Converting Units

A unit is a standard magnitude of a physical quantity, like length, weight, time, temperature, etc. A primitive way to measure a length would be using the span of one's hand. However the span of a hand varies among people, so mankind has developed the concept of a unit to standardize the magnitudes of various physical quantities.

The representative physical quantities are: length, weight, and time. To represent a physical quantity, the meter–kilogram–second (MKS) system is commonly adopted, where meter, kilogram, and second are the units of length, weight, and time, respectively. For example, a unit of speed can be given as meters per second \((\text{m/s}).\)

However, different civilizations have developed unique units of their own, and therefore we may have various units for the same physical quantity. When it comes to length, a unit can be any of meter, inch, foot, yard, mile, light-year, and so on. If the size of a display is given in inches, we may want to convert it in meters or centimeters (\(10^{-2}\) meters). From 1 (in) \(=\) 2.54 (cm), we have 2 (in) \(=\) 5.08 (cm), 3 (in) \(=\) 7.62 (cm), 4 (in) \(=\) 10.16 (cm), \(\dots.\) Therefore, if the size of a display is 152 inches, then the size equals \(152 \times 2.54 = 386.08\) (cm).

Seeing this, we can get a lesson that if we know 1 in one unit \(u_1\) equals some value \(v\) in the other unit \(u_2,\) then we can convert the unit from \(u_1\) to \(u_2\) by multiplying \(v.\) In other words, we can convert a unit by multiplying the conversion factor. The conversion factor for converting inches into centimeters can be obtained by dividing 1 (in) \(=\) 2.54 (cm) by 1 (in), which gives \(1 = \frac{2.54 \text{ (cm)}}{1 \text{ (in)}}.\) Hence if we want to convert 152 (in) to a representation in centimeters, we can multiply 152 (in) by \(\frac{2.54 \text{ (cm)}}{1 \text{ (in)}};\) \[ 152 \text{ (in)} \times \frac{ 2.54 \text{ (cm)} }{ 1 \text{ (in)} } = 386.08 \text{ (cm)} .\]

If 1 inch is 2.54 centimeters, what is 13.3 inches in centimeters?

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We can multiply 13.3 (in) by \( \frac{ 2.54 \text{ (cm)} }{ 1 \text{ (in)} }:\) \[\begin{align} 13.3 \text{ (in)} =& 13.3 \text{ (in)} \times \frac{ 2.54 \text{ (cm)} }{ 1 \text{ (in)} } \\ =& 33.782 \text{ (cm)}.\ _\square \end{align}\]

If 1 inch is 2.54 centimeters, how many inches is 127 centimeters?

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We can multiply 127 (cm) by \( \frac{ 1 \text{ (in)} }{ 2.54 \text{ (cm)} }:\) \[\begin{align} 127 \text{ (cm)} =& 127 \text{ (cm)} \times \frac{ 1 \text{ (in)} }{ 2.54 \text{ (cm)} } \\ =& 50 \text{ (in)}.\ _\square \end{align}\]

If 1 kilogram is 2.2046 pounds, what is 1 pound in kilograms?

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We can multiply 1 (lb) by \( \frac{ 1 \text{ (kg)} }{ 2.2046 \text{ (lb)} }:\) \[\begin{align} 1 \text{ (lb)} =& 1 \text{ (lb)} \times \frac{ 1\text{ (kg)} }{ 2.2046 \text{ (lb)} } \\ \approx& 0.45360 \text{ (kg)}.\ _\square \end{align}\]

If 1 mile is 1609 meters, what is 1 mph (mile per hour) in meters per second?

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\(1 \text{ (h)} = 3600 \text{ (s)} .\) We can multiply 1 (mph) by \( \frac{ 1609 \text{ (m)} }{ 1 \text{ (mile)} } \times \frac{ 1 \text{ (h)} }{ 3600 \text{ (s)} };\)

\[\begin{align} 1 \text{ (mph)}=& 1 \text{ (mph)} \times \frac{ 1609 \text{ (m)} }{ 1 \text{ (mile)} } \times \frac{ 1 \text{ (h)} }{ 3600 \text{ (s)} } \\ =& \frac{1609}{3600} \text{ (m/s)} \\ \approx& 0.4469 \text{ (m/s)} .\ _\square \end{align}\]

If \(1 \text{ (J)} = 2.78 \times 10^{-7} \text{ (kW}\cdot \text{h)} = 2.39 \times 10^{-4} \text{ (kcal)},\) what is \(1 \text{ (kW} \cdot \text{h)} + 1000 \text{ (kcal)} \) in joules?

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We have \[1 \text{ (kW} \cdot \text{h)} \approx 3.60 \times 10^6 \text{ (J)}, \] and \[1000 \text{ (kcal)} \approx 4.18 \times 10^6 \text{ (J)}. \]

Therefore, \[\begin{align} 1 \text{ (kW} \cdot \text{h)} + 1000 \text{ (kcal)} \approx& 3.60 \times 10^6 \text{ (J)} + 4.18 \times 10^6 \text{ (J)}\\ =& 7.78 \times 10^6 \text{ (J)} .\ _\square \end{align}\]

The depth of the troposphere is about 17 km at the equator. What is the volume of the troposphere above a hectare in liters? Assume that the troposphere above a hectare is a cuboid.

