Editing Conversion of units (section)

== Factor–label method ==
{{further|Dimensional analysis}}

The '''factor–label method''', also known as the '''unit–factor method''' or the '''unity bracket method''',<ref name="BodóJones2013">{{cite book |author1=Béla Bodó |url=https://books.google.com/books?id=P46291mjqAsC&q=conversi%C3%B3n+walshaw+methode&pg=SA9-PA129 |title=Introduction to Soil Mechanics |author2=Colin Jones |date=26 June 2013 |publisher=John Wiley & Sons |isbn=978-1-118-55388-6 |pages=9–}}</ref> is a widely used technique for unit conversions that uses the rules of [[algebra]].<ref>{{Cite book |last=Goldberg |first=David |title=Fundamentals of Chemistry |publisher=McGraw-Hill |year=2006 |isbn=978-0-07-322104-5 |edition=5th}}</ref><ref>{{Cite book |last=Ogden |first=James |title=The Handbook of Chemical Engineering |publisher=Research & Education Association |year=1999 |isbn=978-0-87891-982-6}}</ref><ref>{{Cite web |title=Dimensional Analysis or the Factor Label Method |url=http://www.kentchemistry.com/links/Measurements/dimensionalanalysis.htm |website=Mr Kent's Chemistry Page}}</ref>

The factor–label method is the sequential application of conversion factors expressed as fractions and arranged so that any dimensional unit appearing in both the numerator and denominator of any of the fractions can be cancelled out until only the desired set of dimensional units is obtained. For example, 10 [[miles per hour]] can be converted to [[metres per second]] by using a sequence of conversion factors as shown below:
<math display="block"> \frac{\mathrm{10~\cancel{mi}}}{\mathrm{1~\cancel{h}}} \times \frac{\mathrm{1609.344~m}}{\mathrm{1~\cancel{mi}}} \times \frac{\mathrm{1~\cancel{h}}}{\mathrm{3600~s}} = \mathrm{4.4704~\frac{m}{s}}. </math>

Each conversion factor is chosen based on the relationship between one of the original units and one of the desired units (or some intermediary unit), before being rearranged to create a factor that cancels out the original unit. For example, as "mile" is the numerator in the original fraction and {{tmath|1= \mathrm{1~mi} = \mathrm{1609.344~m} }}, "mile" will need to be the denominator in the conversion factor. Dividing both sides of the equation by 1 mile yields {{tmath|1= \frac{\mathrm{1~mi} }{\mathrm{1~mi} } = \frac{\mathrm{1609.344~m} }{\mathrm{1~mi} } }}, which when simplified results in the dimensionless {{tmath|1= 1 = \frac{\mathrm{1609.344~m} }{\mathrm{1~mi} } }}. Because of the identity property of multiplication, multiplying any quantity (physical or not) by the dimensionless 1 does not change that quantity.<ref>{{cite web| title = Identity property of multiplication |url = http://www.basic-mathematics.com/identity-property-of-multiplication.html |access-date = 2015-09-09 }}</ref> Once this and the conversion factor for seconds per hour have been multiplied by the original fraction to cancel out the units ''mile'' and ''hour'', 10 miles per hour converts to 4.4704 metres per second.

As a more complex example, the [[concentration]] of [[nitrogen oxides]] ([[NOx|NO<sub>''x''</sub>]]) in the [[flue gas]] from an industrial [[Industrial furnace|furnace]] can be converted to a [[mass flow rate]] expressed in grams per hour (g/h) of NO<sub>''x''</sub> by using the following information as shown below:
; NO<sub>''x''</sub> concentration := 10 [[parts per notation|parts per million]] by volume = 10&nbsp;ppmv = 10 volumes/10<sup>6</sup> volumes
; NO<sub>''x''</sub> molar mass := 46&nbsp;kg/kmol = 46&nbsp;g/mol
; Flow rate of flue gas := 20 cubic metres per minute = 20&nbsp;m<sup>3</sup>/min
: The flue gas exits the furnace at 0&nbsp;°C temperature and 101.325&nbsp;kPa absolute pressure.
: The [[standard conditions of temperature and pressure#Molar volume of a gas|molar volume]] of a gas at 0&nbsp;°C temperature and 101.325&nbsp;kPa is 22.414&nbsp;m<sup>3</sup>/[[kmol]].

