Engineer's Notebook §2.1

Weight transfer under braking

When a car brakes, the inertia of its mass acts at the centre of gravity and pitches load off the rear axle onto the front. The load transferred is:

ΔW =m·a·hL
ΔW
load transferred
N
m
vehicle mass
kg
a
deceleration
m/s²
h
centre of gravity height
m
L
wheelbase
m
g
gravitational acceleration
m/s²
f
static front share of weight
—
b
CG distance ahead of the rear axle
m
t
track width
m
ay
lateral acceleration
m/s²
Side view of a car facing right, with its centre of gravity marked. Arrows under each wheel show the axle loads while braking. ma rear 4.22 kN front 10.5 kN DIRECTION OF TRAVEL →
Fig. 2.1-1Axle-load arrows to scale with the car's weight; CG height to scale with the wheelbase.
Wfront, static8.09 kN
Wrear, braking4.22 kN
Wfront, braking10.5 kN
ΔW = mah/L2.4 kN

Worked example

Uses the values in Fig. 2.1-1. Change them and the working updates.

Given
  • mvehicle mass1,500 kg
  • adeceleration0.8 g
  • hCG height550 mm
  • Lwheelbase2,700 mm
  • fstatic front share55%
Find

The load transferred to the front axle ΔW, and the axle loads while braking.

Solution
a = 0.8 × 9.81 m/s² = 7.85 m/s² ΔW = mah/L = 1,500 kg × 7.85 m/s² × 0.55 m / 2.7 m = 2.4 kN Wfront = fmg + ΔW = 0.55 × 14.7 kN + 2.4 kN = 10.5 kN Wrear = (1 − f)mg − ΔW = 0.45 × 14.7 kN − 2.4 kN = 4.22 kN
Answer
ΔW = 2.4 kN

The front axle now carries 71.3% of the weight, up from 55%.

Related cases

  • Accelerating

    ΔW = mahLonto the rear axle

  • Cornering, across the track

    ΔW = mayht

  • Static front axle load

    Wfront = mgbL

Assumptions
  • The car is a rigid body. Suspension pitch, which moves the centre of gravity slightly, is ignored.
  • The road is level and aerodynamic forces are ignored.
  • h is the centre of gravity's height above the ground.
  • Total weight is unchanged: load moves between the axles. If the rear load reaches zero, the rear wheels lift.

Derivation
  1. At rest, moments about the rear tyre contact patch give WfrontL = mgb.
  2. Braking adds an inertia force ma, acting forwards at the centre of gravity, height h. The braking forces act at ground level, so they have no moment about the contact patch.
  3. So WfrontL = mgb + mah, and the front axle gains ΔW = mah/L.
  4. Vertical balance: the rear axle loses the same amount.

Transfer grows with CG height and shrinks with wheelbase. That is why long, low cars pitch less under braking.