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If weight was not an issue, would increasing the number of main gear wheels on a jumbo such as the Airbus A380 from 20 to 40 wheels, each equipped with braking disks and braking equipment, considering their extra weight, reduce braking distance.

Would it be any different if the current set up was used but with longer wheelstuds and each wheel was paired with an extra wheel, as with most semi-truck's (articulated lorry's) rear wheels?

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    $\begingroup$ Some aircraft use a drogue parachute to shorten their stopping distance, in case you weren't aware of that option. $\endgroup$ – Andrew Morton Jan 25 '18 at 18:40
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    $\begingroup$ No disrespect to GdD but Sylvian's answer is much better, you know you can change the accepted answer right? $\endgroup$ – Notts90 Jan 26 '18 at 7:59
  • $\begingroup$ Aren't thrust reversers just as -- or more -- important than brakes? More wheels add more weight, complexity and stuff to maintain. You should instead ask whether bigger reversers would reduce landing distances. $\endgroup$ – RonJohn Jan 26 '18 at 19:21
  • $\begingroup$ @RonJohn sure it's obvious more reverse thrust reduces landing distance? That doesn't always make it the best option though. $\endgroup$ – Notts90 Jan 27 '18 at 17:43
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    $\begingroup$ @Notts90 "that’s a separate question entirely" Which I found the answer to here: airspacemag.com/flight-today/… $\endgroup$ – RonJohn Jan 27 '18 at 21:38
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Increasing the number of wheels with brakes will decrease the braking distance as there's more sources of friction to convert the kinetic energy. Doubling the number of wheels per brake (I think what you're asking in the second part of the question) wouldn't automatically decrease braking distance as the brakes are limited to absorb a certain amount of energy, but you could put a more powerful brake on a pair of wheels than a single wheel of the same size as there's more grip.

In reality this wouldn't be done because it's not required, runway length is determined as much by takeoff length than braking distance, you could halve the braking distance of all airplanes and you wouldn't be able to reduce runway length.

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    $\begingroup$ Isn't required takeoff length also affected by braking distance, though? You have to have enough runway left to stop (ideally without blowing tires or setting the brakes on fire) in an RTO. $\endgroup$ – reirab Jan 25 '18 at 15:36
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    $\begingroup$ @reirab RTO>?.. $\endgroup$ – Cloud Jan 25 '18 at 15:51
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    $\begingroup$ @Cloud Sorry. RTO = Rejected Take Off (when you abort a takeoff run.) $\endgroup$ – reirab Jan 25 '18 at 16:03
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    $\begingroup$ @MichaelKjörling Indeed. But there are also reasons why we try to keep them as close as possible. Additionally, if you do have a large gap from V1 to Vr, this will also increase your required takeoff runway length, as you'll need to have sufficient runway remaining after V1 to accelerate from V1 to Vr and climb out safely with an inop engine. $\endgroup$ – reirab Jan 25 '18 at 16:23
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    $\begingroup$ Actually there are two ruling factors. Brake capacity, and adhesion. Increasing the number of wheels won't help adhesion because adhesion is proportional to weight on rubber. Double the rubber surface, half the weight per cubic inch, nets out to same adhesion. What would help is manipulating airflow to push down on main gear, giving more weight-on-gear than the aircraft weighs. That's exactly what an IndyCar spoiler does. $\endgroup$ – Harper - Reinstate Monica Jan 26 '18 at 20:38
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Typically, a little gain is obtained.

Larger airplanes use anti-skid technology. Anti-skid works by modulating brake pressure to ensure the tires never skid. It's important to understand the relation between a tire's load, brake pressure, and actual retarding force. First, look at this image : friction coefficient versus slip

It shows how as you increase pressure and the wheel starts slipping a bit (turning at a rate slightly slower than when free rolling), the friction coefficient rises sharply. Then after the 10% slippage, the peak coefficient is reached and thus the retardation force reduces. Anti-skid aims to keep the slip raitio as close to 10% as posssible. Therefore the retardation force is maximized.

The brakes of airplanes are strong enough to fully lock the tyres, but this is detrimental (results in 100% Slippage, so maybe 20% retardation force loss). So they can and do maintain optimal slip, therefore using the entire available 'grip'.

Now what happens to this picture if you add wheels?

  • If the added wheel is not braked (like the nose wheels are) then you lose braking. Indeed, the braked wheels will have a reduced vertical load, so they will provide less retardation force. It's quite useless.
  • If we add braked wheels then we see the tyiniest bit of gain. Let's see why.

First look at this picture of tire load sensitivity (this one shows lateral force while cornering, but longitudinal force while braking exhibits the same features): tire load sensitivity

What is important to note, is that as a tire gets more loaded, its capacity to generate friction decreases.

Now for a jumbo: When 20 wheels share 500 tons of load, each tire sees 25t. with a default friction coefficient of 0.8, each tire generates 20t retarding force, so total of 400t retarding force.

