Converting between disparate units of measurement is the bane of many science students. The problem is worse for students in the United States of America, who must work with British (“Customary”) units such as the pound, the foot, the gallon, etc. World-wide adoption of the metric system would go a long way toward alleviating this problem, but until then it is important for students to master the art of unit conversions.
It is possible to convert from one unit of measurement to another by use of tables designed expressly for this purpose. Such tables usually have a column of units on the left-hand side and an identical row of units along the top, whereby one can look up the conversion factor to multiply by to convert from any listed unit to any other listed unit. While such tables are undeniably simple to use, they are practically impossible to memorize.
A better way to convert between different units is shown in the next subsection.
An important principle in the physical sciences is to closely track all units of measurement when performing calculations of physical quantities. This practice is generally referred to as dimensional analysis. A brief example of dimensional analysis is shown here, used to analyze the simple formula \(P = IV\) which describes the amount of power dissipated by an electrical load (\(P\)) given its current (\(I\)) and voltage drop (\(V\)):
\[P = IV\]
Substituting units of measurement for each variable in this formula (i.e. Watts for power, Amperes for current, and Volts for voltage), using bracket symbols to denote these as unit abbreviations rather than variables, we get this result:
\[[\hbox{Watts}] = [\hbox{Amperes}] \times [\hbox{Volts}] \hbox{\hskip 20pt or \hskip 20pt} [\hbox{W}] = [\hbox{A}] [\hbox{V}]\]
If we happen to know that “watts” is equivalent to joules of energy dissipated per second, and that “amperes” is equivalent to coulombs of charge motion per second, and that “volts” is equivalent to joules of energy per coulomb of electrical charge, we may substitute these units of measurement into the formula and see that the unit of “coulomb” cancels just like identical variables in the numerator and denominator of multiplied fractions:
\[\left[\hbox{Joules} \over \hbox{Seconds} \right] = \left[\hbox{\sout{Coulombs}} \over \hbox{Seconds}\right] \times \left[\hbox{Joules} \over \hbox{\sout{Coulombs}}\right] \hbox{\hskip 20pt or \hskip 20pt} \left[\hbox{J} \over \hbox{s} \right] = \left[\hbox{\sout{C}} \over \hbox{s} \right] \left[\hbox{J} \over \hbox{\sout{C}} \right]\]
As it so happens, dimensional analysis may be employed in a similar manner to convert between different units of measurement via a technique I like to call unity fractions.
This technique involves setting up the original quantity as a fraction, then multiplying by a series of fractions having physical values of unity (1) so that by multiplication the original value does not change, but the units do. Let’s take for example the conversion of quarts into gallons, an example of a fluid volume conversion:
\[35 \hbox{ qt} = \hbox{??? gal}\]
Now, most people know there are four quarts in one gallon, and so it is tempting to simply divide the number 35 by four to arrive at the proper number of gallons. However, the purpose of this example is to show you how the technique of unity fractions works, not to get an answer to a problem.
To demonstrate the unity fraction technique, we will first write the original quantity as a fraction, in this case a fraction with 1 as the denominator:
\[{35 \hbox{ qt} \over 1}\]
Next, we will multiply this fraction by another fraction having a physical value of unity (1) so that we do not alter the quantity. This means a fraction comprised of equal measures in the numerator and denominator, but having different units of measurement. This “unity” fraction must be arranged in such a way that the undesired unit cancels out and leaves only the desired unit(s) in the product. In this particular example, we wish to cancel out quarts and end up with gallons, so we must arrange a fraction consisting of quarts and gallons having equal quantities in numerator and denominator, such that quarts will cancel and gallons will remain:
\[\left({35 \hbox{ qt} \over 1}\right) \left({1 \hbox{ gal} \over 4 \hbox{ qt}}\right)\]
Now we see how the unit of “quarts” cancels from the numerator of the first fraction and the denominator of the second (“unity”) fraction, leaving only the unit of “gallons” left standing:
\[\left({35 \hbox{ \sout{qt}} \over 1}\right) \left({1 \hbox{ gal} \over 4 \hbox{ \sout{qt}}}\right) = 8.75 \hbox{ gal}\]
The reason this conversion technique is so powerful is it allows one to perform the largest range of unit conversions while memorizing the smallest possible set of conversion factors.
