Home · Articles on balancing and vibration
Tolerances and Residual Unbalance

Residual Unbalance in Practice: g·mm, g·mm/kg and Microns

The report has a line reading “residual unbalance 1200 g·mm,” and that's where the questions start. Is that a lot or a little, which plane does it belong to, and why does the support still read 1.8 mm/s. We go through how the balance quality grade and the speed give you an allowable mass in grams at your radius, why the tolerance gets split between two planes, why it's more convenient to measure large rotors in microns, and which numbers absolutely have to end up in the report.

Updated 27 August 2026 · by AXILINE · Vila Nova de Gaia

In short: Residual unbalance is what's left in the rotor after fitting the correction weights. It's expressed in g·mm per correction plane, and in g·mm/kg per kilogram of rotor mass, where the second figure is numerically equal to the center of mass's offset from the rotation axis, in microns. The allowable value comes from the balance quality grade and the speed: e_per = G/ω, then U_per = e_per·M, then you divide by the actual weight-mounting radius and split it between the planes. This residual doesn't convert directly into mm/s, because the relationship runs through the stiffness of the rotor-supports-foundation system.

Residual unbalance: what's left in the rotor after correction

Before balancing, the rotor carries an initial unbalance. You fit weights, vibration drops, and what's left is the difference between what was there and what you compensated. That difference is the residual unbalance. It's always tied to a specific correction plane: 800 g·mm in plane 1 plus 800 g·mm in plane 2 is nowhere near the same thing as 1600 g·mm in one.

Where you see this number depends on how you're balancing. A balancing machine displays residual unbalance directly in g·mm: the system is calibrated against a known mass at a known radius, and the rotor sits in the machine's own supports. In the field you see something else — the residual running-speed component in mm/s, and its phase. The running-speed component (instruments label it 1x) is vibration at the rotor's rotational frequency, and phase is the angle that ties it to the marker on the shaft. The software calculates the tolerance in g·mm from the grade and the speed, but converting the residual back into g·mm comes out as an estimate. We'll show how to make that estimate honestly below.

There's no such thing as a zero residual. Even if you did everything carefully, quite specific things will leave one behind.

The question isn't whether residual unbalance exists. It always does. The question is whether you know its size, and whether it falls within the tolerance you declared before the job started, not after.

g·mm, g·mm/kg and microns: three views of one residual

Unbalance U in g·mm is the product of the unbalanced mass and the radius, and it has a direction: magnitude and angle. 10 g at a 200 mm radius and 20 g at a 100 mm radius both give the same 2000 g·mm and the same centrifugal force. That's why a mass without a radius means nothing, and a report without a radius can't be read.

Specific residual unbalance in g·mm/kg is U divided by the mass of the rotating part. This quantity is needed because a single tolerance can't fit both a 12 kg impeller and a 1200 kg drum. And it has a nice property: g·mm/kg is numerically equal to the center of mass's eccentricity, in microns. 40 g·mm/kg means the center of mass is offset from the rotation axis by 40 microns.

The balance quality grade G ties the specific quantity to speed: the grade number is e_per multiplied by angular velocity, in mm/s. The logic is simple. Centrifugal force grows with the square of speed, so the same eccentricity is harmless on a slow-running machine and destructive on a fast one. That gives you the main practical consequence: the higher the speed, the less mass you're allowed to leave at the same radius. Look at the spread in the table below — it covers four orders of magnitude.

RotorMass, kgSpeed, rpmGradee_per, µmU_per, g·mmAllowable mass at the radius, g
Grinding spindle812,000G10.860.1 (r 60 mm)
Pump impeller122900G6.3212501.7 (r 150 mm)
Medium-sized fan401450G6.34016005.3 (r 300 mm)
Slow-running drum1200300G16510612,0001200 (r 500 mm)

The numbers are rounded and given as an illustration of the order of magnitude, not as a ready-made acceptance criterion. The grades and the allowable residual unbalance are set by ISO 21940-11 (formerly ISO 1940-1); check the applicable part and edition for your specific machine. There's a separate article on how to choose the grade itself, and why you shouldn't ask for G1 “just in case.”

Sources: ISO 21940-11:2016

How the grade and the speed give you an allowable mass: a rounded example

Let's take a typical field case: a fan wheel between bearings, two correction planes, grade G6.3. The numbers below are rounded so you can see the logic, not so you can copy them straight into a contract.

  1. 01

    Gather four numbers

    The mass of the rotating part, M = 40 kg (not the whole machine — the rotor itself), running speed n = 1450 rpm, grade G6.3, and the actual weight-mounting radius, r = 300 mm at each plane. Take the mass from the datasheet or weigh it, measure the radius with a tape on site.