Let's add 2 wheels in there. Now each wheels sees 22.7t of load. Their friction coefficient could now rise to 0.82, therefore providing 18.6t, for a total of 410 tons. Certainly far of the 40t that could be naively expected from adding 2 braked wheels! Only the load sensitivity nets us a gain.

It then becomes the usual game of trade-offs. How much mass is spent on bigger landing gear and wheels and additional brakes, versus the gain of harder braking (and thus even stronger landing gear legs?)

Apparently the Airbus guys decided the trade stopped at 20 wheels (because 25t of load on a single tire is the maximum the ground could handle) but they didn't even bother to brake them all, if I remember well only 4 of the 6 wheels of the main boggies are braked! So if you wanted to improve brake efficiency:

  1. Install better tires on braked wheels
  2. Brake the non-braked main wheels
  3. Make some smart aero changes to increase aero downforce & drag
  4. Add braked wheels (and prepare to house them in the main body... good luck!)
  5. Brake the nose wheels (but that would be terribly hard because it was never designed for that)
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    $\begingroup$ There's an important effect that doesn't show up in your charts: maximum braking effort of an airliner is limited not by friction, but by heat generation. An airliner braking as hard as possible is right on the edge of blowing tires and/or catching the brakes on fire; more brakes or more braked wheels will let it stop faster by spreading the heat over more mass. $\endgroup$ – Mark Jan 25 '18 at 23:33
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    $\begingroup$ @Mark Or, sometimes, they're a little past the edge... $\endgroup$ – reirab Jan 26 '18 at 3:43
  • $\begingroup$ That's a very elaborate proof of a simpler concept: adhesion is proportional to weight on rubber, not rubber surface area. Increasing surface area decreases weight per surface area, which decreases adhesion per surface area in proportion leaving you right where you started. But you needn't add more gear; just let some air out of the tyres. If you have 10,000 lb on a 100psi tire, your contact patch will be 100 sq.in. Deflate to 80 psi and your contact patch is 125 sq.in. $\endgroup$ – Harper - Reinstate Monica Jan 26 '18 at 20:50
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    $\begingroup$ @Harper, despite what you learned in school, adhesion is not proportional to weight on rubber, it's only approximately so. The actual mechanics of friction are so complicated that the greatly simplified "friction = normal force * coefficient" is what gets taught at the undergraduate level and below. $\endgroup$ – Mark Jan 27 '18 at 2:21
  • $\begingroup$ Sure, I admit that's the rough version of the math, the high school version, and there's some optimizing possible. This for sure: what you say about braking, and how the brakes are able to ride at 10% slip, pretty much shows adhesion is the limiting factor, not braking. Now how to get more adhesion? $\endgroup$ – Harper - Reinstate Monica Jan 27 '18 at 5:01
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Here’s something on Physics.SE that you may find interesting.

It’s all about the force and mass, not the contact area.

More wheels, same brakes = no change

Adding more wheels but keeping the same number of brakes would have negligible affect on stopping distance, as the total force going through the wheels and brakes would be the same and the mass change small.

There would be more wheels but each one takes a smaller load so there’s no additional friction between the wheels and the ground. There maybe some additional forces from the extra bearings and some extra mass from the wheels but that would be negligible in terms of stopping distance.

There’s the same number of brakes so they are providing the same force so they make no difference.

Same wheels, more brakes = shorter distance (possibly)

If we add extra brakes to each wheel (or bigger brakes) that allows us to apply more force to the discs, then we have more braking force and therefore the aircraft will stop in a shorter distance.

There are some issues with this though:

  • Heat - brakes generate a lot of heat that needs to be dissipated, more/bigger brakes will generate more heat. If the brakes get too hot they will fail and then you may not stop in time.
  • Structural limits - putting more force through the structure may cause it to fail which could be catastrophic.
  • Wheels - put to much force/heat through the wheels and they may fail as well. That maybe bursting if the heat/forces are to much or simply skidding if too much force is applied too quick. There are similar systems to the ABS on cars for aircraft the can help with this (Thanks to Michael for the link).

More wheels, more brakes = shorter distance

Adding more wheels with more brakes solves the main issues with the “same wheels, more brakes” scenario.

Why don’t aircraft manufacturers do this? Well all them additional wheels and brakes are mass that’s expensive to transport unnecessarily.

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    $\begingroup$ you missed something- if the brakes are already stronger than the tires ability to stop, for example like most passenger cars in low speeds- you can break hard enough to make the car skid, than adding more wheels (or just better tires) can potentially shorten braking distance $\endgroup$ – Rsf Jan 25 '18 at 15:07
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    $\begingroup$ @Rsf Or use better brakes which don't cause the wheels to lock as easily. Those aren't found just on cars; airplanes have them, too. $\endgroup$ – a CVn Jan 25 '18 at 16:14
  • $\begingroup$ @Rsf I was counting that as failure of the wheels, I’ll be more specific. $\endgroup$ – Notts90 Jan 25 '18 at 16:37
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    $\begingroup$ @MichaelKjörling: Anti-lock brakes work by decreasing the braking force if it would otherwise exceed the available wheel-road frictions. So "better brakes" will not allow you to overcome that limitation. $\endgroup$ – Henning Makholm Jan 25 '18 at 16:41
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    $\begingroup$ @Rsf: But more wheels won’t increase friction. $\endgroup$ – Michael Jan 25 '18 at 16:47
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It's a tradeoff, like just about anything involving aircraft design.

