[mile201]

Originally Posted by arun4242

ABS stopping distance versus Non ABS systems:

.....To all those, who have managed to read this post all the way till here, I would say thanks and hope I have been able to convince you that ABS stopping distances are shorter.

Thanks Arun for the detailed reply. Very helpful!
As you pointed out, I have noticed that the brake force generated when the wheels are moving is more than when they are locked up...this is very apparent on hard ice where after lock up there is not much of decrease in speed (non ABS car). Another thing i noticed was the distinct brake pulse that you feel in paddle is much slower when braking on ice or gravel as compared to asphalt..why is that?

[DerAlte]

Originally Posted by Sutripta

... Can't figure out why braking force should depend on pad area.
More importantly, why should the force at which the wheel starts locking up depend on speed. ...

1. Braking force is directly proportional to the friction exerted on the drum / disc, which is directly proportional to area of contact of pad with drum / dsc -> pad area.

2. Wheel lock happens when force exerted on wheel by braking exceeds the counteracting force (angular momentum of rotating drum/disc). When momentum of the vehicle overcomes friction with road surface (when vehicle is braking), there is a point at which tyre loses adhesion and hence angular momentum on drum/disc becomes 0 suddenly, resulting in brakes locking the wheel. Hence wheel lock is speed (momentum) dependent.

Originally Posted by mile201

... this is very apparent on hard ice where after lock up there is not much of decrease in speed (non ABS car). Another thing i noticed was the distinct brake pulse that you feel in paddle is much slower when braking on ice or gravel as compared to asphalt ...

The modulation (on-off) of the bleed valve is varied due the friction factor of the road surface being different. On low-friction surfaces (wheel slips more on ice and gravel), the valve OPEN time is less than CLOSED time. On asphalt (higher friction) it stays OPEN longer than CLOSED. This difference in modulation is what you notice in the pedal.

[im_srini]

Originally Posted by Sutripta

Can't figure out why braking force should depend on pad area.

Originally Posted by DerAlte

Braking force is directly proportional to the friction exerted on the drum / disc, which is directly proportional to area of contact of pad with drum / dsc -> pad area.

Hi, DetAlte, Sutripta's right, force due to friction (between 2 surfaces) is dependent on only 2 parameters, the normal force acting on the surfaces, & the coefficient of friction - the contact area should not matter.

Originally Posted by Sutripta

More importantly, why should the force at which the wheel starts locking up depend on speed.

Originally Posted by DerAlte

Wheel lock happens when force exerted on wheel by braking exceeds the counteracting force (angular momentum of rotating drum/disc). When momentum of the vehicle overcomes friction with road surface (when vehicle is braking), there is a point at which tyre loses adhesion and hence angular momentum on drum/disc becomes 0 suddenly, resulting in brakes locking the wheel. Hence wheel lock is speed (momentum) dependent.

I side with Sutripta again, if the car's stationary with the brakes engaged to the maximum, & its pulled forward or towed, shouldn't the force the brakes must exert to keep the wheels stationary (so that the car's dragged forward) be the same as the force required to lock the wheels at any speed ? Of course, when in motion, the deceleration would result in weight transfer & hence the braking force would increase. Even then, force required for wheel lock-up would be dependent on the rate of deceleration rather than the absolute velocity of the car, no ?

[DerAlte]

Originally Posted by im_srini

Hi, DetAlte, Sutripta's right, force due to friction (between 2 surfaces) is dependent on only 2 parameters, the normal force acting on the surfaces, & the coefficient of friction - the contact area should not matter.

Sorry sir, I have never been good at physics, but that doesn't explain why skiers use narrower skis for speed skiing than for slower ambling, and ice-skaters have a narrow edge that they skate on rather than using the shoes without. If there was no dependency on pad area, we could have still used the brake shoes from bicycles in motorcycles, assuming that the magnitude of braking would be the same.

Originally Posted by im_srini

Isn't wheel lock-up dependent on the rate of deceleration rather than the absolute velocity of the car ?

