Prep

Section 5.1 Prep

This preparatory section will equip you with the knowledge and skills to wield units conversions with confidence and precision. Through a combination of instructional video, explanatory text, and practice questions, you will become adept at all things unit conversions.

Subsection 5.1.1 Prep Videos: Units Conversions

Watch the following video(s) to get an introduction to unit conversions.

Subsection 5.1.2 Prep Text: Unit Conversions

Read the following text to learn about unit conversions. Don’t forget to try the checkpoint questions along the way!

Subsubsection 5.1.2.1 Conversion Factors as Ratios of One

Definition 5.1.1.

Unit Conversion

Unit conversion is the process of expressing a quantity in a different unit system by multiplying it by one or more conversion factors. A conversion factor is a ratio of two equal quantities such as \(\frac{1\,\text{hr}}{3600\,\text{s}}\) or \(\frac{1000\,\text{m}}{1\,\text{km}}\text{,}\) that has a numerical value of \(1\text{.}\) Applying these ratios changes only the unit in which a quantity is expressed, not the quantity itself. Unit conversions let us translate measurements into the units most useful for a problem without altering their physical meaning.

You’ve measured the mass of a dragon’s scale in ounces, but your spellbook demands grams. You’re plotting a star map in light-years, but your onboard thrusters calculate in meters. What do you do?

You cast a unit conversion spell.

Unit conversions are the dimensional bridges that connect different lands of measurement. They don’t change what you have, they simply describe it in a new tongue. Whether you’re switching from meters to miles, hours to seconds, or joules to electronvolts, you’re not summoning new quantities, you’re just translating between dialects of the universal languages of science.

To convert between units, you multiply by a conversion factor which is a magical ratio equal to one.

Example 5.1.2.

Suppose a star cruiser travels at \(90\, \frac{\text{km}}{\text{hr}}\text{.}\) To express this in meters per second:

\(\frac{90\,\text{km}}{\text{hr}} \left( \frac{1000\,\text{m}}{1\,\text{km}} \right) \left( \frac{1\,\text{hr}}{3600\,\text{s}} \right) = 25\,\text{m/s} \)

Both of the two conversion factors are fractions that equal one and stem from the following: \(1000\, \text{m} = 1\, \text{km}\) and \(3600\, \text{s} = 1\, \text{hr}\text{.}\)

For example, start with the expression \(3600\, \text{s} = 1\, \text{hr} \text{.}\)

If you divide both sides by \(3600\, \text{s}\text{,}\) you get \(1 = \frac{1\, \text{hr}}{3600\, \text{s}}\text{.}\)

Similarly, dividing both sides of \(1000\, \text{m} = 1\, \text{km}\) by \(1\, \text{km}\) gives \(1 = \frac{1000\, \text{m}}{1\, \text{km}}\text{.}\)

This tells us that multiplying any quantity by either of these fraction that equal \(1\) is like casting a harmless illusion; it doesn’t change the true value of a quantity, only the form in which it’s expressed.

In other words, you’re not altering reality, just viewing it through a different unit’s lens.

It’s like casting a chain of transmutation spells, each one carefully crafted so the old unit cancels and the new one takes its place. As long as your units behave, your math will too.

Subsubsection 5.1.2.2 Common Conversions

While there’s no universal conversion for all situations, below are some common conversions that will occur during the first part of of our journey through classical physics:

Length
- \(\displaystyle 2.54\, \text{cm} = 1\, \text{in}\)
- \(\displaystyle 1\, \text{m} = 3.28084\, \text{ft}\)
- \(\displaystyle 1\, \text{m} = 1.09361\, \text{yd}\)
- \(\displaystyle 1.609 \, \text{km} = 1 \, \text{mi}\)

Mass
- \(\displaystyle 1\, \text{kg} = 2.20462\, \text{lb}\)
- \(\displaystyle 1\, \text{kg} = 35.274\, \text{oz}\)
- \(\displaystyle 1\, \text{kg} = 0.0685218\, \text{slug}\)

For derived dimensions like area or volume, you can use the above conversions or use the internet to search for a conversion.

Checkpoint 5.1.3.

