Bearing Duty Cycle and Cubic Mean Load Calculator

The cubic mean load for a varying duty cycle, with each segment’s share of the DAMAGE printed beside its share of the time — and with Miner’s rule run the other way as a check.

Last updated 24 Sep 2026

Duty cycle and cubic mean load

Up to five load, speed and time segments → P_m and the life

Bearing typeThe life exponent, and nothing else on this page. p = 3 for ball bearings and 10/3 for roller bearings, both fitted to test data rather than chosen for tidiness. The larger exponent makes a roller bearing MORE sensitive to a load spike, not less — which is the opposite of what most people expect from the tougher-looking bearing.

Basic dynamic load rating C (kN)Straight off the catalogue page, and the same number the L10 life page wants. It is the load for which the basic rating life is one million revolutions — a definition, not a permissible load, and never a load you should design to.

Target life you want (hours)Used for one thing: the rating C a catalogue must show to reach this life under YOUR duty cycle. That is the number you shop with, and it is the one output of this page that ends in a part number.

Segment 1 — share of the time (%)The longest, most ordinary condition the machine runs in. Time shares are relative — they do not have to add to 100, and a segment left at zero time drops out of the calculation entirely.

… equivalent load P1 (N)The EQUIVALENT dynamic load for this segment, not the radial load. If the segment has an axial component, reduce it to P first on the equivalent load page and bring the answer here.

… speed n1 (rev/min)Speed during this segment. It matters twice: it weights the segment’s damage by how many revolutions it contains, and it sets the mean speed that turns the answer into hours.

Segment 2 — share of the time (%)A second running condition. Load and speed are independent: a slower segment at the same load does less damage, because damage is counted in REVOLUTIONS and not in hours.

… equivalent load P2 (N)The EQUIVALENT dynamic load for this segment, not the radial load. If the segment has an axial component, reduce it to P first on the equivalent load page and bring the answer here.

… speed n2 (rev/min)Speed during this segment. It matters twice: it weights the segment’s damage by how many revolutions it contains, and it sets the mean speed that turns the answer into hours.

Segment 3 — share of the time (%)A heavier condition — a full hopper, a cutting pass, a headwind. This is usually where the damage starts to concentrate.

… equivalent load P3 (N)The EQUIVALENT dynamic load for this segment, not the radial load. If the segment has an axial component, reduce it to P first on the equivalent load page and bring the answer here.

… speed n3 (rev/min)Speed during this segment. It matters twice: it weights the segment’s damage by how many revolutions it contains, and it sets the mean speed that turns the answer into hours.

Segment 4 — share of the time (%)The short heavy spell. This is the segment the page is about, and the one people leave out of a hand average because it is only a few per cent of the time.

… equivalent load P4 (N)The EQUIVALENT dynamic load for this segment, not the radial load. If the segment has an axial component, reduce it to P first on the equivalent load page and bring the answer here.

… speed n4 (rev/min)Speed during this segment. It matters twice: it weights the segment’s damage by how many revolutions it contains, and it sets the mean speed that turns the answer into hours.

Segment 5 — share of the time (%)A fifth segment if you need one. Leave the time share at zero and it is ignored; the fields below it stay whatever you last typed.

… equivalent load P5 (N)The EQUIVALENT dynamic load for this segment, not the radial load. If the segment has an axial component, reduce it to P first on the equivalent load page and bring the answer here.

… speed n5 (rev/min)Speed during this segment. It matters twice: it weights the segment’s damage by how many revolutions it contains, and it sets the mean speed that turns the answer into hours.

Not a circuit: your duty cycle drawn twice. The top row is the LOAD in each segment, every bar scaled to the heaviest one, with the line running across all of them at the cubic mean load. The bottom row is the DAMAGE each segment does, scaled to the worst one. Under each pair of bars are two numbers: the segment's share of the time on the left and its share of the damage on the right. Those two numbers are the page. A segment that is a twentieth of the time can be nearly half the damage, because the life law is cubic and a load that is twice as big does eight times the damage per revolution. Note also where the cubic mean line sits: well above the bars that take up most of the time, and much nearer the tall one. That is why a bearing sized on the average load of a duty cycle comes out too small, and it is why the load you type into the short heavy segment is the one worth measuring rather than estimating: an error there moves the answer by most of itself cubed, while the same error in a long quiet segment barely moves it at all.

