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Library / Automotive and vehicle dynamics / VEH-58

Automotive and vehicle dynamics

Coupled body and wheel natural frequencies

Lower (body) natural frequency of the undisturbed two-mass quarter-car model with a fixed road, from the eigenvalue equation Aλ² − Bλ + C = 0 with A = ms mu, B = ks mu + ms(ks + kt) and C = ks kt.

Ref
VEH-58
Rev
v1
Field
Automotive and vehicle dynamics
Result
f1 [Hz]

For the two-mass model with a fixed road, the natural frequencies come from the eigenvalue equation Aλ² − Bλ + C = 0, with A = ms·mu, B = ks·mu + ms(ks + kt) and C = ks·kt. The lower root is the body bounce and the higher is the wheel hop, and f = √λ/(2π). If the masses are well separated, these reduce to the familiar decoupled values.

Lower (body) natural frequency of the undisturbed two-mass quarter-car model with a fixed road, from the eigenvalue equation Aλ² − Bλ + C = 0 with A = ms mu, B = ks mu + ms(ks + kt) and C = ks kt. The higher (wheel) frequency is in the derived rows. The lower root uses the stable product form.

Calculator

Solve for
Hz
kg
505,000
kg
5500
N/m
1,000200,000
N/m
20,0001,000,000
  • Wheel-hop (higher) frequency, f₂10.56 Hz
  • Body frequency with a rigid tyre, √(ks kt/(ks + kt)/ms)/(2π)1.214 Hz
  • Wheel frequency with a rigid body, √((ks + kt)/mu)/(2π)10.56 Hz

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 natural frequency (Hz) current value 1.213 Hz
ksSuspension spring rateKey driver
0.3172.93
msSprung mass
0.2712.71
ktTyre vertical stiffness
0.9341.26
muUnsprung mass
1.21.21
  • The body bounces at 1.213 Hz and the wheel hops at 10.56 Hz, a ratio of 8.71.
  • Comfort is mostly set by the low frequency, which should stay in the roughly 1 Hz to 1.5 Hz band, while the high one should stay away from the frequencies that road wavelengths excite.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (Hz)Local stepChange in result
ks Suspension spring rate1,000 to 200,000 N/m0.31732.9272.61+10%+4.39 %
ms Sprung mass50 to 5,000 kg0.27142.7052.434+10%−4.65 %
kt Tyre vertical stiffness20,000 to 1,000,000 N/m0.93381.2630.3294+10%+0.427 %
mu Unsprung mass5 to 500 kg1.2021.2140.0117+10%−0.00679 %

Sweep of Suspension spring rate

Suspension spring rate (N/m)Body natural frequency (Hz)
1,0000.3173
25,8801.5
50,7501.966
75,6302.262
100,5002.47
125,4002.627
150,3002.749
175,1002.847
200,0002.927

Sweep one variable

Body natural frequency (Hz)Hover, touch, or use the arrow keys on the chart
0.511.522.53 50k100k150k200k Suspension spring rate (N/m) now 1.213

Variables and how each one contributes

f1

Body natural frequency

Result · Hz

What the equation gives you. Everything else feeds it.

ms

Sprung mass

kg · typical 50 to 5,000

The sprung mass, in kg. A heavier body lowers the body frequency, roughly as 1/√ms. Raising it by 10 % (to 275) moves the result by −4.65 %, from 1.213 to 1.157 Hz.

mu

Unsprung mass

kg · typical 5 to 500

The unsprung mass, in kg. A heavier wheel lowers the wheel-hop frequency, with little effect on the body frequency. Raising it by 10 % (to 44) moves the result by −0.00679 %, from 1.213 to 1.213 Hz.

ks

Suspension spring rate

N/m · typical 1,000 to 200,000

The suspension spring rate, in N/m. Raises the body frequency, roughly as √ks. Raising it by 10 % (to 17,600) moves the result by +4.39 %, from 1.213 to 1.266 Hz.

kt

Tyre vertical stiffness

N/m · typical 20,000 to 1,000,000

The vertical tyre stiffness, in N/m. Raises mainly the wheel-hop frequency. Raising it by 10 % (to 176,000) moves the result by +0.427 %, from 1.213 to 1.218 Hz.

Worked example with the values above

  1. Coefficients: A = 10,000, B = 4.464×107, C = 2.56×109.
  2. Roots: λ₁ = 58.104 rad²/s², λ₂ = 4,405.9 rad²/s².
  3. f₁ = √λ₁/(2π) = 1.2132 Hz, and f₂ = 10.564 Hz.

Assumptions and limits

  • A linear two-mass model with a point-contact tyre that never leaves the road and with no tyre damping.
  • Displacements measured from static equilibrium, so gravity cancels, and a sinusoidal or sampled input as stated.

Common mistakes

  • Using a total displacement instead of one measured from static equilibrium.
  • Mixing displacement and velocity inputs in the force terms.
  • Assuming the body and wheel frequencies decouple when the masses and stiffnesses are not well separated.

Statistics and uncertainty

Tyre stiffness varies by 10 % or more with pressure, load and speed, and damper behaviour is usually nonlinear, so computed responses give trends and frequency estimates rather than exact amplitudes.

Where it is used

  • Ride, comfort and road-holding studies of passenger and off-road vehicles.
  • Choosing spring and damper values for a quarter-car model.

Maths you need

  • Newton’s second law: m z̈ = Σ forces
  • Eigenvalue problem: natural frequencies from the system matrices
  • Complex frequency response: amplitude and phase at a driven frequency

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