# Chapter 3: Measurement

## 3.4 Unit conversions

Sometimes it is necessary to convert between measurements in different units. Conversions within the metric system are straightforward—it designed intentionally for this to be the case. Conversions between metric units and customary units (which is what we use in the US), or between different customary units (such as conversions between fluid ounces and gallons) are just a little more involved.

### 3.4.1 Conversions with only metric units

In the metric system, where everything is based on powers of ten, this can be straightforward: simply use the metric prefixes to write everything in terms of your base units, and you’re good to go.

#### Example 3.6

A beetle has a mass of 23 grams, and a fruit fly has a mass of 20 milligrams. What is the total mass of the two insects in grams?

First we need to convert milligrams to grams. The metric prefix milli- means multiplication by 10-3. So we have

\begin{align*} 23\ \textrm{g} + 20\ \textrm{mg} &= 23\ \textrm{g} + 20\times 10^{-3}\ \textrm{g} \\ &= 23\ \textrm{g} + 0.02\ \textrm{g} \\ &= 23.02\ \textrm{g} \end{align*}

It takes on extra step if you’re going the away from the base unit. Say you wanted to know the total mass from our previous example in units of kilograms. There is actually not a whole lot of math involved, and we can once again start from the definition of our metric prefixes:

\begin{align*} 1\ \textrm{kg} &= 10^3\ \textrm{g} \\ \frac{1\ \textrm{kg}}{10^3} &= \frac{10^3\ \textrm{g}}{10^3} \\ 10^{-3}\ \textrm{kg} &= 1\ \textrm{g} \end{align*}

So we can simply replace grams with ×10-3 kilograms, and our two insects from the previous example have a combined mass of 23.02×10-3 kg.

### 3.4.2 Conversions with metric and customary units

Let’s start by looking at the relationship between a kilogram, and a gram. Specifically, we know that 1 kg measures the exact same amount of mass as 1000 g. So, we could write

$\frac{1\ \textrm{kg}}{10^3\ \textrm{g}} = 1$

(Notice how important it is to include the units! The number $$\frac{1}{1\thinspace000}$$, with no units specified, is certainly not one!) Mathematically, we can multiply anything by the number one, and not have changed anything. So, going back to the previous example, we could say

$23.02\ \textrm{g} \times 1 = \left(23.02\ \textrm{g}\right)\left(\frac{1\ \textrm{kg}}{10^3\ \textrm{g}}\right) = 23.02\times 10^{-3}\ \textrm{kg}$

where we treated the symbol for grams as an algebraic quantity that divided out.

We can apply this technique to a whole series of conversions. To find out how many seconds there are in a year, we do:

\begin{align*} \left(1\ \textrm{year}\right)\left( \frac{365\ \textrm{days}}{1\ \textrm{year}} \right)\left( \frac{24\ \textrm{hours}}{1\ \textrm{day}} \right)\left( \frac{60\ \textrm{min}}{1\ \textrm{hour}} \right)\left( \frac{60\ \textrm{s}}{1\ \textrm{min}} \right) &= (1)(365)(24)(60)(60)\ \textrm{s} \\ &= 3.1536\times10^7\ \textrm{s} \end{align*}

Notice that we took the quantity (1 year) and multiplied it by the number one several times—we just happened to express the number one in a different way at each step, chosen in such a way that all units except seconds divided out. One year and 3.1536×107 seconds measure the exact same amount of time.

#### Example 3.7

The inch is defined as 2.54 cm. How many centimeters are there in 42 feet?

\begin{align*} \left(42\ \textrm{feet}\right)\left(\frac{12\ \textrm{inches}}{1\ \textrm{foot}}\right)\left(\frac{2.54\ \textrm{cm}}{1\ \textrm{in}}\right) &= (42)(12)(2.54)\ \textrm{cm} \\ &= 1.28\times10^3\ \textrm{cm} \end{align*}

#### Example 3.8

One mile and 1.6 kilometers are the same distance. Imagine you are driving in a country that uses only base SI units, and the speed limit is given as 6 m/s. The speedometer in your car says you are driving at 25 mph (miles/hour). Are you speeding?

Let’s convert your speed into meters per second:

\begin{align*} \left(\frac{25\ \textrm{miles}}{\textrm{hour}}\right)\left(\frac{1.6\ \textrm{km}}{1\ \textrm{mi}}\right)\left(\frac{1\thinspace000\ \textrm{m}}{1\ \textrm{km}}\right)\left(\frac{1\ \textrm{hour}}{3\thinspace600\ \textrm{s}}\right) &= \frac{(25)(1.6)(1\thinspace000)}{3\thinspace600}\ \textrm{m/s} \\ &= 11.1\ \textrm{m/s} \end{align*}

Yes, you are driving above the posted speed limit.

#### Practice 3.1

The following practice problems include material from the entire chapter.