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A hectare equals \(10^4\ (\text{m}^2) ,\) and a liter equals \(10^{-3}\ (\text{m}^{3}) .\)

The volume above a hectare is \[\begin{align} 10^4 \times 1.7 \times 10^4\ (\text{m}^3) =& 1.7 \times 10^8\ (\text{m}^3) \\ =& 1.7 \times 10^8\ (\text{m}^{3}) \times \frac{ 1\ (\text{L}) }{10^{-3}\ (\text{m}^{3})} \\ =& 1.7 \times 10^{11}\ (\text{L}) .\ _\square\\ \end{align}\]

There are many different units of measuring, varying from the very large scale to the microscopic, and also varying in systems (for example, metric system vs. imperial system). How to we determine how large one measurement is in different units? Let's start with an example.

Suppose Jane runs a 10 km race. If there are 1.609 km in one mile, how many miles did Jane run?

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Since there are 1.609 km in one mile, Jane ran \[ 10 \text{ km} \times \left( \frac{ 1 \text{ mile} }{ 1.609 \text{ km} } \right) = 6.215 \text{ miles}.\ _\square\]

As seen in the above example, the procedure for converting units is:

• Multiply by a fraction whose numerator and denominator represent equal quantities (in the above example, 1 mile = 1.609 km). How do we decide whether to use the fraction \( \frac{ 1 \text{ mile} }{ 1.609 \text{ km} } \) or the fraction \(\frac{ 1.609 \text{ km} }{1 \text{ mile} }\)? The trick is:

• Choose the fraction that allows us to cancel units.

Since multiplying \(10\text{ km}\) by the fraction \( \frac{ 1 \text{ mile} }{ 1.609 \text{ km} } \) allows us to cancel the unit of \(\text{km}\) appearing in both the top and the bottom, this is the correct fraction to choose.

To convert between different quantities, it is useful to remember a few key conversion factors:

\[ \begin{align} \text{ Distances: } & 1 \text{ km} = 1000 \text{ m}\\ & 1 \text{ m} = 100 \text{ cm}\\ & 1.609 \text{ km} = 1 \text{ mile}\\ & 1 \text{ inch} = 2.54 \text{ cm} \\ & 1 \text{ yard} = 3 \text{ feet}\\ \\ \text{ Time: } & 1 \text{ day} = 24 \text{ hours}\\ & 1 \text{ hour} = 60 \text{ minutes}\\ & 1 \text{ minute} = 60 \text{ seconds}. \end{align}\]

If a car is moving at \(100 \text{ km/h},\) what is its speed in miles per second?

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We have \[\begin{align} \frac{100 \text{ km}}{\text{h}} &=\left(\frac{100 \text{ km}}{\text{h}}\right)\left(\frac{1 \text{ mi}}{1.609 \text{ km}}\right)\left(\frac{1 \text{ h}}{3600 \text{ s}}\right)\\ &=0.017 \text{ mi/s}. \ _\square \end{align}\]

Notice in this example that we have cancelled the units of \(\text{km}\) from numerator and denominator, as well as the units of hour from numerator and denominator. The units that remain are the desired units in the problem (\(\text{mi/s},\) or miles per second).

How many microns are in \(1 \text{ km}?\\ \)

Note: The micron is another name for the micrometer (\(1~ \mu\text{m}\)).

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Solution: Since \(1 \text{ km }=1\times 10^3 \text{ m}\) and \(1 \text{ m }=1\times 10^6 \mu \text{m},\) \[1 \text{ km }=10^3 \text{ m }=\left(10^3\text{ m }\right)\left(10^6 \mu \text{m}/\text{m}\right)=10^9 \mu\text{m}. \] Therefore, there are \(10^9 \mu\text{m}\) microns in 1 km.

Which is larger, one hundred liters or one cranberry barrel? \[\]

Note: The US barrel for cranberries is 5,826 cubic inches.

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Observe that one cranberry barrel is equivalent to \[\left(5826 \text{ in}^3\right) \left(\frac{2.54 \text{ cm}}{1 \text{ in}}\right)^3 \left(\frac{1\text{ L}}{1000 \text{ cm}^3}\right)=95.5\text{ L}<100 \text{ L}.\]

Thus, one hundred liters is larger. \(_\square\)

The speed of light is around \(3.0\times 10^8 \text{ m/s}.\) An astronomical unit (AU) is the average distance between the earth and the sun, which is around \(1.50\times 10^8 \text{ km}.\) Express the speed of light in \(\text{AU/min}.\)

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We have \[\begin{align} 3.0\times 10^8 \text{ m/s} &=\left(\frac{3.0\times 10^8 \text{ m}}{\text{s}}\right)\left(\frac{1 \text{ km}}{1000 \text{ m}}\right)\left(\frac{\text{AU}}{1.50\times 10^8 \text{ km}}\right)\left(\frac{60 \text{ s}}{\text{min}}\right)\\ &=0.12 \text{ AU/min}. \ _\square \end{align}\]