# Concepts in Mass Metrology

## Introduction

There is a misconception that a weighing instrument measures weight. In fact, most accurate weighing instruments measure mass. Concepts in mass metrology states that an object should “weigh” the same, no matter on which weighing instrument and at what location it is being measured. A 10 gm coin of gold must “weigh” 10 gm in Delhi or in Mumbai. Otherwise we could buy a 10 gm coin of gold in Delhi and sell it in Mumbai for a higher price.

Weighing Instruments that are used in trade or commerce are tested by competent authorities for their accuracy in measuring force. Their accuracy is in measuring mass. In order to measure mass, there must be a method to compensate for the acceleration of gravity at a particular location where a spring or load cell scale is used. This process is called “calibration of the weighing instrument”.

To calibrate a weighing instrument, a standard calibration mass is placed on the pan of the weighing instrument. The weighing instrument is then adjusted until it reads the appropriate standard weight. The standard weight is the weight the standard calibrating mass that would have a standard value of the acceleration due to gravity (g). This standard value of g is 9.80665 m/s2.

Once the calibration is performed at a specific location, we have to automatically correct for the acceleration of gravity at that location. No mathematical manipulation of the data will be necessary to correct for latitude or longitude. However, for precise measurements, it is necessary to calibrate the scale just before making a measurement to correct for the variation in gravitational attraction due to the Sun or moon, since this varies with the time of day.

#### Instruments Calibrated At One Location But Used Elsewhere

Problems occur when a load cell type scale is calibrated at one location and then moved to another location to weigh an object. For large scale capacities, it is often not possible to carry a calibration weight to the new site. It may be because this weight is not available, or due to problems of shipping this calibration mass that weighs too much. Therefore, it is necessary to correct for the change in the acceleration of gravity between the site where the scale was calibrated, and the site where the object needs to be measured. This correction may be determined accurately by the following procedure. This is done using an additional small scale and calibration mass.

Let’s say that you have a balance with maximum capacity of 5,000 kg that has been calibrated at its manufacturing factory at Avery India Ballabagarh. This is done using a primary standard set of weights that were calibrated at NPL New Delhi. Now supposing this balance is used at Bangalore to weigh a spacecraft, (where it is not possible to bring the primary standard calibration weights). Now the balance needs to be re-calibrated at this new location

For correcting the balance indication at Bangalore we need a small precision balance and a precisely calibrated standard weight. This precision balance may be a balance of 1 kg capacity and the standard 1 kg calibrated weight.

While in Ballabagarh, to calibrate 1 kg, a small balance that reads exactly 1.000 kilogram is used when the 1 kg standard weight is measured. From this point on we may adjust the zero of this balance but not the entire span.

Now to calibrate this 5000 kg balance, a small 1 kg balance and 1 kg standard weight are brought to Bangalore. On arrival at Bangalore, place the 1 kg weight on the small balance and note the balance indication.  If the indicated mass value is say 0.16% less (i.e. 0.9984 kg), then the acceleration due to gravity is 0.16% lower at this new location. Therefore the measured weight of the large object must be multiplied by 1.0016 to give the correct weight for this change in the acceleration of gravity.

It may be possible in some cases that there is no means of calibrating a weighing scale. It may have been calibrated at the manufacturing unit only. In that case to make a correct measurement we need to know the value of the acceleration due to gravity at the location where the weighing scale was calibrated. Also at the location where we are going to make the measurements. Although anomalies in the earth’s crust will affect local gravity, gravity will mainly differ with latitude and altitude. Gravity at a particular location can be calculated with a precision of about 0.002 % using the following formula:

Where   L is latitude in degrees;    H is altitude in meters;  g is expressed in m/s2: The true mass M can be calculated by the following formula:

M = gc ( Measured mass) / gm

where:

gc   =   acceleration due to  gravity where the weighing scale  was calibrated

gm =  acceleration due to gravity where measurement was made using   weighing scale

#### Concepts In Mass Metrology: Factors affecting mass measurement

There are six main factors that normally affect the measured value of an object using a weighing instrument. They are:

Air Buoyancy

Large objects (low density) weigh less than small objects (high density) of equal mass, because of the buoyancy of an object as it floats in a sea of air. This effect is much more significant than most of us realize. The magnitude of air buoyancy varies due to change in air density, which is affected by the weather and is a function of the atmospheric pressure, relative humidity and temperature. This effect may be compensated by calculating the air density and determining the magnitude of air buoyancy effect for the said volume of the object being weighed. This is discussed in more detail below.

