Tire and Wheel Rotational Speed Calculation
A size P235/75R15 tire rotates ONE complete revolution per second (RPS), or 1 Hz, at a vehicle speed of 8 km/h (5 mph). This means that at 16 km/h (10 mph), the same tire will make TWO complete revolutions in one second, 2 Hz, and so on.
| Tire Size | Tread | Revs/Sec (Hertz) at 8 km/h (5 mph) |
|---|---|---|
| P205/55R16 | AL3 | 1.16 |
| P215/50ZR17 | HW4 | 1.13 |
| Tread Code | ||
| AL3 | All Season Performance | |
| HW4 | Highway Hi Performance | |
- Determine the rotational speed of the tires in revolutions per second (RPS), or Hertz (Hz), at 8 km/h (5 mph), based on the size of the tires. Refer to the preceding Tire Rotational Speed table.
For example: According to the Tire Rotational Speed table, a P205/55R16 tire makes 1.16 revolutions per second (Hz) at a vehicle speed of 8 km/h (5 mph). This means that for every increment of 8 km/h (5 mph) in vehicle speed, the tire's rotation increases by 1.16 revolutions per second (Hz).
- Determine the number of increments of 8 km/h (5 mph) that are present, based on the vehicle speed (km/h, mph) at which the disturbance occurs.
For example: Assume that a disturbance occurs at a vehicle speed of 96 km/h (60 mph). A speed of 96 km/h (60 mph) has 12 INCREMENTS of 8 km/h (5 mph):
96 km/h (60 mph) divided by 8 km/h (5 mph) = 12 increments
- Determine the rotational speed of the tires in revolutions per second (Hz), at the specific vehicle speed (km/h, mph) at which the disturbance occurs.
For example: To determine the tire rotational speed at 96 km/h (60 mph), multiply the number of increments of 8 km/h (5 mph) by the revolutions per second (Hz) for one increment:
12 (increments) X 1.23 Hz = 14.76 Hz (rounded to 15 Hz)
- Compare the rotational speed of the tires at the specific vehicle speed at which the disturbance occurs, to the dominant frequency recorded on the J 38792-A
during testing. See Special Tools and Equipment . If the frequencies match, then a first-order disturbance related to the rotation of the tire/wheel assemblies is present.
If the frequencies do not match, then the disturbance may be related to a higher order of tire/wheel assembly rotation.
- To compute higher order tire/wheel assembly rotation related disturbances, multiply the rotational speed of the tires at the specific vehicle speed at which the disturbance occurs, by the order number:
15 Hz X 2 (for second order) = 30 Hz second-order tire/wheel assembly rotation related
15 Hz X 3 (for third order) = 45 Hz third-order tire/wheel assembly rotation related
If any of these computations match the frequency of the disturbance, a disturbance of that particular order, relating to the rotation of the tire/wheel assemblies and/or driveline components (also rotating at the same speed) is present.