  2. 02

    Convert speed to angular velocity

    ω = 2π·n/60. For 1450 rpm that comes out to roughly 152 rad/s. The easy mental-math version: divide the rpm by 9.55.

  3. 03

    Get the specific tolerance

    e_per = G/ω = 6.3/152 ≈ 0.04 mm. That's 40 microns, which is also 40 g·mm/kg. That's how much you're allowed to leave per kilogram of rotor mass.

  4. 04

    Move to the full tolerance

    U_per = e_per·M = 40 g·mm/kg × 40 kg = 1600 g·mm for the whole rotor. You can already write this number into the report, but you still can't work with it on the rotor: it's not clear how many grams that is, or where to apply them.

  5. 05

    Convert to grams and split between planes

    1600 g·mm at a 300 mm radius is 5.3 g for the whole rotor. For a symmetric rotor between bearings, the tolerance gets split in half: 800 g·mm and about 2.7 g at each plane. Fit the weights at a 150 mm radius instead, and the same 800 g·mm turns into 5.3 g per plane.

The exact calculation, the mapping between grades and machine types, and the rule for splitting the tolerance between planes are all set by the applicable edition of ISO 21940-11. In practice this calculation is usually done by the software: you enter the rotor's mass, speed, radius and grade, and get a tolerance in g·mm and in grams on screen, which the instrument then compares the result against.

Sources: ISO 21940-11:2016

Why the tolerance is split between supports and planes

The standard tolerance applies to the rotor as a whole, but it gets checked at the support, that is, the bearing, planes. That's why two numbers show up in the report instead of one.

For a symmetric rotor between bearings, the total tolerance is split in half, and the arithmetic ends quickly. For an overhung or asymmetric rotor, the split is calculated from how the center of mass distributes the static load across the supports: the support closer to the center of mass gets the larger share. The standard caps the extreme shares here, at roughly 0.3 and 0.7 of the total tolerance, so one plane doesn't take almost everything. Get the exact wording from the applicable edition.

Now, about the temptation to give the whole tolerance to one plane. The first problem is arithmetic: 1600 g·mm at one end of the rotor instead of 800 plus 800 loads the near bearing support almost twice as hard, even though the formal total is the same. The second problem is worse and less obvious. Put 800 g·mm in plane 1 and 800 g·mm in plane 2 at 180°, and you get zero net offset of the center of mass. The calculated eccentricity is zero, the specific unbalance is zero, and yet a couple of forces lives at the supports, growing with the distance between the planes. Checking against a single overall number will let a rotor like this pass, and the machine will shake.

And one more mix-up that costs rework. Correction planes aren't support planes. If you're fitting weights somewhere other than where the tolerance is specified, the numbers need to be recalculated through lever arms, and the instrument can do that on its own. Always write in the report exactly which plane each number belongs to.

For a disc-shaped rotor it's simpler: a single correction plane legitimately takes the whole tolerance, and the splitting rule only applies to two-plane balancing. How to choose the number of planes from the L/D ratio is covered in a separate article.

Sources: ISO 21940-11:2016

Eccentricity in microns: why it's convenient on large rotors

g·mm/kg and microns are the same number, and the second form is often more useful. It doesn't depend on mass, so you can put a 12 kg wheel and a 1200 kg drum on the same scale and immediately see which one demands finer work.

On large rotors, the difference in perception is especially sharp. A line reading “80,000 g·mm” sounds like a huge margin, and the temptation is to fit a hefty weight. Convert it: for a 2 t rotor at 1500 rpm and grade G6.3, that's the same 40 microns of eccentricity, and at an 800 mm radius, only around 100 g. Microns immediately ask the right question: is your rotor's geometry and fit actually tighter than forty microns? If shaft runout or the fit is looser than the tolerance, you'll be chasing your own geometry with weights.

On slow-running machines the picture is reversed. For a 1200 kg drum at 300 rpm, grade G16 allows 510 microns — half a millimeter of center-of-mass offset. Here the tolerance is wider than typical runout, and the right first step isn't weights, it's inspection: the fit, the welds, product build-up, the condition of the shell. Often, after cleaning and re-tightening, there's almost nothing left to balance.

One caveat, without which microns are misleading. Eccentricity only describes the static part of the unbalance. Couple unbalance can't be expressed in microns at all, and needs two planes and two numbers.

Why residual unbalance doesn't convert into mm/s

The same residual in g·mm produces different vibration on different machines, and the difference can be several-fold. There's no conversion factor that works for everyone, because in mm/s you're not measuring the unbalance, you're measuring the system's response to it. Here's what that response depends on.

