Library / Automotive and vehicle dynamics / VEH-56
Automotive and vehicle dynamics
Sprung-mass vertical acceleration
Upward acceleration of the sprung mass in a two-mass quarter-car model from the spring and damper forces between it and the wheel.
- Ref
- VEH-56
- Rev
- v1
- Field
- Automotive and vehicle dynamics
- Result
- z̈s [m/s²]
The sprung mass is moved only by the suspension, so ms·z̈s = ks(zu − zs) + cs(vu − vs): the spring pushes it up when the wheel is above the body and the damper pushes it along with the relative velocity. Displacements are from static equilibrium, so the weight and spring preload cancel and are not in the equation.
Upward acceleration of the sprung mass in a two-mass quarter-car model from the spring and damper forces between it and the wheel. Displacements are from static equilibrium, positive upwards, with a linear suspension and gravity already balanced by preload.
Calculator
- Spring force on the body, ks(zu − zs)200 N
- Damper force on the body, cs(vu − vs)60 N
- Total suspension force, spring plus damper260 N
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.
- The suspension pushes the body up with 260 N, an acceleration of 0.8667 m/s², or 0.0883 g.
- Comfort is judged by this body acceleration, so a lower value for the same road input means a better ride; the damper term here is 23.1 % of the force.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m/s²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| zu Unsprung displacement | −0.2 to 0.2 m | −13.8 | 12.87 | 26.67 | +0.01 m | +76.9 % |
| zs Sprung displacement | −0.2 to 0.2 m | −11.8 | 14.87 | 26.67 | +0.01 m | −76.9 % |
| vu Unsprung velocity | −2 to 2 m/s | −7.533 | 8.467 | 16 | +0.08 m/s | +36.9 % |
| vs Sprung velocity | −2 to 2 m/s | −6.933 | 9.067 | 16 | +0.08 m/s | −36.9 % |
| ks Suspension spring rate | 1,000 to 200,000 N/m | 0.2333 | 6.867 | 6.633 | +10% | +7.69 % |
| ms Sprung mass | 50 to 5,000 kg | 0.052 | 5.2 | 5.148 | +10% | −9.09 % |
| cs Suspension damping | 100 to 10,000 N s/m | 0.6833 | 2.333 | 1.65 | +10% | +2.31 % |
Sweep of Unsprung displacement
| Unsprung displacement (m) | Sprung-mass acceleration (m/s²) |
|---|---|
| −0.2 | −13.8 |
| −0.15 | −10.47 |
| −0.1 | −7.133 |
| −0.05 | −3.8 |
| 0 | −0.4667 |
| 0.05 | 2.867 |
| 0.1 | 6.2 |
| 0.15 | 9.533 |
| 0.2 | 12.87 |
Sweep one variable
Variables and how each one contributes
Sprung-mass acceleration
What the equation gives you. Everything else feeds it.
Suspension spring rate
The suspension spring rate, in N/m. Raises the spring force on the body in direct proportion. Raising it by 10 % (to 22,000) moves the result by +7.69 %, from 0.8667 to 0.9333 m/s².
Unsprung displacement
The unsprung (wheel) displacement, in m. A higher wheel compresses the spring and accelerates the body upwards. Raising it by 10 % (to 0.022) moves the result by +15.4 %, from 0.8667 to 1 m/s².
Sprung displacement
The sprung (body) displacement, in m. A higher body relaxes the spring and reduces the upward force. Raising it by 10 % (to 0.011) moves the result by −7.69 %, from 0.8667 to 0.8 m/s².
Suspension damping
The suspension damping, in N s/m. Raises the damper force in direct proportion to the relative velocity. Raising it by 10 % (to 1,320) moves the result by +2.31 %, from 0.8667 to 0.8867 m/s².
Unsprung velocity
The unsprung velocity, in m/s. A faster upward wheel drags the body up through the damper. Raising it by 10 % (to 0.11) moves the result by +4.62 %, from 0.8667 to 0.9067 m/s².
Sprung velocity
The sprung velocity, in m/s. A faster upward body reduces the damper force. Raising it by 10 % (to 0.055) moves the result by −2.31 %, from 0.8667 to 0.8467 m/s².
Sprung mass
The sprung mass, in kg. A heavier body accelerates less for the same suspension force. Raising it by 10 % (to 330) moves the result by −9.09 %, from 0.8667 to 0.7879 m/s².
Worked example with the values above
- Spring force: ks(zu − zs) = 20,000 × 0.01 = 200 N.
- Damper force: cs(vu − vs) = 1,200 × 0.05 = 60 N.
- z̈s = (200 + 60) / 300 = 0.86667 m/s².
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