Braking at high speed vs low speed.

[Kozy]

I am interested in the forces at play determining wheel lock up at different speeds.

Lets say a car is capable of locking the front wheels up at 130mph.

Does it lock up the wheels significantly easier at 30mph? If so, why?

If we assume it can generate the same acceleration level from both speeds, so the same load transfer, what would cause the tyre to give up grip sooner at low speed than at high speed?

 

[sgb27]

The driver being too scared to hit the brake pedal hard and fast enough at 130mph :-)

 

[cjl]

The braking force from the wheels should be similar at both speeds. Yes, most cars will decelerate faster from 130mph than 30, but this is largely due to the added contribution of aerodynamic drag. If the car makes significant aerodynamic downforce, this could make a difference, but this will not be the case for most normal cars. If the wheels are not skidding though (as with nearly all modern cars, due to ABS), the amount of heat put into the brakes from 130mph will be enormous compared to the heat from 30mph, and a panic stop from 130mph could well overheat certain brake components in a car not designed for high performance.

 

[xxChrisxx]

You can brake much, much harder from high speed due to the increased inertia.

 

[cjl]

You can brake much, much harder from high speed due to the increased inertia.

What makes you say that? Unless the normal force is significantly different (as would be the case if the car is making a lot of downforce), the available friction at the wheels is pretty much the same at 130mph as at 30mph.

 

[russ_watters]

You can brake much, much harder from high speed due to the increased inertia.

I doubt the first half is true, but the second is definitely not or is a misuse of a word intended to be kinetic energy. Inertia is proportional to mass and is a fixed property of the car.

On a first order analysis, I see no reason why braking force should be different at different speeds. There are no basic friction force equations that include speed in them.

But what does matter is the braking energy, which may be what you are getting at. At twice the speed, braking takes 4x the energy and thus at high speed, the brakes get hotter faster, which has its own problems.

 

[xxChrisxx]

What makes you say that? Unless the normal force is significantly different (as would be the case if the car is making a lot of downforce), the available friction at the wheels is pretty much the same at 130mph as at 30mph.

Because I've done it.

If you hold a constant pedal load from a higher speed the wheels will lock as you slow down (or the abs will intervene).

To threshold brake you need to reduce pedal effort as you slow down. It takes a surprisingly large pedal force to lock up at high speed.

The ultimate aim is to have a constant linear deceleration, limited by the grip availaible.


Can't remember the technical explination for it, and it's too late to think it through.

You can switch out the words for any correct terminology you like. I suppose momentum is more correct. Its also got something to do with tyre slip.

 

[Kozy]

The force required on the pedal will vary as you stop from high speed because the temperature of the brakes increases, which changes the co-efficient of friction. Up until the point they fade, most brake pads will increase their friction co-efficient which increasing heat, so that might explain why you need to bleed off pedal force as you slow down.

 

[Highspeed]

Also consider the tire's CoF changes with temperature changes...from higher speeds or braking friction.

 

[cjl]

Because I've done it.

If you hold a constant pedal load from a higher speed the wheels will lock as you slow down (or the abs will intervene).

To threshold brake you need to reduce pedal effort as you slow down. It takes a surprisingly large pedal force to lock up at high speed.

Pedal effort is not necessarily a good indication of braking force available. A log of speed vs time elapsed, or of acceleration, along with knowledge of the aerodynamic properties of the vehicle would be much more reliable.

 

[xxChrisxx]

The problem is more likely is your use of an overly simplistic model.

What was it that Feynman said. If your guess disagrees with experiment. It's wrong.

I experienced this first hand, as part of my driver training. Threshold braking yields a constant deceleration, to do so requires decreasing pedal effort.

If you don't believe me, you can always try it for yourself. Under controlled and safe conditions only! Track day, used airfield, proving ground etc.


I'll try digging out Milliken and Milliken, there's bound to be a bit of this in there.

 

[Damo ET]

Chris is correct, it takes far greater braking pressure to lock wheels at higher speed! Even with the heating of the discs from an extended stop, a constant peddle pressure will lock the wheels at a low speed from a high speed stop (no ABS).

Try it.


Damo

 

[cjl]

Oh, I don't doubt that increased pedal pressure (and probably also brake pressure) is necessary at high speed. That doesn't mean that the available stopping force is greater though, nor does it mean that the tire grips any less at low speed (which is what I thought the original question was asking). Instead, it means that the same brake pressure does not provide the same stopping force at the wheels, so you need a higher brake pressure to generate the same force at high speed (even though the tire grip level is identical). As for why this might be? It could be related to the fact that friction coefficient tends to decrease with sliding velocity, so at higher speed, the friction coefficient between the brake pads and rotors is lower.

