LEMON Manuals: Even more car manuals for everyone
Home >> Chevrolet >> 2002 >> Blazer 4D Utility, 4WD, Part Time >> Repair and Diagnosis >> External Pages >> Different car >> Section 842 (Vibration Symptoms Diagnosis And Correction) >> Diagnostic Information And Procedures >> Component Rotational Speed Calculation >> Tire and Wheel Rotational Speed Calculation

Tire and Wheel Rotational Speed Calculation

WARNING: This page does not describe the selected car, but rather 8 other vehicles, including the 2003 GMC Yukon XL, 2003 GMC Yukon, 2003 Chevrolet Tahoe, 2003 Chevrolet Suburban, and 2003 Chevrolet Avalanche. However, it is still accessible from the selected car via links, so may be relevant.

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 Rotational Speed (at 8 km/h [5 mph]) 

Fig 1: Tire Rotational Speed (at 8 km/h [5 mph])
G02580323Courtesy of GENERAL MOTORS CORP.
  1. 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 Tire Rotational Speed table. For example: According to the Tire Rotational Speed table, a P255/70R16 tire makes 0.96 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 0.96 revolutions per second (Hz).
  2. 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
  3. 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 0.96 Hz = 11.52 Hz (rounded to 12 Hz)
  4. 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. 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.
  5. 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: 12 Hz X 2 (for second order) = 24 Hz second-order tire/wheel assembly rotation related 12 Hz X 3 (for third order) = 36 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 is present.