Turning motion into heat

A stop from 100 km/h puts about 580 kilojoules into your brakes. From 50 it is 145. Same car, half the speed, a quarter of the heat — and that square is the lesson.

Reading time: 8 min read, plus the simulation
Tier: Free, and stays free
Physics review: Awaiting physics review

Rests on

Kinetic energy ½mv²
Conservation of energy
Specific heat capacity
Heat transfer and thermal limits

This lesson assumes

Not required. Read it here, or go down into it and come back.

Where the energy goes — A vehicle never makes energy and never destroys it. It moves energy between forms, and everything it cannot use becomes heat. That is the whole subject.

Your brakes do not stop your car. Your tyres stop your car. The brakes' entire job is to turn motion into heat fast enough that the tyres get a chance — and then to survive the heat they just made.

Hold onto that division of labour, because almost every braking problem an owner ever meets is a failure of the second half, not the first.

How much heat, exactly

A moving car is holding , and there is only one place for it to go when you stop. Not "some of it". All of it. The car arrives at the lights with zero kinetic energy, so every joule it had at the start of the stop has been converted, and friction converts in exactly one direction.

A stop from 100 km/h, and from 50

E = ½mv²

Car mass m = 1500 kg

At 100 km/h: v = 100 ÷ 3.6 = 27.78 m/s, so v² = 771.6 m²/s²

E = 0.5 × 1500 × 771.6 = 578,700 J ≈ 580 kJ

At 50 km/h: v = 13.89 m/s, so v² = 192.9 m²/s²

E = 0.5 × 1500 × 192.9 = 144,700 J ≈ 145 kJ

580 ÷ 145 = 4.00. Half the speed, a quarter of the energy. Exactly, not approximately — because the only thing that changed was v, and v is squared.

Five hundred and eighty kilojoules is an abstraction until you put it next to something domestic. It is enough to take 1.7 litres of water from tap temperature to boiling: 580,000 ÷ (4186 × 80) = 1.73 kg. One stop. One set of lights.

Predict, then run

You stop from 100 km/h instead of 50. Same car, same road, same firm stop. How much more heat do the brakes have to get rid of?

About the same — you are still going from moving to stopped

Twice as much

Three times as much

Four times as much

You learn more from being wrong on purpose than right by accident.

Where 580 kilojoules actually ends up

Into the discs, mostly, and mostly into the front ones. Braking pitches the car forward onto its nose, which loads the front tyres and unloads the rears, so the front brakes are built to take the larger share — typically 60 to 80 per cent of the work in a hard stop.

Temperature rise of a pair of front discs

Assume 70% of the stop goes to the front axle: 0.70 × 580 kJ = 406 kJ

Assume two grey cast-iron discs at 7 kg each: m = 14 kg

Specific heat of grey cast iron: c ≈ 460 J/(kg·K)

ΔT = E ÷ (m × c) = 406,000 ÷ (14 × 460) = 406,000 ÷ 6,440 = 63 K

About sixty-three degrees, from one stop, if the heat spread itself evenly through the whole disc. It does not. In the first second it is concentrated in a few millimetres of the swept face, and the surface runs far hotter than the bulk figure suggests, while the middle of the disc is still cold. That temperature difference across a few millimetres of iron is what makes discs crack, and it is why a disc that has been heat-cycled hard has a different life to one that has not.

Your brakes are the most powerful thing on your car

Energy is only half the story. The other half is how fast you have to move it, and that is power.

Braking power in that same stop

Take a firm stop at 0.8 g: a = 0.8 × 9.81 = 7.85 m/s²

Time to stop from 27.78 m/s: t = v ÷ a = 27.78 ÷ 7.85 = 3.54 s

Average power: P = E ÷ t = 578,700 ÷ 3.54 = 164 kW

Peak power, at the instant you first hit the pedal, where the speed is still highest:

Force F = ma = 1500 × 7.85 = 11,775 N

P = Fv = 11,775 × 27.78 = 327,000 W ≈ 327 kW

Three hundred and twenty-seven kilowatts. That is roughly 440 horsepower of heat production, from a family car whose engine makes perhaps a third of that. The brakes on an ordinary car are, by a wide margin, its highest-powered component — they only ever run flat out for three or four seconds at a time.

Which is the entire design problem. Everything about a brake is a compromise between absorbing an enormous burst and then getting rid of it before the next one.

Fade is two different failures wearing one name

When a driver says the brakes faded, they could mean either of two things, and the difference tells you what actually went wrong.