The dependence is on the momentum of the vehicle, which is composed of the mass of the vehicle (doesn't change while the vehicle is moving) and speed, which overcomes the braking forces at wheel periphery and causes slipping. Wheel lock is a resultant of that, since nothing is acting against the braking force now to cause wheel rolling. Rate of deceleration is only an observation during the process, and it itself depends on these factors

I am not sure what will happen if one has not taken the foot off the gas pedal, clutch is engaged, and engine is still supplying power to the wheel and one has pressed the brake fully at the same time - possibly wheel lock will be delayed or won't happen.

[McLaren Rulez]

Friction is a pretty complex thing. At the heart of it, you're trying to explain how atoms of different bodies interact with each other. Its definitely more than the basic law of friction that one studies in high school

First, I can tell you that the force of friction is sometimes dependent on the surface area of the two bodies in contact. The idea that friction is dependent on only the normal force and the coeffcient of friction is an approximation that works well in many cases but isn't always true. Why do we bother fitting fat tyres on our cars if the grip was indeed independent of the area of contact? We should go with tyres no wider than on bicycles if that was the case.

Second, friction dependence on velocity: I'm not too sure about this but I guess (really guessing here) it has something to do with the fact that at higher speeds, braking causes the temperature of the discs to reach a much higher value. Temperature does affect the properties of a material, including its coefficient of friction. Simple example: Hot rubber sticks better which is why cold tyres on F1 cars are a no-no.

I'm not sure of the details myself but the basic idea is that friction is not just mu*N because that is an ideal case (perfectly rigid bodies, atomic contact increasing linearly with pressure, material properties staying unchanged, etc.).

[Sutripta]

I think we should wait for Arun's clarifications. He after all has formally studied the subject.

Regards
Sutripta

[vivekiny2k]

to start with, mu*N may be right, but if the contact area is small, it will lead to more wearout of the material, and more heating in a small patch of contact(energy dissipated). And this is where we move from science to engineering.

So technically a bicycle brake can give you same braking power in car, it will fail in no time. if you are talking about a 2000 kg car coming to halt from say 40 kmph, imagine the energy dissipated on a tiny brake shoe from a bicycle.

As a whole, deralte is right, wider contact area gives better braking. and so Mclauren is right too.

[arun4242]

Originally Posted by Sutripta

Can't figure out the orientation of the sensor on the wheel rim. Also, the signal is yes/ no in nature?

It would be parallel to the rim and rotate along with the wheel

Originally Posted by Sutripta

Can't figure out why braking force should depend on pad area. Sutripta

Originally Posted by im_srini

Hi, DetAlte, Sutripta's right, force due to friction (between 2 surfaces) is dependent on only 2 parameters, the normal force acting on the surfaces, & the coefficient of friction - the contact area should not matter.

Braking force generated = Frictional force generated

Now Frictional Force = Mu * N where Mu = coefficient friction (considered as standard for two given surfaces) & N = Normal Force applied.

The normal force is defined as the net force compressing two parallel surfaces together; and its direction is perpendicular to the surfaces. In the simple case of a mass resting on a horizontal surface, the only component of the normal force is the force due to gravity, where N= M*G. In this case, the magnitude of the friction force is the product of the mass of the object, the acceleration due to gravity, and the coefficient of friction. However, the coefficient of friction is not a function of mass or volume; it depends only on the material. For instance, a large aluminum block has the same coefficient of friction as a small aluminum block. However, the magnitude of the friction force itself depends on the normal force, and hence the mass of the block.

Larger pads would imply larger mass (an assumption but a very valid one) and hence higher braking force generated for the same amount of pressure applied

Regards
Arun

[arun4242]

Originally Posted by Sutripta

I think we should wait for Arun's clarifications. He after all has formally studied the subject.

After seeing this reply, had no option but to dig up my books (and Wikipedia) for some answers.
and BTW, the practical knowledge of some of the BHPians far exceeds my limited formal knowledge which I can see on this thread

Originally Posted by Sutripta

Interesting post. Some clarifications please.
More importantly, why should the force at which the wheel starts locking up depend on speed.