8,262hExample

A four-segment duty cycle on a bearing rated C = 30 kN: half the time at 2,000 N and 1,500 rev/min, thirty per cent at 3,000 N and 1,200, fifteen per cent at 5,000 N and 900, and five per cent at 9,000 N and 600

Miner’s rule applied to a cubic life law, which is all the cubic mean load is

damage = Σ Nᵢ / N₁₀,ᵢ = 1 at failure · N₁₀,ᵢ = (C/Pᵢ)^p · P_m = (Σ qᵢ (nᵢ/n_m) Pᵢ^p)^(1/p) · n_m = Σ qᵢ nᵢ

the derivation: in four lines. Each segment runs Nᵢ revolutions at load Pᵢ, and the life at that load alone is N₁₀,ᵢ = (C/Pᵢ)^p million revolutions. Miner says the fractions add, so the damage per cycle is ΣNᵢ(Pᵢ/C)^p. Ask what single load P_m run for the same total revolutions ΣNᵢ would do the same damage: ΣNᵢ (P_m/C)^p = ΣNᵢ(Pᵢ/C)^p. The C’s cancel, and P_m^p is the revolution-weighted mean of Pᵢ^p. That is the whole of it — there is nothing in the cubic mean load that is not in those four lines
qᵢ, nᵢ: the time share and the speed of each segment. They enter as a PRODUCT, because what wears a bearing is revolutions, not hours. A segment at half the speed does half the damage for the same time at the same load, and a duty cycle in which the heavy loads happen slowly is genuinely kinder than one in which they happen fast
p: 3 for ball, 10/3 for roller. It is the whole reason this page exists: an average is a p = 1 operation and the damage law is p = 3, so an average of loads is the mean of the wrong quantity. The larger the exponent, the further apart the two answers sit — so a roller bearing suffers MORE from a spiky duty cycle than a ball bearing does
n_m: the mean speed, Σqᵢnᵢ. It is an ordinary time-weighted average and not a cubic one, because it is not doing damage, it is only converting revolutions into hours at the end
the identity: written out, P_m = (ΣPᵢ^p nᵢ tᵢ / Σnᵢ tᵢ)^(1/p) — which is exactly the form NTN prints. The two are the same equation and the page asserts it numerically rather than taking it on faith

Worked example

A four-segment duty cycle on a bearing rated C = 30 kN: half the time at 2,000 N and 1,500 rev/min, thirty per cent at 3,000 N and 1,200, fifteen per cent at 5,000 N and 900, and five per cent at 9,000 N and 600
MEAN SPEED FIRST, because it is the easy one and everything gets weighted by it. n_m = 0.5×1,500 + 0.3×1,200 + 0.15×900 + 0.05×600 = 1,275 rev/min. Note that this one IS an ordinary average: speed does not get cubed
NOW THE DAMAGE WEIGHTS. Each segment contributes in proportion to qᵢ·nᵢ·Pᵢ³. For the first segment that is 0.5 × 1,500 × 2,000³ = 6.00×10¹²; for the last, 0.05 × 600 × 9,000³ = 2.187×10¹³. The short spell is already the biggest term, and it is only one twentieth of the time
THE CUBIC MEAN LOAD. Add the four terms, divide by n_m, take the cube root: P_m = 3,495.7 N. Sanity-check it against the numbers you started with — it has to sit between the smallest and the largest load in the cycle, and it does, but much nearer the top than a glance at the time shares would suggest
THE DAMAGE SHARES, which are the point of the page. Divide each term by the sum: segment 1 does 11.0 per cent of the damage in 50 per cent of the time; segment 4 does 40.2 per cent in 5 per cent of the time. The short heavy spell is doing 8.0 times its share, and it does it while turning slower than anything else in the cycle
THE LIFE. L10 = (C/P_m)³ = (30,000/3,496)³ = 632.1 million revolutions, which at 1,275 rev/min is 8,262 hours. Check it the other way with Miner's rule — work out each segment's own L10 in hours, add qᵢ/Lᵢ and invert — and you get 8,262 hours. The same number, because they are the same calculation
AND WHAT THE AVERAGE WOULD HAVE TOLD YOU. The time-weighted average of the four loads is 3,100 N, which gives 11,847 hours — 1.43 times too long. Weight it by revolutions instead and it gets WORSE, not better: 2,765 N and 16,702 hours, 2.02 times too long, because the heavy segment runs slowly and revolution-weighting discounts it further. Both averages fail for the same reason: the damage law is cubic and an average is not
THE NUMBER TO SHOP WITH. For 20,000 hours at this duty cycle the catalogue must show C = P_m · (L·60n_m/10⁶)^(1/3) = 40,281 N, which is 134 per cent of the rating you have. That is the line to take to the bearing table, and the L10 page will take P_m straight from here

