Library / Automotive and vehicle dynamics / VEH-13
Automotive and vehicle dynamics
Front axle load during braking
Normal load on the front axle while braking: the static front load plus the weight transferred forward by the deceleration acting at the height of the centre of gravity.
- Ref
- VEH-13
- Rev
- v1
- Field
- Automotive and vehicle dynamics
- Result
- Fz,f [kN]
m is the vehicle mass, L the wheelbase, b the distance from the centre of gravity to the rear axle, h the height of the centre of gravity and ab the deceleration in g. Braking shifts a load m ab h / L from the rear axle to the front axle, on top of the static front load m g b / L. The axle loads are totals for both wheels, on a level road with no aerodynamic forces and no pitching acceleration. If the rear load reaches zero, the rear wheels have lifted and the equation no longer applies.
Normal load on the front axle while braking: the static front load plus the weight transferred forward by the deceleration acting at the height of the centre of gravity. Level road, no aerodynamic forces.
Calculator
- Static front axle load, m g b / L7.63 kN
- Load transferred forwards, m ab h / L2.398 kN
- Rear axle load4.687 kN
- Front axle share of the weight68.15 %
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.
- At rest the front axle carries 7.63 kN; braking at 0.8 g moves 2.398 kN forward, taking the front to 10.03 kN (68.1 % of the weight) and unloading the rear to 4.687 kN.
- The rear axle would lift if ab h reached the distance from the centre of gravity to the front axle: at the default geometry, at 2.36 g. Few road vehicles reach it, but short, tall vehicles can.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kN) | Local step | Change in result |
|---|---|---|---|---|---|---|
| m Vehicle mass | 500 to 5,000 kg | 3.343 | 33.43 | 30.08 | +10% | +10 % |
| b Distance from the centre of gravity to the rear axle | 0.9 to 1.8 m | 7.303 | 12.21 | 4.905 | +10% | +7.61 % |
| L Wheelbase | 2.2 to 3.5 m | 7.736 | 12.31 | 4.571 | +10% | −9.09 % |
| ab Braking deceleration | 0.2 to 1.2 g | 8.229 | 11.23 | 2.998 | +10% | +2.39 % |
| h Centre of gravity height | 0.4 to 0.9 m | 9.374 | 11.55 | 2.18 | +10% | +2.39 % |
Sweep of Braking deceleration
| Braking deceleration (g) | Front axle load (kN) |
|---|---|
| 0.2 | 8.229 |
| 0.325 | 8.604 |
| 0.45 | 8.979 |
| 0.575 | 9.354 |
| 0.7 | 9.728 |
| 0.825 | 10.1 |
| 0.95 | 10.48 |
| 1.075 | 10.85 |
| 1.2 | 11.23 |
Sweep one variable
Variables and how each one contributes
Front axle load
What the equation gives you. Everything else feeds it.
Vehicle mass
The vehicle mass, in kg. Directly proportional: both the static load and the transfer scale with the mass. Raising it by 10 % (to 1,650) moves the result by +10 %, from 10.03 to 11.03 kN.
Wheelbase
The wheelbase, in m (the distance between the front and rear axles). A longer wheelbase reduces both the static front load (for a fixed b) and the weight transfer, which is inversely proportional to it. Raising it by 10 % (to 2.97) moves the result by −9.09 %, from 10.03 to 9.116 kN.
Distance from the centre of gravity to the rear axle
The distance from the centre of gravity to the rear axle, in m. A larger b means more weight on the front axle. Raises the static front load in proportion, so a front-heavy vehicle has more load on the front axle even before braking. Raising it by 10 % (to 1.54) moves the result by +7.61 %, from 10.03 to 10.79 kN.
Centre of gravity height
The height of the centre of gravity above the ground, in m: about 0.5 m for a car and 1 m or more for a truck. Raises the weight transferred in direct proportion, so tall vehicles pitch forward more under braking. Raising it by 10 % (to 0.605) moves the result by +2.39 %, from 10.03 to 10.27 kN.
Braking deceleration
The braking deceleration, as a multiple of g. Raises the weight transferred in direct proportion, which is why the front brakes must be larger than the rear brakes. Raising it by 10 % (to 0.88) moves the result by +2.39 %, from 10.03 to 10.27 kN.
Worked example with the values above
- Static front load: m g b / L = 1,500 × 9.81 × 1.4 / 2.7 = 7,630 N.
- Load transfer: m ab h / L = 1,500 × (0.8 × 9.81) × 0.55 / 2.7 = 2,398 N.
- Fz,f = 7,630 + 2,398 = 10,030 N = 10.03 kN. The rear axle carries 4.687 kN.
Assumptions and limits
- A level road, a rigid vehicle and constant deceleration, with no aerodynamic forces and no pitching acceleration.
- The centre of gravity is on the centre line, and the loads are totals for both wheels on an axle.
Common mistakes
- Using the distance to the front axle in place of the distance b to the rear axle.
- Applying the result when the rear wheels have lifted: if the rear load comes out negative, the equation no longer holds.
Statistics and uncertainty
Axle loads follow from the mass and geometry, which are known to 1 to 2 %, but the height of the centre of gravity is hard to measure (typical error 2 to 5 cm, or 5 to 10 %). Since the weight transfer is directly proportional to h, a 10 % error in h gives a 10 % error in the transferred load and about 2 % in the front axle load at the defaults.
Where it is used
- Sizing front and rear brakes and choosing the brake bias.
- Checking front tyre, suspension and wheel-bearing loads in heavy braking.
Maths you need
- Moments about an axle: balance of moments in a statics problem
- Newton’s second law: inertial force m a at the centre of gravity