The ball-bearing problem is a force-and-motion construction: where does the load act, which contacts support it, and about which axis does the bearing rotate? Historical V-groove and angular-contact drawings show how inclined contact forces contain both radial and axial components. This page resolves an ideal pair of opposed contacts and illustrates rolling separately.
Ball-bearing Problem Interactive Calculator
Resolve radial and axial load into two opposed contact reactions. Change the contact angle to see when one contact unloads. The rolling view illustrates the mechanism; the force calculation concerns the selected contact pair.
Equation Used
- Ideal rigid, frictionless opposed contacts.
- Loads refer to the illustrated pair; whole-bearing ball-load distribution is not solved.
- Contact angle is fixed during the calculation.
- No bearing rating, fatigue life, preload or elastic deformation is inferred.
This model resolves bearing contact forces using the load, support and rotation distinction in Hiscox figures 876–877.
Load directions in a ball bearing
A ball transfers load between contacting races. The reaction at an ideal frictionless contact acts normal to the contacting surface. An inclined reaction has radial and axial components. Opposed contact directions can cancel axial components under a radial load, while an applied axial load increases one reaction and reduces the other.
The animation separates the rolling view from an enlarged force construction. In the rolling view the outer race is stationary, the inner race turns and the balls both orbit and spin. The force inset shows two compressive reactions at the selected angle. It is a local equilibrium model, not a detailed prediction of how every ball in a bearing shares the shaft load.
Where this construction is useful
The force construction helps explain V-groove supports, opposed angular contacts and the effect of changing contact angle. It also shows why a contact can unload when the axial component is too large. Actual bearing selection additionally requires manufacturer load factors, contact geometry, preload, speed, lubrication and life calculations.
Contact-force equations
Take radial load Fr downward and axial load Fa to the right. Let each contact normal make angle α with the radial direction. Radial equilibrium gives (Nl + Nr) cos α = Fr; axial equilibrium gives (Nr − Nl) sin α = Fa. Adding and subtracting these relations gives the two reactions shown in the calculator.
Both contacts must remain in compression. Therefore |Fa| must not exceed Fr tan α for this ideal contact pair. Beyond that condition, the calculator shows the unloaded contact rather than inventing a tensile bearing reaction.
Worked example
With Fr = 1000 N, Fa = 0 and α = 45°, each reaction is 707.1 N. Their radial components add to 1000 N and their axial components cancel. With Fa = 300 N at the same angle, the reactions become 495.0 N on the left and 919.2 N on the right. The ideal axial limit for maintaining both contacts is ±1000 N.
What the model does and does not establish
A larger contact angle permits a larger axial component for the same radial load, but raises the required normal reactions. Real rolling contacts deform and distribute load across several balls. The three inputs here are insufficient to establish bearing life, contact stress, assembly preload or an allowable service load.
Common questions
Why does one result say contact unloaded?
The selected forces require a negative normal reaction at that contact. A separated contact cannot pull the ball back, so a different contact set or support arrangement is needed.
Is the rotating illustration a specific bearing product?
No. It is an explanatory rolling view with illustrative proportions and the cage omitted. It is not a dimensioned bearing drawing.
Reference
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances (1901), figures 876–877, printed pages 224–225: load direction, support direction and rotation axes in ball bearings. Read the historical source.
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