Convert \(5.30\) kilograms to pounds.

Hint.

Use the conversion factor \(1\, \text{kg} = 2.20462\, \text{lb}\text{,}\) which can be written as a fraction: \(\frac{2.20462\, \text{lb}}{1\, \text{kg}}\text{.}\)

Answer.

\(11.7\, \text{lb}\)

Solution.

To convert \(5.3\) kilograms to pounds, multiply by the conversion factor:

\(5.3\, \text{kg} \left( \frac{2.20462\, \text{lb}}{1\, \text{kg}} \right) = 11.684\, \text{lb}\)

Using the proper significant figures: \(5.30\) kilograms is equal to \(11.7\) pounds.

Checkpoint 5.1.4.

Convert \(1530\, \text{ft}^2\) to \(\text{m}^2\text{.}\)

Hint.

\(\text{ft}\) is the symbol for feet, and \(\text{m}\) is the symbol for meters. Use the conversion factor \(1\, \text{m} = 3.28084\, \text{ft}\text{,}\) twice.

Answer.

\(142\, \text{m}^2\)

Solution.

Note that our original unit is \(\text{ft}^2\) which can be written as \(\text{ft·ft}\text{.}\)

So we need to multiply by the conversion between feet and meters twice:

\(1530\, \text{ft}^2 \left( \frac{1\, \text{m}}{3.28084\, \text{ft}} \right) \left( \frac{1\, \text{m}}{3.28084\, \text{ft}} \right)= 142.1447\, \text{m}^2\)

...or...

\(1530 \, \text{ft}^2 \left( \frac{1\, \text{m}}{3.28084\, \text{ft}}\right)^2 = 142.1447\, \text{m}^2\)

Using the proper significant figures: \(1530\, \text{ft}^2\) is equal to \(142\, \text{m}^2\text{.}\)

Subsubsection 5.1.2.3 Purpose of Conversions

Humans, robots, magic folk, and aliens tend to use units that make sense in their context. When they want to communicate with each other, they need a way to convert thier measurement’s unit to anothers preferred unit. For example, a human might measure the length of a dragon’s tail in feet, while a robot might use meters. If they want to work together, they need to convert between these units.

Humans in different fields of study even use different units. For example, in physics, we often use the International System of Units (SI), which uses meters, kilograms, and seconds as its base units. While some engineers tend to use feet, slug, and seconds as their base units.

Subsubsection 5.1.2.4 Fermi Problems

So you have made it this far, from scientific notation all the way to unit conversions, despite all the curses we placed upon you when beginning your journe...uh wait nevermind that. Congratulations! Now, you can use your newfound skills to tackle Fermi problems.

A Fermi problem is a type of estimation problem that requires you to make reasonable assumptions and use unit conversions to arrive at an answer. Named after the physicist Enrico Fermi, these problems often involve estimating quantities that are difficult to measure directly, such as the number of jelly beans in a jar or the amount of energy consumed by a city in a day. The key is to break the problem down into smaller, manageable parts, make reasonable assumptions, and use unit conversions to arrive at an answer.

The answer to a Fermi problem is an estimate, not an exact value. Usually the answer is given in scientific notation with the order of magnitude being the most important part of the answer.

Example 5.1.5.

Fermi Problem: How much tea would it take to fill an airship’s cargo bay?

Initial thought 1: The problem didn’t specify the size of the airship’s cargo bay, so we will estimate the size as a volume based on a typical commercial jet’s passenger volume.

Height of the cargo bay.
- The height of a typical commercial jet is tall enough for a person to stand up, so let’s estimate about \(2\) meters.

Width of the cargo bay.
- The width of a typical commercial jet is wide enough for \(6\) people to side shoulder to shoulder, and an aisle for \(1\) person to walk through. So let’s estimate the width of the jet to be about \(7\) people widths. Since we are looking for an estimate, lets round up to \(10\) people widths. A person is probably about \(0.5\) meters wide, so the total width of the jet would be about \(10 \times 0.5 = 5\, \text{m}\text{.}\)

Length of the cargo bay.
- The length of a typical commercial jet is long enough for \(50\) seats to fit in front of or behind each other. Let’s estimate the length for each seat to be about the width of a person. Using \(0.5\) meters as a person width again, that means the total length of the jet’s passenger space is about \(50 \times 0.5\, \text{m} = 25\, \text{m}\text{.}\)

Volume of cargo bay.
- The volume is the product of the cargo bay’s height, width, and length. So the total volume is about \(2\) meters (height) times \(5\) meters (width) times \(25\) meters (length) \(=250\) cubic meters.