The page’s own default duty cycle, worked out

Read the last two columns together. The fourth segment is five per cent of the time and it does 40.2 per cent of the damage — 8.0 times its share, even though it runs at the LOWEST speed in the cycle and therefore contributes the fewest revolutions. The first segment is half the time and does 11.0 per cent. That is what a cubic law does to a duty cycle, and it is why the cubic mean load comes out at 3,496 N when the ordinary average of the same four loads, weighted by time, is 3,100 N. Use the average and this bearing appears to last 11,847 hours instead of 8,262 — 1.43 times too long. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
	Time share (%)	Load (N)	Speed (rev/min)	Share of the REVOLUTIONS (%)	Share of the DAMAGE (%)	Damage share ÷ time share
Segment 1	50	2,000	1,500	58.8	11.0	0.22
Segment 2	30	3,000	1,200	28.2	17.8	0.59
Segment 3	15	5,000	900	10.6	31.0	2.07
Segment 4	5	9,000	600	2.4	40.2	8.03
Whole cycle	100	P_m = 3,496	n_m = 1,275	100.0	100.0	1.00

Ten per cent of the time at a higher load: what it costs

This is the whole page in one table, and the row to look at is 2.081. That is the exact load multiple at which a tenth of the time does as much damage as the other nine tenths, because 0.1·m³ = 0.9 gives m = ∛9 = 2.0801. Below it the long steady spell still governs; above it the short heavy one does, and by three times the load the spell owns three quarters of the damage. Note the second column against the first: the ROLLER bearing, with its larger exponent, is more sensitive to the spike and not less. And note the last column, which is the cost of the mistake: an ordinary time-weighted average promises 28 per cent more life than there is at twice the load and 61 per cent more at three times. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Load of the spell	Its damage share, ball (%)	… roller (%)	Life left, ball (%)	Life the average promised (%)	How optimistic the average is
1.000×	10.0	10.0	100.0	100.0	1.00
1.250×	17.8	18.9	91.3	92.9	1.02
1.500×	27.3	30.0	80.8	86.4	1.07
1.750×	37.3	41.8	69.6	80.5	1.16
2.000×	47.1	52.8	58.8	75.1	1.28
2.080×	50.0	56.1	55.6	73.5	1.32
2.500×	63.5	70.2	40.6	65.8	1.62
3.000×	75.0	81.2	27.8	57.9	2.08
4.000×	87.7	91.9	13.7	45.5	3.32

A continuously varying load: the published shortcuts against the exact answer

NTN publishes three shortcuts for a load that varies continuously rather than in steps: F_m = (F_min + 2F_max)/3 for a monotonic ramp, and 0.75·F_max and 0.65·F_max for two sinusoidal cases. The fourth column here is not quoted from anybody: it is the exact p-mean of each shape, integrated numerically. Two things come out of putting them side by side. The ramp rule is HIGH by three to six per cent at p = 3, which is the safe direction — it over-estimates the load and under-estimates the life. And the 0.75 figure is not a rule of thumb at all: the exact cubic mean of a half-sine hump is ∛(4/3π) = 0.7515, which rounds to 0.75 exactly. NTN’s second sinusoidal case, 0.65, could not be reproduced here from any obvious waveform — the last two rows bracket it at 0.596 and 0.679 — so read it off NTN’s own figure rather than from a description of it. All of these are shortcuts for hand calculation. If you have the shape, put four or five points of it into the calculator above instead. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
How the load varies	Published rule	What the rule gives	Exact cubic mean (ball)	Exact, roller p = 10/3	The rule’s error
Ramp, F_min = 0 to F_max	(F_min + 2F_max)/3	0.667	0.6300	0.6441	5.8
Ramp, 0.2 to 1.0	(F_min + 2F_max)/3	0.733	0.6782	0.6885	8.1
Ramp, 0.5 to 1.0	(F_min + 2F_max)/3	0.833	0.7768	0.7810	7.3
Half-sine hump, zero to F_max and back	0.75·F_max (NTN case a)	0.750	0.7515	0.7630	-0.2
Sinusoid about a mean of F_max/2	—	—	0.6786	0.6952	—
Half-wave: a hump for half the cycle, unloaded for the rest	—	—	0.5965	0.6197	—

Damage is cubic, averages are linear, and that gap is the whole subject

Averaging a duty cycle is a linear operation and bearing life is a cubic law, so the average of the loads is the mean of the wrong quantity. That single sentence is the page. A bearing’s life at one load is (C/P)³ million revolutions; run it at two loads and Miner’s rule says the damage fractions add. Ask what single load would consume life at the same rate and the answer falls out immediately: P_m is the revolution-weighted cube-root-mean of the loads, not their average. The two agree only when the load is constant, and they diverge faster the spikier the cycle gets. On the default cycle here, the difference is a factor of 1.43 on the predicted life.

