1.7: Conversions

I suggest you avoid memorizing lots of conversion factors between SI units and U.S. units. Suppose the United Nations sends its black helicopters to invade California (after all who wouldn't rather live here than in New York City?), and institutes water fluoridation and the SI, making the use of inches and pounds into a crime punishable by death. I think you could get by with only two mental conversion factors:

1 inch = 2.54 cm

An object with a weight on Earth of 2.2 lb has a mass of 1 kg .

The first one is the present definition of the inch, so it's exact. The second one is not exact, but is good enough for most purposes. The pound is a unit of gravitational force, while the kg is a unit of mass, which measures how hard it is to accelerate an object, not how hard gravity pulls on it. Therefore it would be incorrect to say that 2.2 lb literally equaled 1 kg , even approximately.

More important than memorizing conversion factors is understanding the right method for doing conversions. Even within the SI, you may need to convert, say, from grams to kilograms. Different people have different ways of thinking about conversions, but the method I'll describe here is systematic and easy to understand. The idea is that if 1 kg and 1000 g represent the same mass, then we can consider a fraction like

\[\frac{10^3 \mathrm{~g}}{1 \mathrm{~kg}}\notag\]

to be a way of expressing the number one. This may bother you. For instance, if you type 1000/1 into your calculator, you will get 1000, not one. Again, different people have different ways of thinking about it, but the justification is that it helps us to do conversions, and it works! Now if we want to convert 0.7 kg to units of grams, we can multiply 0.7 kg by the number one:

\[0.7 \mathrm{~kg} \times \frac{10^3 \mathrm{~g}}{1 \mathrm{~kg}}\notag\]

If you're willing to treat symbols such as "kg" as if they were variables as used in algebra (which they're really not), you can then cancel the kg on top with the kg on the bottom, resulting in

\[0.7 \mathrm{~kg} \times \frac{10^3 \mathrm{~g}}{1 \mathrm{k} / \mathrm{g}}=700 \mathrm{~g}\notag\]

To convert grams to kilograms, you would simply flip the fraction upside down.

One advantage of this method is that it can easily be applied to a series of conversions. For instance, to convert one year to units of seconds,

\[\begin{array}{l}
1 \mathrm{y} \not \mathrm{dar} \times \frac{365 \mathrm{~d} / \mathrm{ys}}{1 \mathrm{y} / \mathrm{ar}} \times \frac{24 \mathrm{ho} / \mathrm{hrs}}{1 \mathrm{~d} / \mathrm{yy}} \times \frac{60 \mathrm{~m} / \mathrm{n}}{1 \mathrm{~h} / \mathrm{ur}} \times \frac{60 \mathrm{~s}}{1 \mathrm{~m} / \mathrm{in}} \\
=3.15 \times 10^7 \mathrm{~s}
\end{array}\notag\]

Should that exponent be positive or negative?

A common mistake is to write the conversion fraction incorrectly. For instance the fraction

\[\frac{10^3 \mathrm{~kg}}{1 \mathrm{~g}}\notag\]

(incorrect)

does not equal one, because \(10^3 \mathrm{~kg}\) is the mass of a car, and 1 g is the mass of a raisin. One correct way of setting up the conversion factor would be

\[\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~g}}\notag\]

You can usually detect such a mistake if you take the time to check your answer and see if it is reasonable.

If common sense doesn't rule out either a positive or a negative exponent, here's another way to make sure you get it right. There are big prefixes and small prefixes:

\[\begin{array}{llll}
\text { big prefixes: } & \mathrm{k} & \mathrm{M} & \\
\text { small prefixes: } & \mathrm{m} & \mu & \mathrm{n}
\end{array} \notag \]

(It's not hard to keep straight which are which, since "mega" and "micro" are evocative, and it's easy to remember that a kilometer is bigger than a meter and a millimeter is smaller.) In the example above, we want the top of the fraction to be the same as the bottom. Since k is a big prefix, we need to compensate by putting a small number like \(10^{-3}\) in front of it, not a big number like \(10^3\).