 

[xxChrisxx]

Tractive force Ft and braking force Fb are a function of slip ratio. As the slip ratio increases from zero [free rolling condition], the forces rise rapidly to a maximum which usually occurs in the range of 0.1 to 0.15 slip ratio, after which forces fall off. Up to the peak, the forces depend heavily on the elastic properties of the tread and carcass. After the peak, the forces depend on a variety of factors such as tread compsition, road texture, surface moisture, speed, tire temperature, etc.

SR = (wR/V) - 1

w = wheel angualr velocity
R effecive rolling radius
V road speed

Free rolling is when slip ratio = 0 (ie wR/V=1)
Sliding (locked wheel) when SR = -1 (wR/V=0)

Racecar Vehicle Dynamics Section 2.3 - page 37So brake torque is effectively controlling slip ratio. And a constant slip ratio will give constant deceleration.
The faster you are going, the larger the delta wheel speed to road speed required to achieve a given slip ratio.
There is more explanation (maths ***), as the above by itsself doesn't fully answer why larger braking forces are required to achieve the same slip ratio at high speed. But it's making my brain hurt thinking about it.

It's not as simple as the amount of brake torque applied though.EDIT: Or is it...
Looking at it the other way, the road is trying to accelerate the wheel (as the contact patch wants to be traveling at the same speed). A higher road speed would mean a higher 'correcting acceleration'. The brake force input is trying counteracting this.

Hmm I'm not sure.

 

[marellasunny]

One has to arrive at a relationship between slip ratio and braking torque.This is how we were taught in class:
Elaborating on Chris's post:
Slip ratio(in case of braking)=Vehicle speed(V)-Wheel speed(omega)/Vehicle speed(V)
$$ Slip ratio s=\frac{V-\omega}{V}$$

Lets say F_B is the braking force acting on the tyre contact patch.

Then,the 'disk brake' would need to apply a brake torque of magnitude=$F_B * r_{dynamic tyre radius}$

I have not yet arrived at the relation between brake torque F_B and the slip ratio.I go about this in 2 steps.
1.Arrive at a relationship between friction coefficient and slip ratio:

[Note:
Wheel starts slipping when slip-ratio s=1.]
2.Arrive at a relationship between braking force$F_B= \mu F_{normal}$and slip ratio.

You can see that by the time the slip-ratio reaches -100%, you are left with almost negligible traction force being transferred to the road.This means,the wheel has started slipping at -100%.

You say in your question that you want the acceleration to be the same in both cases.So,now make use of this equation to calculate the normal force at the rear tire.
$$F_{normal}=m.g \frac{l_v}{l}+m.a \frac{h}{l}$$
where
h-height of c.g above road
l_v -distance of the cg from front-end of car
The F_{normal} would be the same for both cases since you assume accelerations are the same.
[NOTE:I call $m.a \frac{h}{l}$ the "dynamic load transfer". As long as this is equal for both cases,wheel lock-up will occur at the same time.]

So,my answer to the question would be that the vehicle would lock-up at 30kph or 130kph depending on the "dynamic load transfer". If the "dynamic load transfer" causes the vehicle speed and wheel speed to be equal at say after pressing the pedal 80%($V= \omega$),then the wheels would lock-up-say at the same time.

The way to find-out which case would cause the car to loose grip sooner would be to look at the pedal effort.
In reality,the 130kph car should lock-up sooner because of aerodynamics resistance force.But,say you have the 30kph car load more(than the empty 130 kph car),the greater shift of load to the rear would cause the 30kph car to lock-up sooner.
Correct me if I am wrong.I am also learning. Cheers

 

[Kozy]

Well explained Marella.

Changing the loading would be at odds with the 'all else being equal' aspect of the question.

Very simply, disregarding aerodynamic effects and assuming both cars attaint the same rate of deceleration, I am thinking that there is no real reason why the wheels would lock up easier at 30mph than at 130mph.

 

[xxChrisxx]

Well explained Marella.

Changing the loading would be at odds with the 'all else being equal' aspect of the question.

Very simply, disregarding aerodynamic effects and assuming both cars attaint the same rate of deceleration, I am thinking that there is no real reason why the wheels would lock up easier at 30mph than at 130mph.

