# Braking at high speed vs low speed.

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?

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.

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
Mentor
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.

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.

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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.

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.

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.

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.

Milliken and Milliken said:
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

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 37

So 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 travelling 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.

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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
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

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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.

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
Gold Member
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
Mentor
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.

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 thats 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, ive 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.

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

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...

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