rather common but still unable to find the reason

[Tapshi]

more weight on the landing gear increases brake effictivness. but how?
i have been unable to find any proper reason. isnt coefficient of friction independent of weight?

[777fly]

The coefficient of friction on a given surface is a constant. Its the braking effectiveness which varies with weight . It's a long time since I did A level physics, but I do remember the formula: F=uR, where F= retarding force, u= coefficient of friction, R= weight on tyre. Increase the applied weight and you increase the retarding force.

[Tapshi]

thanks a ton 777fly! :)

[Loose rivets]

The coefficient of friction on a given surface is a constant.

In the real world, wouldn't the increased contact surface have a greater effect?

I'm assuming that's not taken into account in the formula.

[Tapshi]

i think the increased contact area would be negligible practically. and so would the increased friction due to it! not very sure of it though!

[KBPsen]

Friction is (largely) independent of surface area. Force (weight) and material (tire and runway) is what matters.

[EEngr]

Increase the applied weight and you increase the retarding force. Yes, but increase the weight and you increase the retarding force necessary to achieve the same deceleration (F = ma). And that is (I assume) what one means by brake effectiveness. The two cancel each other out.

In reality, the increase in effectiveness is probably a second order effect. Increased tire contact area, etc. Beyond some point (weight), brake effectiveness will begin to drop. One factor is brake fade due to temperature. The more energy the rotors have to dissipate, the hotter they will get, soon reaching some maximum value. The rotor steel will soften for one thing.

Google for a max weight RTO test video. Or perhaps its better not to know what goes on down below. :eek:

[alf5071h]

Also an aspect of anti-skid; see ICAO circular ( http://www.iata.org/iata/RERR-toolkit/assets/Content/Contributing%20Reports/ICAO_Circular_on_Rwy_Surface_Condition_Assessment_Measuremen t_and_Reporting.pdf) on runway friction.
Aircraft aspects in Chapt 6, see paras 6.20, 6.21.

[Loose rivets]

Friction is (largely) independent of surface area.


That's ringing a bell from the very dim and distant past. I find it as hard to believe now as I did then, but the boffins know best.

It leaves me wondering why tyres/tires on my hotter cars are twice as wide as the old days.

[Microburst2002]

Friction coefficient is an entirely empirical constant for each material. And so is the fact that friction force is independent of surface.

The more normal force between the surfaces, the more the friction force. For the same airplane mass, it means that if more weight is on the wheels (when spoilers spoil much of the lift force) there is more friction force, and landing distance or stop distance is reduced.

[deltahotel]

Isn't the important friction, ie the retarding force, between the brake pads and the discs? The weight on wheels thing allows more brake pressure to be applied before the wheels start sliding over the surface. Hence the use of speedbrakes; imagine not using them and the lift still from the wings taking the weight off the wheels - you wouldn't be able to put so much pressure on the brakes without skidding.

[777fly]

Well, whatever the arguments, F=uR always applies. If F is the resistance to motion available ( torsional friction at the wheel hub) at max antiskid brake pressure, its effectiveness can only be influenced by the interaction of surface coefficient of friction with weight on wheels. On a frictionless runway, F will always be zero. If u has any value above zero, any increase in R (the weight) will always lead to a higher effective value of F (retarding force)
EEngr: sorry, but your observation is wrong in the case we are discussing. F=ma applies to linear forces, but F=uR is about a vertical interaction. F=uR will produce a particular value of 'F' in F=ma. Since the aircraft mass is a constant, the deceleration (a) depends on F, which is in turn dependant on uR. Your view that more weight at the wheels would require greater retardation force is incorrect, since we are considering that the brakes are being applied at max antiskid pressure to a fixed mass. The fixed mass of the aircraft still needs to be stopped and it is the value of F that will do it.

[balsa model]

It leaves me wondering why tyres/tires on my hotter cars are twice as wide as the old days.
Speculating here, but perhaps, fatter tires to handle more ponies. Material limits still apply; we don't want to shred this rubber in one go. With lower shear forces per width of rubber material, we also have more choices when picking the rubber material for most desirable coefficient of friction, for example. So it's still all connected.
And the threads don't deform as much in a turn.

[chksix]

A wider tyre can handle a bigger slip angle before losing traction. Rear tyres are often wider than the front ones due to having to cope with cornering forces on top of engine torque.

I have to double check the definitions in the book about car suspensions I have at home though. :8

[Capt Pit Bull]

A few points:

Weight on the tyres, for aircraft, is a variable even when the aircraft mass is constant. Hence the need for spoilers.

Well, whatever the arguments, F=uR always applies.

Hmm. But u is only constant for static friction. Once things are sliding everything gets way more complicated. Materials heat up / liquify / vaporise and this is where the big tyres make a diference.

iirc optimum slip for maximum antiskid braking is about 10% of velocity.

[EEngr]

777fly (http://www.pprune.org/members/104066-777fly): Work out the math(s)

F = uR

F = braking force
u = coefficient of static friction
R = weight (the vertical force due to gravity and other stuff*)

F = ma

a = acceleration (deceleration) resulting from F acting on mass m

R = mg (*see above)

so:

ma = uR = umg

m cancells out:

a = ug

The maximum deceleration available, a (at max antiskid brake pressure) depends on the coefficient of static friction, u. Obviously, this a linear simplification of real life.

