Library / Automotive and vehicle dynamics / VEH-57
Automotive and vehicle dynamics
Unsprung-mass vertical acceleration
Upward acceleration of the unsprung mass in the same quarter-car model, with the tyre stiffness kt acting from the road elevation zr.
- Ref
- VEH-57
- Rev
- v1
- Field
- Automotive and vehicle dynamics
- Result
- z̈u [m/s²]
The wheel feels the tyre force from the road, kt(zr − zu), and the reaction of the suspension on the body: mu·z̈u = kt(zr − zu) − ks(zu − zs) − cs(vu − vs). The suspension terms are exactly the ones that appear in the body equation with the opposite sign, so the pair conserves momentum. It assumes the tyre stays in contact and has no damping.
Upward acceleration of the unsprung mass in the same quarter-car model, with the tyre stiffness kt acting from the road elevation zr. Same coordinates as the sprung-mass equation, continuous tyre contact and no tyre damping.
Calculator
- Tyre force on the wheel, kt(zr − zu)900 N
- Suspension force on the wheel (downward), ks(zu − zs) + cs(vu − vs)260 N
- Tyre deflection, zr − zu5 mm
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 wheel accelerates at 16 m/s² (1.63 g), driven by a net force of 640 N.
- Because the wheel is light, 40 kg against a body of hundreds, a net force of a few hundred newtons gives accelerations of many m/s², which is why wheel-hop is fast.
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 | −884 | 1,116 | 2,000 | +0.01 m | −312 % |
| zr Road elevation | −0.2 to 0.2 m | −996.5 | 803.5 | 1,800 | +0.01 m | +281 % |
| zs Sprung displacement | −0.2 to 0.2 m | −89 | 111 | 200 | +0.01 m | +31.2 % |
| mu Unsprung mass | 5 to 500 kg | 1.28 | 128 | 126.7 | +10% | −9.09 % |
| kt Tyre vertical stiffness | 20,000 to 1,000,000 N/m | −4 | 118.5 | 122.5 | +10% | +14.1 % |
| vu Unsprung velocity | −2 to 2 m/s | −41 | 79 | 120 | +0.08 m/s | −15 % |
| vs Sprung velocity | −2 to 2 m/s | −45.5 | 74.5 | 120 | +0.08 m/s | +15 % |
| ks Suspension spring rate | 1,000 to 200,000 N/m | −29 | 20.75 | 49.75 | +10% | −3.13 % |
| cs Suspension damping | 100 to 10,000 N s/m | 5 | 17.38 | 12.38 | +10% | −0.938 % |
Sweep of Road elevation
| Road elevation (m) | Unsprung-mass acceleration (m/s²) |
|---|---|
| −0.2 | −996.5 |
| −0.15 | −771.5 |
| −0.1 | −546.5 |
| −0.05 | −321.5 |
| 0 | −96.5 |
| 0.05 | 128.5 |
| 0.1 | 353.5 |
| 0.15 | 578.5 |
| 0.2 | 803.5 |
Sweep one variable
Variables and how each one contributes
Unsprung-mass acceleration
What the equation gives you. Everything else feeds it.
Tyre vertical stiffness
The vertical tyre stiffness, in N/m. Raises the tyre force in direct proportion. Raising it by 10 % (to 198,000) moves the result by +14.1 %, from 16 to 18.25 m/s².
Road elevation
The road elevation, in m. A higher road pushes the wheel up through the tyre. Raising it by 10 % (to 0.0275) moves the result by +70.3 %, from 16 to 27.25 m/s².
Unsprung displacement
The unsprung displacement, in m. A higher wheel reduces the tyre force and compresses the suspension, so it lowers the acceleration. Raising it by 10 % (to 0.022) moves the result by −62.5 %, from 16 to 6 m/s².
Suspension spring rate
The suspension spring rate, in N/m. Raises the downward spring force on the wheel. Raising it by 10 % (to 22,000) moves the result by −3.13 %, from 16 to 15.5 m/s².
Sprung displacement
The sprung displacement, in m. A higher body relaxes the spring force on the wheel. Raising it by 10 % (to 0.011) moves the result by +3.13 %, from 16 to 16.5 m/s².
Suspension damping
The suspension damping, in N s/m. Raises the damper force in proportion to the relative velocity. Raising it by 10 % (to 1,320) moves the result by −0.938 %, from 16 to 15.85 m/s².
Unsprung velocity
The unsprung velocity, in m/s. A faster wheel raises the downward damper force. Raising it by 10 % (to 0.11) moves the result by −1.88 %, from 16 to 15.7 m/s².
Sprung velocity
The sprung velocity, in m/s. A faster body lowers the downward damper force. Raising it by 10 % (to 0.055) moves the result by +0.937 %, from 16 to 16.15 m/s².
Unsprung mass
The unsprung mass, in kg. A lighter wheel accelerates more for the same net force, so wheel-hop gets more violent. Raising it by 10 % (to 44) moves the result by −9.09 %, from 16 to 14.55 m/s².
Worked example with the values above
- Tyre force: kt(zr − zu) = 180,000 × 0.005 = 900 N.
- Suspension force: 260 N, acting downward on the wheel.
- z̈u = (900 − 260) / 40 = 16 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