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Mathematics in Trades and Life

Section 1.1 Units

This section addresses the following topics.
  • Interpret data in various formats and analyze mathematical models
  • Communicate results in mathematical notation and in language appropriate to the technical field
This section covers the following mathematical concepts.
  • Unit conversion (skill)
  • Set up and solve proportions (skill)
This book is designed to present mathematics in various contexts including a variety of trades. As a result numbers will frequently be connected to units such as length (feet, meters), time (seconds) or others. These units are part of the arithmetic, and so we must learn what they mean and how to perform this arithmetic. Sometimes the units even suggest to us how to setup the arithmetic.
In this section we will introduce the units so we can understand them when we read technical material (ItemΒ 2) and use them correctly to communicate results to others (ItemΒ 3). We will also learn to convert units (ItemΒ 2.d) which involves our first use of proportions (ItemΒ 2.a).
We measure many things such as distance, time, and weight. We describe these measurements in terms of units like miles, hours, and pounds. But have you ever stopped to think about how these units are defined?
The story of some of these units is lost in history. For example dividing the day into 24 units began with ancient Egyptians. They did not record, that we know of, the reason for choosing 24 units as opposed to 30 or any other number.
Other units, such as the metric (also called the International System of Units or simply SI) are much more modern. Initially many units were based on something physical. For example, one calorie is the amount of heat it takes to raise the temperature of one gram of water \(1^\circ\) C. The meter was originally defined as one ten-millionth of the distance from the equator to the north pole. The problem with this type of measurement is that it is neither fixed (depends on where on the equator your begin) nor easy to measure.
Thus modern definitions were developed. The length of a meter was changed to mean the length of a bar of metal kept in special storage in France. The bar had been carefully constructed and was used to confirm other measurement devices were correctly calibrated. It was changed yet again to be based on wavelengths of radiation. These are uniform no matter where they are done, so they can be used by many people to construct simple measurement tools.

Subsection 1.1.1 Types of Measurement

First we will look at units for different types of measurement in the two major systems. In a specific trade you will need to memorize the units you use most often. For this class you should ask your instructor which units must be memorized and which you may look up when working on problems.
Note that the U.S. Customary system (related to the British Imperial system) is non-uniform, so there are multiple names for some types of measurements. This is in contrast to the SI (metric) which has one name for each property and prefixes to indicate the size. TableΒ 1.1.1 lists names of units for both systems. It is important to be able to recognize which unit (name) goes with which type of measurement (e.g., length, volume, …).
Table 1.1.1. Units of Measure
Measuring US Customary Metric
Length inch (in) meter (m)
foot (ft)
yard (yd)
mile (mi)
nautical mile (nm)
Volume fluid ounce (oz) liter (L or \(\ell\))
cup (c)
pint (pt)
quart (qt)
gallon (g)
Weight ounce (oz) gram (g)
pound (lb)
Temperature degrees Fahrenheit (F) degrees Celsius (C)
Pressure inches of mercury (inHg) bar
pounds per square inch (psi) Pascal (Pa)
Time second (s)
minute (min)
hour (hr)
One name: two meanings Note that fluid ounces and weight ounces are not the same unit. Ten (10) fluid ounces of milk does not weigh ten (10) ounces. You must determine which ounce is referenced by the context. This can be tricky in recipes, which is a good reason to use SI units.
Note a gram is a unit of mass rather than than a unit of weight. Pound and ounce on the other hand are units of weight. Nevertheless gram is often used to describe weight because it is easy to switch between it and weight. Namely, mass can be obtained by dividing by the acceleration due to gravity (see a physics book for more information). The official unit for weight (a force) is a Newton, but we will not use that in this book.

