Library / Automotive and vehicle dynamics / VEH-29

Automotive and vehicle dynamics

Sprung body roll angle

Roll angle of the sprung body in a steady turn: the rolling moment from lateral acceleration divided by the net restoring roll stiffness.

Ref: VEH-29
Rev: v1
Field: Automotive and vehicle dynamics
Result: φ [°]

φ \approx \frac{m_{s} a_{y} h_{s}}{K_{φ} - m_{s} g h_{s}}

m_s is the sprung mass (the body, carried by the springs), a_y the lateral acceleration in g, h_s the height of the sprung mass centre above the roll axis, and K_φ the roll stiffness of the suspension in kN·m per radian. The rolling moment is m_s a_y h_s. The weight of the body, acting above the roll axis as it tilts, reduces the effective restoring stiffness by m_s g h_s. The result is the roll angle in degrees, for small angles. If the effective stiffness is not positive, the body is unstable in roll.

Roll angle of the sprung body in a steady turn: the rolling moment from lateral acceleration divided by the net restoring roll stiffness. The weight of the body acting above the roll axis reduces the effective stiffness.

Calculator

Solve for

φBody roll angleComputed

m_sSprung mass

3004,000

a_yLateral acceleration

0.11

h_sHeight of the sprung mass centre above the roll axis

0.20.9

K_φSuspension roll stiffness

kN·m/rad

40300

Rolling moment, m_s a_y h_s2,869 N·m
Effective restoring stiffness, K_φ − m_s g h_s84.26 kN·m/rad
Roll gradient3.902 °/g

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.

Body roll angle (°) current value 1.951 °

m_sSprung massKey driver

0.4286.99

K_φSuspension roll stiffness

0.5594.8

a_yLateral acceleration

0.393.9

h_sHeight of the sprung mass centre above the roll axis

0.8364.19

At 0.5 g the body rolls 1.951°, a roll gradient of 3.902° per g, which is typical of a road car (about 3 to 6° per g). The weight correction removes 6.38 % of the stiffness.
Roll rises quickly if the stiffness is low: with K_φ = 60 kN·m/rad the angle would be 3.03° at 0.5 g, and the body weight term leaves only 54.26 kN·m/rad of effective stiffness.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (°)	Local step	Change in result
m_s Sprung mass	300 to 4,000 kg	0.4278	6.993	6.565	+10%	+0.21 °
K_φ Suspension roll stiffness	40 to 300 kN·m/rad	0.5587	4.799	4.24	+10%	−0.188 °
a_y Lateral acceleration	0.1 to 1 g	0.3902	3.902	3.512	+10%	+0.195 °
h_s Height of the sprung mass centre above the roll axis	0.2 to 0.9 m	0.8356	4.187	3.352	+10%	+0.21 °

Sweep of Suspension roll stiffness

Suspension roll stiffness (kN·m/rad)	Body roll angle (°)
40	4.799
72.5	2.463
105	1.656
137.5	1.248
170	1.001
202.5	0.8356
235	0.7171
267.5	0.6281
300	0.5587

Sweep one variable

Vary

Body roll angle (°)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Body roll angle

Result · °

What the equation gives you. Everything else feeds it.

m_s

Sprung mass

kg · typical 300 to 4,000

The sprung mass (the mass carried on the springs), in kg: the vehicle mass less the wheels, axles and brakes. Raises the rolling moment and reduces the effective stiffness, so the roll angle grows faster than the mass. Raising it by 10 % (to 1,430) moves the result by +10.8 %, from 1.951 to 2.161 °.

a_y

Lateral acceleration

g · typical 0.1 to 1

The lateral acceleration, as a multiple of g. Directly proportional: roll angle grows in step with lateral acceleration. Raising it by 10 % (to 0.55) moves the result by +10 %, from 1.951 to 2.146 °.

h_s

Height of the sprung mass centre above the roll axis

m · typical 0.2 to 0.9

The height of the sprung mass centre above the roll axis, in m. Raises the rolling moment and lowers the effective stiffness, so the roll angle grows faster than in proportion. Raising it by 10 % (to 0.495) moves the result by +10.8 %, from 1.951 to 2.161 °.

K_φ

Suspension roll stiffness

kN·m/rad · typical 40 to 300

The suspension roll stiffness, in kN·m per radian, from the springs and anti-roll bars. Typical cars are 50 to 150. Lowers the roll angle, a little less than in inverse proportion because of the weight correction. Anti-roll bars are the cheapest way to raise it. Raising it by 10 % (to 99) moves the result by −9.65 %, from 1.951 to 1.763 °.

Worked example with the values above

Rolling moment: m_s a_y h_s = 1,300 × (0.5 × 9.81) × 0.45 = 2,869 N·m.
Effective stiffness: K_φ − m_s g h_s = 90,000 − 5,739 = 84,260 N·m/rad.
φ = 2,869 / 84,260 = 0.03405 rad = 1.951°.

Assumptions and limits

Small roll angles, a fixed roll axis and a linear suspension roll stiffness.
Steady-state cornering with a rigid sprung body, ignoring tyre compliance and the movement of the roll axis.

Common mistakes

Using the total vehicle mass in place of the sprung mass.
Measuring h_s from the ground instead of from the roll axis.

Statistics and uncertainty

The roll angle depends on the roll stiffness and the height of the roll centre, both known to about 10 % at best (springs, bars and bushings all contribute, and the roll axis height changes with suspension travel). The effective stiffness is a difference of two terms, so it carries a larger relative error than the roll stiffness itself, and an error of 10 % in K_φ becomes an error of 10 to 12 % in φ at the defaults.

Where it is used

Setting spring and anti-roll bar stiffness for a target roll angle.
Judging passenger comfort and the camber change of the wheels in cornering.

Maths you need

Moments and torsional stiffness: angle = moment / stiffness
Small angles: tan φ ≈ φ in radians

Related formulas

Practise this formula