A short spell at a high load runs away with the damage. Put a tenth of the cycle at twice the load and that tenth does 47 per cent of the damage — nearly half the wear from a twentieth of the running. At 2.081 times the load, which is the cube root of nine, it crosses over and does more damage than the other ninety per cent of the time put together. At three times the load it does 75 per cent. This is why machines that were specified on their nominal duty eat bearings once a real duty cycle arrives, and it is why the single most useful thing you can measure on a machine with a bearing problem is the peak load rather than the average one.

Revolutions do the damage, not hours. The speed of each segment enters the weighting as a straight multiplier, because fatigue in a rolling contact is counted in stress cycles and a stress cycle is a rolling element passing a point. A duty cycle whose heavy loads happen slowly is genuinely gentler than one whose heavy loads happen fast, by exactly the speed ratio — and a machine that answers a load increase by slowing down is doing something real for its bearings, not just sounding calmer. The mean speed itself, by contrast, is an ordinary time average: it is not doing damage, it is only converting revolutions into hours at the end.

What this page does not do. It does not turn radial and axial loads into an equivalent load — do that segment by segment on the equivalent load page first, because X, Y and the limit e can differ between segments if the axial load does. It does not apply the reliability factor a₁ or the life modification factor a_ISO; take P_m to the L10 page for those. It does not know about lubrication, contamination, temperature or misalignment, which between them end more bearings than fatigue does. And it assumes the duty cycle repeats: five segments in any order give the same answer, because Miner’s rule has no memory of sequence. Real damage does depend on order — a heavy spell on a cold, unlubricated bearing is not the same as the same spell warm — and no linear damage rule can see that.

Where the method comes from. Palmgren applied a linear cumulative damage rule to ball bearings in 1924, twenty-one years before Miner published the general version that carries his name, which is why bearing engineers often call it the Palmgren-Miner rule. It is the oldest working piece of fatigue engineering still in daily use, and it is also the crudest: it assumes damage accumulates linearly, independent of order, with no interaction between load levels and no threshold below which nothing happens. Every one of those assumptions is known to be imperfect. It is used anyway because the alternatives need data nobody has, and because it has been calibrated against eighty years of bearings that failed.

Frequently asked questions

How do you calculate the mean load for a bearing with a varying load?

Not by averaging. Raise each segment’s load to the power p — 3 for a ball bearing, 10/3 for a roller — weight it by the REVOLUTIONS that segment contains, average those, and take the p-th root: P_m = (Σqₓ(nₓ/n_m)Pₓⁿ)¹⁄ⁿ. The revolutions matter because damage is counted per stress cycle, so a segment at half the speed does half the damage for the same duration. The same equation is printed by NTN as (ΣFₓⁿnₓtₓ / Σnₓtₓ)¹⁄ⁿ, which is the same thing written with raw times instead of fractions.

Why can’t I just average the loads?

Because life goes as the cube of load and an average is linear, so the average is the mean of the wrong quantity and it always errs in the unsafe direction. On this page’s own default cycle the time-weighted average load is 3,100 N against a cubic mean of 3,496 N, and that seemingly small gap becomes a factor of 1.43 on the predicted life. The sharper the duty cycle, the worse it gets. Weighting the average by revolutions instead of time does not help — here it makes it worse, because the heavy segment runs slowly.

What is the cubic mean load in bearing selection?

The single constant load that would consume the bearing’s fatigue life at the same rate as the varying load does. It is not an average, a peak, or a root-mean-square; it is the damage-equivalent load, and the cube in its name is the exponent of the life law rather than anything to do with cubing for its own sake. On a roller bearing it is a 10/3-power mean, which is why the phrase “cubic mean” is strictly a ball-bearing term that everyone uses for both.

Does a short overload really matter that much?