Subsection 1.1.2 U.S. Customary

Often we need to convert between units within the U.S. Customary system. This section provides the information needed for conversion and examples of performing them. It is an example of units suggesting how we setup the calculation.
Why would we need to convert units? This can occur because measurements were taken with different scales. For instance, we cannot add 3 inches to 2.2 feet without changing one to make the units match.
Why are there multiple units in the first place? There are different units for different scales (e.g., inches for small lengths and miles for long distances). This is a result of the U.S. Customary system being developed from the British Imperial system which was based on disparate measurements from multiple centuries ago (look up unit names in an etymological dictionary for fun). Converting between units therefore requires remembering special numbers for conversion. Most of these you likely know.
Table 1.1.2. Converting within U.S. Customary
Measuring Unit 1 Unit 2
Length 1 nm 6076 ft
1 mi 5280 ft
1 yd 3 ft
1 ft 12 in
Area 1 acre 43,560 \(\text{ft}^2\)
Volume 1 g 4 qts
1 qt 2 pts
1 pt 2 c
1 c 8 oz
1 g 0.1337 in3
Weight 1 ton 2000 lbs
1 lb 16 oz
Time 1 year 365 days
1 day 24 hrs
1 hr 60 mins
1 min 60 secs
Of course a year is not always the same number of days. In each circumstance it is important to determine whether we can use the common approximation of 365 days without injury or loss.
Review each of these examples to see how to convert from one U.S. Customary unit to another.

Example 1.1.3.

How many quarts is 2.3 gallons?
From TableΒ 1.1.2 we know (or can look up) that each gallon is 4 quarts. This means we have \(\frac{4 \text{ quarts}}{1 \text{ gallon}}\text{.}\) This suggests that we can multiply 2.3 by this ratio, because the gallons will divide out.
\begin{align*} 2.3 \text{ gallons} \cdot \frac{4 \text{ quarts}}{\text{gallon}} & =\\ \frac{2.3 \cancelhighlight{\text{ gallons}}}{1} \cdot \frac{4 \text{ quarts}}{\cancelhighlight{\text{gallon}}} & =\\ 2.3 \cdot 4 \text{ quarts} & = 9.2 \text{ quarts} \end{align*}
We can be confident we set up the conversion correctly because the only units left are quarts.
Sometimes to convert units we need to convert more than once.

Example 1.1.4.

How many cups is 1.5 quarts?
While we do not have an entry for cups per quart in the conversion table, we do have entries for quarts to pints and pints to cups. This suggests we can put these two conversions together to convert quarts to cups.
\begin{align*} 1.5 \text{ quarts} \cdot \frac{2 \text{ pints}}{\text{quart}} & =\\ \frac{1.5 \cancelhighlight{\text{ quarts}}}{1} \cdot \frac{2 \text{ pints}}{\cancelhighlight{\text{quart}}} & =\\ 1.5 \cdot 2 \text{ pints} & = 3.0 \text{ pints}\text{.} \end{align*}
Now we can convert the pints to cups.
\begin{align*} 3.0 \text{ pints} \cdot \frac{2 \text{ cups}}{\text{pint}} & =\\ \frac{3.0 \cancelhighlight{\text{ pints}}}{1} \cdot \frac{2 \text{ cups}}{\cancelhighlight{\text{pint}}} & =\\ 3.0 \cdot 2 \text{ cups} & = 6.0 \text{ cups}\text{.} \end{align*}
When we know that we need multiple conversions we can calculate them all at once.
\begin{equation*} 1.5 \text{ quarts} \cdot \frac{2 \text{ pints}}{\text{quart}} \cdot \frac{2 \text{ cups}}{\text{pint}} = 6.0 \text{ cups}\text{.} \end{equation*}
When we use more than one conversion, we are careful to set up the ratios so that the intermediate units (quarts and pints in this case) are divided out leaving us with only the desired unit.
Nothing restricts this process to two at a time.

Example 1.1.5.

