# Unit Conversion and Dimensional Analysis

Units can be converted from one type to another through dimensional analysis, a method of making comparisons between measurements that are part of different systems or use different units.
Measurements recorded using one set of units often must be converted to another set of units to facilitate comparison of measurements. This conversion is accomplished through dimensional analysis, a technique in which the relationship between units is tracked throughout a calculation. The first step in dimensional analysis is to define the conversion factor, which is the algebraic ratio between two different units. For example,
$1\;\rm L=1,\!000\;\rm{mL}$
Divide each side of the equation by 1 L.
$\frac{1\;\rm L}{1\;\rm L}=\frac{1,\!000\;\rm{mL}}{1\;\rm L}$
Canceling the unit L on the left side of the equation yields
$1=\frac{1,\!000\;\rm{mL}}{1\;\rm L}$
and this is the conversion factor for converting liters to milliliters. For example, a measurement of 2.341 L can be converted to milliliters by multiplying 2.341 L by the conversion factor. This is the same thing as multiplying by 1 because the conversion factor is equal to 1, as demonstrated above.
$2.341\;\rm L\times\left(\frac{1,\!000\;\rm{mL}}{1\;\rm L}\right)=\;2,\!341\;\rm{mL}$
Dimensional analysis can also be used when measurements are squared or cubed, but the conversion factor must be also be squared or cubed to cancel out all the unwanted units so that the two sides of the equation balance. For example, how many cubic centimeters are in 8 m3? The conversion factor is
$1=\left(\frac{100\;\rm{cm}}{1\;\rm m}\right)$
so dimensional analysis reveals:
$(8\;\rm m^3)\left(\frac{100\;\rm{cm}}{1\;\rm m}\right)^3=(8\;\rm m^3)\left(\frac{100\;\rm{cm}}{1\;\rm m}\right)\left(\frac{100\;\rm{cm}}{1\;\rm m}\right)\left(\frac{100\;\rm{cm}}{1\;\rm m}\right)=(8\;\rm m^3)\left(\frac{1\rm{,}000\rm{,}000\;\rm{cm}^3}{1\;\rm m^3}\right)$
The cubic meters cancel out, and the result is
$8\;\rm m^3=8\rm{,}000\rm{,}000\;\rm{cm}^3$
Now consider how to convert a measurement of 8.23 m3 to cubic centimeters. Notice that the measurement has three significant figures, so the measurement expressed in cubic centimeters should also have three significant figures. Following the steps described above, the calculation reveals
$(8.23\;\rm m^3)\left(\frac{1\rm{,}000\rm{,}000\;\rm{cm}^3}{1\;\rm m^3}\right)=8\rm{,}230\rm{,}000\;\rm{cm}^3$
Although the expression 8,230,000 cm3 does have three significant figures, a better way to represent it is with scientific notation. Scientific notation lowers the chance a measurement will be misinterpreted as being more or less precise than it actually is. Scientific notation also makes it easy to compare measurements of different orders of magnitude. In scientific notation, the previous example becomes
$(8.23\;\rm m^3)\left(\frac{1\times10^6\;\rm{cm}^3}{1\;\rm m^3}\right)=8.23\times10^6\;\rm{cm}^3$
The ability to use dimensional analysis to convert units is incredibly useful in every branch of science and engineering. Remember that there are derived SI units defined by their algebraic relationship to SI base units. Through dimensional analysis, it is possible to derive fundamental formulas. Consider the unit of force, the newton.
$1\;\rm N=1\;\frac{\rm{kg}\cdot\rm{m}}{\rm s^2}$
What properties do these base units measure? The kilogram is a measure of mass, and acceleration is measured in meters per second squared, m/s2. So the units of force can separated into units of mass and acceleration.
$\frac{\rm{kg}\cdot\rm{m}}{\rm s^2}=\left(\rm{kg}\right) \!\left(\frac{\rm m}{\rm s^2}\right)$
Unit errors are one of the most common errors in scientific calculations. If a solution to a problem does not make sense, the first thing to check is the units in the calculation. Although different units can be multiplied and divided, it is not possible to add and subtract measurements with different units. Imagine, for example, trying to add 12 g to 30 mL. It doesn't make sense. But 12 g/30 mL is a perfectly reasonable density.

Consider the conversion of atmospheric pressure on Earth of 14.7 pounds per square inch (psi) into pascals (Pa). The purpose of this conversion is to convert the measurement from a unit that is hard to compare to a unit that is more comparable.

The first step is to write the units in terms of their base units:
$1\;\rm{psi}=1\;\frac{\rm{lb}}{\rm{in}^2}\hspace {25pt}1\;\rm{Pa}=1\;\frac{\rm N}{\rm m^2}$
Next, define the conversion factors. Both pounds and newtons are units of force, and square inches and square meters are both units of area. So these are the unit pairs that will have conversion factors. The conversion from pounds to newtons on Earth is $1\;\rm{lb}=4.448\;\rm{N}$, so the conversion factor is
$1=\frac{4.448\;\rm N}{\rm{lb}}$
The conversion from inches to meters is $1\;\rm{in}=0.0254\;\rm{m}$. Both sides of the equation must be squared to find the conversion from square inches to square meters.
$1\;\rm{in}^2=(2.54\times10^{-2}\;\rm m)^2=6.4516\times10^{-4}\;\rm m^2$
$1=\frac{6.4516\times10^{-4}\;\rm m^2}{\rm{in}^2}$
Notice that it is useful to write the values using scientific notation to clearly show the significant figures, but additional digits are carried since this is an intermediate calculation. The two conversion factors can now be used to make the conversion. It is important to write each conversion factor so that the unwanted units cancel.
\begin{aligned}14.7\;{\rm{ psi}}&=\left(\frac{14.7\;\rm{lb}}{\rm{in}^2}\right)\!\left(\frac{4.448\;\rm{N}}{1\;\rm{lb}}\right)\!\left(\frac{1\;\rm{in^2}}{6.4516\times10^{-4}\;\rm{m^2}}\right)\\&=1.01\times10^5\,\frac{\rm N}{\rm m^2}\\&=1.01\times10^5\;\rm {Pa}\end{aligned}
The final answer has three significant figures to match the number of significant figures in the original measurement, 14.7 psi.