More wheels + more brakes means a shorter stopping distance. But...

More wheels means more weight, quite a bit more, to accommodate the extra landing gear, extra retraction/extension gear, extra storage while retracted, extra maintenance for all that extra stuff, etc... that means fewer paying customers and higher operating costs. This also means more things that can go wrong.

More braking means more loads subjected on the paying customer. I recall being on a C130 when the pilots practiced their short field landing: full reverse props just before touchdown, full braking. We hit (and I do mean HIT) the runway with a very loud clang, then we were slammed forward as the plane came to a very quick stop. It was... dramatic... fun when you're a twenty something in military training, but I'm not sure paying customers would like that.

As others have mentioned, the limitation is also takeoff distance. Doesn't do a lot of good to be able to stop in 500 feet, if the takeoff distance is 2000 feet. You'l still need a long runway if you want to get the plane out of the airport.

The one possibility would be safety in the case of an emergency, but the number of aircraft accidents that could have been prevented with more wheels for braking is extremely small.

Current airliner brakes are designed to meet the requirements, while remaining as light and as reliable as possible.

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Absolutely, properly designed, more wheels and more brakes will get you a shorter stopping distance.

However, there is a lot more goes into the design of the landing gear and braking system than you might imagine. Braking distance, and force, is just one factor that is considered as part of that design and is more or less fixed for a particular aircraft style since runway lengths are defined by take-off more than landing.

Other factors which are balanced to achieve that obviously include weight, space lost to stow the landing gear, cost of manufacture, cost of maintenance, reliability and importantly, heat dissipation. Stopping a wide body jet generates A LOT of heat in the brakes which needs to be dissipated quickly to prevent overheating and fire.

Doubling up wheels can, with added stronger brakes, get you more braking, but you also now have to have a stronger carriage and fuselage attachments to absorb that braking force, and again, a larger wheel bay.

How much you want to brake is also dictated by passenger comfort. Brake too hard and you are open to passenger injuries and lawsuits.

That is a complex set of conflicting requirements. Adding more wheels and brakes would make stopping easier, but at a significant cost.

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Nope, and it's because of how tires work.

The limits are tire adhesion and braking capacity. The latter is not the limiting factor - adhesion is.

Adhesion is roughly proportional to weight on tire. Specifically, contact patch area (sq.in.) x weight per square inch, but you notice square inches factor out, leaving only weight on tires. It's more complicated than this, this is the prep school version, and there is some tuning to be done -- but it's been done. And it doesn't help that much.

Take my car (please). It has 600 pounds on a front tire. How big is the contact patch? That's actually easy: the tire pressure is 30 PSI. 600lb divided by 30 lb/sq.in. = 20 sq.in. Contact pressure is, you guessed it, 30 pounds per square inch.

What happens if I reduce my tire pressure to 20 PSI? OK, now 600/20 = 30 square inches of patch area. With 20 PSI contact area.

Which brakes better? It's pretty close to a wash. 30x20 ~= 20x30 =~ 600.

What if I add lead ballast to my front end? Now my weight is 750 lbs/tire (25% more). YES! I get 25% better adhesion and thus braking power. Except I now have 25% more mass to stop! So it doesn't slow any more swiftly.


OK, so you take a 747 and replace its 4-wheel bogeys with magic 6-wheel bogeys that weigh the same. 50% more wheels, right? 50% more contact area, right? Nope. Think about how contact area is determined. Same tire pressure but each tire is carrying 2/3 the weight as before, so the contact patch is 2/3 the size. You end up with exactly the same contact patch (in square inches) and exactly the same pounds per square inch as you had with the 4-wheel bogeys.

Anyway, if you wanted to increase the contact patch 50%, you just reduce the tire pressure 33%. But just like my car, that wouldn't help you either.

All these factors wash out (roughly). The ruling factor in adhesion is the weight on bogeys.

If only there was a way to increase weight on wheels without increasing vehicle mass.

enter image description here

There's the answer. It's an aerodynamic lift surface that pushes down on the tires. those cars go plenty fast enough for the Marlboro to be aerodynamically effective. This works. Weight-on-tires is increased so adhesion increases in proportion. Vehicle mass is not increased so it brakes correspondingly better.

Do the same thing on an airplane. Come up with a way to attach a big wing of some kind directly over the main gear. And set it up to generate downforce.

Not sure how to do it, though. Maybe this? ;-)

enter image description here

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  • $\begingroup$ The lift dumpers do help. $\endgroup$ – Koyovis Jan 27 '18 at 4:37

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