For large commercial jets is it possible to land and slow sufficiently to leave the runway without using reverse thrust or brakes

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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 :)