: <math chem="">
\frac{1000\ \ce{g\ NO}_x}{1 \cancel{\ce{kg\ NO}_x}} \times
\frac{46\ \cancel{\ce{kg\ NO}_x}}{1\ \cancel{\ce{kmol\ NO}_x}} \times
\frac{1\ \cancel{\ce{kmol\ NO}_x}}{22.414\ \cancel{\ce{m}^3\ \ce{NO}_x}} \times
\frac{10\ \cancel{\ce{m}^3\ \ce{NO}_x}}{10^6\ \cancel{\ce{m}^3\ \ce{gas}}} \times
\frac{20\ \cancel{\ce{m}^3\ \ce{gas}}}{1\ \cancel{\ce{minute}}} \times
\frac{60\ \cancel{\ce{minute}}}{1\ \ce{hour}} =
24.63\ \frac{\ce{g\ NO}_x}{\ce{hour}}
</math>

After cancelling any dimensional units that appear both in the numerators and the denominators of the fractions in the above equation, the NO<sub>''x''</sub> concentration of 10&nbsp;ppm<sub>v</sub> converts to mass flow rate of 24.63&nbsp;grams per hour.

=== Checking equations that involve dimensions ===
The factor–label method can also be used on any mathematical equation to check whether or not the dimensional units on the left hand side of the equation are the same as the dimensional units on the right hand side of the equation. Having the same units on both sides of an equation does not ensure that the equation is correct, but having different units on the two sides (when expressed in terms of base units) of an equation implies that the equation is wrong.

For example, check the [[Ideal gas law|universal gas law]] equation of {{nowrap|1=''PV'' = ''nRT''}}, when:
* the pressure ''P'' is in pascals (Pa)
* the volume ''V'' is in cubic metres (m<sup>3</sup>)
* the amount of substance ''n'' is in moles (mol)
* the [[universal gas constant]] ''R'' is 8.3145&nbsp;Pa⋅m<sup>3</sup>/(mol⋅K)
* the temperature ''T'' is in kelvins (K)

<math display="block">\mathrm{Pa{\cdot}m^3} = \frac{\cancel{\mathrm{mol}}}{1} \times
\frac{\mathrm{Pa{\cdot}m^3}}{\cancel{\mathrm{mol}}\ \cancel{\mathrm{K}}} \times \frac{\cancel{\mathrm{K}}}{1}
</math>

As can be seen, when the dimensional units appearing in the numerator and denominator of the equation's right hand side are cancelled out, both sides of the equation have the same dimensional units. Dimensional analysis can be used as a tool to construct equations that relate non-associated physico-chemical properties. The equations may reveal undiscovered or overlooked properties of matter, in the form of left-over dimensions – dimensional adjusters – that can then be assigned physical significance. It is important to point out that such 'mathematical manipulation' is neither without prior precedent, nor without considerable scientific significance. Indeed, the [[Planck constant]], a fundamental physical constant, was 'discovered' as a purely mathematical abstraction or representation that built on the [[Rayleigh–Jeans law]] for preventing the [[ultraviolet catastrophe]]. It was assigned and ascended to its quantum physical significance either in tandem or post mathematical dimensional adjustment – not earlier.

=== Limitations ===
The factor–label method can convert only unit quantities for which the units are in a linear relationship intersecting at 0 ([[Level of measurement#Ratio scale|ratio scale]] in Stevens's typology). Most conversions fit this paradigm. An example for which it cannot be used is the conversion between the [[Celsius scale]] and the [[Kelvin scale]] (or the [[Fahrenheit scale]]). Between degrees Celsius and kelvins, there is a constant difference rather than a constant ratio, while between degrees Celsius and degrees Fahrenheit there is neither a constant difference nor a constant ratio. There is, however, an [[affine transform]] ({{tmath|1= x \mapsto ax+b }}, rather than a [[linear transform]] {{tmath|1= x \mapsto ax }}) between them.