It seems I was right about wheel lock depending on speed but my reasons were wrong. (DerAlte has got his facts correct intuitively)
Some guy called Couloomb has formulated a law which says kinetic friction is independent of velocity. Apparently friction involves atomic level interactions and normal physics of more speed=more force does not hold good for friction.
This implies that braking force generated by friction at 100 kmph or 20 kmph will be the same if brake pressure applied is equal at both speeds. However the car's momentum/ forward force would be dependent on the speed and would effect itself by the rolling velocity at the wheels.
Now wheel lock would occur when the braking force > momentum of car.
At high speeds with full brakes, the momentum may be> braking force and wheel lock would not occur.
At low speeds with full brakes, the reverse would occur resulting in braking force > momentum causing wheel lock. (This is because braking force is equal for both speeds while momentum varies)
Hope this clarifies - regards

[vivekiny2k]

Originally Posted by arun4242

Now wheel lock would occur when the braking force > momentum of car.
At high speeds with full brakes, the momentum may be> braking force and wheel lock would not occur.
At low speeds with full brakes, the reverse would occur resulting in braking force > momentum causing wheel lock. (This is because braking force is equal for both speeds while momentum varies)
Hope this clarifies - regards

I would replace "momentum" with "momentum that can be transferred via the wheel's contact patch". So at high speed a wheel lock will not occur at a dry F1 track but it will occur at icy roads because it can not be transferred to the brake pad via the tyres. (I am just following you statement, not sure about the speed factor).

Second, wheel lock occurs when kinetic friction changes to static friction (at brake pad, while at contact patch it change from static to kinetic). this will take longer at a high speed. This might explain the previous point better.

EDIT: An example I remember, when I used to ride my uncle's scooter, I could brake hard and sometimes even skid a little. But when I had a pillion (my late uncle), no matter how much I pressed the pedal, I could not get it to skid, or even stop in time. And then he would scold me. His mass was not only increasing the total momentum, it as also providing extra traction at rear wheel so all of the momentum was getting transferred to the brake assembly, and the brake pad could just not deal with that much force. I also think the pads should have been changed, but the inadequate brakes did teach me a little about friction.

His stance was I should drive sensibly so I should never have to brake that hard , so brakes never got fixed.

[DerAlte]

Originally Posted by arun4242

... facts correct intuitively ...

Baap re baap, this one has reached too high levels of maths and physics for my comfort now (I am getting disturbing visions of equations with Σαβγθ) *running away fast*

[Sutripta]

Originally Posted by arun4242

It would be parallel to the rim and rotate along with the wheel

I expect the tube is curved, and not straight.
It will discriminate between just locked, and not locked. It cannot detect 'about to lock'.

Braking force generated = Frictional force generated

Now Frictional Force = Mu * N where Mu = coefficient friction (considered as standard for two given surfaces) & N = Normal Force applied.

The normal force is defined as the net force compressing two parallel surfaces together; and its direction is perpendicular to the surfaces. In the simple case of a mass resting on a horizontal surface, the only component of the normal force is the force due to gravity, where N= M*G. In this case, the magnitude of the friction force is the product of the mass of the object, the acceleration due to gravity, and the coefficient of friction. However, the coefficient of friction is not a function of mass or volume; it depends only on the material. For instance, a large aluminum block has the same coefficient of friction as a small aluminum block. However, the magnitude of the friction force itself depends on the normal force, and hence the mass of the block.

Correct. Because, as you have said, the normal force is the weight of the block.

Larger pads would imply larger mass (an assumption but a very valid one) and hence higher braking force generated for the same amount of pressure applied

Regards
Arun

Normal force in this case is a function of the hydraulic/ mechanical system. Mass/ weight of pads does not figure in it.

Regards
Sutripta

[Sutripta]

Originally Posted by arun4242

Some guy called Couloomb has formulated a law which says kinetic friction is independent of velocity. Apparently friction involves atomic level interactions and normal physics of more speed=more force does not hold good for friction.
This implies that braking force generated by friction at 100 kmph or 20 kmph will be the same if brake pressure applied is equal at both speeds. However the car's momentum/ forward force would be dependent on the speed and would effect itself by the rolling velocity at the wheels.
Now wheel lock would occur when the braking force > momentum of car.
At high speeds with full brakes, the momentum may be> braking force and wheel lock would not occur.
At low speeds with full brakes, the reverse would occur resulting in braking force > momentum causing wheel lock. (This is because braking force is equal for both speeds while momentum varies)
Hope this clarifies - regards

No. Not clarification. More confusion.