How many cups is 1.7 gallons?
Solution.
The conversion table does not include the number of cups per gallon. However, we can start by converting gallons to quarts. Then looking at the table again, we can convert quarts to pints, and finally we can convert pints to cups.
As with the previous example we can treat each conversion as a ratio of units. We setup the units so that multiplying will result in units dividing out.
\begin{equation*} 1.7 \text{ gallons} \cdot \frac{4 \text{ quarts}}{\text{gallon}} \cdot \frac{2 \text{ pints}}{\text{quart}} \cdot \frac{2 \text{ cups}}{\text{pint}} = 27.2 \text{ cups} \end{equation*}

Example 1.1.6.

How many days is 17 hours?
Solution.
Here we are going from a small unit (hours) to a bigger one (days). This does not change our process. We can still multiply the amount by the unit conversion. Because we want to end up with days we use \(\frac{1 \text{ day}}{24 \text{ hours}}\)
\begin{equation*} 17 \text{ hours} \cdot \frac{1 \text{ day}}{24 \text{ hours}} \approx 0.7083 \text{ days} \end{equation*}
Check that you can perform a unit conversion using this Checkpoint.

Subsection 1.1.3 Metric (SI)

Just as with U.S. Customary units, we often need to convert between SI units. This section provides the information needed for conversion and examples of performing them. Again, units suggest how we setup the calculations.
Rather than have different names for different scales, metric uses one name of the unit (e.g., liter) and then uses prefixes to indicate size. These can be converted easily, because each prefix is a power of ten (uniform).
You will need to memorize a few of the prefixes. As with units which ones depends on your work. Ask your instructor which prefixes you should memorize for this course.
Table 1.1.8. Metric Prefixes
Multiple Prefix
\(10^{12}\) tera (T)
\(10^9\) giga (G)
\(10^6\) mega (M)
\(10^3\) kilo (k)
\(10^2\) hecto (h)
\(10\) deka (da)
\(1/10\) deci (d)
\(1/{10^2}\) centi (c)
\(1/{10^3}\) milli (m)
\(1/{10^6}\) micro (\(\mu\))
\(1/{10^9}\) nano (n)
\(1/{10^{12}}\) pico (p)
This next table illustrates these prefixes in the context of length (meters). Notice how it is easier to avoid the fractions (last three entries).
Table 1.1.9. Metric Conversion
Scaled Unit Base Unit
kilometer (km) \(10^3=1000\) meters
hectometer \(10^2=100\) meters
10 decimeters meter
\(10^2=100\) centimeters meter
\(10^3=1000\) millimeters meter
Review each of these examples to see how to convert from one SI unit to another.

Example 1.1.10.

How many centimeters is 3.8 meters?
From TableΒ 1.1.8 we know \(10^2=100\) centimeters is 1 meter, that is, \(\frac{100 \text{ cm}}{1 \text{ m}}\text{.}\) This suggests that we can multiply 3.8 by the ratio which will cause the meters units to divide out.
\begin{align*} 3.8 \text{ m} \cdot \frac{100 \text{ cm}}{\text{m}} & =\\ \frac{3.8 \cancelhighlight{\text{ m}}}{1} \cdot \frac{100 \text{ cm}}{\cancelhighlight{\text{m}}} & =\\ 3.8 \cdot 100 \text{ cm} & = 380 \text{ cm}\text{.} \end{align*}
Note, because centi is a power of ten (\(10^2\)) the result is shifting the decimal place two positions. 3.8 meters becomes 380 centimeters.
Using this idea we can convert 0.76 meters to 76 centimeters by just shifting the decimal (no additional process necessary).

Example 1.1.11.