Initial thought 2: We now have the volume of the cargo bay in meters cubed, thus we also know much volume of tea we need in meters cubed: \(250\) cubic meters.

While this is an answer, it might not be the most human friendly answer. In other words, do you immediately have a mental visual of how much \(250\) cubic meters of tea is? Let’s convert it to a more human friendly unit, cups. We will use cups since is reasonable to assume folks drink tea a cup at a time.

Cups of tea.
- A quick internet search tells us that \(1\) cubic meter is equal to \(4226.75\) cups. But remember, Fermi problems are about estimates, so we will round that to \(4000\) cups. Or in other words, \(1\, \text{m}^3 = 4000\) cups.
- So the total amount of tea needed to fill the airship’s cargo bay is: \(250\, \text{m}^3 \times \frac{4000\, \text{cups}}{1\, \text{m}^3} = 1000000\, \text{cups}\text{.}\)

That’s \(1 \times 10^6\, \text{cups}\text{.}\) Or one million cups of tea. Enough to host a tea party for every elf in the eastern forest, twice.

A Parting Word

In general, when working on a project, if you start in SI units, and stay in SI units unless you have a solid reason (like a prophecy, or a stubborn client from another dimension).

Subsection 5.1.3 Prep Questions: Unit conversions

Unit Conversions

Checkpoint 5.1.6.

Convert \(5.91\) feet to meters.

\(1.80\) meters
Correct!
\(0.555\) meters
Your targeting system glitched, try again. Looks like you might have inverted the answer. Start with \(5.92\) meters and multiply it by one obtained from the conversion factor: \(1\, \text{m} = 3.28084\, \text{ft}\)
\(19.4\) meters
Your enchanted click missed the mark, try again. It looks like you might have multiplied instead of divided. Start with \(5.92\) meters and multiple it by one obtained from the conversion factor: \(1\, \text{m} = 3.28084\, \text{ft}\)
\(0.549\) meters
Your digital blade missed the mark, try again. It looks like you might have applied the conversion factor twice instead of once. Since the given value is "meters", we only need to apply the conversion factor once.

Checkpoint 5.1.7.

Convert \(55.0\) miles per hour to meters per second.

\(24.6\, \text{m/s}\)
Correct!
\(0.0246\, \text{m/s}\)
Your targeting system glitched, try again. This is \(55.0\, \text{mi/hr} = 0.0246\, \text{km/s}\text{.}\) Continue working on the conversion to meters per second.
\(88.5\, \text{m/s}\)
Your enchanted click missed the mark, try again. This is \(55.0\, \text{mi/hr} = 88.5\, \text{km/hr}\text{.}\) Continue working on the conversion to meters per second.
\(88500\, \text{m/s}\)
Your digital blade missed the mark, try again. This is \(55.0\, \text{mi/hr} = 88500\, \text{m/hr}\text{.}\) Continue working on the conversion to meters per second.

Checkpoint 5.1.8.

Convert \(55\underline{0}0\, \text{yd}^2\) to \(\text{m}^2\)

\(46\underline{0}0\, \text{m}^2\)
Correct!
\(6580\, \text{m}^2\)
Your targeting system glitched, try again. It looks like you used the correct conversion factor but inverted the conversion fraction.
\(5030\, \text{m}^2\)
Your enchanted click missed the mark, try again. It looks like you are using the correct conversion factor, but only applied it once. Apply the same conversion factor twice since the original unit is in yards squared.
\(6010\, \text{m}^2\)
Your digital blade missed the mark, try again. It looks like you used the correct conversion factor but inverted the conversion fraction. You also probably only applied the conversion factor once instead of twice.

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