Judged on the static equivalent load P₀ = 0.6Fr + 0.5Fa (never less than Fr), which is a different quantity from the dynamic P that drives the life calculation above.
Selection usually runs the other way round from the life formula: you know the load, the speed and how long the machine has to last, and you need to know which bearing clears it. Rearranging L10 = (C/P)³ gives the dynamic rating you must buy — then the catalogue tells you which parts have it.
C is fatigue under rotation, and feeds the life formula; C₀ is denting of a stationary raceway. Both matter; they answer different questions.
L10 = (C/P)³ × 10⁶ revolutions: the load under which 90% of a batch survive. Double the load, an eighth of the life.
One number, P = XFr + YFa (ISO 281). For deep-groove balls X and Y come from Fa/C₀ alone, so no per-part lookup.
Below Fa/Fr = e axial load is free; above it every extra newton costs life at the cube.
L10 scales with the cube of load, so small overloads cost disproportionate life. Running a bearing at 120% of its design load costs about 42% of the life; a 26% increase already halves it. Margins on load matter enormously.
6200 and 6300 series bearings sharing a bore are not interchangeable: their outer diameters differ, and the heavier 6300 uses larger balls in a bigger ring for much higher ratings at the cost of a larger housing. Check both bore and series before ordering a replacement.
A statistical rating: 90% of an identical batch should reach it, 10% will fail sooner, some run far longer. It assumes clean lubrication, correct alignment and normal temperature; contamination, misalignment or heat can cut real life to a fraction. Demanding 99% reliability instead of 90% means designing for the unluckiest bearings: the a1 factor (0.25 at 99%) prices that in, and multiplies the rating needed by about 1.59.
Below e the contact angle already carries the thrust and P is just the radial load, so people over-design for small thrusts. Above it P = 0.56Fr + YFa and life falls with the cube; Y can be as high as 2, so a thrust equal to the radial load can more than double P and cut life to under a fifth. Find out which side of e you are on before sizing anything.
You start from a load, a speed and a life the machine must reach, and solve C = P·L10⅓ for the rating to buy. If a lighter series cannot clear a comfortable margin, stepping up a series usually buys far more life per millimetre of housing than any other change.
s₀ = C₀/P₀: at least 1 for ordinary rotating duty, 1.5 where shock is expected, 2 where the bearing must turn quietly or precisely afterwards: a static overload leaves a permanent dent the bearing runs over once per revolution for the rest of its life. The static load P₀ = 0.6Fr + 0.5Fa (floored at Fr by ISO 76) differs from the dynamic P: at equal radial and axial load P is about 1.9× P₀, and comparing P against C₀ (as this page did until v5.60) reports nearly twice the static utilisation that exists.
They share a bore but not a cross-section. The 6300 (heavy) series uses larger balls in a bigger outer ring, giving substantially higher load ratings for the same shaft — at the cost of a larger housing bore and more radial space.
Because they are different equivalent loads, computed for different failure modes. The dynamic equivalent load P = 0.56Fr + YFa is what fatigues a turning bearing, and it is what the cube law uses. The static equivalent load P₀ = 0.6Fr + 0.5Fa is what dents a stationary one, and ISO 76 floors it at Fr because a small thrust cannot make a bearing safer than its radial load alone.
They agree exactly under pure radial load, which is why the difference is easy to miss. Put equal radial and axial load on a bearing and P is about 1.9× P₀ — so comparing P against C₀ reports nearly twice the static utilisation that exists. This page did exactly that until v5.60.
s₀ = C₀ / P₀, and the usual guidance is s₀ ≥ 1 for ordinary rotating duty, ≥ 1.5 where shock loads are expected, and ≥ 2 where the bearing must turn quietly or precisely afterwards — because the damage a static overload does is a permanent dent in the raceway, and the bearing runs over it once per revolution for the rest of its life.
C (dynamic) governs fatigue life under rotation — it's what the L10 formula uses. C₀ (static) is the load a stationary bearing can take before the balls permanently dent the raceway. A bearing can be adequately sized for running duty and still be damaged by a heavy shock load while parked.
No — it's a statistical rating. 90% of an identical batch, under identical conditions, should reach or exceed this life; 10% will fail sooner, and some will run far longer. It also assumes clean lubrication and normal temperature — contamination or heat cuts real life well below the catalogue number.
Switch the load input to Radial + axial and enter both. The tool computes the equivalent dynamic
load P the ISO 281 way and shows the factors it used. For single-row deep-groove ball
bearings the factors e and Y depend only on the ratio Fa/C₀,
not on the individual part, so they are the same across manufacturers.
An earlier version of this page said combining loads needed per-size catalogue factors and declined to do it. That is true of angular-contact and taper roller bearings, but not of the deep-groove bearings this tool covers — so the calculation is now included.
Only once Fa/Fr exceeds the threshold e (typically
0.22–0.44). Below that, the bearing's existing contact angle carries the thrust and the equivalent load
is simply the radial load — the axial component costs you nothing. Above it,
P = 0.56Fr + YFa and life starts falling with the cube of the result.
Rearrange the life formula: C = P × (L10)^⅓, with L10 in millions of revolutions. The
Size for a target life section does this and then names the smallest bearing in each series that
meets it. Note that demanding higher reliability raises the required C — a 99% requirement needs
about 1.59× the rating of a 90% one for the same life.
Basic rating life is L10 = (C / P)p million revolutions, where C is the dynamic load rating, P the equivalent load and p = 3 for ball bearings (10/3 for roller bearings in the standard). In hours, L10h = 106 / (60 n) × (C / P)3 with n in rpm. A load of one fifth of the rating at 1,500 rpm gives about 1,390 hours, and halving the load makes the life eight times longer.
This page has changed in 194 archived releases, but each of those was a site-wide pass, so none is attributable to this tool on its own and none is listed here. That is not a claim that the tool never changed — a release that reworked many pages at once may well have altered this one too. The changelog has them.