure nobody is behind you so the next crew doesn’t have to squeeze their muscles in anticipation waiting if you vacate the runway in time or not :)

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  • $\begingroup$ thank you for your answer @busdriver - I have to admit my gut feel was that this could be possible - although you do say this is for a medium size jet - would this include, jets in the class 757, 767, A330 or A321/A320/737 ? Am I correct in thinking you are a professional airline pilot ? $\endgroup$
    – robbie70
    Oct 20, 2023 at 9:01
  • $\begingroup$ This information is incorrect. With no braking, you aren't going to be close to taxi speed before the end of a normal runway, unless something very unusual (massive headwinds, extraordinary upslope, etc) is going on. $\endgroup$
    – Ralph J
    Oct 20, 2023 at 16:07

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