Here is a set of six equal volumes, each one expressed in a different unit of measurement:
1 gallon (gal) = 231.0 cubic inches (in\(^{3}\)) = 4 quarts (qt) = 8 pints (pt) = 128 fluid ounces (fl. oz.) = 3.7854 liters (l)
Since all six of these quantities are physically equal, it is possible to build a “unity fraction” out of any two, to use in converting any of the represented volume units into any of the other represented volume units. Shown here are a few different volume unit conversion problems, using unity fractions built only from these factors (all canceled units shown using strike-out lines):
40 gallons converted into fluid ounces (using 128 fl. oz. = 1 gal in the unity fraction):
\[\left({40 \hbox{ \sout{gal}} \over 1}\right) \left({128 \hbox{ fl. oz.} \over 1 \hbox{ \sout{gal}}}\right) = 5120 \hbox{ fl. oz}\]
5.5 pints converted into cubic inches (using 231 in\(^{3}\) = 8 pt in the unity fraction):
\[\left({5.5 \hbox{ \sout{pt}} \over 1}\right) \left({231 \hbox{ in}^3 \over 8 \hbox{ \sout{pt}}}\right) = 158.8 \hbox{ in}^3\]
1170 liters converted into quarts:
\[\left({1170 \hbox{ \sout{l}} \over 1}\right) \left({4 \hbox{ qt} \over 3.7854 \hbox{ \sout{l}}}\right) = 1236 \hbox{ qt}\]
By contrast, if we were to try to memorize a 6 \(\times\) 6 table giving conversion factors between any two of six volume units, we would have to commit 30 different conversion factors to memory! Clearly, the ability to set up “unity fractions” is a much more memory-efficient and practical approach.
This economy of conversion factors is very useful, and may also be extended to cases where linear units are raised to powers to represent two- or three-dimensional quantities. To illustrate, suppose we wished to convert 5.5 pints into cubic feet instead of cubic inches: with no conversion equivalence between pints and cubic feet included in our string of six equalities, what do we do?
We should know the equality between inches and feet: there are exactly 12 inches in 1 foot. This simple fact may be applied by incorporating another unity fraction in the original problem to convert cubic inches into cubic feet. We will begin by including another unity fraction comprised of 12 inches and 1 foot,just to see how this might work:
5.5 pints converted into cubic feet (our first attempt!):
\[\left({5.5 \hbox{ \sout{pt}} \over 1}\right) \left({231 \hbox{ in}^3 \over 8 \hbox{ \sout{pt}}}\right) \left(1 \hbox{ ft} \over 12 \hbox{ \sout{in}}\right) = 13.23 \hbox{ in}^2 \cdot \hbox{ft}\]
Unfortunately, this yields a non-sensical unit of square inch-feet. Even though \({1 \hbox{ ft} \over 12 \hbox{ in}}\) is a valid unity fraction, it does not completely cancel out the unit of cubic inches in the numerator of the first unity fraction. Instead, the unit of “inches” in the denominator of the unity fraction merely cancels out one of the “inches” in the “cubic inches” of the previous fraction’s numerator, leaving square inches (in\(^{2}\)). What we need for full cancellation of cubic inches is a unity fraction relating cubic feet to cubic inches. We can get this, though, simply by cubing the \({1 \hbox{ ft} \over 12 \hbox{ in}}\) unity fraction:
5.5 pints converted into cubic feet (our second attempt!):
\[\left({5.5 \hbox{ pt} \over 1}\right) \left({231 \hbox{ in}^3 \over 8 \hbox{ pt}}\right) \left(1 \hbox{ ft} \over 12 \hbox{ in}\right)^3\]
Distributing the third power to the interior terms of the last unity fraction:
\[\left({5.5 \hbox{ pt} \over 1}\right) \left({231 \hbox{ in}^3 \over 8 \hbox{ pt}}\right) \left(1^3 \hbox{ ft}^3 \over 12^3 \hbox{ in}^3\right)\]
Calculating the values of \(1^3\) and \(12^3\) inside the last unity fraction, then canceling units and solving:
\[\left({5.5 \hbox{ \sout{pt}} \over 1}\right) \left({231 \hbox{ \sout{in}}^3 \over 8 \hbox{ \sout{pt}}}\right) \left(1 \hbox{ ft}^3 \over 1728 \hbox{ \sout{in}}^3\right) = 0.0919 \hbox{ ft}^3\]
Now the answer makes sense: a volume expressed in units of cubic feet.
Once again, this unit conversion technique shows its power by minimizing the number of conversion factors we must memorize. We need not memorize how many cubic inches are in a cubic foot, or how many square inches are in a square foot, if we know how many linear inches are in a linear foot and we simply let the fractions “tell” us whether a power is needed for unit cancellation.