How many kilotons is 2.3 megatons?
We know one kiloton is \(10^3\) tons and one megaton is \(10^6\) tons. Because we have two units we can convert twice.
\begin{align*} 2.3 \text{ megatons} \cdot \frac{10^6 \text{ tons}}{\text{megatons}} \cdot \frac{\text{kilotons}}{10^3 \text{ tons}} & =\\ \frac{2.3 \cancelhighlight{\text{ megatons}}}{1} \cdot \frac{10^6 \secondcancelhighlight{\text{ tons}}}{\cancelhighlight{\text{megatons}}} \cdot \frac{\text{kilotons}}{10^3 \secondcancelhighlight{\text{ tons}}} & = \\ \frac{2.3 \cdot 10^6}{10^3} \text{ kilotons} & =\\ 2.3 \cdot 10^3 \text{ kilotons} & = 2,300 \text{ kilotons}\text{.} \end{align*}
Just as in the previous example, we can pay attention to the powers of ten to find a shortcut. Kilo and mega are three powers apart (\(6-3=3\)), which means we shift the decimal position three places (we see this in the last two steps above). Because we are converting from a large unit to a smaller unit, we move the decimal place to the right (make the number bigger). 2.3 megatons is 2,300 kilotons.

Example 1.1.12.

How many centiliters is 13.6 milliliters?
Solution.
We know \(10^2=100\) centiliters is 1 liter and \(10^3=1000\) milliliters is 1 liter. This means we shift the decimal \(2-(3)=-1\text{.}\) Because we are moving from a smaller unit to a larger unit, we move the decimal place to the left (shown by the negative) to make the number smaller. 13.6 milliliters is 1.36 centiliters.

Subsection 1.1.4 Converting between Systems

Commonly we end up with measurements in both U.S. Standard system and SI. We will need to convert all units to one system before using them together. This process is the same as converting one Standard unit to another (e.g., converting miles to feet). This section provides the information needed for conversion and examples of performing them. It is an example of units suggesting how we setup the calculation.
Table 1.1.14. U.S. Customary to SI
Measuring Standard SI
Length 1 nm 1.85200 km
Length 1 mi 1.60934 km
1 ft 0.304800 m
1 in 2.54000 cm
Volume 1 gal 3.78541 L
1 oz 29.5735 mL
Weight 1 lb 0.453592 kg
1 oz 28.3495 g
Table 1.1.15. SI to U.S. Customary
Measuring SI Standard
Length 1 km 0.621371 mi
1 m 3.28084 ft
1 cm 0.393701 in
Volume 1 L 0.264172 gal
1 mL 0.0338140 oz
Weight 1 kg 2.20462 lb
1 g 0.0352740 oz
Review each of these examples to see how to convert between U.S. Customary units and SI units.

Example 1.1.16.

How many kilometers is 26.2 miles?
Solution.
From TableΒ 1.1.14 we know each mile is 1.60934 km; this means there is
\begin{equation*} \frac{1.60934 \text{ km}}{1 \text{ mi}}\text{.} \end{equation*}
The ratio suggests that we can multiply 23.6 miles by the ratio, because the miles will divide out.
\begin{equation*} 26.2 \text{ mi} \cdot \frac{1.60934 \text{ km}}{\text{mi}} = 42.164708 \text{ km} \approx 42.2 \text{ km} \end{equation*}
We have rounded here to 3 digits to match the original number (26.2). A reason to do so is presented in SectionΒ 1.2.

Example 1.1.17.

How many inches is 15 centimeters?
Solution.
From TableΒ 1.1.15 we know each centimeter is 0.393701 inches; this means there is
\begin{equation*} \frac{0.393701 \text{ in}}{1 \text{ cm}}\text{.} \end{equation*}
The ratio suggests that we can multiply 15 centimeters by the ratio, because the centimeters will divide out.
\begin{equation*} 15 \text{ cm} \cdot \frac{0.393701 \text{ in}}{\text{cm}} = 5.905515 \text{ in} \approx 5.9 \text{ in} \end{equation*}
We have rounded here to 2 digits to match the original number (15). A reason to do so is presented in SectionΒ 1.2.

Example 1.1.18.