Except that they do...

We haven't found the answer yet.

 

[Ranger Mike]

it takes 996,424 (lb/ft) energy to slow down a 1760 pound formula car from 130 mph to 0.
It takes 53,064 (lb/ft) energy to slow down a 1760 pound formula car from 30 mph to 0.
So if you don’t want to wipe out and lock up the rotors it will take a lot longer to use up the 996,424 (lb/ft) energy in the pads/rotors.
I got the formula ifin anyone cares to see it

 

[russ_watters]

The energy is different because of the distances traveled (energy isn't force): those two numbers imply a nearly identical braking force (actually a slightly lower force at lower speed).

My guess would be that as you brake, the rotor heats up and expands, increasing the braking force unless you take care to lift your foot a little. But I don't know for sure - there are a lot of things that happen as you brake.

 

[xxChrisxx]

The energy is different because of the distances traveled (energy isn't force): those two numbers imply a nearly identical braking force (actually a slightly lower force at lower speed).

My guess would be that as you brake, the rotor heats up and expands, increasing the braking force unless you take care to lift your foot a little. But I don't know for sure - there are a lot of things that happen as you brake.

I don't think that's the answer. Its plausible for heavy braking from high speed, but the same phenomenon exists on the initial bite too. Which would be before any significant heat related expansion.

The same phenomenon happens with carbon brakes too I belive, which wouldn't expand in the same way. I have no practical experience of it, I've only ever used steel road brakes.



The net retarding force at the wheels must be the same, as threshold braking gives a linear deceleration.

Just why the hell can you push the pedal so much harder without locking? I'm still reading through the tyre dynamics books, but it's just confusing me.

 

[marellasunny]

it takes 996,424 (lb/ft) energy to slow down a 1760 pound formula car from 130 mph to 0.
It takes 53,064 (lb/ft) energy to slow down a 1760 pound formula car from 30 mph to 0.
So if you don’t want to wipe out and lock up the rotors it will take a lot longer to use up the 996,424 (lb/ft) energy in the pads/rotors.
I got the formula ifin anyone cares to see it

Formula please.

 

[cjl]

Just why the hell can you push the pedal so much harder without locking? I'm still reading through the tyre dynamics books, but it's just confusing me.

My guess would be that the friction coefficient between the brake pads and the brake rotor is lower at higher sliding velocity. A quick googling seems to indicate that reduction of kinetic friction at high velocity is a known (and common) phenomenon...

 

[marellasunny]

The energy is different because of the distances traveled (energy isn't force): those two numbers imply a nearly identical braking force (actually a slightly lower force at lower speed).

My guess would be that as you brake, the rotor heats up and expands, increasing the braking force unless you take care to lift your foot a little. But I don't know for sure - there are a lot of things that happen as you brake.

There is a relation between the energy(kinetic energy in our case) and force(braking force).
$$E_{Kinetic}=\int_{0}^{s}F_{braking}ds$$

where s-braking distance

I hope someone could suggest how we could inculcate 'brake slip' into the above equation and we might have a mathematical answer to the question.

To make things more simple,I could find the braking force by using the equation:
$$F_{braking}=m_{vehicle} a_{braking}$$

acceleration(/deceleration)=f(velocity)=g(slip)

If someone could be kind enough to derive the above,we might have an answer..I think..

 

[Ranger Mike]

formula you asked for

 

[sgb27]

If you lock the brakes, the brake system only needs to absorb the energy to stop the wheels rotating, the linear kinetic energy of the car itself (given in the formula above) is then absorbed by the tyre and the road. This is why the rubber melts onto the road if you lock up at 130mph.

 

[Ranger Mike]

The original question is -
1. forces at play determining wheel lock up at different speeds.
2. is it easier to lock up the brakes at 30 mph vs. 130 MPH

Typical passenger cars with hydraulic master cylinder and wheel cylinders, drum / disc brakes has a mechanical linkage at brake pedal to the Master Cylinder of 6:1 ratio. This means you push the pedal with 100 lbs. leg force and the master cylinder squeezes the caliper and brake pads at 600 lbs. force on the rotors.

Assumption- Once the brake system is locked up ( the rotor can not turn anymore) you now have all tires not rotating. Tires are sliding on one small contact patch per tire. You have effectively lost control.