Pad fade. The friction material has a temperature range it was designed for. Past it, the resins and binders holding the material together start to break down and gas off, and the coefficient of friction falls away. The symptom: the pedal still feels firm, and the car will not slow. You are pushing just as hard and getting less back.

Fluid fade, or vapour lock. Heat conducts out of the disc, through the pad, into the caliper piston, and reaches the brake fluid sitting behind it. Boil that fluid and you have vapour in a system designed around a liquid — and liquid does not compress, while vapour does. The symptom: the pedal goes long and soft. Your foot travels further and the pads move less.

Why brake fluid is changed on a calendar, not an odometer

Glycol-ether brake fluid is — it pulls water out of the air, through the reservoir cap and through the walls of the flexible hoses, whether the car is driven or parked.

The specification numbers show what that costs. DOT 4 fluid must boil no lower than 230 °C when dry and no lower than 155 °C when wet — and "wet" in the standard means a fluid humidified to a few per cent water content, which a real system reaches in a couple of years of ordinary life.

Seventy-five degrees of margin, given away to time rather than distance. That is why the interval is two years and not forty thousand kilometres.

The descent, which is where this really bites

A single stop is a burst. A long descent is a tap left running, and the arithmetic is unkind.

Coming down 500 metres of hill

Gravitational potential energy: E = mgh

E = 1500 × 9.81 × 500 = 7,357,500 J ≈ 7.36 MJ

Against one stop from 100 km/h at 580 kJ:

7,360,000 ÷ 578,700 = 12.7

Nearly thirteen full-speed stops, delivered back to back with no gap in which to cool. Thirteen stops' worth of heat is roughly 800 K of temperature rise on those two front discs if none of it escapes — and while plenty does escape to the air, the balance runs one way.

Note where the failure actually happens, though. The fluid does not need the disc to reach 800 °C. It needs the caliper to reach 155 °C, and the caliper is bolted to the thing making the heat.

This is the whole reason for the advice to select a lower gear on a long descent, and it is not folklore. In a lower gear the engine is being driven by the wheels and pumping air against a closed throttle, so it takes a share of that 7.36 MJ and turns it into heat in the cooling system and the exhaust instead — a system built to shed heat continuously, which the brakes are not.

Two levers, and only one of them is linear

Look at ½mv² as a design equation and it tells you what to do about brakes.

Mass, m, is linear. Add 50% to the mass and you add 50% to the energy. A heavier car needs bigger brakes in direct proportion.
Speed, v, is squared. Add 50% to the speed and you add 125% to the energy.

That is why a 2200 kg electric SUV needs more brake than a 1500 kg hatchback — 0.5 × 2200 × 771.6 = 849 kJ from 100 km/h, about 47% more — and why the same hatchback needs more brake on a country road than it does in the suburbs, by a much larger factor.

What the electric version changes, and what it does not

runs the traction motor backwards as a generator, so some of that kinetic energy returns to the pack as electricity instead of leaving as heat. It is the single largest reason an electric car is more efficient in traffic than on the highway — the exact opposite of a petrol car — because braking stops being pure waste.

It is a partial counterpart, not a replacement, and the arithmetic above shows why. That stop demanded 327 kW at its peak. An electric car that can push 100 kW back into its pack under braking — a generous figure for a mainstream one — covers under a third of the peak. The rest has to come from friction. Regen also acts only through the driven axle, fades out as the car slows toward walking pace, and backs off when the pack is cold or nearly full.

So electric cars keep their discs, and two honest consequences follow:

They are heavier, so the total energy in a stop is larger.
They use the friction brakes less, so discs corrode rather than wear, and a seized caliper slide pin is now a more likely fault than a worn pad.

The one everyone gets wrong

If your pedal pulses under braking, you will be told the rotors are warped. Almost none of them are. What you have is nearly always — a disc that is a few hundredths of a millimetre thicker in some places than others, so the pads squeeze harder once per revolution. The cause is usually uneven pad transfer film or a run-out that let one part of the disc wear differently, not a bent casting.

The distinction is worth money, because the fix is different, and resurfacing a disc without measuring the thickness variation first is a coin flip. If your pedal has gone long or your car pulses under brakes and you are in Melbourne's west, we will measure it before we quote you. If it is within spec, we will tell you that.

What you now know

Where the energy went, how much of it there was, why the number goes up with the square, what heat does to the friction material and to the fluid, and why the answer on a long descent is a lower gear rather than a firmer foot.

That is most of what a brake fault ever is.