How many inches is 1 meter?
Solution.
From TableΒ 1.1.15 we know each meter is 3.28084 feet. From TableΒ 1.1.2 that each foot is 12 inches. We use the method of setting up a product of ratios so that the units divide out. We start with meters, so the first ratio must be feet per meters. We want to end with inches so the second ratio must be inches per feet.
\begin{equation*} 1 \text{ m} \cdot \frac{3.28084 \text{ ft}}{1 \text{ m}} \cdot \frac{12 \text{ in}}{1 \text{ ft}} = 39.37008 \text{ in} \approx 39 \text{ in} \end{equation*}
We have rounded here to 2 digits to match the original number (15). A reason to do so is presented in SectionΒ 1.2.

Subsection 1.1.5 Converting Compound Units

Some units, such as speeds, are compound. For example speed is distance per time. This section provides examples of converting compound units.

Example 1.1.20.

How many meters per second is 15 miles per hour?
We start with \(\frac{15 \text{ mi}}{1 \text{ hr}}\text{.}\) We can convert miles to feet and feet to meters (multi-step conversion like ExampleΒ 1.1.4). The conversion ratios suggest we can multiply the 15 mi/hr by the conversion ratios.
\begin{equation*} \frac{15 \text{ mi}}{1 \text{ hr}} \cdot \frac{5280 \text{ ft}}{1 \text{ mi}} \cdot \frac{0.3048 \text{ m}}{1 \text{ ft}} \approx \frac{24140 \text{ m}}{1 \text{ hr}} \end{equation*}
We can use the same method (multiplying by conversion ratios to divide out units) to also conver hours to seconds.
\begin{equation*} \frac{24140 \text{ m}}{1 \text{ hr}} \cdot \frac{1 \text{ hr}}{60 \text{ min}} \cdot \frac{1 \text{ min}}{60 \text{ sec}} \approx 6.706 \frac{\text{m}}{\text{sec}}\text{.} \end{equation*}
Note, we could perform that conversion in on step by multiplying all the conversion ratios at once.

Example 1.1.21.

How many pounds does a tablespoon of water weigh? Note that one gallon of pure water weighs 8 lbs. Also a tablespoon is a half fluid ounce.
Solution.
We need to convert the gallons into tablespoons. Because conversions are ratios, we multiply 8 lbs by the necessary conversions. The ounces (oz) in this conversion are fluid (volume) ounces: we are not converting pounds to ounces here.
\begin{align*} \frac{8 \text{ lbs}}{1 \text{ gal}} \cdot \frac{1 \text{ gal}}{4 \text{ qts}} \cdot \frac{1 \text{ qt}}{2 \text{ pints}} \cdot \frac{1 \text{ pint}}{2 \text{ cups}} \cdot \frac{1 \text{ cup}}{16 \text{ oz}} \cdot \frac{1 \text{ oz}}{2 \text{ tbs}} & =\\ \frac{8 \text{ lbs}}{1 \cancelhighlight{\text{ gal}}} \cdot \frac{1 \cancelhighlight{\text{ gal}}}{4 \secondcancelhighlight{\text{ qts}}} \cdot \frac{1 \secondcancelhighlight{\text{ qt}}}{2 \cancelhighlight{\text{ pints}}} \cdot \frac{1 \cancelhighlight{\text{ pint}}}{2 \secondcancelhighlight{\text{ cups}}} \cdot \frac{1 \secondcancelhighlight{\text{ cup}}}{16 \cancelhighlight{\text{ oz}}} \cdot \frac{1 \cancelhighlight{\text{ oz}}}{2 \text{ tbs}} & =\\ \frac{8 \text{ lbs}}{4 \cdot 2 \cdot 2 \cdot 16 \cdot 2 \text{ tbs}} & = 0.015625 \frac{\text{lbs}}{\text{tbs}}\text{.} \end{align*}
Another kind of compound unit is square units such as square feet or seconds squared. When converting these we must account for the square.

Example 1.1.23.

