---
title: "Air helps hold up the object on a balance"
description: "Follow a 100 g sample to see how air changes a balance reading and why replacing the sample on the same pan removes an arm-length error."
canonical_url: "https://sandboxphysics.com/stories/instruments/precision-balance/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/instruments/precision-balance/"
last_modified: "2026-10-28"
---

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Inside the instrument

# Air helps hold up the object on a balance

Air pushes upward on the object on a balance. It also pushes upward on the reference weights, but the two objects often displace different volumes. A precise mass comparison has to account for the part of the weight that the air supports.

Sandbox Physics · 2026-10-28

## The larger object gets more help from the air

Picture two objects with the same mass of 100 grams: a compact steel weight and a much larger sample. Air presses on every surface of both objects. The pressure is slightly greater underneath than above, so the pressures leave a small upward force. This is buoyancy, the same effect that helps a block float in water.

The upward force equals the weight of the air displaced by the object. A larger volume displaces more air. The effect is easy to overlook in everyday weighing, but a laboratory balance can resolve changes much smaller than a grain of rice. Air buoyancy can then shift several digits.

A balance compares the downward forces delivered to its pans. Of two equal masses, the larger-volume object presses down less in air. When that object is compared with compact steel weights, its indicated mass comes out a little low. Here, indicated mass means the result of comparing the object with known true-mass standards in the air present during the weighing.

Sources: [1]

## A 100 gram sample can read about 105 milligrams low

Choose round numbers for a teaching example. Give the sample a mass of 100 grams and a density of 1,000 kilograms per cubic metre. It occupies 100 cubic centimetres. Set the density of the steel standard to 8,000 kilograms per cubic metre; the same mass occupies only 12.5 cubic centimetres. Take the air density as 1.2 kilograms per cubic metre.

The sample displaces 120 milligrams of air. A 100 gram steel standard displaces just 15 milligrams. The difference is about 105 milligrams. The amount of steel that balances the sample has a mass of about 99.895 grams. Correcting for buoyancy restores the original 100 gram sample mass. This worked example uses calculated values for the stated teaching densities.

The correction uses the densities of the sample, the reference weights and the air. The relation below converts the indication into sample mass.

$$
m_x=m_{\mathrm{ind}}\frac{1-\rho_a/\rho_r}{1-\rho_a/\rho_x}
$$

The left side is sample mass. The right side uses indicated mass, air density, reference density and sample density, respectively. Equal sample and reference densities give the same fractional buoyancy, so the correction factor becomes one.

Sources: [1]

## Put the sample and standard on the same pan

There is another small difference to consider: the two arms of a balance may have slightly different lengths. A downward force turns the beam according to its distance from the pivot. A slightly longer left arm gives the same load more turning effect on that side. A direct comparison can therefore include a mass error from the geometry.

Substitution weighing changes the comparison. Leave the same counterpoise on the right pan. First put the sample on the left pan and read the detector. Then replace it with a standard and read again. Both weighings use the same suspension point and arm. Subtracting the signals removes the fixed right-pan contribution. A small known calibration weight establishes the sensitivity needed to convert that signal difference into a mass difference.

The detector zero can also drift with time. One useful sequence weighs the sample, then the standard, then the sample again, with equal time intervals. If the zero changes at a constant rate during the sequence, the mean of the two sample readings has the zero appropriate to the middle reading. This cancels linear drift. The buoyancy difference between sample and standard still needs its own correction.

Sources: [1]

## An electronic balance holds the load with a current

An electronic balance can replace the opposite weight with a coil in a magnetic field. Current through the coil produces a force. When a sample pushes the pan down, a sensor detects the displacement and a control circuit adjusts the current to bring the beam back. The current needed to hold that position tells the instrument the balancing force. This is electromagnetic force compensation.

Air continues to support part of the sample even when the readout is electronic. Calibration weights establish the relation between current and indicated mass; the sample may displace a different amount of air from those weights. A more precise weighing needs better knowledge of the densities and the surrounding air. That is the practical limit of the correction: uncertainty in sample density also leaves uncertainty in the corrected mass.

Sources: [2] [1]

## [Compare readings of the same sample in air](https://sandboxphysics.com/labs/balance-beam/)

Compare Air lifts the sample with Equal densities, then weigh the sample with Unequal arms and Replace on the same pan.

This quasi-static model derives mass indications from moments and simulated detector signals, with stored zero, sensitivity and substitution observations.

## Sources and further reading

1. [NIST SOP 2, Applying Air Buoyancy Corrections (2019)](https://www.nist.gov/system/files/documents/2019/05/13/sop-2-applying-air-buoyancy-20190506.pdf)

   Equation 3 gives true mass from a reference and a buoyancy-corrected difference; section 5.1 defines same-pan substitution. Equation 5 distinguishes conventional mass.

2. [Arthur Reichmuth, METTLER TOLEDO, Adverse Influences and Their Prevention in Weighing (2001)](https://www.mt.com/mt_ext_files/Editorial/Generic/2/Weigh_Uncertain_Number1_0x0003d6750003db6700091746_files/Adverse_Influences.pdf)

   Printed pages 1–2 explain electrodynamic force compensation; page 6 defines sensitivity as the slope between two load/readout points.

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