Unity fractions are also useful when we need to convert more than one unit in a given quantity. For example, suppose a flowmeter at a wastewater treatment facility gave us a flow measurement of 205 cubic feet per minute but we needed to convert this expression of water flow into units of cubic yards per day. Observe the following unit-fraction conversion to see how unity fractions serve the purpose of converting cubic feet into cubic yards, and minutes into days (by way of minutes to hours, and hours to days):
\[\left(205 \hbox{ \sout{ft}}^3 \over \hbox{\sout{min}} \right) \left(1^3 \hbox{ yd}^3 \over 3^3 \hbox{ \sout{ft}}^3\right) \left(60 \hbox{ \sout{min}} \over 1 \hbox{ \sout{hr}}\right) \left(24 \hbox{ \sout{hr}} \over 1 \hbox{ day}\right) = 10933.3 \hbox{ yd}^3\hbox{/day}\]
Note how the only units left un-canceled on the left-hand side of the “equals” symbol are cubic yards (yd\(^{3}\)) and days, which therefore become the units of measurement for the final result.
A major caveat to this method of converting units is that the units must be directly proportional to one another, since this multiplicative conversion method is really nothing more than an exercise in mathematical proportions. Here are some examples (but not an exhaustive list!) of conversions that cannot be performed using the “unity fraction” method:
The following subsections give sets of physically equal quantities, which may be used to create unity fractions for unit conversion problems. Note that only those quantities shown in the same line (separated by \(=\) symbols) are truly equal to each other, not quantities appearing in different lines!
Note: all of the conversion factors given for temperature are exact, not approximations.
\(^{o}\)F = (\(^{o}\)C)(9/5) + 32
\(^{o}\)C = (\(^{o}\)F \(-\) 32)(5/9)
\(^{o}\)R = \(^{o}\)F + 459.67
K = \(^{o}\)C + 273.15
Note: all of the conversion factors given for distance are exact, not approximations.
1 inch (in) = 2.54 centimeters (cm)
1 foot (ft) = 12 inches (in)
1 yard (yd) = 3 feet (ft)
1 mile (mi) = 5280 feet (ft)
Note: all conversion factors shown in bold type are exact, not approximations.
1 gallon (gal) = 231.0 cubic inches (in\(^{3}\)) = 4 quarts (qt) = 8 pints (pt) = 16 cups = 128 fluid ounces (fl. oz.) = 3.7854 liters (l)
1 milliliter (ml) = 1 cubic centimeter (cm\(^{3}\))
Note: all conversion factors shown in bold type are exact, not approximations.
1 mile per hour (mi/h) = 88 feet per minute (ft/m) = 1.46667 feet per second (ft/s) = 1.60934 kilometer per hour (km/h) = 0.44704 meter per second (m/s) = 0.868976 knot (knot – international)
1 pound-mass (lbm) = 0.4535924 kilogram (kg) = 0.031081 slugs
1 pound-force (lbf) = 4.448222 newtons (N)
1 kilogram-force (kgf) = 9.80665 newtons (N)
Note: all conversion factors shown in bold type are exact, not approximations.
1 acre = 43560 square feet (ft\(^{2}\)) = 4840 square yards (yd\(^{2}\)) = 4046.86 square meters (m\(^{2}\))
Note: all conversion factors shown in bold type are exact, not approximations.
1 pounds per square inch (PSI) = 2.03602 inches of mercury at 0 \(^{o}\)C (in. Hg) = 27.6799 inches of water at 4 \(^{o}\)C (in. W.C.) = 6.894757 kilopascals (kPa) = 0.06894757 bar
1 bar = 100 kilopascals (kPa) = 14.504 pounds per square inch (PSI)
1 meter of water at 4 \(^{o}\)C (m W.C.) = 9.80665 kilopascals (kPa)
Note: all conversion factors shown in bold type are exact, not approximations.
1 standard atmosphere (Atm) = 14.7 pounds per square inch absolute (PSIA) = 101.325 kilopascals absolute (kPaA) = 1.01325 bar absolute = 760 millimeters of mercury absolute (mmHgA) = 760 torr (torr)
1 British thermal unit (Btu – “International Table”) = 251.996 calories (cal – “International Table”) = 1055.06 joules (J) = 1055.06 watt-seconds (W-s) = 0.293071 watt-hour (W-hr) = 1.05506 x 10\(^{10}\) ergs (erg) = 778.169 foot-pound-force (ft-lbf)
Note: all conversion factors shown in bold type are exact, not approximations.