Capt Pit Bull (http://www.pprune.org/members/8103-capt-pit-bull):
Hmm. But u is only constant for static friction. Once things are sliding everything gets way more complicated. Materials heat up / liquify / vaporise and this is where the big tyres make a diference.
An approximation of static friction is what you have as long as you don't skid. Its an approximation because under braking forces, the tires begin to flex and there is some scuffing between them and the pavement. This results in heating and a resulting change in u (sometimes an increase as the tires warm up and get a bit softer).

I'll stand behind my statement that, for the simplified case, braking 'effectiveness' doesn't depend on aircraft mass. Mass does affect factors such as the tire contact patch area, tire heating and change in u, weight shifting due to braking (a much larger effect in automobiles) and a number of other factors. Some of these have highly nonlinear relationships to weight (downward force on the contact patch) and depend on which side of the peak on some graph an engineer has placed the braking characteristics.

Look at the case of a truck: More tires (and wider). Heavier (greater R per tire), and yet they tend to have longer stopping distances (at maximum no-skid braking) than a passenger car. That is because the engineers (due to economics) have designed closer to, or on the other side of, the point of maximum braking effectiveness. Which is why you see more 'road alligators' (tire carcasses) from trucks than cars on the highway.:mad:

[Yankee Whisky]

What about increased weight causing increased inertia to be dissipated

for a residual constant max brake parts efficiency ?

Or are you chaps just talking about normal landing weight after a long

flight and dumping the lift ? I believe glider type airbrakes would provide

better lift dumping than spoilers. This is true in gliders, so why not big

jobs ! I found retracting the flaps on power planes effective as well, though

less than airbrakes on a glider.

I guess that runway length covers a lot of shortcomings in this

department ! :):)

[tightcircuit]

You are confusing weight, mass and downforce. Weight is the product of mass and gravity. You cannot increase it aerodynamically but you can add aerodynamic downforce to it. The mass of the a/c is not affected by increasing the downforce and so inertia is not affected. By increasing the downforce you are increasing the surface area in contact with the runway. The statement that friction does not increase with increased contact area is very simplistic and only applies if no other factor changes. It occurs because any increase in area must also be accompanied by a decrease in contact pressure. In the case of adding aerodynamic downforce to weight then the tyres deform more adding extra contact area but this is acheived at the same contact pressure. This increases the total friction available without a change in inertia.

[FullWings]

I think the OPs question may be more related to the effect of lift dumping on deceleration, rather than the complicated area of friction vs. tyre pressure.

If you didn't have spoilers, on touchdown there would still be a significant amount of lift generated by the wing. This means the maximum braking force would be limited by the weight on the wheels being a lot less than the mass of the aircraft, leading to a slower deceleration.

A lot of overruns have late/no spoiler deployment as a prime cause, especially on wet/contaminated runways.

[Owain Glyndwr]

If I may jump in ...

To get any braking at all, there must be some slippage between tyre and runway surface. It is usual to express this as a ratio (slip ratio) which is a percentage of the unbraked wheel speed.
The braking friction starts out being proportional to the slip ratio up to a maximum at about 20% slip and thereafter it deteriorates as the wheel goes into a true skid. The aim of the ABS is to keep the braked wheel as close as possible to this optimum slip ratio. To do this (at least in the systems I am familiar with) the actual wheel rotational speed is compared with an unbraked wheel speed derived from a measured nosewheel rotational speed.
The resulting braking friction coefficient is not constant; on dry concrete it varies from (about) 0.8 at low speed through 0.7 around 75 kts down to 0.65 at about 150 kts.
The braking force will be the appropriate coefficient times the wheel reaction at that speed. Obviously, immediately after touch down the braking will be at a minimum on both counts. The wheel reaction will be the aircraft weight less the aerodynamic lift. Even with all wheels on the ground this can be a considerable proportion of the actual weight - maybe as high as 50% with full flap.
Deploying lift dumpers drops this to something close to zero and at the same time just about doubles the aerodynamic drag - both very useful at a time when runway distance is being gobbled up.
When the aircraft gets to lower speeds the wheel reaction is close to the actual weight and the coefficient of braking friction is also near its maximum. In these conditions the torque needed to maintain the optimum slip may become greater than than available from the brakes. The braking force is then limited by maximum brake capability.

On the question of a landing distance monitor, I agree that it should be possible, but I'm not sure how reliable it might be given the possibility that the braking characteristics of any given runway can vary with time - how for example do you know how much friction you are going to get on that final bit of runway that is contaminated by rubber smears from repeated touchdowns and maybe also wet?

[Owain Glyndwr]

msbbarrett

The Formula 1 guys know a thing or two about tyres and brakes. They got to the point where the tyres started losing grip on the wheel rim, not the road. They had to go to toothed rims to stop this happening. Does a similar problem exists on aircraft?

Not so far as I know.

[Mr Optimistic]

Is it really this complicated ? Did the OP want to know about the benefits of reducing lift off the wing when braking ? The earlier math(s) neglected that R= mg-Lift, so dumping lift improves braking action.

F = mu x R is fine for this, surely ?