Tidal Variation

An object on the surface of the earth is attracted to every celestial body, but most of these masses are too far away to have any significance on their weight. But the Sun and the moon have a significant effect. If we have a weighing scale whose accuracy is 0.003 % or better, we can notice the variation in the weight of an object as a function of the time of day. This effect is most pronounced during spring and fall when the sun and moon align. This produces the “neap tides” that often cause flooding of marinas at these critical dates.

Measured weight varies slightly throughout the day due to changing attraction to the Sun and moon

As we can see from the above figure, the gravitational mass attraction to the Sun and moon varies during the year. It may reach 0.003% of the acceleration of earth’s gravity at certain dates during the year when the Sun and moon align.

Condensation

In areas with high humidity, sudden changes in temperature will produce condensation on the object being measured, which adds to the weight of the object. This usually occurs when an object is brought from an air conditioned ambience to a non-air-conditioned one. Condensation can be minimized by allowing an object’s temperature to equalize its surrounding before weighing it.

Electrostatic attraction

It’s remarkable how dramatic this effect can be if you are weighing a large lightweight object such as a Mylar decoy reentry cone. If the object is covered by a clear plastic draft shield during measurement, the attraction between object and shield can differ by as much as 2% of the weight of the object.

Magnetic attraction

If the object that needs to be weighed contains permanent magnets, there will be an attraction to any magnetic material near or on the scale. There will also be a small attraction force to the magnetic north of the earth. Often an object has been magnetized by the magnetic chuck used in the machining of the object. It may be necessary to demagnetize the object before weighing it. Generally you can detect magnetic errors by repositioning the object on the weighing pan of the scale. If the readings are very sensitive to position on the pan, then the problem may be magnetic attraction. But it might also be the corner loading error of the scale. A Boy Scout compass is also a good way to test for magnetic attraction.

Drafts or air currents

Generally, drafts or air currents effect will be quite obvious, since drafts introduce a random variation in the readings. However, there are instances where drafts can produce a relatively steady downward or upward force. For example, if sunlight heats up an object, then the updraft from the surface will produce an upward force, reducing the weight of the measured object. Conversely, if an object is brought in from an unheated storage area to be weighed, the cooler object will cause a downdraft, thereby increasing the measured weight. These effects can be minimized by making sure that the object is at the same temperature as the surrounding and by avoiding direct sunlight or bright lights.

Note: This method does not correct for differences in air buoyancy at the two locations. See later discussion for more sophisticated method to correct for both gravity and buoyancy.

Concept of Mass

The property of a body due to which it requires force to change its state of motion is called inertia, and mass is the numerical measure of this property. In physical terms, mass is a property of matter existing in a body.

And this exerts

• Inertia to change its condition of motion – Inertial Mass
• Attraction of other bodies – Gravitational Mass.

According to the Principle of Equivalence, under experimental conditions, the inertial mass is equal to the gravitational mass.

In order to measure a physical quantity, a reference quantity and a measuring instrument are required. This reference quantity is called the unit of that quantity.

According to Maxwell “every physical quantity can be expressed as the product of a number and a unit. Where the unit is a selected reference quantity in terms of which all quantities of the same kind can be expressed.”

The Unit of Mass

The fundamental unit of mass is the international kilogram and in the International System of Units (SI) the unit of mass, the kilogram is defined as:

The kilogram is the unit of mass and it is equal to the mass of the international prototype of the kilogram.

This kilogram is kept at the Pavilion de Britoil, which is the location of the International Bureau of Weights & Measures (BIPM) at Paris in France.

This international prototype of the kilogram is a cylinder made of an alloy of 90% platinum and 10% iridium. Its height is (39 mm) and it is equal to its diameter so that its surface area in minimum. Its density is of the order of 21,500 kg/m3.

Originally this platinum-iridium artifact was designed to have the mass of 1 cubic decimeter of pure water at the temperature of the maximum density of water, 4 0C. Later on, determination of the density of pure water removing air at 4 0C, under standard atmospheric pressure (101.325 Pascal), yielded the present value of 1.000028 cubic decimeter for the volume of one kilogram of water

Measurement of Mass

The mass of a body is determined by weighing i.e. the comparing the mass of an object to be weighed with the mass of a standard weight of known mass on a balance after allowing for necessary corrections, particularly, buoyancy correction.