We agree that
A) Two values of friction, static and kinetic.
B) Each of these is independent of velocity.

These values determine the maximum braking (Or tractional) force which can be generated by the tyre, when rolling, and when scrubbing. Thus if we exceed this value when the tyre is rolling, it will start to lock, and almost immediately (by human time standards) lock.

Once again velocity does not appear in the maximum force which can be generated.

(NB. We are not talking of weight transfer to the front wheels during braking.)

Regards
Sutripta

[DerAlte]

Originally Posted by Sutripta

... These values determine the maximum braking (Or tractional) force which can be generated by the tyre, when rolling, and when scrubbing. Thus if we exceed this value when the tyre is rolling, it will start to lock, and almost immediately (by human time standards) lock.

Once again velocity does not appear in the maximum force which can be generated.

(NB. We are not talking of weight transfer to the front wheels during braking.)...

Errrr... we are confusing some terms here.

a. Braking force is not generated by the tyre - it is generated by the brake shoes rubbing against the brake rotor (drum or disc attached to the wheel). Though this is tangential to the rotor rotation, it can be modeled as angular force with a -ive sign (opposite to rotor rotation)
b. The momentum of the vehicle (speed dependent) is being transferred as angular force to the brake rotor, +ive since it is along the direction of rotation of the rotor
c. As long as sum of the above angular forces is +ive, the brake rotor, and hence the wheel, keeps rotating
d. As soon as this sum is 0 or -ive, there is no further rotation (wheel lock). This is either
- the brake force (a above) exceeding - especially when brake is pressed real hard - the force transferred from the vehicle momentum (b above; speed dependent) or
- b suddenly becomes 0 because tyre loses adhesion so momentum is no longer transferred (depends on coeffiecient of tyre friction with road surface; skidding if incidentally there was an sandy, icy or oily patch at the time of braking)
- the brake rotor deceleration is rapid when the sum is reaching 0, i.e. this is not a linear decline (as you rightly pointed out)

The above is not a simple interaction by any means. We tend to simplify the model so that it is easier to understand. Whether the wheel lock happened before the skidding or after is simplified as 'skid because of wheel lock', forgetting that coeff. of friction between tyre and surface is variable either because of road surface or the tyre chemistry.

Don't disregard the transfer of forces to the front wheel during braking. Cheaper 3-point 3-sensor ABS systems for small & medium trucks rely on this, i.e. they concentrate the action on the front wheels and save a channel from the rear saving cost.

[Sutripta]

Originally Posted by DerAlte

Errrr... we are confusing some terms here.

a. Braking force is not generated by the tyre - it is generated by the brake shoes rubbing against the brake rotor (drum or disc attached to the wheel).

Used the term the way it is popularly used.

Though this is tangential to the rotor rotation, it can be modeled as angular force with a -ive sign (opposite to rotor rotation)
b. The momentum of the vehicle (speed dependent) is being transferred as angular force to the brake rotor, +ive since it is along the direction of rotation of the rotor

Linear Momentum is M.V. (=P). Angular Momentum is M.w. F= dP/dt. Different dimensions.
For this particular case, I find it easier to think in terms of F=M.Acc, rather than F= dP/dt.

The maximum decelaration (of the body = car) that we can achieve (by braking a wheel) is limited by the maximum force (result of static friction) between tyre and ground, and cannot exceed this value. This maximum value is not dependent on speed (AFAIK) (This is the crux of the discussion). The maximum deceleration we will get at 100 kmph will be the same as the maximum deceleration we will get at 10 kmph.

Or the deceleration we will get with wheels locked at 100 kmph will be the same as the maximum deceleration we will get at 10 kmph (again with wheels locked).

Because both the mass remains the same, and the force remains the same, not being a function of speed. (Incidentally the deceleration with wheels locked will be less than the first case)

I thought that that is what we were discussing/ debating. First order terms/ approximations. Not a full modelling of braking behaviour in which we start factoring in higher order terms. That I'm sure can be handled in another thread.

Regards
Sutripta