Brake heat budget

Enter the vehicle mass and its speeds before and after one braking event. Choose the percentage of the removed kinetic energy that reaches the discs being considered, and enter the combined mass of exactly those discs.

With speeds converted from km/h to m/s, the removed translational kinetic energy is ΔE = ½m(v₀² − v₁²). The heat in the selected discs is Q = (f/100)ΔE. Their idealized mean temperature rise is ΔT = Q/(M c), where M is their total mass and c their constant specific heat in J/(kg·K). A rise of 1 K equals a temperature difference of 1 °C.

For two discs of 4 kg each, enter 8 kg and the energy share going to that pair. Do not divide again by the number of discs. The share is your assumption; the tool does not estimate heat partition, regeneration or other losses. The default 100% assigns all removed translational kinetic energy to the chosen discs. The default specific heat of 460 J/(kg·K) is an editable example, not a material specification.

Example: 1200 kg, 108 → 72 km/h, a 60% share, 12 kg of discs and 500 J/(kg·K) give ΔE = 300 kJ, Q = 180 kJ and ΔT = 30 K. Equal speeds or a zero share give a valid zero result.

The model assumes uniform heating and constant specific heat. It omits cooling, rotational energy and changes in elevation. It gives neither peak surface temperature nor an absolute final temperature, and cannot establish braking safety or suitability. Input ranges are calculation limits, not approved operating ranges: vehicle mass 1–100000 kg; speeds 0–500 km/h with end ≤ start; share 0–100%; total disc mass 0.01–1000 kg; specific heat 1–5000 J/(kg·K).

Sources: NASA Glenn: conservation of energy; MIT: specific heats; OpenStax: heat and specific heat.