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Automotive and vehicle dynamics

Rear axle load during acceleration

Normal load on the rear axle while accelerating: the static rear load plus the weight transferred rearward.

Ref
VEH-14
Rev
v1
Field
Automotive and vehicle dynamics
Result
Fz,r [kN]

m is the vehicle mass, L the wheelbase, a the distance from the centre of gravity to the front axle, h the height of the centre of gravity and ax the forward acceleration in g. Accelerating shifts a load m ax h / L from the front axle to the rear axle, on top of the static rear load m g a / L. The loads are totals for the axle, on a level road with no aerodynamic forces. If the front load reaches zero the front wheels have lifted and the equation no longer applies.

Normal load on the rear axle while accelerating: the static rear load plus the weight transferred rearward. Level road, no aerodynamic forces.

Calculator

Solve for
kN
kg
5005,000
m
2.23.5
m
0.91.8
m
0.40.9
g
0.050.8
  • Static rear axle load, m g a / L7.085 kN
  • Load transferred rearwards, m ax h / L1.199 kN
  • Front axle load6.431 kN
  • Rear axle share of the weight56.3 %

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.

Rear axle load (kN) current value 8.284 kN
mVehicle massKey driver
2.7627.6
aDistance from the centre of gravity to the front axle
6.111
LWheelbase
6.3910.2
axForward acceleration
7.239.48
hCentre of gravity height
7.969.05
  • At rest the rear axle carries 7.085 kN; accelerating at 0.4 g moves 1.199 kN rearwards, taking it to 8.284 kN (56.3 % of the weight).
  • This is why rear-wheel drive launches harder than front-wheel drive at high acceleration: the load on the driven wheels increases. A front-wheel-drive car loses load on its driven axle in the same acceleration, which limits its launch.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kN)Local stepChange in result
m Vehicle mass500 to 5,000 kg2.76127.6124.85+10%+10 %
a Distance from the centre of gravity to the front axle0.9 to 1.8 m6.10411.014.905+10%+8.55 %
L Wheelbase2.2 to 3.5 m6.39110.173.776+10%−9.09 %
ax Forward acceleration0.05 to 0.8 g7.2359.4832.248+10%+1.45 %
h Centre of gravity height0.4 to 0.9 m7.9579.0471.09+10%+1.45 %

Sweep of Forward acceleration

Forward acceleration (g)Rear axle load (kN)
0.057.235
0.14377.516
0.23757.797
0.33128.078
0.4258.359
0.51888.64
0.61258.921
0.70639.202
0.89.483

Sweep one variable

Rear axle load (kN)Hover, touch, or use the arrow keys on the chart
7.588.599.5 0.10.30.50.7 Forward acceleration (g) now 8.284

Variables and how each one contributes

Fz,r

Rear axle load

Result · kN

What the equation gives you. Everything else feeds it.

m

Vehicle mass

kg · typical 500 to 5,000

The vehicle mass, in kg. Directly proportional: both the static load and the transfer scale with mass. Raising it by 10 % (to 1,650) moves the result by +10 %, from 8.284 to 9.112 kN.

L

Wheelbase

m · typical 2.2 to 3.5

The wheelbase, in m. A longer wheelbase lowers the static rear load (for a fixed a) and reduces the transfer. Raising it by 10 % (to 2.97) moves the result by −9.09 %, from 8.284 to 7.531 kN.

a

Distance from the centre of gravity to the front axle

m · typical 0.9 to 1.8

The distance from the centre of gravity to the front axle, in m. A larger a puts more weight on the rear axle. Raises the static rear load in proportion, so a rear-heavy vehicle starts with more rear load. Raising it by 10 % (to 1.43) moves the result by +8.55 %, from 8.284 to 8.992 kN.

h

Centre of gravity height

m · typical 0.4 to 0.9

The height of the centre of gravity above the ground, in m. Raises the weight transferred in direct proportion, so taller vehicles squat more. Raising it by 10 % (to 0.605) moves the result by +1.45 %, from 8.284 to 8.404 kN.

ax

Forward acceleration

g · typical 0.05 to 0.8

The forward acceleration, as a multiple of g. Raises the weight transferred in direct proportion; a harder launch loads the rear tyres more. Raising it by 10 % (to 0.44) moves the result by +1.45 %, from 8.284 to 8.404 kN.

Worked example with the values above

  1. Static rear load: m g a / L = 1,500 × 9.81 × 1.3 / 2.7 = 7,085 N.
  2. Load transfer: m ax h / L = 1,500 × (0.4 × 9.81) × 0.55 / 2.7 = 1,199 N.
  3. Fz,r = 7,085 + 1,199 = 8,284 N = 8.284 kN. The front axle carries 6.431 kN.

Assumptions and limits

  • A level road, a rigid vehicle and constant acceleration, 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 b (to the rear axle) in place of the distance a to the front axle.
  • Using the result when the front wheels have lifted: if the front load is negative, the equation no longer holds.

Statistics and uncertainty

As in braking, the weakest input is the height of the centre of gravity, which is known only to about 5 to 10 %. The weight transfer is directly proportional to it, so a 10 % error in h gives a 10 % error in the transferred load and a smaller error in the rear axle load. Suspension squat, tyre compliance and aerodynamic lift also move the real loads by a few per cent.

Where it is used

  • Traction calculations for rear-wheel-drive vehicles and launch analysis.
  • Choosing the weight distribution and driven axle for a performance or off-road vehicle.

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

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