Forces effecting vehicle braking
Aerodynamics –the faster you go the more effort it takes to overcome aero drag. This works for you when braking.
Tire grip- any slip angle advantage due to down force will go totally away once you lock the brakes.
Again let us look at the mechanics of the situation. You have managed to motivate a race car to 130 MPH having overcome aero drag et al..
If we apply the brakes and manage not to lock them we have to deal with the forward momentum being reduced by the brake system. In this case momentum energy is converted to heat caused by friction of the tires in contact with the pavement working thru the brake rotors being pinched by the brake calipers as actuated by the wheel cylinders via the pedal on the master cylinder. This HEAT is the big difference between a 30 mph stop and a 130 mph stop.

it takes 996,424 (lb/ft) energy to slow down a 1760 pound formula car from 130 mph to 0.
It takes 53,064 (lb/ft) energy to slow down a 1760 pound formula car from 30 mph to 0.

Now the brake system has to deal with 18 times the energy and it does not do it well. Momentum energy changed to heat energy can heat things up in the brake system to 1000 degrees F.
Heat reduces the ability of the brake system to “ lock up”. The coefficient of friction of the brake pads is severely reduced. The calipers squeeze the brake pads but the pads just can not slow the rotors.. They are cooked. Pads just can not cope. And..the brake fluid can boil. The higher quality racing brake fluids are very expensive but have a 600 degree boil point. Cheaper convectional brake fluid can turn to a gaseous state at the caliper/ brake pad interface so you got no pedal...very spongy. These factors contribute to what we in racing call Brake Fade.
Now evolution in racing has brought about super good brake fluid and ceramic hi tech brake pads and rotors and cooling ducts to counter this problem. But the fact remains you have to deal with 18 times the energy and its not that you need to hit the brakes “ harder” to brake from 130 MPH...it is that the brake system itself is not functioning at any where near the effectiveness as it is at 30 MPH for obvious reasons.

At 30 mph the brake pads and rotors are pretty cool. Brake fade is nil in that the system can easily absorb the energy without boiling the fluid or causing brake pad fade. So you can lock up the brake system easier at the lower speed. Or something like that and I'm out of beer..

 

[sgb27]

The original question is -
1. forces at play determining wheel lock up at different speeds.
2. is it easier to lock up the brakes at 30 mph vs. 130 MPH
...
Now the brake system has to deal with 18 times the energy and it does not do it well. Momentum energy changed to heat energy can heat things up in the brake system to 1000 degrees F.
Heat reduces the ability of the brake system to “ lock up”.

The brake system only needs to deal with 18x the energy if you *don't* lock up. Assuming you lock up quickly (before the car has slowed much) then the rest of the car's kinetic energy goes into the tyre/road and not the brake system, so the discs/pads don't heat up and the friction doesn't change.

A quick back-of-the-envelope calculation shows that energy needed to be dissipated to stop 4 wheels rotating from 130mph is about the same as needed to stop the entire car from 25mph. So locking the wheels at 130mph is not going to generate any more heat *in the brake system* than a normal stop from 25mph (or thereabouts). And as I said before, the rest of the KE from the car (the 18x factor you mentioned) goes into melting the tyres onto the road.

 

[Geegee]

Hello, I am new but I have some technical knowledge of vehicles that I can share that might assist you in finding the answer. The mechanical advantage of the brake pedal is applied to the brake booster, a vacuum is stored on the engine side of a diaphragm and atmospheric pressure is allowed into the cabin side of the diaphragm when you apply the pedal. You would then have 14.7 psia on one side and -5.3 psia on the other, depending on the size and number of diaphragms you would multiply the pressure by the square inches. Then the hydraulic system generally has a .5" piston inside the master cylinder which would force fluid against about a 2" piston in the caliper if it's just one piston, thus multiplying force once again, the drums wheel cylinder has a relatively small piston area about equal to the master cylinder. The smaller piston will travel farther than the larger piston. You will have a wedging affect taking place on a vehicle with duo-servo style shoes used on larger vehicles, and no wedging affect on non-servo drum systems used on smaller vehicles, which also affects braking force. Then you have the size of the friction surface. In a typical disc/drum system we also use a metering valve which builds pressure on the rear drum system first to allow them to engage because they have to travel further and to keep the vehicles rear end from trying to pass the front end. A proportioning valve is used on disc/drum or disc/disc, it allows the pressure to the front brakes to increase or overcome rear brake pressure as braking force increases past a given pressure. The brakes tend to work better once they are warm but start to outgas as the temperature gets too high. The fluid can also begin to boil. Then you have weight transfer and tire grip. I believe the explanation for why a brake rotor will lock up easier at 30 than 130 is an object in motion will tend to stay in motion, as the speed of the rotor increases it's energy increases, thus it takes more energy to stop it. Much the same way a flywheel hybrid system has more energy storage at 50,000 RPM than it does at 20,000 RPM. I tend to think of rotational mass the same as torque so I would like to say the rotor has more torque at 130.