For example, the freezing point of water is 0&nbsp;°C and 32&nbsp;°F, and a 5&nbsp;°C change is the same as a 9&nbsp;°F change. Thus, to convert from units of Fahrenheit to units of Celsius, one subtracts 32&nbsp;°F (the offset from the point of reference), divides by 9&nbsp;°F and multiplies by 5&nbsp;°C (scales by the ratio of units), and adds 0&nbsp;°C (the offset from the point of reference). Reversing this yields the formula for obtaining a quantity in units of Celsius from units of Fahrenheit; one could have started with the equivalence between 100&nbsp;°C and 212&nbsp;°F, which yields the same formula.

Hence, to convert the numerical quantity value of a temperature ''T''[F] in degrees Fahrenheit to a numerical quantity value ''T''[C] in degrees Celsius, this formula may be used:
: ''T''[C] = (''T''[F] − 32) × 5/9.

To convert ''T''[C] in degrees Celsius to ''T''[F] in degrees Fahrenheit, this formula may be used:
: ''T''[F] = (''T''[C] × 9/5) + 32.

=== Example ===
Starting with:
: <math>Z = n_i \times [Z]_i</math>
replace the original unit {{tmath|1= [Z]_i }} with its meaning in terms of the desired unit {{tmath|1= [Z]_j }}, e.g. if {{tmath|1= [Z]_i = c_{ij} \times [Z]_j }}, then:
: <math>Z = n_i \times (c_{ij} \times [Z]_j) = (n_i \times c_{ij}) \times [Z]_j</math>

Now {{tmath|1= n_i }} and {{tmath|1= c_{ij} }} are both numerical values, so just calculate their product.

Or, which is just mathematically the same thing, multiply ''Z'' by unity, the product is still ''Z'':
: <math>Z = n_i \times [Z]_i \times ( c_{ij} \times [Z]_j/[Z]_i )</math>

For example, you have an expression for a physical value ''Z'' involving the unit ''feet per second'' ({{tmath|1= [Z]_i }}) and you want it in terms of the unit ''miles per hour'' ({{tmath|1= [Z]_j }}):

{{ordered list
|1= Find facts relating the original unit to the desired unit:
: 1 mile = 5280 feet and 1 hour = 3600 seconds

|2= Next use the above equations to construct a fraction that has a value of unity and that contains units such that, when it is multiplied with the original physical value, will cancel the original units:
: <math>1 = \frac{1\,\mathrm{mi}}{5280\,\mathrm{ft}}\quad \mathrm{and}\quad 1 = \frac{3600\,\mathrm{s}}{1\,\mathrm{h}}</math>

|3= Last, multiply the original expression of the physical value by the fraction, called a ''conversion factor'', to obtain the same physical value expressed in terms of a different unit. Note: since valid conversion factors are [[dimensionless]] and have a numerical value of [[one]], multiplying any physical quantity by such a conversion factor (which is 1) does not change that physical quantity.
: <math> 52.8\,\frac{\mathrm{ft}}{\mathrm{s}} =
 52.8\,\frac{\mathrm{ft}}{\mathrm{s}}
 \frac{1\,\mathrm{mi}}{5280\,\mathrm{ft}}
 \frac{3600\,\mathrm{s}}{1\,\mathrm{h}} =
 \frac {52.8 \times 3600}{5280}\,\mathrm{mi/h}
 = 36\,\mathrm{mi/h}</math>
}}

Or as an example using the metric system, you have a value of fuel economy in the unit ''litres per 100 kilometres'' and you want it in terms of the unit ''microlitres per metre'':
: <math> \mathrm{\frac{9\,\rm{L}}{100\,\rm{km}}} =
 \mathrm{\frac{9\,\rm{L}}{100\,\rm{km}}}
 \mathrm{\frac{1000000\,\rm{\mu L}}{1\,\rm{L}}}
 \mathrm{\frac{1\,\rm{km}}{1000\,\rm{m}}} =
 \frac {9 \times 1000000}{100 \times 1000}\,\mathrm{\mu L/m} =
 90\,\mathrm{\mu L/m}</math>