Yes, and the arithmetic is unforgiving. Ten per cent of the time at twice the load does 47 per cent of the damage; at 2.081 times the load that tenth does more damage than the other ninety per cent combined; at three times it does 75 per cent. The practical consequence is that the peak is where the accuracy of your input matters most: a twenty per cent error in a short heavy segment moves the answer far more than the same error in a long quiet one. Be careful costing the fix, though. On this page’s default cycle, cutting the peak segment’s load by fifteen per cent buys 18 per cent more life, while a fifteen per cent bigger rating buys 52 per cent, because the rating applies to every segment and the peak is only one of them. Cutting every load by fifteen per cent buys 63 per cent.

Is a roller bearing better than a ball bearing for a shock duty cycle?

It usually wins, but not for the reason people give, and it wins by less than they think. The exponent for a roller bearing is 10/3 rather than 3, which makes its life MORE sensitive to load, not less — on this page’s default cycle the short heavy spell’s damage share rises from 40.2 to 47.4 per cent when you switch the type. What wins the argument is that a roller bearing’s basic dynamic load rating C is much larger for the same envelope, because line contact carries far more load than point contact. The exponent is working against it; the rating is working for it, harder.

How do I handle a load that varies continuously rather than in steps?

Either integrate it, or use one of the published shortcuts. NTN gives F_m = (F_min + 2F_max)/3 for a load that ramps monotonically and 0.75·F_max for a sinusoidal one. The ramp rule turns out to be three to six per cent high at p = 3 — the safe direction — and the 0.75 figure is not an approximation at all: the exact cubic mean of a half-sine hump is the cube root of 4/3π, which is 0.7515. In practice, splitting the real curve into four or five steps and putting them into the calculator above is both easier and more accurate than either.

Does the order of the segments matter?

Not to this calculation, and somewhat to the real bearing. Miner’s rule adds damage fractions with no memory of sequence, so five segments in any order give exactly the same answer here. Real damage is not quite so obliging: a heavy spell on a cold bearing with a thin oil film is harder on it than the same spell at running temperature, load sequence affects residual stresses in the raceway, and a machine that always starts under full load is doing something a duty-cycle table cannot show. None of that is in the model, and none of it is usually large enough to change a bearing selection.

Related calculators

References

M. A. Miner, Cumulative Damage in Fatigue, Journal of Applied Mechanics 12 (1945), A159–A164, and Arvid Palmgren, Die Lebensdauer von Kugellagern, VDI-Zeitschrift 68 (1924), 339–341. The linear damage rule, and its original application to ball bearings, which came first. Everything this page does follows from one sentence of it: each spell of running consumes a fraction of life equal to the revolutions run divided by the revolutions that load alone would give, and the part fails when the fractions add to one. The cubic mean load is what you get when you ask which single load consumes the fractions at the same rate.
NTN, Bearing load calculation, technical section 4 of catalogue CAT. No. 2203-E/A (ntnglobal.com, read 29 September 2026). The named source for three published approximations: the stepped duty cycle mean F_m = [ΣF_i^pn_it_i / Σn_it_i]^1/p, the monotonically varying load F_m = (F_min + 2F_max)/3, and the two sinusoidal cases at 0.75 F_max and 0.65 F_max. The stepped form and the form used on this page are the same equation written two ways, which is asserted rather than assumed.
ISO 281:2007, Rolling bearings — Dynamic load ratings and rating life. Cited by number and not reproduced. It is the document behind L₁₀ = (C/P)^p and behind the exponents 3 and 10/3, and it is the reason this page’s output is a load rather than a life: the standard’s life equation takes ONE load, and a duty cycle has to be reduced to one before it can be used. The reduction itself — the damage-equivalent mean — is Palmgren and Miner’s, not ISO 281’s.
SKF, Size selection based on rating life and Equivalent dynamic bearing load, P (skf.com). Cited as the second publisher of the same duty-cycle method, in the form P = (ΣU_iP_i^p)^1/p over time fractions U at constant speed. NOT verified by fetch: the page would not load from this sandbox on 29 September 2026. What was checked instead is numerical — SKF’s constant-speed form and NTN’s speed-weighted form are the same equation when every n_i is equal, and the code asserts it.
American Roller Bearing, Bearing Life Calculation — Bearing Loads and Speeds (amroll.com). The source this plugin’s L₁₀ page already uses for the 0.02 C minimum load rule, cited here for the same rule: a duty cycle with an idle or very lightly loaded segment is the commonest way a bearing ends up below its minimum load, and no amount of mean-load arithmetic will show it.