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Automotive and vehicle dynamics

Brake torque from clamp force

Braking torque produced by a disc brake: pad friction coefficient times the total clamp force on both pad faces times the effective friction radius.

Ref
VEH-16
Rev
v1
Field
Automotive and vehicle dynamics
Result
Tb [N·m]

μp is the friction coefficient between pad and disc, Fclamp the total normal force of the pads on the disc (the sum over both faces, so the hydraulic force on one piston when it acts on both sides), and Reff the effective radius at which the friction force acts, close to the mid-radius of the pad. The friction force on the two faces is μp Fclamp, and multiplying by the radius gives the torque on the disc, in N·m. The result is the torque on one disc, which the wheel converts to a force at the tyre.

Braking torque produced by a disc brake: pad friction coefficient times the total clamp force on both pad faces times the effective friction radius.

Calculator

Solve for
N·m
0.20.7
kN
180
mm
50250
  • Friction force on the disc, μp Fclamp8 kN
  • Torque per kN of clamp force48 N·m/kN
  • Heat dissipated per disc revolution, 2π Tb6.032 kJ

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.

Brake torque (N·m) current value 960 N·m
FclampTotal clamp force (sum of both pad faces)Key driver
483,840
ReffEffective pad friction radius
4002,000
μpPad friction coefficient
4801,680
  • With μp = 0.4, 20 kN of clamp force and a 120 mm effective radius, one disc gives 960 N·m. At the tyre with Re = 0.3 m this is a braking force of 3,200 N from that wheel.
  • Pad fade is the reason for large margins. If hot pads lose 25 % of their friction coefficient, the torque falls by exactly 25 %, with the same pedal force.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (N·m)Local stepChange in result
Fclamp Total clamp force (sum of both pad faces)1 to 80 kN483,8403,792+10%+10 %
Reff Effective pad friction radius50 to 250 mm4002,0001,600+10%+10 %
μp Pad friction coefficient0.2 to 0.7 4801,6801,200+10%+10 %

Sweep of Total clamp force (sum of both pad faces)

Total clamp force (sum of both pad faces) (kN)Brake torque (N·m)
148
10.88522
20.75996
30.631,470
40.51,944
50.382,418
60.252,892
70.133,366
803,840

Sweep one variable

Brake torque (N·m)Hover, touch, or use the arrow keys on the chart
01000200030004000 10305070 Total clamp force (sum of both pad faces) (kN) now 960

Variables and how each one contributes

Tb

Brake torque

Result · N·m

What the equation gives you. Everything else feeds it.

μp

Pad friction coefficient

· typical 0.2 to 0.7

The friction coefficient between the pad and the disc: about 0.35 to 0.45 for road pads, 0.5 or more for race pads when hot. Directly proportional, which is why pad fade (a fall in μp when hot) directly reduces braking torque. Raising it by 10 % (to 0.44) moves the result by +10 %, from 960 to 1,056 N·m.

Fclamp

Total clamp force (sum of both pad faces)

kN · typical 1 to 80

The total clamp force of the pads on the disc (the sum of the normal forces on both faces), in kN. Directly proportional to the brake line pressure and the piston area, so doubling the pressure doubles the torque. Raising it by 10 % (to 22) moves the result by +10 %, from 960 to 1,056 N·m.

Reff

Effective pad friction radius

mm · typical 50 to 250

The effective friction radius of the pad on the disc, in mm: roughly the mid-radius of the pad. Directly proportional: a larger disc diameter gives more torque for the same force. Raising it by 10 % (to 132) moves the result by +10 %, from 960 to 1,056 N·m.

Worked example with the values above

  1. Friction force: μp Fclamp = 0.4 × 20 kN = 8 kN.
  2. Effective radius: Reff = 120 mm = 0.12 m.
  3. Tb = 8,000 N × 0.12 m = 960 N·m.

Assumptions and limits

  • A constant friction coefficient and a clamp force spread evenly over the pad faces, with the force acting at the effective radius.
  • Fclamp is the sum on both pad faces. For a floating calliper it equals twice the piston force.

Common mistakes

  • Using the force on one pad face alone for Fclamp, which gives half the torque if the disc has two pad faces.
  • Entering the outer disc radius in place of the effective pad radius, which overstates the torque.

Statistics and uncertainty

The torque is a product of three terms. The clamp force is known from the pressure and piston area to about 3 to 5 %, the effective radius to better than 2 %, but the pad friction coefficient varies by ±10 to 20 % with temperature, speed, wear and moisture, so it dominates. A quoted brake torque is only valid for the stated pad temperature.

Where it is used

  • Sizing callipers, pads and discs to a braking requirement.
  • Converting hydraulic pressure and piston area into braking torque at the wheel.

Maths you need

  • Friction: F = μ N
  • Torque: T = F r

Related formulas

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