If Mw is the mass of the test weight which is compared with a standard S of known mass Ms, on a suitable balance. And if Dm is the difference in mass values of the two weights as observed by the balance, then the condition of equilibrium is:  Mw. (1-ρaw) = Ms. (1-ρas) + Dm      ………..    (1)

Where ρw and ρs are densities of the material of the test weight W and the standard weight S respectively. ρa is the density of air during the comparison.

From the above relation mass of the test weight is: –

Mw = (MS + Dm) + (Vow -Vs). ρa               ……… …(2)

Where Vw= Mww and Vs= Mss are volumes of the weights W and S respectively

Calibration of a weight

Calibration of a weight involves assigning a mass value to the weight by comparing it against a reference standard of known mass whose nominal mass is equal to that of the weight under calibration.

This comparison of the weights is done on a suitable weighing instrument. The indication of the weighing instrument is used only to determine the difference Dm between the reference standard and the test weight.

The comparison of two weights is always carried out according to the substitution method using ABBA or ABA weighing cycle to eliminate linear drift.

## Substitution Weighing:

The Reference Weight of known mass MR and the Test Weight of unknown mass MT are put one after another on the same weighing pan and their weighing difference Dm is determined as:

Dm     = (MT    –  MR ) (3)

Weighing  Cycles

For a single comparison, three internationally accepted weighing cycles are employed to eliminate linear drift of the Weighing Instrument. These are:

1. ABBA Weighing Cycle
2. ABA Weighing Cycle
3. AB1B2 … BnA Weighing Cycle

A     :   Represents Weight of the Reference Weight

B     :   Represents Weight of the Test Weight

ABBA  &  ABA  are used for  calibration of  E and F Class  of Weights and  AB1, B2,  BnA  for  M  Class  of  Weights

#### ABBA Weighing Cycle

 Weight on Pan Balance Indication First Diff.(B – A) Second Diff.(B – A) Mean Diff.(B – A) = Dm A I1 (I2 – I1) (I3 – I4) [(I2 – I1) + (I3 – I4)]/2 B I2 B I3 B  =  A + [(I2 – I1) + (I3 – I4)] / 2 A I3

#### ABA Weighing Cycle

 Weight on Pan Balance Indication First Diff.(B – A) Second Diff.(B – A) Mean Diff.(B – A) = Dm A I1 (I2 – I1) (I2 – I3) [(I2 – (I1 + I3) /2] B I2 A I3 B  =  A + [(I2 – (I1 + I3) /2]

Remark :

1.  The time interval between weighing’s should be kept constant.

2. The above sequence may be repeated n times ( n  ³ 1 ).

#### AB1…BnA Weighing Cycle

 Weight on Pan A B1 B2 B3 B4 B5 A B6 B7 B8 ….……. Balance Indication Ia1 Ia1 Ia1 Ia1 Ia1 Ia1 Ia1 Ia1 Ia1 Ia1 ….…….

( B1 – A )   = [Ib1 – ( Ia1+ I a2 ) / 2 ]  = Dm1

( B2 – A )    =    [Ib2 – ( Ia1 + Ia2  ) / 2 ]=Dm2

( B3 – A )   =     [Ib3 – ( Ia1 + Ia2 ) / 2 ]=Dm3

## Mass Measuring Instruments

For high accuracy mass determination, analytical balances and comparator balances are utilized.  An analytical balance is characterized by a small-scale interval (d) in comparison to the maximum capacity (Max). The ratio d/Max generally is lesser than 10-5 and the maximum capacity usually does not exceed 10 kg.  Analytical balances are grouped into weighing instrument of special accuracy (class I) and of high accuracy (class II). For weighing instruments of special accuracy, the following classification is generally used:

Macro balance                Max.  >  100 g     d < 10-5 Max   (e.g. d = 0.1 mg)

Half-Micro balance          Max.  £  100 g     d < 10-5 Max   (e.g. d = 0.01 mg)

Micro balance                  Max.  £    50 g     d < 10-5 Max   (e.g. d = 0.001 mg)

Ultra micro balance       Max   £      5 g     d < 10-6 Max   (e.g. d = 0.0001 mg)

For weighing instruments higher than these, the term comparator balance or mass comparator is used. The mass comparator allows only differential weighing whereas analytical balances can be used for both differential weighing as well as for proportional weighing.  Differential weighing stands for mass comparison using mass standards or other suitable objects. Whereas proportional weighing is the simple process of mass comparison without having recourse of reference standards.