The consequence cuts both ways. Your 0.7 mm/s at the support doesn't confirm grade G6.3, and hitting grade G6.3 doesn't guarantee landing in the best zone under the overall vibration-level limits. These are different tolerances, different units and different control points; we covered them together in a separate article on the three tolerances in balancing.

Support and foundation stiffness

The same force produces a small support displacement on a rigid foundation, and a noticeably larger one on a compliant frame. The response is set by the whole chain: rotor, bearings, supports, frame, foundation.

Closeness to resonance

If running speed approaches the structure's natural frequency, the amplitude grows several-fold for the same residual. Here, mm/s is telling you about resonance, not about the quality of the balancing.

Mass ratio

A light rotor on a heavy bed rocks it weakly. The same residual in a light housing with a belt drive will give noticeably more mm/s.

Measurement point and direction

Horizontal and vertical at the same bearing support can differ several-fold. A number without a stated point and direction can't be compared to anything.

Not just unbalance

The running-speed frequency isn't home to just one defect. A bent shaft, misalignment, looseness and electrical causes all contribute to 1x too, and that contribution isn't removed by weights.

Estimating the residual in the field: the influence coefficient as an exchange rate

You don't have a calibrated machine on site, but there is one honest way to get the residual in g·mm. The influence coefficient the instrument calculated on the trial run is exactly a local exchange rate between g·mm and mm/s: for this machine, this speed, this point and this direction. Take it and work backward.

  1. 01

    Calculate what you added

    A trial weight of 15 g at a 300 mm radius is 4500 g·mm. Use the actual mass from the scale and the actual radius, not what you meant to fit.

  2. 02

    See how much 1x changed

    Say the running-speed vector changed by 2.0 mm/s. Strictly, that's a vector difference accounting for phase, and the software calculates it on its own; for an order-of-magnitude estimate, the magnitudes alone are enough.

  3. 03

    Get the sensitivity

    2.0 mm/s over 4500 g·mm gives roughly 0.45 mm/s per 1000 g·mm. Write that number down: it'll come in useful on this machine next time too.

  4. 04

    Convert the residual

    The residual 1x after correction is 0.35 mm/s. Divide by the sensitivity and you get around 800 g·mm, referred to this plane. The tolerance from the example above is also 800 g·mm per plane. So you're standing right on the boundary, there's no margin, and one trim step is justified here.

This is an estimate, not acceptance under the standard. It works as long as the system is linear, the speed is the same, and the sensors haven't been moved, and it only applies to the plane, point and direction where you measured. If part of the residual 1x comes from something other than unbalance, the estimate will come out inflated. Confirming a grade under the standard means work on a balancing machine with a calibrated measuring system. There are separate articles with more detail on influence coefficients themselves and on trim balancing.

What limits the residual: weight increment and angle accuracy

A 2.7 g tolerance per plane and a set of weights in 5 g steps don't work together: you'll keep jumping back and forth across the boundary. Keep the correction increment no coarser than a third of the tolerance, or the last run turns into a lottery.

The angle costs you more than the mass does, and it's worth calculating this once. An angle error of Δ leaves behind roughly 2·sin(Δ/2) of the correction vector. Ten degrees is 17%. For a 4500 g·mm correction, a 10° error leaves about 780 g·mm behind — that is, your entire tolerance. A 5% mass error leaves only 5% behind, about 220 g·mm. The conclusion is simple: nail the angle first, then fine-tune the mass.

If you're removing metal, work out the volume in advance. Removing 3 g of steel takes about 0.38 cm³: an 8 mm diameter hole to a depth of roughly 7.5 mm. Drill it by eye, and you'll take off too much and end up with a new unbalance on the opposite side. There's a separate article on measuring the angle, fixed positions, and splitting a weight between neighboring blades.

What to write in the report

A year from now, someone else will be reading your report, quite possibly in a dispute over the cause of a failure. Without these lines, they won't be able to either repeat the measurement or argue the substance of it.

Write a separate line stating what the report doesn't confirm. On-site balancing in the rotor's own bearings gives you a residual 1x, a phase, and an estimate of the residual in g·mm. Confirming a grade under the standard means work on a balancing machine with a calibrated system. An honest statement heads off half of the future disputes with the client.

Sources: ISO 21940-11:2016

When there's no time to calculate the tolerance and defend the report

The Balanset-1A calculates the tolerance by G grades from the rotor's mass, speed and radius, and displays the residual running-speed component and phase at each support. The instrument stores influence coefficients for repeat visits, does trim balancing and recalculates weights for other planes, supports fixed positions and the drilling calculation, keeps an archive, and prints reports. There's a version without the case for building into machines and rigs.

