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Library / Automotive and vehicle dynamics / VEH-28

Automotive and vehicle dynamics

Lateral load transfer

The load shifted from the inside to the outside wheels in a steady turn.

Ref
VEH-28
Rev
v1
Field
Automotive and vehicle dynamics
Result
ΔFz [kN]

m is the vehicle mass, ay the lateral acceleration in g, h the height of the centre of gravity and t the track width (assumed equal at the front and rear). The lateral inertial force m ay acts at height h, producing an overturning moment m ay h that is resisted by a vertical force couple ΔFz t. ΔFz is the load added to the outside wheels and removed from the inside wheels, so the outside minus inside difference is 2ΔFz. Springs and anti-roll bars decide how it is split front to rear, not the total.

The load shifted from the inside to the outside wheels in a steady turn. It depends on the lateral acceleration, the height of the centre of gravity and the track width, and not on springs or roll stiffness.

Calculator

Solve for
kN
kg
5005,000
g
0.11.5
m
0.31.2
m
1.22.2
  • Total load on the outside wheels, m g / 2 + ΔFz11.4 kN
  • Total load on the inside wheels, m g / 2 − ΔFz3.311 kN
  • Share of the static side load transferred, 2ΔFz / m g55 %

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.

Load transferred to the outside (kN) current value 4.047 kN
mVehicle massKey driver
1.3513.5
ayLateral acceleration
0.5067.59
hCentre of gravity height
2.218.83
tTrack width
2.945.4
  • Cornering at 0.8 g moves 4.047 kN to the outside, so the outside wheels carry 11.4 kN against 3.311 kN on the inside, almost three and a half times as much.
  • The inside wheels lift when ΔFz reaches half the weight, at ay = t / (2h) = 1.45 g for this vehicle. A truck with h = 1.2 m and t = 1.8 m has a limit of only 0.75 g.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kN)Local stepChange in result
m Vehicle mass500 to 5,000 kg1.34913.4912.14+10%+10 %
ay Lateral acceleration0.1 to 1.5 g0.50587.5877.082+10%+10 %
h Centre of gravity height0.3 to 1.2 m2.2078.8296.622+10%+10 %
t Track width1.2 to 2.2 m2.9435.3962.453+10%−9.09 %

Sweep of Lateral acceleration

Lateral acceleration (g)Load transferred to the outside (kN)
0.10.5058
0.2751.391
0.452.276
0.6253.161
0.84.047
0.9754.932
1.155.817
1.3256.702
1.57.587

Sweep one variable

Load transferred to the outside (kN)Hover, touch, or use the arrow keys on the chart
02468 0.20.40.60.811.21.4 Lateral acceleration (g) now 4.047

Variables and how each one contributes

ΔFz

Load transferred to the outside

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. Raising it by 10 % (to 1,650) moves the result by +10 %, from 4.047 to 4.451 kN.

ay

Lateral acceleration

g · typical 0.1 to 1.5

The lateral acceleration, as a multiple of g. A road car in a hard corner reaches 0.8 to 1 g. Directly proportional: doubling the lateral acceleration doubles the transfer. Raising it by 10 % (to 0.88) moves the result by +10 %, from 4.047 to 4.451 kN.

h

Centre of gravity height

m · typical 0.3 to 1.2

The height of the centre of gravity above the ground, in m. Directly proportional: it is the main reason that tall vehicles roll over more easily. Raising it by 10 % (to 0.605) moves the result by +10 %, from 4.047 to 4.451 kN.

t

Track width

m · typical 1.2 to 2.2

The track width, in m: the lateral distance between the left and right wheel centres. Inversely proportional: a wide track resists the overturning moment and reduces the transfer. Raising it by 10 % (to 1.76) moves the result by −9.09 %, from 4.047 to 3.679 kN.

Worked example with the values above

  1. Lateral force: m ay = 1,500 × (0.8 × 9.81) = 11,770 N, acting at h = 0.55 m.
  2. Transfer: ΔFz = m ay h / t = 11,770 × 0.55 / 1.6 = 4,047 N.
  3. ΔFz = 4.047 kN. Outside wheels carry 11.4 kN and inside wheels 3.311 kN.

Assumptions and limits

  • Quasi-static cornering on a flat road with equal front and rear track widths, a centred centre of gravity and a rigid vehicle.
  • Suspension roll and the movement of the centre of gravity during roll are ignored.

Common mistakes

  • Treating ΔFz as the whole outside–inside difference. The difference is 2ΔFz.
  • Adding the lateral acceleration in m/s² into an input that expects g.

Statistics and uncertainty

The load transfer is directly proportional to the height of the centre of gravity, which is typically known to 5 to 10 %, and to the lateral acceleration, which is measured to a few per cent. A combined error of 10 % is common. The total transfer is independent of the suspension, but how it divides between front and rear depends on the roll stiffness, which is known to about ±15 %, and affects the handling balance.

Where it is used

  • Predicting wheel lift and tyre load differences in a corner.
  • Setting anti-roll bar stiffness and judging rollover risk.

Maths you need

  • Moments: overturning moment and resisting couple
  • Lateral acceleration: a = U² / R

Related formulas

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