Air Buoyancy and Concept of Conventional Mass

Like water, air has mass. A certain volume of air will have a certain mass depending on the density of air. And like water, displaced air results in a buoyant or lifting force. An object in water experiences an upward force equal to the weight of the water displaced by the object. Likewise, an object in air experiences an upward force equal to the weight of the air displaced by the object.

If an object is placed at rest on a balance in vacuum. It will then experience a downward gravimetric force equal to the mass (or true mass) of the object times the acceleration due to gravity. The object will press down on the balance with this force and the balance will indicate the object’s true mass.

Now imagine this same object at rest on a balance in air. The object will experience the downward gravimetric force as well as the upward force of air buoyancy.

The resulting force with which the object presses down on the balance is less than the gravimetric force on the object by an amount equal to the air buoyant force. The balance does not know whether the object is in a vacuum or in air. And thus it will indicate that the object is less massive than before. The mass of the object will appear to be less

The magnitude of the mass by which the object appears to be less is equal to the volume of the object multiplied by the density of air displaced by it and is called buoyancy correction. When a weight of mass MT and volume VT is compared against a standard weight of mass MS and Volume VS in air of density ra then the buoyancy correction term will be expressed as (VT – VR).ra, see the second term in eq.(2) above.

Imagine this same object on an equal arm balance in vacuum. It is perfectly balanced by a standard of 8.0 g/cm3 density. Thus, the standard and the object have the same “true mass”

Conventional Mass

Now imagine that same object and standard balance at 20 0C with air of density 1.2 mg/cm3. If the object has a density different from 8.0 g/ cm3 then for it to have the same “true mass” as the standard, the volume has to be the same. Thus, the lifting force on the object and that on the standard are not the same and the object will no longer appear to have the same mass as the standard

Both the object and the standard will experience the downward force of gravity equal to their true mass times the acceleration due to gravity. However, since their volumes are different, the amount of air displaced by the object is different from that displacement by the standard, and hence the weight of air displaced by the object is different from the weight of air displaced by the standard.

The mass that the object now is called the “conventional mass“.

According to OIML Recommendation No.R.33 (now it is Document No. D-28), the Conventional Mass is defined as follows:

For a weight at 20 °C, the conventional mass is the mass of a reference weight of a density of 8,000 kg/m3 that it balances in air of density 1.2 kg/m3.

If the mass of a weight is known then the conventional mass can be calculated according to this definition using the formula

MT = MC + (VS – VT) x 1.2   (5)

Where   VS = Volume of the reference weight of density 8000 kg/m3

VT = Volume of the test weight in m3

1.2 is the air density in kg/m3

In terms of the densities

MC = MT (1-ρ0/ρ)/(1-ρ0/ρs)            (6)

If the mass of a weight is known, then its conventional mass can be calculated according to the above formula.

This definition shows that the conventional mass mc of a weight made of stainless steel which has a density ρ  close to ρ = 8000 kg/m3 deviates from its mass m by only a small amount.  For other materials the relative deviation of mc from m, ranges from about – 3 x10-4, for aluminum to +10-4 for platinum. Nevertheless, the difference between the conventional mass value of weight and its true mass can be significant even for weights made of stainless steel.

Method to calculate buoyancy, using calculated air density

•  Calibrate the scale and adjust its output to read the exact value of the calibration weight.
• Determine the density of the calibration weight (brass = 8570 kg/m3; stainless steel = 7870 kg/m3; cast iron = 7100 kg/m3)
• Measure the weight of the object.
• Determine the volume of the air displaced by the object (often not an easy task).
• Determine the approximate density of the object by dividing the measured weight by the volume of air displaced.
• Determine the density of the air by using the following formula:

Where

p = atmospheric pressure in mbar (hPa)

h = relative humidity in %

t = temperature in degrees Celsius

d = air density in kg/m3

Calculate the buoyancy correction using formula:

where :

s = density of calibration weight in kg/m3

u = density of unknown in kg/m3

d = density of air in kg/m3

Example

Air buoyancy correction using following data :

p = atmospheric pressure = 1000 mbar ;

h = relative humidity = 75% ;   t = temperature air = 23 oC

s = density of calibration weight   = 8400 kg/m3

u = density of unknown in            =   500 kg/m3

Solution

Air density is:

Buoyancy correction = (1-1.167/8400) / (1-1.167/500)  = 1.0022

True mass = Measured mass * 1.0022

The measured weight will be 0.22% lesser if no correction is made for air buoyancy.

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