The instruments are designed and manufactured by engineers who balance with them on site themselves, which is why our conversations about tolerances tend to be short and concrete. If you need not just a number but a decision on what to do with it, AXILINE comes to your site, takes the reading, balances in the rotor's own bearings, and puts together a report with the caveats that will hold up under scrutiny. Advisory support on tolerance calculation and reading a report is part of the job: send us the rotor's mass, speed, weight-mounting radius and the current mm/s at the supports, and we'll tell you what's realistically achievable on site, and what needs a machine.

Frequently asked questions

How do you calculate residual unbalance in g·mm?

The order is this: convert the speed to angular velocity, ω = 2π·n/60, divide the grade number by ω to get the specific tolerance e_per in mm (equal to microns, equal to g·mm/kg), multiply by the mass of the rotating part to get U_per in g·mm for the whole rotor, then divide by the actual weight-mounting radius and split it between the planes. That's how you calculate the tolerance. The residual itself is read directly in g·mm on a balancing machine, and estimated in the field through the influence coefficient obtained on a trial run.

What's the difference between g·mm and g·mm/kg?

g·mm is the total unbalance, the product of the unbalanced mass and the radius. g·mm/kg is the same thing divided by the rotor's mass — a specific quantity. It's needed because a single tolerance can't suit both a 12 kg wheel and a 1200 kg drum. A convenient feature: g·mm/kg is numerically equal to the center of mass's eccentricity, in microns.

Can you find residual unbalance from vibration in mm/s?

Strictly speaking, no. The same residual gives 0.5 mm/s on a rigid bed, and several mm/s on a compliant frame near resonance. You can estimate the value if you have an influence coefficient from a trial run: it acts as a local exchange rate between g·mm and mm/s, for this machine, this speed and this point. That's an estimate, not acceptance under the standard.

Why can't you give the whole tolerance to one plane?

For two reasons. First, putting the whole tolerance at one end of the rotor loads the near bearing support almost twice as hard, even though the formal sum is the same. Second, two equal residuals in opposite phase form a couple: the calculated eccentricity comes out at zero, while the supports shake, and it gets worse the further apart the planes are. Checking against a single overall number will let a rotor like this pass.

Is zero residual unbalance ever achieved?

No. It's left behind by the discreteness of the weights, error in the angle and radius, measurement error, and the part of the running-speed vibration that doesn't come from unbalance. On top of that, the rotor itself changes from heating, deposits and erosion. The question isn't whether a residual exists, it's whether you know its size and whether it falls within the tolerance declared before the work started.

What residual unbalance counts as normal for a fan?

Calculate it from the grade and the speed, not from a number you're used to. For a 40 kg wheel at 1450 rpm and grade G6.3, the tolerance works out to around 40 microns of eccentricity, roughly 1600 g·mm for the whole rotor, about 800 g·mm for each of the two planes, and on the order of 2.7 g at a 300 mm radius. Double the speed, and the allowable mass changes by half in the other direction.

Related content

Vibration Measurement Units: mm/s, g, µm, and What RMS Means

Vibration is described by three quantities: displacement in µm (usually peak-to-peak), velocity in mm/s (usually RMS — root-mean-square value), and acceleration in m/s² or g. Displacement works at low frequencies and on shafts, velocity gives a universal condition assessment for housings over the 10–1000 Hz band, and acceleration shows up high frequencies, bearings and impacts. Converting between quantities is only possible for a single component at a known frequency, and the 1.41 factor between RMS and peak only holds for a sine wave. That's why a figure with no stated quantity, amplitude type, frequency band and measurement point simply has nothing to be compared against.

Open page

Savings from balancing without removal: how to calculate the full cost and avoid overpaying in downtime

On-site balancing wins not on the price of the work itself, but on what it doesn't include: disassembly, rigging, transport, reassembly, shaft alignment after reassembly, and calendar days of downtime. Calculate both options against the same boundary: from the moment the machine stops to the moment it's back within tolerance and running again. In the overwhelming majority of cases, the outcome comes down to two line items: your hourly downtime rate, and whether you have a standby unit. Removal stays cheaper wherever a weight physically can't be fitted on site, where the rotor is flexible, where the geometry is damaged, or where acceptance requires a report against a balance quality grade G.

Open page

On-site balancing of a forest mulcher rotor, without removing it from the host machine

Yes, we balance forest mulcher rotors on site, in their own bearing housings, without removing the drum. The conditions: the tooth set is complete and matched, the drum has been washed clean of soil and wood pulp, the drive holds a stable speed, and the drum ends are reachable once the flap guard or access ports are opened. If the drum is bent from an impact or the teeth are inconsistent, we'll say so first, because weights won't fix that.

Open page

Describe your equipment and the problem

We'll answer your questions, clarify the details, and let you know what's needed for an estimate and a visit.

Submit a Request WhatsApp Pricing