1 horsepower = 550 foot-pounds per second (ft-lbf/s) = 745.7 watts (W) = 2544.43 British thermal units per hour (Btu/h) = 0.0760181 boiler horsepower (hp – boiler)
Acceleration of gravity at sea level = 9.806650 meters per second per second (m/s\(^{2}\)) = 32.1740 feet per second per second (ft/s\(^{2}\))
Atmospheric pressure = 14.7 pounds per square inch absolute (PSIA) = 760 millimeters of mercury absolute (mmHgA) = 760 torr (torr) = 1.01325 bar (bar)
Atmospheric gas concentrations (by volume, not mass):
Density of dry air at 20 \(^{o}\)C and 760 torr = 1.204 mg/cm\(^{3}\) = 1.204 kg/m\(^{3}\) = 0.075 lb/ft\(^{3}\) = 0.00235 slugs/ft\(^{3}\)
Absolute viscosity of dry air at 20 \(^{o}\)C and 760 torr = 0.018 centipoise (cp) = 1.8 \(\times\) \(10^{-5}\) pascal-seconds (Pa\(\cdot\)s)
Freezing point at sea level = 32 \(^{o}\)F = 0 \(^{o}\)C
Boiling point at sea level = 212 \(^{o}\)F = 100 \(^{o}\)C
Density of water at 4 \(^{o}\)C = 1000 kg/m\(^{3}\) = 1 g/cm\(^{3}\) = 1 kg/liter = 62.428 lb/ft\(^{3}\) = 1.94 slugs/ft\(^{3}\)
Specific heat of water at 14 \(^{o}\)C = 1.00002 calories/g\(\cdot\)\(^{o}\)C = 1 BTU/lb\(\cdot\)\(^{o}\)F = 4.1869 joules/g\(\cdot\)\(^{o}\)C
Specific heat of ice \(\approx\) 0.5 calories/g\(\cdot\)\(^{o}\)C
Specific heat of steam \(\approx\) 0.48 calories/g\(\cdot\)\(^{o}\)C
Absolute viscosity of water at 20 \(^{o}\)C = 1.0019 centipoise (cp) = 0.0010019 pascal-seconds (Pa\(\cdot\)s)
Surface tension of water (in contact with air) at 18 \(^{o}\)C = 73.05 dynes/cm
pH of pure water at 25 \(^{o}\)C = 7.0 (pH scale = 0 to 14)
Note: all constants shown in bold type are exact, not approximations. Parentheses show one standard deviation (\(\sigma\)) of uncertainty in the last digits: for example, Avogadro’s number given as 6.02214179(30) \(\times\) \(10^{23}\) means the center value (\(6.02214179 \times 10^{23}\)) plus or minus \(0.00000030 \times 10^{23}\).
Avogadro’s number (\(N_A\)) = 6.02214179(30) \(\times\) \(10^{23}\) per mole (mol\(^{-1}\))
Boltzmann’s constant (\(k\)) = 1.3806504(24) \(\times\) \(10^{-23}\) joules per Kelvin (J/K)
Electronic charge (\(e\)) = 1.602176487(40) \(\times\) \(10^{-19}\) Coulomb (C)
Faraday constant (\(F\)) = 9.64853399(24) \(\times\) \(10^{4}\) Coulombs per mole (C/mol)
Gravitational constant (\(G\)) = 6.67428(67) \(\times\) \(10^{-11}\) cubic meters per kilogram-seconds squared (m\(^{3}\)/kg-s\(^{2}\))
Molar gas constant (\(R\)) = 8.314472(15) joules per mole-Kelvin (J/mol-K) = 0.08205746(14) liters-atmospheres per mole-Kelvin
Planck constant (\(h\)) = 6.62606896(33) \(\times\) \(10^{-34}\) joule-seconds (J-s)
Stefan-Boltzmann constant (\(\sigma\)) = 5.670400(40) \(\times\) \(10^{-8}\) Watts per square meter-Kelvin\(^{4}\) (W/m\(^{2} \cdot\)K\(^{4}\))
Speed of light in a vacuum (\(c\)) = 299792458 meters per second (m/s) = 186282.4 miles per second (mi/s)
All constants taken from NIST data “Fundamental Physical Constants – Extensive Listing”, published 2006.
All density figures approximate for samples at standard temperature and pressure.