Unit Conversion Tutorial

The conversion of numbers from one system of units to another often puzzles students, but if it is treated as just another problem in arithmetic using arithmetic's rules the problems disappear.

For example: Knowing that there are \(2.54\; cm\) in \(1.0\) inch, how many \(cm\) are there in \(15\; inches\)?

Of course the answer is simple  \(15\; in = 15.0 \times 2.54 = 38.1 \;cm\)

But what actually was being done here? The complete solution is as follows:

\(15.0 in \times 2.54 \frac {cm}{in} = 38.1 cm\)

Notice that the units were treated just like arithmetic quantities and the "\(in\)" were canceled.

Let's do one that is not quite as obvious! How many \((mm)^2\) are there in \(4.0 (in)^2\)? The solution is:

\(4.0(in)^2 \times \biggl( 2.54 \frac {cm}{in} \biggr)^2 \times \biggl( 10 \frac {mm}{cm} \biggr)^2\)

\(= 4.0(in)^2 \times (2.54)^2 \frac {(cm)^2}{(in)^2} \times 10^2 \frac {(mm)^2}{(cm)^2} = 2.58 \times 10^3 (mm)^2\)

Notice that the quantities \(2.54\; cm/in\) and \(10 \;mm/cm\) were used but, because the units had to be squared, then the numbers that accompanied them had also to be squared.

Another example: Convert \(30 \;mi/hr\) into \(m/s.\) This is a common conversion. It is necessary to know that there are \(1.6 \;km\) in \(1 \;mile\) and \(3600 \;s\) in \(1\;hr (60 \times 60).\)

\(30 \frac {mi}{hr} \times 1.6 \frac{km}{mi} \times \frac{1}{3600} \frac {hr}{s} \times 1000 \frac {m}{km} = 13.3 m/s\)

Notice that the only tricky part here is the time conversion \(3600\; s/hr\) which is the wrong way up for our conversion but of course it is equally true that there are \((1/3600)\; hr/s.\)

So long as the relevant relations between the quantities are assembled in advance, then any conversion can be performed using these strict arithmetic rules.

Try this more complicated conversion:

An old (and ridiculous) unit of thermal conductivity sometimes still encountered in building materials is

\(Btu.hr^{-1}.in.F^{-1}.ft^{-2}\)

where \(1 Btu = 1054.8 J\)

\(1 in = 2.54 cm\)

\(1 ft^2 = 0.0929 m^2\)

\(9 F = 5 C\)
 

What is 1 \((Btu.hr^{-1}.in.F^{-1}.ft^{-2})\) in proper units \((W.m^{-1}.C^{-1})\)?

Just take the quantities one at a time:
  1. Convert \(Btu\) to \(J\) and cancel \(Btu\)
  2. Convert \(in\) to \(cm\) and then \(m\) canceling \(in\)
  3. Convert \(ft^2\) to \(m^2\) and cancel \(ft^2\)
  4. Convert \(F\) to \(C\) and cancel \(F\)
  5. Convert \(hr\) to \(s\) and cancel \(hr\)

 

\(1 \frac {Btu}{hr} \frac {in}{^ \circ F}\frac {1}{ft^2} \times 1054.8 \frac {J}{Btu} \times 2.54 \frac {cm}{in} \times \frac {1}{100} \frac {m}{cm} \times 10.7643 \frac {ft^2}{m^2} \times \frac {9}{5} \frac {^ \circ F}{^ \circ C} \times \frac {1}{3600} \frac {hr}{s}\)
\(= 0.144 W \cdot m^{-1} \cdot ^ \circ C^ {-1}\)