Library / Automotive and vehicle dynamics / VEH-17
Automotive and vehicle dynamics
Brake disc temperature rise
Bulk temperature rise of a brake disc when a share χ of the vehicle kinetic energy lost in a stop goes into it, with no cooling during the stop.
- Ref
- VEH-17
- Rev
- v1
- Field
- Automotive and vehicle dynamics
- Result
- ΔTd [K]
The kinetic energy lost in the stop is ½m(U₀² − U₁²), with speeds converted from km/h to m/s. A fraction χ of it goes into the one disc considered; the rest goes to the other discs and to the air. The disc heats uniformly (a bulk, adiabatic estimate) so the rise is the energy it receives divided by its heat capacity md cd. It excludes the cooling during the stop, the temperature gradient through the disc, and the energy in the spinning wheels. The final speed must be below the initial speed.
Bulk temperature rise of a brake disc when a share χ of the vehicle kinetic energy lost in a stop goes into it, with no cooling during the stop. It is the heating level to compare with fade limits.
Calculator
- Kinetic energy lost in the stop578.7 kJ
- Energy entering this disc, χE86.81 kJ
- Heat capacity of the disc, md cd3.68 kJ/K
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- One stop from 100 km/h heats the disc by only 23.59 K, but the rise adds up: after repeated stops without cooling, ten of them would raise it by about 235.9 K.
- The rise grows with the speed squared: from 200 km/h the same stop gives 94.35 K, four times the rise. Mountain descents and track use are the cases where the disc temperature, and brake fade, become the limit.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| md Disc mass | 1 to 30 kg | 6.29 | 188.7 | 182.4 | +10% | −9.09 % |
| U0 Speed at the start of braking | 20 to 250 km/h | 0.9435 | 147.4 | 146.5 | +10% | +21 % |
| U1 Speed at the end of braking | 0 to 200 km/h | −70.77 | 23.59 | 94.35 | +10% | 0 % |
| χ Fraction of the lost energy entering this disc | 0.02 to 0.5 | 3.145 | 78.63 | 75.48 | +10% | +10 % |
| m Vehicle mass | 500 to 5,000 kg | 7.863 | 78.63 | 70.77 | +10% | +10 % |
| cd Specific heat capacity of the disc | 400 to 900 J/(kg·K) | 12.06 | 27.13 | 15.07 | +10% | −9.09 % |
Sweep of Speed at the start of braking
| Speed at the start of braking (km/h) | Disc temperature rise (K) |
|---|---|
| 20 | 0.9435 |
| 48.75 | 5.606 |
| 77.5 | 14.17 |
| 106.3 | 26.63 |
| 135 | 42.99 |
| 163.8 | 63.25 |
| 192.5 | 87.41 |
| 221.3 | 115.5 |
| 250 | 147.4 |
Sweep one variable
Variables and how each one contributes
Disc temperature rise
What the equation gives you. Everything else feeds it.
Vehicle mass
The vehicle mass, in kg. Directly proportional: heavier vehicles have more energy to dissipate. Raising it by 10 % (to 1,650) moves the result by +10 %, from 23.59 to 25.95 K.
Speed at the start of braking
The speed at the start of braking, in km/h. The energy grows with the square of the speed, so the heating rises sharply with speed. Raising it by 10 % (to 110) moves the result by +21 %, from 23.59 to 28.54 K.
Speed at the end of braking
The speed at the end of braking, in km/h. Zero for a stop. A higher final speed means less energy removed, so the disc gets cooler. Raising it by 20 km/h (to 20 km/h) moves the result by −4 %, from 23.59 to 22.64 K.
Fraction of the lost energy entering this disc
The fraction of the lost energy that enters this one disc: about 0.15 for each front disc of a car that brakes 60 % at the front. Directly proportional, so rear discs, which take less, heat less. A higher front bias means hotter front discs. Raising it by 10 % (to 0.165) moves the result by +10 %, from 23.59 to 25.95 K.
Disc mass
The mass of the disc, in kg. Inversely proportional: a heavier disc stores more heat per degree. Raising it by 10 % (to 8.8) moves the result by −9.09 %, from 23.59 to 21.44 K.
Specific heat capacity of the disc
The specific heat capacity of the disc material, in J/(kg·K): about 460 for cast iron and 800 or more for carbon–ceramic. Inversely proportional, but nearly fixed for a given material. Raising it by 10 % (to 506) moves the result by −9.09 %, from 23.59 to 21.44 K.
Worked example with the values above
- Kinetic energy lost: ½m(U₀² − U₁²) = 0.5 × 1,500 × (27.78² − 0²) = 578,700 J.
- Energy into this disc: χE = 0.15 × 578,700 = 86,810 J. Heat capacity: md cd = 8 × 460 = 3,680 J/K.
- ΔTd = 86,810 / 3,680 = 23.59 K.
Assumptions and limits
- The disc takes a fixed fraction χ of the energy and heats uniformly, with no cooling during the stop.
- All kinetic energy is lost in the brakes (no drag, rolling resistance, engine braking or regeneration), and the rotating inertia is neglected.
Common mistakes
- Using the bulk rise as the surface temperature. The friction surface is much hotter than the disc average during the stop.
- Entering the whole vehicle energy share as χ = 1. That puts all the heat in one disc.
Statistics and uncertainty
The weak point is the share χ of the energy entering the disc, which depends on the brake bias, the weight transfer and the cooling, and is known only to ±20 %. The mass and specific heat of the disc are known to a few per cent, though the specific heat rises with temperature, so the estimate is conservative at high temperature. Use the result as an order-of-magnitude guide, not a prediction of the disc surface temperature.
Where it is used
- Estimating brake fade risk for repeated stops, steep descents and track use.
- Sizing the mass and material of brake discs.
Maths you need
- Kinetic energy: ½ m U²
- Heat capacity: Q = m c ΔT