 

[sgb27]

I believe the explanation for why a brake rotor will lock up easier at 30 than 130 is an object in motion will tend to stay in motion, as the speed of the rotor increases it's energy increases, thus it takes more energy to stop it.

I think this is the key, it's not that the *force* required to lock up the wheel is any different, just that because it is rotating faster it takes a longer *time* to lock up with the same force. So if you want to lock up the wheels at 130mph *in the same time* as you can lock up wheels at 30 mph, you will need to apply a larger force. This gives the impression of the wheels being "easier" to lock up at 30.

 

[Kozy]

The original question is -
1. forces at play determining wheel lock up at different speeds.
2. is it easier to lock up the brakes at 30 mph vs. 130 MPH

Typical passenger cars with hydraulic master cylinder and wheel cylinders, drum / disc brakes has a mechanical linkage at brake pedal to the Master Cylinder of 6:1 ratio. This means you push the pedal with 100 lbs. leg force and the master cylinder squeezes the caliper and brake pads at 600 lbs. force on the rotors.

Not quite right Mike. The pedal ratio might be 6:1, but you have boosters and hydralics in the mix too. A typical production car is more like 30:1 on pedal to brake force ratio.

Assumption- Once the brake system is locked up ( the rotor can not turn anymore) you now have all tires not rotating. Tires are sliding on one small contact patch per tire. You have effectively lost control.

Forces effecting vehicle braking
Aerodynamics –the faster you go the more effort it takes to overcome aero drag. This works for you when braking.
Tire grip- any slip angle advantage due to down force will go totally away once you lock the brakes.
Again let us look at the mechanics of the situation. You have managed to motivate a race car to 130 MPH having overcome aero drag et al..
If we apply the brakes and manage not to lock them we have to deal with the forward momentum being reduced by the brake system. In this case momentum energy is converted to heat caused by friction of the tires in contact with the pavement working thru the brake rotors being pinched by the brake calipers as actuated by the wheel cylinders via the pedal on the master cylinder. This HEAT is the big difference between a 30 mph stop and a 130 mph stop.

it takes 996,424 (lb/ft) energy to slow down a 1760 pound formula car from 130 mph to 0.
It takes 53,064 (lb/ft) energy to slow down a 1760 pound formula car from 30 mph to 0.

Now the brake system has to deal with 18 times the energy and it does not do it well. Momentum energy changed to heat energy can heat things up in the brake system to 1000 degrees F.
Heat reduces the ability of the brake system to “ lock up”. The coefficient of friction of the brake pads is severely reduced. The calipers squeeze the brake pads but the pads just can not slow the rotors.. They are cooked. Pads just can not cope. And..the brake fluid can boil. The higher quality racing brake fluids are very expensive but have a 600 degree boil point. Cheaper convectional brake fluid can turn to a gaseous state at the caliper/ brake pad interface so you got no pedal...very spongy. These factors contribute to what we in racing call Brake Fade.
Now evolution in racing has brought about super good brake fluid and ceramic hi tech brake pads and rotors and cooling ducts to counter this problem. But the fact remains you have to deal with 18 times the energy and its not that you need to hit the brakes “ harder” to brake from 130 MPH...it is that the brake system itself is not functioning at any where near the effectiveness as it is at 30 MPH for obvious reasons.

At 30 mph the brake pads and rotors are pretty cool. Brake fade is nil in that the system can easily absorb the energy without boiling the fluid or causing brake pad fade. So you can lock up the brake system easier at the lower speed. Or something like that and I'm out of beer..

A subtle difference in the wording, 'locking up at 130mph' vs 'from 130mph'. For heat soak to affect lockup, it would have had to absorb a significant amount of energy, and thus will have slowed the car down significantly. I agree that braking from 130mph, you are less likely to be locking up when you get down to 30mph due to heat soak.

At the moment you touch the brakes at 130mph though, there is no heat soak.

 

[Ranger Mike]

agreed and yes I did not go to production psi as power booster not used on our race cars..good points though!