Convert 2 acres to units of square miles.
Solution.
First we note that an acre is \(43,560 \text{ ft}^2\text{.}\) From TableΒ 1.1.2 we know that there are 5280 ft per mile. These conversion ratios suggest that we can multiply the 2 acres by the ratios to obtain the result in square miles.
\begin{align*} 2 \text{ acres} \cdot \frac{43,560 \text{ ft}^2}{\text{ acre}} \cdot \frac{\text{mi}}{5280 \text{ ft}} \cdot \frac{\text{mi}}{5280 \text{ ft}} & =\\ 2 \text{ acres} \cdot \frac{43,560 \text{ ft}^2}{\text{ acre}} \cdot \left(\frac{\text{mi}}{5280 \text{ ft}}\right)^2 & =\\ 2 \text{ acres} \cdot \frac{43,560 \text{ ft}^2}{\text{ acre}} \cdot \frac{\text{mi}^2}{5280^2 \text{ ft}^2} & = \frac{1}{320} \text{ mi}^2\\ & = 0.003125 \text{ mi}^2 \end{align*}
It is not necessary to write all of the steps above if you understand how the final conversion line is obtained. The steps are included here to show how the squares show up in the final conversion.

Example 1.1.25.

Calculate the value and resulting (simplified) units for
\begin{equation*} \frac{1}{2} \cdot \frac{0.002309 \text{ slugs}}{\text{ft}^3} \cdot (174 \text{ ft}^2)(3.1067)\left(\frac{75.95 \text{ ft}}{\text{s}}\right)^2\text{.} \end{equation*}
Note, the unit slug is defined as
\begin{equation*} \text{slug} = \frac{\text{lbs} \text{ s}^2}{\text{ft}}\text{.} \end{equation*}
\begin{align*} \frac{1}{2} \cdot \frac{0.002309 \text{ slugs}}{\text{ft}^3} \cdot (174 \text{ ft}^2)(3.1067)\left(\frac{75.95 \text{ ft}}{\text{s}}\right)^2 \amp = \text{ First, substitute slugs}\\ \frac{1}{2} \cdot \frac{0.002309 \text{ lbs} \text{ s}^2}{\text{ft} \cdot \text{ft}^3} \cdot (174 \text{ ft}^2)(3.1067)\left(\frac{75.95 \text{ ft}}{\text{s}}\right)^2 \amp = \text{ Second, handle powers}\\ \frac{1}{2} \cdot \frac{0.002309 \text{ lbs} \text{ s}^2}{\text{ft}^4} \cdot (174 \text{ ft}^2)(3.1067)\left(\frac{75.95 \text{ ft}^2}{\text{s}^2}\right) \amp = \text{ Now, collect units}\\ \frac{1}{2}(0.002309)(174)(3.1067)(75.95) \frac{ \text{ lbs} \text{ s}^2 \text{ ft}^2 \text{ ft}^2}{\text{ft}^4 \text{s}^2} \amp = \text{ Combine units}\\ \frac{1}{2}(0.002309)(174)(3.1067)(75.95) \frac{ \text{ lbs} \text{ s}^2 \text{ ft}^4}{\text{ft}^4 \text{s}^2} \amp = \text{ Calculate the result}\\ 35\underline{9}9.963183 \frac{ \text{ lbs} \text{ s}^2 \text{ ft}^4}{\text{ft}^4 \text{s}^2} \amp = \text{ Divide units}\\ 35\underline{9}9.963183 \frac{ \text{ lbs} \cancelhighlight{\text{ s}^2} \cancelhighlight{\text{ ft}^4}}{\cancelhighlight{\text{ft}^4} \cancelhighlight{\text{s}^2}} \amp = \text{ Divide units}\\ 35\underline{9}9.963183 \text{ lbs} \amp = \text{ Round using sigfigs}\\ 3600 \text{ lbs} \end{align*}

Exercises 1.1.6 Exercises

Dimensional Analysis.

Identify the correct units to make these work.

Converting Units.

Convert the units as directed. Use the conversion tables (TableΒ 1.1.2, TableΒ 1.1.9, or TableΒ 1.1.14) as needed.

Applications.

Answer these questions by converting units as needed.

Quantitative Literacy.

Use your knowledge of these items to decide on appropriate units.